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Taxes_Smuggling_Prelude_to_Revolution_Crash_Course_US_History_6.txt

Hi, I'm John Green, this is Crash Course: US History, and today we begin discussing the American Revolution. So two things to keep in mind here: One, the American Revolution and the American War for Independence are not the same thing. And two, while I know this will upset some of you, the American Revolution was not really about taxes. Mr. Green, Mr. Green! It was about tea, right? Also, it was not about tea. The Boston Tea party was about taxes and our God-given right to smuggle. It's a little confusing, me from the past, but that's why Crash Course is here! [Theme Music] So as you'll recall, the Seven Years War ended with the Treaty of Paris in 1763 which made the colonists cranky because it limited their ability to take land from the Indians, and it also left them holding the bag for a lot of war debt. Wars, as you may have noticed, are expensive, and the British government had to borrow 150,000,000 pounds, and the interest payments on that money ate up half of the national budget. So in order to pay for the war, the British decided to raise taxes. And since the primary beneficiaries of the war had been the American colonists, the British government felt it was only fair if some of the burden fell to them. Now, taxes on colonial trade were nothing new. The British government had placed taxes on a bunch of items in order to reduce competition with Britain, including wool and hats and “mole-asses.” Why did they place a tax on mole-asses? It doesn’t seem like that would be a huge market. Oh, molasses. Right, of course. But those taxes were about trying to regulate trade in a mercantilist way more than trying to pay back war debt. And also they were easy to avoid via smuggling, which we did because this is America! [Patriotic Rock Music] But mostly the colonists were angry because they didn’t have any say about the new taxes that Britain was imposing. I mean after all, by 1760, some colonies had been setting their own taxes through their own legislatures for 100 years. So the taxes themselves weren’t really the problem; it was their lack of Parliamentary representation. The first purportedly oppressive tax, the Sugar Act of 1764, extended the Molasses Act by changing the tax on imports from the Caribbean from 6 cents per gallon all the way up to 3 cents per gallon. So they actually cut the tax, but they decided to start enforcing it by stamping out smuggling. And to that end, the Act also gave British courts the right to try colonial smugglers, taking that power away from colonial courts, which had been notoriously lenient when it came to smuggling on account of how they enjoyed smuggled rum as much as the next guy. But those initial acts weren’t nearly as annoying as the Stamp Act passed in 1765. The Stamp Act declared that all printed material had to carry a stamp. Unsurprisingly, that stamp was not free. This was purely to gain revenue for Britain, and it mostly affected people who used a lot of paper. You know, like newspaper printers and lawyers. Just the kind of people you want to anger about taxes! So in October, protesters organized the Stamp Act Congress, which after a meeting, decided to boycott British goods. And this was the first major coordinated action by the colonies together, and it might be the first time that we can speak of the colonies acting in a united way. Almost like, say, a government. Committees of correspondence, which had been created to encourage opposition to earlier acts, now grew to coordinate the boycott efforts, and they helped people become aware of their "liberties." And they also spurred street actions that occasionally became violent. These direct actions were organized by groups calling themselves the Sons of Liberty, and guess what? Coordinated action worked! The British Parliament repealed the Stamp Act, but they did pass the Declaratory Act which was all like, “Listen, you’re not the boss of us. We can tax you. We don’t want to tax you right now as it happens, but we could if we wanted to. But we won’t, but we could!” So the repeal of the Stamp Act was seen by many in the colonies as a huge victory, but most of the people organizing the protests were elites. You know, the kind of people who use paper. But once you start talking about the idea of representation, everybody wants in. Meanwhile, Great Britain still needed money, so Chancellor of the Exchequer Charles Townshend got Parliament to pass new taxes in 1767. The so called “Townshend Acts” also created a new board of customs to stop smuggling which we didn’t like one bit. You don’t like it when I say we? Well tough luck, I’m an American. Bring back the libertage, Stan! [Patriotic Rock Music] Many colonists again responded with a boycott, and women got in on the act this time, with the Daughters of Liberty, encouraging homespun clothes to replace British ones. But not all the states were on board, like artisans loved the boycotts because they got more money. But merchants from cities like Philadelphia and New York weren’t so happy, because they made their livings by importing and selling the very goods that were now being boycotted. On occasion, protests did get out of hand as in the Boston Massacre of March 5, 1770, which, while it was not much of a massacre, was definitely the worst outcome of a snowball fight in American history. I mean five colonists were killed, including most famously Crispus Attucks, a sailor of mixed race ancestry. And then of the nine British soldiers put on trial, 7 were acquitted and 2 were convicted only of manslaughter, thanks to the top-notch lawyering of one John Adams. Don’t worry though, that guy comes around to the American cause. But overall boycotts and protests were effective, and British merchants pushed for the repeal of these acts, leaving only a tax on tea. Let’s go to the Thought Bubble. The 1773 Tea Act offered tax exemptions and rebates for tea coming in from the British-East India Company, which allowed them to dump cheap tea on the colonies, which actually lowered the price of tea. So why were the colonists so mad that on December 16, 1773 they dressed up as Indians and dumped enough tea into Boston Harbor to cause the modern equivalent of a $4,000,000 loss? Some colonists were upset that cheap tea would cut into the profits of smugglers and established tea merchants, but most were just angry on principle. To our great national shame, tea was at the time as important a beverage in the colonies as it was to Brits living in Britain, and to allow the British to tax a near universal product set a precedent that Britain could tax whatever they wanted. But the tea partiers miscalculated, thinking that the British would back down in response to their protest. Instead, the British responded by passing a series of acts that colonists came to call the “Intolerable Acts.” The Massachusetts Government Act curtailed self-government there. The Quartering Act forced colonists to house British soldiers in their homes when ordered to. The Quebec Act extended the southern boundary of Quebec and granted religious toleration to Catholics, which was none too popular with the Great Awakening crowd, having recently awoken. The colonial response to these acts is really the start of the American Revolution. First Massachusetts passed a set of resolutions calling for colonists to: one, disobey the Intolerable Acts, two, stop paying taxes, and three, prepare for war. And in September 1774, a group of delegates from 12 of the 13 colonies – Georgia! – met in Philadelphia to coordinate the resistance of the Intolerable Acts. This was the First Continental Congress, which in setting up the continental association to police the boycott and encourage domestic manufacturing, was the first real colony-wide government in British America. Thanks, Thought Bubble. So this sort of phenomenon is known by historians as kind of a big deal. I mean coordinating action to achieve some end is what governments do, so it’s not an exaggeration to say that the First Continental Congress was the first government of America. You might even say that sending delegates to Philadelphia in 1774 was the first truly revolutionary act of the American Revolution, but it was not a call for independence. However, there was a change in attitude among many colonists because rather than seeing themselves as standing up for their rights as English people, they began to make claims based on abstract ideas about freedom and natural rights. In this respect, the Continental Congress was between worlds, because it justified its actions as liberties of free and natural-born subjects within the realm of England. But it also talked about immutable laws of nature. And this idea that all humans have certain rights derived from natural law has become a pretty big deal, not just in the United States, but also in the green lands of “not-America.” I mean these days “human rights” is a phrase that we bandy about with – Putin! Do you show up every time the words “human rights” are mentioned to make sure that we’re talking about China and not you? And this brings me back to an important point: although we tend to equate the two, the American Revolution and the American War for Independence were not the same thing. I mean for one thing, the fighting started 15 months before the Declaration of Independence. For another, simply declaring independence does not make you an independent nation, as I will now demonstrate. I hereby declare this studio the independent nation of John Green-sylvania! Yeah, see nothing happened. The war between colonists and Britain began in 1775 – on April 19th to be exact – when fighting broke out between the British soldiers and Massachusetts militiamen, the minutemen, at Concord and Lexington. Or Lexington and Concord, depending on whether you live in Lexington or Concord. This was the famous “shot heard around the world” immortalized in Longfellow’s poem "The Midnight Ride of Paul Revere." So while the colonists actually lost the famous battle of Bunker Hill, which was technically fought on Breeds Hill, the British suffered such heavy casualties that soon thereafter they were forced to abandon Boston. But then they got some revenge by taking over New York, which they held for most of the rest of the war. But in thinking about the war, it’s very important to understand that not all colonists were pro-independence. Like elites in colonies like New York and Pennsylvania were very nervous about all this revolutionary fervor that was whipping up artisans and small-time farmers to think that they deserve to have say in the political process. Oh it’s time for the Mystery Document? Awesome, I love getting shocked. The rules here are simple: if I guess the author of the Mystery Document correctly, I do not get the shock pen. If I’m wrong, I do get the shock pen. All right, let’s see what we got here. [clears throat] "The Americans are properly Britons. They have the manners, habits, and ideas of Britons; and have been accustomed to a similar form of government. But Britons could never bear the extremes, either of monarchy or republicanism. Some of their kings have aimed at despotism; but always failed. Repeated efforts have been made towards democracy, and they equally failed. If we may judge of future events by past transactions, in similar circumstances, this would most probably be the case in America, were a republican form of government adopted in our present ferment." Hmm. All right, so we’ve got an educated person who thinks that Americans are Britons who will inevitably want to walk a middle path between republicanism and monarchy and therefore that the revolution is not a good idea. I know it’s a colonist, because of the reference to “our present ferment.” All right, I’m going to guess that it is Ben Franklin’s son William Franklin. [Buzzing Noise] Ahhh! Dang it! Who is it? Who the hell is Charles Ingles? Charles Ingles? Charles freaking Ingles? I’ve never even heard of that guy! It’s not fair! [Sigh] [Shock Noise] Ahh! Oh I hate that. Apparently he’s a bishop or something. Anyway, people like Ingles reminds us that not everyone in the colonies was all fired up to be an independent nation. In fact, in July of 1775, the Continental Congress sent the Olive Branch Petition to King George III suggesting that Americans were loyal British subjects who wanted reconciliation with the mother country. But then along came Thomas Paine’s pamphlet Common Sense. Why couldn’t that have been the Mystery Document? Common Sense appeared in January of 1776 and it was like the Harry Potter of its time, only with liberty instead of wizard school. Written in relatively straightforward English, the pamphlet contains many powerful rhetorical arguments like: “of more worth is one honest man to society and in the sight of God than all the crowned ruffians that ever lived.” Others were just common sense, like this nugget: “There is something absurd in supposing a continent to be perpetually governed by an island.” Pow! Also there is the beautiful sentiment: "The weapon we have is love.” Oh that’s from Harry Potter? I told you they were similar. Ultimately Paine’s arguments all contributed to the idea that America is somehow special, even exceptional. I mean, talking about independence and freedom he said, “The cause of America is in great measure: the cause of all mankind.” That’s powerful stuff, and Paine’s pamphlets sold 150 thousand copies, and it was extremely widely read. By the way, he still managed to die penniless, and only 8 people attended his funeral because of his vitriolic ridicule of Christianity. But anyway America eventually declared independence for many reasons, but Paine’s persuasive arguments were one important reason. And it marks a moment when the pen truly was, if not more powerful, then at least more important, than the sword. I mean, within 6 months of the publication of Common Sense, the Second Continental Congress had declared independence and signed one of the most important documents in the history of the world. Which is where we’ll pick up next week. Thanks for watching! Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, and myself. And our graphics team is Thought Bubble. If you have questions about today’s video or anything from American history, good news! You can ask them in comments, where they will be answered by our team of historians. Thanks for watching Crash Course. If you liked today's video make sure you subscribe to our channel. And as we say in my hometown, don't forget to be awesome!

US_History

The_Market_Revolution_Crash_Course_US_History_12.txt

Hi, I'm John Green, this is Crash Course US History and today we return to one of my favorite subjects: economics. Mr. Green, Mr. Green, I don't wanna brag, but economics is actually my best subject. Like, I got the bronze medal at the state academic decathlon tournament...among C students. Yeah, I remember, Me from the Past. By the way, thanks for getting that picture into our show. It just goes to show you: aptitude is not destiny. Anyway, economics is about much more than, like, supply and demand curves. Ultimately, it's about the decisions people make and how those decisions shape their lives and the world. So today we're going to turn to one of the least-studied but most interesting periods in American history: the Market Revolution. There weren't any fancy wars or politically charged debates, but this discussion shaped the way that most Americans actually live their lives and think about work on a daily basis. Like, if you or someone you know goes to work, well, then you have the Market Revolution to thank, or possibly to curse. [Theme Music] The Market Revolution, like the Industrial Revolution, was more of a process than an event. It happened in the first half of the 19th century, basically, the period before the Civil War. This was the so-called "Era of Good Feelings," because between 1812 and 1836, there was really only one political party, making American politics, you know, much less contentious. Also, more boring. The Market Revolution saw many Americans move away from producing stuff largely for themselves on independent farms – that Jeffersonian ideal – and toward producing goods for sale to others, often others who were very far away, with prices set by competition with other producers. This was closer to Hamilton's American dream. In the end, buddy, you didn't get to be president, but you did win. In many ways, this was the beginning of the modern commercial industrial economy, not just in the United States, but in the world. The first thing that enabled this massive economic shift was new technology, specifically in transportation and communication. Like, in the 18th century, it was very difficult to bring goods to markets, and that meant that markets were local and small. Most trade was over land, and transporting goods 30 miles over land in the United States literally cost as much as shipping them to England. So, to get something from Cincinnati to New York, for instance, the most efficient way was to go down the Mississippi River, through the Gulf of Mexico, around Florida, and then up the Atlantic coast, which took three months, but that was still less time and less money than more direct overland routes. But new transportation changed this. First came better roads, which were largely financed by tolls. Even the federal government got in on the act, building the so-called National Road, which reached all the way from the massive city of Cumberland, Maryland, across our great nation to the equally metropolitan Wheeling, West Virginia. Mr. Green! Mr – Mr. Green, Mr. Green! I know, Me from the Past, West Virginia did not yet exist, argh, shut up! More important than roads were canals, which made transport much cheaper and more efficient, and which wouldn't have been possible without the steam boat. Robert Fulton's steam boat Clermont first sailed from New York to Albany in 1807, demonstrating the potential of steam-powered commerce. And by 1811, there were steam boats on the Mississippi. The introduction of steam boats set off a mania for canal building. Between 1800 and the depression of 1837, which put a halt to most construction, more than 3,000 miles of canals were built. And no state was more instrumental in the canal boom than New York, which in 1825 completed the 363-mile-long Eerie Canal, linking the Great Lakes with the Hudson River, which made New York the nation's premier port. Other cities like Buffalo, Rochester, and Syracuse grew up along the canals. So much so that Nathaniel Hawthorne once said, "The canal is like fertilizer, causing cities to spring up alongside it." That's such a good simile, Nathaniel Hawthorne. It's almost like the United States didn't have any good writers until Mark Twain, but we need to read somebody from the early 19th century, so I guess it's you. But from a long-term perspective, the most important new transportation? Railroads. The first commercial railroad, the Baltimore & Ohio, was begun in 1828 and by 1860, there were more than 30,000 miles of rails in the United States. And on the communication side, we got the telegraph, so no longer would Andrew Jackson fight battles two weeks after the end of a war. Telegraphs allowed merchants to know when to expect their shipments and how much they could expect to sell them for. And then, as now, more information meant more robust markets. But perhaps the most important innovation of the time was the factory. Now, when you think of factories, you might think of, like, Chinese political prisoners making smartphones, but early factories looked like this. More than just a technological development, the factory was an organizational innovation. Like, factories gathered workers together in one place and split up tasks among them, making production much faster and also more efficient. The first factories relied on water power, which is the reason they were all east of the fall line – the geographic reason why there are so many waterfalls and rapids on the east coast. But after 1840, steam power was introduced, so factories could be located in other places, especially near the large cities that were sprouting up in what we now know as the Midwest. So the American system of manufacturing, which centered on mass-production of interchangeable parts, grew up primarily in New England, but then it moved to the Midwest, where it spent its adolescence and its adulthood, and now its tottering decline into senility. So, all these new economic features – roads, canals, railroads, telegraphs, factories – they all required massive up-front capital investment. Like, you just can't build a canal in stages as it pays for itself. So, without more modern banking systems and people willing to take risks, none of this would have happened. Some of these investments were facilitated by new business organizations, especially the Limited Liability Corporation, which enabled investors to finance business ventures without being personally responsible for losses other than their own. In other words, corporations can fail without, like, ruining their stockholders and directors. People don't always like that, by the way, but it's been very good for economic growth in the last 180 years or so. So having angered a bunch of people by talking about the important role that big businesses played in growing the American Economy in the 19th century, I will now anger the rest of you by talking about the important role that the state played. In the 1830s, states began passing general incorporation laws, which made it easier to create corporations, and the Supreme Court upheld them and protected them from further interference in cases like Gibbons vs. Ogden, which struck down a monopoly that New York had granted to one steamboat company. And the Charles River Bridge case, which said that building a second bridge over the Charles River did not infringe upon the charter of the first bridge. In both those cases, the court was using its power to encourage competition. And this brings up something really important about the growth of American capitalism: government helped. The federal government built roads and canals and its highest court protected businesses. And states issued bonds to build canals and offered sweetheart deals to companies that built railroads. And despite what we may believe about the heroic risk-taking entrepreneurs building the American economy through solitary efforts, without the government protecting their interests, they wouldn't have been able to do much. All right, let's go to the Thought Bubble. The Market Revolution changed the landscape of work, which, for most of the prior 200 years, happened at home. Small-scale production of clothes and other goods had been done in the home, largely by women, and initially, this is how industrial production worked as well. Factory owners would produce some of the products, like patterns for shoes, and then farm the finishing out to people working in their houses. Eventually, they realized that it would be more efficient to gather the workers together in one place, although the older, "putting-out system" continued in some industries, especially in big cities. After the Market Revolution, more and more Americans went to work instead of working from home. The Market Revolution also changed the way we imagined work and leisure time. Like, on farms, the seasons and hours of daylight regulated the time for work, but in factories, work is regulated by the clock. Which, by the way, was one of the first products to be manufactured using the American system of manufacturing. Railroads and shipping timetables further required the standardization of time. Factories also made it possible for more people to do industrial work. At first, this meant women. The workers in the early textile mills at Lowell, Massachusetts, for instance, were primarily young, New England farm girls who worked for a few years in the mills before returning home to get married. Women were cheaper to employ, because it was assumed that they would not be a family's sole breadwinner. At least, this was the excuse for not paying them more at the time. I can't remember what excuse we have now, but I'm sure it's a great one. Anyway, all of this meant that the nature of work had changed. In colonial America, artisans worked for what they called their "price," which was linked to what they produced. In a factory, however, workers were paid a wage according to the number of hours they worked, regardless of how much they produced. This may not sound like a big deal, but working for wages with one's livelihood defined by a clock and the whims of an employer was a huge change, and it undermined the idea of freedom that was supposedly the basis of America. Thomas Jefferson had worried that men working in factories, dependent upon their employers, were inherently un-free, and that this would make them unfit to be proper American citizens. And as it happens, many factory workers agreed with him. Thanks, Thought Bubble. So, one reaction to the restrictions of the wage worker was to engage in the great American pastime of lighting out for the territories. With less and less farmland available in New England, young men had been migrating west for decades. And, after the War of 1812, this flood of migration continued and even grew. Between 1790 and 1840, 4.5 million people crossed over the Appalachian Mountains, and six new states were created between 1815 and 1821. Ohio's population grew from 231,000 in 1810 to over 2 million by 1850. People even took up the motto 'Malaria isn't going to catch itself!' and moved to Florida after we purchased it from Spain in 1819. Moving out West was a key aspect of American freedom, and the first half of the 19th century became the age of "manifest destiny": the idea that it was a God-given right of Americans to spread out over the North American continent. The term was coined by a New York journalist, John L. O'Sullivan, who wrote that the people living out West – i.e, the Native Americans – must succumb to quote, "our manifest destiny to overspread and to possess the whole of the continent which providence has given us for the development of the great experiment in liberty." Stan, he actually wrote "overspread"! One thing I love about providence is that it has like a 100% rate of giveth-ing unto us and taketh-ing away from them. One of the results of this migration was that it was really difficult for factory owners to find men who could work in their factories. First, they looked to Yankee women to fill the factories, but increasingly, those jobs were filled by immigrants. Fortunately, the US had lots of immigrants, like the more than one million Irish people that came here fleeing poverty, especially after the potato famine of 1845 to 1851. Lastly, let's turn to the intellectual responses to the Market Revolution. Oh, it's time for the Mystery Document? The rules here are simple. If I fail to guess the author of the Mystery Document, I get shocked with the shock pen. And yes, this is a real shock pen! Lots of people are commenting, saying I am faking the shocks. I am not faking the shocks! I am in the business of teaching you history, not in the business of faking pain! All right, let's do this thing. "They do not yet see, and thousands of young men as hopeful now crowding to the barriers of the career, do not yet see, that if the single man plant himself indomitably on his instincts, and there abide, the huge world will come round to him. Patience - patience; - with the shades of all the good and great for company; and for solace the perspectives of your own infinite life; and for work, the study and the communication of principles, the making those instincts prevalent, the conversion of the world. Is it not the chief disgrace in the world, not to be a unit; - not to be reckoned one character; - not to yield that peculiar fruit which each man was created to bear, but to be – " Oh, god, Stan, I can't bear it anymore. It's Emerson. [dinging noise] It's definitely Emerson. It is unreadably Emerson. Indeed, the most linguistically convoluted of the Transcendentalists, which is really saying something. Anyway, I don't get punished, but I did kind of get half punished, because I had to read that. The Transcendentalists – like Margaret Fuller, Henry David Thoreau, Walt Whitman – were trying to redefine freedom in a changing world. Work was increasingly regimented. Factory workers were as interchangeable as the parts that they made. But the Transcendentalists argued that freedom resided in an individual's power to remake oneself, and maybe even the world. But there would be a reaction to this in American literature as it became clear that escaping drudgery to reinvent yourself was no easy task for wage workers. So, the early 19th century saw a series of booms and busts, sometimes called business cycles. And with those business cycles came a growing disparity in wealth. To protect their interests, workers began forming political organizations called Working Man's Parties, that eventually morphed into unions, calling for higher wages and better working conditions. And we'll have more to say about that in coming weeks, but for now, it's important to remember that as America grew more prosperous, many people – women and especially slaves, but also free, wage-working men – recognized that the Market Revolution left them with much less freedom than they might have enjoyed 50 or 100 years earlier. My favorite commentary on the Market Revolution actually comes from the author Herman Melville in his short story "Bartleby the Scrivener." Melville worked at the customs house in New York, so he knew all about world markets first-hand. In "Bartleby," he tells the story of a young clerk who works for a lawyer in New York City. Now, when you're a farmer, your work has an intrinsic meaning. When you work, you have food, and when you don't work, you don't. But when you're a copyist like Bartleby, it's difficult to find meaning in what you do every day. You know that anyone else could do it, and you suspect that if your work doesn't get done, it won't actually matter very much. And in light of this, Bartleby just stops working, saying, "I prefer not" when asked, well, pretty much anything. Seeing his boss and society's reaction to someone who simply doesn't buy into the market economy is comic, and then ultimately tragic. And it tells us a lot about the Market Revolution beyond the famous people and inventions and heroic individualism. Now, most people read "Bartleby" as an existentialist narrative, and it definitely is that, but, for me, the story's subtitle proves that it's also about the market economy. The full title of the story is, "Bartleby the Scrivener: A Story of Wall Street." I'll see you next week. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our associate producer is Danica Johnson. The show is written by my high school history teacher Raoul Meyer and myself. And our graphics team is Thought Cafe. If you have questions about today's video, please ask them in comments, where they'll be answered by our team of historians. Also, suggest Libertage captions. Thanks for watching Crash Course World History. If you enjoy Crash Course, make sure you're subscribed, and as they say in my hometown, don't forget to be awesome. Just kidding, thanks for watching Crash Course US History! DFTBA!

US_History

American_Imperialism_Crash_Course_US_History_28.txt

Episode 28: American Imperialism Hi, I’m John Green, this is CrashCourse U.S. History and today we’re gonna talk about a subject near and dear to my white, male heart: imperialism. So, here at CrashCourse we occasionally try to point out that the U.S., much as we hate to admit it, is actually part of a larger world. Mr. Green, Mr. Green, you mean like Alaska? No, Me from the Past, for reasons that you will understand after your trip there before your senior year of college, I do not acknowledge the existence of Canada’s tail. No, I’m referring to all of the Green Parts of Not-America and the period in the 19th century when we thought, “Maybe we could make all of those green parts like America, but, you know, without rights and stuff.” Intro So, the late 19th and early 20th centuries were a period of expansion and colonization in Asia and Africa, mostly by European powers. As you’ll know if you watched Crash Course World History, imperialism has a long, long history pretty much everywhere, so this round of empire building is sometimes called, rather confusingly, New Imperialism. Because the U.S. acquired territories beyond its continental boundaries in this period, it’s relatively easy to fit American history into this world history paradigm. But there’s also an argument that the United States has always been an empire. From very early on, the European settlers who became Americans were intent on pushing westward and conquering territory. The obvious victims of this expansion/imperialism were the Native Americans, but we can also include the Mexicans who lost their sovereignty after 1848. And if that doesn’t seem like an empire to you, allow me to draw your attention to the Russian Empire. Russians were taking control of territory in Central Asia and Siberia and either absorbing or displacing the native people who lived there, which was the exact same thing that we were doing. The empires of the late 19th and early 20th centuries were different because they were colonial in their own special way. Like, Europeans and Americans would rule other places but they wouldn’t settle them and more or less completely displace the native people there. (Well, except for you, Australia and New Zealand.) American historians used to try to excuse America’s acquisitions of a territorial empire as something of an embarrassing mistake, but that’s misleading because one of the primary causes of the phenomenon of American imperialism was economics. We needed places to sell our amazing new products. And at the time, China actually had all of the customers because apparently it was opposite day. It’s also not an accident that the U.S. began pursuing imperialism in earnest during the 1890s, as this was, in many ways, a decade of crisis in America. The influx of immigrants and the crowded cities added to anxiety and concern over America’s future. And then, to cap it all off, in 1893 a panic caused by the failure of a British bank led the U.S. into a horrible economic depression, a great depression, but not The Great Depression. It did however feature 15,000 business failures and 17% unemployment, so take that, 2008. According to American diplomatic historian George Herring, imperialism was just what the doctor ordered to help America get out of its Depression depression. Other historians, notably Kristin Hoganson, imply that America embarked on imperial adventures partly so that American men could prove to themselves how manly they were. You know, by joining the Navy and setting sail for distant waters. In 1890, Captain Alfred Thayer Mahan published “The Influence of Seapower upon History” and argued that, to be a great power like Great Britain, the U.S. needed to control the seas and dominate international commerce. Tied into this push to become a maritime power was the obsession with building a canal through Central America and eventually the U.S. decided that it should be built in Panama because you know how else are we gonna get malaria. In order to protect this canal we would need a man, a plan, a canal. Panama. Sorry, I just wanted to get the palindrome in there somewhere. No we would actually need much more than a man and a plan. We would need ships and in order to have a functioning two-ocean navy, we would need colonies. Why? Because the steamships at the time were powered by coal and in order to re-fuel they needed coal depots. I mean, I suppose we could have, like, rented harbor space, but why rent when you can conquer? Also, nationalism and the accompanying pride in one’s “country” was a worldwide phenomenon to which the U.S. was not immune. I mean, it’s no accident that the 1890s saw Americans begin to recite the pledge of allegiance and celebrate Flag Day, and what better way to instill national pride than by flying the stars and stripes over … Guam. So pre-Civil War attempts to expand beyond what we now know as the continental United States included our efforts to annex Canada, which were sadly unsuccessful, and also filibustering, which before it meant a senator talking until he or she had to stop to pee was a thing where we tried to take over Central America to spread slavery. But, the idea of taking Cuba persisted into the late 19th century because it is close and also beautiful. The Grant administration wanted to annex it and the Dominican Republic, but Congress demurred. But we did succeed in purchasing Canada’s tail. You can see how I feel about that. To be fair, discovery of gold in the Yukon made Seward’s icebox seem like less of a Seward’s folly and it did provide coaling stations in the Pacific. But we could have had rum and Caribbean beaches. Ugh, Stan, all this talk about how much I hate Alaska has me overheated, I gotta take off my shirt. Ughhh. Waste of my life. So hard to take off a shirt dramatically. I’m angry. Anyway, coal stations in the Pacific were important because in 1854 we “opened” Japan to American trade by sending a flotilla of threatening black ships under Matthew Perry. No Stan, not that Matthew Perry. You know better. By far, America’s best piece of imperial business before 1898 was Hawaii. Like, I like oil and gold as much as the next guy but Hawaii has pineapples and also had sugar, which was grown on American owned plantations by Chinese, Japanese, Filipino, and native workers. Treaties between the U.S. and the Hawaiian governments exempted this sugar from tariffs, and America also had established a naval base at Pearl Harbor, which seemed like a really good idea...then. We eventually annexed Hawaii in 1898 and this meant that it could eventually become a state, which it did in 1959, two years before Barack Obama was born in Kenya. And this leads us nicely to the high tide of American imperialism, the Spanish-American-Cuban-Fillipino War. The war started out because native Cubans were revolting against Spain, which was holding on to Cuba for dear life as the remnant of a once-great empire. The Cubans’ fight for independence was brutal. 95,000 Cubans died from disease and malnutrition after Spanish general Valeriano Weyler herded Cubans into concentration camps. For this Weyler was called “Butcher” in the American yellow press, which sold a lot of newspapers on the backs of stories about his atrocities. And at last we come to President William McKinley who responded cautiously, with a demand that Spain get out of Cuba or face war. Now Spain knew that it couldn’t win a war with the U.S. but, as George Herring put it, they “preferred the honor of war to the ignominy of surrender.” Let that be a lesson to you. Always choose ignominy. Oh, it’s time for the Mystery Document? The rules here are simple. I guess the author of the Mystery Document. I’m either right or I get shocked. Alright, let’s see what we’ve got today. With such a conflict waged for years in an island so near us and with which our people have such trade and business relations; when the lives and liberty of our citizens are in constant danger and their property destroyed and themselves ruined; where our trading vessels are liable to seizure and are seized at our very door by warships of a foreign nation, the expeditions of filibustering that we are powerless to prevent altogether -- all these and others that I need not mention, with the resulting strained relations, are a constant menace to our peace, and compel us to keep on a semiwar footing with a nation with which we are at peace. Thank you, Stan. This is obviously President William McKinley’s war message to Congress. You can tell it’s a war message because it includes the word “peace” more than the word “war.” By the way, it’s commonly thought that the President McKinley asked Congress for a declaration of war, he didn’t; he let Congress take the lead. That’s the only time that’s ever happened in all of American history, which would be more impressive if we had declared war more than 5 times. So, the document shows us that, at least according to McKinley, we officially went to war for American peace of mind and to end economic uncertainty. It was not to gain territory, at least not in Cuba. How do we know? Because Congress also passed the Teller Amendment, which forswore any U.S. annexation of Cuba, perhaps because representatives of the U.S. sugar industry like Colorado’s Senator Henry Teller feared competition from sugar produced in an American Cuba. Or maybe not. But probably so. Also not the cause of the war was the sinking of the USS Maine. The battleship which had been in Havana’s harbor to protect American interests sank after an explosion on February 15, 1898 killing 266 sailors. Now, most historians chalk up the sinking to an internal explosion and not to Spanish sabotage, but that didn’t stop Americans from blaming the Spanish with their memorable meme: “Remember the Maine, to hell with Spain.” Let’s go to the Thoughtbubble. The actual war was one of the most successful in U.S. history, especially if you measure success by brevity and relative paucity of deaths. Secretary of State John Hay called it a “splendid little war” and in many ways it was. Fighting lasted about 4 months and fewer than 400 Americans were killed in combat, although 5,000 died of, wait for it, disease. Stupid disease, always ruining everything. There weren’t a ton of battles but those that happened got an inordinate amount of press coverage, like the July attack on San Juan Hill at the Cuban city of Santiago, led by future president Theodore Roosevelt. While it was a successful battle, the real significance is that it furthered Roosevelt’s career. He returned a hero, promptly became Governor of New York and by 1900 was McKinley’s vice president. Which was a good job to have because McKinley would eventually be assassinated. A more important battle was that of Manila Bay in which commodore George Dewey destroyed a tiny Spanish fleet and took the Philippines. This battle took place in May of 1898, well before the attack on Cuba, which strongly suggests that a war that was supposedly about supporting Cuban independence was really about something else. And what was that something else? Oh right. A territorial empire. As a result of the war, the U.S. got a bunch of new territories, notably the Philippines, Puerto Rico and Guam. We also used the war as an opportunity to annex Hawaii to protect our ships that would be steaming toward the Philippines. We didn’t annex Cuba, but we didn’t let it become completely independent, either. The Platt Amendment in the Cuban Constitution authorized American military intervention whenever it saw fit and gave us a permanent lease for a naval base at Guantanamo Bay. Thanks Thoughtbubble. So, Cuba and Puerto Rico were gateways to Latin American markets. Puerto Rico was particularly useful as a naval station. Hawaii, Guam, and especially the Philippines opened up access to China. American presence in China was bolstered by our contribution of about 3,000 troops to the multinational force that helped put down the Boxer Rebellion in 1900. But in the Philippines, where Americans had initially been welcome, opinion soon changed after it became clear that Americans were there to stay and exercise control. Emiliano Aguinaldo, leader of the Filipino rebellion against Spain, quickly turned against the U.S. because his real goal was independence and it appeared the U.S. would not provide it. The resulting Philippine War lasted 4 years, from 1899-1903. And 4,200 Americans were killed as well as over 100,000 Filipinos. The Americans committed atrocities, including putting Filipinos in concentration camps, torturing prisoners, rape, and executing civilians. And much of this was racially motivated and news of these atrocities helped to spur anti-imperialist sentiment at home, with Mark Twain being one of the most outspoken critics. Now, there was some investment in modernization in the Philippines, in railroads, schools, and public health, but the interests of the local people were usually subordinated to those of the wealthy. So, American imperialism in short looked like most other imperialism. So Constitution nerds will remember that the U.S. Constitution has no provision for colonies, only territory that will eventually be incorporated as states. Congress attempted to deal with this issue by passing the Foraker Act in 1900. This law declared that Puerto Rico would be an insular territory; its inhabitants would be citizens of Puerto Rico, not the United States and there would be no path to statehood. But this wasn’t terribly constitutional. Congress did extend U.S. citizenship to Puerto Ricans in 1917. Now it’s a commonwealth with its own government that has no voice in U.S. Congress or presidential elections and no control over its own defense or environmental policy. The Philippines were treated similarly to Puerto Rico, in a series of cases between 1901 and 1904 collectively called the Insular Cases. But Hawaii was treated differently. Because it had a sizeable population of American settlers who happened to be white. Ergo, it became a traditional territory with a path to statehood because white people and also pineapples. Now let’s briefly talk about anti-imperialism. There were lots of people who objected to imperialism on racial grounds, arguing that it might lead to, like, diversity. But there were also non-racist anti-imperialists who argued that empire itself with its political domination of conquered people was incompatible with democracy, which, to be fair, it is. The Democratic Party, which had supported intervention in Cuba, in 1900 opposed the Philippine War in its platform. Some Progressives opposed imperialism too because they believed that America should focus on its domestic problems. Yet those who supported imperialism were just as forceful. Among the most vocal was Indiana Senator Albert Beveridge who argued that imperialism was benevolent and would bring “a new day of freedom.” But, make no mistake, underneath it all, imperialism was all about trade. According to Beveridge, America’s commerce “must be with Asia. The Pacific is our ocean … Where shall we turn for consumers of our surplus? Geography answers the question. China is our natural customer.” In the end, imperialism was really driven by economic necessity. In 1902, Brooks Adams predicted in his book The New Empire that the U.S. would soon “outweigh any single empire, if not all empires combined.” Within 20 years America would be the world’s leading economic power. We didn’t have the most overseas territory, but ultimately that didn’t matter. Now, the reasons for imperialism, above all the quest for markets for American goods, would persist long after imperialism became recognized as antithetical to freedom and democracy. And we would continue to struggle to reconcile our imperialistic urges with our ideals about democracy until...now. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, Rosianna Rojas, and myself. And our graphics team is Thought Café. Every week there’s a new caption for the libertage. You can suggest captions in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome. This is the part where Stan gets nervous, like, is he gonna go this way or this way or this way? I’m going this way. Imperialism -

US_History

George_HW_Bush_and_the_End_of_the_Cold_War_Crash_Course_US_History_44.txt

Hi, I’m John Green, this is CrashCourse U.S. history and we’ve finally done it we have reached the moment where we get to talk about the presidency of George HW Bush. The 2nd most important man named George Bush ever to be President of the United States. A man so fascinating that we did not give him a face. Mr. Green, Mr. Green, so we’re almost in the present? Well we’re never really gonna get to the present Me From The Past because we’re always in the past. But you are like 20 years in the past which is soon going to create a time paradox that I can not possibly deal with. So I’m just going to let Hank deal with that over on the science shows. Intro Anyway despite like calendars and everything, the 1990s really began in 1988 with the election of George Herbert Walker Bush, who had probably the best resume of any presidential candidate since Teddy Roosevelt. I mean he was a war hero, having enlisted in the Navy upon graduating from high school and then going on to become the youngest pilot in Navy history. He flew 58 missions in the Pacific during WWII and received the Distinguished Flying Cross for completing a mission in a burning plane before ditching into the sea. So just consider that the next time you complete a heroic mission in Call of Duty 4. After the war Bush went to college at Yale, and then moved to Texas where he made millions in the oil industry. Then, he became a Congressman, and then ambassador to the UN, and then director of the CIA, and then Vice President. The guy had more careers than Barbie! Plus like every great American politician George Bush grew up in hardscrabble poverty working his way through the Depression… just kidding he was the son of Connecticut Senator Prescott Bush. But I guess after like 20 years of peanut farmers and former actors, America was ready to have an aristocrat at the helm again, as long as he pretended to be from Texas. Like certain Crash Course teachers wearing striped polo shirts George HW Bush was an old school Episcopalion so he was never totally comfortable with like public professions of faith. So when it came down to pick his vice presidential candidate, Bush chose J Danforth Quayle aka Dan. A young, family values, senator from right here in Indiana. Now these days of course Dan Quayle is primarily known for getting in an argument with a fictional television character named Murphy Brown, and also for not being able to spell the word potato, but once upon a time he was a promising young Republican. Bush’s opponent in that 1988 was Massachusetts Governor Michael Dukakis, who was perceived as competent but kinda heartless and weak and a little bit clueless. As this famous picture of him in a tank indicates he was not a war hero. But at the beginning of the Democratic primary the leading contender was actually the reverend, Jesse Jackson, who had a legitimate shot at being the first African American Democratic Presidential nominee. That would have to wait. Because instead the Democrats chose the northern, liberal governor Dukakis and paired him with Texas Senator Lloyd, I’m gonna make Dan Quayle look good, Bentsen. Which I bring up primarily to point out that Texas actually used to have Democrats. So negative campaign ads had existed before 1988 but the 1988 election took it to an entirely new level and ushered in an era of going negative in politics. Like everybody says they hate negative ads, but they also work like the Bush campaign’s efforts to make Dukakis look weak on defense and crime were brutally effective. The most infamous ad featured Willie Horton who while on furlough from prison committed rape and murder. And even though Dukakis’ Republican predecessor had actually started the furlough program, the Horton crime occurred while Dukakis was governor. The ad featured a terrifying photo of Horton and prisoners walking through a revolving door and it worked. Dukakis was regarded as a liberal who was weak on crime. In fact, it was George HW Bush who was the first to use the word liberal as an insult in American politics, which represents the larger shifts that were happening. So in retrospect, possibly the most important thing about the 1988 campaign was George Bush’s famous pledge at the Republican convention: “Read my lips, No New Taxes!” No way that’s gonna come back to bite him. So once he was President, it’s not surprising that Bush focused more on foreign policy than domestic concerns. I mean that was his background with the UN and at the CIA. But it also makes sense in the larger historical context because the Cold War actually ended during Bush’s presidency. Even though no one ever gives him credit for it. I mean the Berlin Wall came down, Poland’s military rule ended, the Velvet Revolution happened in Czechoslovakia during Bush’s watch. Let’s go to the Thought Bubble The end of the Cold War was really a failure on the part of the USSR rather than the result of successful American policies. But it left the U.S. in something of a policy limbo. I mean after all, the idea of a super-powerful malevolent Bowser Boss Soviet Union poised to destroy the American Way of Life provided a comfortable structure for all our foreign relations for almost 50 years as well as providing the reason for massive military build up and all the jobs that came with it. One positive result of the end of the Cold War was a reduction in nuclear weapons. Under Bush the U.S. and USSR negotiated and implemented the START I and START II treaties, which limited the number of warheads each country could possess to between 3,000 and 3,500. I mean that was still enough to end human life on Earth several times over but it was amazing progress. The collapse of the Soviet Union and end of the Cold War led the president to declare the dawn of a New World Order, but calling it a New World Order didn’t make foreign policy any easier. Without the Cold War to orient us foreign policy issues were much more confusing and messy. So for example, Bush kept the United States out of Yugoslavia, which disintegrated in 1991, turning into a bloodbath. But he sent troops into Somalia to help deliver food aid, resulting in the botched operation described in the movie, and book, Black Hawk Down. And then there was the foreign policy crisis that Bush handled decisively: Saddam Hussein’s invasion of Kuwait in August 1990. Bush brought the issue to the UN and ushered through a Security Council resolution that set a deadline for Saddam to leave Kuwait. When he didn’t meet the January 15, 1991 deadline, the U.S. had already put together a coalition of 34 nations ready to make him leave. America first launched a spectacular air war that destroyed much of the Iraqi defense capability. And then our technological prowess was on display for the world on CNN, which featured coverage of “smart bombs” blowing stuff up. When ground troops led by General Norman z finally moved in, they were able to defeat the Iraqi army in just 100 hours. Thanks Thought Bubble. So the Iraq war, I guess we now have to say first Iraq war, was a huge military success. America lost fewer than 300 soldiers. Iraq suffered somewhere between 1500 and 9500 killed in action.[1] And the US’s military objectives had been achieved clearly and quickly. And Bush claimed that the victory had forever banished the so called “Vietnam Syndrome,” the reluctance to use American military power for fear of becoming bogged down in another “quagmire.” Now in hindsight, if the Americans had supported Iraqi efforts to topple Saddam Hussein and build a new Iraq, we might have achieved that objective as well, but the mission under the UN resolution was to get Iraqis out of Kuwait and so that’s what we did. Bush didn’t want to take it any further. Oh it’s time for the Mystery Document? The rules here are simple I read the mystery document. I either guess the author correctly or I get shocked. “Five of the seven agree with President Bush that the war is just or at least necessary. But not one wants to fight in it. All are opposed to a draft, though a few said one might be necessary as a last resort. They said they would gladly serve in non-military public service jobs. “This might sound selfish, but I think it would be a shame to put America’s best minds on the front line,” said Jason Bell, 20, a junior English major from Elizabethtown, KY. “If we have to go, we have to go, but I think it would be a shame.” [2] Yeah, Jason Bell, that does sound selfish. Alright Stan this is from like a newspaper or magazine. I assume that you are using it to call attention to the fact that this was really the first big American military initiative without a draft. And it also reminds us that the war was not universally popular, I mean at least before it was fought, after it was fought it pretty much was. But I have no idea who actually wrote the piece in the mystery document, how would I know that? Is it a famous journalist? Is it like David Halberstam? No? David Maraniss? Who the hell is that? Does he have a Wikipedia article? Meredith does he have a Wikipedia article? Alright apparently he does have a Wikipedia article. He even won a Pulitzer Prize so congratulations sir. Ahhh! So the Gulf lifted the President Bush’s approval rating to an unheard of 89%. And in April 1991 it looked like there was no way that George HW Bush would lose his re-election bid, but he didn’t consider the domestic issues that were kind of important to Americans. We are very happy to talk about all the wars that we are fighting unless and until someone raises our taxes! So Bush wasn’t much interested in putting together a domestic agenda – he once called it “the vision thing” – and anyway he would have had a hard time getting anything through the Democratically controlled Congress. So Congress continued to pass New Deal style “liberal” legislation including expanded funding for Head Start and welfare, as well as a Family and Medical Leave bill (which Bush vetoed twice) but eventually passed nonetheless. With the Family Medical Leave Act of course America joined every other country in the world in offering paid maternity and paternity leave to new parents. What’s that? We didn’t? We still don’t? We still don’t have that? We still don’t have paid leave? Oh god.. However you are no longer allowed to be fired for 12 whole weeks while you take unpaid leave to care for your child. That’s why Stan couldn’t replace me with text-to-voice software after my daughter, Alice, was born. But in news that actually was sort of cutting edge Congress also enacted the Americans With Disabilities Act in 1991. Before I talk about the recession that ended George Bush’s presidency I want to talk about Rodney King. Because this revealed huge fissures in the American population and called into question the achievements of the rights revolution. In April 1992 an all white jury in Simi Valley found three of four policemen not guilty of beating black motorist Rodney King, even though the incident had been recorded on videotape. After the verdict, Los Angeles erupted into the deadliest riots seen in America since the New York City Draft Riots. 52 people were killed and 2,300 injured in rioting that caused $1billion in property damage. So obviously race remained a volatile issue in the U.S. It was also an issue that Bush seemed unprepared to deal with like he toured burned out LA neighborhoods but had little in the way of real comfort to offer, contributing to the perception that he was this millionaire, Ivy League-educated, Washington insider who was out of touch with regular Americas. But the biggest issue to most Americans was money. America fell into recession in 1990 and the slump lasted until 1992. It might have been caused by the end of the Cold War and the subsequent reductions in defense spending, or by Fed Chairman Alan Greenspan’s sluggish refusal to lower interest rates, or maybe the economy just needed to reset at a lower number after growing every year since 1982, or maybe macroeconomics is more complicated than who is President and sometimes people unjustly get blamed or credited for things that they had very little to do with. Regardless, 4.5 million Americans lost their jobs and the unemployment rate rose from 5.3% in 1989 to 7.5% in 1992, its highest level in almost a decade. Along with the many thousands of manufacturing workers who lost their jobs in America’s continuing de-industrialization, white-collar workers were thrown out of work, too, and college graduates, of whom there were record numbers, couldn’t find work as they came out of school. Stop me if any of that sounds familiar. One person who struggled to find a job after graduating during the Bush Recession was none other than CrashCourse writer Raoul Meyer, who after sending 100 resumes out got 3 job interviews and ended up working at a small independent school in Alabama, where he became the teacher of … Me From the Past. Now the recession was certainly bad for Bush politically, but what probably destroyed Bush’s re-election hopes was the whole taxes thing. In 1991, with tax receipts dropping and spending not slowing very much, President Bush did something that now seems unthinkable: he authorized a tax increase. And in doing this he called his conservative credentials into question. Especially in the eyes of small-government-wanting-libertarian-leaning republicans. They had never really trusted the faux Texan Bush anyway, but he had said, “Read my lips..” and they believed him but it turned out he had no lips! Now when coupled with Bush’s lukewarm support of the evangelical wing of the republican party and his running mate’s inability to spell the word potato it all prompted a primary challenge from conservative commentator Pat Buchanan. Which he beat back easily, however some of the GOP voter base, especially the evangelical Christians, stayed home on Election Day. Then there was also a third party candidate, Texas Billionaire and muppet impersonator H. Ross Perot, who won 19% of the vote (the best third-party performance since Teddy Roosevelt in 1912) All of this came together to open the door for a pudgy lad from Hope, Arkansas, who had never inhaled marijuana, and didn’t cheat on his wife except for sometimes, named William Jefferson Clinton. Looking back from today the fascinating thing about the George HW Bush administration is that it seems like a weird interruption in a larger narrative. For a couple decades we had seen increasing conservatism and rising partisanship and then suddenly George HW Bush comes along and everybody kind of works together. They didn’t always make good decisions when working together, but they did make decisions! But what’s really fascinating to me is that if you’re from Eastern Europe or China this period was one of the most important in history. Whereas if you’re American arguably the most important thing the leader of this era ever did was raise George W Bush. For better and for worse America didn’t really change that much as a result of the end of the Cold War. But we’re creeping up now on the growth of the Internet which would change the way that Americans and everyone else imagines history and everything else forever. Thanks for watching. I’ll see you next week. Crash Course is made with the help of all of these nice people and it exists because of your support at Subbable, a voluntary subscription platform where you can give whatever you want to help make Crash Course awesome. There’s also great perks and stuff, so check out Subbable and, thank you for watching Crash Course, thanks for making it possible, and as we say in my hometown, “Don’t forget to be awesome.” ________________ [1] source for these numbers is Patterson, Restless Giant p. 235. [2] From “It’s Their War, Too” by David Maraniss Washington Post, 2/11/91

US_History

Who_Won_the_American_Revolution_Crash_Course_US_History_7.txt

Hi, I’m John Green; this is Crash Course U.S. History. There are two kinds of revolutions: those where things DO change and those where things don’t change. Like, not to get all Crash Course Mathematics on you or anything, but a Revolution is a 360 degree turn, which leaves you back where you started. That’s what happened with the French Revolution, basically they just exchanged a Bourbon for a Bonaparte. What? I don’t have to say it all French-y. This is American history. And shut up French people about how if it weren’t for your support in the American Revolution, this would be the History of Southern Canada. But other revolutions, like the Industrial Revolution, actually change things. So, which was the American Revolution? Well, little of column A, little of Column B. Mr. Green, Mr. Green! Yeah, we went from a bunch of rich white guys running the show all the way to a bunch of rich white guys running the show. You’re not wrong, Me from the Past. But the 1700s were a pretty good century for rich white guys everywhere. I mean, they were running the show in Holland and Portugal and Spain, but only the United States became the country that invented baseball, the Model T, and competitive eating. So you’re right, Me from the Past, but even if the US didn’t live up to its rhetoric, that rhetoric was still powerful. And in the end whether you care more about ideas or policy defines whether you think the American Revolution really was Revolutionary. [Theme Music] All right, let’s start with the War for Independence. If you’ve been watching Crash Course, you’ll know that we’re not big on gratuitous war details. But we’re obligated to tell you something about it. The main strategy of the British in the Revolutionary war was to capture all the cities and force the colonists to surrender. And the first part of that strategy pretty much worked. They captured Boston and New York and Charleston, but all the colonists had to do was NOT QUIT. I mean, they had home-field advantage, knowledge of the terrain, easier supply lines, and Mr. Creepy Eyes down here. So while the British took the cities, the Americans, or Continentals, held onto the countryside. The most famous battle of the war was probably the battle of Trenton, where Washington was like, “I’m gonna cross the Delaware on Christmas morning.” He had a funny voice. Everybody knows he had a funny voice. It’s famous. That’s a made up fact! Don’t put it on your AP test. “What do I know about Washington? Well, I know he had a funny voice.” Washington surprised a bunch of Hessians, which was a pretty impressive victory especially since he had just come off of a string of defeats. But he wasn’t able to turn it into an all out rout, and ended up having to spend a miserable winter at Valley Forge. But remember, generals always get to eat. But the most important battle, at least in the North, was not Trenton, but Saratoga. This was a major defeat for the British, and while it’s often put forth as an example of the superiority of the Continental fighting man, the British mostly lost because of terrible generalling. The French would eventually bankrupt themselves helping us, which would lead to their own Revolution. As thanks, we named our most important food after them. In the South the country-city trend continued with the British taking Charleston but then continuing to lose smaller scale battles and be harassed by guerrilla style tactics. The key battle of the war in the south – because it was the one where the British surrendered – was at Yorktown in 1781. Lord Cornwallis made the brilliant tactical decision to station his troops on a peninsula, surrounded on three sides by water filled with French ships, and the British lost the war. So what did this all mean for actual people? Well, Americans like to think that we all pitched in together and got rid of British tyranny, and lived happily ever after. Also that the Continental army was the bravest, most loyal, and most effective fighting force in human history thanks to the leadership of George Washington. [Patriotic Rock Music] But actually, well, yeah, let’s go to the Thought Bubble. Morale among continental soldiers was often pretty low. Rations were poor and soldiers went unpaid. As Joseph Plumb Martin, a soldier from Connecticut, wrote, they felt they were, “starving in detail for an ungrateful people who did not care what became of us.” And many other colonists didn’t fight for independence; they fought with the British. Others were pacifists, like the Quakers, who often had their property confiscated when they refused to fight. And in colonial America, of course, losing property also meant losing rights. And for slaves, the so-called fight for freedom was very different than it was for Continental soldiers, because loyalty to Britain in the war could mean freedom. In 1775, British governor Lord Dunmore issued a proclamation that granted freedom to any slave who deserted his master and fought for the British. Something like 5,000 slaves took him up on the offer. And in addition, many slaves saw the revolution as chance to escape. Boston King left a cruel master and later wrote, “I determined to go to Charles-Town and throw myself into the hands of the English. They received me readily, and I began to feel the happiness of liberty, of which I knew nothing before.” 100,000 slaves are estimated to have fled to the British. Now, many slaves were returned to their masters, but more than 15,000 left the U.S. when the British did. And it’s worth remembering that the British empire abolished slavery in all of its territory by 1843 and without a civil war. Thanks, Thought Bubble. So, Native Americans were also profoundly affected by the Revolutionary War. Generally, they wanted to stay out of it, and the Colonists mostly wanted them to remain neutral, too. Like, the Continental Congress was eager to remind the Iroquois of their history of neutrality, writing: “This is a family quarrel between us and Old England. You Indians are not concerned in it. We don’t wish you to take up the hatchet against the king’s troops. We desire you to remain at home, and not join on either side, but keep the hatchet buried deep.” Right, well, many of the Iroquois fought for the British, anyway. The Oneidas joined the Patriots, fighting against the Iroquois. Sometimes there were divisions within tribes themselves. Like, with the Cherokees; younger chiefs tended to side with the British, older ones with the Americans. And it should be mentioned that, unsurprisingly, American troops were particularly brutal to American Indians who fought for the British, burning their villages and enslaving prisoners, contrary to the accepted rules of war. And, if the American revolution was really about, as Thomas Jefferson would have it, "the inalienable rights of life, liberty, and the pursuit of happiness," then the Indians were definitely the losers because they didn’t get any of those rights. So, we know slaves and Indians didn’t get much out of the Revolutionary War. How did it go for women? Yeah, not great. Some colonial women fought in the war: Deborah Sampson dressed up as a man and fought at several battles, once even pulling a bullet out of her own leg. But women didn’t get much out of the Revolution – they were basically still considered wards of their husbands. Or, if they were unmarried, saleable assets of their fathers. However, the idea of Republican Motherhood became really important. It held that for the republic to survive, it was necessary to have a well-educated citizenry. And since women were the primary educators, they themselves needed to be educated so they could, to quote Founding Father Benjamin Rush, “instruct their sons in the principles of liberty and government.” But not vote or own property. So the war didn’t end slavery, it didn’t much change the roles of women. And it didn’t displace the elite, land-owning, pasty white guy leadership of America. So what was revolutionary? Well, the ideas. A lot of which are summed up in a single sentence of the Declaration of Independence: “We hold these truths to be self evident, that all men are created equal, that they are endowed by their creator with certain inalienable rights and that among these are life, liberty, and the pursuit of happiness.” So, when the colonies became states, they all created constitutions, which opened voting to more people. While most states still had property qualifications for voting, the bar was lowered, so there were far more voters than there had been. Although they were mostly white and male, but still. Another aspect of the American revolution that was pretty revolutionary was the beginning of true religious freedom. Like, with independence, the Church of England ceased to be the Church of America. And some founders, like Jefferson, were Deists, believing that God had created the world, but then stepped away to, like, create other universes or try to build a boulder too big for him to lift. Jefferson called for a “wall of separation” between Church and State that’s best embodied in the Bill for Establishing Religious Freedom in Virginia, which Jefferson was so proud of that he had it mentioned on his tombstone. And the American revolution profoundly changed the economy, too. Like, all these new ideas of liberty led to a decline in apprenticeship and indentured servitude. And, immediately after the war, you began to see the split between the North, with its reliance on paid labor, and the South, with its reliance on slavery. Slavery was actually on the decline in the South until Eli Whitney went and invented the cotton gin in 1793, which: A. made it possible to turn a profit growing inferior American cotton, and B. reinvigorated slavery. Yay, innovation. Oh, no. It’s time for the Mystery Document! The rules are simple: Mystery Document. Get it wrong: shock pen. Get it right: WOOO. “An equality of property constantly operating to destroy combinations of powerful families, is the very soul of a republic – While this continues, the people will inevitably possess both power and freedom; when this is lost, power departs, liberty expires, and a commonwealth will inevitably assume some other form. Let the people have property, and they will have power – a power that will for ever be exerted to prevent a restriction of the press, and the abolition of trial by jury, or the abridgment of any other privilege.” Stan, why did you put communism in my Mystery Document? All right, so we’ve got a fan of wealth distribution. But, it isn’t Marx because: a) he’s not American, b) he wasn’t born. There were a bunch of far-left hippies in early America with their hemp-growing and their liberty-espousing. Ugh, I hate the shock pen. All right, I’m gonna guess that it is noted lexicographer, Noah Webster. [Dinging Noise] AH HA YES! Yes, yes, yes! Yes! NAILED IT! Yes. Stan, that was the best one ever. That was my biggest victory to date. So it’s worth remembering that some early Americans proposed a vision of liberty that sprung out of the idea of equality of property, which is very different from the way we imagine liberty today. But ideas of liberty – as diverse as they were – are really at the heart of what makes the American Revolution revolutionary. And that brings us back to slavery. The most common complaint among American high school students is that the Revolution was deeply hypocritical. I mean, how could this guy write that “All men were created equal” when he himself held slaves? And had kids with one of them. And, even crazier, American colonists, often referred to themselves as slaves because they were denied the right to have a vote in parliament about their taxation. Now, some people recognized that it was a smidge hypocritical to claim to be enslaved by British taxation while they themselves were ACTUALLY enslaving people. But very few made the leap to say that liberty should mean freedom for the slaves. One exception was James Otis of Massachusetts who wrote, concerning America’s slaves, that unless they were free, there could be no liberty: “What man is or ever was born free if every man is not?” But most of the Founders, including this guy and this guy, were the cream of the colonial elites, and so they held slaves and made arguments against abolition. Like, many historians now argue that Jefferson was trying to condemn slavery in the Declaration of Independence, but without slavery he wouldn’t have had his amazing life. I mean, if he’d been working, he couldn’t have designed Monticello or stolen all of those ideas from John Locke. And speaking of Locke, Locke equated liberty with property. And a revolution based on securing property against tyranny couldn’t very well turn around and take slaves, who after all were considered property. I mean, Jefferson once calculated that his slaves gave him a better financial return than his real estate investments. That being said, there were many and frequent protests against slavery. The most vociferous protesters were often African Americans, and in the northern states at least, their pleas were heard. Between 1777 and 1804, all states north of Maryland got rid of slavery, although most did so at a very slow pace and were careful not to deprive slave owners of the value of their property. Like, as late as 1830, there were still about 3,500 slaves in the North; and on the eve of the Civil war there were still 18 in New Jersey. NEW JERSEY. So, the number of free people of color in the U.S. skyrocketed. There were fewer than 10,000 in 1776; by 1810, there were nearly 200,000 free black Americans. So, in the end, real change came, as it usually does, not through a revolutionary event but through a revolutionary process. To me, the really novel idea that emerged from the American Revolution was of American equality. Now obviously this was (and remains) a vastly unequal social order, but I’m talking about the kind of equality that Gordon Wood described in his famous book “The Radicalism of the American Revolution”: The idea that no one American is inherently better than any other. Prior to the revolution, and certainly in Europe, there were definitely classes of superior people, usually determined by birth. I mean, people knew their place and they were expected to be deferential to their “betters.” But all that talk of freedom and inalienable rights introduced the idea that birth wasn’t destiny, and that all people should be treated with respect. And the idea that no one should be denied the opportunity to succeed because of who their parents were catalyzed change not just in America but around the world. And while the U.S. no longer leads in equality of opportunity, that early American idea that we are all equal in our capacity to reason and to work became the foundation not just for the American Revolution, but for many others that would come afterward. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller, edited by Stan and Mark Olsen. Our associate producer is Danica Johnson. The script supervisor is Meredith Danko. The show is written by my high school history teacher, Raoul Meyer, and myself. And our graphics team is Thought Bubble. If you have questions about today’s video, good news! There are historians waiting for you in comments. So ask away. Thanks for watching Crash Course and as we say in my hometown, Don’t Forget To Be Awesome.

US_History

The_Election_of_1860_the_Road_to_Disunion_Crash_Course_US_History_18.txt

Hi I’m John Green; this is Crash Course US History and today we discuss one of the most confusing questions in American history: What caused the Civil War? Just kidding it’s not a confusing question at all: Slavery caused the Civil War. Mr. Green, Mr. Green, but what about, like, states rights and nationalism, economics – Me from the Past, your senior year of high school you will be taught American Government by Mr. Fleming, a white Southerner who will seem to you to be about 182 years old, and you will say something to him in class about states rights. And Mr. Fleming will turn to you and he will say, “A state’s rights to what, sir?” And for the first time in your snotty little life, you will be well and truly speechless [Theme Music] The road to the Civil War leads to discussions of states rights...to slavery, and differing economic systems... Specifically whether those economic systems should involve slavery, and the election of Abraham Lincoln. Specifically how his election impacted slavery, but none of those things would have been issues without slavery. So let’s pick up with the most controversial section of the Compromise of 1850, the fugitive slave law. Now, longtime Crash Course viewers will remember that there was already a Fugitive Slave Law written into the United States Constitution, so what made this one so controversial? Under this new law, any citizen was required to turn in anyone he or she knew to be a slave to authorities. And that made, like, every person in New England into a sheriff, and it also required them to enforce a law they found abhorrent. So, they had to be sheriffs and they didn’t even get little gold badges. Thought Bubble, can I have a gold badge? Oh. Awesome. Thank you. This law was also terrifying to people of color in the North, because even if you’d been, say, born free in Massachusetts, the courts could send you into slavery if even one person swore before a judge that you were a specific slave. And many people of color responded to the fugitive slave law by moving to Canada, which at the time was still technically an English colony, thereby further problematizing the whole idea that England was all about tyranny and the United States was all about freedom. But anyway the most important result of the fugitive slave law was that it convinced some Northerners that the government was in thehands of a sinister “slave power.” Sadly, slave power was not a heavy metal band or Britney Spears’s new single or even a secret cabal of powerful slaves, but rather a conspiracy theory about a secret cabal of pro-slavery congressmen. That conspiracy theory is going to grow in importance, but before we get to that let us discuss Railroads. Underrated in Monopoly and underrated in the Civil War. Let’s go to the Thought Bubble. Railroads made shipping cheaper and more efficient and allowed people to move around the country quickly, and they had a huge backer (also a tiny backer) in the form of Illinois congressman Stephen Douglas, who wanted a transcontinental railroad because 1. he felt it would bind the union together at a time when it could use some binding, and 2. he figured it would go through Illinois, which would be good for his home state. But there was a problem: To build a railroad, the territory through which it ran needed to be organized, ideally as states, and if the railroad was going to run through Illinois, then the Kansas and Nebraska territories would need to become state-like. So Douglas pushed forward the Kansas Nebraska Act in 1854. The Kansas-Nebraska Act formalized the idea of popular sovereignty, which basically meant that (white) residents of states could decide for themselves whether the state should allow slavery. Douglas felt this was a nice way of avoiding saying whether he favored slavery; instead, he could just be in favor of letting other people be in favor of it. Now you’ll remember that the previously bartered Missouri Compromise banned slavery in new states north of this here line. And since in theory Kansas or Nebraska could have slavery if people there decided they wanted it under the Kansas-Nebraska Act despite being north of that there line, this in practice repealed the Missouri Compromise. As a result, there was quite a lot of violence in Kansas, so much so that some people say the Civil War really started there in 1857. Also, the Kansas Nebraska Act led to the creation of a new political party: The Republicans. Yes, those Republicans. Thanks, Thought Bubble. So, Douglas’s law helped to create a new coalition party dedicated to stopping the extension of slavery. It was made of former Free-Soilers, Northern anti-slavery Whigs and some Know-Nothings. It was also a completely sectional party, meaning that it drew supporters almost exclusively from the free states in the North and West, which, you’ll remember from like, two minutes ago, were tied together by common economic interests, and the railroad. I’m telling you, don’t underestimate railroads. By the way, we are getting to you, Dred Scott. And now we return at last to “slave power.” For many northerners, the Kansas Nebraska Act which repealed the Missouri Compromise, was yet more evidence that Congress was controlled by a sinister “slave power” group doing the bidding of rich plantation owners. Which, as conspiracy theories go, wasn’t the most far-fetched. In fact, by 1854, the North was far more populous than the South – it had almost double the South’s congressional representation – but in spite of this advantage, Congress had just passed a law extending the power of slave states, and potentially – because two new states meant four new senators – making the federal government even more pro-slavery. And to abolitionists, that didn’t really seem like democracy. The other reason that many northerners cared enough about Kansas and Nebraska to abandon their old party loyalties was that having them become slave states was seen as a threat to northerner’s economic self-interest. Remember the west was seen as a place where individuals – specifically white individuals – could become self-sufficient farmers. As Lincoln wrote: “The whole nation is interested that the best use be made of these territories. We want them for the homes of free white people. They cannot be, to any considerable extent, if slavery is planted within them. New Free States are places for poor people to go to and better their condition.” So, the real question was: Would these western territories have big slave-based plantations like happened in Mississippi? Or small family farms full of frolicking free white people, like happened in Thomas Jefferson’s imagination? So the new Republican party ran its first presidential candidate in 1856 and did remarkably well. John C. Fremont from California picked up 39% of the vote, all of it from the North and West, and lost to the Democrat James Buchanan, who had the virtue of having spent much of the previous decade in Europe and thus not having a position on slavery. I mean, let me take this opportunity to remind you that James Buchanan’s nickname was The Old Public Functionary. Meanwhile, Kansas was trying to become a state by holding elections in 1854 and 1855. I say trying because these elections were so fraudulent that they would be funny, except that everything stops being funny like 12 years before the Civil War and doesn’t get really funny again until Charlie Chaplin. Ah, Charlie Chaplin, thank you for being in the public domain and giving us a much-needed break from a nation divided against itself, discovering that it cannot stand. Right so part of the Kansas problem was that hundreds of so called border ruffians flocked to Kansas from pro-slavery Missouri to cast ballots in Kansas elections, which led to people coming in from free states and setting up their own rival governments. Fighting eventually broke out and more than 200 people were killed. In fact, in 1856, pro-slavery forces laid siege to anti-slavery Lawrence, Kansas with cannons. One particularly violent incident involved the murder of an entire family by an anti-slavery zealot from New York named John Brown. He got away with that murder but hold on a minute, we’ll get to him. Anyway, in the end Kansas passed two constitutions because, you know, that’s a good way to get started as a government. The pro-slavery Lecompton Constitution was the first that went to the U.S. Congress and it was supported by Stephen Douglas as an example of popular sovereignty at work, except that the man who oversaw the voting in Kansas called it a “vile fraud.” Congress delayed Kansas’ entry into the Union (because Congress’s primary business is delay) until another, more fair referendum took place. And after that vote, Kansas eventually did join the U.S. as a free state in 1861, by which time it was frankly too late. All right so while all this craziness was going on in Kansas and Congress, the Supreme Court was busy rendering the worst decision in its history. Oh, hi there, Dred Scott. Dred Scott had been a slave whose master had taken him to live in Illinois and Wisconsin, both of which barred slavery. So, Scott sued, arguing that if slavery was illegal in Illinois, then living in Illinois made him definitionally not a slave. The case took years to find its way to the Supreme Court and eventually, in 1857, Chief Justice Roger B. Taney, from Maryland, handed down his decision. The Court held that Scott was still a slave, but it went even further, attempting to settle the slavery issue once and for all. Taney ruled that black people “had for more than a century before been regarded as beings of an inferior order, and altogether unfit to associate with the white race, either in social or political relations; and so far inferior that they had no rights which the white man was bound to respect; and that the negro might justly and lawfully be reduced to slavery for his benefit.” So...that is an actual quote from an actual decision by the Supreme Court of the United States of America. Wow. I mean, Taney’s ruling basically said that all black people anywhere in the United States could be considered property, and that the court was in the business of protecting that property. This meant a slave owner could take his slaves from Mississippi to Massachusetts and they would still be slaves. Which meant that technically, there was no such thing as a free state. At least that’s how people in the north, especially Republicans saw it. The Dred Scott decision helped convince even more people that the entire government, Congress, President Buchanan, and now the Supreme Court, were in the hands of the dreaded “Slave Power.” Oh, we’re going to do the Mystery Document now? Stan, I am so confident about today’s Mystery Document that I am going to write down my guess right now. And I’m going to put it in this envelope and then when I’m right I want a prize. All I ever get is punishment, I want prizes. OK. The rules here are simple. I guess the author of the Mystery Document. I already did that. And then I get rewarded for being right. Alright total confidence. Let’s just read this thing. And then I get my reward. “I look forward to the days when there shall be a servile insurrection in the South, when the black man … shall assert his freedom and wage a war of extermination against his master; when the torch of the incendiary shall light up the towns and cities of the South, and blot out the last vestige of slavery. And though I may not mock at their calamity, nor laugh when their fear cometh, yet I will hail it as the dawn of a political millennium.” I was right! Right here. Guessed in advance. John Brown. [buzzing] What? STAN! Ohio Congressman Joshua Giddings? Seriously, Stan? AH! Whatever. I’m gonna talk about John Brown anyway. In 1859, John Brown led a disastrous raid on the federal arsenal at Harpers Ferry, hoping to capture guns and then give them to slaves who would rise up and use those guns against their masters. But, Brown was an awful military commander, and not a terribly clear thinker in general, and the raid was an abject failure. Many of the party were killed and he was captured. He stood trial and was sentenced to death. Thus he became a martyr to the abolitionist cause, which is probably what he wanted anyway. On the morning of his hanging, he wrote, “I, John Brown, am now quite certain that the crimes of this guilty land will never be purged away but with blood.” Well, he was right about that, but in general, any statement that begins “I-comma-my-name” Meh. And, so the stage was set for one of the most important Presidential elections in American history. Dun dun dun dun dun dahhhhh. In 1860, the Republican Party chose as its candidate Abraham Lincoln, whose hair and upper forehead you can see here. He’d proved his eloquence, if not his electability, in a series of debates with Stephen Douglas when the two were running for the Senate in 1858. Lincoln lost that election, but the debates made him famous, and he could appeal to immigrant voters, because he wasn’t associated with the Know Nothings. The Democrats, on the other hand, were – to use a historian term – a hot mess. The Northern wing of the party favored Stephen Douglas, but he was unacceptable to voters in the deep South. So Southern Democrats nominated John C. Breckinridge of Kentucky, making the Democrats, the last remaining truly national party, no longer truly a national party. A third party, the Constitutional Union Party, dedicated to preserving the Constitution “as it is” i.e. including slavery, nominated John Bell of Tennessee. Abraham Lincoln received 0 votes in nine American states. But he won 40% of the overall popular vote, including majorities in many of the most populous states, thereby winning the electoral college. So, anytime a guy becomes President who literally did not appear on your ballot, there is likely to be a problem. And indeed, Lincoln’s election led to a number of Southern states seceding from the Union. Lincoln himself hated slavery, but he repeatedly said that he would leave it alone in the states where it existed. But the demographics of Lincoln’s election showed Southerners and Northerners alike that slave power – to whatever extent it had existed –was over. By the time he took office on March 1, 1861, seven states had seceded and formed the Confederate States of America. And the stage was set for the fighting to begin, which it did, when Southern troops fired upon the Union garrison at Fort Sumter in Charleston harbor on April 12, 1861. So, that’s when the Civil War started, but it became inevitable earlier – maybe in 1857, or maybe in 1850, or maybe in 1776. Or maybe in 1619, when the first African slaves arrived in Virginia. Because here’s the thing: In the Dred Scott decision, Chief Justice Taney said that black Americans had “no rights which the white man was bound to respect.” But this was demonstrably false. Black men had voted in elections and held property, including even slaves. They’d appeared in court on their own behalf. They had rights. They’d expressed those rights when given the opportunity. And the failure of the United States to understand that the rights of black Americans were as inalienable as those of white Americans is ultimately what made the Civil War inevitable. So next week, it’s off to war we go. Thanks for watching. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The show is written by my high school history teacher, Raoul Meyer, and myself. Our associate producer is Danica Johnson. And our graphics team is Thought Café. Usually every week there’s a libertage with a caption, but there wasn’t one this week because of stupid Chief Justice Roger Taney. However, please suggest captions in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course US History and as we say in my hometown of Nerdfighteria, don’t forget to be awesome.

US_History

The_Civil_War_Part_I_Crash_Course_US_History_20.txt

Hi I’m John Green this is Crash Course US History and today we come at last to the Civil War, the conflict that in many ways created a nation. So here’s what you won’t be getting today. We will not be describing battles and tactics. If that’s your bag, might I suggest Ken Burns or if you prefer books, like 1000 authors, my favorites being James McPherson and Shelby Foote. And 2. We won’t be bashing and/or praising Abraham Lincoln very much, although we do have multiple Lincolns here because we’ve heard that’s good for ratings. I mean, to watch or read certain accounts, you would think that the Civil War was a lengthy chess game played by Abraham Lincoln against his cunning opponent Abraham Lincoln, but of course there were other people involved. We are going to quote a fair bit of Lincoln, though, because, you know, that won Tony Kushner an Academy Award nomination. 3. We won’t be claiming that the Civil War was somehow secretly about something other than slavery, because that is just so early 20th century. And 4. There will not be a lot of jokes today because hahaha 700,000 people died. Mr. Green, actually only 680,000 people died. Yeah, it depends on how you count, you sniveling little ghoul. But recent estimates are between 680,000 and 800,000 total casualties. Deadlier for Americans than the American Revolution, World War I, World War II, and Vietnam combined. [Theme Music] So let’s start with some basic facts about the American Civil War. 1861 to 1865, which corresponded with the presidency of Abraham Lincoln. The Union, or more colloquially the North, fought against the forces of the Confederate States of America, or the South. Sometimes people call the Union ‘the blue” and the confederates “the gray,” but in fact the uniforms weren’t very uniform, they were all different kinds of color. And also, with all that dirt and blood, they were all just brown. Alright, let’s go to the Thought Bubble. You’ll notice from this map that not all the states that held slaves were part of the Confederacy. The border states of Kentucky, Missouri, Delaware and Maryland allowed slavery and never left the United States. All of these border states were critical to the Union--Maryland was north of the nation’s capitol in Washington D.C.; Kentucky controlled the Ohio River; Missouri was the gateway to the West; Delaware actually wasn’t that important. So none of that should be particularly controversial, unless you’re from Delaware, but the causes of the war, that’s another story. The Civil War was about slavery--actual historians will back me up on this, like David Goldfield, who wrote, “Both Northerners and Southerners recognized slavery as the immediate cause of the Civil War.” Also, Lincoln said in his second inaugural address, “One eighth of the whole population were colored slaves, not distributed generally over the Union, but localized in the southern part of it. These slaves constituted a peculiar and powerful interest. All knew that this interest was, somehow, the cause of the war.” That said, in comments lots of people will be like, the war was about agriculture versus industry, or the states’ rights to protect themselves from the tyranny of a big federal government, but if it were REALLY about that, the Civil War would’ve started during the Nullification crisis in the 1830s, when-- as I’m sure you’ll remember--Andrew Jackson said that South Carolina couldn’t declare a federal tariff null in their state. Why didn’t that cause a Civil War? The Confederate government passed the first conscription act in American history, implemented national taxes, created a national currency, and had a government bureaucracy of about 70,000 people, more than the federal bureaucracy in Washington D.C. Thanks, Thought Bubble. That said, in the beginning of the war, Lincoln deliberately tried to downplay the slavery angle, arguing that the war was only about preserving “the Union.” But the war was also about religion, for both sides. As David Goldfield put it, “In protecting the Revolutionary ideals, northerners would preserve God’s plan to extend democracy and Christianity across an unbroken continent and around the world. Southerners welcomed a war to create a nation more perfect in its fealty to God than the one they had left.” But it’s also important to remember that regular soldiers often had more prosaic reasons for going off to fight, as you will eventually learn when you are forced to read The Red Badge of Courage, Goldfield tells the story of one Alabamian who enlisted only after his girlfriend mailed him a dress and told him he should start wearing if he wasn’t willing to go fight. And for Northerners, Union, religion and an end to slavery mixed together to form a potent rationale for war. It’s summed up nicely by Julia Ward Howe’s words to the song that would become famous as the Battle Hymn of the Republic: “As he died to make men holy, let us die to make men free.” You thought I was going to sing, but you were wrong. So spoiler alert the Union won the war, which in a sense was unsurprising, because they had massive advantages: For starters, they had many more people, approximately 22 million as compared to 9 million in the South, of whom 3.5 million were slaves and therefore unlikely to be sympathetic to the Southern cause. Also, the north manufactured more than 90% of all goods in America; its factories turned out 17 times more textiles than the South, 30 times more shoes and boots, 13 times more iron, and 32 times more firearms. Plus, at the outbreak of the war the North had twenty thousand miles of railroad compared with the South’s ten thousand. This made it easier for the Union to move its army, which over the course of the war enlisted more than 2 million men, compared with 900,000 for the Confederacy. Even northern agriculture was also more productive, taking greater advantage of mechanization than southern farmers did. Really the only advantage the south had was better leaders, like most of the tactically famous generals of the Civil War, Robert E. Lee, Stonewall Jackson, J. E. B. Stewart, etc., were Southerners. And also, by the way, they all had great last words. Lee said “Strike the tent,” Stonewall Jackson said “Let us cross over the river and rest under the shade of those trees,” and JEB Stuart after being mortally wounded in battle said to his close friend and lieutenant, “Honey-bun, how do I look in the face?” Famous Union general Ulysses Grant’s last word was “Water,” which isn’t near so good, but he said that last word after having survived the war and getting to be, like, President of the United States and stuff. Right but anyway, this all raises an interesting question: Was the result of the war a foregone conclusion? The Confederacy had to create a nation from scratch and build national unity among people who were committed to the autonomy of their individual home states. So that’s a problem. And, then there was the issue of overcoming class conflicts, especially when the ruling class was often exempted from actually fighting in the war. But when you put aside all that nation-building stuff and just focus on the actual fighting of the war, the question of the union’s inevitable win becomes much trickier. Some have argued that all the Confederacy really to do was outlast the Northern efforts to bring them back into the Union, like Washington had to do against the British. And the idea was that the war of attrition would eventually wear down northern resolve. But, there were two problems with this theory. First, the North had such superiority in its resources that it would take a long time to wear down. Secondly, fighting a war of attrition would be costly to the South, as well and their resources would be depleted long before the North’s. Oh it’s time for the Mystery Document? The rules here are simple. Woah! That was intense. I try to identify the author of the Mystery Document. If I am right, I do not get shocked, but I’m never right because Stan makes it too hard. Alright, let’s see what we’ve got today. “I therefore determined, first, to use the greatest number of troops practicable against the armed force of the enemy, preventing him from using the same force at different seasons [and] second, to hammer continuously against the armed force of the enemy and his resources, until by mere attrition … there should be nothing left to him but submission.” Okay so the strategy of attrition was a Confederate strategy. But, Stan is a jerk. But it talks about the enemy AND HIS RESOURCES, which was kind of a Union focus. And more importantly, it talks about preventing him from using the same force at different seasons. That makes me think it is a Union general. Final answer Ulysses S. Grant. OH HOW DO YOU LIKE THEM APPLES. Grant was different from previous Union generals in that he was willing to sustain enormous casualties in pursuit of his goal to wear down the South. Because of this, Grant was branded a butcher, like he was willing to weather incredible losses including the 52,000 men -- 41% of his army -- who were injured or killed at the battles of the Wilderness and Cold Harbor. But his grim determination not just to defeat but to destroy his opponent is what made Grant one of the first truly modern generals and also the most successful leader the Union found. So, Grant’s brutal strategy coupled with the vast superiority in Northern resources suggests that the outcome of the Civil war really was inevitable, but it also points to some of the reasons to be cautious about that conclusion. First off, it took three years before the Union actually fully adopted Grant’s strategy, and between 1861 and 1864 it was possible that Southern victories would eventually force the Union to give in. I mean, the Union lost a lot of battles in the first two years, largely due to ineffective General-ing and nothing saps a nation’s motivation for war like losing. Now, some argue that the North had superior motivation to prosecute the war because they had God on their side and they were against slavery, but that’s also pretty problematic. I mean, for many men who joined the federal army, a war to end slavery had very little appeal, especially poor enlistees who might be afraid that newly-freed slaves would compete with them for jobs. Also, while we are correct in considering slavery unjust, southerners who took up arms for the Confederacy saw themselves as engaged in a fight for their own freedom, rather than a fight to protect slavery. The truth is, when it comes to fighting, motivation is a very tricky business, and I’m most comfortable agreeing with James McPherson who argued that motivation waxes and wanes with victory, and that the outcome of the war was contingent on a number of turning points. And we’re just gonna discuss two of the most important: July 1863 and August 1864. July 1863 saw two of the most important Union victories in the whole war. In the western theater, General Grant laid siege to and captured Vicksburg Mississippi, thus giving the federals control of the lower Mississippi river. I mean, by then, the North already had New Orleans, which made it pretty much impossible for the Confederates to ship cotton or anything else along the Mississippi River. After that, Grant was able to turn his attention to the east with the aforementioned hammering of the enemy and their resources. More famously, especially in the eastern part of the United States, the first three days of July 1863 saw the battle of Gettysburg in Pennsylvania. This was General Lee’s furthest major offensive in the north and had he won the battle it is likely that panic would have set in in places like Philadelphia and maybe even New York. Actually panic did overcome New York in draft riots that killed more than 100 people and only ended after troops from Gettysburg were called in. I’m not going to go into detail about either of these battles, but they shifted the tide of the war in favor of the North, although not enough to bring the war to a quick end. Confederate forces would never again threaten a northern city. August 1864 saw another turning point that really spelled the doom of the Confederacy, and that was when Union general Sherman took Atlanta. Atlanta was a railroad hub and manufacturing center but its capture was more significant politically than militarily because it happened close to the election of 1864. And that American election was really the last time that the Confederate states of America could have won the Civil War. It’s easy to forget this, but Lincoln actually had to run for reelection during the Civil War, and by the summer of 1864 the war was pretty unpopular and it looked like Lincoln might lose. The capture of Atlanta changed public opinion about Lincoln and meant it that his Democratic opponent and former top general George McClellan didn’t stand a chance of winning, which was really significant for the war because Lincoln was committed to ending it with a Union victory and McClellan, meh. I think it says a lot about American history that in the end the war’s outcome was insured not just by military victories but by a political one. Next week, we’ll examine the effects of the Civil War and the enduring questions that have arisen out of it, such as who, exactly, freed the slaves? But, until then, thanks for watching. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our show is written by my high school history teacher, Raoul Meyer, and myself. Our graphics team is Thought Café. And our associate producer is Danica Johnson, also responsible for felt Abraham Lincoln. If you want to suggest captions for the libertage, you can do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

History_of_the_4th_of_July_Crash_Course_US_History_Special.txt

Hi, I'm John Green, this is Crash Course US History, and today we're gonna talk about July 4th, which in the United States is known as Independence Day. This is the day that Americans celebrate our Independence from Great Britain by doing what we do best: blowing stuff up, offering significant discounts on mattresses, driving long distances for uncomfortable family interactions, and eating a lot of grilled meat. [intro music] Right, so the story goes that the founders of this nation signed the Declaration of Independence on July 4th 1776, parting ways with King George to found the freest, finest nation on the face of the Earth. [Libertage: America: It's our birthday!] Yeah, except the continental congress actually approved a resolution of independence on July 2nd. The Lee Resolution was proposed by Richard Henry Lee of Virginia in June of 1776, and was a simple legal declaration of separation from England. John Adams got so excited about it that he wrote to his wife Abigail, The second day of July 1776 will be the most memorable epic in the history of America. I am apt to believe that it will be celebrated by succeeding generations as the great anniversary festival. It ought to be solemnized with pomp and parade, with shows, games, sports, guns, bells, bonfires, and illuminations from one end of this continent to the other, from this time forward forevermore. So what happened on the fourth? Well, not that much, actually. The Declaration of Independence was the formal announcement of independence and its text was approved on July 4th 1776. The fancy parchment version with all the pretty calligraphy wouldn't get drawn up until July 19th, and most members of the Congress signed the Declaration on August 2nd. Adams may have been wrong about the date, but he was definitely right about the celebration. Americans started celebrating the 4th of July as early as 1777, and, as Adams predicted, the holiday was observed with feasts, 13 gun salutes, and fireworks. Why don't we call them illuminations anymore, Stan? Y'know, people can say, like, "oh, we put a man on the moon and we can refrigerate our meat now" but I miss the old days! Anyway, in 1778 George Washington celebrated the Fourth by giving his soldiers a double ration of rum, and also there was much more shooting than usual. But while the people celebrated the anniversary from the beginning, the federal government took its sweet time in formalizing the holiday. Independence day became an unpaid holiday for federal employees in 1870 and became a paid day off in 1938. 4th of July observances have evolved over the years, but they generally involve patriotic displays, including decorations, fire, and explosives. Essentially, we celebrate our independence by having a one-day fake war each summer. Huzzah! [pops party popper] Aw, freedom-tenitis. Early observances were marked by huge bonfires, but the litigious nature of modern American society has greatly reduced the number of bonfires. However, we still have a lot of fireworks on the 4th of July. Many cities and towns across the country sponsor fireworks displays on the Fourth; New York's fireworks display is the largest in the nation. Despite the grandeur of these officially sanctioned displays, many, many individuals across the nation feel the need to blow stuff up on their own in their yards, because, y'know, they have all of their fingers and that seems like too many. Many states have restricted the sale and personal use of fireworks freedom haters! but would-be patriots find their way to marginal neighborhoods every year to buy fireworks out of the backs of vans. In my neighborhood, these personal pyrotechnic celebrations started about mid-May and continue well into August, causing my dog to have a very stressful summer. John Adams's prediction about gunfire holds true as well. On most military bases, 50 gunshots -- one for each state -- are fired at noon on July 4th as a salute to the Union. And in Stan's neighborhood, where you go to buy the fireworks out of the backs of vans, celebratory gunfire is common throughout the year. Here is an actual bullet hole in Stan's actual window. Stan, can we get the Libertage again? [Libertage: America - seriously, a bullet came into my house] In the 19th century, many ex-presidents celebrated the 4th of July by dying. Both Thomas Jefferson and John Adams died July 4th 1826, and James Madison died on the 4th of July in 1831. Calvin "Silent Cal" Coolidge was born on July 4th 1876, but as he was never demonstrably alive, no one really cares that much. Finally, lest we forget, Americans also celebrate the 4th of July by eating. Millions of Americans host cook-outs to celebrate independence, and the greatest spectacle in professional sports happens every year on the 4th of July. I am referring of course to the Nathan's Hot Dog Eating Contest. The current world record is 68 HDBs, hot dogs and buns, in ten minutes. That means the world record holder, Joey Chestnut, consumed about 28,500 calories in ten minutes, but don't worry -- he threw it up shortly thereafter. So happy 4th of July from all of us here at Crash Course. We may not celebrate American independence on the right day, but I would argue that, in spite of it all, it's still worth celebrating. Thanks for watching; I'll see you next week. [pops party popper] Boop! Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our associate producer is Danica Johnson. Today's show was written by Stan. And our graphics team is Thought Cafe. Every week, there's a new caption for the Libertage. If you'd like to suggest one, you can do so in comments, where you can also ask questions about today's video that will be answered by our team of historians. Thank you for watching Crash Course, and as we say in my home town, don't forget to be awesome.

US_History

Progressive_Presidents_Crash_Course_US_History_29.txt

Hi, I’m John Green, this is CrashCourse U.S. History and today we’re going to finish our discussion of Progressivism, and indulge in a bit of “great man” history. Mr. Green, Mr. Green! Great man history, huh? Well I was born on a sunny, summer morning in 197-- Yeah you’re not great, Me from the Past. Also, you’re a boy not a man, and the only historically significant thing you ever participated in was a brilliant senior prank that wasn’t even your idea. However, 39 of our 43 presidents were, at least arguably, great men and today we’ll be talking about three of them. It will be kind of like a Jefferson vs. Hamilton for the 20th century, except not like that at all. But there will be a canal, and TWO people get shot. Intro So, as we saw in CrashCourse World History, national governments were on the rise from the middle of the 19th century until basically now. And in the U.S., Corporations became national and then, by the twentieth century, international. Like, the British East India Company was kind of an international corporation, but it wasn’t the same as Coca-Cola, although they did both deal in narcotics. And this mania for nationalization even affected sports. Like, in baseball, the National league and the American league were formed and in 1903 they played the first inaccurately named World Series. I’m sorry, was Botswana invited? Then it’s not a World Series. Anyway, the rise of a strong, national government was seen as an alternative to people’s lives being controlled by provincial city and state governments or by ever-growing corporations. Like, Herbert Croly, editor of the New Republic, thought that to achieve the Jeffersonian democratic self-determination ideal of individual freedom, the country needed to employ Hamiltonian government intervention in the economy. And he wasn’t the only one who believed that. Okay, so in 1901, 42-year-old Theodore Roosevelt became the youngest American president ever after William McKinley was assassinated by Leon Czolgosz. Czol--? Czolg--? Czol--? Hold on. “Czolgosz. Polish.” Czolgosz? Czolgosz? His name was Leon Chuckles? Man, Leon Chuckles was a real barrel of laughs for an anarchist. Usually they’re very serious. Right, so Leon Chuckles paved the way for Teddy Roosevelt, who in many ways the model of the 20th century president. He was very engaged in both domestic and foreign policy and he set the political agenda for the whole country. His political program, the Square Deal, aimed to distinguish good corporations that provided useful products and services at fair prices from evil corporations that existed just to make money. That is hilarious. A corporation that doesn’t exist just to make money. That’s fantastic, Teddy. Everybody knows that corporations are just inherently greedy people, but they are people. Roosevelt felt it was the federal government’s responsibility to regulate the economy directly and to break up power of wealthy corporations, and he used the Sherman Act to prosecute bad trusts such as the Northern Securities Company, which was a holding company created by J.P. Morgan that directed three major railroads and monopolized transport. And that did not make J.P. Morgan a happy bunny. Thank you for that, Stan. That’s, that’s wonderful. Shockingly, the legislative and executive branches managed to work together and Congress passed some actual legislation, including the Hepburn Act of 1906, which gave the Interstate Commerce Commission the power to regulate railroad rates and examine their company books. And Roosevelt was also a conservationist. He wanted to preserve the environment from economic exploitation, probably so that there would be plenty of animals for him to hunt with his big stick while he walked softly. Having appointed noted progressive Gifford Pinchot head of the forest service, millions of acres were set aside for new, highly managed national parks reflecting the progressive idea that experts could manage the world. But then in 1908, Teddy Roosevelt decided to go elephant hunting instead of running for re-election and he picked William Howard Taft to be his successor, but the man who became our largest president massively disappointed Roosevelt. When I say “our largest,” by the way, I don’t mean our greatest. I mean our largest. Taft was a pretty hard-core trust-buster who ordered the prosecution that broke up Standard Oil in 1911, but he didn’t see big business as bad unless the corporations stifled competition. He also supported the 16th amendment, allowing Congress to pass an income tax, and that paved the way for the 18th amendment, Prohibition, because with an income tax, the federal government didn’t have to rely on liquor excise taxes. So, why didn’t Roosevelt like Taft? Well, not only was Taft more conservative than most progressives, he also fired Pinchot in 1910. And Roosevelt was so frustrated with Taft that he actually challenged the incumbent president for the Republican nomination in 1912. Which Roosevelt lost, but he didn’t let it drop. He founded his own Progressive Party, called the Bull Moose Party so that he could run again. So, the election of 1912 featured four candidates: Taft; Teddy Roosevelt for the Bull Moose Party; Eugene Debs, for the Socialist Party; and Democrat Woodrow Wilson. It’s worth noting that in contemporary American political discourse, all four of these people would have been seen as somewhere between crazy liberals and actual communists. So Eugene Debs, from right here in my home state of Indiana, did not support the Socialist Party’s goal of abolishing capitalism, but he ran on a platform that included public ownership of railroads and banks, and laws limiting work hours. And running on the socialist ticket, Debs won 6% of the vote, which was, to quote another president, “not bad.” But the election of 1912 turned out to be a contest between Wilson and Roosevelt’s competing views over the dangers of increasing government power and economic concentration. Wilson claimed, “Freedom today is something more than being let alone. The program of government must in these days be positive, not negative merely.” That’s just not good grammar, sir. His program, called New Freedom, was supposed to reinvigorate democracy by restoring market competition and preventing big business from dominating government. It included stronger anti-trust laws and policies to encourage small businesses. Roosevelt’s answer to New Freedom was a program he called New Nationalism, because, of course, in election years all things are new. Roosevelt recognized the inevitability of big business and hoped to use government intervention to stop its abuses. New Nationalism included heavy taxes on personal and corporate fortunes and greater federal regulation of industries. So, the Bull Moose Party platform was in some ways a vision of a modern welfare state, it called for: Women’s suffrage Federal regulation National labor and health legislation for women and children Eight hour days and living wage for all workers National systems of social insurance for health, unemployment, and old age What are we, Canada? God, I wish we were Canada...You weren’t recording that, were you, Stan? Roosevelt thought his party’s platform was one of the most important documents in the history of mankind, and Americans agreed, they supported him and elected him in a landslide. Oh wait, no they didn’t. Instead, he lost. And also, a guy shot him at one of his campaign stops, that’s shooting #2. Roosevelt however survived and even went on to make the speech after being shot. What happened in the election is that Taft and Roosevelt split the Republican vote, leaving Woodrow Wilson president with a mere 42% of the popular vote, giving us our only democratic president between 1896 and 1932. Oh, it’s time for the mystery document? The rules here are simple. I guess the author of the mystery document. If I’m wrong, I get shocked by the shock pen, which many of you insist is fictional, but I promise, it’s not. “The two things we are fighting against, namely, excessive tariffs and almost universal monopoly, are the very things that these two branches of the Republican party both decline to combat. (...) They intend to accept these evils and stagger along under the burden of excessive tariffs and intolerable monopolies as best they can through administrative commissions. I say, therefore, that it is inconceivable that the people of the United States, whose instinct is against special privilege and whose deepest convictions are against monopoly, should turn to either of these parties for relief when these parties do not so much as pretend to offer them relief.” Alright, it’s definitely about the 1912 election. It talks about the Republican party being split into two parts, so it’s by a democrat. Or a socialist, but probably a democrat judging from the Mystery Document itself. You always make it hard, Stan. So it’s not going to be Woodrow Wilson because that would be obvious, but I do not know the names of any other prominent democrats, so I am going to guess Woodrow Wilson. YES? Get in! So, with its stirring anti-tariff, anti-monopoly, do not pass GO, do not collect $200 stance, New Freedom won out, and because the Democrats also controlled Congress, Wilson was able to implement this program. The Underwood Tariff reduced import duties and after the ratification of the 16th amendment, Congress imposed a graduated income tax on the richest 5% of Americans. Other legislation included the Clayton Act of 1914, which exempted unions from antitrust laws and made it easier for them to strike; the Keating-Owen Act, which outlawed child labor in manufacturing; and the Adamson Act which mandated an eight hour workday for railroad workers. If Wilson’s New Freedom sounds a lot like Roosevelt’s New Nationalism, that’s because they ended up being pretty similar. Wilson engaged in less trust busting than expected, and more regulation of the economy. Wilson didn’t institute a national system of health and unemployment insurance, but he did expand the powers of the Federal Trade Commission to investigate and prohibit unfair monopolistic practices. He also supported the founding of the Federal Reserve System in 1913, which gave the government much more control over banks in response to the Panic of 1907 where the U.S. had to be bailed out by J.P. Morgan. Fear not, big banks, the government will bail you out in due time. So, under Roosevelt, and Wilson, and to a lesser extent Taft, Progressivism flourished domestically, but it also became an international phenomenon as presidents expanded national government power outside the country’s border, mostly in the Western Hemisphere. Like, between 1901 and 1920, U.S. marines landed in Caribbean countries over 20 times, usually to create a more friendly environment for American businesses, but sometimes just to hang out on the beach. And this points to an interesting contradiction, Progressive presidents were very concerned about big business as a threat to freedom in the United States, but in Latin America and the Caribbean, they weren’t that concerned about freedom at all. Teddy Roosevelt especially was much more active in international diplomacy than his predecessors. He was the first president to win the Nobel Peace prize, for instance, for helping to negotiate the Treaty of Portsmouth that ended the Russo Japanese War in 1905. You may be familiar with his motto, “Speak Softly and Carry a Big Stick” – which essentially meant “the U.S. will intervene in Latin America whenever we want.” And probably the most famous such intervention was the building of the Panama Canal. It featured feats of engineering and succeeding where the French had failed...Stan, these are my favourite things! Let’s go to the Thought Bubble. The way we got the 10 mile wide canal zone wasn’t so awesome. In 1903, Panama was part of Colombia but the U.S. encouraged Philippe Bunau-Varilla to lead an uprising. Bunau-Varilla, a representative of the Panama Canal Company, was so grateful after the U.S. sent a gunboat to ensure that the Colombian army couldn’t stop him that he signed a treaty giving the U.S. the right to build and operate the canal and sovereignty over newly independent Panama’s Canal Zone, which we gave up in 2000 after enjoying nearly 100 Years of sovereignty thanks to Carter’s stupid altruism. Roosevelt also added the Roosevelt Corollary to the Monroe Doctrine, the 1823 statement that the U.S. would defend independent Latin American states from European intervention. Now, according to Roosevelt, we would wield our big stick like a policeman waving around a nightstick exercising an “international police power” over the western hemisphere. In practice, this meant using American troops to ensure that Latin American countries were stable enough for Americans to invest there. Like, in 1904 we seized the customs house in the Dominican Republic to make sure that they paid their debts to investors, then by “executive agreement” American banks got control of the DR’s finances. Roosevelt also encouraged investment by the United Fruit Company in Honduras and Costa Rica, helping to turn those nations into Banana Republics. No, not the store, Thought Bubble. Yes. Taft, on the other hand, maybe because of his experiences as governor of the Philippines, was less eager to wave America’s Big Stick. He emphasized loans and economic investment as the best way to spread American influence in a policy that came to be known as Dollar Diplomacy. Ultimately, Dollar Diplomacy was probably more effective, but it seemed weak to many people in contrast to Roosevelt’s strategy of SEND ALL THE TROOPS RIGHT NOW. Thanks, Thought Bubble. I wore my Banana Republic shirt just for this occasion. So, we’ve discussed Roosevelt and Taft’s foreign policy. Let’s move on to Wilson, who was, of course, an isolationist. No. Woodrow Wilson. Okay. Woodrow Wilson was not a volleyball. He was the son of a Presbyterian Minister, a former American history professor and once had been governor of New Jersey, so he understood moral indecency. Wilson thought the best way to teach other countries about the greatness of America was to export colossal amounts of American products. Like, in 1916, he instructed a group of businessmen, “Sell goods that will make the world more comfortable and happy, and convert them to the principles of America.” In short, Woodrow Wilson believed correctly that the the essence of democracy is the freedom to choose among hundreds of brightly coloured breakfast cereals. Still, Wilson intervened in Latin America more than any other U.S. President and his greatest moral triumph was in Mexico, where he wanted to teach the Mexicans “to elect good men”. To do this, Wilson sent troops to stop weapons from flowing to the military dictator Victoriano Huerta but the Americans, who landed at Veracruz were not welcomed with open arms, and 100 Mexicans and 19 Americans were killed. And then in 1916, having learned his lesson (just kidding), Wilson sent 10,000 troops into northern Mexico to chase after revolutionary bandit Pancho Villa. Villa had killed 17 Americans in New Mexico. And everyone knows that the proper response to such a criminal act is to send 10,000 troops into a foreign country. Pershing’s expedition was a smashing success fortunately…except that he actually did not capture Pancho Villa. But all of that was a prelude to Wilson’s leading America to our first international moral crusade, our involvement in the Great War. So, this period of American history is important because Roosevelt, Taft, and Wilson oversaw the expansion of the power of the federal government both at home and abroad, and in doing so they became the first modern American presidents. I mean, these days, we may talk about small government and large government, but really, we’re always talking about large government. Roosevelt, Taft, and Wilson recognized that the national government was going to have to deal with big business, and that it would have to get big to do that. And also that it had a role to play in ensuring that Americans would retain some freedom in this new industrial era. And they also built neo-imperialistic foreign policies around the idea that the safer the world was for American business, the better it was for Americans. As our old friend Eric Foner wrote: “The presidents who spoke the most about freedom were likely to intervene most frequently in the affairs of other countries.” Sometimes for good and sometimes for ill, we’ll see an extreme and ambiguous case of that next week when we look at America in World War I. Thanks for watching. I’ll see you then. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. Our show is written by my high school history teacher, Raoul Meyer, Rosianna Rojas, and myself. And our graphics team is Thought Café. Every week, there’s a new caption for the libertage. If you’d like to suggest one, you can do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

Growth_Cities_and_Immigration_Crash_Course_US_History_25.txt

Hi, I’m John Green, this is CrashCourse U.S. History and today we’re going to continue our extensive look at American capitalism. Mr. Green, Mr. Green, I’m sorry are you saying that I grow up to be a tool of the bourgeoisie… Oh not just a tool of the bourgeoise, Me from the Past, but a card-carrying member of it. I mean, you have employees whose labor you can exploit because you own the means of production, which in your case includes a chalkboard, a video camera, a desk, and a xenophobic globe. Meanwhile Stan, Danica, Raoul, and Meredith toil in crushing poverty – STAN, DID YOU WRITE THIS PART? THESE ARE ALL LIES. CUE THE INTRO. [Theme Music] So, last week we saw how commercial farming transformed the American west and gave us mythical cowboys and unfortunately not-so-mythical Indian reservations. Today we leave the sticks and head for the cities, as so many Americans and immigrants have done throughout this nation’s history. I mean we may like to imagine that the history of America is all “Go west young man,” but in fact from Mark Twain to pretty much every hipster in Brooklyn, it’s the opposite. So, population was growing everywhere in America after 1850. Following a major economic downturn in the 1890s, farm prices made a comeback, and that drew more and more people out west to take part in what would eventually be called agriculture’s golden age. Although to be fair agriculture’s real golden age was in like 3000 BCE when Mesopotamians were like, “Dude, if we planted these in rows, we could have MORE OF IT THAN WE CAN EAT.” So it was really more of a second golden age. But anyway, more than a million land claims were filed under the Homestead Act in the 1890s. And between 1900 and 1910 the populations of Texas and Oklahoma together increased by almost 2 million people. And another 800,000 moved into Kansas, the Dakotas, and Nebraska. That’s right. People moved TO Nebraska. Sorry, I just hadn’t yet offended Nebraskans. I’m looking to get through the list before the end of the year. But one of the central reasons that so many people moved out west was that the demand for agricultural products was increasing due to...the growth of cities. In 1880, 20% of the American population lived in cities and there were 12 cities with a population over 100,000 people. This rose to 18 cities in 1900 with the percentage of urban dwellers rising to 38%. And by 1920, 68% of Americans lived in cities and 26 cities had a population over 100,000. So in the 40 years around the turn of the 20th century, America became the world’s largest industrial power and went from being predominantly rural to largely urban. This is, to use a technical historian term, a really big deal. Because it didn’t just make cities possible, but also their products. It’s no coincidence that while all this was happening, we were getting cool stuff like electric lights and moving picture cameras. Neither of which were invented by Thomas Edison. I don’t know if you’ve noticed, but suddenly there are a lot more photographs in Crash Course U.S. History b-roll. So the city leading the way in this urban growth was New York, especially after Manhattan was consolidated with Brooklyn (and the Bronx, Queens and Staten Island) in 1898. At the turn of the century, the population of the 23 square miles of Manhattan Island was over 2 million. And the combined 5 boroughs had a population over 4 million. But, while New York gets most of the attention in this time period, and all time periods since, it wasn’t alone in experiencing massive growth. Like, my old hometown of Chicago, after basically burning to the ground in 1871, became the second largest city in America by the 1890s. Also, they reversed the flow of the freaking Chicago River. Probably the second most impressive feat in Chicago at the time. The first being that the Cubs won two World Series. Even though I’m sorely tempted to chalk up the growth of these metropolises to a combination of better nutrition and a rise in skoodilypooping, I’m going to have to bow to stupid historical accuracy and tell you that much of the growth had to do with the phenomenon that this period is most known for: immigration. Of course, by the end of the 19th century, immigration was not a new phenomenon in the United States. After the first wave of colonization by English people, and Spanish people, and other Europeans, there was a new wave of Scandinavians, French people, and especially the Irish. Most of you probably know about the potato famine of the 1840s that led a million Irish men and women to flee. If you don’t know about it, it was awful. And the second largest wave of immigrants was made up of German speakers, including a number of liberals who left after the abortive revolutions of 1848. All right, let’s go to the Thought Bubble. The Irish had primarily been farmers in the motherland, but in America, they tended to stay in cities, like New York and Boston. Most of the men began their working lives as low-wage unskilled laborers, but over time they came to have much more varied job opportunities. Irish immigrant women worked too, some in factories or as domestic servants in the homes of the growing upper class. Many women actually preferred the freedom that factory labor provided and one Irish factory woman compared her life to that of a servant by saying: “Our day is ten hours long, but when it’s done, it’s done, and we can do what we like with the evenings. That’s what I’ve heard from every nice girl that’s tried service. You’re never sure that your soul is your own except when you’re out of the house.” Most German speakers had been farmers in their home countries and would remain farmers in the US, but a number of skilled artisans also came. They tended to stay in cities and make a go of entrepreneurship. Bismarck himself saw emigration from Germany as a good thing saying, “The better it goes for us, the higher the volume of emigration.” And that’s why we named a city in North Dakota after him. Although enough German immigrants came to New York that the lower east side of Manhattan came to be known for a time as Kleindeutschland (little Germany), many moved to the growing cities of the Midwest like Cincinnati and St. Louis. Some of the most famous German immigrants became brewers. And America is much richer for the arrival of men like Frederick Pabst, Joseph Schlitz, and Adolphus Busch. And by richer, I mean drunker. Hey. Thanks for not ending on a downer, Thought Bubble. I mean, unless you count alcoholism. So, but by the 1890s, over half of the 3.5 million immigrants who came to our shores came from southern and eastern Europe, in particular Italy and the Russian and Austro Hungarian empires. They were more likely than previous immigrants to be Jewish or Catholic, and while almost all of them were looking for work, many were also escaping political or religious persecution. And by the 1890s they also had to face new “scientific” theories, which I’m putting in air quotes to be clear because there was nothing scientific about them, which consigned them to different “races” whose low level of civilization was fit only for certain kinds of work and predisposed them to criminality. The Immigration Restriction League was founded in Boston in 1894 and lobbied for national legislation that would limit the numbers of immigrants, and one such law even passed Congress in 1897 only to be vetoed by President Grover Cleveland. Good work, Grover! You know, his first name was Stephen, but he called himself Grover. I would have made a different choice. But before you get too excited about Grover Cleveland, Congress and the President were able to agree on one group of immigrants to discriminate against: the Chinese. Chinese immigrants, overwhelmingly male, had been coming to the United States, mostly to the West, since the 1850s to work in mines and on the railroads. They were viewed with suspicion because they looked different, spoke a different language, and they had “strange” habits, like regular bathing. By the time the Chinese Exclusion Act went into effect in 1882, there were 105,000 people of Chinese descent living in the United States, mainly in cities on the West Coast. San Francisco refused to educate Asians until the state Supreme Court ordered them to do so. And even then the city responded by setting up segregated schools. The immigrants fought back through the courts. In 1886, in the case of Yick Wo v. Hopkins the United States Supreme court ordered San Francisco to grant Chinese-operated laundries licenses to operate. Then in 1898 in United States v. Wong Kim Ark, the Court ruled that American born children of Chinese immigrants were entitled to citizenship under the 14th Amendment, which should have been a duh but wasn’t. We’ve been hard on the Supreme Court here at Crash Course, but those were two good decisions. You go, Supreme Court! But despite these victories Asian immigrants continued to face discrimination in the form of vigilante-led riots like the one in Rock Springs, Wyoming that killed 26 people. And congressionally approved restrictions, many of which the Supreme Court did uphold, so, meh. Also it’s important to remember that this large-scale immigration – and the fear of it – was part of a global phenomenon. At its peak between 1901 and the outbreak of World War 1 in 1914, 13 million immigrants came to the United States. In the entire period touched off by the industrialization from 1840 until 1914, a total of 40 million people came to the U.S. But at least 20 million people emigrated to other parts of the Western Hemisphere, including Brazil, the Caribbean, Canada (yes, Canada) and Argentina. As much as we have Italian immigrants to thank for things like pizza (and we do thank you), Argentina can be just as grateful for the immigrant ancestors of Leo Messi. Also the Pope, although he has never once won La Liga. And there was also extensive immigration from India to other parts of the British Empire like South Africa; Chinese immigration to South America and the Caribbean; I mean, the list goes on and on. In short, America is not as special as it fancies itself. Oh it’s time for the Mystery Document? The rules here are simple. I guess the author of the Mystery Document. I get it wrong and then I get shocked with the shock pen. Sorry I don’t mean to sound defeatist, but I don’t have a good feeling about this. All right. “The figure that challenged attention to the group was the tall, straight, father, with his earnest face and fine forehead, nervous hands eloquent in gesture, and a voice full of feeling. This foreigner, who brought his children to school as if it were an act of consecration, who regarded the teacher of the primer class with reverence, who spoke of visions, like a man inspired, in a common classroom. I think Miss Nixon guessed what my father’s best English could not convey. I think she divined that by the simple act of delivering our school certificates to her he took possession of America.” Uhh, I don’t know. At first I thought it might be someone who worked with immigrants, like Jane Addams, but then at the end suddenly it’s her own father. [buzz] Jane Addams’s father was not an immigrant. Mary Antin? Does she even have a Wikipedia page?! She does? Did you write it, Stan? Stan wrote her Wikipedia page. AH. So, this document, while it was written by someone who should not have a Wikipedia page, points out that most immigrants to America were coming for the most obvious reason: opportunity. Industrialization, both in manufacturing and agriculture, meant that there were jobs in America. There was so much work, in fact, that companies used labor recruiters who went to Europe to advertise opportunities. Plus, the passage was relatively cheap, provided you were only going to make it once in your life, and it was fast, taking only 8 to 12 days on the new steam powered ships. The Lower East Side of Manhattan became the magnet for waves of immigrants, first Germans, then Eastern European Jews and Italians, who tended to re-create towns and neighborhoods within blocks and sometimes single buildings. Tenements, these 4, 5 and 6 story buildings that were designed to be apartments, sprang up in the second half of the 19th century and the earliest ones were so unsanitary and crowded that the city passed laws requiring a minimum of light and ventilation. And often these tenement apartments doubled as workspaces because many immigrant women and children took in piecework, especially in the garment industry. Despite laws mandating the occasional window and outlawing the presence of cows on public streets, conditions in these cities were pretty bad. Things got better with the construction of elevated railroads and later subways that helped relieve traffic congestion but they created a new problem: pickpockets. “Pickpockets take advantage of the confusion to ply their vocation… The foul, close, heated air is poisonous. A healthy person cannot ride a dozen blocks without a headache.” So that’s changed! This new transportation technology also enabled a greater degree of residential segregation in cities. Manhattan’s downtown area had, at one time, housed the very rich as well as the very poor, but improved transportation meant that people no longer had to live and work in the same place. The wealthiest, like Cornelius Vanderbilt and J.P. Morgan, constructed lavish palaces for themselves and uptown townhouses were common. But until then, one of the most notable feature of gilded age cities like New York was that the rich and the poor lived in such close proximity to each other. And this meant that with America’s growing urbanization, the growing distance between rich and poor was visible to both rich and poor. And much as we see in today’s megacity, this inability to look away from poverty and economic inequality became a source of concern. Now one way to alleviate concern is to create suburbs so you don’t have to look at poor people, but another response to urban problems was politics, which in cities like New York, became something of a contact sport. Another response was the so-called progressive reform movement. And in all these responses and in the issues that prompted them – urbanization, mechanization, capitalism, the distribution of resources throughout the social order – we can see modern industrial America taking shape. And that is the America we live in today. Thank you for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. The show is written by my history teacher, Raoul Meyer, Rosianna Halse Rojas, and myself. Our associate producer is Danica Johnson. And our graphics team is Thought Café. Every week, there’s a new caption for the libertage. If you’d like to suggest one, you can do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

Ford_Carter_and_the_Economic_Malaise_Crash_Course_US_History_42.txt

Hi, I’m John Green, this is Crash Course U.S. History and today we are going to talk about one of the most important periods in American history, the mid-to-late 1970s. Stan why is there nothing on the chalkboard? We can’t find a picture of Gerald Ford somewhere around here? Don’t worry Crash Course fans we got one. Thanks for your support through Subbable. It paid for this 90 cent Gerald Ford photograph. These really are the years where everything changed in the United States and amidst all that turmoil something wonderful was born. Mr. Green? Mr. Green? Strong with the force, this episode is. No, me from the past, Yoda doesn’t show up until Empire Strikes Back which came out in 1980. I’m referring of course to the fact that we were born! It’s the beginning of the John Green era! From here on out, almost everything we discuss will have happened in my lifetime. Or as most Crash Course viewers refer to it, “that century before I was born”. But it wasn’t just the birth of me and the death of Elvis, the late 1970s were truly a period of momentous change, and for most Americans it sucked. Intro So how Americans reacted to those no good very bad years really has shaped the world in which we find ourselves. The big story of the 1970s is economics. Twenty-five years of broad economic expansion and prosperity came to a grinding halt in the 1970s meaning the our party was over. And what did we get instead? Inflation and extremely slow growth. The worst hangover ever. Just kidding, the worst hangover was The Depression. The 2nd worst hangover was the 2008 recession, and then the 3rd worst hangover was Hangover Part III. It was the 4th worst hangover in American history. Narrowly beating out America’s 5th worst hangover the Hangover Part II. What happened to the American economy in the 1970s was the result both of long-term processes and unexpected shocks. The long-term process was the gradual decline of manufacturing in the U.S. in relation to competing manufacturing in the rest of the world. Part of this was due to American policy; after World War II, you’ll remember that we promoted the economic growth of Japan, Germany, South Korea and Taiwan, ignoring the tariffs that they set up to protect nascent industries, and effectively subsidizing them by providing for their defense. And not having to build nuclear arsenals of their own really allowed them to invest in their domestic economies. And then one day, a bunch of Toyotas and Mercedes showed up, and you could drive them up to like 40 thousand miles before they would break down and we were like, “wait a second”. In 1971, for the first time in the 20th century, America experienced an export trade deficit, importing more goods than it exported, which is the same problem that my aunt has with QVC. I mean, they hardly import anything from her. One reason for this deficit was because the dollar was linked to gold, making it a strong currency but also making American products more expensive abroad. So Nixon took the U.S. off the gold standard, hoping to make American goods cheaper overseas and reduce imports, but that didn’t really work. Because the U.S. was also competing against cheaper labor costs, and cheaper raw materials, and more productive economies. And in many cases this growing global competition put American firms that couldn’t compete out of business. This was especially true in manufacturing. In 1960, 38% of Americans worked in manufacturing. In 1980, it was 28%. Today, it’s nine. Not 9%, nine people. Stan wants me to tell you that was a joke. It actually is 9%. Unionized workers were hit particularly hard. In the 1940s and 1950s unions had won generous concessions from corporate employers including paid vacation, and health benefits, and especially pensions, which employers would agree to as a kind of deferred compensation so that they wouldn’t have to pay higher w ages to people while they were working. And this worked great, until people started to retire. So by 1970, competition led employers to either eliminate high-paying manufacturing jobs, or else to increase automation, or to shift workers to lower wage regions of the U.S. or even overseas. The American South benefitted from this trend because its anti-union stance was attractive to manufacturers. But then, non-union industries that were already in the South found that they had no way to find new workers so the only way to survive was to move production overseas. And also as industries moved production to the Sunbelt that increased the political influence of the region, and because the South and Southwest are generally conservative politically, the nation’s politics continued to move to the right. Meanwhile the northern industrial cities, particularly the Rust Belt of the Midwest, were becoming the empty urban playgrounds that we know and love today. Detroit and Chicago had lost half of their manufacturing jobs by 1980 and smaller cities fared even worse. As industry moved away, they found their tax bases dried up, and they were unable to provide even basic services to their citizens. I mean with the world of Wall Street fat cats this is hard to imagine today, but in 1975 New York City faced bankruptcy. In addition to these long term structural changes to the American economy and our demographics, the 1970s saw two oil shocks that sent the economy into a tailspin. In 1973, in response to Western support of Israel, Middle Eastern Arab states suspended oil exports to the U.S which led to the price of oil quadrupling. This resulted in long lines for gasoline, dramatically higher oil prices, and Americans deciding to purchase smaller, more fuel efficient cars, which is to say Japanese cars. Also, prices of everything else went up because oil is either used for the production of or transportation of just about everything. I mean with 70’s inflation, this 90 cent portrait of Gerald Ford would have cost at least $1.10. The paint that covers the green parts of not-America, oil based. The plastic that comprises the DVD’s of Crash Course World History, available now at DFTBA.com, oil based. Those were a fantastic bargain and they would have been way more expensive if the price of oil was higher. And then, in 1979, a second oil shock hit the United States after the Iranian revolution. Wait Stan, did we say 1979? We’ve got to put up a picture of Jimmy Carter. Bam! Sorry, Gerald Ford there’s a peanut farmer in town. So during the 1970s inflation soared to 10% a year and economic growth slowed to 2.4%, resulting in what came to be known as stagflation. Unemployment rose, and a new economic statistic was born: the misery index, the combination of unemployment and inflation. At the beginning of the decade it was 10.8, by 1980 it had doubled. If you’re looking for the roots of America’s contemporary economic inequality, the 1970s are a good milestone, since according to our old friend Eric Foner, “beginning in 1973, real wages essentially did not rise for twenty years.” [1] Americans got to experience the joy of two years of Gerald Ford before poor Jimmy Carter had a chance to fail at improving the economy. The only president never to have been elected even to the vice presidency, Gerald Ford was so insignificant to American history that we already replaced him on the chalkboard. One of Ford’s first acts was to pardon Nixon making him immune from prosecution for obstruction of justice. That very unpopular decision probably made it impossible for Ford to win in 1976. Coincidentally, WIN was the only memorable domestic program that Ford proposed. It stood for Whip Inflation Now and it was basically a plea for Americans to be better shoppers, spend less, and wear WIN buttons. Thirty-five years later Charlie Sheen would turn winning into an incredibly successful social media campaign, but sadly at the time there was no Twitter. Inflation did drop, but unemployment went up, especially during the recession of 1974-75 where it topped 9%. Now, Ford would have liked to cut taxes and reduce government regulation, but the Democratic Congress wouldn’t let him. So that’s Ford, probably best known today as the first president to be satirized on Saturday Night Live. Then, in 1976, we got a new president: Jimmy Carter. Now Jimmy Carter is generally considered by historians to have been a failure as president. Although, he is often seen as a really good ex-president. He tried to fight the inflation part of stagflation, but to do it he acted in some rather un-New Deal Democrat ways. He cut government spending, deregulated the trucking and airline industries, and he supported the Fed’s decision to raise interest rates. Oh, it’s time for the mystery document? The rules here are simple... I read the mystery document, I guess the author, and if I’m wrong I get shocked. Alright, let’s see what we’ve got today. “I want to speak to you first tonight about a subject even more serious than energy or inflation. I want to talk to you right now about a fundamental threat to American democracy. I do not mean our political and civil liberties. They will endure. And I do not refer to the outward strength of America, a nation that is at peace tonight everywhere in the world, with unmatched economic power and military might. The threat is nearly invisible in ordinary ways. It is a crisis of confidence. It is a crisis that strikes at the very heart and soul and spirit of our national will. We can see this crisis in the growing doubt about the meaning of our own lives and in the loss of a unity of purpose for our nation. The erosion of our confidence in the future is threatening to destroy the social and political fabric of America.” It’s Jimmy Carter’s “Crisis of Confidence” speech, my favorite speech ever made that also cost a president 20 points of approval rating. So Carter says that Americans have lost their ability to face the future and some of their can-do spirit. The rest of the speech talks about how Americans’ values are out of whack, how Americans are wasteful, and need a new more vibrant approach to the energy crisis. Let me tell you a lesson from history Jimmy Carter, you don’t get reelected by telling Americans how to do more with less. You get reelected by telling Americans, “more, more, always more, more for you. More. More. More. I promise.” The speech ultimately called for a renewal of spirit, but all people remember is the part where Jimmy Carter was criticizing them, and it’s gone down as a great example of Carter’s political ineptitude. Domestically, Carter paid lip-service to liberal ideas like energy conservation, even installing solar panels on the White House, but his bigger plan to solve the energy crises was investment in nuclear power. And nuclear power did grow, although never to the extent we saw in certain European countries, partly because of the accident at Three Mile Island in 1979 when radioactive vapor was released into the air. This of course spurred public fears of a nuclear meltdown and drove a huge anti-nuclear energy movement. But some of Carter’s more conservative policies did ultimately have an impact, like his support for deregulation of the airlines. Before airline deregulation, prices were fixed, so airlines had to compete by offering better service. Now, of course, flights are much cheaper and also so much more miserable. In many ways, Carter was more important as a foreign policy president, but as with his energy initiatives, he’s mostly remembered for his failures. Aiming to make Human Rights a cornerstone of America’s foreign policy, Jimmy Carter tried to turn away from the Cold War framework and focus instead on combatting 3rd world poverty and reducing the spread of nuclear weapons. Let’s go to the Thought Bubble. Carter’s notable changes included cutting off aid to Argentina during its “Dirty War” and signing a treaty in 1978 that would transfer the Panama Canal back to Panama. His greatest accomplishment was probably brokering the Camp David Accords. This historic peace agreement between Egypt and Israel has, as we all know, led to a lasting peace in the Middle East, just kidding, but it has been a step in the right direction and one that’s lasted. But the U.S. continued to support dictatorial regimes in Guatemala, the Philippines and South Korea. Carter’s most significant failure in terms of supporting international bad guys, though, is the Shah of Iran. Iran had oil and was a major buyer of American arms, but the Shah was really unpopular and our support of him fuelled anti-American sentiments in Iran. Those boiled over in the 1979 Iranian Revolution, especially after Carter allowed the Shah to get cancer treatments in America, which in turn prompted the storming of the American embassy in Tehran and the capture of 53 American hostages. The Iranian hostage crisis lasted 444 days and although Carter’s secretary of state did negotiate their release, it didn’t happen until the day Carter’s successor Ronald Reagan was inaugurated. The inability to free the hostages and the botched rescue attempt -- Affleck’s ARGO notwithstanding -- added to the impression that Carter was weak. Events in the Middle East also increased Cold War tensions especially after 1979, when the USSR invaded Afghanistan. Carter claimed that the invasion of Afghanistan was the greatest threat to freedom since World War II and proclaimed the Carter Doctrine, which was basically said that the U.S. would use force, if necessary, to protect its interests in the Persian Gulf region. In direct response to the Soviets, the U.S. put an embargo on grain shipments and organized the boycott of the 1980 Olympics in Moscow. Thanks for another dose of good news Thought Bubble. So despite focusing on Carter, I’ll again stress that the real story of the 1970s was the economy. High inflation and high unemployment had monumental effects in shaping America. And no president could have dealt with it effectively. Not Carter, not Gerald Ford, not anyone. The truth is, history isn’t about individuals. Oil shocks and inevitable systemic changes led to the poor economy and that weakened support for New Deal liberalism and increased the appeal of conservative ideas like lower taxes, reduced regulation, and cuts in social spending. All of which, for the record, started under the Democrat Jimmy Carter, not the Republican Ronald Reagan. More abstractly, the economic crisis of the 1970’s dealt a serious blow to the Keynesian consensus that Government action could actually solve macro-economic problems. I mean according to the economic theory that had prevailed for the previous 50 years, unemployment and inflation were supposed to be inversely proportional, the so-called Phillips Curve. When that relationship broke down and we had both high inflation and high unemployment it undermined the entire idea of government intervention. And that opened the door for a different way of thinking about economics that emphasized the economy as an aggregate of individual economic decisions. Now that might sound like a small thing, but whether you think of individual choices or governmental policies really make economies work or not work turns out to be pretty freaking important. And this has come to really shape the contemporary American political landscape especially when it comes to taxes. Which we’ll talk about more next week. Thanks for watching. Crash Course is made with all the help from these nice people and it exists because of your support through Subbable and also because so many of you are buying Crash Course World History on DVD. Thank you! Our mission here at Crash Course is to make educational content freely available to everyone forever and you can help us in that mission, if you’re able, by subscribing at Subbable. Subbable is a voluntary subscription platform where you can get amazing perks liked signed posters and lots of other stuff so check it out. Thank you for supporting Crash Course, thanks for watching, and as we say in my hometown, “don’t forget to be awesome.” ________________ [1] Foner. Give me Liberty ebook version p. 1097

US_History

에이지_오브_잭슨_크래쉬_코스_미국_역사_14.txt

Hi I’m John Green. This is Crash Course U.S. History and today, after last week’s bummer on slavery, we turn to a happier topic: the rise of democratization in the U.S. This was also known as the Age of Jackson, no Stan, not that Jackson. No, no, Stan, come’on seriously. No not, no, no, no, no, no, no, no, no, no, no, no. YES. That Jackson. Andrew Jackson. [Theme Music] ...Sorry, I just had to check my collar. Right, so you’ll recall that the initial democracy of the United States wasn’t terribly democratic—almost all voters were white male land owners. Mr. Green, Mr. Green. That’s just radically unfair. Exactly, Me from the Past. But, between 1820 and 1850, this started to change. State legislatures lowered, or else eliminated, the property qualifications for voting, which allowed many more people to vote, so long as they were, you know, both white and male. Mr. Green, Mr. Green. So, I’d be in, right? Yeah, that seems reasonable. Yeah, Me from the Past, quick privilege check. One of the reasons we study history is so that you can learn that people like you are not actually at the center of history, even though, you know, you’ve been taught that. But, anyway, the whole idea of owning land as a prerequisite for voting is sort of Jeffersonian— an individual who works his own land can be truly independent, because he doesn’t need to rely upon markets to acquire stuff or, God forbid, wages to give him money with which to buy stuff. No, he makes his own stuff and he doesn’t need anybody...except for slaves and also women to make shoes and clothes and to cook food and also make children. But, in light of the Market Revolution, the idea of excluding wage workers seemed very outdated. The idea of excluding women and non-white people, though, still quite popular. But, this defining characteristic of the Age of Jackson really had very little to do with Andrew Jackson himself because, by the time he became President in 1829, every state except for North Carolina, Virginia, and Rhode Island had already gotten rid of their property requirements. In fact, that’s probably why he got elected. Right so you’ll recall that America’s mostly fake victory in the War of 1812 and the subsequent collapse of the Federalist party ushered in the “Era of Good Feelings” which was another way of saying that there was basic agreement on most domestic policies. The American System was a program of economic nationalism built on (1) federally financed internal improvements, like roads and canals, what we would now call “infrastructure” (2) tariffs, to protect new factories and industries, and (3) a national bank that would replace the First Bank of the United States whose charter expired in 1811. You’ll never guess what we called this second bank, unless you guessed that we called it “The Second Bank of the United States.” The main supporters of this American System were our old friend John C. Calhoun and our new friend Henry Clay. Both were Jeffersonian Republicans, which isn’t surprising because that was the only political party, but it’s kind of surprising because the American System had nothing to do with the Agrarian Republic that Jefferson had championed. But whatever, this was the Era of Good Feelings, so we’re gonna go with it. By the way, this nationalism also extended to foreign affairs. After Latin America won its independence from Spain, President Monroe made a speech proposing that Europe shouldn't try to retake colonies in the Western hemisphere. And if they did, we would, like, do stuff. This so called “Monroe Doctrine” also said that the U.S. would stay out of European wars. Hahahaha that is hilarious! But, we did live up to the other end of it, you’ll remember that when the British came for the Falkland Islands, we were like, “This shall not stand.” Just kidding. We were like, “Go ahead.” The last Good Feelings era president was John Quincy Adams, who was quite the diplomat and expansionist. He actually wrote the Monroe Doctrine, for instance. But in fact, it turns out that all feelings were not good. There was significant disagreement over three main issues. First, many people felt that the federal government shouldn’t invest in infrastructure. Like, James Madison, who’d initially supported those bills, ended up vetoing one of them that included a big spending increase to finance roads and canals. Now, the roads and canals did get built, but, in the end, most of the financing fell to the states. There were also big problems with the Second Bank of the United States, which you known is why you can’t visit a branch of it these days. But we’ll get to that in a minute! And, lastly, there was the perennial issue of slavery. In this case the problem started, as so many problems do, in Missouri. So, in 1819 Missouri had enough people in it to become a state, but despite the fact that there were already more than 10,000 slaves there, a New York congressman, named James Tallmadge, made a motion to prohibit the introduction of further slaves into the proposed state. It took almost two years to work out the John C. Calhoun storm that blew up after this. Actually, it took more than that. It took until the end of the Civil War basically. But in the short run, Missouri was allowed to enter the union as a slave state, while Maine was carved out of Massachusetts to keep the balance of things. But the Missouri Compromise also said that no state admitted above the 36 30 line of latitude would be allowed to have slaves, except, of course, for Missouri itself, which as you can see, is well above the line. Anyway, this solution to westward expansion worked out magnificently provided that you enjoy Civil Wars. So, Thomas Jefferson, who was by the way was still alive, which gives you some context for how young the nation truly was, wrote that the Missouri Compromise was “like a fire bell in the night that awakened and filled me with terror. I considered it at once the death knell of the union.” Eventually, almost. But in the short term, it did mean the rise of political parties. So, America was becoming more democratic, but if there was only one political party, that democratic spirit had nowhere to go. Fortunately, there was a tiny little magician named Martin Van Buren. They really did call him the “Little Magician,” by the way. Also “The red fox of Kinderhook,” but we remember him as the worst-haired president. So, despite having been President of the United States, Van Buren is arguably more important for having invented the Democratic Party. He was first to realize that national political parties could be a good thing. So, I mentioned that Martin Van Buren was known as the “Little Magician, and I know this sounds a little bit silly, but I think it’s telling. You see, Van Buren was only the second American president with a well-used nickname. And the first was his immediate predecessor, Andrew Jackson, or Old Hickory. Why does this matter? Well when you’re actually having to campaign for office, as all presidential candidates did after the election of 1828, and you’re trying to appeal to the newly enfranchised “common man” what better way to seem like a regular guy than to have a nickname? I mean, if you think this is crazy, just think of the nicknames of some some of our most popular presidents. “Honest Abe,” “The Bull Moose,” “The Gipper.” Even our lesser known presidents had nicknames. “Young Hickory,” “Handsome Frank;” “Old Rough and Ready,” “Big Steve.” James Buchanan, and I am not making this up, was “Old Public Functionary.” Who’re you gonna vote for? Oh, I think the “Old Public Functionary.” He seems competent. As it happens, he wasn’t. So, by now you’re probably wondering, where does Andrew Jackson fit into all of this? When we last caught up with Jackson, he was winning the battle of New Orleans shortly after the end of the War of 1812. He continued his bellicose ways, fighting Indians in Florida, although he was not actually authorized to do so, and became so popular from all of his Indian killing that he decided to run for president in 1824. The election of 1824 was very close. And it went to the House, where John Quincy Adams was eventually declared the winner. And Jackson denounced this as “a corrupt bargain.” So, in 1828, Jackson ran a much more negative campaign—one of campaign slogans was “Vote for Andrew Jackson who can fight, not John Quincy Adams who can write.” Adams’ supporters responded by arguing that having a literate president wasn’t such a bad thing and also by accusing Jackson of being a murderer, which given his frequent habit of dueling and massacring, he sort of was. So as you can see, the quality of discourse in American political campaigns has come a long way. Anyway, Jackson won. Jackson ran as the champion of the common man and in a way he was. I mean, he had little formal schooling and in some ways he was the archetypal self made man. Jackson’s policies defined the new Democratic party, which had formerly been known as the Jeffersonian Democratic Republicans. So who were these new Democrats? Well generally, they tended to be lower to middle class men, usually farmers, who were suspicious of the widening gap between the rich and the poor that was one of the results of the Market Revolution. And they were particularly worried about bankers, merchants and speculators, who seemed to be getting rich without actually producing anything. Stop me if any of this sounds familiar. This vision probably would have carried the day except a new party arose in response to Jackson’s election: the Whigs. No, Stan, the Whigs. Yes. The American Whigs took their name from the English Whigs, who were opposed to absolute monarchy. And the American Whigs felt that Andrew Jackson was grabbing so much power for the executive branch that he was turning himself into “King Andrew.” So, the Whigs were big supporters of the American System and its active federal government. You know, tariffs, infrastructure, etc. Their greatest support was in the Northeast, especially from businessmen and bankers who benefitted from those tariffs and the stability provided by a national bank. And they also thought the government should promote moral character because that was necessary for a person to act as a truly independent citizen. So Jackson’s policies must have been pretty egregious for them to spawn an entire new political party. What did he actually do as president? Well, let’s go to the Thought Bubble. Let’s start with Nullification. So, in 1828, Congress passed the Tariff of 1828 because they were not yet in the habit of marketing their bills via naming them with funny acronyms. Jackson supported this in spite of the fact that it benefitted manufacturers. The tariff raised prices on imported manufactured goods made of wool and iron, which enraged South Carolina because they’d put all their money into slavery and none into industry. Unlike northerners, who could avoid the higher prices by manufacturing sweaters and pants and such at home, South Carolinians would have to pay more. They were so angry at this “Tariff of Abominations” that the South Carolina legislature threatened to nullify it. Jackson didn’t take kindly to this affront to federal power, but South Carolina persisted, and when Congress passed a new tariff in 1832– one that actually lowered the duties -- the Palmetto State’s government nullified it. Jackson responded by getting Congress to pass the Force Act, which authorized him to use the army and navy to collect taxes. A full blown crisis was averted when Congress passed a new tariff in 1833 and South Carolina relented. This smelled a bit of dictatorship – armed tax collectors and all – and helped to cement Jackson’s reputation as a tyrant, at least among the Whigs. And then we have the Native Americans, much of Jackson’s reputation there was based on killing them, so it’s no surprise that he supported southern states’ efforts to appropriate Indian lands and make the Indians move. This support was formalized in the Indian Removal Act of 1830, which Jackson supported. The law provided funds to re-locate the Cherokees, Chickasaws, Choctaws, Creek and Seminole Indians from their homes in Georgia, North Carolina, Florida, Mississippi, and Alabama. In response, these tribes adopted a novel approach, and sued the government. And then, the Supreme Court ruled that Georgia’s actions in removing the Cherokees violated their treaties with the federal government and that they had a right to their land. To which Jackson supposedly responded by saying, “John Marshall has made his decision. Now let him enforce it.” So, Jackson set the stage for the forced removal of the Cherokees from Georgia to Oklahoma, but it actually took place in the winter of 1838-1839 under Jackson’s successor Van Buren. At least ¼ of the 18,000 Indians died during the forced march that came to be known as the Trail of Tears. Boy, Thought Bubble, you do know how to end on a downer. But, thank you. But Andrew Jackson also changed our banking system. Just as today, banks were very important to the industrial and mercantile development of the U.S. And at the beginning of Jackson’s Presidency, American banking was dominated by the Second National Bank, which you’ll remember, had been established by Congress as part of the American system. Oh it’s time for the Mystery Document? The rules here are simple. When I inevitably fail to guess the author of the Mystery Document, I get shocked with the shock pen. “The powers, privileges, and favors bestowed upon it in the original charter, by increasing the value of the stock far above its par value operated as a gratuity of many millions to its stockholders … Every monopoly and all exclusive privileges are granted at the expense of the public which ought to receive a fair equivalent. The many millions which this act proposes to bestow on the stockholders of the existing bank must come directly or indirectly out of the earnings of the American people … Stan, I know this one! Is it not conceivable. It is not conceivable how the present stockholders can have any claim to the special favor of Government. Should [the bank’s] influence become concentrated, as it may under the operation of such an act as this, in the hands of a self-elected directory… will there not be cause to tremble for the purity of our elections[?]” It is Andrew Jackson’s veto of the charter of the Second Bank of the United States. YES. So in 1832 bank leader Nicholas Biddle persuaded Congress to pass a bill extending the life of the Second US Bank for 20 years. Jackson thought that the Bank would use its money to oppose his reelection in 1836, so he vetoed that bill. In fact, the reason I knew that was from the veto message is because it talks about the bank as an instrument to subvert democracy. Jackson set himself up as a defender of the lower classes by vetoing the bank’s charter. Now, Whigs took exception to the idea that the president was somehow a more democratic representative of the people than the legislature, but in the end Jackson’s view won out. He used the veto power more than any prior president, turning it into a powerful tool of policy. Which it remains to this day, by the way. So the Second Bank of the U.S. expired in 1836, which meant that suddenly we had no central institution with which to control federal funds. Jackson ordered that money should be disbursed into local banks, unsurprisingly preferencing ones that were friendly to him. These so-called “pet banks” were another version of rewarding political supporters that Jackson liked to call “rotation in office.” Opponents called this tactic of awarding government offices to political favorites the spoils system. Anyway, these smaller banks proceeded to print more and more paper money because, you know, free money. Like, between 1833 and 1837 the face value of banknotes in circulation rose from $10 million to $149 million, and that meant inflation. Initially, states loved all this new money that they could use to finance internal improvements. But, inflation is really bad for wage workers. And also, eventually, everyone. So all this out-of-control inflation, coupled with rampant land-speculation eventually lead to an economic collapse, the Panic of 1837. The subsequent depression lasted until 1843. And Jackson’s bank policy proved to be arguably the most disastrous fiscal policy in American history, which is really saying something. It also had a major effect on American politics because business-oriented Democrats became Whigs, and the remaining Democrats further aligned with agrarian interests, which meant slavery. So the Age of Jackson was more democratic than anything that came before and it gave us the beginnings of modern American politics. I mean, Jackson was the first president to really expand executive power and to argue that the president is the most important democratically elected official in the country. One of the things that makes Andrew Jackson’s presidency so interesting and also so problematic is that he was elected via a more democratic process, but he concentrated more power in the executive in a thoroughly undemocratic way. In the end, Andrew Jackson probably was the worst American president to end up on currency, particularly given his disastrous fiscal policies. But the Age of Jackson is still important. And it’s worth remembering that all that stuff in American politics started out with the expansion of democracy. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, and myself. And our graphics team is Thought Cafe. If you have libertage caption suggestions, please leave them in comments, where you can also leave questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome...WHAT.

US_History

The_Constitution_the_Articles_and_Federalism_Crash_Course_US_History_8.txt

Hi, I’m John Green, this is Crash Course U.S. History, and today we’re going to talk about the United States Constitution. And, in doing so, we’re going to explore how the American style of government became the envy of the entire world, so much so that everyone else copied us. What’s that, Stan? We’re not gonna talk about other countries stealing our form of government? Because no other country stole our form of government? That – that doesn’t seem possible, Stan. [Patriotic Rock Music] No, Stan, not the Libertage, cue the intro! [Theme Music] So, today we’re going to learn why the green areas of not-America didn’t copy us. All right, so as Americans may dimly remember from history classes, the Constitutional system we’ve been living under since 1788, the year of the first Presidential election, was not the original American government. The first government set up by the Continental Congress was called the Articles of Confederation and it was, in a word: Bad. In two words, it was not good, which is why it only lasted 10 years. The problem with the confederation is that it wasn’t so much a framework for a national government as it was a “firm league of friendship,” which unfortunately only sounds like a team of Care Bear Superheroes. The Articles set up a “government” that consisted of a one-house body of delegates, with each state having a single vote, who, acting collectively, could make decisions on certain issues that affected all the states. There was no president and no judiciary. You can try to tell me that John Hanson, the president of the congress, was the first American president, but it’s just not true. Any decision required 9 of the 13 congressional votes, which pretty much guaranteed that no decisions would ever be made. Ahh, super majorities: Always so efficient. But besides the 2/3rds requirement, the Congress was very limited in what it could actually do. The government could declare war, conduct foreign affairs and make treaties – basically, the stuff you need to do to go to war with England. It could coin money, but it couldn’t collect taxes; that was left to the states. So if you needed money to, say, go to war with Britain, you had to ask the states politely. The articles could be amended, but that required a unanimous vote, so zero amendments were ever passed. The government was deliberately weak, which followed logically from Americans’ fear of tyrannical governments taxing them and quartering soldiers in their houses and so on. But here’s the thing, weak government is like nonalcoholic beer: It’s useless. That said, the Articles government did accomplish a couple things. First, it won the war, so, yay – unless you were a slave or a Native American, in which case, you know, probable boo. Second, the government developed rules for dealing with one of the most persistent problems facing the new nation: Ohio. Which was called the northwest, presumably because it is north and west of Virginia. Getting control of the land meant taking it from the Indians who were living there, and the Articles government was empowered to make treaties, which it did. Crash Course World History fans will remember the Athenians telling the Melians that “the strong do as they can and the weak suffer what they must,” and the Americans definitely went to the Athenian School of Treaty-Making. Through treaties signed at Fort Stanwix and Fort McIntosh, the Indians surrendered land north of the Ohio River. The biggest accomplishment of the Articles government was the Northwest Ordinance of 1787, which set up a process to create 5 new states between the Ohio and Mississippi rivers. Two things to know about this: first, it acknowledged that American Indians had a claim to the land and that they had to be treated better if settlers wanted to avoid violence. And second, Stan, can I get the foreshadowing filter? Yes, perfect. The ordinance outlawed slavery in all five of the new states. Still, the Articles government was a complete disaster for exactly one reason: It could not collect taxes. Both the national government and the individual states had racked up massive debt to pay for the war, and their main source of revenue became tariffs, but because Congress couldn’t impose them, states had to do it individually. And this made international trade a total nightmare, a fact worsened by the British being kinda cranky about us winning the war and therefore unwilling to trade with us. In 1786 and 1787, the problem got so bad in Massachusetts that farmers rose up and closed the courts to prevent them from foreclosing upon their debt-encumbered farms. This was called Shays’ Rebellion, after Revolutionary War veteran and indebted farmer Daniel Shays. The uprising was quelled by the state militia, but for many, this was the sign that the Articles government, which couldn’t deal with the crisis at all, had to go. But not for everyone; Thomas Jefferson, for instance, was a fan of Shay’s Rebellion. “A little rebellion now and then is a good thing. The tree of liberty must be refreshed from time to time with the blood of patriots and tyrants.” Which is all fine and good, I mean, unless you’re the bleeding patriots or tyrants. But to most elites, Shays’ Rebellion showed that too much democratic liberty among the lower classes could threaten private property. Also, people who held government bonds were nervous, because without tax revenue, they were unlikely to get paid back. And when rich people feel like something has to be done, something is usually done. Let’s go to the Thought Bubble. The first attempt to do something was a meeting in Annapolis in 1786 aimed at better regulating international trade. Only six states sent delegates, but they agreed to meet the next year in Philadelphia to “revise” the Articles of Confederation. The delegates who met in Philly the next year had a funny definition of “revision,” though. Rather than make tweaks to the articles, they wrote a new charter of government, the Constitution, which is, with some significant alterations, the same one that Americans live under and argue about today. Despite what some seem to believe, the 55 men who met in Philadelphia and hammered out a new form of government were not gods, but they were far from ordinary, especially for the time. Most were wealthy, some very much so. More than half had college educations, which was super rare since .001% of Americans attended college at the time. About 40% had served in the army during the war. But, one thing they all shared was a desire for a stronger national government. The delegates agreed on many things: the government should have executive, legislative, and judicial branches; and should be republican, with representatives, rather than direct democracy. But the devil appeared in the details. Alexander Hamilton, probably the biggest proponent of very strong government, wanted the President and Senate to serve life terms, for example. That idea went nowhere because the overarching concern of almost all the delegates was to create a government that would protect against both tyranny by the government itself and tyranny by the people. They didn’t want too much government, but they also didn’t want too much democracy, which is why our Presidents are still technically elected not directly by regular people but by 538 members of the Electoral College. This system is so byzantine and strange that when American politicians speak of spreading democracy through the world, they never actually advocate for American-style elections. Thanks, Thought Bubble. Yes, I know, you have fantastic elections in Canada. Yeah, right, OK. All that too. I get it, OK? It’s U.S. History, Thought Bubble. So conflicts between competing interests arose quickly at the Constitutional Convention in Philadelphia. The first being between states with big populations and those with small populations. Large states supported James Madison’s Virginia Plan, which called for a two-house legislature with representation in both proportional to a state’s population. And smaller states, fearing that the big boys would dominate, rallied behind the New Jersey plan. [muttered] New Jersey. This called for a single legislative house with equal representation for each state, as with the Articles of Confederation. But, of course, coming from New Jersey, it had no chance of succeeding, and sure enough, it didn’t. Instead we got the Great Compromise, brokered by Connecticut’s Roger Sherman, which gave us two houses: a House of Representatives with representation proportional to each state’s population, and a Senate with two members from each state. House members, also called Congressmen, served two year terms while Senators served six year terms, with 1/3 of them being up for election in every 2 year cycle. The House was designed to be responsive to the people, while the Senate was created to never pass anything and it was so masterfully designed that it still works to this day. However, this solution created another problem: Who should be counted in terms of representation? Slaveholding states wanted slaves to count toward their population, even though of course they could not vote, because they were property. States with few slaves argued that slaves shouldn’t be counted as people because, just to be clear, none of these dudes were not racist. This issue was solved with the notorious 3/5ths compromise. For the purpose of determining the population, the total number of white people plus 3/5ths the population of “other persons” – the word “slave” was never used – would be the basis for the calculation. So yeah, that’s still in the Constitution. The Constitution also contains a fugitive slave clause requiring any escaped slave to be returned to their master. And this meant that a slave couldn’t escape slavery by moving to a state where slavery was outlawed, which meant that on some level some states couldn’t enforce their own laws. Spoiler alert: this becomes problematic. But except for the tyranny of slavery, the framers really hated tyranny. To avoid tyranny of the government, the Constitution embraced two principles: Separation of powers and federalism. The government was divided into three branches – legislative, executive, and judiciary, and the Constitution incorporated checks and balances: each branch can check the power The legislature can make laws, but the president can veto those laws. The judiciary can declare laws void, too, but that’s a power they had to grant themselves. You won’t find it in the Constitution – I promise, you can look for it. And federalism is the idea that governmental authority rests both in the national and the state governments. As an American, I am a citizen both of the United States and of the state of Indiana. And the national government, the one set up by the Constitution, is supposed to be limited in scope to certain enumerated powers. Most other powers, especially the protection of health, safety and morals, are left to the states. But the Constitution also seeks to protect against the radicalism that too much democracy can bring. The mostly rich framers worried that the people, many of whom were poor and indebted, might vote in congress people, or God forbid a President, in favor of, like, redistribution of property. To hedge against this, senators were elected by the states, usually by state legislatures, and they were supposed to be, like, leading citizen types. You know, the kind of good Americans who take bribes and have adulterous affairs in airport bathrooms and patronize prostitutes and shoot Alexander Hamilton. Anyway, the other hedge against too much democracy is the aforementioned Electoral College, which many Americans hate because it has the potential to elect a president who did not win the popular vote, but that’s kind of the point. The electors were supposed to be prominent, educated men of property who were better able to elect a president than, like, the rabble. But, the Constitution of the United States is a really impressive document, especially when you consider its longevity. I mean, as Crash Course World History fans will remember, the nation-state is pretty new on the historical scene, and the United States established by the Constitution, is actually one of the oldest ones. But the Constitution would be meaningless if it hadn’t been ratified, which it was, but not without a fight that helped clarify America’s political ideology. 9 out of the 13 states were required to ratify the Constitution in special conventions called for the purpose. In order to convince the delegates to vote for it, three of the framers, Alexander Hamilton, James Madison, and John Jay wrote a series of 85 essays that together are known as the Federalist Papers. Taken together, they’re a powerful and ultimately persuasive argument for why a strong national government is necessary and ultimately not a threat to people’s liberty. Oh, it’s time for the Mystery Document? The rules here are simple. If I name the author of the Mystery Document, shock as in surprise. If I don’t shock as in [gurgling noise] All right, Stan, let’s see what we’ve got here. “If circumstances should at any time oblige the government to form an army of any magnitude that army can never be formidable to the liberties of the people while there is a large body of citizens, little, if at all, inferior to them in discipline and the use of arms, who stand ready to defend their own rights and those of their fellow-citizens. This appears to me the only substitute that can be devised for a standing army, and the best possible security against it, if it should exist.” Federalist Papers. Alexander Hamilton. [dinging noise] YES. Too easy, Stan, although I appreciate the opportunity for a rant. The whole idea of the Second Amendment was that the people could protect themselves from a standing army by being equally well-armed. Which, these days, would mean not that citizens should have the right to buy assault rifles, but that they should have the right to buy, like, unmanned drones. And arguably, suitcase nukes. And by the way, in the Constitution, this is not listed as a privilege, it is listed as a right. And, as a right, if I can’t afford my own predator drone, I guess the government should buy one for me. It’s almost as if Alexander Hamilton had no way of knowing that weaponry would one day advance past the musket. P.S. You know how Alexander Hamilton died? GUNSHOT. Sorry, I just, I had to. I am on a roll. So, it would be easy to ignore the people who opposed the Constitution because, you know, they lost. But some of the ideas of these so-called Anti-Federalists were particularly powerful, and they deserve a bit of attention. Anti-Federalists, unlike the mostly wealthy federalists, were usually supported by common people, small farmers who weren’t as involved in commercial activity. They saw less need for a strong national government that would foster trade and protect creditors. And, the Anti-Federalists were very afraid of a strong government, especially one dominated by the wealthy. Writers like James Winthrop held that a large group of united states would be like an empire and “that no extensive empire can be governed upon Republican principles.” As evidence, he could point to Britain, or all the way back to Rome. Smaller, more local governments, are more responsive to the people and better able to protect their rights. To the Anti-Federalists, that meant state governments. And while ultimately the Federalists won out and the Constitution was ratified, the issue of how large government should be did not go away. So, the Constitution was really only a starting point. It’s a vague document, and the details would be worked out in the political process. And then on the battlefield. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The show is written by my high school history teacher, Raoul Meyer, and myself. Edited by Stan and Mark Olsen. The associate producer is Danica Johnson. And our graphics team is Thought Bubble. If you have questions about today’s video, or anything about American history, good news: there are historians in comments, so ask away. Thanks for watching Crash Course and as we say in my hometown, Don’t Forget To Be Awesome.

US_History

The_Cold_War_Crash_Course_US_History_37.txt

Hi I’m John Green; this is Crash Course U.S. history and today we’re gonna talk about the Cold War. The Cold War is called “Cold” because it supposedly never heated up into actual armed conflict, which means, you know, that it wasn’t a war. Mr. Green, Mr. Green, but if the War on Christmas is a war and the War on Drugs is a war… You’re not going to hear me say this often in your life, Me from the Past, but that was a good point. At least the Cold War was not an attempt to make war on a noun, which almost never works, because nouns are so resilient. And to be fair, the Cold War did involve quite a lot of actual war, from Korea to Afghanistan, as the world’s two superpowers, the United States and the U.S.S.R., sought ideological and strategic influence throughout the world. So perhaps it’s best to think of the Cold War as an era, lasting roughly from 1945 to 1990. Discussions of the Cold War tend to center on international and political history and those are very important, which is why we’ve talked about them in the past. This, however, is United States history, so let us heroically gaze--as Americans so often do--at our own navel. (Libertage.) Stan, why did you turn the globe to the Green Parts of Not-America? I mean, I guess to be fair, we were a little bit obsessed with this guy. So, the Cold War gave us great spy novels, independence movements, an arms race, cool movies like Dr. Strangelove and War Games, one of the most evil mustaches in history. But it also gave us a growing awareness that the greatest existential threat to human beings is ourselves. It changed the way we imagine the world and humanity’s role in it. In his Nobel Prize Acceptance Speech, William Faulkner famously said, “Our tragedy today is a general and universal physical fear so long sustained by now that we can even bear it. There are no longer problems of the spirit. There is only the question: When will I be blown up?” So, today we’re gonna look at how that came to be the dominant question of human existence, and whether we can ever get past it. intro So after WWII the U.S. and the USSR were the only two nations with any power left. The United States was a lot stronger – we had atomic weapons, for starters, and also the Soviets had lost 20 million people in the war and they were led by a sociopathic mustachioed Joseph Stalin. But the U.S. still had worries: we needed a strong, free-market-oriented Europe (and to a lesser extent Asia) so that all the goods we were making could find happy homes. The Soviets, meanwhile, were concerned with something more immediate, a powerful Germany invading them. Again. Germany--and please do not take this personally, Germans--was very, very slow to learn the central lesson of world history: Do not invade Russia. Unless you’re the Mongols. (Mongoltage.) So at the end of World War II, the USSR “encouraged” the creation of pro-communist governments in Bulgaria, Romania, and Poland--which was a relatively easy thing to encourage, because those nations were occupied by Soviet troops. The idea for the Soviets was to create a communist buffer between them and Germany, but to the U.S. it looked like communism might just keep expanding, and that would be really bad for us, because who would buy all of our sweet, sweet industrial goods? So America responded with the policy of containment, as introduced in diplomat George F. Kennan’s famous Long Telegram. Communism could stay where it was, but it would not be allowed to spread. And ultimately this is why we fought very real wars in both Korea and Vietnam. As a government report from 1950 put it the goals of containment were: 1. Block further expansion of Soviet power 2. Expose the falsities of soviet pretensions 3. Induce a retraction of the Kremlin’s control and influence, and 4. In general, foster the seeds of destruction within the Soviet system. Harry Truman, who as you’ll recall, became President in 1945 after Franklin Delano Prez 4 Life Roosevelt died, was a big fan of containment, and the first real test of it came in Greece and Turkey in 1947. This was a very strategically valuable region because it was near the Middle East, and I don’t know if you’ve noticed this, but the United States has been just, like, a smidge interested in the Middle East the last several decades because of oil glorious oil. Right, so Truman announced the so-called Truman Doctrine, because you know why not name a doctrine after yourself, in which he pledged to support “freedom-loving peoples” against communist threats, which is all fine and good. But who will protect us against “peoples,” the pluralization of an already plural noun? Anyway, we eventually sent $400 million in aid to Greece and Turkey, and we were off to the Cold War races. The Truman Doctrine created the language through which Americans would view the world with America as free and communists as tyrannical. According to our old friend Eric Foner, “The speech set a precedent for American assistance to anticommunist regimes throughout the world, no matter how undemocratic, and for the creation of a set of global military alliances directed against the Soviet Union.”[1] It also led to the creation of a new security apparatus – the National Security Council, the Central Intelligence Agency, the Atomic Energy Commission, all of which were somewhat immune from government oversight and definitely not democratically elected. And the containment policy and the Truman Doctrine also laid the foundations for a military build-up – an arms race – which would become a key feature of the Cold War. But it wasn’t all about the military, at least at first. Like, the Marshall Plan was first introduced at Harvard’s Commencement address in June 1947 by, get this, George Marshall, in what turned out to be, like, the second most important commencement address in all of American history. Yes, yes, Stan, okay. It was a great speech, thank you for noticing. Alright, let’s go to the Thought Bubble. The Marshall Plan was a response to economic chaos in Europe brought on by a particularly harsh winter that strengthened support for communism in France and Italy. The plan sought to use US Aid to combat the economic instability that provided fertile fields for communism. As Marshall said “ our policy is not directed against any country or doctrine, but against hunger, poverty, desperation and chaos.” [2] Basically it was a New Deal for Europe, and it worked; Western Europe was rebuilt so that by 1950 production levels in industry had eclipsed pre-war levels and Europe was on its way to becoming a U.S. style-capitalist-mass-consumer society. Which it still is, kind of. Japan, although not technically part of the Marshall Plan, was also rebuilt. General Douglas MacArthur was basically the dictator there, forcing Japan to adopt a new constitution, giving women the vote, and pledging that Japan would foreswear war, in exchange for which the United States effectively became Japan’s defense force. This allowed Japan to spend its money on other things, like industry, which worked out really well for them. Meanwhile Germany was experiencing the first Berlin crisis. At the end of the war, Germany was divided into East and West, and even though the capital, Berlin, was entirely in the east, it was also divided into east and west. This meant that West Berlin was dependent on shipments of goods from West Germany through East Germany. And then, in 1948, Stalin cut off the roads to West Berlin. So, the Americans responded with an 11-month-long airlift of supplies that eventually led to Stalin lifting the blockade in 1948 and building the Berlin Wall, which stood until 1991, when Kool Aid Guy--no, wait, wait, wait, wait, that wasn’t when the Berlin Wall was built. That was in 1961. I just wanted to give Thought Bubble the opportunity to make that joke. Thanks, Thought Bubble. So right, the Wall wasn’t built until 1961, but 1949 did see Germany officially split into two nations, and also the Soviets detonated their first atomic bomb, and NATO was established, AND the Chinese Revolution ended in communist victory. So, by the end of 1950, the contours of the Cold War had been established, West versus East, Capitalist Freedom versus Communist totalitarianism. At least from where I’m sitting. Although now apparently I’m going to change where I’m sitting because it’s time for the Mystery Document. The rules here are simple. I guess the author of the Mystery Document and about 55% of the time I get shocked by the shock pen. “We must organize and enlist the energies and resources of the free world in a positive program for peace which will frustrate the Kremlin design for world domination by creating a situation in the free world to which the Kremlin will be compelled to adjust. Without such a cooperative effort, led by the United States, we will have to make gradual withdrawals under pressure until we discover one day that we have sacrificed positions of vital interest. It is imperative that this trend be reversed by a much more rapid and concerted build-up of the actual strength of both the United States and the other nations of the free world.” I mean all I can say about it is that it sounds American and, like, it was written in, like, 1951 and it seems kind of like a policy paper or something really boring so I...I mean... Yeah, I’m just going to have to take the shock. AH! National Security Council report NSC-68? Are you kidding me, Stan? Not-not 64? Or 81? 68? This is ridiculous! I call injustice. Anyway, as the apparently wildly famous NSC-68 shows, the U.S. government cast the Cold War as a rather epic struggle between freedom and tyranny, and that led to remarkable political consensus--both democrats and republicans supported most aspects of cold war policy, especially the military build-up part. Now, of course, there were some critics, like Walter Lippmann who worried that casting foreign policy in such stark ideological terms would result in the U.S. getting on the wrong side of many conflicts, especially as former colonies sought to remove the bonds of empire and become independent nations. But yeah, no, nothing like that ever happened. Yeah, I mean, it’s not like that happened in Iran or Nicaragua or Argentina or Brazil or Guatemala or Stan are you really going to make me list all of them? Fine. Or Haiti or Paraguay or the Philippines or Chile or Iraq or Indonesia or Zaire or, I’m sorry, THERE WERE A LOT OF THEM, OKAY? But these interventions were viewed as necessary to prevent the spread of communism, which was genuinely terrifying to people and it’s important to understand that. Like, national security agencies pushed Hollywood to produce anticommunist movies like “The Red Menace,” which scared people. And the CIA funded magazines, news broadcasts, concerts, art exhibitions, that gave examples of American freedom. It even supported painters like Jackson Pollack and the Museum of Modern Art in New York because American expressionism was the vanguard of artistic freedom and the exact opposite of Soviet socialist realism. I mean, have you seen Soviet paintings? Look at the hearty ankles on these socialist comrade peasants. Also because the Soviets were atheists, at least in theory, Congress in 1954 added the words “under God” to the pledge of allegiance as a sign of America’s resistance to communism. The Cold War also shaped domestic policy--anti-communist sentiment, for instance, prevented Truman from extending the social policies of the New Deal. The program that he dubbed the Fair Deal would have increased the minimum wage, extended national health insurance and increased public housing, Social Security and aid to education. But the American Medical Association lobbied against Truman’s plan for national health insurance by calling it “socialized” medicine, and Congress was in no mood to pay money for socialized anything. That problem goes away. But the government did make some domestic investments as a result of the Cold War--in the name of national security the government spent money on education, research in science, technology like computers, and transportation infrastructure. In fact we largely have the Cold War to thank for our marvelous interstate highway system, although part of the reason Congress approved it was to set up speedy evacuation routes in the event of nuclear war. And, speaking of nuclear war, it’s worth noting that a big part of the reason the Soviets were able to develop nuclear weapons so quickly was thanks to espionage, like for instance by physicist and spy Klaus Fuchs. I think I’m pronouncing that right. Fuchs worked on the Manhattan Project and leaked information to the Soviets and then later helped the Chinese to build their first bomb. Julius Rosenberg also gave atomic secrets to the Soviets, and was eventually executed--as was his less-clearly-guilty wife, Ethel. And it’s important to remember all that when thinking about the United States’s obsessive fear that there were communists in our midst. This began in 1947 with Truman’s Loyalty Review System, which required government employees to prove their patriotism when accused of disloyalty. How do you prove your loyalty? Rat out your co-workers as communists. No seriously though, that program never found any communists. This all culminated of course with the Red Scare and the rise of Wisconsin senator Joseph McCarthy, an inveterate liar who became enormously powerful after announcing in February 1950 that he had a list of 205 communists who worked in the state department In fact, he had no such thing, and McCarthy never identified a single disloyal American, but the fear of communism continued. In 1951’s Dennis v. United States, the Supreme Court upheld the notion that being a communist leader itself was a crime. In this climate of fear, any criticism of the government and its policies or the U.S. in general was seen as disloyalty. There was only one question--when will I be blown up--and it encouraged loyalty, because only the government could prevent the spread of communism and keep us from being blown up. We’ve talked a lot about different ways that Americans have imagined freedom this year, but this was a new definition of freedom--the government exists in part to keep us free from massive destruction. So, the Cold War changed America profoundly: The U.S. has remained a leader on the world stage and continued to build a large, powerful, and expensive national state. But it also changed the way we imagine what it means to be free, and what it means to be safe. Thanks for watching. I’ll see you next week. Crash Course is created by all of these nice people and it is possible because of you and your support through Subbable.com. Subbable is a crowdfunding website that allows you to support the stuff you love on a monthly basis. Our Subbable subscribers make this show possible. Thanks to them. If you value Crash Course, please check out our Subbable. There are great perks there. And thanks to all of you for watching. As we say in my hometown, don’t forget to be awesome...Wait, wait, wait, Stan, is that music copyrighted? All right! It's not! Woo! That saved us a thousand dollars.

US_History

When_is_Thanksgiving_Colonizing_America_Crash_Course_US_History_2.txt

Hi I’m John Green, this is Crash Course US History, and today we're going to tell the story of how a group of plucky English people struck a blow for religious freedom, and founded the greatest, freest and fattest nation the world has ever seen. [Libertage] These Brits entered a barren land containing no people, and quickly invented the automobile, baseball and Star Trek and we all lived happily ever after. Mr. Green, Mr. Green, if it is really that simple, I am so getting an A in this class. Oh, me from the past, you're just a delight. [Theme Music] So most Americans grew up hearing that the United States was founded by pasty English people who came here to escape religious persecution. And that's true of the small proportion of people who settled in the Massachusetts Bay and created what we now know is New England. But these Pilgrims and Puritans, there's a difference, weren’t the first people or even the first Europeans to come to the only part of the globe we didn't paint over. In fact they weren’t the first English people. The first English people came to Virginia. Off topic but how weird is it that the first permanent English colony in the Americas was named not for Queen Elizabeth’s epicness but for her supposed chastity. Right anyway, those first English settlers weren't looking for religious freedom, they wanted to get rich. So the first successful English colony in America was founded in Jamestown, Virginia in 1607. I say "successful" because there were two previous attempts to colonize the region. They were both epic failures. The more famous of which was the colony of Roanoke Island set up by Sir Walter Raleigh, which is famous because all the colonists disappeared leaving only the word "Croatoan" on carved into a tree. Jamestown was a project of the Virginia Company, which existed to make money for its investors, something it never did. The hope was that they would find gold in the Chesapeake region like the Spanish had in South America, so there were a disproportionate number of goldsmiths and jewelers there to fancy up that gold which of course did not exist. Anyway, it turns out that jewelers dislike farming -- so much so, that Captain John Smith who soon took over control of the island once said that they would rather starve than farm. So in the first year, half of the colonists died. 400 replacements came, but, by 1610, after a gruesome winter called "The Starving Time," the number of colonists had dwindled to 65. And eventually word got out that the new world’s 1 year survival rate was like 20% and it became harder to find new colonists. But 1618, a Virginia company hit upon a recruiting strategy called the headright system which offered 50 acres of land for each person that a settler paid to bring over. And this enabled the creation of a number of large estates, which were mostly worked on and populated by indentured servants. Indentured servants weren't quite slaves, but they were kind of temporary slaves. Like they could be bought and sold and they had to do what their masters commanded. But after seven to ten years of that, if they weren't dead, they were paid their freedom dues which they hoped would allow them to buy farms of their own. Sometimes that worked out, but often either the money wasn't enough to buy a farm, or else they were too dead to collect it. Even more ominously in 1619, just 12 years after the founding of Jamestown, the first shipment of African slaves arrived in Virginia. So the colony probably would have continued to struggle along, if they hadn't found something that people really loved: tobacco. Tobacco had been grown in Mexico since at least 1000 BCE, but the Europeans had never seen it and it proved to be kind of a "thank you for the small pox; here's some lung cancer” gift from the natives. Interestingly King James hated smoking. He called it “a custom loathsome to the eye and hateful to the nose" but he loved him some tax revenue, and nothing sells like drugs. By 1624 Virginia was producing more than 200,000 pounds of tobacco per year. By the 1680s, more than 30 million pounds per year. Tobacco was so profitable the colonists created huge plantations with very little in the way of towns or infrastructure to hold the social order together, a strategy that always works out brilliantly. The industry also structured Virginian society. First off, most of the people who came in the 17th century, three-quarters of them, were servants. So Virginia became a microcosm of England: a small class of wealthy landowners sitting atop a mass of servants. That sounds kind of dirty but it was mostly just sad. The society was also overwhelmingly male, because male servants were more useful in the tobacco fields, they were the greatest proportion of immigrants. In fact they outnumbered women 5 to 1. The women who did come over were mostly indentured servants, and if they were to marry, which they often did because they were in great demand, they had to wait until their term of service was up. This meant delayed marriage which meant fewer children which further reduced the number of females. Life was pretty tough for these women, but on the upside Virginia was kind of a swamp of pestilence, so their husbands often died, and that created a small class of widows or even unmarried women who, because of their special status, could make contracts and own property, so that was good, sort of. OK. A quick word about Maryland. Maryland was the second Chesapeake Colony, founded in 1632, and by now there was no messing around with joint stock companies. Maryland was a proprietorship: a massive land grant to a single individual named Cecilius Calvert. Calvert wanted to turn Maryland into like a medieval feudal kingdom to benefit himself and his family, and he was no fan of the representational institutions that were developing in Virginia. Also Calvert was Catholic, and Catholics were welcome in Maryland which wasn't always the case elsewhere. Speaking of which, let's talk about Massachusetts. So Jamestown might have been the first English colony, but Massachusetts Bay is probably better known. This is largely because the colonists who came there were so recognizable for their beliefs and also for their hats. That’s right. I’m talking about the Pilgrims and the Puritans. And no, I will not be talking about Thanksgiving ...is a lie. I can’t help myself. But only to clear up the difference between Pilgrims and Puritans and also to talk about Squanto. God I love me some Squanto. Let's go to the Thought Bubble. Most of the English men and women who settled in New England were uber-Protestant Puritans who believed the Protestant Church of England was still too Catholic-y with its kneeling and incense and extravagantly-hatted archbishops. The particular Puritans who, by the way did not call themselves that -- other people did, who settled in new England were called Congregationalists because they thought congregations should determine leadership and worship structures, not bishops. The Pilgrims were even more extreme. They wanted to separate more or less completely from the Church of England. So first they fled to the Netherlands, but the Dutch were apparently too corrupt for them, so they rounded up investors and financed a new colony in 1620. They were supposed to land in Virginia, but in what perhaps should have been taken as an omen, they were blown wildly off course and ended up in what's now Massachusetts, founding a colony called Plymouth. While still on board their ship the Mayflower, 41 of the 150 or so colonists wrote and signed an agreement called the Mayflower Compact, in which they all bound themselves to follow "just and equal laws" that their chosen representatives would write-up. Since this was the first written framework for government in the US, it's kind of a big deal. But anyway, the Pilgrims had the excellent fortune of landing in Massachusetts with 6 weeks before winter, and they had the good sense not to bring very much food with them or any farm animals. Half of them died before winter was out. The only reason they didn't all die was that local Indians led by Squanto gave them food and saved them. A year later, grateful that they had survived mainly due to the help of an alliance with the local chief Massasoit, and because the Indians had taught them how to plant corn and where to catch fish, the Pilgrims held a big feast: the first Thanksgiving. Thanks Thought Bubble! And by the way, that feast was on the fourth Thursday in November, not mid-October as is celebrated in some of these green areas we call Not America. Anyway Squanto was a pretty amazing character and not only because he helped save the Pilgrims. He found that almost all of his tribe, the Patuxet had been wiped out by disease and eventually settled with the Pilgrims on the site of his former village and then died... of disease because it is always ruining everything. So the Pilgrims struggled on until 1691 when their colony was subsumed by the larger and much more successful Massachusetts Bay Colony. The Massachusetts Bay colony was chartered in 1629 by London merchants who, like the founders of the Virginia Company, hoped to make money. But unlike Virginia, the board of directors relocated from England to America, which meant that in Massachusetts they had a greater degree of autonomy and self-government than they did in Virginia. Social unity was also much more important in Massachusetts than it was in Virginia. The Puritans' religious mission meant that the common good was, at least at first, put above the needs or the rights of the individual. Those different ideas in the North and South about the role of government would continue...until now. Oh God. It's time for the mystery document? The rules are simple. I read the mystery document which I have not seen before. If I get it right, then I do not get shocked with the shock pen, and if I get it wrong I do. All right. "We must be knit together in this work as one man, we must entertain each other in brotherly affection, we must be willing to abridge ourselves of our superfluities (su-per-fluities? I don't know), for the supply of others necessities, we must uphold a familiar commerce together in all meekness, gentleness, patience and liberality, ...for we must consider that we shall be as a city upon a hill, the eyes of all people are upon us; so that if we shall deal falsely with our god in this work we have undertaken and so cause him to withdraw his present help from us, we shall be made a story and a byword through the world." Alright, first thing I noticed: the author of this document is a terrible speller or possibly wrote this before English was standardized. Also, a pretty religious individual. And the community in question seems to embrace something near socialism: abridging the superfluous for others' necessities. Also it says that the community should be like a city upon a hill, like a model for everybody. And because of that metaphor, I know exactly where it comes from: the sermon "A Model of Christian Charity" by John Winthrop. Yes! Yes! No punishment! This is one of the most important sermons in American history. It shows us just how religious the Puritans were, but it also shows us that their religious mission wasn't really one of individualism but of collective effort. In other words, the needs of the many outweigh the needs of the few or the one. But this city on a hill metaphor is the basis for one kind of American exceptionalism: the idea that we are so special and so godly that we will be a model to other nations, at least as long, according to Winthrop, as we act together. Lest you think Winthrop’s words were forgotten, they did become the centerpiece of Ronald Reagan’s 1989 farewell address. Okay so New England towns were governed democratically, but that doesn't mean that the Puritans were big on equality or that everybody was able to participate in government because no. The only people who could vote or hold office were church members, and to be a full church member you had to be a “visible saint", so really, power stayed in the hands of the church elite. The same went for equality. While it was better than in the Chesapeake Colonies or England, as equality went...eh, pretty unequal. As John Winthrop declared, "Some must be rich, and some poor. Some high, an eminent in power, and dignity; others mean and in subjection." Or as historian Eric Foner put it "Inequality was considered an expression of God's will and while some liberties applied to all inhabitants, there were separate lists of rights for freemen, women, children and servants." There was also slavery in Massachusetts. The first slaves were recorded in the colony in 1640. However, Puritans really did foster equality in one sense. They wanted everyone to be able to read the Bible. In fact, parents could be punished by the town councils for not properly instructing their children in making them literate. But when Roger Williams called for citizens to be able to practice any religion they chose, he was banished from the colonies. So was Ann Hutchinson who argued the church membership should be based on inner grace and not on outward manifestations like church attendance. Williams went on to found Rhode Island, so that worked out fine for him, but Hutchinson, who was doubly threatening to Massachusetts because she was a woman preaching unorthodox ideas, was too radical and was further banished to Westchester, New York where she and her family were killed by Indians. Finally, somebody who doesn't die of disease or starvation. So Americans like to think of their country as being founded by pioneers of religious freedom who were seeking liberty from the oppressive English. We've already seen that's only partly true. For one thing, Puritan ideas of equality and representation weren't particularly equitable or representational. In truth, America was also founded by indigenous people and by Spanish settlers, and the earliest English colonies weren't about religion; they were about money. We'll see this tension between American mythology and American history again next week and also every week. Thanks for watching; I’ll see you next time. Crash Course is produced and directed by Stan Muller, our script supervisor is Meredith Danko, the associate producer is Danica Johnson, the show is written by my high school history teacher, Raoul Meyer and myself, and our graphics team is Thought Bubble. If you have questions about today's video or really about anything about American history, ask them in comments; the entire Crash Course team and many history professionals are there to help you. Thanks for watching Crash Course. Please make sure you are subscribed and, as we say in my home town, "Don't forget to be awesome."

US_History

Westward_Expansion_Crash_Course_US_History_24.txt

Hi, I’m John Green, this is Crash Course U.S. History and today we leave behind the world of industry and corporations to talk about the Wild Wild West. Spoiler Alert: You You have died of dysentery. And in the process, we’re going to explore how all of us, even those of us who are vegan or eat sustainably-produced food benefit from massive agrobusiness that has its roots in the Wild Wild West. The West still looms large in American mythology as the home of cowboys and gunslingers and houses of ill repute and freedom from pesky government interference. But in fact: It was probably not as wild as we’ve been told. Ugh, Mr. Green, why can’t America live up to its myths just once? Because this is America, Me from the Past, home to Hollywood and Gatsby and Honey Boo Boo. We are literally in the mythmaking business. [Theme Music] So, before the Hollywood western, the myth of the Frontier probably found its best expression in Frederick Jackson Turner’s 1893 lecture, “the Significance of the Frontier in American History.” Turner argued that the West was responsible for key characteristics of American culture: beliefs in individualism, political democracy, and economic mobility. Like, for 18th and 19th century Americans, the western frontier represented the opportunity to start over, and possibly to strike it rich by dint of one’s own individual effort, even back when the West was, like, Ohio. In this mythology, the west was a magnet for restless young men who lit out for the uncorrupted, unoccupied, untamed territories to seek their fortune. But, in reality, most western settlers went not as individuals but as members of a family or as part of an immigrant group. And they weren’t filling up unoccupied space either because most of that territory was home to American Indians. Also, in addition to Easterners and migrants from Europe, the West was settled by Chinese people and by Mexican migrant laborers and former slaves. Plus, there were plenty of Mexicans living there already who became Americans with the treaty of Guadalupe Hidalgo. And the whole west as “a place of rugged individualism and independence” turns out to be an oversimplification. I mean, the federal government, after all, had to pass the law that spurred homesteading, then had to clear out American Indians already living there, and had to sponsor the railroads that allowed the West to grow in the first place. About as individualistic as the government buying Walden Pond for Henry David Thoreau. What’s that? It’s a state park now? The government owns it? Well, there you go. Now, railroads didn’t create the desire to settle the west but they did make it possible for people who wanted to live out west to do so, for two reasons. First, without railroads there would be no way to bring crops or other goods to market. I mean, I guess you could dig a canal across Kansas, but, if you’ve ever been to Kansas that is not a tantalizing proposition. Second, railroads made life in the west profitable and livable because they brought the goods that people needed, such as tools for planting and sowing, shoes for wearing, books for putting on your shelf and pretending to have read. Railroads allowed settlers to stay connected with the modernity that was becoming the hallmark of the industrialized world in the 19th century. Now, we saw last week that the Federal government played a key role in financing the transcontinental railroad, but state governments got into the act too, often to their financial detriment. In fact, so many states nearly went bankrupt financing railroads that most states now have constitutional requirements that they balance their budgets. But perhaps the central way that the Federal government supported the railroads, and western settlement and investment in general, was by leading military expeditions against American Indians, rounding them up on ever-smaller reservations, and destroying their culture. Let’s go to the Thought Bubble. There was an economic as well as a racial imperative to move the Native Americans off their land: white people wanted it. Initially it was needed to set down railroad tracks, and then for farming. But eventually it was also exploited for minerals like gold and iron and other stuff that makes industry work. I mean, would you really want a territory called the Badlands unless it had valuable minerals? Early western settlement, of the Oregon Trail kind, did not result in huge conflicts with Native Americans, but by the 1850s, a steady stream of settlers kicked off increasingly bloody conflicts that lasted pretty much until 1890. Even though the fighting started before the Civil War, the end of the “war between the states” meant a new, more violent phase in the warring between American Indians and whites. General Philip H. “Little Phil” Sheridan set out to destroy the Indians’ way of life, burning villages and killing their horses and especially the buffalo that was the basis of the plains tribes’ existence. There were about 30 million buffalo in the U.S. in 1800; by 1886 the Smithsonian Institute had difficulty finding 25 “good specimens.” In addition to violent resistance, some Indians turned to a spiritual movement to try to preserve their traditional way of life. Around 1890 the Ghost Dance movement arose in and around South Dakota. Ghost Dancers believed that if they gathered together to dance and engage in religious rituals, eventually the white man would disappear and the buffalo would return, and with them the Indians’ traditional customs. But even though a combined force of Sioux and Cheyenne warriors completely destroyed George Custer’s force of 250 cavalrymen at Little Bighorn in 1876, and Geronimo took years to subdue in the Southwest, western Native Americans were all defeated by 1890, and the majority were moved to reservations. Thanks, Thought Bubble. Boy, this Wild West episode sure is turning out to be loads of fun! It’s just like the Will Smith movie! All right, Stan, this is about to get even more depressing, so let’s look at, like, some pretty mountains and western landscapes and stuff, while I deliver this next bit. So in 1871 the U.S. government ended the treaty system that had since the American Revolution treated Native American land as if they were nations. And then with the Dawes Act of 1887, the lands set aside for the Indians were allotted to individual families rather than to tribes. Indians who “adopted the habits of civilized life,” which in this case meant becoming small scale individualistic Jeffersonian farmers, would be granted citizenship and there were supposed to be some protections to prevent their land from falling out of Native American possession. But, these protections were not particularly protective and much of the Indian land was purchased either by white settlers or by speculators. After the passage of the Dawes Act “Indians lost 86 million of the 138 million acres of land in their possession.” Oh boy, it’s time for the Mystery Document. The rules here are simple. I guess the author of the Mystery Document. And then you get to see me get shocked when I’m wrong. All right. I have seen the Great Father Chief the Next Great Chief the Commissioner Chief; the Law Chief; and many other law chiefs and they all say they are my friends, and that I shall have justice, but while all their mouths talk right I do not understand why nothing is done for my people. I have heard talk and talk but nothing is done Words do not pay for my dead people. They do not pay for my country now overrun by white men. They do not protect my father's grave. Good words will not give my people a home where they can live in peace and take care of themselves. I am tired of talk that comes to nothing. It makes my heart sick when I remember all the good words and all the broken promises. I mean that could be almost any American Indian leader. This is totally unfair, Stan. All I really know about this is that the Great Father Chief is the President. I mean it could be any of a dozen people. How bout if I say the name in 10 seconds I don’t get punished? Aaaand start. Sitting Bull, Crazy Horse, Geronimo, Chief Big Foot, um, Keokuk, Chief Oshkosh, Chief Joseph Ch-OH YES YES! And now let us move from tragedy to tragedy. So if you’re thinking that it couldn’t get worse for the Native Americans: it did. After killing off the buffalo, taking their land and forcing Indians onto reservations, the Bureau of Indian Affairs instituted a policy that amounted to cultural genocide. It set up boarding schools, the most famous of which was in Carlisle, PA, where Indian children were forcefully removed from their families to be civilized. This meant teaching them English, taking away their clothes, their names, and their family connections. The idea put succinctly, was to “kill the Indian, save the man.” Now, the U.S. wasn’t the only nation busy subjugating its indigenous inhabitants and putting them on reservations in the late 19th century. Like, something similar was happening in South Africa, in Chile, and even to First Peoples in Canada. And you’re usually so good, Canada. Although the slower pace of western settlement meant that there was much less bloodshed, so, another point to Canada. And as bad as the American boarding school policy was, at least it was short lived compared with Australia’s policy of removing Aboriginal children from families and placing them with white foster families, which lasted until the 1970s. All right, Stan, we need to cheer this episode up. Let’s talk about cowboys! The Marlboro Man riding the range, herding cows and smoking, solitary in the saddle, alone in his emphysema. Surely that is the actual West, the men and women but mostly men who stood apart from the industrializing country as the last of Jefferson’s rugged individuals. But, no. Once again, we have the railroad to thank for our image of the cowboy. Like, those massive cattle drives of millions of cows across open range Texas? Yeah, they ended at towns like Abilene, and Wichita, and Dodge City – because that’s where the railheads were. Without railroads, cowboys would have just driven their cattle in endless circles. And without industrial meat processing, there wouldn’t have been a market for all that beef. And it was a lot of beef. You know what I’m talking about. I’m actually talking about beef. By the mid 1880s the days of open range ranching were coming to an end, as ranchers began to enclose more and more land and set up their businesses closer to, you guessed it, railroad stations. There are also quite a few things about western farming that just fly in the face of the mythical Jeffersonian yeoman farmer ideal. Firstly, this type of agricultural work was a family affair; many women bore huge burdens on western farms, as can be seen in this excerpt from a farm woman in Arizona: “Get up, turn out my chickens, draw a pail of water … make a fire, put potatoes to cook, brush and sweep half inch of dust off floor, feed three litters of chickens, then mix biscuits, get breakfast, milk, besides work in the house and this morning had to go half mile after calves.” These family-run farms were increasingly oriented towards production of wheat and corn for national and even international markets rather than trying to eke out subsistence. Farmers in Kansas found themselves competing with farmers in Australia and Argentina, and this international competition pushed prices lower and lower. Secondly, the Great Plains, while remarkably productive agriculturally, wouldn’t be nearly as good for producing crops without massive irrigation projects. Much of the water needed for plains agriculture comes a massive underground lake, the Oglala Aquifer. Don’t worry, by the way, the Aquifer is fed by a magic and permanent H20 factory in the core of the earth that you can learn about in Hank’s show, Crash Course Chemis– What’s that? It’s going dry. MY GOD THIS IS A DEPRESSING EPISODE. Anyway, large-scale irrigation projects necessitate big capital investments, and therefore large, consolidated agricultural enterprises that start to look more like agri-business than family farms. I mean, by 1900, California was home to giant commercial farms reliant on irrigation and chemical fertilizers. Some of them were owned, not by families, but by big corporations like the Southern Pacific Railroad. And they were worked by migrant farm laborers from China, the Philippines, Japan, Mexico. As Henry George, a critic of late 19th century corporate capitalism, wrote “California is not a country of farms, but … of plantations and estates.” When studying American history, it’s really easy to get caught up in the excitement of industrial capitalism with its robber barons, and new technologies, and fancy cities because that world looks very familiar to us, probably because it’s the one in which we live. After all, if I was running a farm like that Arizona woman I talked about earlier, there’s no way I could be making these videos because I’d be chasing my calves. I don’t even know what a litter of chickens is. Is it 4 chickens? 12? 6? It’s probably 12 because eggs do come in dozens. The massive agricultural surplus contemporary farms create, and the efficient transportation network that gets that surplus to me quickly, makes everything else possible – from YouTube to Chevy Volts. And no matter who you are, you benefit from the products that result from that massive surplus. That’s why we’re watching YouTube right now. So, agriculture and animal husbandry did change a lot in late 19th century America, as we came to embrace the market driven ethos that we either celebrate or decry these days. And in the end, the Wild West ends up looking a lot more like industrial capitalism than like a Larry McMurtry novel. The Wild West, like the rest of the industrialized world, was incentivized to increase productivity and was shaped by an increasingly international economic system. And it’s worth remembering that even though we think of the Oregon Trail and the Wild West being part of the same thing. In fact, they were separated by the most important event in American history: the Civil War. I know that ain’t the mythologizing you’ll find in Tombstone, but it is true. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer of the show is Danica Johnson. The show is written by my high school history teacher Raoul Meyer, Rosianna Halse Rojas, and myself. And our graphics team is Thought Café. Every week, there’s a new caption for the libertage. If you’d like to suggest one you can do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course. If you enjoy it, make sure you subscribe. And as we say in my hometown, don’t forget to be awesome. OH, ahh I didn’t get a good push.

US_History

The_Civil_War_Part_2_Crash_Course_US_History_21.txt

Hi, I’m John Green; this is Crash Course U.S. History and today we return to...wait, what are we talking about today, Stan? Ah, the Civil War! I can tell because Lincoln’s here. But this week we’re not gonna talk about casualty counts or battles or its generals with their heroic and probably fictional dying declarations. Mr. Green, Mr. Green, wait did that one guy not really say “Honeybun how do I look in the face?” because that was the best part of this whole class. Jeb Stewart did say that, Me from the Past, but it probably wasn’t his last words. But anyway today we’re going to try to focus on what’s really important. In the end the really vital stuff isn’t, like, Pickett’s Charge or Lee saying “It is well that war is so terrible - otherwise we would grow too fond of it” or the surrender at the Appomattox Court House. That stuff matters and I don’t want to deny it, but the Civil War and the way we remember it is still shaping the world today, and that’s what I want to focus on, because it’s the stuff that might actually change the way you think about your own life in your own country, whether it’s the United States or the Green Parts of Not America. [Theme Music] So let’s start with one of the big questions historians still ask about the Civil War: Did Lincoln free the slaves? The answer, as with so much here on Crash Course is yes … and also no. Let’s go straight to the Thought Bubble today. So Lincoln’s reputation as the Great Emancipator rests largely on his Emancipation Proclamation, an executive order which went into effect on January 1, 1863. This order ostensibly freed all the slaves in territory currently rebelling against the United States, i.e. in areas where the U.S. government had no authority to free slaves. This is rather like the United States announcing that from here on out, North Korea will be ruled by Lady Gaga. Sure, it’s a great idea, but it’s not really your jurisdiction. In areas where the U.S. did have the authority to free slaves, the border states and some of the areas of the Confederacy that had been effectively conquered and occupied by federal troops, those slaves were NOT freed. So Lincoln didn’t free the slaves that he actually had the power to free. Many historians argue that, in fact, slaves freed themselves. How? By running away to union lines and becoming “contrabands.” Because this was a time of war and slaves were seen as a valuable resource to the enemy, when they escaped and sought refuge with Union troops, Union commanders wouldn’t give them back, despite fugitive slave laws still being on the books. So many slaves escaped, the argument goes, that Lincoln was basically forced to issue the Emancipation Proclamation, because until he did so, those contraband slaves were still technically property of their Southern masters, and the Union generals were breaking American laws by not returning them. The Emancipation Proclamation then had the added bonus of encouraging more slaves to come over to the Union lines, many of whom joined the army, which eventually included about 180,000 former slaves and free black men. Thanks, Thought Bubble. So Lincoln may also have issued the proclamation in order to shift the focus of the war from union to slavery to prevent the British from recognizing the Confederacy. Arguably the Confederacy’s best chance to win the Civil War was to get some kind of foreign patron, and Britain was the likeliest choice as it was very dependent on Confederate textiles. But as you’ll remember from all those people going to Canada, Britain had already abolished slavery and it was the historic source of abolitionist sentiment, and so it was very shrewd of Lincoln to make the war about slavery. Off-topic, but if I may put on my world historian hat for a moment. Thank you, Stan. The fact that the British did not recognize the South had profound effects on the whole world, because it meant that the British shifted their focus to Egypt and India as sources of cotton for their textile mills. All that noted, I think Lincoln does deserve some credit for freeing the slaves for two reasons. First, he pushed for the Thirteenth amendment which actually ended slavery in the United States. And perhaps more importantly, he continued the war to its conclusion and demanded that the end of slavery and the return of the Southern states to the Union be conditions for peace. This may seem obvious today, but in 1864 it wasn’t. In fact, there were numerous calls in the North for an end to the war that would allow the South to exist as a separate country and leave slavery intact. Now, of course, the rest of world history indicates that at some point slavery would have ended, but by prosecuting the war to its end, Lincoln brought about slavery’s end sooner. But the Civil War didn’t just end slavery. If it had gone differently, Me from the Past might have been annoying teachers in a different country from the one in which I now live. I might’ve needed a passport to visit my parents in North Carolina and slavery might have survived for decades--Brazil didn’t fully abolish slavery until 1888. And the South would be covered in green as part of Not-America. Or, the North, depending on where you’re watching this video, I guess. And, the people who lived through the Civil War knew it was momentous. In his famous Gettysburg Address, Abraham Lincoln fostered the idea that the Civil War was a kind of second American Revolution, or at least a culmination and reaffirmation of the first one. “From these honored dead we take increased devotion to that cause for which they gave the last full measure of devotion--that we here highly resolve that these dead shall not have died in vain— that this nation, under God, shall have a new birth of freedom—and that government of the people, by the people, for the people, shall not perish from the earth.” We tried to hire Daniel Day-Lewis for that, but he was unavailable. That phrase “new birth” of freedom had religious significance as well because it was, like, the 19th century equivalent to “born again.” So, the Civil War was the first modern war in terms of its scale and its destruction. Like, others have waged war on civilians to break the spirit of their enemies (STAN! Mongoltage OPPORTUNITY!) [Mongoltage] But new technologies made this one of the most destructive wars yet recorded. And, yes, I know the Taiping rebellion took more lives, and in terms of percentage of population killed, the contemporaneous war in Paraguay was worse, but bear with me. Rifles, and toward the end of the Civil War, machine guns shifted the way that people fight. It became easier to defend a line, so cavalry charges and huge waves of attacks started to be just slaughtery although it would take World War I for the rest of the world to figure that out. And the incredible numbers of dead and wounded really changed Americans’ relationship with death itself. Like, the Gettysburg address was given to dedicate a new national cemetery, and the Civil War helped to create a culture of meditation on mortality itself that led to cemeteries replacing churchyards as the final resting places for most Americans. And the sight of slaughter and the sheer weight of it had profound existential effects on a generation of American intellectuals from Walt Whitman and Oliver Wendell Holmes. Oh, it’s time for the Mystery Document? The rules here are simple. I guess the Mystery Document and usually I am shocked. Oh my gosh, today’s Mystery Document is on an iPad! This appears to be a photograph of wounded soldiers in hospital. I’m gonna go ahead and call it as being by Mathew Brady. What? I already got it? But I didn’t get to say the name... Oh, it’s called Wounded Soldiers in Hospital. Thank you for an easy one, Stan. So, Mathew Brady was a prolific photographer during the Civil War, although, like a lot of prolific people, he often took credit for work done by his employees. And Brady really changed the way that people thought about war. He and his staff created some 10,000 images during the Civil War. And it was the first time that an event had been photographically documented so thoroughly. By the way, lest you think that the unreliability of images began with Photoshop, many of Brady’s photographs were staged. He would move bodies, sometimes soldiers were apparently told to act dead. But of course, at the time, photographs felt inherently authentic and written accounts of battles could now be accompanied by actual images of the fighting and its aftermath. But, perhaps the most important impact of the Civil War was the new nation that it created. Like, the American Civil War fits right in with the global phenomenon of nation-building that was happening. Soon we would have places on the map like Italy and Germany, and older places like Greece would be re-born as nation states. And then all of these places would be known to Americans as Not-America. But, by the way, congratulations to Italy on the recent election of their 732nd Prime Minister in just 180 years of existing. By far, the most successful of these new nation states were the ones that embraced industrialization and modern ideas of organization and centralized government. Northern victory in the Civil War meant the United States would follow the path that the North laid down. It would become an industrial rather than agrarian nation, with a national government pre-eminent over those of individual states. It would become a nation. And its not a coincidence that over the course of the 19th century, people stopped pluralizing the United States; they stopped saying, “The United States are a great place to live,” and began saying, “The United States is a great place to live.” The Civil War helped singularize what had been until then a plural nation. And Abraham Lincoln was the first president to truly expand the power of the executive. He ordered blockades and suspended habeas corpus, in addition to emancipating the slaves. But the Republican dominated congress played a role in this federalization too. Congress passed the Homestead Act in 1862 that encouraged settlement of the west by basically giving away land to anyone who had $18 and was willing to live on it and farm it for five years. Meanwhile, the Morrill Land Grant Act financed colleges to offer training in new scientific agricultural techniques. The Department of Agriculture was created to generate statistics and share best practices in farming. Congress also helped unify the country with the massive land grants in the Pacific Railway Act of 1862. And during the war the Lincoln administration gave away 158 million acres to railroads to tie the nation together. Get it? Tie? Railroad ties? The nation toge-? I’ll take my coat and go. Plus, as you may have noticed, wars are expensive. And in order to finance the Civil War, Congress passed the first progressive income tax in American history, as well as floating huge bond issues to the public. And when that wasn’t enough, the administration began printing federal money on green paper called “greenbacks.” These, along with notes issued by banks under the National Bank Act of 1863 became the first national currency in the United States. Altogether, the total cost of the war for the Union was $6.7 billion. Interestingly, if in 1860 the federal government had purchased every slave and granted a 40-acre farm to each family, the total cost would have been $3.1 billion. But a) it would have been hard to get that bill through Congress, and b) at the time the federal government had no way to raise that kind of money. The federal government also actively promoted the industrial economy that was to become dominant in the United States after the war. In fact, industrialization was so healthy that visitors to cities in the North during the Civil War would have been hard pressed to notice that they were even in a war. So, ultimately, the Civil War was a victory for Alexander Hamilton’s federalist vision of what America should be. I mean, Thomas Jefferson could never have imagined the United States that emerged from the Civil War, a government that supported an army of a million men, carried a $2.5 billion national debt, distributed public lands, printed a national currency, and collected an array of internal taxes. It sounds like Britain! So, the Civil War wasn’t just a victory of North over South or of freedom over slavery. It created the nation that the United States of America has become. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. Too far! Our associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, and myself. And our graphics team is Thought Café. Every week, there’s a new caption for the Libertage. You can suggest some in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thank you for watching and as we say in my hometown, don’t forget to be awesome.

US_History

Battles_of_the_Civil_War_Crash_Course_US_History_19.txt

Hi, I’m John Green and this is Crash Course US History. Starting next week, we’re going to be talking about the Civil War. As you may have noticed, Crash Course doesn’t usually focus on military history, because we’re more interested in causes and effects and that kind of stuff, but because some in our audience are likely to insist that a series on American History has to include the battles of the Civil War, I am now going to tell you about EVERY SINGLE fight of the war. Oh. Stan, this says there were 8,000 instances of violence between the Union and the Confederate States of America between 1861 and 1865. Can that be right? Slight change of plan. I’m going to tell you about the MAJOR battles of the Civil War. [Theme Music] The shooting started in 1861. In April the first shots of the war were fired at the Battle of Fort Sumter, South Carolina, which the South won. Next the Battle of Rich Mountain went to the North. First Bull Run happened in Manassas, VA in July. The South won, General Jackson got the nickname Stonewall, and the North realized this war was going to be serious business. The South had another victory at Wilson’s Creek in August, but lost to the Union at Carnifex Ferry in September. The North got another win at the Battle of Cheat Mountain, but the South finished the year strong with wins at Ball’s Bluff and Chustenahlah. The North came roaring back in 1862 with wins at the battles of Mill Springs, Fort Henry, Roanoke Island, and Fort Donelson, where the Confederate general was named Simon Bolivar Buckner. What!? There was fighting in the west, with the South winning at Valverder, NM, and Pea Ridge, AR. The Union won at New Madrid, MO. The Battle of Hampton Roads, VA in March was a draw, and featured the first fight between two ironclad warships, the USS Monitor and the CSS Virginia. The Union won the Battle of Bern, and the first battle of Kernstown, and then draws at Glorieta Pass, NM, and Yorktown. The Union won the Battle of Shiloh, where future Ben Hur author Lew Wallace was accused of incompetence and cowardice. The Union also won at the Battle of Fort Pulaski and the Battles of Forts Jackson and St. Philip, which gave the North control of New Orleans. The North then won the Siege of Corinth, and Union general McClellan fought one of the most indecisive battles he would be involved in at Williamsburg, Virginia, which is really saying something, because he was pretty indecisive. Then Stonewall Jackson had a great run, winning battles at McDowell, Front Royal, and Winchester. Union forces captured Memphis, TN in May, but then lost a couple more to Jackson at Cross Keys and Port Republic. In Virginia that June, Robert E. Lee and George McClellan fought a series of six battles in seven days, which were called the seven days battles, because historians are so good at naming things. There was a draw at Oak Grove, Union victory at Beaver Dam Creek, a win for Lee at Gaines' Mill, ties at Garnett's and Golding's Farms, The Battle of Savage's Stations, and the Battle for Glendale. The Union finally won the Seven Days at Malvern Hill, but McClellan withdrew after the battle, allowing Lee and the remaining confederates to escape. In July 1862, one of the least consequential battles of the war took place in Stan’s hometown, Newburgh, IN. A force of 35 Confederate irregulars built some fake cannons out of stovepipes that they called Quaker guns, crossed the Ohio River, captured some weapons and a hospital full of wounded Union soldiers, and then abandoned the town later the same day. Later that summer came the Battles of Baton Rouge, Cedar Mountain, Mannassas Station, the Second Battle of Bull Run, Richmond, KY, Chantilly, and Harper’s Ferry, all of which the Confederates won. The Union won at South Mountain, but lost at Munfordville. On September 17 McClellan ended Lee’s invasion of the north at Anteitam, MD. This was the bloodiest single day of the war with 22,717 dead, wounded, or missing. The rest of 1862 saw the battles of Perryville, Prairie Grove, Fredericksburg, and Chickasaw Bayou. By the way, if this is starting to sound like baseball box scores, maybe you're getting a glimpse of why we don't usually do the military history. Right, but back to the Civil War: 1863 started in Tennessee with a Union victory at the Battle of Stones River, but also with a simultaneous Confederate victory at Galveston, TX. There were lots of smallish skirmishes during the winter and early Spring, until the Battle of Chancellorsville in May, where Lee defeated Hooker, but Stonewall Jackson was mortally wounded. Jackson lost his left arm and then Lee said "I have lost my right arm." Jackson then proceeded to famously say "Let us cross over the river and rest under the shade of those trees," and then died 8 days later. His arm, by the way, is buried with its own headstone near Chancellorsville. The North won a bunch of battles in Mississippi at Port Gibson, Raymond, Jackson, Champion Hill, Big Black River Ridge, and finally, Vicksburg. That victory, along with the victory at Port Hudson, effectively ended the Confederates' ability to use the Mississippi River. Fighting in June 1863 in Virginia was inconclusive with draws in Brandy Station, Aldie and Upperville, and then July brought the Battle of Gettysburg, a major Union victory. This battle featured Pickett's famous charge, it was the end of Lee's second invasion of the North, it was the costliest battle of the war in terms of casualties and it led to, surprisingly, the Gettysburg Address. The rest of that summer brought split results to Confederate victories at Fort Wagner, South Carolina. They also won at Fort Sumter again, holding the fort against heavy Union bombardment, and they perpetrated a massacre of civilians in Warrens, Kansas. The rest of 1863 saw battles at the Bayou of Force, Arkansas; a major Confederate win at Chickamauga; the battles of Bristoe Station, Wauhatchie, Rappahannock Station, Chattanooga again, Ringgold Gap, Fort Sanders and Mossy Creek. Stan, are you making some of these places up? A creek cannot be mossy. Oh man, we’re only to 1864. Gotta keep it moving. Stan, can you just indicate who won these on screen? Awesome. The spring saw the Battles of Mansfield; Pleasant Hill, Louisiana and Fort Pillow, Tennessee. Then there was the Battles of the Wilderness, Spotsylvania Courthouse, New Market, North Anna, Old Church and the Battle of the Cold Harbor, all in Virginia. Summer brought the battles of Marietta, Georgia and Petersburg were all fought in Virginia as well. And there was Kennesaw Mountain, Georgia; Monocacy, Maryland and Fort Stevens in the District of Columbia. By the way, Abraham Lincoln himself went to observe that battle and the guy standing next to him got shot. Then there were the Battles of Peach Tree Creek, Georgia and Atlanta, Georgia - a major Union victory won by General Sherman. Also fought were the Battles of Ezra Church, Georgia; the Battle of the Crater in Virginia; the Battles of Mobile Bay, Alabama; Deep Bottom Globe Tavern and Jonesboro, Georgia, which solidified Union control of Atlanta. Okay, autumn 1864 we are really in the home stretch. The Battles of Opequon, Fisher's Hill, Chaffin's Farm and Cedar Creek were all fought in Virginia. Johnsonville, Tennessee; the Sand Creek Massacre; the Battle of Franklin, Tennessee; Fort McAllister, Georgia and Nashville, Tennessee finished out the year. Alright, 1865 here we go: The Battles of Fort Fisher, North Carolina; Hactcher's Run, Virginia; Bentonville, North Carolina; Fort Stedman, Virginia; Five Forks, Virginia; the Third Battle of Petersburg, Virginia; Fort Blakely, Alabama; Sailor's Creek, Virginia, Appomattox Station, Virginia; and, finally, the decisive battle, at Appomattox Courthouse, Virginia, on April 8th, 1865, which resulted in Lee's surrender to Grant - there were a few more minor skirmishes - but the war was over! [Libertage: American wins! And America loses?] Ugh, so there you have it, an episode of Crash Course entirely about battles. What did we learn? Very little in the end. And I know I missed many battles of the war, but I also didn't miss many. There's no big finish today because I am exhausted from all of those battles and all of that fighting and death - thanks for watching, see you next week. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our Associate Producer is Danica Johnson. Today's show was written - by Stan. And our Graphics Team is Thought Café. Every week there's a new caption for the libertage. If you'd like to suggest one you can do so in comments or you can also ask questions about today's video that will be answered by our team of historians. Thank you for watching Crash Course and as we say in my hometown: Don't Forget to be Awesome.

US_History

The_Industrial_Economy_Crash_Course_US_History_23.txt

Episode 23: The Rise of the Industrial Economy Hi I’m John Green this is Crash Course U.S. History and today we’re going to discuss economics and how a generation of- Mr. Green, Mr. Green, is this going to be one of those boring ones no wars or generals who had cool last words or anything? Alright, Me From The Past, I will give you a smidge of Great Man history. But only a smidge. So today we’re gonna discuss American industrialization in the decades after the Civil War, during which time the U.S. went from having per capita about a third of Great Britain’s industrial output to becoming the richest and most industrialized nation on earth. Libertage Meh, you might want to hold off on that Libertage, Stan because this happened mostly thanks to the Not Particularly Awesome Civil War, which improved the finance system by forcing the introduction of a national currency and spurred industrialization by giving massive contracts to arms and clothing manufacturers. The Civil War also boosted the telegraph, which improved communication, and gave birth to the transcontinental railway via the Pacific Railway Act of 1862, all of which increased efficiency and productivity. So thanks, Civil War! Intro If you want to explain America’s economic growth in a nutshell chalk it up to G, D, and L: Gerard, Depardieu, and Lohan. No, Geography, Demography and Law. However, while we’re on the topic, when will Gerard, Depardieu, and Lindsay Lohan have a baby? Stan, can I see it? Yes. Yes. Geographically, the U.S. was a huge country with all the resources necessary for an industrial boom. Like, we had coal, and iron and, later, oil. Initially we had water to power our factories, later replaced by coal. And we had amber waves of grain to feed our growing population which leads to the Demography. America’s population grew from 40 million in 1870 to 76 million in 1900 and 1/3 of that growth was due to immigration. Which is good for economies. Many of these immigrants flooded the burgeoning cities, as America shifted from being an agrarian rural nation to being an industrial, urban one. Like, New York City became the center of commerce and finance and by 1898 it had a population of 3.4 million people. And the industrial heartland was in the Great Lakes region. Chicago became the second largest city by 1900, Cleveland became a leader in oil refining, and Pittsburgh was a center of iron and steel production. And even today, the great city of Pittsburgh still employs 53 Steelers. Last but not least was the Law. The Constitution and its commerce clause made the U.S. a single area of commerce – like a giant customs union. And, as we’ll see in a bit the Supreme Court interpreted the laws in a very business friendly way. Also, the American constitution protects patents, which encourag4B-es invention and innovation, or at least it used to. And despite what Ayn Rand would tell you, the American government played a role in American economic growth by putting up high tariffs, especially on steel, giving massive land grants to railroads and by putting Native Americans on reservations. Also, foreigners played an important role. They invested their capital and involved Americans in their economic scandals like the one that led to a depression in 1893. The U.S. was at the time was seen by Europeans as a developing economy; and investments in America offered much higher returns than those available in Europe. And the changes we’re talking about here were massive. In 1880, for the first time, a majority of the workforce worked in non-farming jobs. By 1890 2/3 of Americans worked for wages, rather than farming or owning their own businesses. And, by 1913 the United States produced 1/3 of the world’s total industrial output. NOW bring out the Libertage, Stan. Libertage Awesome. And even better, we now get to talk about the perennially underrated railroads. Let’s go to the Thought Bubble. Although we tend to forget about them here in the U.S., because our passenger rail system sucks, railroads were one of the keys to America’s 19th century industrial success. Railroads increased commerce and integrated the American market, which allowed national brands to emerge, like Ivory Soap and A&P Grocery Stores. But railroads changed and improved our economy in less obvious ways, too: For instance, they gave us time zones, which were created by the major railroad companies to make shipping and passenger transport more standard. Also because he recognized the importance of telling time, a railroad agent named Richard Warren Sears turned a $50 dollar investment in watches into an enormous mail order empire, and railroads made it possible for him--and his eventual partner Roebuck--to ship watches, and then jewelry, and then pretty much everything, including unconstructed freaking houses throughout the country. Railroads were also the first modern corporations. These companies were large, they had many employees, they spanned the country. And that meant they needed to invent organizational methods, including the middle manager--supervisors to supervise supervisors. And for the first time, the owners of a company were not always day-to-day managers, because railroads were among the first publicly traded corporations. They needed a lot of capital to build tracks and stations, so they sold shares in the company in order to raise that money, which shares could then be bought and sold by the public. And that is how railroads created the first captains of industry, like Cornelius “They Named a University after Me” Vanderbilt and Andrew “Me Too” Carnegie (Mellon) and Leland “I Named a University After My Son” Stanford. The Railroad business was also emblematic of the partnership between the national government and industry. The Transcontinental Railroad, after all, wouldn’t have existed without Congressional legislation, federal land grants, and government sponsored bond issues. Thanks, Thought Bubble. Apparently it’s time for the Mystery Document. The rules here are simple. I guess the author of the Mystery Document and if I’m wrong, which I usually am, I get shocked. Alright. “The belief is common in America that the day is at hand when corporations far greater than the Erie – swaying such power as has never in the world’s history been trusted in the hands of mere private citizens, controlled by single men like Vanderbilt...– will ultimately succeed in directing government itself. Under the American form of society, there is now no authority capable of effective resistance.” Corporations directing government? That’s ridiculous. So grateful for federal ethanol subsidies brought to you by delicious Diet Dr. Pepper. Mmm I can taste all 23 of the chemicals. Anyway, Stan, I’m pretty sure that is noted muckraker Ida Tarbell. No! Henry Adams? HOW ARE THERE STILL ADAMSES IN AMERICAN HISTORY? That makes me worry we’ll never escape the Clintons. Anyway, it should’ve been Ida Tarbell. She has a great name. She was a great opponent of capitalism. Whatever. AH! Indeed industrial capitalists are considered both the greatest heroes and the greatest villains of the era, which is why they are known both as “captains of industry” and as “robber barons,” depending on whether we are mad at them. While they often came from humble origins, took risks and became very wealthy, their methods were frequently unscrupulous. I mean, they often drove competitors out of business, and generally cared very little for their workers. The first of the great robber barons and/or captains of industry was the aforementioned Cornelius Vanderbilt who rose from humble beginnings in Staten Island to make a fortune in transportation, through ferries and shipping, and then eventually through railroads, although he once referred to trains as “them things that go on land.” But the poster boy of the era was John D. Rockefeller who started out as a clerk for a Cleveland merchant and eventually became the richest man in the world. Ever. Yes, including Bill Gates. The key to Rockefeller’s success was ruthlessly buying up so many rivals that by the late 1880s Standard Oil controlled 90% of the U.S. oil industry. Which lack of competition drove the price of gasoline up to like 12 cents a gallon, so if you had one of the 20 cars in the world then, you were mad. The period also saw innovation in terms of the way industries were organized. Many of the robber barons formed pools and trusts to control prices and limit the negative effects of competition. The problem with competition is that over time it reduces both prices and profit margins, which makes it difficult to become super rich. Vertical integration was another innovation – firms bought up all aspects of the production process – from raw materials to production to transport and distribution. Like, Philip Armour’s meat company bought its own rail cars to ship meat, for instance. It also bought things like conveyor belts and when he found out that animal parts could be used to make glue, he got into the glue-making business. It was Armour who once proclaimed to use “everything but the squeal.” Horizontal integration was when big firms bought up small ones. The best example of this was Rockefeller’s Standard Oil, which eventually became so big incidentally that the Supreme Court forced Standard Oil to be broken up into more than a dozen smaller oil companies. Which, by the way, overtime have slowly reunited to become the company known as Exxon-Mobil, so that worked out. U.S. Steel was put together by the era’s giant of finance, J.P. Morgan, who at his death left a fortune of only $68 million – not counting the art that became the backbone of the Metropolitan Museum of Art – leading Andrew Carnegie to remark in surprise, “And to think he was not a rich man.”[1] Speaking of people who weren’t rich, let us now praise the unsung heroes of industrialization: workers. Well, I guess you can’t really call them unsung because Woody Guthrie. Oh! Your guitar! And my computer! I never made that connection before. Anyway, then as now, the benefits of economic growth were shared...mmm shall we say...a smidge unevenly. Prices did drop due to industrial competition, which raised the standard of living for the average American worker. In fact, it was among the highest in the world. But due to a growing population, particularly of immigrant workers, there was job insecurity. And also booms and busts meant depressions in the 1870s and 1890s, which hit the working poor the hardest. Also, laborers commonly worked 60 hours per week with no pensions or injury compensation, and the U.S. had the highest rate of industrial injuries in the world: an average of over 35,000 people per year died on the job. These conditions and the uncertainty of labor markets led to unions, which were mostly local but occasionally national. The first national union was the Knights of Labor, headed by Terence V. Powderly which grew from 9 members in 1870 to 728,000 by 1884. The Knights of Labor admitted unskilled workers, black workers, and women, but it was irreparably damaged by the Haymarket riot in 1886. During a strike against McCormick Harvesting Company, a policeman killed one of the strikers and in response there was a rally in Chicago’s Haymarket Square at which a bomb killed seven police officers. Then, firing upon the crowd, the police killed four people. Seven anarchists were eventually convicted of the bombing, and although Powderly denounced anarchism, the public still associated the Knights of Labor with violence. And by 1902, its membership had shrunk considerably--to 0. The banner of organized labor however was picked up by the American Federation of Labor under Samuel L. Gompers. Do all of these guys have great last names? They were more moderate than the anarchists and the socialist International Workers of the World, and focused on bread and butter issues like pay, hours, and safety. Founded in 1886, the same year as the Haymarket Riot, the AFL had about 250,000 members by 1892, almost 10% of whom were iron and steel workers. And now we have to pause to briefly mention one of the most pernicious innovations of the era: Social Darwinism: a perversion of Darwin’s theory that would have made him throw up. Although to be fair, almost everything made him throw up. Social Darwinists argued that the theory of survival of the fittest should be applied to people and also that corporations were people. Ergo, big companies were big because they were fitter and we had nothing to fear from monopolies. This pseudoscience was used to argue that government shouldn’t regulate business or pass laws to help poor people. It assured the rich that the poor were poor because of some inherent evolutionary flaw, thus enabling tycoons to sleep at night. You know, on a big pile of money, surrounded by beautiful women. But, despite the apparent inborn unfitness of workers, unions continued to grow and fight for better conditions, sometimes violently. There was violence at the Homestead Steel Strike of 1892 and the Pullman Rail strike of 1894 when strikers were killed and a great deal of property was destroyed. To quote the historian Michael Lind: “In the late 1870s and early 1880s, the United States had five times as many unionized workers as Germany, at a time when the two nations had similar populations.”[2] Unions wanted the United States and its citizens to imagine freedom more broadly, arguing that without a more equal economic system, America was becoming less, not more, free, even as it became more prosperous. If you’re thinking that this free-wheeling age of fast growth, uneven gains in prosperity, and corporate heroes/villains resembles the early 21st century, you aren’t alone. And it’s worth remembering that it was only 150 years ago that modern corporations began to form and that American industry became the leading driver in the global economy. That’s a blink of an eye in world history terms, and the ideas and technologies of post Civil War America gave us the ideas that still define how we--all of us, not just Americans--think about opposites like success and failure, or wealth and poverty. It’s also when we people began to discuss the ways in which inequality could be the opposite of freedom. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, Rosianna Halse Rojas, and myself. And our graphics team is Thought Café. Each week there’s a new caption for the Libertage. You can suggest captions in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course. Make sure you’re subscribed. And as we say in my hometown, don’t forget to be awesome. Industrial Economy - ________________ [1] Brands, American Colossus p 6. [2] Lind, Land of Promise 171

US_History

America_in_World_War_I_Crash_Course_US_History_30.txt

Episode 30: America and World War I Hi I’m John Green, this is Crash Course U.S. history and today we’re finally going to make the military history buffs happy. That’s right, today we’re going to talk about how the United States with its superior technology, innovative tactics and remarkable generalship turned the tide of World War I. Mr. Green, Mr. Green. Finally. I’ve been waiting for months to learn about tanks and airplanes and Ernest Hemingway. Well that’s a shame, Me from the Past, because I was kidding about this being an episode full of military details. But I do promise that we will mention Ernest Hemingway. And in a few weeks I will tell you about how he liberated the martinis of Paris. intro Americans were only involved in the Great War for 19 months and, compared with the other belligerents, we didn’t do much fighting. Still, the war had profound effects on America at home, on its place in the world and it also resulted in an amazing number of war memorials right here in Indianapolis. So, The Great War, which lasted from 1914 until 1918, and featured a lot of men with hats and rifles, cost the lives of an estimated 10 million soldiers. Also the whole thing was kind of horrible and pointless, unless you love art and literature about how horrible and pointless World War I was in which case, it was a real bonanza. So, when the war broke out, America remained neutral, because we were a little bit isolationist owing to the fact that we were led, of course, by President Wilson. But many Americans sided with the British because by 1914 we’d pretty much forgotten about all the bad parts of British rule, like all that tea and monarchy. Plus, they’re so easy to talk to with their English. But there were a significant number of Progressives who worried that involvement in the war would get in the way of social reforms at home. In fact, Wilson courted these groups in the 1916 presidential campaign running on the slogan “He kept us out of War.” And will continue to keep us out of war until we reelect him and then he gets us into war. But, for that slogan to make sense, there had to have been some way in which war was avoided, which brings me to one of the classic errors made by American history students. What? I haven’t even said anything yet. But you were about to, Me From the Past, because if I had asked you what event led the U.S. to enter World War I, you would have surely told me that it was the sinking of the cruise ship Lusitania by German submarines. 124 American passengers died when the ship, which had been carrying arms and also guns, was torpedoed off the coast of Ireland. Even though Secretary of State William Jennings Bryan had warned Americans not to travel on British, French, or German ships, Wilson refused to ban such travel because, you know: freedom. Bryan promptly resigned. So how do I know it wasn’t the immediate cause of our involvement in the war? Because the United States declared war on Germany and the Central powers on April 2, 1917, almost two years after the sinking of the Lusitania. So why did the United States declare war for only the fourth time in its history? Was it the Germans’ decision to resume unrestricted submarine warfare in early 1917? Was it the interception and publication of the Zimmerman Telegram in which the German Foreign Secretary promised to help Mexico get back California if they joined Germany in a war against the U.S? Or was it the fall of the Tsarist regime in Russia, which made Wilson’s claims that he wanted to fight to make the world safe for democracy a bit more plausible? Yes, yes, and yes. Also there was our inclination to help Britain, to whom we had loaned a $2 billion. That’s the thing about wars. They never start for easy, simple reasons like Lusitania sinkings. Stupid truth, always resisting simplicity. Oh, it’s time for the Mystery Document? The rules here are simple. I guess the author of the mystery document. I’m either right or I get shocked I. [or possibly “one”] Open covenants of peace, openly arrived at, after which there shall be no private international understandings of any kind but diplomacy shall proceed always frankly and in the public view. II. [I’m starting to think these are Roman numerals] Absolute freedom of navigation upon the seas, outside territorial waters, alike in peace and in war, except as the seas may be closed in whole or in part by international action for the enforcement of international covenants III. The removal, so far as possible, of all economic barriers and the establishment of an equality of trade conditions among all the nations consenting to the peace and associating themselves for it’s maintenance. [And] XIV. [I’m going to guess we skipped some.] A general association of nations must be formed under specific covenants for the purpose of affording mutual guarantees of political independence and territorial integrity of great and small states alike. Stan, thank you for throwing me a softball. That’s my favorite kind of ball. Other than you, Wilson. With its mention of self-determination, freedom of the seas, open diplomacy, and liberal use of Roman numerals, I know it is Woodrow Wilson’s Fourteen Points. Our second consecutive Woodrow Wilson week and my second consecutive non-shock. Given all of his quasi-imperialism, there’s something a little bit ideologically inconsistent about Wilson, but his Fourteen Points are pretty admirable as a statement of purpose. Most of them deal specifically with colonial possessions, and were pretty much ignored, but I suppose if we have learned anything, it’s that in American history, it’s the thought that counts. [Libertage] America’s primary contribution to the Entente powers winning the war was economic as we sent all sorts of arms and money “over there.” Troops didn’t arrive until the spring of 1918 and eventually over 1 million American doughboys served under General John J. Pershing. Not all of these people saw combat. They were much more likely to die of flu than bullet wounds, but their sheer numbers were enough to force the defeat of the exhausted Germans. And now, as promised, I will mention Ernest Hemingway. He served as an ambulance driver, which gave him a close up view of death and misery and led to his membership in the so called Lost Generation of writers who lived in Paris in the 1920s and tried to make sense of everything. Turns out, it’s pretty hard to make sense of and you’re just going to end up with a lot of six-toed cats and then eventually suicide. Okay, so I said earlier than a lot of American Progressives were anti-war, but certainly not all of them. Like, according to Randolph Bourne, “War is the health of the state.” And for progressives like him, “the war offered the possibility of reforming American society along scientific lines, instilling a sense of national unity and self-sacrifice, and expanding social justice.” Let’s go to the ThoughtBubble. World War I made the national government much more powerful than it had ever been. Like, in May of 1917, Congress passed the selective service act, which required 24 million men to register for the draft and eventually increased the size of the army from 120,000 to 5 million. The government also commandeered control of much of the economy to get the country ready to fight, creating new agencies to regulate industry, transportation, labor relations, and agriculture. The War Industries Board took charge of all elements of wartime production setting quotas and prices and establishing standardized specification for almost everything, even down to the color of shoes. The Railroad Administration administered transportation, and the Fuel Agency rationed coal and oil. This regulation sometimes brought about some of the progressives’ goals. Like, the War Labor Board, for instance, pushed for a minimum wage, eight hour days and the rights of workers to form unions. Wages rose substantially in the era, working conditions improved and union membership skyrocketed. But then so did taxes, and the wealthiest Americans ended up on the hook for 60% of their income. Also, in World War I as never before, the government used its power to shape public opinion. In 1917 the Wilson administration created the Committee on Public Information, which only sounds like it’s from an Orwell novel. Headed by George Creel, the CPI’s team created a wave of propaganda to get Americans to support the war, printing pamphlets, making posters and advertising in swanky motion pictures. The best known strategies were the speeches of 75,000 four minute men, who in that amount of time delivered messages of support for the war in theaters, schools, and other public venues. The key concepts in the CPI propaganda effort were democracy and freedom. “Creel believed that the war would accelerate movement towards solving the “age old problems of poverty, inequality, oppression, and unhappiness,” because, obviously, war is the most effective antidepressant. Thanks, Thoughtbubble. So the aforementioned Randolph Bourne might have had good things so say about war, but he was also correct when he suggested that the war would encourage and empower the “least democratic forces in American life.” World War I may have been a war to make the world safe for democracy but according to one historian “the war inaugurated the most intense repression of civil liberties the nation has ever known.” War suppressing civil liberties, eh? I’m glad those days have passed. Speaking of the repression of civil liberties, the NSA is about to start watching this video because I’m about to use the word “espionage.” The Espionage act of 1917 prohibited spying, interfering with the draft and “false statements” that might impede military success. Even more troubling was the Sedition Act passed in 1918, which criminalized statements that were intended to cast “contempt, scorn or disrepute” on our form of government or that advocated interference with the war effort. So basically these laws made it a crime to criticize either the war or the government. In fact, Eugene Debs, the Socialist who ran for president in 1912, was one of those convicted for giving an anti-war speech. He was sentenced to 10 years in prison and he served three of them, but he ran for president from prison and got 900,000 votes. Fortunately, thanks to checks and balances, you can turn to the courts. Unfortunately, they weren’t very helpful. Like in Schenck v. the U.S., the Supreme Court upheld the conviction of a guy named Schenck for encouraging people to avoid the draft and ruled that the government can punish critical speech when it presents a “clear and present danger,” to the state and its citizens. This was when Justice Oliver Wendell Holmes introduced the famous exception to free speech, that it is not okay to “shout fire in a crowded theater.” Nor apparently is it okay to shout, “We shouldn’t be in this war, I don’t think. Just my opinion.” But, some went even further. The 250,000 strong American Protective League helped the Justice Department identify radicals by harassing people in what were called “slacker raids.” Good thing those stopped before you got to high school, right Me from the Past? Slacker. In Bisbee, Arizona vigilantes went so far to put striking copper miners in boxcars, shipped them out to the middle of the desert and left them there. The war also raised the question of what it meant to be a ‘real American.’ Like, public schools “Americanized” immigrants and sought to “implant in their children, so far as can be done, the Anglo-Saxon conceptions of righteousness, law and order, and popular government.” Many cities sponsored Americanization pageants, especially around the Fourth of July, which the CPI in 1918 re-christened “Loyalty Day”. Hamburgers, a German word, became liberty sandwiches. World War I certainly didn’t create anti-immigrant feeling in the United States, but it was used to justify it. Like, IQ tests, introduced to screen army applicants, were soon used to argue that certain immigrant groups were inferior to white protestants and could never be fully assimilated into the United States. Now, of course, those tests were tremendously biased, but no matter. But, to return to the questions of dissent and free speech, the suppression continued after the war with the 1919 Palmer Raids, for instance, named after Attorney General A. Mitchell Palmer and headed up by a young J. Edgar Hoover. To be fair, someone did try to blow up Palmer. So there was some dissent related to the suppression of dissent. Also, more than 4 million workers engaged in strikes in the United States in 1919 but that didn’t legally justify the arrest of more than 5,000 suspected radicals and labor organizers. Most of them were arrested without warrants and held without charge, sometimes for months. And it’s difficult to imagine that all of this would have happened without the heightened sense of patriotism that always accompanies war. However, there were a handful of good things to come out of the Great War, and not just the stylings of Irving Berlin. Like, students are often taught that the war led directly to the passage of the 19th amendment, although a number of states had actually granted the franchise to women before the war. In Montana, for instance, women didn’t just vote, they held office. Congresswoman Jeannette Rankin voted against the declaration of war in 1917, and was the only member of the House to vote against the declaration of war against Japan in 1941. New opportunities in wartime industry also provided incentives for African Americans to move north, thus beginning the so-called great migration and the growth of black populations in northern cities like Chicago and New York. The biggest gain was in Detroit where between 1910 and 1920 the black population rose from 5,741 to 40,838, a 611% increase. So it’s true that World War I provided some new opportunities for African Americans and women, but if World War I was supposed to be an opportunity for America to impose its progressive ideas on the rest of the world, it failed. The Versailles peace conference where Wilson tried to implement his 14 Points raised hope for a new diplomatic order. But, the results of the treaty made the 14 points look hypocritical. I mean, especially when Britain and France took control of Germany’s former colonies and carved up the Arabian provinces of the Ottoman Empire into new spheres of influence. Wilson’s dream of a League of Nations was realized, but the U.S. never joined it largely because Congress was nervous about giving up its sovereign power to declare war. And disappointment over the outcome of World War I led the U.S. to, for the most part, retreat into isolationism until World War II. And therein lies the ultimate failure of World War I. It’s not called “The World War,” it’s called “World War I,” because then we had to go and have a freaking other one. We’ll talk about that in a few weeks, but next week we get to talk about suffrage. Yes! We finally did something right. I’ll see you then. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher Raoul Meyer, Rosianna Rojas, and myself. And our graphics team is Thought Café. Every week, there’s a new caption for the Libertage. If you’d like to suggest one, you can do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome. Stan, can you do some movie magic to get me out of here? Perfect.

US_History

The_Natives_and_the_English_Crash_Course_US_History_3.txt

Hi I'm John Green, this is Crash Course US History and today we're going to talk about one of the worst relationships in American history. No Thought Bubble, not my college girlfriend and me. Mr. Green, Mr. Green! Your relationship with your high school girlfriend? Oh Me From The Past, you and I both know that I didn't have a high school girlfriend. No, I'm talking about the relationship between Native Americans and English Settlers. [Theme Music] So as you'll no doubt remember from last week, the first English settlers came to the Chesapeake area – now Virginia – in 1607. The land the English found was, of course, already inhabited by Indian tribes unified under the leadership of Chief Wahunsenacawh, and I will remind you that mispronouncing things is my thing! The English called this Chief Powhatan because, of course, mispronouncing things was also their thing. Powhatan was actually his title and the name of his tribe, but to say that the English lacked cultural sensitivity would be an understatement. So Powhatan didn't get to be leader of over 30 tribes by being a dummy and he quickly realized that: 1. The English were pretty clueless, when it came to not dying of starvation, and 2. They were useful – because they had guns. So he decided to help them and the English were indeed grateful. In fact, colony leader John Smith went so far as to order the colonists to stop stealing food from the Indians. Aaauugh, in the book business this is known as foreshadowing. So as previously noted, relationships, whether between individuals or collectives, tend to go well when they are mutually beneficial, and for a while, both the English and the Indians were better off for these interactions. I mean, you know, post-smallpox. The Virginia Company existed to make money, and since the Chesapeake lacked gold or silver, making money required trade. OK, let's go to the Thought Bubble: We tend to think of trade between Europeans and Natives as being a one-way exchange, like savvy, exploitative Europeans tricking primitive, pure, indigenous people into unfair deals. But that isn't quite accurate. Both sides traded goods that they had in surplus for those they did not. The English were happy to give up iron utensils, tools, guns, woven cloth in exchange for furs and, especially in the early days, food, which the Indians could easily part with because they had plenty. Soon, though, there were problems. In order to keep up trade relations, Indian men devoted more time to hunting and less to agriculture which upset traditional gender balance in their society. And European ideas about land use started to overcome traditional Indian ways of life, and that led to conflict. The English liked to fence in some of their land, which kept the Indians off it, and also the English let their pigs and cattle roam freely and the animals would eat Natives' crops. And as Europeans' appetite for furs grew, Indian tribes began to fight with each other over access to the best hunting grounds, leading to inter-tribal warfare, which suddenly included guns. But this was still a relatively calm time. Yes, at one point John Smith was captured by the Indians and had to be "saved" by Powhatan's daughter, Pocahontas, but this was probably all a ritual planned by Powhatan to demonstrate his dominance over the English. Pocahontas never married John Smith by the way, but she was kidnapped by the English and held for ransom in 1613, and she did eventually marry another Englishman, John Rolfe. She converted to Christianity and went to England, where she became a sensation and died of disease. Stupid disease always deciding the course of human history. Anyway, despite not marrying Pocahontas, John Smith is still important to this story because when he left Virginia for England after being injured in a gunpowder explosion, things between the Native Americans and the English immediately began to deteriorate. How? Well, the English went back to stealing Indians' crops and also began stealing their lives via massacres. Thanks, Thought Bubble – man, you guys sure know how to end on a downer. Although to be fair, there are not a lot of uppers in this story. So after a period of peace following Pocahontas' marriage to John Rolfe in 1614, dramatized here, things finally came to a head in 1622, when Chief Opechancanough led a rebellion against the English. It had become abundantly clear that more and more English were going to show up and they weren't just there to trade. They wanted to take Indian land. But the English struck back, as empires will, and the uprising of 1622 ultimately failed. And after another failed uprising in 1644, the 2,000 remaining Native Americans were forced to sign a treaty that consigned them to reservations in the West. Well, the west of Virginia, at least. But the 1622 uprising was the final nail in the coffin of the Virginia Company, which was a failure in every way. It never turned a profit, and despite sponsoring 6,000 colonists, by 1644 when Virginia became a royal colony, only 1,200 of those people were still alive, proving once again that governments are better at governing than corporations. Up in New England, you'll recall that the Pilgrims probably wouldn't have survived their first winter without help from the Native Americans, which of course led to the first Thanksgiving, and then centuries of mutually beneficial trade and generosity– just kidding. While some of the Puritans that settled in New England – notably Roger Williams – tried to treat the Indians fairly, in general it was very similar to what we saw in the Chesapeake. Settlers thought Native Americans could be replaced because they weren't "properly using the land." Now John Winthrop, who you'll remember from last week, at least realized that it was better to buy land from Indians than just take it. But Puritan land purchases usually came with strings attached. The main string being that the Native Americans had to submit to English authority. Now, the Puritans had a rather conflicted view of the Indians. On the one hand, they saw natives as heathens in need of salvation, as evidenced by the Massachusetts seal, which features an Indian saying "Come over and help us." On the other hand, they recognized that the Native American way of life – with its relative abundance and equality, especially when it came to women – might be tempting to some people, who might want to go native. This was such a concern that in 1642, the Massachusetts General Court prescribed a sentence of three years hard labor for anyone who left the colony and went to live with the indigenous people. There was even anti-Indian propaganda in the form of books. Captivity narratives, in which Europeans recounted their desire to return to Christian society after living with the Indians, were quite popular. Even though some, like the famous Sovereignty and Goodness of God by Mary Rowlandson, did admit that the Indians often treated their European captives quite well. New England's native population lacked an overarching leader like Powhatan, but by 1637, the inevitable conflict between the English and the Indians did happen. It was called the Pequot War. After some Pequots killed an English fur trader, soldiers from Massachusetts, the newly-formed colony of Connecticut, and some Narragansett Indians, who saw an opportunity to gain an upper hand over the Pequots, attacked a Pequot village at Mystic, burning it and massacring over 500 people. The war continued for a few months after this, but to call it a war is, in a way, to give it too much credit. The Indians were over-matched from the beginning, and by the end, almost all of them had been massacred or sold into slavery in the Caribbean. The War opened up the Connecticut River to further settlement. It also showed that Native Americans were going to have a tough time resisting, because they were outnumbered and they had inferior weapons. But the brutality of the massacre in Mystic shocked even some Puritans, like William Bradford, who wrote, "It was a fearful sight to see them frying in the fire." But despite the odds, New England natives continued to resist the English. In 1675, Native Americans launched their biggest attack on New England colonists in what would come to be known as King Philip's War. It was led by a Wampanoag chief named Metacom, which was why it is also sometimes called Metacom's War. The English called Metacom "King Philip" due to their fantastic cultural sensitivity. The conflict was marked by brutality on both sides and it nearly ended English settlements in the northeast. The fighting itself lasted 2 years. Indians attacked half of the 90 towns the English had founded, and 12 of those towns were destroyed. About 1,000 of the 52,000 Europeans and 3,000 of the 20,000 Indians involved died in the War. As I mentioned before, the War was particularly brutal. The Battle of the Great Swamp was really just a massacre of Indians by the English. And when King Philip was finally killed, ending the War, his decapitated head was placed on a stake in the Plymouth town square, where it remained for decades. And on the other side? Well, to quote Nathaniel Saltonstall, who lived through the war, "The heathen rarely [gave] quarter to those that they take, but if they were women, they first forced them to satisfy their filthy lusts and then murdered them." Saltonstall went on to describe a particularly brutal way that natives would kill colonists' cows: by cutting "their bellies and letting them go several days trailing their guts after them." That indigenous people would reserve such brutality for livestock says something really important about this war. The Indians correctly saw European colonization as a threat to their way of life, and that included the animals who trampled Indians' land and whose grazing patterns required the English to take more and more territory. Some of the stories told about Native American brutality also suggest the symbolic nature of this war. Like, one English colonist was disemboweled and had a Bible stuck in his body cavity. Supposedly, the natives who buried him explained, "You English, since you came into this country have grown exceedingly above the ground. Let us see how well you grow when planted into the ground." But it wasn't just the Indians who felt their way of life being threatened. It's time for this week's Mystery Document! The rules here are simple. I read the Mystery Document. I try to guess its author. If I'm right, I don't get shocked with the shock pen. If I'm wrong, I do. "The righteous god hath heightened our calamity and given commission to the barbarous heathen to rise up against us and to become a smart rod and a severe scourge to us in burning and depopulating several hopeful plantations, murdering many of our people of all sorts and seeming as it were to cast us off hereby speaking aloud to us to search and try out our ways and turn again unto the Lord our God from whom we have departed with a great backsliding." OK, I don't know this one, so I'm going to have to piece it together. Uh, we have a plural narrator, that's important. Seemingly monotheistic, feels like the heathens in this context – likely the Native Americans – have been sent as a scourge [scorge], or scourge [scurge], as it is apparently properly pronounced. What, I'm from Alabama, I don't know how to say a ton of words. I mean, I just recently learned that you don't check your Ya-HOO! mail, you check your YA-hoo! mail, and Yahoo!'s over already! All right so plural narrator, scourge, great backsliding uhh Stan, you're going to get to shock me this time, who is it? [Buzzing Sound] The Laws of War passed by the General Court of Massachusetts in 1675. Are you kidding? From now on, the Mystery Document must always be written by a single human person! I hate this. I hate this so much. It's worse now, because I've had it before, so I know it's gonna – GAHHHHHH!!! This shows us the way the Puritans understand the world, but it also show us that within 50 years of its founding, Puritans already felt that the mission of their colony – to be a great Christian community – was already kind of a failure. If they'd been as righteous as they were supposed to be, God wouldn't have sent the Indians to burn their homes and kill them. So it's important to understand that this was a war to preserve a way of life for both the Indians and the English. And that brings us to another question: What's the point of even telling these bloody stories about massacres and atrocities. One point is to remind ourselves that much of what we learn about American history, like all history, has been cleaned up to conform to our mythological view of ourselves. Native Americans have been so successfully marginalized, both geographically and metaphorically, that it's easy to either forget about them or else to view them merely as people to be pitied or reviled. But it's important to know the ways that they resisted colonization, because it reminds us that Native Americans were people who acted in history, not just people who were acted upon by it. And it also reminds us that the history of Indigenous people on this land mass isn't separate from American history; it's an essential part of it. Thanks for watching. I'll see you next week. Crash Course is produced and directed by Stan Muller, our script supervisor is Meredith Danko, the associate producer is Danica Johnson, and the show is written by my high school history teacher, Raoul Meyer and myself. Our graphics team is Thought Bubble. If you have questions about today's video, please ask them in comments. They will be answered by our team of crack historians. By the way, our team of crack historians is a team of excellent historians, not a team of historians who study crack cocaine. Thanks for watching. As we say in my hometown, don't forget to be awesome.

US_History

The_Black_Legend_Native_Americans_and_Spaniards_Crash_Course_US_History_1.txt

Hi, I’m John Green and this is Crash Course U.S. History— No, Stan, that’s not gonna work actually. I mean, we’re talking about the 16th century today, when this was neither United nor States. By the way, this globe reflects the fact that I believe that Alaskan statehood is illegitimate. In fact, we’re gonna call this whole show US History, but inevitably it’s going to involve other parts of the world. And also, not to brag, a small part of the moon. Sorry, we can be a little bit self- aggrandizing sometimes here in America. [Patriotic Rock Music] So to begin U.S. History, we’re not going to talk about the United States or this guy. We’re going to talk about the people who lived here before any Europeans showed up. [Theme Music] North America was home to a great variety of people, so it’s difficult to generalize but here’s what we can say: 1. When the Europeans arrived, there were no “classical style civilizations” with monumental architecture and empires like the Aztec or the Incas; And 2. Native North Americans had no metal work, no gunpowder, no wheels, no written languages, and no domesticated animals. However, they did have farming, complex social and political structures, and widespread trade networks. Mr. Green, Mr. Green! So they were pretty backwards, huh? Well, I mean, or at least primitive. Primitive is a funny word, Me from the Past, because it implies a romanticization— the simple people who never used more than they needed and had no use for guns— and it also implies an infantalization. It’s like you believe that just because you have an beeper and they didn’t, they were somehow less evolved humans. But you can’t see the human story as one that goes from primitive to civilized. That’s not just Eurocentric; that’s contemporary-centric. The idea that we’re moving “forward” as a species implies a linear progression that just does not reflect the reality of life on this planet. I get that you like to imagine yourself as the result of millennia of advancement and very pinnacle of humanness. But from where I’m sitting, that worldview is a lot more backwards than living without the wheel. So no one knows exactly how many people lived in North America before the Europeans got here. Some estimates are as high as 75 million, but in the present US borders the guesses are between 2 and 10 million. And like other Native Americans, their populations were decimated by diseases, such as smallpox and influenza. Actually, it was much worse than decimation. As many of you have pointed out “decimation” means 1 in 10. This was much worse than that. It was closer maybe to 8 in 10, which would be an oct-icimation. So, there had been civilizations in North America, but they peaked before the Europeans arrived. The Zuni and Hopi civilization, roundabout here, peaked around 1200 CE. They had large multiple family dwellings in canyons which they probably left because of drought. Crash Course World history fans will remember that environmental degradation often causes the decline of civilizations. I’m looking at you, Indus Valley, and also you, Entire Future Earth. But complex civilizations weren’t the rule in North America. And now we’re about to begin generalizing, a bad habit historians have, partly because there’s a limited historical record. But also because Eurocentric historians have a bad habit of primitivizing and simplifying others. So I want to underscore that there was huge diversity in the pre-Columbus American experience, and that talking about someone who lived here in 1,000 BCE, and talking about someone who lived here 2,000 years later is just inherently problematic. That said, let’s go to the Thought Bubble. Most native groups in most places organized as tribes, and their lives were dominated by the natural resources available where they lived. So, west coast Indians primarily lived by fishing, gathering, and hunting sea mammals. Great plains Indians were often Buffalo hunters. These tribal bands often united into loose confederacies or leagues, the best known of which was probably the Iroquois Confederacy, also called the Great League of Peace. This was kind of like an upstate New York version of NATO, but without nuclear weapons or the incessant international meddling or Latvians. Okay, it was nothing like NATO actually. Religion usually involved the vibrant spiritual world with ceremonies geared toward the tribe’s lifestyle; hunting tribes focused on animals, agricultural tribes on good harvests. And most Indian groups believed in a single Creator-god who stood above all the other deities, but they weren’t monotheistic in the way that Christians who came to the new world were. American Indians also saw property very differently from Europeans. To first peoples, land was a common resource that village leaders could assign families to use— but not to own— and most land was seen as common to everyone. As Black Hawk, a leader of the Sauk tribe said: “The Great Spirit gave it to his children to live upon and cultivate as far as necessary for their subsistence; and so long as they occupy and cultivate it, they have a right to the soil.” Thanks, Thought Bubble. So, many of us tend to romanticize American Indians as being immune from greed and class, but in fact there were class distinctions in Indian tribes. Rulers tended to come from the same families, for instance. That said, wealth was much more evenly distributed than it was in Europe. And while most tribal leaders were men, many tribes were matrilineal, meaning that children become member's of their mothers family. Also, women were often important religious leaders. Women also often owned dwellings and tools although not land because again that idea did not exist. Also, in many tribes, women engaging in premarital skoodilypooping wasn’t taboo. In general, they were just much less obsessed with female chastity than Europeans were. I mean, I will remind you, the first English settlement in America was called “Virginia.” The idea that Native Americans were noble savages, somehow purer than Europeans, and untouched by their vices is not a new one. Like, some of the earliest Europeans saw the Indians as paragons of physical beauty and innocent of European’s worst characteristics. But for most Europeans, there was little noble about what they saw as pure Indian savagery. I mean, Indians didn’t have writing, they suffered from the terrible character flaw of being able to have sex without feeling ashamed, and most importantly they weren’t Christians. The Spanish were the first Europeans to explore this part of the world. Juan Ponce de Leon arrived in what is now Florida in 1513 looking for gold and the fabled fountain of youth. In 1521, he encountered a Calusa Brave’s poison-tipped arrow and died before discovering that the Fountain of Youth is, of course, delicious Diet Dr. Pepper. Mmm, oooh, I can taste all 23 flavors. There were many more Spanish explorers in the first half of the 16th century, including one Alvar Nunez Cabeza de Vaca who wandered through the American southwest looking for gold, which I mention entirely because I think that guy’s last name means cow head. Of course, none of these people found any gold, but they did make later European colonization easier by bringing over the microbes that wiped out most native populations. So, the Spanish wanted to colonize Florida to set up military bases to thwart the pirates who preyed on silver-laden Spanish galleons coming out of Mexico. But Spanish missionaries also came over, hoping to convert local native populations. This of course worked out magnificently. Just kidding. It went terribly. And many of the missions were destroyed by an uprising of Guale Indians in 1597. And I will remind you, mispronouncing things is my thing. In general, colonizing Florida sucked, because it was hot and mosquito-ey. Spain was much more successful at colonizing the American Southwest. In 1610, Spain established its first permanent settlement in the Southwest at Santa Fe, New Mexico, and you couldn’t really say that it flourished, since Santa Fe’s population never got much above 3000. But it had a great small town feel. And New Mexico is really important because it’s the site of the first large scale uprising by Native Americans against Europeans. I mean, the native people, who the Spanish called Pueblos, had seen their fortunes decline significantly since the arrival of Europeans. How much decline? Well, between 1600 and 1680, their population went from about 60,000 to about 17,000. Also, the Franciscan friars who came to convert the indigenous people, became increasingly militant about stamping out all native religion. The Spanish Inquisition just wasn’t very keen on the kind of cultural blending that made early conversion efforts successful. So, while the Spanish saw all the Pueblos as one people, they also knew there were tribal differences that made it difficult for the Indians to unite and rise up against Spanish. But nothing unites like a common enemy, and in 1680, a religious leader called Pope organized an uprising to drive the Spaniards out. Pope organized about 2,000 warriors who killed 400 Spanish colonists and forced the rest to leave Santa Fe. So, the Spanish colony in New Mexico was effectively destroyed. The Pueblos tore down all the Christian churches and replaced them with “kivas,” their places of worship. But, like most awesome uprisings, it didn’t last. But after the revolt, the Spanish were much more tolerant of indigenous religion and they also abandoned the forced labor practice called encomienda. Oh, it’s time for the new Crash Course feature the Mystery Document? How mysterious. The rules here are simple: I read and attempt to identify the mystery document. If I am right, I do not get shocked by this shock pen. And if I am wrong, I do. OK, what do we have here? The Indians … were totally deprived of their freedom and were put in the harshest, fiercest most horrible servitude and captivity which no one who has not seen it can understand. Even beasts enjoy more freedom when they are allowed to graze in the fields. But our Spaniards gave no such opportunity to Indians and truly considered them perpetual slaves… I sometimes came upon dead bodies on my way, and upon others who were grasping and moaning in their death agony repeating, “Hungry, hungry.” And this was the freedom, the good treatment and the Christianity the Indians received.” Well, that’s nice. OK, so the mystery document is always a primary source and since the writer refers to “Our” Spaniards, I’m gonna guess that he or she -- probably he -- is European. And a Spaniard sympathetic to the Indians, which narrows the list of suspects considerably. So, it probably wasn’t de Sepulveda, for instance, who argued that the Indians might not even be human. OK, Stan, I’m actually pretty confident here. I believe it is from A Short Account of the Destruction of the Indies by Bartolome de las Casas. No? DANG IT! Stan just told me I have the author right, but the book wrong. It’s A History of the Indies. Ugh, I hate shocks, both literal and metaphorical. [Buzzing nosie] Gah! So we’ve focused a lot on the brutality of the Spanish toward the Indians. But at least one Spaniard, de las Casas, recognized that his countrymen were…terrible. This realization is a good thing obviously, but it leads us to one of the big problems when it comes to studying this time and place. The Black Legend is the tale that the Spanish unleashed unspeakable cruelty on the Indians. Now that tale is true. But, that idea was used by later settlers, especially the English to justify their own settlements. Like, part of the reason they needed to expand their empire was to save the Indians from the awful Spanish. But were the English so much better? Yeah, probably not. As we mentioned at the beginning of today’s episode, American Indians didn’t have writing, so we don’t have records of their perspective. Now, some Europeans, like de las Casas, were critical of the Spaniards, but most considered the Indians heathens, and implied— or even outright said— that they deserved whatever horrible things befell them. So, at the beginning of our series, I want to point out something that we need to remember throughout. One of the great things about American history is that we have a lot of written sources. This is the advantage of the US coming onto the scene so late in the game, historically speaking. But every story we hear comes from a certain point of view, and we always need to remember who is speaking, why they are speaking, and, especially which voices go unheard— and why. Thanks for watching, I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, and myself. And our graphics team is Thought Bubble. If you have questions about today’s video, you should ask them in comments. Everybody who works on Crash Course, as well as a team of historians, will be there to answer them. Thanks for watching. Please make sure you’re subscribed to Crash Course. And as we say in my hometown, Don't Forget To Be Awesome.

US_History

The_Quakers_the_Dutch_and_the_Ladies_Crash_Course_US_History_4.txt

Hi, I'm John Green, this is Crash Course US History, and today we're going to cram 150 years of American history into one video. Why? Well, many American history classes don't cover the colonial period at all, because most major American history tests have, like, one question about it. Mr. Green, Mr. Green, so this isn't going to be on the test? That's awesome because I have some flirtatious notes to exchange with Jessica Alvarez. Yeah, me from the past. So listen, would you rather do well on one test or lead a richer, more productive life as a result of having a better understanding of the complicated factors that led to the creation of the greatest nation in history? Stan, can I get a Libertage? [Patriotic Rock Music] So listen up me from the past! It's time to bask in our own greatness, and by greatness I mean morally dubious dominance over people who would have been just fine without us. [Theme Music] So, contrary to popular mythology Colonial America was more than just Jamestown and Massachusetts. There was, for instance, New Amsterdam. The tale goes, the Dutch traders bought the island of Manhattan from Lenape Indians for $24 in 1624 – that isn't quite true, but it contains a truth. The Dutch traders who founded their colony were businessmen and New Amsterdam was, above everything else, a commercial venture. This is still true in New York, actually. I mean, Manhattan is all about Wall Street. In fact, Crash Course writer and history teacher Raoul Meyer is believed to be the last person living on the island of Manhattan who does not work for an investment bank. So the Dutch let anyone into New Amsterdam who could help them turn a profit, including Jews and even Quakers. But they didn't like Indians very much, in fact they drove them out of the colony. But anyway the $24 the Lenapes supposedly got for New England was $24 more than the Dutch got when the English took over the colony in 1664, by sailing four frigates into the harbor and asking for the colony in a threatening voice. So New Amsterdam became New York which was a mixed blessing. The population doubled in the decade after the English takeover, but English rule meant less economic freedom for women who, under the Dutch were able to inherit property and conduct business for themselves. And under the English, free black people lost a lot of the jobs they had been able to hold under the Dutch. Things were better in Pennsylvania. So much so that it was known as the "best poor man's country," which admittedly in the 17th century was a low bar to jump over. Given by Charles II to this guy William Penn in 1681, Pennsylvania was a huge tract of land round about here. The land included contemporary Pennsylvania and Delaware and New Jersey. But we've made an editorial decision NOT to talk about New Jersey here on Crash Course, due to my long-standing anti-New Jersey bias. So Penn wanted his colony to be a haven for Quakers because he was a Quaker, as you know if you have ever seen a container of Quaker Oats. Quakers were a pretty tolerant bunch, except when it came to slavery which they opposed vehemently. And under Penn's leadership, the colony showed remarkable religious toleration and also an amazing respect for Indian communities but then, after Penn was gone – yeah the usual. In 1737, Pennsylvania colonists perpetrated one of the most famous frauds of colonial America: the Walking Purchase. Indians agreed to cede a tract of land bound by the distance a man could walk in 36 hours. But the clever governor James Logan hired a bunch of fast runners who marked out territory much larger than the Indians anticipated. Quakers had to resort to such tricks because they were pacifists. I should also mention that they weren't particularly fond of loose living. The government prevented swearing and drunkenness for instance but, you know, it was still pretty great compared to the other colonies. More than half of the male population was eligible to vote and Pennsylvania's dual promise of religious freedom and cheap land attracted a lot of German immigrants. Well I should say German-speaking immigrants. There was of course not a Germany at the time, as many viewers of Crash Course World History have pointed out to me. And now let us venture south, where we will find many mosquito-borne illnesses and somewhat less abolitionist sentiment. In 1663 English King Charles II gave eight English proprietors the right to set up a colony just North of the Spanish-controlled Florida to serve as a buffer. This became South Carolina and its original settlers came from the sugar colony of Barbados, which helps to explain why they were so awesome at slavery. They tried to enslave the Indians and ship them to the Caribbean but when that didn't work out they began to import African slaves. We're going to talk a lot more about slavery in future episodes but for now just bear in mind: it sucked. OK, so in the last quarter of the 17th century the British colonies and the Americas experienced this series of crises. Oh it's time for the mystery document? The rules here are simple: I guess the author of the document. I get it right no shock, I get it wrong shock. OK. "We accuse Sir William Berkeley as guilty of each and every one of the same, and as one who hath traitorously attempted, violated and injured His Majesties interest here, by a loss of a great part of his colony and many of his faithful loyal subjects, by him betrayed and in a barbarous and shameful manner exposed to the incursions and murther of the heathen, and we do further declare these ensuing persons in this list, to have been his wicked and pernicious counselors." Both wicked and pernicious – those are some terrible counselors. OK so this guy clearly hated William Berkeley, who I happen to know as governor of Virginia. Particularly upset about the colonists being incurred upon and murdered by the heathen (that is the Native Americans). Uhhh I mean I have a guess but I'm not brimming with confidence. Ah, the one person I know who hated William Berkeley was Nathaniel Bacon? Yes!!! Yes! Yes! No shock for me and no pleasure for you, you schadenfreudic Crash Course viewers. So Nathaniel Bacon arrived in Virginia in 1673 and led an armed uprising against Governor Berkeley just three years later. And just to be clear, he was mad not because Berkeley did a poor job protecting colonists from Indians, but because Berkeley wouldn't allow them to kill more Indians and take more land. Berkeley had already given all the really good land to his cronies, those aforementioned "wicked and pernicious counsellors," leaving men like Bacon with serious beef. I hate myself. Before the rebellion was quelled by the arrival of English warships, Bacon burned Jamestown and made himself ruler of Virginia and looted Berkeley's supporters' land. 23 of the rebels were hanged, but not Bacon, who died shortly after taking control from – you guessed it – dysentery! Dang it, dysentery, it's called "history" not "dysentery-story!" Bacon's rebellion is sometimes portrayed as an early example of lower-class artisans and would-be farmers rising up against the corrupt British elite, which I guess kind of. But the biggest effects of the rebellion were: 1. A shift away from indentured servants to slaves, and 2. A general desire by the English crown to control the colonies more. OK, so in 1686 King James II really tried to put the hammer down by consolidating Connecticut, Plymouth, Massachusetts, New Hampshire, Rhode Island, New York, and East and West Jersey into one big mega-colony called The Dominion of New England. It's near dictatorial ruler was former New York governor Edmund Andros, who proceeded to appoint his own officials and lay his own taxes without even consulting any of the elected assemblies. Luckily – or unluckily, depending on your perspective – a major event in British history reversed this policy: the Glorious Revolution. Now thankfully, this isn't Crash Course: British History or it would quickly turn into Crash Course: John is Bored History. But the upshot is that Britain got a fancy new royal family from Holland, which sparked uprisings in the colonies, and Andros was thrown into a Boston jail as the colonies re-asserted their independence. And these new guys imposed the English Toleration Act of 1690, which decreed that all Protestants could worship freely. As Toleration Acts go, this one wasn't that tolerant – I mean, it still discriminated against Jews – but it did mark the end of the Puritan Experiment. No longer would membership at a Congregationalist church be a requirement for voting in general-court elections, property ownership would now be the determining factor. And Massachusetts would now have a governor from England, not from a company board residing inside the colony itself. This was the context for one of the most talked-about events of colonial history: the Salem Witch Trials. A lot of ink has been expended on this incident, and the interpretations of it are numerous and controversial, so I'm just gonna point out that when the Witch Trials – which claimed the lives of fourteen of the nearly 150 women and men accused of witchcraft – happened in 1691, New England as a colony had basically just failed in its religious mission. The Tolerance Act meant that people in Massachusetts would have to accept even Quakers as virtual equals! Quakers! So it's not surprising that colonists would look for scapegoats, or that their male leaders would seek to re-assert their gender dominance. OK, to talk about Colonial American economics, let's go to the Thought Bubble. Most colonists were farmers, or worked on farms, and they were mostly small, unlike the giant plantations that predominated in the Caribbean. Since New England contains relatively little in the way of tropical diseases, and was increasingly free of Native Americans, the colonial population there skyrocketed – so fast that families began to run out of land, so second and third sons increasingly had to go make their way in growing coastal cities. We'll talk about this more in future episodes, but for now let's just note the idea of a person owning a small farm and the idea of freedom are pretty closely intertwined in the early part of American history. It's more of an "amber waves of grain" place than a "behold this metropolis" place. In fact, they were richer than any other colonial elites. Were they rich enough to dominate the Constitutional Convention? Time will tell. Now not everyone was a farmer or slave. There were growing numbers of artisans in the colonies. Although British colonial policy discouraged local manufacturing, the growing population in America meant that there was certain to be a market for locally produced goods, especially clothing and metalwork. Remember, one of the heroes of the American Revolution – Paul Revere – was a silversmith. Thanks, Thought Bubble. So that variety of jobs leads us nicely to our last topic today: colonial society. Although Americans like to think of themselves as class-less, pun intended, that's not really the case. I mean, the colonies definitely had an elite ruling class, especially in the South, that did what it could to perpetuate itself. George Washington's father and grandfather were both justices of the peace, an important role in colonial times, meaning that George Washington had deeply elite roots. So that was the top of colonial society and at the bottom was a growing number of poor people. While it's never good to be poor, it was much better to be poor in the colonies than it was in England or much of the rest of Europe, which is why people kept indenturing themselves to get here. America had lots of food, and there was the possibility of maybe, someday, getting some land – provided you didn't die of dysentery. OH, and also provided you weren't a woman. Married women in 18th-century colonial America generally couldn't own property, and husbands usually willed their land to their sons and their personal items to their daughters, meaning that almost all landowners were male. In the earliest days of colonization, when everyone was needed to ensure their survival of the colonies, women had a greater role in the economy, although they were still expected to be wives and mothers above all else. Male dominance was written into law and solidified in practice. Women's work was mostly confined to the home, and especially for lower-class women, it involved a lot of drudgery. As one woman, Mary Cooper, wrote in her diary in 1769, "I am dirty and distressed, almost wearied to death. This day is forty years since I left my father's house and come here, and here have I seen little else but hard labor and sorrow." Aaand that's actually a good place to end, because it reminds us that history is about much more than the lives of kings like James II and rebels like Nathaniel Bacon. And while history classes – and exams – tend to focus on those kinds of men – and they were mostly men – the real story of history is about regular people trying to take care of their families and not die. The colonial era often gets skipped for its lack of large-scale drama, but those small scale dramas can be found in abundance. Next week we'll go back to all that great men and dramatic events crap, when we start talking about the American Revolution. I'll see you then. Thanks for watching. Crash Course is produced and directed by Stan Muller; our script supervisor is Meredith Danko; the associate producer is Danica Johnson. The show is written by my high school history teacher, Raoul Meyer, and myself, and our graphics team is Thought Bubble. Last week's phrase of the week – oh wait, we don't do phrase of the week anymore. If you have questions about today's video, you can ask them in comments where they will be answered by our team of crack historians. Thanks for watching Crash Course, and as they say in my hometown: Don't Forget to Be Awesome. [off screen] CRASH COURSE!! Everything is fine.

US_History

Slavery_Crash_Course_US_History_13.txt

Hi, I'm John Green, this is Crash Course U.S. History, and today, we're going to talk about slavery, which is not funny. Yeah, so we put a lei on the eagle to try and cheer you up, but let's face it, this is going to be depressing. With slavery, every time you think, like, "Aw, it couldn't have been that bad," it turns out to have been much worse. Mr. Green, Mr. Green! But what about – Yeah, Me from the Past, I'm going to stop you right there, because you're going to embarrass yourself. Slavery was hugely important to America. I mean, it led to a civil war and it also lasted what, at least in U.S. history, counts as a long-ass time, from 1619 to 1865. And yes, I know there's a 1200-year-old church in your neighborhood in Denmark, but we're not talking about Denmark! But slavery is most important because we still struggle with its legacy. So, yes, today's episode will probably not be funny, but it will be important. [Theme Music] So the slave-based economy in the South is sometimes characterized as having been separate from the Market Revolution, but that's not really the case. Without southern cotton, the North wouldn't have been able to industrialize, at least not as quickly, because cotton textiles were one of the first industrially products. And the most important commodity in world trade by the nineteenth century, and 3/4 of the world's cotton came from the American South. And speaking of cotton, why has no one mentioned to me that my collar has been half popped this entire episode, like I'm trying to recreate the Flying Nun's hat. And although there were increasingly fewer slaves in the North as northern states outlawed slavery, cotton shipments overseas made northern merchants rich. Northern bankers financed the purchase of land for plantations. Northern insurance companies insured slaves who were, after all, considered property, and very valuable property. And in addition to turning cotton into cloth for sale overseas, northern manufacturers sold cloth back to the South, where it was used to clothe the very slaves who had cultivated it. But certainly the most prominent effects of the slave-based economy were seen in the South. The profitability of slaved-based agriculture, especially King Cotton, meant that the South would remain largely agricultural and rural. Slave states were home to a few cities, like St. Louis and Baltimore, but with the exception of New Orleans, almost all southern urbanization took place in the upper South, further away from the large cotton plantations. And slave-based agriculture was so profitable that it siphoned money away from other economic endeavors. Like, there was very little industry in the South. It produced only 10% of the nation's manufactured goods. And, as most of the capital was being plowed into the purchase of slaves, there was very little room for technological innovation, like, for instance, railroads. This lack of industry and railroads would eventually make the South suck at the Civil War, thankfully. In short, slavery dominated the South, shaping it both economically and culturally, and slavery wasn't a minor aspect of American society. By 1860, there were four million slaves in the U.S., and in the South, they made up one third of the total population. Although in the popular imagination, most plantations were these sprawling affairs with hundreds of slaves, in reality, the majority of slaveholders owned five or fewer slaves. And, of course, most white people in the South owned no slaves at all, though, if they could afford to, they would sometimes rent slaves to help with their work. These were the so-called yeoman farmers who lived self-sufficiently, raised their own food, and purchased very little in the Market Economy. They worked the poorest land and, as a result, were mostly pretty poor themselves. But even they largely supported slavery, partly, perhaps, for aspirational reasons, and partly because the racism inherent to the system gave even the poorest whites legal and social status. And southern intellectuals worked hard to encourage these ideas of white solidarity and to make the case for slavery. Many of the founders, a bunch of whom you'll remember, held slaves, saw slavery as a necessary evil. Jefferson once wrote, quote, "As it is, we have the wolf by the ear, and we can neither hold him, nor safely let him go. Justice is on one scale, and self-preservation in the other." The belief that justice and self-preservation couldn't sit on the same side of the scale was really opposed to the American idea, and, in the end, it would make the Civil War inevitable. But as slavery became more entrenched in these ideas of liberty and political equality were embraced by more people, some southerners began to make the case that slavery wasn't just a necessary evil. They argued, for instance, that slaves benefited from slavery. Because, you know, because their masters fed them and clothed them and took care of them in their old age. You still hear this argument today, astonishingly. In fact, you'll probably see asshats in the comments saying that in the comments. I will remind you, it's not cursing if you are referring to an actual ass. This paternalism allowed masters to see themselves as benevolent and to contrast their family-oriented slavery with the cold, mercenary Capitalism of the free-labor North. So yeah, in the face of rising criticism of slavery, some southerners began to argue that the institution was actually good for the social order. One of the best-known proponents of this view was John C. Calhoun, who, in 1837, said this in a speech on the Senate floor: "I hold that, in the present state of civilization, where two races of different origin and distinguished by color and other physical differences as well as intellectual, are brought together, the relation now existing in the slave-holding states between the two is, instead of an evil, a good. A positive good." Now, of course, John C. Calhoun was a fringe politician, and nobody took his views particularly seriously. Stan: Well, he was Secretary of State from 1844 to 1845. John: Well, I mean, who really cares about the Secretary of State, Stan? Danica: Eh, he was also Secretary of War from 1817 to 1825. John: All right, but we don't even have a Secretary of War anymore, so... Meredith: And he was Vice President from 1825 to 1832. John: Oh my god, were we insane?! We were, of course, but we justified the insanity with Biblical passages and with the examples of the Greeks and Romans, and with outright racism, arguing that black people were inherently inferior to whites. And that not to keep them in slavery would upset the natural order of things. A worldview popularized millennia ago by my nemesis, Aristotle. God, I hate Aristotle. You know what defenders of Aristotle always say? "He was the first person to identify dolphins." Well, ok, dolphin identifier. Yes, that is what he should be remembered for, but he's a terrible philosopher! Here's the truth about slavery: It was coerced labor that relied upon intimidation and brutality and dehumanization. And this wasn't just a cultural system, it was a legal one. I mean, Louisiana law proclaimed that a slave "owes his master... a respect without bounds, and an absolute obedience." The signal feature of slaves' lives was work. I mean, conditions and tasks varied, but all slaves labored, usually from sunup to sundown, and almost always without any pay. Most slaves worked in agriculture on plantations, and conditions were different, depending on which crops are grown. Like, slaves on the rice plantations of South Carolina had terrible working conditions, but they labored under the task system, which meant that once they had completed their allotted daily work, they would have time to do other things. But lest you imagine this is like how we have work and leisure time, bear in mind that they were owned and treated as property. On cotton plantations, most slaves worked in gangs, usually under the control of an overseer, or another slave who was called a "driver." This was back-breaking work done in the southern sun and humidity, and so it's not surprising that whippings – or the threat of them – were often necessary to get slaves to work. It's easy enough to talk about the brutality of slave discipline, but it can be difficult to internalize it. Like, you look at these pictures, but because you've seen them over and over again, they don't have the power they once might have. The pictures can tell a story about cruelty, but they don't necessarily communicate how arbitrary it all was. As, for example, in this story, told by a woman who was a slave as a young girl: "[The] overseer... went to my father one morning and said, "Bob, I'm gonna whip you this morning." Daddy said, "I ain't done nothing," and he said, "I know it, I'm going to whip you to keep you from doing nothing," and he hit him with that cowhide – you know it would cut the blood out of you with every lick if they hit you hard." That brutality – the whippings, the brandings, the rape – was real, and it was intentional, because, in order for slavery to function, slaves had to be dehumanized. This enabled slaveholders to rationalize what they were doing, and it was hoped to reduce slaves to the animal property that is implied by the term "chattel slavery." So the idea was that slaveholders wouldn't think of their slaves as human, and slaves wouldn't think of themselves as human. But it didn't work. Let's go to the Thought Bubble. Slaves' resistance to their dehumanization took many forms, but the primary way was by forming families. Family was a refuge for slaves and a source of dignity that masters recognized and sought to stifle. A paternalistic slave owner named Bennet H. Barrow wrote in his rules for the Highland Plantation: "No rule that I have stated is of more importance than that relating to Negroes marrying outside of the plantation... It creates a feeling of independence." Most slaves did marry, usually for life, and, when possible, slaves grew up in two-parent households. Single-parent households were common, though, as a result of one parent being sold. In the upper South, where the economy was shifting from tobacco to different, less labor-intensive cash crops, the sale of slaves was common. Perhaps one-third of slave marriages in states like Virginia were broken up by sale. Religion was also an important part of life in slavery. While masters wanted their slaves to learn the parts of the Bible that talked about being happy in bondage, slave worship tended to focus on the stories of Exodus, where Moses brought the slaves out of bondage, or Biblical heroes, who overcame great odds, like Daniel and David. And, although most slaves were forbidden to learn to read and write, many did anyway. And some became preachers. Slave preachers were often very charismatic leaders, and they roused the suspicion of slave owners, and not without reason. Two of the most important slave uprisings in the South were led by preachers. Thanks, Thought Bubble. Oh, it's time for the Mystery Document? We're doing two set pieces in a row? All right. [buzzing noise] [music] The rules here are simple. I wanted to re-shoot that, but Stan said no. I guess the author of the Mystery Document. If I am wrong, I get shocked with the shock pen. "Since I have been in the Queen's dominions I have been well contented, yes well contented for sure, man is as God intended he should be. That is, all are born free and equal. This is a wholesome law, not like the southern laws which puts man made in the image of God on level with brutes. O, what will become of the people, and where will they stand in the day of judgment. Would that the 5th verse of the 3rd chapter of Malachi were written as with a bar of iron, and the point of a diamond upon every oppressor's heart that they might repent of this evil, and let the oppressed go free..." All right, it's definitely a preacher, because only preachers have read Malachi. Probably African American, probably not someone from the South. I'm going to guess that it is Richard Allen, the founder of the African Methodist Episcopal Church? [buzzing noise] DAAAH, DANG IT! It's Joseph Taper, and Stan just pointed out to me that I should have known it was Joseph Taper because it starts out, "Since I have been in the Queen's dominions..." He was in Canada. He escaped slavery to Canada. The Queen's dominions! All right, Canadians, I blame you for this, although, thank you for abolishing slavery decades before we did. [electric sounds] AHHH! So, the Mystery Document shows one of the primary ways that slaves resisted their oppression: by running away. Although some slaves like Joseph Taper escaped for good by running away to northern free states, or even to Canada, where they wouldn't have to worry about fugitive slave laws, even more slaves ran away temporarily, hiding out in the woods or the swamps, and eventually returning. No one knows exactly how many slaves escaped to freedom, but the best estimate is that a thousand or so a year made the journey northward. Most fugitive slaves were young men, but the most famous runaway has been hanging out behind me all day long: Harriet Tubman. Harriet Tubman escaped to Philadelphia at the age of 29, and over the course of her life, she made about 20 trips back to Maryland to help friends and relatives make the journey north on the Underground Railroad. But a more dramatic form of resistance to slavery was actual, armed rebellion, which was attempted. Now, individuals sometimes took matters into their own hands and beat or even killed their white overseers or masters. Like Bob, the guy who received the arbitrary beating, responded to it by killing his overseer with a hoe. But that said, large-scale slave uprisings were relatively rare. The four most famous ones all took place in a 35-year period at the beginning of the 19th century. Gabriel's Rebellion in 1800 – which we've talked about before – was discovered before he was able to carry out his plot. Then, in 1811, a group of slaves upriver from New Orleans seized cane, knives, and guns, and marched on the city before militia stopped them. And in 1822, Denmark Vesey, a former slave who had purchased his freedom, may have organized a plot to destroy Charleston, South Carolina. I say "may have" because the evidence against him is disputed and comes from a trial that was not fair. But regardless, the end result of that trial was that he was executed, as were 34 slaves. But the most successful slave rebellion, at least in the sense that they actually killed some people, was Nat Turner's in August 1831. Turner was a preacher, and with a group of about 80 slaves, he marched from farm to farm in South Hampton County, Virginia, killing the inhabitants, most of whom were women and children, because the men were attending a religious revival meeting in North Carolina. Turner and 17 other rebels were captured and executed, but not before they struck terror into the hearts of whites all across the American South. Virginia's response was to make slavery worse, passing even harsher laws that forbade slaves from preaching, and prohibited teaching them to read. Other slave states followed Virginia's lead and, by the 1830s, slavery had grown, if anything, more harsh. So, this shows that large-scaled armed resistance was – Django Unchained aside – not just suicidal, but also a threat to loved ones and, really, to all slaves. But, it is hugely important to emphasize that slaves did resist their oppression. Sometimes this meant taking up arms, but usually it meant more subtle forms of resistance, like intentional work slowdowns or sabotaging equipment, or pretending not to understand instructions. And, most importantly, in the face of systematic legal and cultural degradation, they re-affirmed their humanity through family and through faith. Why is this so important? Because too often in America, we still talk about slaves as if they failed to rise up, when, in fact, rising up would not have made life better for them or for their families. The truth is, sometimes carving out an identity as a human being in a social order that is constantly seeking to dehumanize you, is the most powerful form of resistance. Refusing to become the chattel that their masters believed them to be is what made slavery untenable and the Civil War inevitable, so make no mistake, slaves fought back. And in the end, they won. I'll see you next week. Crash Course is produced and directed by Stan Muller. The script supervisor is Meredith Danko. Our associate producer is Danica Johnson. The show is written by my high school history teacher Raoul Meyer and myself. And our graphics team is Thought Cafe. Every week, there's a new caption to the Libertage, but today's episode was so sad that we couldn't fit a Libertage in... UNTIL NOW! [Libertage Rock Music] Suggest Libertage caption in comments, where you can also ask questions about today's video that will be answered by our team of historians. Thanks for watching Crash Course, and as we say in my home town, don't forget to be abolitionist.

US_History

The_Great_Depression_Crash_Course_US_History_33.txt

Hi, I'm John Green, this is Crash Course U.S. history and Herbert Hoover's here, which is never a good sign. Today we're gonna return to two of my favorite topics: economics and inaccurate naming conventions. That's right, we're gonna be talking about the Great Depression, which was only great if you enjoy, like, being a hobo or selling pencils. Now some of you might get a bit frustrated today because there's no real consensus about the Great Depression, and simple, declarative statements about it really say much more about you than they do about history. Why are you looking at me, Mr Green? I didn't say anything. I thought it. Because, Me From the Past, you always want things to fit into this simplistic narrative: she loves me, she loves me not, the Great Depression was caused by x or was caused by y. It's complicated! (Intro Music) Many people tell you that the Great Depression started with the stock market crash in October 1929, but a) that isn't true and b) it leads people to mistake correlation with cause. What we think of as the Great Depression did begin AFTER the stock market crash, but not because of it. Like, as we saw last week, the underlying economic conditions in the U.S. before the stock market crash weren't all moonshine and rainbows. The 1920s featured large-scale domestic consumption of relatively new consumer products, which was good for American industry. But much of this consumption was fueled by credit and installment buying which, it turned out, was totally unsustainable. The thing about credit is that it works fine unless and until economic uncertainty increases at which point POW. That's a technical historian term, by the way. Meanwhile the agricultural sector suffered throughout the 1920s and farm prices kept dropping for two reasons. First, American farms had expanded enormously during World War I to provide food for all those soldiers, and second, the expansion led many farmers to mechanize their operations. As you'll know if you've ever bought a tractor, that mechanization was expensive, and so many farmers went into debt to finance their expansion. And then a combination of overproduction and low prices meant that often their farms were foreclosed upon . And other signs of economic weakness appeared throughout the decade. Like by 1925, the growth of car manufacturing slowed, along with residential construction. And, worst of all was what noted left wing radical Herbert Hoover labeled "an orgy of mad speculation" in the stock markets that began in 1927. By the way I'm kidding about him being a left wing radical. Just look at him. According to historian David Kennedy, "By 1929, commercial bankers were in the unusual position of loaning more money for stock market and real estate investments than for commercial ventures."[1] I wonder if we would ever find ourselves in that position again. Oh right we did in 2008. Anyway, it's tempting to see the stock market crash as the cause of the depression, possibly because it turns American economic history into morality play, but the truth is that the stock market crash and the depression were not the same thing. A lot of rich people lost money in the market, but what made the Great Depression the Great Depression was massive unemployment and accompanying hardship, and this didn't actually begin until, like, 1930 or 1931. The end of 1929 was actually okay. Unless you were a farmer. Or a stockbroker obviously. So what did actually cause the Depression? Well that's a big question and it's one that economists have struggled with ever since. They want to find out so they can keep it from ever happening again. No pressure, economists. Only 3% of Americans actually owned stock, and the markets recovered a lot of their value by 1930, although they did then go down again because, you know, there was a depression on. And even though big banks and corporations were buying a lot of stock, much of it was with borrowed money, known as margin buying, and all of that still was not nearly a big enough iceberg to sink the world's economy. But if I had to name a single cause of the Great Depression, it might be America's weak banking system. Alright. Let's go to the ThoughtBubble. Although the Federal Reserve system had been created in 1913, the vast majority of America's banks were small, individual institutions that had to rely on their own resources. When there was a panic and depositors rushed to take the money out of the bank -- like they do in the obscure arthouse movie Mary Poppins -- the bank went under if it didn't have enough money on reserve. So in 1930, a wave of bank failures began in Louisville that then spread to Indiana, Illinois, Missouri, and eventually Arkansas and North Carolina. As depositors lined up to take their money out before the banks went belly up, banks called in loans and sold assets. Ultimately this meant that credit froze up, which was what really destroyed the economy. A frozen credit system meant that less money was in circulation, and that led to deflation. Now you're probably thinking, "Big deal, deflation, can't be as bad as inflation right?" No. Deflation is much worse, as anyone who has ever slept on an air mattress knows. When prices drop, businesses cut costs, mainly by laying off workers. These workers then can't buy anything so inventories continue to build up and prices drop further. Banks weren't lending money, so employers couldn't borrow it to make payroll to pay their workers and more and more businesses went bankrupt leaving more and more workers unable to purchase the goods and services that would keep the businesses open. So if we have to lay the blame for the Great Depression on someone we can blame the banks, which isn't completely wrong, and it gives us a chance to shake our fists at Andrew Jackson whose distrust of central banking got us into this mess in the first place. That's probably too simple, but the Federal Reserve does deserve a good chunk of the blame for not rescuing the banks and not infusing money into the economy to combat this deflationary cycle. Thanks, Thoughtbubble. So, economics fans out there might be saying, "Why didn't the Hoover administration engage in some good old fashioned Keynesian pump priming?" The thinking there is that if governments do large-scale economic stimulus and a bunch of infrastructure projects, it can kind of create a bottom that stops the deflationary cycle. And that does often work, but unfortunately the Hoover Administration did not have a TARDIS. John Maynard Keynes' great work The General Theory of Employment, Interest and Money (he wasn't very good at titles) wasn't published until 1936, when the Depression was well under way. Venturing into the green nightmare of not-America for a moment, Herbert Hoover offered a global explanation in his memoirs for the global phenomenon that was the Great Depression. He claimed that its primary cause was World War One. And to be fair, the war did set the stage for a global economic disaster because of the web of debts and reparations that it created. Like, under the Versailles Treaty, Germany had to pay $33 billion in reparations mostly to France and Britain, which it couldn't pay without borrowing money from ... American banks. In addition the U.S. itself was owed $10 billion by Britain and France, some of which those countries paid back with German reparations. But then once American credit dried up, as it did in the wake of the stock market crash and the American bank failures, the economies of Germany, France, and Britain also fell off a cliff. And then with the largest non-U.S. industrial economies in total turmoil, fewer people abroad could buy American products, or French wine, or Brazilian coffee, and world trade came to a halt. And then when what the world really needed was more trade, America responded by raising tariffs to their highest levels ever with the Hawley Smoot tariff, a law that was as bad as it sounds. The idea of the high tariff was to protect American industry, but since Europe responded with their own high tariffs, that just meant that there were fewer buyers for American goods, less trade, fewer sales, and ultimately fewer jobs. So what did Hoover do? Not enough. It's important to remember that the American government is not just the President. Hoover couldn't always get Congress to do what he wanted but his political ineptitude was not particularly surprising because the first elected office that he ever held in his life was President of the United States. Like, let's take the foreign debt issue. Hoover proposed a moratorium on intergovernmental debt payments and he actually got Congress to go along with it, but it wasn't enough, mainly because the central bankers in Europe and America refused to let go of the gold standard, which would have allowed the governments to devalue their currency and pump needed money into their economies. And when Britain, rather heroically I might add, did abandon the gold standard in 1931 and stopped payments in gold, the U.S. did not follow suit, which meant that world financial markets froze up even further. Like this is a little bit complicated, but if you and I have always used Cheetos as currency to exchange goods and services and one day I announce that we can't do that anymore because it doesn't give us the flexibility that we need to pull ourselves out of this deflationary spiral. If I don't also agree to abandon Cheetos, then it's going to be a total disaster, which it was. And then, even worse, the Fed raised its discount rate, making credit even harder to come by. By the end of 1931, 2,294 American banks had failed, double the number that had gone under in 1930. Now, it's easy to criticize poor Herbert Hoover for not doing enough to stop the Great Depression, and he probably didn't do enough, but part of that is down to our knowledge of what happened afterward: the New Deal. That FDR at least tried to do something about the Depression makes us forget that when Hoover was president, orthodox political and economic theory counseled in favor of doing nothing. And at least Hoover didn't follow the advice of his treasury secretary who, according to Hoover anyway, argued that that the solution was to "liquidate labor, liquidate stocks, liquidate the farmers, liquidate real estate," which sounds like the worst milkshake ever. Instead, Hoover believed that the best course of action was to "use the powers of government to cushion the situation"[2] and in a White House meeting he persuaded a large number of industrialists to agree to maintain wage rates. He also got the Federal Farm Board to support agricultural production, and got Congressional approval for $140 million in new public works. Overall, he nearly doubled the federal public works expenditures between 1929 and 1931. It just wasn't nearly enough. Because what Hoover didn't allow was for the federal government to take over the situation completely. He relied primarily on private businesses and state and local governments to stimulate the economy, and that was insufficient. It's not surprising when you consider that in 1929 Federal expenditures accounted for 3% of our gross domestic product. Today it's more like 20%. So, it was just really hard to imagine the Federal government doing anything on such a large scale to address a national problem because it had never really done that much before. Hoover also hiked taxes as part of a plan to stabilize the banks by balancing the federal budget, providing confidence for foreign creditors, and stopping them from buying American gold. This would support bonds and also keep the federal government out of competition with private borrowers. The Revenue Act of 1932 passed Congress, but it didn't do much to stop the Depression. In fact, arguably it made it worse. Though ultimately, this dire situation forced Hoover into a truly radical move. In January 1932 he and Congress created the Reconstruction Finance Corporation, which was basically a federal bailout program that borrowed money to provide emergency loans to banks, building-and-loan societies, railroads, and agricultural corporations. The problem was that by 1932 bailing out the banks wasn't enough and the Great Depression started to take shape. By early 1932 well over 10 million people were out of work, 20% of the labor force. And in big cities the numbers were even worse, especially for people of color. Like, in Chicago, 4% of the population was African American, but they made up more than 16% of the unemployed. Although Hoover famously claimed that no one starved, which was a little bit let-them-eat-cake-y, people did search trash cans for food. And many Americans were forced to ask for relief. Hoover's response was to try to encourage private charity through the unfortunately acronymed President's Organization on Unemployment Relief. Or "POUR." New York City's government relief programs rose from $9 million in 1930 to $58 million in 1932, and private charitable giving did increase from $4.5 million to $21 million, and that sounds great until you realize that the total of $79 million that New York City spent on relief in 1932 was less than ONE MONTH's lost wages for the 800,000 people who were unemployed.[3] Oh, it's time for the Mystery Document? I hope it's a break from the unrelenting misery. Probably not. The rules here are simple. I guess the author of the Mystery Document and then usually fail and get shocked with the shock pen, which is a real shock pen no matter what you people say. Alright, what do we got here? "We sit looking at the floor. No one dares think of the coming winter. There are only a few more days of summer. Everyone is anxious to get work to lay up something for that long siege of bitter cold. But there is no work. Sitting in the room we all know it. This is why we don't talk; much. We look at the floor dreading to see that knowledge in each other's eyes. There is a kind of humiliation in it. We look away from each other. We look at the floor. It's too terrible to see this animal terror in each other's eyes." I mean, Stan, unemployment was 25% and this could be literally any of those people. I'm gonna guess that it's a woman, because men were usually on the road trying to find work while women would go to these offices to look. I - I mean it could be many - I have no idea. Ummm Janet Smith. Meridel Le Sueur? She's a good writer. Maybe we should hire her. AH! So, often at Crash Course we try to show how conventional wisdom about history isn't always correct. But in the case of the hardships experienced during the Great Depression, it really is. The pictures of Dorothea Lange and Walker Evans, and Steinbeck's description in Grapes of Wrath of Okies leaving the dust bowl in the usually vain hope of a better life in California, they tell the story better than I can. Thousands of Americans took to the road in search of work and thousands more stood in breadlines. There were shantytowns for the homeless called Hoovervilles, and there were protests, like the Bonus March on Washington by veterans seeking an early payment of a bonus due to them in 1945. A lot of the debate around the Great Depression revolves around the causes, while still more concerns the degree to which the federal government's eventual response, the New Deal, actually helped to end the Depression. Those questions are controversial because they're still relevant. We're still talking about how to regulate banking. We're still talking about what the government's role in economic policy should be and whether a strong federal government is ultimately good for an economy or bad for it. And how you feel about the government's role in the Great Depression is going to depend on how you feel about government in general. That said, we shouldn't let our ideological feelings about markets and governments and economics obscure the suffering that millions of Americans experienced during the Great Depression. For generations of Americans, it was one of the defining experiences of their lives. Thanks for watching. I'll see you next week. Crash Course is produced and directed by Stan Muller, written by Raoul Meyer, and made with the help of all of these nice people. And it is possible because of your support through Subbable. These videos are only possible because of the support Crash Course viewers give the show on a monthly basis through Subbable. There's a link in the video info if you'd like to join those subscribers. Cool perks and stuff, but mostly educational video available for free to everyone forever. Thank you for watching and supporting Crash Course and as we say in my hometown, don't forget to be awesome...I'm gonna hit the globe! Nailed it.

US_History

The_Roaring_20s_Crash_Course_US_History_32.txt

Hi, I'm John Green, this is Crash Course US History, and today we're gonna learn about one of the best eras ever: the 1920s. The 20s gave us jazz, movies, radio, making out in cars, illegal liquor, and the 20s also gave us prosperity--although not for everybody-- and gangsters, and a consumer culture based on credit, and lots of prejudice against immigrants, and eventually the worst economic crisis the US has ever seen. Mr. Green, Mr. Green, but what about Gatsby? Yeah, me from the past, it's true that Gastby turned out all right in the end, but what preyed on Gatsby, what foul dust trailed in the wake of his dreams, did temporarily close out my interest in the aborted sorrows and short-winded elations of men. *theme music* So there's a stereotypical view of the 1920s as "The Roaring 20s," a decade of exciting change and new cultural touchstones, as well as increased personal freedom and dancing. And it really was a time of increased wealth-- for some people. The quote of the decade has to go to our famously taciturn president from Massachusetts, Calvin Coolidge, who said, Jay-Z would later update this for the 21st century noting, But anyway, during the 1920s, the government helped business grow like gangbusters, largely by not regulating it much at all. This is known as “laissez-faire” capitalism. Or “laissez-faire” capitalism if you’re good at speaking French. The Republican Party dominated politics in the 1920s, with all the presidents elected in the decade being staunch conservative Republicans. The federal government hewed to the policies favored by business lobbyists, including lower taxes on personal income and business profits, and efforts to weaken the power of unions. Presidents Harding, Coolidge, and Hoover stocked the boards of the Federal Reserve and the Federal Trade Commission with men who shared their pro-business views, shifting the country away from the economic regulation that had been favored by Progressives. And that was very good for the American economy, at least in the short run. The 1920s were also marked by quite a bit of government corruption, most of which can be pinned to the administration of Warren G. Harding. Now, Harding himself wasn't terribly corrupt, but he picked terrible friends. They included Attorney General Harry Daugherty who accepted money to not prosecute criminals, and Interior Secretary Albert fall, who took half a million dollars from private business in exchange for leases to government oil reserves at Teapot Dome. Fall later became the first cabinet member ever to be convicted of a felony, but on the other hand, business, man! Productivity rose dramatically largely because older industry's adopted Henry Ford's assembly line techniques and newer industries like aviation, chemicals, and electronics grew up to provide Americans with new products and new jobs. During the 1920s annual production of cars tripled to 4.8 million, and automobile companies were gradually consolidated into the big three that we know today: Ford, Chrysler, and Harley-Davidson. What? General Motors. By 1929 half of all American families owned a car and thus began the American love affair with the automobile, which is also where love affairs were often consummated, which is why in the 1920s cars came to be known as Scootaloo pooping chariots. What's that? They were called brothels on wheels? And the economy also grew because American corporations were extending their reach overseas, and American foreign investment was greater than that of any other country. The dollar replaced the pound as the most important currency for trade and by the end of the decade America was producing eighty-five percent of the world's cars and forty percent of its overall manufactured goods. Stan can I get a Libertage? And companies turned out all kinds of labor-saving devices like vacuum cleaners, toasters, refrigerators, and not having to spend all day washing your clothes, or turning over your own toast like some kind of common or meant that Americans had more time for leisure. And this was provided by radios and baseball games boxing matches vacations dance crazes. I mean before Gangnam style there was the windy and the Charleston but probably the most significant leisure product was movies and I'm not just saying that because I'm staring into a camera. The American film industry moved out to Hollywood before World War one because land was cheap and plentiful all that sunshine meant that you could shoot outside all year round and it was close to everything: desert, mountains, ocean, plastic surgeons. And by 1925 the American film industry had eclipsed all of its competitors and become the greatest in the world, especially if you count by volume and not quality, and more and more people had money to go see those movies thanks to consumer debt. The widespread use of credit and lay away buying plans meant that it was acceptable to go into debt to maintain what came to be seen as the American standard of living and this was a huge change in attitude. These days we don't even think of credit cards as debt, really. But they are. And that was a relatively new idea as was another feature of American life in the 20s that is still with us: celebrity. Opera singer Enrico Caruso has often been called the first modern celebrity but now he's a lot less famous than Charlie Chaplin or Rudolph Valentino or Babe Ruth but probably the biggest celebrity of the decade was Charles Lindbergh whose claim to fame was flying across the Atlantic Ocean by himself without stopping although he did use an airplane which makes it slightly less impressive. Now Lindbergh wasn't a truly contemporary celebrity in the sense of being famous for being famous, but he was a business more than a businessman. High culture also flourished. This was the age of the lost generation of American writers, many of whom lived and worked in Europe but America had its own version of Paris in New York. The decade of the 1920s saw continued migration of African American people from the South to cities in the nNorth, and Harlem became the capital of Black America. And speaking of migration, let us now migrated to the chair for the Mystery Document. The rules here are simple: I guess the author of the mystery document, I'm either right or I get shocked with the shock pen. Alright let's see we got here. “If we must die would it not be like hogs hunted and penned in an inglorious spot, while round us bark the mad and hungry dogs, making their market are a curse a lot... Like men we'll face the murderous, cowardly pack, pressed to the wall, dying but fight back.” Stan thank you for the poetry I appreciate that it's not some obscure document from 18th century blah blah blah It's Claude McKay Harlem Renaissance poet, the poem is called “If We Must Die.” Ah, it's the only thing in the world I'm actually good at. Now I know this from the imagery alone, especially the line about “mad and hungry dogs” that would figuratively and literally make up the mobs at the lynchings, but the giveaway here is the ultimate sentiment that we will fight back. This was part of the spirit of the Harlem Renaissance which rejected stereotypes and prejudice and sought to celebrate African-American experience. Meanwhile, things for changing for women as well, as they found new ways to express autonomy. Flappers kept their hair and skirts short, smoked and drank illegally in public, and availed themselves of birth control. And marketers encouraged them to buy products like cigarettes christened torches of freedom by Edward Bernays. Liberation had its limits though; most women were still expected to marry, have children, and find their freedom at home through the use of washing machines, but the picture of prosperity is as usual more complicated than it at first appears. The fact that so many Americans were going into debt in order to pursue the American dream meant that if the economy faltered, and it did, there was going to be lots of trouble. Let's go to the Thought Bubble. Prosperity in the 1920s wasn't equally distributed through the population. Real industrial wages rose by a quarter between 1922 and 1929 but corporate profits rose at twice that rate. By 1929, one percent of the nation's banks controlled fifty percent of the nation's financial resources and the wealthiest five percent of Americans share of national income exceeded that of the bottom sixty percent. An estimated forty percent of Americans lived in poverty. Now many Americans celebrated big business, and Wall Street was often seen as heroic possibly because by 1920 about 1.5 million Americans owned some kind of stock, but big business also meant that smaller businesses disappeared. During the 1920s the number of manufacturing workers declined by 5%, the first time this class of workers had seen its numbers drop, but not the last. Now some of these jobs were made up for by new jobs in retail finance and education, but as early as the 1920s New England was beginning to see unemployment in deindustrialization as textile companies moved their operations to the south where labor was cheaper and working-class people still made up the majority of Americans and they often couldn't afford these newfangled devices, like in 1930, seventy-five percent of american homes didn't have a washing machine, and only forty percent of them had a radio. Farmers were even worse off many had prospered during World War One when the government subsidized farm prices in order to keep farms producing for the war effort, but when the subsidies ended, production didn't subside, largely due to mechanization and increased use of fertilizer. Farmers incomes dropped steadily and many saw banks foreclose upon their property. For the first time in American history the number of farms declined during the 1920s. For farmers the Great Depression began early. Thanks, Thought Bubble. So in general the federal government did little to nothing to help farmers or workers. The Supreme Court was the only segment of the government that kept any progressive ideas alive as they began to craft a system of ideas that we call the jurisprudence of civil liberties. Now the court still voted to uphold convictions of left-wing critics of the government but gradually began to embrace the idea that people had the right to express dissonant views in what Oliver Wendell Holmes called the “Marketplace of ideas.” In Near vs. Minnesota, the Supreme Court struck down censorship of newspapers and by 1927 Justice Brandeis was writing that “Freedom to think as you will and to speak as you think are indispensable to the discovery and spread of political truth.” But despite increased free speech and torches of liberty and flappers and the Harlem Renaissance the 1920s was in many ways a reactionary period in American history. For instance the decade saw the resurgence of the Ku Klux Klan in a new and improved form and by improved I mean much more terrible. Spurred on by the hyper patriotism that was fostered during World War One, the Klan denounced immigrants and Jews and Catholics as less than one hundred percent American, and by the mid 20s the Klan claimed more than 3 million members and it was the largest private organization right here in my home state of Indiana. And with more immigrants coming from Southern and Eastern Europe who were often Catholic and Jewish, White Protestants became more and more concerned about losing their dominant position in the social order. Spoiler alert: it turns out okay for you, White Protestants The first immigration restriction bill was passed in 1921, limiting the number of immigrants from Europe to 357,000. In 1924, a new immigration law dropped that number to 150,000 and established quotas based on national origin. The numbers of immigrants allowed from Southern and Eastern Europe were drastically reduced and Asians except for Filipinos were totally forbidden. The quota for Filipinos was set at 50 per year although they were still allowed to emigrate to Hawaii because their labor was needed there. There were no restrictions, however, on immigration from the Western Hemisphere because California's large-scale farms were dependent upon seasonal laborers from Mexico. These immigration restrictions were also influenced by fear of radical anarchists and pseudo scientific ideas about race; whites were seen as scientifically superior to people of color and as President Coolidge himself declared when he signed the 1924 immigration law, “America must be kept American” Tell me Calvin Coolidge about how American you are. Are you Cherokee, or Cree, or Lakota? The 1920s also saw increased tension between science education in the United States and religious beliefs. The best known example is of course the trial of John Scopes in Tennessee in 1925. Scopes was tried for breaking the law against teaching evolution which he had been encouraged to do by the ACLU as a test case for freedom of speech. Scopes was prosecuted by William Jennings Bryan whom you will remember as having recently resigned as Secretary of State and who had become a leader of the Fundamentalist Movement. And Scopes was defended by Clarence Darrow, that famous defense attorney who contemporary defense attorneys always point to to argue that defense attorneys aren't all scum. Scopes and Darrow actually lost the trial but the case drew national attention and ultimately led to evolution being taught in more American schools. The Scopes trial is often seen as a victory for free thinking and science and modernism, and I suppose it was, but for me it's more a symbol of the contradictions of the 1920s. This is the decade that gave us mass consumer culture and celebrity worship, which are important and very complicated legacies. And it also saw the birth of modern conceptions of civil liberties. It was a period when tolerance became an important value, but at the same time it saw a rise in lynchings. Immigrants were necessary for the economic boom of the 1920s, but at the same time their numbers were restricted, as they were seen as a threat to traditional American value, and that raises a question that we're still struggling with today: What are those values? I don't mean that rhetorically let me know in comments. Thanks for watching. I'll see you next week. Crash Course is produced and directed by Stan Muller, our script supervisor is Meredith Danko, the Associate Producer is Danica Johnson to show is written by my high school history teacher Raoul Meyer Rosianna Rojas and myself and our graphics team is Thought Cafe. I nailed that. Every week there's a new caption for the Libertage. You can suggest your own in comments or ask questions about today's video that will be answered by our team of historians. Thank you for watching Crash Course, If you enjoyed today's episode make sure you subscribe. And as we say in my hometown: Don't Forget to be Awesome.

US_History

19th_Century_Reforms_Crash_Course_US_History_15.txt

Hi I’m John Green. This is Crash Course U.S. History and today we finally get to talk about sex. Also some other things. Today we’re gonna discuss religious and moral reform movements in 19th century America, but I promise there will be some sex. Mr. Green, Mr. Green. Is it gonna be about real sex or is it gonna be able people who are obsessed with not having sex? You got me there, Me from the Past. But how (and whether) we skoodilypoop ends up saying a lot about America, and also people in general. [Theme Music] So, one response to the massive changes brought about by the shift to an industrialized market economy was to create utopian communities where people could separate themselves from the worst aspects of this brave new world. The most famous at the time, and arguably still, were the Shakers, who were famous for their excellent furniture, so you can’t say that they really fully withdrew from the market system. Still Shaker communities did separate themselves from the competition that characterized free markets, especially in terms of the competition for mates. They were celibate, and therefore only able to increase their numbers by recruitment, which was made a little bit difficult by celibacy. But they did do a lot of dancing to sublimate their libidinous urges, they embraced equality of the sexes, and at their peak they had more than 6,000 members. Today, they are still one of the most successful utopian communities to have emerged in the 19th century. They have three members. Much more successful in the long run were the Latter Day Saints, also called Mormons, although at the time their ideas were so far out of the mainstream that they were persecuted and chased from New York all the way to Utah. In addition to the Bible, The LDS Church holds the Book of Mormon as a holy scripture, which tells of the resurrected Jesus’s visits to the Americas. And while it was subject to widespread persecution, and even some massacres, the LDS Church continued to grow, and in fact continues to today. So, while some of these communities were based in religion, others were more worldly attempts to create new models of society, like Brook Farm. Founded in 1841 by a group of transcendentalists, is a dependent clause that always ends in failure, Brook Farm tried to show that manual labor and intellectual engagement could be successfully mixed. This community drew on the ideas of the French socialist Charles Fourier, who as you may recall from Crash Course World History believed—no joke—that socialism would eventually turn the seas to lemonade. And much like Fourier’s planned communities, Brook Farm did not work out, largely because—and I can say this with some authority—writers do not enjoy farming. Nathaniel Hawthorne, for instance, complained about having to shovel horse manure. But if he’d only kept shoveling horse manure, he might not have shoveled The Blithedale Romances onto an unsuspecting reading public. I’m sorry, Nathaniel Hawthorne. I do like The Scarlet Letter, but I feel like the only reason you’re read is because you were, like, the only author in pre-Civil War America. So either we have to pretend that America began with Huck Finn’s journey on the Mississippi or else we’re stuck with you. It was just, like, you, Thomas Paine, Mary Rowlandson, a bunch of printed sermons, and James Fenimore Pooper. Anyway, the most utopian of the utopian communities were set up at Utopia, Ohio and Modern Times, New York by Josiah Warren. Everything here was supposed to be totally unregulated and voluntary including marriage, which, as you can imagine worked out brilliantly. But, without any laws to regulate behavior, Warren’s communities were individualism on steroids, so they collapsed spectacularly and quickly. But these utopian communities were relatively rare; many more 19th century Americans participated in efforts to reform society rather than just withdraw from it. And behind most of those reform movements was religion, particularly a religious revival called the 2nd Great Awakening. This series of revival meetings reached their height in the 1820s and 1830s with Charles Grandison Finney’s giant camp meetings in New York. And in a way the 2nd Great Awakening made America a religious nation. The number of Christian ministers in the United States went from 2,000 in the 1770s to 40,000 by 1845. And western New York was the center of this revivalism. That’s where Joseph Smith had his revelations. It’s also where John Humphrey Noyes founded his Oneida Community, in which postmenopausal women introduced teenage boys to sex, and which eventually ceased being a religious community and evolved into—wait for it—one of the world’s largest silverware companies. That’s right, every time you take a bite of food with Oneida cutlery, you’re celebrating free love and May-December relationships. Well, more like February-December relationships. (Libertage: Turning Free Love into Fancy Forks) So, yes, religious fervor burned so hot in upstate New York that it became known as the “burned-over district,” and New York remains the heartland of conservative Christianity to this day. Or not. The Awakening stressed individual choice in salvation and a personal relationship with Jesus Christ, and it was deeply influenced by the market revolution. So, like, while many preachers criticized the selfish individualism inherent in free market competition, there was sort of a market for new religions and preachers, who would travel the country drumming up business. Awakening ministers also preached the values of sobriety, industry and self-discipline, which had become the essence of both the market economy and the impulse for reform. There are three points I want to make about the religious nature of all these 19th century reform movements. First, it was overwhelmingly Protestant. Like, all these “new” religions were Protestant denominations, which meant that they wouldn’t have a lot of appeal to immigrants from Ireland and Germany who started to pour into the United States in the middle of the 19th century because A. those people were mostly Catholic, and B. reasons we’ll get to momentarily. Secondly, many of these reformers believed in perfectionism, the idea that individuals and society were capable of unlimited improvement. And third, many of the reform movements were based ultimately on a different view of freedom than we might be used to. And this is really important to understand, for 19th century reformers, freedom was the opposite of being able to do whatever you wanted, which they associated with the word license. They believed that true freedom was like an internal phenomenon that came from self-discipline and the practice of self control. Essentially, instead of being free to drink booze, you would be free from the temptation to drink booze. According to Philip Schaff, a minister who came to Pennsylvania in the 1840s, “true national freedom, in the American view [is] anything but an absence of restraint … [It] rests upon a moral groundwork, upon the virtue of self possession and self control in individual citizens.” Members of the fastest growing Protestant denominations like Methodists and Baptists were taught that it wasn’t enough to avoid sin themselves; they also needed to perfect their communities. And that leads us to America’s great national nightmare, temperance. Now you’re not going to see me advocate for prohibition of alcohol, but to be fair, Americans in the first half of the 19th century were uncommonly drunk. In fact, in 1830, per capita liquor consumption was 7 gallons per year, more than double what it is now. And that doesn’t even count wine, beer, hard cider, zima, pruno. By the way, some people like to have home breweries or whatever, but at our office, Stan’s been making pruno under the couch. The growing feeling among reformers that we should limit or even ban alcohol appealed to those protestant ideas of restraint and perfecting the social order. And that’s also precisely why it was so controversial, especially among Catholic immigrants, who A. came largely from Germany and Ireland, two nations not known for their opposition to strong drink, and B. were Catholic and the Catholic church’s morality didn’t view alcohol or dancing as inherently sinful the way that so many Protestant denominations did. And then we have the widespread construction of asylums and other homes for outcasts. Anyone who’s ever done a bit of urban exploring knows that these places were built by the hundreds in the 19th century—jails, poorhouses, asylums for the mentally ill— and while they might not seem like places of freedom, to reformers they were. Remember, freedom was all about not having the choice to sin so you could be free of sin. Bear in mind, of course, that the crusading reformers who built these places usually chose not to live in them. And speaking of places you’re forced to go regardless of whether you want to, the mid 19th century saw the growth of compulsory state-funded education in the United States. These new schools were called common schools, and education reformers like Horace Mann hoped that they would give poor students the moral character and body of knowledge to compete with upper-class kids. And that worked out great. Just look at where we are on the equality of opportunity index. Now, this may seem like an obvious win for all involved, but many parents opposed common schools because they didn’t want their kids getting moral instruction from the government. That said, by 1860, all northern states had established public schools. But they were far less common in the South, where the planter class was afraid of education falling into the wrong hands, like for instance, those of poor whites and especially slaves. Which brings us to abolition. Let’s go to the Thought Bubble. Abolitionism was the biggest reform movement in the first half of the 19th century, probably because—sorry alcohol and fast dancing—slavery was the worst. In the 17th and 18th centuries, the only challengers to slavery were slaves themselves, free blacks, and Quakers. But in the early 19th century, colonizationists began to gain ground. Their idea was to ship all former slaves back to Africa, and the American Colonization Society became popular and wealthy enough to establish Liberia as an independent homeland for former slaves. While the idea was impractical, and racist, it appealed to politicians like Andrew Jackson and Henry Clay. And some black people, who figured that America’s racism would never allow them to be treated as equals, did choose to emigrate to Liberia. But most free blacks opposed the idea; in fact in 1817, 3,000 of them assembled in Philadelphia and declared that black people were entitled to the same freedom as whites. By 1830, advocates for the end of slavery became more and more radical, like William Lloyd Garrison, whose magazine The Liberator was first published in 1831. Known for being “as harsh as truth and as uncompromising as justice,” Garrison once burned the Constitution, declaring it was a pact with the devil. Radical abolitionism became a movement largely because it used the same mix of pamphleteering and charismatic speechifying that people saw in the preachers of the Second Great Awakening, which in turn brought religion and abolition together in the North, preaching a simple message: Slavery was a sin. By 1843, 100,000 Northerners were aligned with the American Anti-Slavery Society. What made the radical abolitionists so radical was their inclusive vision of freedom. It wasn’t just about ending slavery but about equality—the extension of full citizens’ rights to all people, regardless of race. By the way, it was abolitionists who re-christened the Old State House Bell in Philadelphia the “Liberty Bell.” Why does all this awesome stuff happen in Philadelphia? Thanks, Thought Bubble. So, needless to say, not all Americans were quite so thrilled about abolitionism, which is why slavery remained unabolished. Often, resistance to abolitionism was violent—like, in 1838, a mob in Philadelphia burned down Pennsylvania Hall because people were using it to hold abolitionist meetings. And you were doing so well, Philadelphia! A year later, a mob in Alton, Illinois murdered antislavery editor Elijah P. Lovejoy when he was defending his printing press. This was the fifth time, by the way, that a mob had destroyed one of his newspapers. Even Congress got in on the “let’s suppress free speech and the press” act by adopting the gag rule in 1836. The gag rule prohibited members of congress from even reading aloud or discussing calls for the emancipation of slaves. Seriously. And you thought the filibuster was dysfunctional. The best known abolitionist was Frederick Douglass, a former slave whose life story was well known because he wrote the brilliant Narrative of the Life of Frederick Douglass, An American Slave. But he wasn’t the only former slave to write about the evils of slavery: Josiah Henderson’s autobiography was probably the basis for the most famous anti-slavery novel ever, Uncle Tom’s Cabin. Uncle Tom’s Cabin sold more than a million copies between 1851 and 1854. And despite the unreadable, heavy-handed prose drenched in sentimentality, the book is a great reminder that bad novels can also change the world, which is why it was so widely banned in the South. But while based on a black man’s story, Uncle Tom’s Cabin was written by a white woman, which shows us that black abolitionists were battling not just slavery but near ubiquitous racism. Like Pat Boone rerecording Little Richard to make it safe for the white kids at the sockhop. They had to fight the pseudoscience arguing that black people were physically inferior to white people or just born to servitude, and they had to counter the common conception—still common, I’m afraid—that there was no such thing as African civilization. Oh, it’s time for the mystery document? The rules here are simple. If I guess the author of the mystery document, I do not get shocked. Let’s see what we got today. “Beloved brethren – here let me tell you, and believe it, that he lord our God, as true as he sits on his throne in heaven, and as true as our Savior died to redeem the world, will give you a Hannibal, and when the Lord shall have raised him up, and given him to you for your possession, O my suffering brethren! remember the divisions and consequent sufferings of Carthage and of Haiti … But what need have I to refer to antiquity, when Haiti, the glory of the blacks and terror of tyrants, is enough to convince the most avaricious and stupid of wretches?” Alright Stan, this is going to take some serious critical thinking skills so let’s break this down. So the author’s clearly African American, and an admirer of the Haitian Revolution, which means this was written after 1800. Plus, he references Hannibal, who Crash Course World History fans will remember almost conquered the Romans using freaking elephants! And Hannibal was from Carthage which, I don’t need to tell you, is in Africa. He also warns that Haiti is the terror of tyrants, referencing the widespread massacring of white people after the revolution. Okay that’s what we know. And now we shall make our guess. Henry Highland Garnett? UGH I HATE MYSELF. It’s David Walker? I’m not gonna lie to you, Stan, I don’t even know who that is, so I probably deserve this. AH! That’s how you learn, fellow students. It’s not about positive reinforcement. It’s about shocking yourself when you screw up. I got a 3 on the AP American History test, so I should know. So black abolitionists like Frederick Douglass, Henry Highland Garnett and apparently David Walker were the most eloquent spokesmen for the ideal of equal citizenship in the United States for black and white people. In his 1852 Independence Day Address. By the way, international viewers, our Independence Day is July 4th, so he gave this speech on July 4th. Frederick Douglass said: “Would you argue with me that man is entitled to liberty? That he is the rightful owner of his own body? You have already declared it. Must I argue the wrongfulness of slavery? … There is not a man beneath the canopy of heaven that does not know that slavery is wrong for him.” And in the end, the sophistication and elegance of the black abolitionists’ arguments became one of the strongest arguments for abolition. If black people were better off enslaved, and inherently inferior, how could anyone account for a man like Frederick Douglass? Abolitionism—at least until after the Civil War—pushed all other reform movements to the edges. But I just want to note here at the end that it’s no coincidence that so many abolitionist voices, like Harriet Beecher Stowe for instance, were female. And their work toward a more just social order for others transformed the way that American women imagined themselves as well, which is what we’ll be discussing next week. I’ll see you then. Thanks for watching. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher Raoul Meyer and myself. And our graphics team is Thought Café. If you have questions about today’s video, you can ask them in comments where they’ll be answered by our team of historians. You can also suggest captions for the libertage. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

The_Cold_War_in_Asia_Crash_Course_US_History_38.txt

Hi, I’m John Green, this is CrashCourse US History, and today we’re going to talk about the Cold War again. Really less about the “cold,” more about the “war.” As usual, we’re not going to focus so much on the generals and the tactics, but instead on why the wars were fought and what it all meant. And today we get to visit a part of the world that we haven’t seen much on this series: [spins] Asia. Not my best work. intro So, we’re gonna start today with the place where the Cold War really heated up, at least as far as America’s concerned. Mr Green, it’s Vietnam. Close, Me from the Past, but like all your romantic endeavors, unsuccessful. The correct answer is of course Korea. Like MFTP, many Americans have forgotten about the Korean War, which lasted three years from 1950 to 1953 and is sometimes called the Forgotten War. But it was real. The Korean War was the first real like shooting war that Americans were involved in after World War II and it was the only time that American troops directly engaged with an honest to goodness Communist power. I’m referring not to North Korea, but to China, which became communist in 1949 and qualifies as a major world power because it was, and also is, huge. We love you China. Just kidding, you’re not watching. Because of the Great Fire Wall. So the end of WWII left Korea split between a Communist north led by Kim Il crazypants Sung and an anti-communist but hardly democratic South led by Syngman Rhee. The two were supposed to reunite, but that was impossible because they were constantly fighting that cost around 100,000 lives. The civil war between the two Koreas turned into a full-fledged international conflict in June of 1950 when Kim Il Sung invaded the South, and the US responded. Truman thought that Kim’s invasion was being pushed by the Soviets and that it was a challenge to the “Free World.” Truman went to the United Nations and he got authorization, but he didn’t go to Congress and never called the Korean War a “war.” Insisting instead that American troops were leading a UN “police action” but that was kind of a misleading statement. General Douglas MacArthur was in command of this tiny little police force at the start of the war because he was the highest ranking general in the region. He was also really popular, at least with the press, although not so much with other generals, or with the president. Under MacArthur, UN forces – which basically meant American and South Korean forces -- pushed the North Koreans back past the 38th parallel where the two countries had been divided, and then Truman made a fateful decision: The United States would try to re-unify Korea as a non-communist state. Which, if you’ve looked at a map recently, you’ll notice went swimmingly. America’s allies and the UN all agreed to this idea, so up north they went, all the way to the northern border with China at the Yalu river. At that point, Chinese forces, feeling that American forces were a smidge too close to China, counter-attacked on November 1, 1950 and by Christmas the two sides were stalemated again at the 38th parallel, right where they started. The war dragged on for two more years with the U.S. pursuing a “scorched earth” policy and dropping more bombs on Korea than had been dropped in the entire Pacific theater during WWII. The two sides tried to negotiate a peace treaty, but the sticking point was the repatriation of North Korean and Chinese prisoners who didn’t want to go back to their communist homelands. Meanwhile, at home, Americans were growing tired of a war that they weren’t winning, which helped to swing the election of 1952 for Dwight Eisenhower. Also he was also running against perennial presidential loser Adlai Stevenson, who was perceived as an egghead intellectual because his name was Adlai Stevenson. In addition to helping get Ike elected, the Korean War had a number of profound effects. First and most importantly, it was expensive, both in terms of lives and money. In 3 years of fighting 33,629 Americans were killed, 102,000 were wounded and nearly 4 million Koreans and Chinese were killed, wounded, or missing. The majority of Korean casualties were civilians. The Korean War also further strengthened executive power in the United States – Truman went to war without a declaration and Congress acquiesced – this doesn’t mean that the war was initially was unpopular, it wasn’t. People wanted to see America do something about Communism and allowing Kim to take the south and possibly threaten Japan was unacceptable. However, the whole idea that you don’t really need to declare war to go to war, while not new in America, sure has been important the last 60 years. And the Korean War also strengthened the Cold War mentality. I mean, this was the height of the Red Scare. Also, the Korean War set the stage for America’s longer, more destructive, and more well known engagement in Asia, the Vietnam War. Oh it’s time for the Mystery Document? The rules here are simple. I guess the author of the Mystery Document. I’m either right or I get shocked. Alright, let’s see what we’ve got. "The Declaration of the French Revolution made in 1791 on the Rights of Man and the Citizen also states: "All men are born free and with equal rights, and must always remain free and have equal rights." Those are undeniable truths. Nevertheless, for more than eighty years, the French imperialists, abusing the standard of Liberty, Equality, and Fraternity, have violated our Fatherland and oppressed our fellow citizens. They have acted contrary to the ideals of humanity and justice. In the field of politics, they have deprived our people of every democratic liberty. Well Stan that sounds like a Frenchman who really doesn’t want to be French anymore. So it’s somebody who’s very disappointed by the way the France has been running their colonies. I’m going to guess that it’s North Vietnamese leader and Crash Course chalk board person: Ho Chi Minh. Wait Stan says he needs his real name. It’s Nguyen Sinh Cung. Yes! So, this document it points out that, at least rhetorically, Ho Chi Minh was fighting for liberation from a colonial power as much as, if not more than, he was trying to establish a communist dictatorship in Vietnam. But because of the Cold War and its prevailing theories, the United States could only see Ho as a communist stooge, a tool of the Kremlin. Under the so-called “domino theory” Vietnam was just another domino that had to be propped up or else the rest of South East Asia would fall to communism like a row of, dominos. That wasn’t my best work. Now, in retrospect, this was a fundamental misunderstanding, but it’s important to remember that at the time, people felt that they didn’t want the Soviet Union to expand the way that, say, Nazi Germany had. America’s involvement in Vietnam, like most things Cold War, dates back to World War II, but it really picked up in the 1950s as we threw our support behind the French in their war to maintain their colonial empire. Wait, Stan, how why would we fight with the French to maintain a colonial empire? Oh right, because we were blinded by our fear of communism. Now, Eisenhower wisely refused to send troops or use atomic weapons to help the French. Really good call. And the Geneva Accords were supposed to set up elections to reunite North and South, which had been divided after WWII, but then we didn’t let that happen. Because sometimes democracies don’t vote for our guy. Instead, the U.S. began supporting the repressive, elitist regime of Ngo Dinh Diem as a bulwark against communism. Diem was a Catholic in a majority Buddhist country and his support of landowners didn’t win him any fans. But he was against communism, which was good enough for us. The first major involvement of American troops, then called advisors, began in the early 1960s. Technically, their role was to advise the Army of the Republic of Vietnam, also called ARViN. It was doomed. How did they not know this was doomed? Let’s fight for Arvin. Against this guy. You are scary. Seriously. Anyway, pretty quickly this advising turned into shooting, and the first American advisors were killed in 1961, during John Kennedy’s presidency. However, most Americans consider Vietnam to be Lyndon Johnson’s war, and they aren’t wrong. The major escalation of American troops started under Johnson, especially in 1965 after the Gulf of Tonkin incident. This is one of the great incidents in all of American history. So, in August 1964, North Korean patrol boats attacked US warships in the Gulf of Tonkin. As a result Johnson asked Congress to authorize the president to take “all necessary measures to repel armed attack” in Vietnam, which Congress dutifully did with the Gulf of Tonkin Resolution. So why is this one of the great incidents in American history? Because the whole patrol boats attacking warships thing? That didn’t happen. None of that stuff happened except we did actually go to war. Now, in retrospect, this seems like a terrible idea but it was very popular at the time because to quote the historian James Patterson, “Preventing Communism, after all, remained the guiding star of American policy.”[1] Wait a second, did I just say to quote historian James Patterson, like the crime novelist? Oh it’s a different guy apparently. That’s a bummer. He doesn’t write his own books because he’s so busy with his secret career - being a historian. So, the number of American troops began a steady increase and so did the bombing. The frightfully named Operation Rolling Thunder began in the spring of 1965. And in March of that year two Marine battalions arrived at Danang airbase authorized to attack the enemy. No advising about it. But, Johnson didn’t actually tell the American public that our troops had this authorization, which was part of a widening credibility gap between what the government told Americans about the war and what was really happening. Let’s go to the ThoughtBubble. By 1968 there were about half a million American soldiers in Vietnam and the government was confidently saying that victory was just around the corner. But then in January Vietnamese forces launched the Tet Offensive and while it was eventually repelled, the fact that the North Vietnamese were able to mount such an offensive cast doubts on the claims that U.S. victory was imminent. The Vietnam War itself was particularly brutal, with much of the ground fighting taking place in jungles. Rather than large-scale offensives, troops were sent on search and destroy missions and often it was difficult to tell enemy from civilians. Capturing territory wasn’t meaningful, so commanders kept track of body counts. Like, if more enemy were killed than Americans, we were winning. In addition to jungle fighting, there was a lot of bombing. Like, more bombs were dropped on North and South Vietnam than both the Axis and Allied powers used in all of World War II. The U.S. used chemical defoliants like Agent Orange to get rid of that pesky jungle, and also napalm, which was used to burn trees, homes, and people. Television coverage meant that Vietnam was the first war brought into American living rooms. And people were horrified at what they saw. They were especially shocked at the My Lai massacre, which took place in 1968 but was only reported a year later, in 1969. These draftees were young, and disproportionately from the lower classes because enrollment in college or grad school earned you a deferment. So unlike previous American wars, the burden of fighting did not fall evenly across socioeconomic class. Thanks, ThoughtBubble. So, as Americans at home became increasingly aware of what was going on in Vietnam, protests started. But it’s important to remember that the majority of Americans were not out in the streets or on college campuses burning their draft cards. Right up through 1968 and maybe even 1970, most Americans supported the Vietnam War. During the 1968 presidential campaign, Richard Nixon promised that he had a secret plan to end the war and appealed to the silent majority of Americans who weren’t on board with the anti-war movement. So, the first part of Nixon’s secret plan was “vietnamization” -- gradually withdrawing American troops and leaving the fighting to the Vietnamese. The second part involved more bombing and actually escalating the war by sending American troops into Cambodia in order to cut off the so-called Ho Chi Minh, named for this guy, a supply line that connected north to south. Not only did this not work, it also destabilized Cambodia and helped the Khmer Rouge to come to power. The Khmer Rouge represented the absolute worst that Communism had to offer, forcing almost all Cambodians into communes and massacring one third of the country’s population. So, not a great secret plan. By 1970 the anti-protests had grown and discontent within the armed forces was enormous. Vietnam veterans, including future almost-president John Kerry, were participating in protests. And things got even worse when in 1971 the New York Times published the Pentagon Papers, classified documents that showed that the government had been misleading the public about the war for years. Congress eventually responded by passing the War Powers Act in 1973 which was supposed to limit the president’s ability to send troops overseas without their approval and prevent another Gulf of Tonkin Resolution. John: How does that work out, Stan? Stan: Not great. John: Yeah. I’ll say. After five years of negotiations, Nixon and his secretary of state Henry Kissinger were able to end America’s involvement in Vietnam. In 1973 the Paris Peace Agreement made it possible for America to withdraw its troops, although it left North Korea in control of some of South Vietnam. The war between North and South Vietnam however continued until 1975 when the North finally conquered the South and created a single, communist Vietnam. The Vietnam War cost the United States more than $100 billion spent, 58,000 Americans died as well as between 3 and 4 million Vietnamese people And Vietnam was the first war in America that we definitively lost. We lost it because we didn’t understand the Vietnamese and we didn’t understand why they were fighting. To return to the historian James Patterson, “the unyielding determination of the enemy … wore down the American commitment, which proved to be far less resolute.”[2] America expected that its superior technology and wealth would eventually wear down the Vietnamese and they’d just give up communism. But the Vietnamese weren’t fighting for communism. They were fighting for Vietnam. This fundamental misunderstanding combined with the government’s dishonesty changed American’s relationship with their leaders. Before Vietnam, most Americans trusted their government, even when they knew it did horrible things. But, after the war, and largely because of it, that trust was gone. Thanks for watching. I’ll see you next week. Crash Course is made with the assistance of all of these nice people and also through your support at Subbable. Subbable is a voluntary subscription service that allows you to support crash course at the monthly price of your choice including 0 dollars. So if you enjoy and value Crash Course, I hope you’ll consider supporting us through Subbable. If you can’t afford to do so, that’s fine. We’re just glad that you’re watching. You can click my face or there’s also a link in the video info. Thanks again for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

Women_in_the_19th_Century_Crash_Course_US_History_16.txt

Hi, I’m John Green; this is Crash Course U.S. History and today we’re going to talk about wonder women. Mr. Green, Mr. Green, finally we get to the history of the United States as seen through the lens of Marvel comic superheroes. Oh, Me from the Past, you sniveling little idiot. Wonder Woman is from the DC Universe. Also this is the study of history, which means a constant reexamination and redefinition of what it means to be a hero, and in the case of this episode, it’s about taking the first steps towards acknowledging that not all heroes worthy of historical recognition are men. So we’re going to talk about how women transformed pre-Civil War America as they fought to improve prisons, schools, decrease public drunkenness, and end slavery. And while fighting for change and justice for others, American women discovered that the prisoners, children, and slaves they were fighting for weren’t the only people being oppressed and marginalized in the American democracy. [Theme Music] So in the colonial era, most American women of European descent lived lives much like those of their European counterparts: They were legally and socially subservient to men and trapped within a patriarchal structure. Lower and working class women were actually more equal to men of their own classes, but only because they were, like, equally poor. As usual, it all comes back to economics. In general, throughout world history, the higher the social class, the greater the restrictions on women— although high class women have traditionally had the lowest mortality rates, which is one of the benefits of you know doors and extra lifeboats and whatnot. So at least you get to enjoy that oppression for many years. As previously noted, American women did participate in the American Revolution, but they were still expected to marry and have kids rather than, like, pursue a career. Under the legal principle of “coverture” actually husbands held authority over the person, property and choices of their wives. Also since women weren’t permitted to own property and property ownership was a precondition for voting, they were totally shut out of the political process. Citizens of the new Republic were therefore definitionally male, but women did still improve their status via the ideology of “Republican Motherhood.” Women were important to the new Republic because they were raising children—ESPECIALLY MALE CHILDREN—who would become the future voters, legislators, and honorary doctors of America. So women couldn’t themselves participate in the political process, but they needed to be educated some because they were going to potty train those who would later participate in the political process. What’s that? There were no potties? Really? Apparently instead of potties they had typhoid. Actually it was a result of not having potties. So even living without rights in a pottyless nation, the Republican Mother idea allowed women access to education, so that they could teach their children. Also women—provided they weren’t slaves--were counted in determining the population of a state for representation purposes, so that was at least an acknowledgement that they were at, like, five fifths human. And then the market revolution had profound effects on American women, too, because as production shifted from homes to factories, it shifted away from women doing the producing. This led to the so-called “cult of domesticity,” which like most cults, I am opposed to. That’s right, Stan, I’m opposed to the Blue Oyster Cult, The Cult, The Cult of Personality by In Living Color, and the three remaining Shakers. Sorry, Shakers. But who are we kidding? You’re not watching. You’re too busy dancing. The cult of domesticity decreed that a woman’s place was in the home, so rather than making stuff, the job of women was to enable their husbands to make stuff, by providing food and a clean living space, but also by providing what our favorite historian Eric Foner called “non-market values like love, friendship, and mutual obligation,” which is the way we talk about puppies these days. And indeed that’s in line with actual story titles from early 19th century American women’s magazines, like “Woman, a Being to Come Home To” and “Woman: Man’s Best Friend.” Oh, it’s time for the Mystery Document? I hope it’s from “Woman: Man’s Best Friend.” The rules here are simple. I either get the author of the Mystery Document right...oh, hey there, eagle...or I get shocked. Let’s see what we’ve got. “Woman is to win everything by peace and love; by making herself so much respected, esteemed and loved, that to yield to her opinions and to gratify her wishes, will be the free-will offering of the heart. … But the moment woman begins to feel the promptings of ambition, or the thirst for power, her aegis of defense is gone. All the sacred protection of religion, all the generous promptings of chivalry, all the poetry of romantic gallantry, depend upon woman’s retaining her place as dependent and defenseless, and making no claims, and maintaining no right but what are the gifts of honor, rectitude and love.” Well it was definitely a dude and I have no idea which dude, so I’m just going to guess John C. Calhoun because he’s a bad person. No? Well, what can you do? It wasn’t a dude? It was apparently Harriet Beecher Stowe’s sister Catharine who was an education reformer and yet held all of those opinions, so aaaaAAAAH. So I assume Stan brought up Harriet Beecher Stowe’s sister to point out that it wasn’t just men who bought into the Cult of Domesticity. The idea of true equality between men and women was so radical that almost no one embraced it. Like, despite the economic growth associated with the market economy, women’s opportunities for work were very limited. Only very low paying work was available to them and in most states they couldn’t control their own wages if they were married. But, still poor women did find work in factories or as domestic servants or seamstresses. Some middle class women found work in that most disreputable of fields, teaching, but the cult of domesticity held that a respectable middle class woman should stay at home. The truth is, most American women had no chance to work for profit outside their houses, so many women found work outside traditional spheres in reform movements. Okay, let’s go to the Thought Bubble. Reform movements were open to women partly because if women were supposed to be the moral center of the home, they could also claim to be the moral conscience of the nation. Thus it didn’t seem out of the ordinary for women to become active in the movement to build asylums for the mentally ill, for instance, as Dorothea Dix was, or to take the lead in sobering the men of America. Many of the most famous advocates for legally prohibiting the sale of alcohol in the US were women, like Carry Nation attacked bars with a hatchet and not because she’d had a few too many. The somewhat less radical Frances Willard founded the Women’s Christian Temperance Union in 1874, which would be one of the most powerful lobbying groups in the United States by the end of the 19th century. And women gave many temperance lectures featuring horror stories of men who, rather than seeking refuge from the harsh competition of the market economy and the loving embrace of their homes, found solace at the bottom of a glass or at the end of a beer hose. And by the way, yes, there were bars that allowed you to drink as much beer as you could, from a hose, for a nickel. Today, these establishments are known as frat houses. These temperance lectures would tell of men spending all their hard earned money on drink, leaving wives and children—there were always children—starving and freezing, because in the world of the temperance lecture, it was always winter. Now don’t get me wrong: Prohibition was a disaster, because 1. Freedom, and 2. It’s the only time we had to amend the constitution to be like, “Just kidding about that other amendment,” but it’s worth remembering that back then people drank WAY more than we do now, and also that alcohol is probably a greater public health issue than some recreational drugs that remain illegal. But regardless, the temperance movement made a huge difference in American life because eventually, male and female supporters of temperance realized that women would be a more powerful ally against alcohol if they could vote. Thanks Thought Bubble. So, in 1928, critic Gilbert Seldes wrote that if prohibition had existed in 1800, “the suffragists might have remained for another century a scattered group of intellectual cranks.” And to quote another historian, “the most urgent reasons for women to want to vote in the mid-1800s were alcohol related: They wanted the saloons closed down, or at least regulated. The wanted the right to own property, and to shield their families’ financial security from the profligacy of drunken husbands. They wanted the right to divorce those men, and to have them arrested for wife beating, and to protect children from being terrorized by them. To do all these things they needed to change the laws that consigned married women to the status of chattel. And to change those laws, they needed the vote.” Many women were also important contributors to the anti-slavery movement, although they tended to have more subordinate roles. Like, abolitionist Maria Stewart was the first African American woman to lecture to mixed male and female audiences. Harriet Beecher Stowe wrote the terrible but very important Uncle Tom’s Cabin. Sarah and Angelina Grimke, daughters of a South Carolina slaveholder, converted to Quakerism and became outspoken critics of slavery. Sarah Grimke even published the Letters on the Equality of the Sexes in 1838, which is pretty much what the title suggests. By the way, Stan, you could have made Sarah Grimke’s letters the Mystery Document. I would have gotten that. But I want to say one more thing about Harriet Beecher Stowe. There’s a reason we read Uncle Tom’s Cabin in history classes and not in literature ones, but Uncle Tom’s Cabin introduced millions of Americans to the idea that African American people were people. At least in 19th century readers, Uncle Tom’s Cabin humanized slaves to such a degree that it was banned throughout most of the south. So many women involved in the abolitionist movement, when studying slavery, noticed that there was something a little bit familiar. Now, some male abolitionists, notably Frederick Douglass and William Lloyd Garrison became supporters of women’s rights, but ultimately the male leaders of the anti-slavery movement denied women’s demands for equality, believing that any calls for women’s rights would undermine the cause of abolition. And they may have had a point because slavery only existed in parts of the country whereas women existed in all of it. In fact, one of the arguments used by pro-slavery forces was that equality under the law for male slaves might lead to a slippery slope ending with, like, equality for WOMEN. And out of this emerging consciousness of their own subordinate position, the movement for women’s rights was born. The most visible manifestation of it was the issue of woman’s suffrage, raised most eloquently at the Seneca Falls Convention of 1848 where Elizabeth Cady Stanton, Lucretia Mott, Susan B. Anthony and many others wrote and published the Declaration of Sentiments, modeled very closely on the Declaration of Independence. Except, in some ways this declaration was much more radical than the Declaration of Independence because it took on the entire patriarchal structure. Okay, so there are three things I want to quickly point out about the 19th century movement for women’s rights. First, like abolitionism, it was an international movement. Often American feminists travelled abroad to find allies, prefiguring the later transatlantic movement of other advocates for social justice like Florence Kelley and W.E.B DuBois. Secondly, for the most part, like other reform movements, the women’s movement was primarily a middle-class or even upper class effort. Most of the delegates at Seneca Falls, for instance, were from the middle class. There were no representatives of, like, cotton mills, but this didn’t mean that 19th century feminists didn’t acknowledge the needs of working women. Like, Sojourner Truth, probably the most famous black woman abolitionist, spoke eloquently of the plight of working class women, especially slaves, since she’d been one until 1827. And other women recognized that women needed to be able to participate in the market economy to gain some economic freedom. Now, of course all the women who wrote about the moral evils of 19th century America or spoke out or took hatchets to saloons were doing what we would now recognize as work. But they were not being paid. Amelia Bloomer got paid, though, because she recognized that it was impossible for women to easily participate in economic activities because of their crazy clothes. So she popularized a new kind of clothing featuring a loose fitting tunic, trousers, and eponymous undergarments. But then Bloomer and her pants were ridiculed in the press and in the streets, and this brings up the third important thing to remember about the 19th century women’s movement. It faced strong resistance. Patriarchy, like the force, is strong, which is why Luke and Yoda and Darth Vader and Obi-Wan and whoever Samuel Jackson played...all dudes. By the way, why did they train Luke up and not Princess Leia who was cooler and had more to fight for and was less screwed up? Patriarchy. Many women’s rights advocates were fighting to overturn not just laws, but also attitudes. Some of those goals, such as claiming greater control over the right to regulate their own sexual activity and whether or not to have children were twisted by critics to claim that women advocated “free love.” It’s interesting to note that the United States ended slavery more than 50 years before it granted women the right to vote and that although much of the march towards equality between the sexes has been slow and steady, the Equal Rights Amendment, despite being passed by Congress, was never ratified. But by taking leading roles in the reform movements in the 19th century, not just when it came to temperance and slavery, but also prisons and asylums, women were able to enter the public sphere for the first time. And these great women changed the world for better and for worse, just as great men do. And along the way, they made “the woman question” part of the movement for social reform in the United States. And in doing so, American women chipped away at the idea that a woman’s place must be in the home. That might not have been a presidential election or a war, but it is still bringing real change to our real lives on a daily basis. Thanks for watching. I’ll see you next week. Crash Course is produced and directed by Stan Muller. Our script supervisor is Meredith Danko. The associate producer is Danica Johnson. The show is written by my high school history teacher Raoul Meyer, and myself. And our graphics team is Thought Café. If you want to suggest captions for the libertage, please do so in comments where you can also ask questions about today’s video that will be answered by our team of historians. Thanks for watching Crash Course and as we say in my hometown, don’t forget to be awesome.

US_History

Terrorism_War_and_Bush_43_Crash_Course_US_History_46.txt

Hi, I’m John Green, this is CrashCourse U.S. history and today we’ve done it! WE’VE FINALLY REACHED THE 21st CENTURY! Today, we boldly go where no history course has gone before, because your teacher ran out of time and never made it to the present. Also, if you’re preparing for the AP test it’s unlikely that today’s video will be helpful to you because, you know, they never get to this stuff. Mr. Green, Mr. Green? Awesome, free period. Yeah, Me From the Past, there’s no such thing as a free period. There’s only time, and how you choose to use it. Also, Me From the Past, we’re in your future, hold on I’ve got to take this stuff off it’s hard to take me seriously with that. We’re in the future for you which means that you are learning important things about the you who does not yet exist. You know about Lady GaGa, Kanye and Kim, Bieber, well you’re not going to find out about any of those things because this is a history class, but it’s still going to be interesting. INTRO So the presidency of George W. Bush may not end up on your AP exam, but it’s very important when it comes to understanding the United States that we live in today The controversy starts with the 2000 Election. Democratic presidential candidate Al “I invented the Internet” Gore was sitting Vice President, and he asked Bill Clinton not to campaign much because a lot of voters kind of hated Bill Clinton. The republican candidate was George W. Bush, governor of Texas and unlike his father a reasonably authentic Texan. You know, as people from Connecticut go. Bush was a former oil guy and baseball team owner and he was running as a Compassionate Conservative, which meant he was organizing a coalition of religious people and fiscal conservatives. And that turned out to be a very effective coalition and George W Bush got a lot of votes. He did not however get as many votes as Al Gore. But as you’ll no doubt remember from earlier in Crash Course US History, in the United States presidential elections are not decided by popular vote. They are decided by the Electoral College. So the election was incredibly close. It solidified the Red-Blue divide that has become a trope for politicians since. And in the end Gore won the popular vote by about 500,000 votes. However, Al Gore did not have the necessary electoral votes to become president. Unless he won Florida. Did he win Florida? I don’t even want to go there… In Florida the vote was ridiculously close, but George W Bush had a gigantic advantage which is that his brother, Jeb Bush, was the governor of Florida. So when it came time to certify the election Jeb was like, “Yeah. My brother won. No big deal.” But then the Gore campaign sued to have a recount by hand which is allowed under Florida law. But then Bush’s lawyers asked the Supreme Court to intervene and they did. Their decision in Bush v. Gore remains rather controversial. They ruled that the recount should be stopped, interfering with a state law and also a state’s electoral process, which is a weird decision for strict constructionists to make. However, one of the strong points of the United States these past couple centuries has been that sometimes we have the opportunity to go to war over whether this person or that person should be president and we chose not to. So regardless of whether you think the recount should have gone on, or George W Bush should have been elected, he was, and he set to work implementing his campaign promises, including working on a missile defence system that was very similar to Star Wars. And that was Ronald Reagan’s Star Wars, not George Lucas’ Star Wars. Man if we could get a federally funded new Star Wars trilogy that doesn’t suck that would be awesome. Anyway, in the first 100 days of his presidency Bush also barred federal funding for stem cell research, and he supported oil drilling in the Arctic National Wildlife Refuge. And speaking of environmental policy, the Bush administration announced that it would not abide by the 1997 Kyoto Protocol on carbon emissions and that didn’t go over well with environmentalists in the U.S. or in all of these green parts of not-America because they were like, “You guys made all the carbon.” To which we said, “This is America.” Libertage Bush also attempted education reform with the No Child Left Behind Act, which mandated that states implement “rigorous” standards and testing regimes to prove that those standards were being met. The No Child Left Behind Act is especially controversial with teachers who are great friends of Crash Course US History so we will say nothing more. Most importantly, George W Bush pushed through the largest tax cut in American history in 2001. Claiming that putting more money in Americans’ pockets would stimulate growth in an economy that had stumbled after the bursting of the dot-com bubble in 2000. Oh, it’s time for the Mystery Document? The rules here are simple. I guess the author of the Mystery Document, I either get it right, or I get shocked with the shock pen. Alright, what have we got here today. I’ve got a feeling it’s going to be a sad one. “It was a beautiful fall day, with a crisp, blue sky. I was coming in to work late that day; I guess I didn’t have first period class. It was only the second or third day of school. When I emerged from the subway, Union Square was strangely quiet, which only added to the beauty of the day. People were standing still, which is weird in New York under any circumstances, and looking down University Place towards lower Manhattan. Before I even looked I asked a passerby what had happened. She, or he, I really don’t remember, said that a plane had crashed into the Trade Center. Then I looked and saw the smoke coming billo wing out of the South Tower. I thought it was an accident, but I knew that this was not going to be an easy day. Well it’s obviously someone who was in New York City on September 11, 2001, but that only narrows it down to like 10 million people. However, I happen to know that it is Crash Course historian and my high school history teacher Raoul Meyer who wrote that account. This is the saddest I have ever been not to be shocked. So whether George Bush’s domestic policy would have worked is up for debate, but the events of September 11, 2001 ensured that foreign policy would dominate any discussion of the opening decade of the 21st century. That morning terrorists affiliated with al Qaeda hijacked 4 airliners. Two planes were flown into Manhattan’s World Trade Center, a third was crashed into the Pentagon in Washington and a fourth, also headed for Washington DC crashed in Pennsylvania when passengers overpowered the hijackers. Almost 3,000 people died including almost 400 policemen and firefighters. As Americans rushed to help in the search for survivors and to rebuild a devastated city, a shared sense of trauma and a desire to show resolve really did bring the country together. President Bush’s popularity soared in the wake of the attacks. In a speech on September 20, the president told Americans watching on television that the terrorists had targeted America “Because we love freedom […]. And they hate freedom.” This is another critical moment in American history where the definition of freedom is being reimagined. And we were reminded in the wake of September 11th that one of the central things that government does to keep us free is to keep us safe. But at the same time ensuring our safety sometimes means impinging upon our freedoms. And the question of how to keep America safe while also preserving our civil liberties is one of the central questions of the 21st century. At any rate, in the September 20th speech, the president announced a new guiding principle in foreign policy that became known as the Bush Doctrine. America would go to war with terrorism making no distinction between the terrorists and nations that harbored them. Bush laid out the terms for the world that night: “Either you are with us or you are with the terrorists.” But that dichotomy of course would prove to be a bit of an oversimplification. So on October 7, the United States launched its first airstrikes on Afghanistan, which at the time was ruled by a group of Islamic fundamentalists called the Taliban who were protecting Osama bin Laden, al Qaeda’s leader. This was followed by American ground troops supporting the anti-Taliban Northern Alliance in chasing out the Taliban and setting up a new Afghan government that was friendly to the United States. This new government did undo many of the worst Taliban policies, for instance allowing women and girls to go to school, and even to serve in the parliament. More women than girls in the parliament naturally. But by 2007 the Taliban was beginning to make a comeback and although fewer than 100 Americans died in the initial phase of the war, a sizeable force remained and in the ensuing 12 years the number of Americans killed would continue to rise. And then, by January 2002, Bush had expanded the scope of the Global War on Terror by proclaiming that Iran, Iraq and North Korea were an “axis of evil” that harbored terrorists, even though none of those nations had direct ties to the September 11 attacks. The ultimate goal of Bush Doctrine was to make the world safe for freedom and also to spread it and freedom was defined as consisting of political democracy, free expression, religious toleration, free trade and free markets. These freedoms, Bush said, were, “right and true for every person, in every society”. And there’s no question that the Saddam Hussein led Iraq of 2003 was not, by any of those definitions, free. But the 2003 invasion of Iraq by the United States was predicated on two ideas. First, that Iraq had weapons of mass destruction - chemical and biological weapons that they were refusing to give up. And second, that there was, or at least may have been, a link between Saddam Hussein's Iraq and the Al Qaeda attacks of 9-11. So in March 2003 the United States, Britain, and a coalition of other countries, invaded Iraq. Within a month Baghdad was captured, Saddam Hussein was ousted, Iraq created a new government that was more democratic than Saddam’s dictatorship, and then descended into sectarian chaos. After Baghdad fell, President Bush declared the end of major combat operations in Iraq, but troops soon found themselves trying to manage an increasingly organized insurgency that featured attacks and bombings. And by 2006 American intelligence analysts concluded that Iraq had become a haven for Islamist terrorists, which it hadn’t been, before the invasion. In fact, Saddam Hussein’s socialist government, while it occasionally called upon religion to unify people against an enemy, was pretty secular. Although fewer than 200 Americans had died in the initial assaults, by the end of 2006, more than 3,000 American soldiers had been killed and another 20,000 wounded. Hundreds of thousands of Iraqis had died in the conflict and the costs of the war which were promised to be no more than $60 billion had ballooned to $200 billion dollars. So that, and we try really hard here at Crash Course to be objective was a bit of a disaster. But let’s now go back to the domestic side of things and jump back in time to the passage of the USA PATRIOT act. Which believe it or not is an acronym for the Uniting and Strengthening America by Providing Appropriate Tools Required to Intercept and Obstruct Terrorism act of 2001. Oh, Congress you don’t pass many laws these days but when you do… mmhm…. there’s some winners. The PATRIOT act gave the government unprecedented law enforcement powers to combat domestic terrorism including the ability to wiretap and spy on Americans. At least 5000 people connected to the Middle East were called in for questioning and more than 1200 were arrested, many held for months without any charge. The administration also set up a camp for accused terrorists in Guantanamo Bay, in Cuba, but not the fun kind of camp, the prison kind, it housed more than 700 suspects. The president also authorized the National Security Agency to listen in to telephone conversations without first obtaining a warrant, the so-called warrantless wiretapping. In 2013 Americans learned that NSA surveillance has of course gone much farther than this with surveillance programs like PRISM which sounds like it’s out of an Orwell novel - I mean both like the name and the actual thing it refers to. Meredith would like us to point out that Prism is also the name of a Katy Perry album proving that we here at Crash Course are young and hip and with it. Who is Katy Perry? Oh right, she has that song in Madagascar 3. Sorry, I have little kids. The Supreme Court eventually limited the executive branch’s power and ruled that enemy combatants do have some procedural rights. Congress also banned the use of torture in a 2005 defense appropriations bill sponsored by Republican John McCain who himself had been a victim of torture in Vietnam. But the Defense Department did condone the continued use of so-called “enhanced interrogation techniques” like waterboarding. Which most countries do consider torture. But George W Bush won re-election in 2004, defeating the surprisingly weak John Kerry, who was characterized as a “waffler” on a number of issues including the Iraq war. Kerry’s history as a Vietnam protester and also terrible windsurfer probably didn’t help him much. Bush’s victory is still a bit surprising to historians admittedly at that moment the Iraq war seemed to be going pretty well. But during Bush’s first term, the economy, which is usually what really drives voters, wasn’t that great at all. A recession began during 2001 and the September 11 attacks made it much worse. And while the GDP did begin to grow again relatively quickly, employment didn’t recover, hence all the description of it as a “jobless recovery.” 90% of the jobs lost in the 2001-2002 recession were in manufacturing, continuing a trend that we had been seeing for 30 years. The number of steelworkers dropped from 520,000 in 1970 to 120,000 in 2004. And in his first term George W Bush actually became the first president since Herbert Hoover to oversee a net loss of jobs. Now I want to be clear that that’s not necessarily his fault as I have said many times before - economics are complicated. And presidents do not decide whether economies grow. But at any rate George W Bush was re-elected and went on to have an extremely controversial second term. Let’s go to the thoughtbubble. In 2005 several events undermined the public’s confidence in the Bush administration. First, Vice President Dick Cheney’s chief of staff was indicted for perjury and then House Majority Leader Tom “The Hammer” DeLay was indicted for violating campaign finance laws. Then in August 2005, Hurricane Katrina slammed into the gulf coast near New Orleans submerging much of the city, killing nearly 1500 people, and leaving thousands stranded without basic services. Disaster preparation and response was poor on the state, local, and federal levels, but the slow response of the Department of Homeland Security and Federal Emergency Management Agency was particularly noticeable as thousands of mostly African American New Orleans residents suffered without food or water. Damage to the city was estimated at around $80 billion dollars. And the Katrina disaster exposed the persistent poverty and racial divisions in the city. While the Katrina response probably contributed to the reversal of fortune for Congressional Republicans in the 2006 mid-terms, it was more likely the spike in gasoline prices that resulted from the shutting down of refining capacity in the gulf and increased demand for oil from rapidly growing China. Voters gave Democrats majorities in both houses, and Nancy Pelosi of California became the first woman Speaker of the House in American history. And then, in 2007, the country fell back into recession as a massive housing bubble began to deflate, followed by the near collapse of the American banking system in 2008. Thought Bubble, thank you once again for the tremendous downer. So, the Bush years are still in the recent past, and it’s impossible to tell just what their historical significance is without some distance. But the attacks on September 11 had far ranging effects on American foreign policy but also on the entire world. Under the leadership of George W Bush the United States began a global fight against terrorism and for freedom. But as always, what we mean by the words is evolving and there’s no question that in trying to ensure a certain kind of freedom we have undermined other kinds of freedom. We’ll get to the even messier and murkier world of the 2008 financial collapse next week. Until then, thanks for watching. Crash Course is made with the help of all these nice people and it exists because of your support through Subbable.com - a voluntary subscription service that allows you to subscribe monthly to Crash Course for the price of your choosing. There are great perks over at Subbable, but the biggest perk of all is knowing that you helped make Crash Course possible so please check it out, thank you for watching, thanks for supporting Crash Course, and as we say in my hometown, “Don’t forget to be awesome.”

MIT_804_Quantum_Physics_I_Spring_2016

Photons_and_the_loss_of_determinism.txt

PROFESSOR: Determinism. And it all begins with photons. Einstein reluctantly came up with the idea that light was made of quanta-- quanta of light called photons. Now when you think of photons, we think of a particle. So everybody knew that light was a wave. Maxwell's equations had been so successful. Nevertheless, photoelectric effect-- Planck's work-- all were leading to the idea that, in some ways, photons were also particles. So when you think of a particle, however, there is an important difference between a particle in the sense of Newton, which is an object with zero size that carries energy and has a precise position and velocity at any time, and the quantum mechanical idea of particle, which is just some indivisible amount of energy or momentum that propagates. So light was made of photons-- packets of energy. And a photon is a particle-- a quantum mechanical particle. Not in the sense that maybe it has position and velocity determined or it's a point particle, but more like a packet that is indivisible. You cannot decompose it in further packets. So Einstein realized that for a photon, the energy was given by h nu. Where nu is the frequency of the light that this photon is helping build up. So if you have a beam of light, you should think it's billions of photons. And according to the frequency of that light that is related to the wavelength-- by the equation frequency times wavelength is velocity of light-- you typically know, for light, the wavelength, and you know the frequency, and then you know the energy of each of the photons. The photons have very, very little energy. We have very, very little energy, but your eyes are very good detectors of photons. If you're in a totally dark room, your eye, probably, can take as little as five photons if they hit your retina. So it's a pretty good detector of photons. Anyway, the thing that I want to explain here is what happens if a beam of light hits a polarizer. So what is a polarizer? It's a sheet of plastic or some material. It has a preferential direction. Let me align that preferential direction with the x-axis, and that's a polarizer. And if I send light that is linearly polarized along the x-axis, it all goes through. If I send light linearly polarized along the y-axis, nothing goes through. It all gets absorbed. That's what a polarizer does for a living. In fact, if you send light in this direction, the light that comes out is identical to the light that came in. The frequency doesn't change. The wavelength doesn't change. It's the same light, the same energy. So far, so good. Now let's imagine that we send in light linearly polarized at some angle alpha. So we send an electric field E alpha, which is E0 cosine alpha x hat plus E0 sine alpha y hat. Well, you've studied electromagnetism, and you know that this thing, basically, will come around and say, OK, you can go through because you're aligning the right direction, but you are orthogonal to my preferential direction, or orthogonal I absorbed, so this disappears. So after the polarizer, E is just E0 cosine alpha x hat. That's all that is left after the polarizer. Well here is something interesting-- you know that the energy on electromagnetic field is proportional to the magnitude of the electric field square, that's what it is. So the magnitude of this electric field-- if you can notice, it's the square root of the sum of the squares will give you E0 as the magnitude of this full electric field. But this electric field has magnitude E0 cosine alpha. So the fraction of power-- fraction of energy through is cosine alpha squared. The energy is always proportional to the square. So the square of this is E0 squared cosine squared alpha. And for this one, the magnitude of it is E0, so you divide by E0 and cosine alpha is the right thing. This is the fraction of the energy. If alpha is equal to 0, you get cosine of 01. You get all the energy 1. If alpha is equal to pi over 2, the light is polarized along the y direction, nothing goes through-- indeed, cosine of pi over 2 is 0, and nothing goes through. So the fraction of energy that goes through is cosine squared alpha. But now, think what this means for photons. What it means for photons is something extraordinarily strange. And so strange that it's almost unbelievable that we get so easily in trouble. Here is this light beam over here, and it's made up of photons. All identical photons, maybe billions of photons, but all identical. And now, think of sending this light beam over there-- a billion identical photons-- you send them one by one into the state, and see what happens. You know what has to happen, because classical behavior is about right. This fraction of the photons must go through, and 1 minus that must not go through. You see, it cannot be there comes a photon and half of it goes through, because there's no such thing as half of it. If there would be half of it, it would be half the energy and, therefore, different color. And we know that after a polarizer, the color doesn't change. So here is the situation. You're sending a billion photons and, say, one-third has to get through. But now, the photos are identical. How can that happen in classical physics? If you send identical photos, whatever happens to a photon will happen to all, but the photon either gets absorbed or goes through. And if it gets absorbed, then all should get absorbed. And if it goes through, all should go through because they are all identical. And now you have found a situation which identical set of experiments with identically prepared objects sometimes gives you different results. It's a debacle. It's a total disaster. What seems to have happened here-- you suddenly have identical photons, and sometimes they go through, and sometimes they don't go through. And therefore, you've lost predictability. It's so simple to show that if photons exist, you lose predictability. And that's what drove Einstein crazy. He knew when he entered these photons that he was getting in trouble. He was going to get in trouble with classical physics. So possible ways out-- people speculate about it-- people said, well, yes, the photos are identical, but the polarizer has substructure. If it hits in this interatomic part, it goes through, and in that interatomic part, it doesn't go through. People did experiments many times. It's not true. The polarizer is like that. And then came a more outrageous proposition by Einstein and others-- that there are hidden variables. You think the photons are identical, but a photon has a hidden variable-- a property you don't know about. If you knew that property about the photon, you would be able to tell if it goes through or it doesn't go through. But you don't know it, so that's why you're stuck with probabilities. It's because the quantum theory is not complete. There are hidden variables. And once you put the hidden variables, you'll discover the photon has more something inside it, and they are not the same, even though they look the same. And that's a hidden variable theory. And it sounds so philosophical that you would think, well, if you don't know about them, but they are there, these properties, how could you ever know they are there? And the great progress of John Bell with the Bell inequalities is that he demonstrated that that would not fix the problem. Quantum mechanics cannot be made deterministic with hidden variables. It was an unbelievable result-- the result of John Bell. So that's something we will advance towards in this course but not quite get there. 805 discusses this subject in detail. So at the end of the day, we've lost determinism. We can only predict probabilities. So photons either gets through or not, and can only predict probabilities. Now we write, in classical physics, a beam like that. But how do we write the wave function of a photon? Well, this is quite interesting. We think of states of a particle as wave functions. And I will call them, sometimes, states; I will call them, sometimes, wave functions; and I sometimes will call them vectors. Why vector? Because the main thing you do with vectors is adding them or multiplying them by numbers to scale them. And that's exactly what you can do with a linear equation. So that's why people think of states, or wave functions, as vectors. And Dirac invented a notation in which to describe a photon polarized in the x direction, you would simply write something like this. Photon colon x and this object-- you think of it as some vector or wave function, and it represents a photon in the x direction. And we're not saying yet what kind of vector this is, but it's some sort of vector. It's not just a symbol, it represents a vector. And that's a possible state. This is a photon polarized along x. And you can also have, if you wish, a photon polarized along y. And linearity means that if those photos can exist, the superposition can exist. So there can exist a state called cos alpha photon x plus sine alpha photon y, in which I've superposed one state with another-- created a sum-- and this I call the photon state polarized in the alpha direction. So this is how, in quantum mechanics, you think of this-- photons-- we will elaborate that and compare with this equation. It's kind of interesting. What you lose here is this ease. There's no ease there because it's one photon. When you have a big electric field, I don't know how many photons there are. I would have to calculate the energy of this beam and find the frequency that I didn't specify, and see how many photons. But each photon in this beam quantum mechanically can be represented as this superposition. And we'll talk more about this superposition now because our next subject is superpositions and how unusual they are. Well the hidden variable explanation failed because Bell was very clever, and he noted that you could design an experiment in which the hidden variables would imply that some measurements would satisfy an inequality. If the existed hidden variables and the world was after all classical, the results of experiments would satisfy a Bell inequality. And then a few years later, the technology was good enough that people could test the Bell inequality with an experiment, and they figured out it didn't hold. So the hidden variables lead to Bell inequalities that are experimentally shown not to hold. And we will touch a little bit on it when we get to untangle them. After the polarizer, the photon is in the state photon x. It's always polarized along the x direction, so it's kind of similar that this doesn't go through. This goes through, but at the end of the day, as we will explain very soon, the cosine alpha is not relevant here. When it goes through, the whole photon goes through. So there's no need for a cosine alpha. So that's what goes out of the polarizer.

MIT_804_Quantum_Physics_I_Spring_2016

Widths_and_uncertainties.txt

PROFESSOR: So we go back to the integral. We think of k. We'll write it as k naught plus k tilde. And then we have psi of x0 equal 1 over square root of 2pi e to the ik naught x-- that part goes out-- integral dk tilde phi of k naught plus k tilde e to the ik tilde x dk. OK. So we're doing this integral. And now we're focusing on the integration near k naught, where the contribution is large. So we write k as k naught plus a little fluctuation. dk will be dk tilde. Wherever you see a k, you must put k naught plus k tilde. And that's it. And why do we have to worry? Well, we basically have now this peak over here, k naught. And we're going to be integrating k tilde, which is the fluctuation, all over the width of this profile. So the relevant region of integration for k tilde is the range from delta k over 2 to minus delta k over 2. So maybe I'll make this picture a little bigger. Here is k naught. And here we're going to be going and integrate in this region. And since this is delta k, the relevant region of integration-- integration-- for k tilde is from minus delta k over 2 to delta k over 2. That's where it's going to range. So all the integral has to be localized in the hump. Otherwise, you don't get any contribution. So the relevant region of integration for the only variable that is there is just that one. Now as you vary this k tilde, you're going to vary the phase. And as the phase changes, well, there's some effect [? on ?] [? it. ?] But if x is equal to 0, the phase is stationary, because k tilde is going to very, but x is equal to 0. No phase is stationary. And therefore, you will get a substantial answer. And that's what we know already. For x is going to 0 or x equal to 0, we're going to get a substantial answer. But now think of the phase in general. So for any x that you choose, the phase will range over some value. So for any x different from 0, the face in the integral will range over minus delta k over 2x and to delta k over 2x. You see, x is here. The phase is k tilde x. Whatever x is, since k tilde is going run in this range, the phase is going to run in that range multiplied by x. So as you do the integral-- now think you're doing this integral. You have a nice, real, smooth function here. And now you have a running phase that you don't manage to make it stationary. Because when x is different from 0, this is not going to be stationary. It's going to vary. But it's going to vary from this value to that value. So the total, as you integrate over that peak, your phase excursion is going to be delta k times x-- total phase excursion is delta k times x. But then that tells you what can happen. As long as this total phase excursion is very small-- so if x is such that delta k times x is significantly less than 1-- or, in fact, I could say less than 1-- there will be a good contribution if x is such that-- then you will get a contribution. And the reason is because the phase is not changing much. You are doing your integral, and the phase is not killing it. On the other hand, if delta k times x-- delta k times x is much bigger than 1, then as you range over the peak, the phase has done many, many cycles and is going to kill the integral. So if k of x is greater than 1, the contribution goes to 0. So let's then just extract the final conclusion from this thing. So psi of x 0 will be sizable in an interval x belonging from minus x0 to x0. So it's some value here minus x0 to x0. If, even for values as long as x0, this product is still about 1-- if for delta k times x0, roughly say of value 1, we have this. And therefore the uncertainty in x would be given by 2x0. So x0 or 2x0, this x0 is basically the uncertainty in x. And you would get that delta k times delta x is roughly equal to 1-- so delta k delta x roughly equal to 1. So I'm dropping factors of 2. In principle here, I should push a 2. But the 2s, or 1s, or pi's at this moment are completely unreliable. But we got to the end of this argument. We have a relation of uncertainties is equal to 1. And the thing that comes to mind immediately is, why didn't Fourier invent the uncertainty principle? Where did we use quantum mechanics here? The answer is nowhere. We didn't use quantum mechanics. We found the relation between wave packets, known to Fourier, known to electrical engineers. The place where quantum mechanics comes about is when you realize that these waves in quantum mechanics, e to the ikx represent states with some values of momentum. So while this is fine and it's a very important intuition, the step that you can follow with is-- it's interesting. And you say that, well, since p, the momentum, is equal to h bar k and that's quantum mechanical-- it involves h bar. It's the whole discussion about these waves of matter particles carrying momentum. You can say-- you can multiply or take a delta here. And you would say, delta p is equal to h bar delta k. So multiplying this equation by an h bar, you would find that delta p, delta x is roughly h bar. And that's quantum mechanical. Now we will make the definitions of delta p and delta x precise and rigorous with precise definitions. Then there is a precise result, which is very neat, which is that delta x times delta p is always greater than or equal than h bar over 2. So this is really exact. But for that, we need to define precisely what we mean by uncertainties, which we will do soon, but not today. So I think it's probably a good idea to do an example, a simple example, to illustrate these relations. And here is one example. You have a phi of k of the form of a step that goes from delta k over 2 to minus delta k over 2, and height 1 over square root of delta k. That's phi of k. It's 0 otherwise-- 0 here, 0 there. Here is 0. Here is a function of k. What do you think? Is this psi of x, the psi x corresponding to this phi of k-- is it going to be a real function or not? Anybody? AUDIENCE: This equation [? is ?] [? true, ?] [? but-- ?] PROFESSOR: Is it true or not? AUDIENCE: I think it is. PROFESSOR: OK. Yes, you're right. It is true. This phi of k is real. And whenever you have a value at some k, there is the same value at minus k. And therefore the star doesn't matter, because it's real. So phi is completely real. So phi of k is equal to phi of minus k. And that should give you a real psi of x-- correct. So some psi of x-- have to do the integral-- psi of x0 is 1 over square root of 2 pi minus delta k over 2 to delta k over 2. The function, which is 1 over delta k in here-- that's the whole function. And the integral was supposed to be from minus infinity to infinity. But since the function only extends from minus delta k over 2 to plus delta k over 2, you restrict the integral to those values. So we've already got the phi of k and then e to the ikx dx. Well, the constants go out-- 2 pi delta k. And we have the integral is an integral over x-- no, I'm sorry. It's an integral over k. What I'm writing here-- dk, of course. And that gives you e to the ikx over ix, evaluated between delta k over 2 and minus delta k over 2. OK, a little simplification gives the final answer. It's delta k over 2pi sine of delta kx over 2 over delta kx over 2. So it's a sine of x over x type function. It's a familiar looking curve. It goes like this. It has some value-- it goes down, up, down, up like that-- symmetric. And here is psi of x and 0. Here is 2 pi over delta k, and minus 2 pi over delta k here. Sine of x over x looks like that. So this function already was defined with the delta k. And what is the delta x here? Well, the delta x is roughly 2 pi over delta k. No, it's-- you could say it's this much or half of that. I took [? it half ?] of that. It doesn't matter. It's approximate that at any rate now. So delta x is this. And therefore the product delta x, delta k, delta x is about 2 pi.

MIT_804_Quantum_Physics_I_Spring_2016

Phase_shift_for_a_potential_well.txt

PROFESSOR: So let's do an example where we can calculate from the beginning to the end everything. Now, you have to get accustomed to the idea of even though you can calculate everything, your formulas that you get sometimes are a little big. And you look at them and they may not tell you too much unless you plot them with a computer. So we push the calculations to some degree, and then at some point, we decide to plot things using a computer and get some insight on what's happening. So here's the example. We have a potential up to distance, a, to 0. The wall is always there, and this number is minus v naught. So it's a well, a potential well. And we are producing energy. Eigenstates are coming here. And the question now is to really calculate the solution so that we can really calculate the phase shift. We know how the solutions should read, but unless you do a real calculation, you cannot get the phase shift. So that's what we want to do. So for that, we have to solve the Schrodinger equation. Psi of x is equal to what? Well, there's a discontinuity. So we probably have to write an answer in which we'll have a solution in one piece and a solution in the other piece. But then we say, oh, we wrote the solution in the outside piece already. It is known. It's always the same. It's universal. I don't have to think. I just write this. E to the i delta. I don't know what delta is, but that's the answer, E to the i delta sine kx plus delta should be the solution for x greater than a. You know if you were not using that answer, it has all the relevant information for the problem, time delays, everything, you would simply write some superposition of E to the i kx and E to the minus i kx with two coefficients. On the other hand, here, we will have, again, a wave. Now, it could be maybe an E to the i kx or E to the minus i kx. Neither one is very good because the wave function must vanish at x equals 0. And in fact, the k that represents the kinetic energy here, k is always related to E by the standard quantity, k squared equal to mE over h squared or E equal the famous formula. On the other hand, there is a different k here because you have different kinetic energy. There must be a k prime here, which is 2m E plus v naught. That's a total kinetic energy over h squared. And yes, the solutions could be E to the ik prime x equal to minus ik prime x minus ik prime x, but since they must vanish at 0, should be a sine function. So the only thing we can have here is a sine of k prime x for x less than a and a coefficient. We didn't put the additional normalization here. We don't want to put that, but then we must put the number here, so I'll put it here. That's the answer, and that's k and k prime. Now we have boundary conditions that x equals a. So psi continues at x equals a. What does it give you? It gives you a sine of ka is equal to E to the i delta sine of ka plus delta. And psi prime continues at x equals a will give me ka cosine ka equal-- I have primes missing; I'm sorry, primes-- equals k E to the i delta cosine ka plus delta. What do we care for? Basically we care for delta. That's what we want to find out because delta tells us all about the physics of the scattering. It tells us about the scattering amplitude, sine squared delta. It tells us about the time delay, and let's calculate it. Well, one way to calculate it is to take a ratio of these two equations so that you get rid of the a constant. So from that side of the equation, you get k cotangent of ka plus delta is equal to k prime cotangent of k prime a. Or cotangent of ka plus delta is k prime over k. We'll erase this. And now you can do two things. You can display some trigonometric wizardry, or you say, OK, delta is arc cotangent of this minus ka. That is OK, but it's not ideal. It's better to do a little bit of trigonometric identities. And the identity that is relevant is the identity for cotangent of a plus b is cot a cot b minus 1 over cot a plus cot b. So from here, you have that this expression is cot ka cot delta minus 1 over cot ka plus cot delta. And now, equating left-hand side to this right-hand side, you can solve for cotangent of delta. So cotangent of delta can be solved for-- and here is the answer. Cot delta is equal to tan ka plus k prime over k cot k prime a over 1 minus k prime over k cot k prime a tan ka. Now, who would box such a complicated equation? Well, it can't be simplified any more. Sorry. That's the best we can do.

MIT_804_Quantum_Physics_I_Spring_2016

The_nature_of_superposition_MachZehnder_interferometer.txt

PROFESSOR: Superposition is very unusual and very interesting. Now we've said about superposition that in classical physics, when we talk about superposition we have electric fields, and you add the electric fields, and the total electric field is the sum of electric fields, and it's an electric field. And there's nothing strange about it. The nature of superposition in quantum mechanics is very strange. So nature of superposition-- I will illustrate it in a couple of different ways. One way is with a device that we will get accustomed to. It it's called the Mach-Zehnder interferometer, which is a device with a beam splitter in here. You send in a beam of light-- input- beam splitter and then the light-- indeed half of it gets reflected, half of it gets transmitted. Then you put the mirror here-- mirror 1, you put the mirror 2 here, and this gets recombined into another beam splitter. And then if there would be just a light going in, here there would be two things going out. There's another one coming from the bottom. There will be two. There will be interference. So you put a detector D0 here and a detector E1 here to detect the light. So that's the sketch of the Mach-Zehnder interferometer-- beam splitters and mirrors. Take a beam, spit the light, go down, up, and then recombine it and go into detectors. This was invented by these two people, independently, in the 1890s-- '91 to '92 apparently. And people did this with light-- beams of light before they realized they're photons. And what happens with a beam of light-- it's interesting-- comes a beam of light. The beam splitter sends half of the light one way half of the light the other way. You already know with quantum mechanics that's going to be probabilistic some photons will go up maybe some photons will go down or something more strange can happen. If you have a superposition, some photons may go both up and down. So that's what can happen in quantum mechanics. If you send the beam, classical physics, it divides half and half and then combines. And there's an interference effect here. And we will design this interferometer in such a way that sometimes we can produce an interference that everything goes to D0 or everything goes to D1 or we can produce suitable interferences that we can get fractions of the power going into D1 and D2-- D0 and D1. So we can do it in different ways, but we should think of this as a single photon. Single photos going one at a time. You see, whatever light you put in here, experimentally, the same frequency goes out here. So what is interference? You might think, intuitively, that interference is one photon interfering with another one, but it can't be. If two photos would interfere in a canceling, destructive interference, you will have a bunch of energy. It goes into nothing. It's impossible. If they would interfere constructively, you would add the electric fields and the amplitude would be four times as big because it's proportional to the square. But two photos are not going to go to four photons. It cannot conserve energy. So first of all, when you get light interference, each photon is interfering with itself. It sounds crazy, but it's the only possibility. They cannot interfere with each other. You can send the photons one at a time and, therefore, each photon will have to be in both beams at the same time. And then, each photon as it goes along, there will be an interference effect, and the photon may end up here or end up there in a probabilistic way. So you have an example of superposition. Superposition. A single photon state a single photon is equal to superposition of a photon in the upper beam and a photon in the lower beam. It's like two different states-- a little different from here, you had photons in two different polarizations states superposed. Here you have photons in two different beams-- a single photon is in both beams at the same time. And unless you have that, you cannot get a superposition and an interference that is consistent with experiment. So what does that mean for superpositions? Well, it means something that we can discuss, and I can say things that, at this moment, may not make too much sense, but it would be a good idea that you think about them a little bit. We associated states with vectors. States and vectors are the same thing. And it so happens that when you have vectors, you can write them as the sum of other vectors. So the sum of these two vectors may be this vector. But you can also write it as the sum of these two vectors-- these two vectors add to the state. And you can write any vector as a sum of different vectors, and that's, actually, quite relevant. You will be doing that during the semester-- writing a state a superposition of different things. And in that way you will understand the physics of those states. So for example, we can think of two states-- A and B. And you see, as I said, states wave functions, vectors-- we're all calling them the same thing. If you have a superposition of the states A and B, what can happen? All right, we'll do it the following way. Let's assume if you measure some property on A, you always get value A. So you measure something-- position, momentum, angular momentum, spin, energy, something-- on A, it states that you always get A. Suppose you measured the same property on B. You always get B as the value. And now suppose you have a quantum mechanical state, and the state is alpha A plus beta B-- it's a superposition. This is your state. You superimpose A and B. And now you measure that property. That same property you could measure here, you measure it in your state. The question is, what will you get? You've now superimpose those states. On the first state, you always get A; on the second state, you always get B. What do you get on the superimposed states, where alpha and beta are numbers-- complex numbers in general? Well the most, perhaps, immediate guess is that you would get something in-between maybe alpha A plus beta B or an average or something. But no, that's not what happens in quantum mechanics. In quantum mechanics, you always get A or you always get B. So you can do the experiment many times, and you will get A many times, and you may get B many times. But you never get something intermediate. So this is very different than in classical physics. If a wave has some amplitudes and you add another wave of different amplitudes, you measure the energy you get something intermediate. Here not! You make the superposition and as you measure you will either get the little a or the little b but with different probabilities. So roughly speaking, the probability to get little a is proportional to the number in front of here is alpha squared, and the probability to measure little b is proportional to beta squared. So in a quantum superposition, a single measurement doesn't yield an average result or an intermediate result. It leads one or the other. And this should connect with this. Think of the photon we were talking about before. If you think of the photon that was at an angle alpha in this way, you could say that the polarizer is measuring the polarization of the object. And therefore, what is the possible result it may measure the polarizations say oh, if it's in the x direction you get it right, and what is the probability that you get it to be in the x direction is proportional to cosine squared alpha-- the coefficient here squared. So the probability that you find the photon after measuring in the x direction is closer in squared alpha, and the probability that you'll find that here is sine squared alpha. And after you measure, you get this state which is to say the following thing. The probability to get the value A is alpha squared, but if you get A, the state becomes A because this whole state of the system becomes that. Because successive measurements will keep giving you the value A. If you get B, the state becomes B. So this is what is called the postulate of measurement and the nature of superposition. This is perhaps the most sophisticated idea we've discussed today, in which in a quantum superposition the results are not intermediate. So when you want to figure out what state you have, you have to prepare many copies of your state in this quantum system and do the experiment many times. Because sometimes you'll get A, sometimes you'll get B. After you've measured many times, you can assess the probabilities and reconstruct the state.

MIT_804_Quantum_Physics_I_Spring_2016

Commuting_observables_for_angular_momentum.txt

BARTON ZWIEBACH: We want to understand now our observables. So we said these are observables, so can we observe them? Can we have a state in which we say, what is the value of Lx, the value of Ly, and the value of Lz. Well, a little caution is necessary because we have states and we have position and momentum operator and they didn't commute and we ended up that we could not tell simultaneously the position and the momentum of a state. So for this angular momentum operators, they don't commute, so a similar situation may be happening. So I want to explain, for example, or ask, can we have simultaneous eigenstates of Lx, Ly, and Lz? And the answer is no. And let's see why that happens. So let's assume we can have simultaneous eigenstates and let's assume, for example, that Lx on that eigenstate phi nought is some number lambda x phi nought, and Ly and phi nought is equal to lambda y phi nought. Well, the difficulty with this is essentially-- well, we could even say that Lz on phi nought is equal to lambda z phi nought. So what is the complication? The complication are those commutators. If you do Lx, Ly and phi nought, you're supposed to get i h-bar Lz and phi nought. And therefore, you're supposed to get i h-bar lambda z times phi nought, because it's supposed to be an eigenstate. But how about the left hand side? The left hand side is LxLy and phi nought minus LyLx and phi nought. When Ly acts, it produces a lambda y, but then phi nought, and then when Lx acts, it produces a lambda x, so this produces lambda x lambda y phi nought minus lambda y lambda x phi nought, which is the same thing, so the left hand side is 0. 0 is equal to lambda z phi nought, so you get a-- lambda z must be 0. If you have a non-trivial state, lambda z should be 0. By the other commutators-- this can be attained or applied to phi nought-- would be 0 again, because each term produces a number and the order doesn't matter. But then it would show that lambda x is 0, and this will show that lambda y is 0. So at the end of the day, if these three things hold, then all of them are 0. Lambda x equals lambda y equals lambda z equals 0. So you can have something that is killed by all of the operators, but you cannot have a non-trivial state with non-trivial eigenvalues of these things. So we cannot have-- we cannot tell what is Lx on this state and Ly on this state simultaneously. Any of those two is too much. So if we can't tell that, what can we tell? So what is the most we can tell about this state Is our question now. We can tell maybe what is its value of Lx, but then Ly and Lz are undetermined. Or we can tell what is Lz and then Lx and Ly are undetermined, incalculable, impossible in principle to calculate them. So let's see what we can do, and the answer comes from a rather surprising thing, the fact that if you think about what could commute with Lx, Ly, Lz, it should be a rotationally invariant thing, because Lx, Ly, and Lz do rotations. So the only thing that could possibly commute with this thing is something that is rotationally invariant. The thing that could work out is some thing that is invariant and there are rotations. Now we said, for example, the magnitude of the vector R is invariant under rotation. You rotate the vector, the demanded is invariant. So we can try the operator L squared, which is proportionate to the magnitude squared, so we define it to be LxLx plus LyLy plus LzLz. And we tried, we tried to see if maybe Lx commutes with L squared. Remember, we had a role for L squared in this differential operator that had the Laplacian, the angular part of the Laplacian was our role for L squared, so L squared is starting to come back. So let's see here-- this is Lx-- now, I'll write the whole thing-- LxLx plus LyLy plus LzLz. Now, Lx and Lx commute, so I don't have to bother with this thing, that's 0. But the other ones don't commute. So let's do the distributive law. So this would be an Lx, Ly Ly plus Ly Lx, Ly-- this is from the first-- plus Lx, Lz Lz plus Lz Lx, Lz. You know, if you don't put these operators in the right order, you don't get the right answer. So I think I did. Yes. It's correct. Now you use the commutators and hope for the best. So Lx, Ly is i h-bar LzLy. Lx, Ly is plus i h-bar LyLz. So far, no signs of canceling, these two things are very different from each other. They don't even appear with a minus sign, so this is not a commutator, but anyway, what is this? Lx with Lz. Well, you should always think cyclically. Lz with Lx is i h-bar, so this would be minus i h-bar LyLz, and this is again Lz with Lx would have been i h-bar Ly, so this is minus i h-bar LzLy, and it better cancel-- yes. This term cancels with the first and this term cancels with this and you get 0. That's an incredible relief, because now you have a second operator that is measurable simultaneously. You can get eigenstates that are eigenstates of one of the L's-- for example, Lx and L squared, because they commute, and you won't have the problems you have there. In fact, it's a general theorem of linear algebra that-- we'll see a little bit of that in this course and you'll see it more completely in 805-- that if you have two Hermitian operators that commute, you can find a simultaneous eigenstates of both operators. I mean, eigenstates that are eigenstates of 1, and eigenstates of the second. Simultaneous eigenstates are possible. So we can find simultaneous eigenstates of these operators, and in fact, you could find simultaneous eigenstates of Lx and L squared, but given the simplicity of all this, it also means that Ly commutes with L squared, and that Lz also commutes with L squared. So you have a choice-- you can choose Lx, Ly, or Lz and L squared and try to form simultaneous eigenstates from all these operators. Two of them. Let's study those operators as differential operators a little bit. So x, y, and z are your spherical coordinates and they are r sin theta cos phi, r sin theta sin phi, and r cos theta. We're trying to calculate the differential operators associated with angular momentum using spherical coordinates. So r is x squared plus y squared plus z squared. Theta is cosine minus 1 of z over r and 5 is tan minus 1 y over z. And there's something very nice about one angular momentum operator in spherical coordinates, there is only one angular momentum that is very simple-- its rotations about z. Rotations about z don't change the angle theta of spherical coordinates, just change the angle phi. r doesn't change. The other rotation, the rotation about x messes up phi and theta and all the others are complicated, so maybe we can have some luck and understand what is d/d-phi, the d/d-phi operator. Well, the d/d-phi operator is d/dy dy/d-phi plus d/dx dx/d-phi-- the rules of chain rule for partial derivatives-- plus d/dz dz/d-phi. But z doesn't depend on phi. On the other hand, dy/d-phi is what? dy/d-phi, this becomes a cos phi-- it's x. X d/dy. And dx/d-phi is minus y. And you say, wow, x d/dy is like x py minus y px, that's a z component of angular momentum! So indeed, Lz, which is h-bar over i x d/dy minus y d/dx, h-bar over i is because of the p's-- x py minus y px. And this thing is d/d-phi, so Lz, we discover, is just h-bar over i d/d-phi. A very nice equation that tells you that the angular momentum in the z direction is associated with its operator. So I have left us exercises to calculate the other operators that are more messy, and to calculate Lx Ly as well in terms of d/d-theta's and d/d-phi's. And as you remember, angular momentum has units of h-bar and angles have no units, so the units are good and we should find that. So that calculation is left as an exercise, but now you probably could believe that L squared, which is LxLx plus LyLy plus LzLz is really minus h squared 1 over sin theta d/d-theta. No, it's-- not 1 over sin theta-- uh, yep. 1 sin theta d/d-theta-- sin theta d/d-theta plus 1 over sin squared theta d second d phi squared. So the claim that they had relating the angular momentum operator to the Laplacian is true. But, you know, you now see the beginning of how you calculate these things, but it will be a simple and nice exercise for you to do it.

MIT_804_Quantum_Physics_I_Spring_2016

Translation_operator_Central_potentials.txt

PROFESSOR: The fact is that angular momentum is an observable, and as such it deserves attention. There is an active way of thinking of observables, and we have not developed it that much in this course. But for example, with a momentum operator you've learned that the momentum operator can give you the differential operator. It's a derivative, and derivatives tell you how to move, how a function varies. So with the momentum operator, for example, you have the momentum operator p hat, which is h bar over i d dx. And you could ask the question of, OK, so the momentum operator moves or takes a derivative, does the momentum operator move a function? Does it generate a translation? And the answer is, yes. That's another way of thinking of the momentum operator as a generator of translations. But how does it do it? This is a Hermitian operator, and it takes a derivative. It doesn't translate the function. But there is a universal trick that if you exponentiate i times a Hermitian operator, you get a new kind of operator that actually, in this case, moves things. So we could think of exponentiating e to the i p hat, and for purposes of units I have to put a constant with units of length, and an h bar here. And now you have the exponential of an operator. That's good. That's a very interesting operator, and we can ask what does it do when you act on a wave function? It's an operator. And look, simplify by putting what p is going to do. P is h over i d dx. So this is like a d dx exponentiated acting on psi of x. And as an exponential, it can be expanded in a Taylor series with this funny object there, but it would be the sum from n equals 0 to infinity 1 over n factorial a d dx. I will write it this as normal derivatives, because we just have a function of x, a d dx to the n psi of x. And you see that, of course, this is psi of x plus a d psi dx plus 1 over 2 a squared, d second psi dx squared. But this is nothing else but the Taylor series for this. And there it is, the miracle. The e to the i momentum generated translation. It really moves the wave function. So that in a sense is a deeper way of characterizing the momentum operator as a generator of translations. With the angular momentum operators, we will have that they generate rotations. So I need a little bit more mathematics here, because I have to deal with three dimensions, a vector, and produce an exponential that rotates the vector, so that it gives you the wave function at a rotated point. But this will be the same story. Angular momentum will generate rotations the same way as momentum generates translations. And there is yet another story that when you will appreciate the abstract properties of angular momentum that some of them will appear today, you will realize that in addition of angular momentum that represent rotations of objects doing things, there is another way of having angular momentum. And that's spin angular momentum. That mysterious property of particles that have-- even though they have 0 size, they behave as if they were little balls rotating and spinning. That spin angular momentum has no ordinary wave functions associated to it, and it's fractional sometimes. And the study of angular momentum inspired by orbital angular momentum associated with normal rotations, will lead us to understand where spin angular momentum comes about. So it's a gigantic interesting subject, and we're beginning with it today. So it's really quantum mechanics in three dimensions, central potentials, and angular momentum. And let's begin by mentioning that if we are in three dimensions-- and many things with it so far in this course, we always took the time to write them in three dimensions. So we wrote this, for example, as a generalization of the derivative form of the momentum operator. Meaning there is a Px, which is h bar over i d dx, Py h bar over i d dy, and Pz equal h bar over i d dz. And we had commutators within Px and x, Py and y, and Pz and z. There were always the same commutators of the form x Px equal i h bar. Similar things here. With this we wrote the three dimensional Schrodinger equation, which was minus h squared over 2m, and instead of p squared three dimensional, he would have a derivative if you were doing in one dimension. For three dimensions you have the Laplacian. And this time you have a wave function that depends on the vector x plus v of r-- v of x. Should I write r? Let me write r vector. V of r psi of r equal e psi of r. This is our time independent Schrodinger equation. This corresponds to the energy eigenstate, but in three dimensions. So this is the equation we wish to understand, and our ability to understand that equation in a simple and nice way rests on a simplification. That is not always true, but it's true under so many circumstances that it's worth studying by itself. And it's the case when you have a central potential, and by that we mean that the potential is not quite the vector function of r, but is just a function of the magnitude of r. That's a little bit funny way of writing it, because I'm using the same letter v, but I hope there's no confusion. I mean that the potential just depends on the value of r. So what this means physically is that over concentric spheres, the potential is constant. All over the surface of spheres of constant radius, the potential is constant, because it only depends on the radius. And this potential is there for a spherically symmetric. You can rotate the world, and the potential still looks the same, because rotations don't change the magnitude of vectors. If you have a vector of some length, you rotate it, it's the same length, and therefore you remain on this sphere. So the central potential are spherically symmetric. By that we mean they're invariant under rotations. So this is the reason why angular momentum will play an important role, because precisely the angular momentum operators, in the fashion we discussed a minute ago, generate rotation. So they will have a nice interplay, to be developed in the following lectures, with the Hamiltonian. So at this moment we have a central potential, and let's assume that's the case. And we need to understand a little more of this differential equation. So let me write the formula for the Laplacian of a function. It has a radial contribution. You know it's second order derivatives. And it has a radial part, and an angular part. The units are 1 over length squared. So you need, if you have an angular part, all over here is going to be angular, you still need the 1 over on r squared here for the units to work out. So here it is, it's slightly complicated. d d theta sine theta d d theta of-- well, put the psi, plus 1 over sine squared theta, d second d phi squared all acting on psi. It's a complicated operator, and here is some radial derivatives, and here there are some angular derivatives. So you see, today's lecture will have many steps, and you have to keep track of where we're going. And what we're going to do is, build up a structure that allows us pretty much to forget about all this thing. That's our goal. And angular momentum will play a role in doing this. So there are in fact two things I want to justify, two facts to be justified. So I will erase this. The first fact is the relation between this differential operator and angular momentum. So two facts to justify. The first is that minus h squared 1 over sine theta d d theta sine theta d d theta plus 1 over sine squared theta d second d phi squared. This whole thing can be viewed as the differential operator version of angular momentum. Remember, d dx was a differential operator version of momentum. So maybe this has to do with angular momentum, and indeed this whole thing, remember, units of angular momentum is h bar. Angular momentum is length times momentum. And from the certainty principle, you know that x times p has units of h bar. So angular momentum has units of h bar. So there's h bar squared here. So this must be angular momentum squared. In fact, if you think about that, angular momentum is x times p. So x times a derivative. So it's a first order differential operator, but this is a second order one. So this could not be just angular momentum. Anyway, angular momentum is a vector. So this will turn out to be, and we will want to justify L squared. The quantum version of the angular momentum operator squared. And the other thing I want to justify if I write-- call this equation one. So this is fact one, and fact two, is that equation one is relevant, when-- let me wait a second to complete this. This equation is an equation for a particle moving in a potential, a spherically symmetric potential. It turns out that is relevant under more general circumstances. If you have two particles whose potential energy-- if you have two particles you have a potential energy between them, maybe it's a electromagnetic-- if the potential energy just depends on the distance that separates them, this two body problem can be reduced to a one body problem of this form. This is a fairly non-trivial fact, and an absolutely interesting one. Because if you want to really solve the hydrogen atom, you have an electron and a proton. Now it turns out that the proton is almost 2,000 times heavier than the electron. And therefore, you could almost think that the proton creates a potential in which the electron moves. But similar analysis is valid for neuron orbiting a nucleus. And in that case, the neuron is still lighter than the proton, but not that much lighter. Or maybe for a quark and an anti-quark orbiting each other. Or an electron and a positron orbiting each other, and this would be valid and useful. So we need to somehow explain that as well. If you really want to understand what's going on, is that equation one is relevant when we have a two body problem with a potential function v of x1 x2. The potential energy given that configuration, x1 and x2 of the first and second particle, is a function of the separation only. The absolute value or the length of the vector, it's 1 minus x2. This far we'll get through today. This will be next lecture still.

MIT_804_Quantum_Physics_I_Spring_2016

Uncertainty_and_eigenstates.txt

PROFESSOR: This definition in which the uncertainty of the permission operator Q in the state psi. It's always important to have a state associated with measuring the uncertainty. Because the uncertainty will be different in different states. So the state should always be there. Sometimes we write it, sometimes we get a little tired of writing it and we don't write it. But it's always implicit. So here it is. From the analogous discussion of random variables, we were led to this definition, in which we would have the expectation value of the square of the operator minus the square of the expectation value. This was always-- well, this is always a positive quantity. Because, as claim 1 goes, it can be rewritten as the expectation value of the square of the difference between the operator and its expectation value. This may seem a little strange. You're subtracting from an operator a number, but we know that numbers can be thought as operators as well. Operator of minus a number acting on a state is well defined. The operator acts on the state, the number multiplies a state. So this is well defined. And claim 1 is proven by direct computation. You certainly indeed prove. You can expand what is inside the expectation value, so it's Q hat squared. And then the double product of this Q hat and this number. Now, the number and Q hat commute, so it is really the double product. If you have A plus B times A plus B, you have AB plus BA, but if they commute it's 2AB, so this is minus 2 Q hat Q. Like that. And then, the last term is the number squared, so it's plus Q squared. And sometimes I don't put the hats as well. And all this is the expectation value of the sum of all these things. The expectation value of a sum of things is the expectation value of the first plus the expectation value of the second, plus the expectation value of the next. So we can go ahead and do this, and this is therefore expectation value of Q squared minus the expectation value of this whole thing. But now the expectation value of a number times an operator, the number can go out. And this is a number, and this is a number. So it's minus 2 expectation value of Q, number went out. And then you're left with expectation value of another Q. And the expectation value of a number is just the number, because then you're left within the world of psi star psi, which is equal to 1. So here is plus Q hat squared. And these two terms, the second and the third, are the same really. They are both equal to expectation value of Q squared. They cancel a little bit, and they give you this. So indeed, this is equal to expectation value of Q squared minus expectation value of Q squared. So claim 1 is true. And claim 1 shows in particular that this number, delta Q squared, in the expectation value of a square of something, is positive. We'll see more clearly in a second when we have claim number 2. And claim number 2 is easily proven. That's another expression for uncertainty. For claim number 2, we will start with the expectation value of Q minus Q squared, like this, which is the integral dx psi star of x and t, Q minus expectation value of Q, Q minus expectation value of Q, on psi. The expectation value of this thing squared is psi star, the operator, and this. And now, think of this as an operator acting on all of that. This is a Hermitian operator. Because Q hat is Hermitian, and expectation value of Q is real. So actually this real number multiplying something can be moved from the wave function to the starred wave function without any cost. So even though you might not think of a real number as a Hermitian operator, it is. And therefore this whole thing is Hermitian. So it can be written as dx. And now you have this whole operator, Q minus Q hat, acting on psi of x and t. And conjugate. Remember, the operator, the Hermitian operator, moves to act on psi, and the whole thing [INAUDIBLE]. And then we have here the other term left over. But now, you see that you have whatever that state is and the state complex conjugated. So that is equal to this integral. This is the integral dx of the norm squared of Q hat minus Q hat psi of x and t squared, which means that thing, that's its complex conjugate. So this completes our verification that these claims are true, and allow us to do the last step on this analysis, which is to show that if you have an eigenstate of Q, if a state psi is an eigenstate of Q, there is no uncertainty. This goes along with our measurement postulate that says an eigenstate of Q, you measure Q and you get the eigenvalue of Q and there's no uncertainty. In particular, we'll do it here I think. If psi is an eigenstate of Q, so you'll have Q psi equal lambda psi, where lambda is the eigenvalue. Now, this is a nice thing. It's stating that the state psi is an eigenstate of Q and this is the eigenvalue, but there is a little bit more than can be said. And it is. It should not surprise you that the eigenvalue happens to be the expectation value of Q on the state psi. Why? Because you can take this equation and integrate dx times psi star. If you bring that in into both sides of the equation then you have Q psi equals integral dx psi star psi, and the lambda goes up. Since my assumption whenever you do expectation values, your states are normalized, this is just lambda. And by definition, this is the expectation value of Q. So lambda happens to be equal to the expectation value of Q, so sometimes we can say that this equation really implies that Q hat psi is equal to expectation value of Q psi times psi. It looks a little strange in this form. Very few people write it in this form, but it's important to recognize that the eigenvalue is nothing else but the expectation value of the operator of that state. But if you recognize that, you realize that the state satisfies precisely Q hat minus Q on psi is equal to 0. Therefore, if Q hat minus Q on psi is equal to 0, delta Q is equal to 0. By claim 2. Q hat minus Q expectation value kills the state, and therefore this is 0. OK then. The other way is also true. If delta Q is equal to 0, by claim 2, this integral is 0. And since it's the sum of squares that are always positive, this state must be 0 by claim 2. And you get that Q minus Q hat psi is equal to 0. And this means that psi is an eigenstate of Q. So the other way around it also works. So the final conclusion is delta Q is equal to 0 is completely equivalent of-- I'll put in the psi. Psi is an eigenstate of Q. So this is the main conclusion. Also, we learned some computational tricks. Remember you have to compute an expectation value of a number, uncertainty, you have these various formulas you can use. You could use the first definition. Sometimes it may be the simplest. In particular, if the expectation value of Q is simple, it's the easiest way. So for example, you can have a Gaussian wave function, and people ask you, what is delta of x of the Gaussian wave function? Well, on this Gaussian wave function, you could say that delta x squared is the expectation value of x squared minus the expectation value of x squared. What is the expectation value of x? Well, it would seem reasonable that the expectation value of x is 0. It's a Gaussian centered at the origin. And it's true. For a Gaussian it would be 0, the expectation value of x. So this term is 0. You can also see 0 because of the integral. You're integrating x against psi squared. Psi squared is even, x is odd with respect to x going to minus x. So that integral is going to be 0. So in this case, the uncertainty is just the calculation of the expectation value of x squared, and that's easily done. It's a Gaussian integral. The other good thing about this is that even though we have not proven the uncertainty principle in all generality. We've only [? multivated ?] it. It's precise with this definition. So when you have the delta x, delta p is greater than or equal to h bar over 2, these things are computed with those definitions. And then it's precise. It's a mathematically rigorous result. It's not just hand waving. The hand waving is good. But the precise result is more powerful.

MIT_804_Quantum_Physics_I_Spring_2016

The_wave_for_a_free_particle.txt

PROFESSOR: So what are we trying to do? We're going to try to write a matter wave. We have a particle with energy e and momentum p. e is equal to h bar omega. So you can get the omega of the wave. And p is equal to h bar k. You can get the k of the wave. So de Broglie has told you that's the way to do it. That's the p and the k. But what is the wave? Really need the phase to-- how does the wave look like? So the thing is that I'm going to do an argument based on superposition and very basic ideas of probability to get-- to find the shape of the wave. And look at this possibility. Suppose we have plane waves-- plane waves in the x plus direction. A particle that is moving in the plus x direction. No need to be more general yet. So what could the wave be? Well, the wave could be sine of kx minus omega t. Maybe that's the de Broglie wave. Or maybe the de Broglie wave is cosine of kx minus omega t. But maybe it's neither one of them. Maybe it is an e to ikx minus i omega t. These things move to the right. The minus sign is there. So with an always an e to the minus i omega t. Or maybe it's the other way around. It's e to the minus ikx plus i omega t. So always an e to the i omega t. And then you have to change the sine of the first term in order to get a wave that is moving that way. And now you say, how am I ever going to know which one is it? Maybe it's all of them, a couple of them, none of them. That's we're going to try to understand. So the argument is going to be based on superposition and just the rough idea that somehow this has to do with the existence of particles having a wave. And it's very strange. In some sense, it's very surprising. To me, it was very surprising, this argument, when I first saw it. Because it almost seems that there's no way you're going to be able to decide. These are all waves, so what difference can it make? But you can decide. So my first argument is going to be, it's all going to be based on superposition. Use superposition-- --position. Plus a vague notion of probability-- --bility. So I'm going to try to produce with these waves a state of a particle that has equal probability to be moving to the right or to the left. I'm going to try to build a wave that has equal probability of doing this thing. So in case 1, I would have to put a sine of kx minus omega t. That's your wave that is moving to the right. I have to change one sine here. Plus sine of kx. Say, plus omega t. And that would be a wave that moves to the right. Just clearly, this is the wave that moves to the left. And roughly speaking, by having equal coefficients here, I get the sense that this would be the only way I could produce a wave that has equal probability to move to the left and a particle that moves to the right. On the other hand, if I expand this you get twice sine of kx cosine omega t. The fact is that this is not acceptable. Why it's not acceptable? Because this wave function vanishes for all x at t omega t equal to pi over 2, 3pi over 2, 5pi over 2. At all those times, the wave is identically 0. The particle has disappeared. No probability of a particle. That's pretty bad. That can't be right. And suddenly, you've proven something very surprising. This sort of wave just can't be a matter particle. Again, in the way we're trying to think of probabilities. Same argument for 2 for same reason. 2-- So this is no good. No good. The wave function cannot vanish everywhere at any time. If it vanished everywhere, you have no particle. You have nothing. With 2, you can do the same thing. You have a cosine plus another cosine. Cosine omega t minus-- kx minus omega t plus cosine of kx plus omega t. That would be 2 cosine kx cosine omega t. It has the same problems. Let's do case number 3. Case number 3 is based on the philosophy that the wave that we have-- e to the ikx minus i omega t always has an e to the minus i omega t as a phase. So to get a wave that moves in the opposite direction, we have to do minus ikx minus i omega t. Because I cannot change that phase. Always this [INAUDIBLE]. Now, in this case, we can factor the time dependence. You have e to the ikx e to the minus ikx e to the minus i omega t. And be left with 2 cosine kx e to the minus i omega t. But that's not bad. This way function never vanishes all over space. Because this is now a phase, and this phase is always non-zero. The e to the minus i omega t is never 0. The exponential of something is never 0, unless that something is real and negative. And a phase is never 0. So this function never vanishes for all x-- vanishes for all x. So it can vanish at some point for all time. But those would be points where you don't find the particle. The function is nonzero everywhere else. So this is good. Suddenly, this wave, for some reason, is much better behaved than these things for superposition. Let's do the other wave, the wave number 4. And wave number 4 is also not problematic. So case 4, you would do an e to the minus ikx e to the i omega t plus an e to the ikx e to the i omega t. Always the same exponential. This is simply 2 cosine of kx e to the i omega t. And it's also good. At least didn't get in trouble. We cannot prove it is good at this point. We can only prove that you are not getting in trouble. We are not capable of producing a contradiction, so far. So actually, 3 and 4 are good. And the obvious question that would come now is whether you can use both of them or either one at the same time. So the next claim is that both cannot be true at the same time. You cannot use both of them at the same time. So suppose 3 and 4 are good. Both 3 and 4-- and 4 are both good-- both right, even. Then remember that superimposing a state to itself doesn't change the state. So you can superimpose 3 and 4-- e to the ikx minus i omega t. That's 3. You can add to it 4, which is e to the minus ikx-- minus omega t. I factor a sine. And that's 4. And that should still represent this same particle moving to the right. But this thing is twice cosine of kx minus omega t. So it would mean that this represents a particle moving to the right. And we already know that if this represents a particle moving to the right, you get in trouble. So now, we have to make a decision. We have to choose one of them. And it's a matter of convention to choose one of them, but happily, everybody has chosen the same one. So we are led, finally, to our matter wave. We're going to make a choice. And here is the choice. Psi of x and t equal to the ikx minus i omega t. The energy part will always have a minus sign. Is the mother wave or wave function for a particle with p equal hk and e equal h bar omega according to de Broglie. You want to do 3 dimensions, no problem. You put e to the i k vector, x vector, minus i omega t. On p, in this case, is h bar k vector. So it's a plane wave in 3 dimensions. So that's the beginning of quantum mechanics. You have finally found the wave corresponding to a matter particle. And it will be a deductive process to figure out what equation it satisfies, which will lead us to the Schrodinger equation.

MIT_804_Quantum_Physics_I_Spring_2016

Delta_function_potential_I_Preliminaries.txt

PROFESSOR: Delta function potential. So it's still a one-dimensional potential-- potential is a function of x. We'll write it this way-- minus alpha delta of x, where alpha is positive. So this is a delta function in a negative direction. So if you want to draw the potential-- there's no way to draw really nicely a delta function. So you just do a thick arrow with it pointing down. It's a representation of a potential that, somehow, is rather infinite at x equals zero-- but infinite and negative. It can be thought of as the limit of a square well that is becoming deeper and deeper. And, in fact, that could be a way to analytically calculate the energy levels-- by taking carefully the limit of a potential. It is becoming thinner and thinner, but deeper and deeper, which is the way you define or regulate the delta function. You can imagine the delta function as a sequence of functions, in which it's becoming more and more narrow-- but deeper at the same time. So that the area under the curve is still the same. So, at any rate, the delta function potential is a potential that should be understood as 0 everywhere , else except at the delta function where it becomes infinite. And there are all kinds of questions we can ask. OK. Are there bound states? What are bound states in this case? They are energy eigenstates with energy less than zero. So bound states, which means e less than zero. Do they exist? Does this potential have bound states? And, if it does, how many bound states? 1, 2, 3? Does It depend on the intensity of the delta function? When you get more bound states, the deeper the potential is. Well, we'll try to figure out. In fact, there's a lot that can be figured out without calculating, too much. And it's a good habit to try to do those things before you-- not to be so impatient that you begin, and within a second start writing the differential equation trying to solve it. Get a little intuition about how any state could look like, and how could the answer for the energy eigenstates-- the energies-- what could they be? Could you just reason your way and conclude there's no bound states? Or one bound state? Or two? All these things are pretty useful. So one way, as you can imagine, is to think of units. And what are the constants in this problem? In this problem we'll have three constants. Alpha, the mass of the particle, and h-bar. So with alpha, the mass and the particle, and h-bar you can ask, how do I construct the quantity with units of energy? If there there's only one way to construct the quantity with units of energy, then the energy of a bound state will be proportional to that quantity-- because that's the only quantity that can carry the units. And here, indeed, there's only one way to construct that quantity with units of energy-- from these three. That's to be expected. With three constants that are not linearly dependent-- whatever that is supposed to mean-- you can build anything that has units of length, mass, or time. And from that you can build something that has units of energy. So you can now decide, well, what are the units of alpha? The units of alpha have give you energy, but the delta function has units of one over length. This has one over length. , Remember if you integrate over x the delta function gives you 1. So this has units of 1 over length. And, therefore, alpha has to have units of energy times length. So this is not quite enough to solve the problem, because I want to write e-- think of finding how do you get units of energy from these quantities? But l-- we still don't have a length scale either. So we have to do a little more work. So from here we say that units of energy is alpha over l. There should be a way to say that this is an equality between units. I could put units or leave it just like that. So in terms of units, it's this. But in terms of units, energy-- you should always remember-- is p squared over m. And p is h-bar over a length. So that's p squared and that's m. So that's also units of energy From these two you can get what has units of length. Length. You pass the l to this side-- the l squared to this left-hand side. Divide. So you get l is h squared over m alpha. And if I substitute back into this l here, e would be alpha over l, which is h squared, alpha squared, m. So that's the quantity that has units of energy. M alpha squared over h squared has units of energy. If this has units of energy-- the bound state energy. Now, if you have a bounce state here, it has to decay in order to be normalizable. In order to be normalizable it has to decay, so it has to be in the forbidden region throughout x. So the energy as we said is negative, energy of a bound state-- if it exists. And this bound state energy would have to be negative some number m alpha squared over h squared. And that's very useful information. The whole problem has been reduced to calculating a number. It better be and the answer cannot be any other way. There's no other way to get the units of energy. So if a bound state exists it has to be that. And that number could be pi, it could be 1/3, 1/4, it could be anything. There's a naturalness to that problem in that you don't expect that number to be a trillion. Nor do you expect that number to be 10 to the minus 6. Because there's no way-- where would those numbers appear? So this number should be a number of order one, and we're going to wait and see what it is. So that's one thing we know already about this problem. The other thing we can do is to think of the regulated delta function. So we think of this as a potential that has this form. So here is v of x, and here is x. And for this potential-- if you have a bound state-- how would the wave function look? Well, it would have to-- suppose you have a ground state-- it's an even potential. The delta function is even, too. It's in the middle. It's symmetric. There's nothing asymmetric about the delta function. So if it's an even potential the ground states should be even, because the ground state is supposed to have no nodes. And it's supposed to be even if the potential is even. So how will it look? Well, it shouldn't be decaying in this region. So, presumably, it decays here. It decays there-- symmetrically. And in the middle it curves in the other direction. It is in an allowed region-- and you remember that's kind of allowed this way. So that's probably the way it looks. Now, if that bound state exists, somehow, as I narrow this and go down-- as it becomes even more narrow, very narrow now, but very deep. This region becomes smaller. And I would pretty much expect the wave function to have a discontinuity. You basically don't have enough power to see the curving that is happening here. Especially because the curving is going down. The distance is going down. So if this bound state exists, as you approach the limit in which this becomes a delta function the energy moves a little, but stays finite at some number. And the curvature that is created by the delta function is not visible, and the thing looks just discontinuous in its derivative. So this is an intuitive way to understand that the wave function we're looking for is going to be discontinuous on its derivative. Let's write the differential equation, even though we're still not going to solve it. So what is the differential equation? Minus h squared over m, psi double prime, is equal to E psi. And, therefore-- and I write this, and you say, oh, what are you writing? I'm writing the differential equation when x is different from 0. No potential when x is different from 0. So this applies for positive x and negative x. It doesn't apply at x equals 0. We'll have to deal with that later. So then, no potential for x different from 0. And this differential equation becomes psi double prime equals minus 2m e over h squared psi. And this is equal to kappa squared psi, where kappa squared is minus 2me over h squared. And it's positive. Let's make that positive. It's positive because the energy is negative and we're looking for bound states. So we're looking for bound states only. Kappa squared is positive. And this differential equation is just this. I'll copy it again here. Kappa squared psi. And the solutions of this equation are-- solutions-- are e to the minus kappa x and e to the kappa x. Or, if you wish, cosh kappa x and sinh kappa x-- whichever you prefer. This is something we now have to use in order to produce a solution. But now, let's see if I can figure out how many bound states there are. If there is one bound state, it's going to be even. It's the ground state. It has no nodes. It has to be even, because the potential is even. If I have the first excited state after the ground state, it will have to be odd. It would have to vanish at x equals 0, because it's odd. There is it's node-- it has to have one node. For an odd bound state-- or first excited state-- you'd have to have psi equals 0 at x equals 0. And the way to do that would be to have a sinh, because this doesn't vanish at zero. This doesn't vanish at zero. And cosh doesn't vanish at zero. So you would need psi of x equals sinh of kappa x. But that's not good. sinh of kappa x is like this and blows up. Blows down. It has to go like this. It is in a forbidden region, so it has to be convex towards the axis. And convex here. But it blows up. So there's no such solution. No such solution. You cannot have an odd bound state. So since the bound states alternate-- even, odd, even, odd, even, odd-- you're stuck. You only will have a ground state-- if we're lucky-- but no excited state that is bound, while a finite square will. You remember this quantity z0 that tells you how many bound states you can have. Probably you're anticipating that in the case of the delta function potential, you can only have one bound state, if any. The first excited state would not exist. So, enough preliminaries. Let's just solve that now.

MIT_804_Quantum_Physics_I_Spring_2016

Recursion_relation_for_the_solution.txt

PROFESSOR: This was a differential equation for the energy eigenstates phi. Supposed to be normalizable functions. We looked at this equation and decided we would first clean out the constant. We did that by replacing x by a unit-free coordinate u. For that we needed a constant that carries units of length, and that constant is given by this combination of the constants of the problem. H-bar, m, and omega-- the frequency of the oscillator. We also defined the unit-free energy-- calligraphic e. In terms of which the real energies are given by multiples of h omega over 2. So the problem has now become-- and this whole differential equation turns into this simple differential equation. Simple looking-- let's say properly-- differential equation for phi-- as a function of u-- which is the new rescaled coordinate, and where the energy shows up here. And for some reason, this equation doesn't have normalizable solutions. Unless those energies are peculiar values that allow a normalizable solution to exist. We looked at this equation for u going to infinity, and realized that e to the plus minus u squared over 2 are the possible dependencies. So we said-- without loss of generality-- that phi could be written as some function of u to be determined times this exponential. And we hope for a function that may be a polynomial. So that the dependence at the infinity is governed by this factor. So with this for the function phi, we substitute back into the differential equation. Now the unknown is h. So you can take the derivatives and find this differential equation for h-- a second order differential equation. And that's the equation. It may look a little more complicated than the equation we started with, but it's much simpler, actually. There would be no polynomial solution of this equation, but there may be a polynomial solution of the second equation. So we have to solve this equation now. And the way to do it is to attempt a serious expansion. So we would try to write h of u equals the sum over j. Equals 0 to infinity. Ak, u to the k. P equals 0 to infinity. Now, one way to proceed with this is to plot this expansion into the differential equation. You will get three sums. You will have to shift indices. It's kind of a little complicated. Actually, there's a simpler way to do this in which you think in the following way. You have this series and you imagine there's a term aj, u to the j, plus aj plus 1, u to the j plus 1, plus aj plus 2. U to the j plus 2. And you say, let me look at the terms with u to the j in the differential equation. So just look at the terms that have a u to the power j. So from this second-- h vu squared-- what do we get? Well, to get a term that has a u to the j, you must start-- if you take two derivatives and to end up with u to the j-- you must have started with this. U to the j plus 2. So this gives you j plus 2, j plus 1-- taking the two derivatives-- aj plus 2, u to the j. From the series, the term with u to the j from the second hvu squared is this one. How about for minus 2u dh, du? Well, if I start with h and differentiate and then multiply by u, I'm going to get u to the j starting from u to the j. Because when I differentiate I'll get u to the j minus 1, but the u will bring it back. So this time I get minus 2 a j-- or minus 2. One derivative j. That's aj, u to the j. So it's from this [? step. ?] Minus 2. You differentiate and you get that. From the last term, e minus 1 times h, it's clearly e minus 1 times aj u to the j. So these are my three terms that we get from the differential equation. So at the end of the day, what have we gotten? We've gotten j plus 2, j plus 1, aj plus 2, minus 2jaj, plus e minus 1 aj. All multiplied by u to the j. And that's what we get for u to the j. So if you wish, for the whole differential equation-- all of the differential equation-- you get the sum from j equals 0 to infinity of these things. And that should be equal to zero. So this is the whole left-hand side of the differential equation. We calculated what is the term u to the j. And there will be terms from u to the zeroth to u to the infinity. So that's the whole thing. And we need this differential equation to be solved. So this must be zero. And whenever you have a function of u like a polynomial-- well, we don't know if it's a polynomial-- and it stops. But if you have a function of u like this, each coefficient must be 0. Therefore, we have that j plus 2, times j plus 1, times aj plus 2, is equal to 2 j plus 1 minus e, aj. I set this whole combination inside brackets to 0. So this term is equal to this term and that term on the other side. You get a plus 2j. A plus 1 and minus e. So basically this is a recursion relation. Aj plus 2 is equal to 2j plus 1 minus e, over j plus 2, j plus 1 times aj. And this is perfectly nice. This is what should have happened for this kind of differential equation-- a second-order linear differential equation. We get a recursion that jumps one step. That's very nice. And this should hold for j equals 0, 1, 2-- all numbers. So when you start solving this, there's two ways to solve it. You can decide, OK, let me assume that you know a0-- you give it. Give a0. Well, from this equation-- from a0-- you can calculate a2. And then from a2 you can calculate a4. And successively. So you get a2, a4-- all of those. And this corresponds to an even solution of the differential equation for h. Even coefficients. Even solution for h. Or-- given this recursion-- you could also give a1-- give it-- and then calculate a3, a5-- and those would be an odd solution. So you need two conditions to solve this. And those conditions are a0 and a1, which is the same as specifying the value of the function h at 0-- because the value of the function h at 0 is a0. And the value of the derivative of the function at 0, which is a1. [INAUDIBLE] h of mu is a0 plus a1u plus a2u squared. So the derivative at 0 is [? h1. ?] And that's what you must have for solving a differential equation-- a second-order differential equation for h. You need to know the value of the function at zero and the value of the function at the derivative of the function-- at zero, as well. And then you can start integrating it. So this first gives you a solution a0 plus a2u squared, plus a4u to the fourth. And the second is an a1u plus a3u cubed, plus these ones. So all looks pretty much OK.

MIT_804_Quantum_Physics_I_Spring_2016

Motion_of_a_wavepacket.txt

PROFESSOR: Let me demonstrate now with plain doing the integral that, really, the shape of this wave is moving with that velocity. So in order to do that, I basically have to do the integral. And of course, if it's a general integral, I cannot do it. So I have to figure out enough about the integral. So here it is. We have psi of x and t. It's integral dk phi of k e to the ikx minus omega of kt. OK. It's useful for us to look at this wave at time equals 0 so that we later compare it with the result of the integral. So phi psi at time equals 0 is just dk phi of k e to the ikx. Only thing you know is that phi has peaked around k0. You don't know more than that. But that's psi of x and time equals 0. Let's look at it later. So we have this thing here. And I cannot do the integral unless I do some approximations. And I will approximate omega. Omega of k, since we're anyway going to integrate around k0, let's do a Taylor series. It's omega of k0 plus k minus k0, the derivative of omega with respect to k at k0 plus order k minus k0 squared. So let's-- do this here. So if I've expanded omega as a function of k, which is the only reasonable thing to do. k's near k0 are the only ones that contribute. So omega of k may be an arbitrary function, but it has a Taylor expansion. And certainly, you've noted that you get back derivative that somehow is part of the answer, so that's certainly a bonus. So now we have to plug this into the integral. And this requires a little bit of vision because it suddenly seems it's going to get very messy. But if you look at it for a few seconds, you can see what's going on. So psi of x and t, so far, dk phi of k e to the ikx. So far so good. I'll split the exponential so as to have this thing separate. Let's do this. e to the minus i. I should put omega of k times t. So I'll begin. Omega of k0 times t. That's the first factor. e to the minus i, the second factor. k-- k d omega dk. k0 times t. And the third factor is this one with the k0. e to the minus-- it should be e to the plus. i k not d omega dk. k0 t. Plus order-- higher up. So e to the negligible-- negligible until you need to figure out distortion of wave patterns. We're going to see the wave pattern move. If you want to see the distortion, you have to keep that [INAUDIBLE]. We'll do that in a week from now. This is the integral. And then, you probably need to think a second. And you say, look. There's lots of things making it look like a difficult integral, but it's not as difficult as it looks. First, I would say, this factor-- doesn't depend on k. It's omega evaluated at k0. So this factor is just confusing. It's not-- doesn't belong in the integral. This factor, too. k0 is not a function of k. d omega dk evaluated at k0 is not a function of k. So this is not really in the integral. This is negligible. This is in the integral because it has a k. And this is in the integral. So let me put here, e to the minus i omega of k0 t e to the minus-- to the plus i k0 d omega dk-- at k0 t. Looks messy. Not bad. dk. And now I can put phi of k. e to the i k x minus these two exponentials, d omega dk at k0 times t. And I ignore this. So far so good. For this kind of wave, we already get a very nice result because look at this thing. This quantity can be written in terms of the wave function at time equals 0. It's of the same form at 5k integrated with ik and some number that you call x, which has been changed to this. So to bring in this and to make it a little clearer-- and many times it's useful. If you have a complex number, it's a little hard to see the bump. Because maybe the bump is in the real part and not in the imaginary part, or in the imaginary part and not in the real part. So take the absolute value, psi of x and t, absolute value. And now you say, ah, that's why. This is a pure phase. The absolute value of a pure phase is that. So it's just the absolute value of this one quantity, which is the absolute value of psi at x minus d omega dk k0 t comma 0. So look what you've proven. The wave function-- the norm of the wave function-- or the wave. The new norm of the wave at any time t looks like the wave looked at time equals 0 but just displaced a distance. If there was a peak at x equals 0, at time equals 0. If at time equals 0, psi had a peak when x is equal to zero, it will have a peak-- This function, which is the wave function at time equals 0, will have a peak when this thing is 0, the argument. And that corresponds to x equals to d omega dk times t, showing again that the wave has moved to the right by d omega dk times t. So I've given two presentations, basically, of this very important result about wave packets that we need to understand.

MIT_804_Quantum_Physics_I_Spring_2016

Parseval_identity.txt

PROFESSOR: What we want to understand now is really about momentum space. So we can ask the following question-- what happens to the normalization condition that we have for the wave function when we think in momentum variables? So yes, I will do this first. So let's think of integral dx psi of x star psi of x. Well, this is what we called the integral, the total integral for x squared, the thing that should be equal to 1 if you have a probability interpretation, for the wave function. And what we would like to understand is what does it say about phi of k? So for that, I have to substitute what [? size ?] r in terms of k and try to rethink about this integral how to evaluate it. So for example, here, I can make a little note that I'm going to use a variable of integration that I call k for the first factor. And for this factor, I'm going to use k prime. You should use different variables of integration. Remember, psi of x is an integral over k. But we should use different ones not to get confused. So here we go-- dx 1 over square root of 2 pi-- integral-- we said this one is over k. So it would be phi star-- let me put the dk first-- dk phi star of k e to the minus ikx and dk. That's the first psi. This is psi star. And now we put psi. So this is the dk prime-- phi of x dk prime phi of k prime e to the ik prime x. It's the same x in the three places. OK, at this moment, you always have to think, what do I do next? There are all these many integrals. Well, the integrals over k, there's no chance you're going to be able to do them apparently-- not to begin with, because they are abstract integrals. So k integrals have no chance. Maybe the integral that we have here, the dx, does have a chance. So in fact, let me write this as integral dk phi star of k integral dk prime phi of k prime. And then I have 1 over 2 pi integral dx e to the ik prime minus kx. I think I didn't miss any factor. And now comes to help this integral representation of the delta function. And it's a little opposite between the role of k and x. Here the integration variable is over x. There is was over k. But the spirit of the equality or the representation is valid. You have the 1 over 2 pi, a full integral over a variable, and some quantity here. And this is delta of k prime minus k. And finally, I can do the last integral. I can do the integral, say, over k prime. And that will just give me-- because a delta function, that's what it does. It evaluates the integrand at the value. So you integrate over k prime-- evaluates phi at k. So this is equal to integral dk phi star of k phi of k. And that's pretty neat. Look what we found. We've found what is called Parseval's theorem, which is that integral dx of psi of x squared is actually equal to integral dk phi of k squared. So it's called Parseval theorem-- Parseval's theorem. Sometimes in the literature, it's also called Plancherel's theorem. I think it depends on the generality of the identity-- so Plancherel's theorem. But this is very nice for us, because it begins to tell us there's, yes indeed, some more physics to phi of k. Why? The fact that this integral is equal to 1 was a key thing. Well, the fact that it didn't change in time thanks to showing the equation was very important. It's equal to 1. We ended up with a probabilistic interpretation for the wave function. We could argue that this could be a probability, because it made sense. And now we have a very similar relation for phi of k. Not only phi of k represents as much physics as psi of k, as psi of x, and it not only represents the weight with which you superimpose plane waves, but now it also satisfies a normalization condition that says that the integral is also equal to this integral, which is equal to 1. It's starting to lead to the idea that this phi of k could be thought maybe as a probability distribution in this new space, in momentum space. Now I want to make momentum space a little bit more clear. And this involves a little bit of moving around with constants, but it's important. We've been using k all the time. And momentum is h bar k. But now let's put things in terms of momentum. Let's do everything with momentum itself. So let's put it here. So let's go to momentum space, so to momentum language-- to momentum language. And this is not difficult. We have p is equal to h bar k. So dp over h bar is equal to dk in our integrals. And we can think of functions of k, but these are just other functions of momentum. So I can make these replacements in my Fourier relations. So these are the two equations that we wrote there-- are the ones we're aiming to write in a more momentum language rather than k, even though it's going to cost a few h bars here and there. So the first equation becomes psi of x equal 1 over square root of 2 pi integral. Well, phi of k is now phi of p with a tilde. e to the ikx is e to the ipx over h bar. And dk is equal to dp over h bar. Similarly, for the second equation, instead of phi of k, you will put phi tilde of p 1 over square root of 2 pi integral psi of x e to the minus ipx over h bar. And that integral doesn't change much [? dx. ?] So I did my change of variables. And things are not completely symmetric if you look at them. Here, they were beautifully symmetric. In here, you have 1 over h bar here and no h bar floating around. So we're going to do one more little change for more symmetry. We're going to redefine. Let phi tilde of p be replaced by phi of p times square root of h bar. You see, I'm doing this a little fast. But the idea is that this is a function I invented. I can just call it a little different, change its normalization to make it look good. You can put whatever you want. And one thing I did, I decided that I don't want to carry all these tildes all the time. So I'm going to replace phi tilde of p by this phi of p. And that shouldn't be confused with the phi of k. It's not necessarily the same thing, but it's simpler notation. So if I do that here, look-- you will have a phi of p, no tilde, and 1 over square root of h bar. Because there will be e square root of h bar in the numerator and h bar there. So this first equation will become psi of x 1 over square root of 2 pi h bar integral phi of p e to the ipx over h bar dp. And the second equation, here, you must replace it by a phi and a square root of h bar, which will go down to the same position here so that the inverse equation is phi of p, now 1 over square root of 2 pi h bar integral psi of x e to the minus ipx over h bar dx. So this is Fourier's theorem in momentum notation, in which you're really something over momenta. And you put all these h bars in the right place. And we've put them symmetrically. You could do otherwise. It's a choice. But look at the evolution of things. We've started with a standard theorem with k and x, then derived a representation for the delta function, derived Parseval's theorem, and finally, rewrote this in true momentum language. Now you can ask what happens to Parseval's theorem. Well, you have to keep track of the normalizations what will happen. Look, let me say it. This left-hand side, when we do all these changes, doesn't change at all. The second one, dk, gets a dp over h bar. And this is becomes phi tilde of p. But phi tilde of p then becomes a square root, then it's phi of p. So you get two square roots in the numerator and the dk that had a h bar in the denominator. So they all disappear, happily. It's a good thing. So Parseval now reads, integral dx psi squared of x equal integral dk phi of k-- phi of p, I'm sorry. I just doing p now. And it's a neat formula that we can use.

MIT_804_Quantum_Physics_I_Spring_2016

Waves_on_the_finite_square_well.txt

PROFESSOR: Today's lecture continues the thing we're doing with scattering states. We send in a scattering state. That is an energy eigenstate that cannot be normalized into a step barrier. And we looked at what could happen. And we saw all kinds of interesting things happening. There was reflection and transmission when the energy was higher than the barrier. And there was just reflection and a little exponential decay in the forbidden region if the energy was lower than the energy of the barrier. We also observed when we did the packet analysis that a wave packet sent in would have a delay in coming back out. It doesn't come out immediately. And that's the property of those complex numbers that entered into the reflection coefficient. Those complex numbers were a phase that had an energy dependence. And by the time you're done with analyzing how the wave packet is moving, there was a delay. So today we're going to see another effect that is sometimes called resonant transmission. And it's a rather famous. Led to the so-called Ramsauer-Townsend effect. So we'll discuss that. And then turn for the last half an hour into the setup where we can analyze more general scattering problems. So let's begin with this Ramsauer-Townsend effect. And Ramsauer Townsend effect. Before discussing phenomenologically what was involved in this effect, let's do the mathematical and physics analysis of a toy problem relevant to this effect. And for that toy problem we have the finite square well. We've normalized this square well. Having width to a. So it extends from minus a to a. That's x equals 0. And there's a number minus v0. v0 we always define it to be positive. Therefore, v0 is the depth of this potential. And the question, is what happens if you send in a wave? So you're sending in a particle. And you want to know what will happen to it. Will it get reflected? Will it get transmitted? What probabilities for reflection, what probabilities for transmission? So of course, we would have to send a wave packet to represent the physical particle. But we've learned that dealing with energy eigenstates teaches us the most important part of the story. If a particle has some energy, well roughly, it will tend to behave the way an energy eigenstate of that energy does, as far as reflection and as far as transmission is concerned. So we'll set up a wave. And there's a coefficient A. So there's an Ae to the ikx. That is our wave. And it's moving to the right. Because you remember the time factor, e to the minus iet over h bar. And when you put them together, you see that it's moving to the right. But presumably, there will be a reflection here. The wave comes in, and some gets reflected and some gets transmitted. And the part that gets transmitted probably will get partially reflected back here and partially transmitted forward. But the part that is partially reflected will get again reflected here and partially transmitted back. It seems like a never ending process of which, if you think physically what's happening, there's a reflection at the first barrier. Now you say, wait a moment. Why would there be a reflection? Classically, there would never be a reflection. If the potential goes down, the particle would just be able to continue. We had reflection when we had a barrier. Well in quantum mechanics, any change in the potential is bound to produce a reflection. So yes, if you have a potential like this, like a jumping board, and you come in here, there's a tiny probability that you will be reflected as you come into this potential drop. So OK, so this will be reflection. So at the end of the day, there might be many bouncings. If you imagine a particle doing this. Some probability of reflection, some of transmission. But the end of the day, there will be some wave moving to the left here. So we'll represent it by Be to the minus ikx. So in this region, we have A and a wave with amplitude B going this way. In the middle region, the same will be true. There will be some wave going here, and some wave that bounces due to this reflection. So there will be a-- I don't know what letters I used. I'd better keep the same. C in this direction and D in this direction. And now I would have Ce to the ikx. Well, that e to the ikx is not quite right. Because k here, k squared represents the energy. k is the momentum and k squared is energy. So if you have an energy eigenstate-- oh, my picture is very crowded. So I'll do it anyway. A is here. Maybe I'll put C and D here. And now with A and B here I can write this as the energy. You have a particle with some energy coming in. And there is this wave here and k squared. It's 2mE over h squared. E is positive scattering states. And indeed, E from that equation is h squared k squared over 2m. What do you know? But at this point, the total energy, kinetic energy of the particle is bigger. If e is replaced by e plus v0, which is the magnitude of this drop. So here there will be a k2x plus De to the minus ik2x. And k2 refers because it's region two, presumably. People use that name. k2 squared will be 2me plus v0 over h squared. And finally, to the right of the potential square well there will be just one wave. Because intuitively, we should be able to interpret this as some wave that goes through, but has nothing to make it bounce or reflect. So we try to get the solution, which will have just some wave going in this direction. And it's called Fe to the ikx. And I can go back to the label k because you have the same energy available as kinetic energy you had to the left of the barrier. OK, so we've set up the problem. The wave function I would have to write it as three expressions. One for x less than a, one for x in between a and minus a, and one for x greater than a. And those are this one, two, and three formulas. Any questions about this setup so far? OK. Well, at this moment you will eventually have some practice on that. The thing that you want to do is relate the various coefficients and define some reflection and transmission coefficients. We saw we had to think in terms of probability current. That's the better way to get an idea of what you should call reflection or transmission coefficient. So we have to be careful with the case of the step potential when we compared the meaning of the wave that was moving to the right. The amplitude divided by the incoming wave was not quite the transmission coefficient. But in this case, the nice thing is that the wave to the left and the wave to the right are experiencing the same potential. So they can be compared directly. So I will be able to conjecture, it's reasonable to conjecture that we can define reflection and transmission coefficients as follows. We'll have a reflection coefficient should be B over A squared. B represents a reflected wave, A the incoming wave. The transmission coefficient you may guess that it's F over A squared. And this all will make sense if we have the reflection plus the transmission is equal to 1. So we have current conservation, conservation. And the current, which is the net probability flow to the left of the barrier or the depression over here, is J on the left is proportional to A squared minus B squared. You remember, we computed it last time. If you compute the probability current to this Ae to the ikx plus Be to the minus ikx, you get two contributions. Essentially A squared minus B squared. There's a factor of h bar k over m in front. At any rate, this should be equal to the current that is flowing out in this direction. You see, whatever current is coming in to the left must be the current going out to the right. So this is f squared. So current conservation really tells you that A squared minus B squared is equal to F squared. And if you pass the B to the other side, you get A squared equal B squared plus F squared. And dividing by A squared you get 1 is equal to B over A squared plus F over A squared. And that's the reflection plus the transmission So the way we've defined things makes sense. Reflection and transmission are properly defined. And this is because current conservation works well. So reflection coefficient is essentially the flux in the reflected, the probability current or in the flux in the reflected wave compared to the flux in the incoming wave. The transmission is the probability current or flux of probability in the transmitted wave compared with the one incoming wave. So you've done all of this set up. Now you can't avoid, however, doing a little bit of calculation, which is boundary conditions. You have one, two, three, four, five variables. And somehow, you want to calculate these ratios. So you have to say that the wave function and the derivative is continuous at this point. And the wave function and the derivative is continuous and this point. That would give you four conditions. And that's reasonable. We have five variables. But you know, the overall normalization could never be determined. So things can be determined in terms of A. So you can expect that with four equations you can solve for B, C, D, and F in terms of A. But the overall scale of this total wave function is undetermined. Its boundary conditions will give you constraints, but it will never determine the magnitude, the overall magnitude of an energy eigenstate.

MIT_804_Quantum_Physics_I_Spring_2016

Entanglement.txt

BARTON ZWIEBACH: Let's talk now about entanglement. So we talk about entanglement when we have two non-interacting particles. You don't need a strong interaction between particles to produce entanglement, the particles can be totally non-interacting. Suppose particle 1 can be in any of these states-- u 1, u 2. Let's assume just u 1 and u 2. And particle 2 can be in states v 1 and v 2. And you have these two particles flying around, these are possible states of particle 1 and possible states of particle 2. Now you want to describe the full system, the quantum state of the two particles. States of the two particles. Two particles. Well, it seems reasonable that to describe the state of the two particles that are not interacting, I should tell you what particle 1 is doing and what particle 2 is doing. OK, so particle 1 could be doing this. Could be u 1. And particle 2 could be doing v 1. And in a sense, by telling you that, we've said what everything is doing. Particle 1 is doing u 1, particle 2 is doing u 2. And mathematically, we like to make this look like a state and we want to write it in a coherent way. And we sort of multiply these two things, but we must say sort of multiply, because this strange multiplication, this, you know, we think of them as vectors or states, so how do you multiply states? So you put something called the tensor product, a little multiplication like this. So you could say, don't worry, it's kind of like a product, and it's the way we do it. We don't move things across, the first state here, the second state here, and that's a possible state. Now, I could have a different state. Because particle 1, in fact, could be doing something a little different. Could be doing alpha 1 u 1 plus alpha 2 u 2, and maybe particle 2 is doing beta 1 v 1 plus beta 2 v 2. And this would be all right. I'm telling you what particle 1 is doing and I'm telling you what particle 2 is doing and the rules of tensor multiplication or this kind of multiplication to combine those states are just like a product, except that as I said, you never move the states across. So you just distribute, so you have alpha 2, beta 1, the number goes out, u 1 v 1-- that's the first factor-- plus alpha 1 beta 2 u 1 v 2 plus alpha 2 beta 1 u 2 v 1 plus alpha 2 beta 2 u 2 v 2. I think I got it right. Let me know. I just multiplied and got the numbers out. The numbers can be move out across this product. OK, so that's a state and that's a superposition of states, so actually, I could try to write a different state now. You see, we're just experimenting, but here is another state. u 1 v 1 plus u 2 v 2. Now this is a state that actually seems different. Quite different. Because I don't seem to be able to say that what particle 1 is doing and what particle 2 is doing separately. You see, I can say when particle 1 is doing u 1, particle 2 is doing v 1. And if when particle 1 is doing u 2, this is v 2. But can I write this as some state of the first particle times some state of the second particle? Well, let's see. Maybe I can and can write it in this form. This is the most general state that you can say, particle 1 is doing this, and particle 2 is doing that. So can they do that? Well, I can compare these two terms with those and they conclude that alpha 1 beta 1 must be 1. Alpha 2 beta 2 must be also 1. But no cross products exist, so alpha 1 beta 2 must be 0 and alpha 2 beta 1 must be 0. And that's a problem because either alpha 1 is 0, which is inconsistent, or beta 2 is 0, which is inconsistent with that, so now this state is un-factorizable. It's a funny state in which you cannot say that this quantum state can be described by telling what the first particle is doing and what the second particle is doing. What the first particle is doing depends on the second and what the second is doing depends on the first. This is an entangled state. And then we can build entangled states and our very strange states. So with two particles with spins, for example, we can build an entangled states of 2 spin 1/2 particles. And this state could look like this-- the first particle is up along z and the second particle is down along z, plus a particle that is down along z for the first particle, but the second is up along z. And these are 2 spin 1/2 particles and in the usual notation, these experiments in quantum mechanics and black hole physics, people speak of Alice and Bob. Alice has one particle, Bob has the other particle. Maybe Alice is in the moon and has her electron and Bob is on earth and has his electron, and the two electrons, one on the earth and in the moon, are in this state. So then we say that Alice and Bob share an entangled pair. And all kinds of strange things happen. People can do those things in the lab-- not quite one in the earth and one in the moon, but one photon at one place and another photon entangled with it at 100 kilometers away, that's pretty doable. And they are in this funny state in which their properties are currently that in surprising ways. So what happens here? Suppose Alice goes-- or let's say Bob goes along and measures his spin and he finds his spin down. So-- oh, you look here, oh, here is down for Bob. So at this moment, the whole state collapses into this. Because up with Bob didn't get realized. So once Bob measures and he finds down, the whole state goes into this. So if Alice-- on the moon or in another galaxy-- at that instant looks at her spin, she will find it's up before light has had time to get there. Instantaneously. It will go into this state. People were sure somehow this violates special relativity. It doesn't. You somehow when you think about this carefully, you can't quite send information, but the collapse is instantaneous in quantum mechanics. Somehow, Bob and Alice cannot communicate information by sharing this entangled pair, but it's an interesting thing why it cannot happen. Einstein again objected to this. And he said, this is a fake thing. You guys are going to share-- and now, of course, they have to share many entangled pairs to do experiments, so maybe 1,000 entangled pairs. And Einstein would say, no, that's not what's happening. What's happening is that some of your entangled pairs are this. That is, Bob is down, Alice is up, some of them are this-- and there's no such thing as this entanglement and indeed, if you measure and you find down, she will find up, and if you measure and you find up, she will find down, and there's nothing too mysterious here. But then came John Bell in 1964 and discovered his Bell inequalities that demonstrated that if Alice and Bob can measure in three different directions, they will find correlations that are impossible to explain with classical physics. It took a lot of originally for Bell to discover this, that you have to measure in three directions, and therefore, the kind of correlations that appear in entangled states are very subtle and pretty difficult to disentangle. So that's why entanglement is a very peculiar subject. People think about it a lot because it's very mysterious. It somehow violates classical notions, but in a very subtle way.

MIT_804_Quantum_Physics_I_Spring_2016

Potentials_that_satisfy_Vx_Vx.txt

PROFESSOR: Now we prove the other thing that we used in order to solve the square well. So this is property number 3. It's so important that I think I should do it here. If a potential is even-- here comes again the careful statement-- the energy eigenstates can be chosen to be even or odd under x goes to minus x. So that's analog of the first sentence in property number 2. But then comes the second sentence, that you can imagine what it is. For 1D potentials the bound states are either even or odd. So look again at this freedom. You have a general problem-- you're not talking bound states. You have a wave function that solves a problem of a potential of the symmetric around a mid-point. Then you find an arbitrary solution, no need to work with that solution. You can work with a solution that is even, and a solution that is odd. You can always choose to be even or odd. But if you have one-dimensional potential, there is no such solution that is neither even nor odd. You cannot find it. It will be, automatically, even or odd-- which is kind of remarkable. It's sufficiently subtle that in the general exam at MIT for graduate quantum mechanics, the professor that invented the problem forgot this property and the problem had to be cancelled. It's a very interesting. So let's just try to prove it. So complete proof in this case. So what is the equation? Proof. What is the equation we have to solve? Sine double prime of x, plus 2M over h squared. E minus v of x. Equals 0. Now, the proof, actually, is very simple. I just do it and I elaborate on it, because it's possible to get a little confused about it. I think it's kind of interesting. So here's equation one. And sine double prime of x notation means the second derivative of psi evaluated at x. You see, what you want to do is to show that psi of minus x solves the same equation. Right? It's kind of clear. Well, you sort of put psi of minus x here, minus x here. Well, you would do minus x here. But if the potential is even it will solve the same equation. Now, the only complication here is that there are a few X's in the derivatives here. But whether there's a complication or not there's two derivatives, so the [? signs ?] should not matter. But I want to make this a little clearer, and in order to do that I will just define phi of x to be equal to psi at minus x. So if you have that, the derivative of phi-- with respect to x-- you must differentiate this with respect to the argument. You evaluate at the argument, and then differentiate the argument with respect to x. So that leaves you a minus 1. On the second derivative of phi with respect to x squared-- at x-- will be yet another derivative. So you now get a second derivative evaluated at the thing. And then differentiate the thing inside again. So minus 1 plus another minus 1. So this is just psi double prime of x-- of minus x. Now, evaluate equation one at minus x. Well, it would be the second derivative of psi evaluated at minus x, plus 2m h squared, E minus V of minus x-- but that's the same as v of x-- psi at minus x equals 0. And then you'll realize that this thing is just the second phi of x. [? h squared ?] of x, plus 2m over h squared, e minus v of x, phi of x equals 0. So actually you've proven that phi defined this way solved the same Schrodinger equation with the same energy. So if one is true-- this thing-- I guess we could call it [? three ?] or [? two. ?] So you've proven that both psi of x and phi of x-- which is equal to psi of minus x are solutions of the Schrodinger equation with the same energy. And, therefore, if you have two solutions-- and now I emphasize this psi of minus x, and the psi of x. If you have two solutions then you can form the symmetric part of the wave function, which is 1/2 psi of x, plus psi of minus x. And the anti-symmetric part of the wave function, which would be 1/2 of psi of x minus psi of minus x. And notice that by definition psi s of minus x is indeed psi s of x. It's symmetric. If you change x for minus x on the left hand side, this goes into this, this goes into that. So it's unchanged. Here it's changed by a psi. So psi a of minus x is equal to psi minus psi a of x. And these two are solutions with the same energy-- psi s and psi a. You see, if you have-- remember that key fact-- if you have two solutions of the Schrodinger equation with the same energy, any linear combination of them is a solution with the same energy. So we form two linear combinations and they have the same energy. And, therefore, the theorem has been proven. The first part of the theorem-- the wave functions that you work with-- can be chosen to be even or odd. And that's pretty nice. But now we go to the second part of the statement. So for one-dimensional bound states-- 1-D bound states. Again, there cannot be two solutions. So it cannot be that there are two degenerate solutions. So after all, psi of x and psi of minus x, we have two solutions. This and psi of minus x. But if you're in a one-dimensional bound state you cannot have two solutions. So they must be proportional to each other. Now, if you started with a solution-- I want to say this. You start with a solution from there, from the beginning you can assume now-- because of property two-- that the solution is real. [? You can work those ?] with real solutions. So in here, I can assume that psi is real. Just simpler. So these two solutions-- that would be two real solutions-- would be degenerating energy-- there's no degeneracy for bound states. Therefore, these two must be the same up to a constant, that again-- because psi is real c-- is real. There cannot be two solutions. Let x goes to minus x in this equation. So you would get psi of x equals c of psi of minus x. But psi of minus x uses this equation again. You get c times c psi of x. But from comparing these two sides, you get that c squared must be equal to 1. But c is real. Therefore, there's only two solutions. Two options. C is equal to plus 1-- in which case-- psi is even-- automatically. Or c is equal to minus 1 and psi is odd. You have no option. You may think that the general solution of a bound state-- of a symmetric potential-- could be arbitrary. But no. The solutions come out automatically symmetrical or anti-symmetrical. And that's why-- when we decided to search for all the solutions of the finite square-- well, we could divide it into two cases. Let's find the symmetric solutions and the anti-symmetric solutions. There is no other solution of the Schrodinger equation. But what if you add a symmetric to an anti-symmetric solution? Don't you get the general solution? Well you cannot add them, because for bound states they are different energies. And adding two solutions with different energies is pointless. It's not a energy eigenstate anymore. So a very powerful theorem. We'll be using it a lot, and I thought you really ought to see it.

MIT_804_Quantum_Physics_I_Spring_2016

Particle_on_the_forbidden_region.txt

PROFESSOR: We have found in this solution with some energy like this, that there's a decaying exponential over this side. And the question is often asked, well what happens if you tried to measure the particle in the forbidden region? Must be a problem. If you find the particle in the forbidden region, it has energy E that is less than v0, so you you have found the particle with negative kinetic energy. How does it look? How can it happen? What's going on? Can you really find the particle in the forbidden region? And then how does this negative kinetic energy look like? The answer is that it's kind of funny what happens here. You can make two statements. It would be contradictory, contradictory if you could make-- could say the following things. One, that the particle is in the forbidden region, forbidden region. And two, that the particle has energy less than v0. Because then it would mean negative kinetic energy. So if you can say these two things, it seems contradictory. So quantum mechanics evades this problem. Now, this is not discussed as far as I can see, except in some lecture notes of Gordon [? Boehme. ?] And because the argument is not 100% precise, but they think the spirit of the argument is clear. So I want to share it with you. So here is the catch. This particle, remember it's governed by e to the minus kappa x is the forbidden region. So the length scale here where you can find it, the particle. The length scale is, this forbidden region stretches to about x of the order 1 over kappa. If you are going to find it, it is in the region of a distance 1 over kappa. At 10 1 over kappa you're not going to find it. The exponential is too small. But remember, what was kappa? Kappa squared was 2m v0 minus E over h squared. That's what it was. Now if you want to see and declare that you have this particle, you would have to be able to measure position with some precision, with a precision a little smaller than this. Otherwise if you measure with precision 10 times that, well maybe it's to the left, maybe it's somewhere else. So you need to measure position with delta x a little smaller than 1 over kappa, otherwise you cannot really tell it's inside the forbidden region. But now the problem is that if you do a position measurement, and you localize the wave function, there is some momentum uncertainty. The particle that you're looking at, as opposed to the particle to the left, has no momentum. It's a different kind of wave function. There's no momentum really associated or well-defined momentum to it. So because you make a position, you're localizing x, whatever wave function you have. You're going to have some uncertainty, and some momentum that is going to be kind of bigger than h bar over delta x. So a momentum that is bigger than, or a little bigger, than h bar kappa. If delta x is less than that inequality, it goes in the same direction. So there's going to be an uncertainty P. And therefore, this particle has now some kinetic energy due to this uncertain momentum. So uncertainty in the kinetic energy is how much? It's P squared over 2m, where P is this uncertain momentum. So this is equal to h bar kappa squared over 2m, which is equal to v0 minus E. So actually, if you think about it, here is v0. This difference is v0 minus E. And you were going to say, oh, I found the particle, it has negative kinetic energy. But no. The uncertainty principle says, you found it localized? OK. Your kinetic energy, I'm sorry, no. There's an uncertainty. How much? v0 minus E. So whatever you wanted to prove, it has been disproved. You can't do it. The total energy, total energy is now E plus the uncertainty in the energy, which is E plus v0 minus E. And it's therefore greater than or equal to v0. And no real contradiction. So the uncertainty principle sort of conspires to prevent you from finding a particle with negative kinetic energy. And if you do detect a particle in the forbidden region, it will have total energy 0, or total kinetic energy 0. It will be a normal particle. Nothing strange about it.

MIT_804_Quantum_Physics_I_Spring_2016

Expectation_values_on_stationary_states.txt

PROFESSOR: How about the expectation value of the Hamiltonian in a stationary state? You would imagine, somehow it has to do with energy ion states and energy. So let's see what happens. The expectation value of the Hamiltonian on this stationary state. That would be integral dx stationary state Hamiltonian stationary state. And we're going to see this statement that we made a few minutes ago become clear. Well what do we get here? dx psi star of x e to the i Et over h bar H e to the minus i Et over h bar psi of x. And H hat couldn't care less about the time dependence, that exponential is irrelevant to H hat. That exponential of time can be moved across and cancelled with this one. And therefore you get that this is equal to dx psi star of x H hat psi of x, which is a nice thing to notice. The expectation value of H on the full stationary state is equal to the expectation value of H on the spatial part of the stationary state. That's neat. I think it should be noted. So it's equal to the H of little psi of x. But this one, we can evaluate, because if we are in a stationary state, H hat psi of x is E times psi of x. So we get an E integral the x psi star of psi, which we already show that integral is equal to one, so we get the energy. So two interesting things. The expectation value of this quantity of H in the stationary state is the same as it's quotation value of H in the spatial part, and it's manually equal to the energy. By the way, you know, these states are energy eigenstates, these psi of x's, so you would expect zero uncertainty because they are energy eigenstates. So the zero uncertainty of the energy operator in an energy eigenstate. There's zero uncertainty even in the whole stationary state. If you have an H squared here, it would give you an E squared, and the expectation value of H is equal to E, so the expectation value of H squared minus the expectation value of H squared would be zero. Each one would be equal to E squared. Nothing would happen, no uncertainties whatsoever. So let me say once more, in general, being so important here is the comment that the expectation value of any time independent operator, so comments 1, the expectation value of any time-independent operator Q in a stationary state is time-independent. So how does that go? It's the same thing. Q hat on the psi of x and t is general, now it's integral dx capital Psi of x and t Q hat psi of x and t equals integral dx-- you have to start breaking the things now. Little psi star of x E to the i et over H bar. And I'll put the whole thing here. Q hat Psi of x E to the minus i et over H. So it's the same thing. Q doesn't care about time So this factor just moves across and cancels this factor. The time dependence completely disappears. And in this case, we just get-- this is equal to integral dx psi star Q psi, which is the expectation value of Q on little psi of x, which is clearly time-independent, because the state has no time anymore and the operator has no time. So everybody loves their time and we're in good shape. The second problem is kind of a peculiarity, but it's important to emphasize superposition. It's always true, but the superposition of two stationary states is or is not a stationary state? STUDENT: No. PROFESSOR: No, good. It's not a stationary state in general because it's not factorizing. You have two stationary states with different energies, each one has its own exponential, and therefore, the whole state is not factorized between space and time. One time-dependence has one space-dependence plus another time-dependence and another space-dependence, you cannot factor it. So it's not just a plain fact. So the superposition of two stationary states of different energy is not stationary. And it's more than just saying, OK, it's not stationary. What it means is that if you take the expectation value of a time-independent operator, it may have time-dependence, because you are not anymore guaranteed by the stationary state that the expectation value has no time-dependence. That's how, eventually, these things have time-dependence, because these things are not [INAUDIBLE] on stationary states. On stationary states, these things would have no time-dependence. And that's important, because it would be very boring, quantum mechanics, if expectation values of operators were always time-independent. So what's happening? Whatever you measure never changes, nothing moves, nothing changes. And the way it's solved is because you do have those stationary states that will give you lots of solutions. And then we combine them. And as we combine them, we can get time-dependence and we can get the most [INAUDIBLE] equation.

MIT_804_Quantum_Physics_I_Spring_2016

Angular_momentum_operators_and_their_algebra.txt

PROFESSOR: So angular momentum, we need to deal with angular momentum, and the inspiration for it is classical. We have L is r cross p. Classically. So let's try to just use that information and write the various operators. And in fact, we're lucky in this case. The operators that we would write inspired by the classical definition are good operators and will do the job. So what do we have? Lx, if you remember the cross-product rule, that would be y Pz minus z Py. Now you can think of this thing as a cyclic, like a circle when you have x and Px, y and Py, and z and Pz. Things are cyclically symmetric. There's no real difference between this core. And so you can go cyclically here. So you say, let's go cyclical on this index. Ly is equal to the next cyclic to y in that direction is z Px minus x Pz. And Lz is equal to x Py minus y Px. And these things, I'll think of them as the operators. Let's put hats to everything. The first thing I can wonder with a little bit of trepidation is maybe I got the ordering wrong. Should I have written-- here classically, you put r cross p, and then the order of these two terms doesn't matter. Does it matter quantum mechanically? Happily, it doesn't matter because y and Pz commute. z and Py commute, so you could even have written them the other way, and they are good. All of them are ambiguous. You could have even written them the other way, and they would be fine. But now these are operators. And moreover, they are Hermitian operators. Hermitian. Let's see. Lx dagger. Well, the dagger of two operators you would do Pz dagger y hat dagger-- recall the dagger changes the order-- minus Py dagger z dagger. Now, p and x's are all for Hermitian operators, so this is Pz y minus Py z. And we use, again, that y and Pz commute, and z and Py commute to put it back in the standard form. And that's, again, Lx. So it is an Hermitian operator. And so is Ly and Lz. That means these operators are observables. That's all you need for the operator to be an observable. And that's a very good thing. So these operators are observables. Li's are observable. But they're funny properties. With these operators, they're not all that simple in some ways. So next we have these operators. Whenever you have quantum operators, the thing you do next is compute their commutators Just like we did with x and p, we wanted to know what that commutator is. We want to know what is the commutator of this L operator. So we'll do Lx with Ly. Try to compute the commutators. So Lx is y Pz. Let me forget the hat, so basically minus z Py. And Ly is z Px minus x Pz. Here is a y. The y commutes with everything here, so the y doesn't get. The Pz gets stuck with the z and doesn't care about this. So this term just talks to that term. And here the z Py, the Py doesn't care about anybody here, but this z, well, doesn't care about that z, but it does care about this Pz. So the only contribution, there could have been four terms out of this commutator, but only two are relevant. So let's write them down. y Pz with z Px and minus, it's a plus z Py x Pz. Well, you can start peeling off things. You can think of this as a single operator with this too, and it will fail to commit with the first. So you have y Pz z Px. That's all this commutator gets. And the same thing here. This fails to commute just with Pz, so the x can go out, x z Py Pz. And then here the y actually can go out, doesn't care about this z, goes out on the left. Not that it matters much here, but that's how using the commutator identities does. And this Py can go out and let's go out on the right, z Pz. And basing this on this identity, we just have A BC commutator and then AB C commutators, how things distribute. Now, this is minus i h bar, and this is i h bar. So here we get i h bar x Py minus y Px. See everything came out in the right position. And you recognize that operator as Lz. So this commutator here has given you Lx with Ly equal i h bar Lz. It's a very interesting and fascinating property that somehow you're doing this commutator, it could have been a mess, but it combined to give you another angular momentum operator. Now, it looks like a miracle, but physically, it's not that miraculous. It actually has to do with the concept of symmetry. Symmetry transformations. If you have a symmetry transformation and you do commutators within those symmetry operators, you must get an operator that corresponds to that symmetry, or you must get a symmetry at the very least. So if we say that the potential has very close symmetry, that suggests that when you do operations with these operators that generate rotation, you should get some rotation here. And alternatively, although, again, this is suggestive, it can be made very precise, when you do rotations in different order, you don't get the same thing at the end. Everybody knows if you have a page and you do one rotation and then the other as opposed to the other and then the first one, you don't get the same thing. Rotations do not commute. A single rotation does commute in one direction, but rotations in different directions don't commute. That is the reason for this equation. And this equation, as we said, everything is cyclic. so you don't have to work again to argue that then Ly Lz, going cyclic, must be equal to i h bar Lx. And that Lz Lx must be i h bar Ly. And this is called the quantum algebra of angular momentum. In fact, it is so important that this algebra appears in all fields of physics and mathematics, and all kinds of things show up. This algebra is related to the algebra of generators of the group SU2, Special Unitary Transformations in Two Dimensions. It is related to the orthogonal group in three dimensions where you rotate things in three-dimensional space. It is here, the algebra of operators and in a sense, it's a deeper result than the derivation. It is one of those cases when you start with something very concrete and you suddenly discover a structure that is rather universal. Because we started with very concrete representation of L's in terms of y P's and all these things. But then they form a consistent unit by themselves. So sometimes there will be operators that satisfy these relations, and they don't come from x's and P's, but still they satisfy that. And that's what happens with spin angular momentum. The spin angular momentum operators will be denoted with Sx, for example, and Sy will have i h bar spin in the z direction, and the others will follow. But nevertheless nobody will ever be able to write spin as something like that because it's not, but spin exists. And it's because this structure is more general than the situation that allowed us to discover it. It's a lot more general and a lot more profound. So in fact, mathematicians don't even mention angular momentum. They say, let's study. The subject of Lie algebra is the subject of classifying all possible consistent commutation relations. And this is the first non-trivial example they have, and they studied the books on this algebra.

MIT_804_Quantum_Physics_I_Spring_2016

Interpretation_of_the_wavefunction.txt

PROFESSOR: interpretation of the wave function. --pretation-- the wave function. So you should look at what the inventor said. So what did Schrodinger say? Schrodinger thought that psi represents particles that disintegrate. You have a wave function. And the wave function is spread all over space, so the particle has disintegrated completely. And wherever you find more psi, more of the particle is there. That was his interpretation. Then came Max Born. He said, that doesn't look right to me. If I have a particle, but I solve the Schrodinger equation. Everybody started solving the Schrodinger equation. So they solved it for a particle that hits a Coulomb potential. And they find that the wave function falls off like 1 over r. OK, the wave function falls off like 2 over r. So is the particle disintegrating? And if you measure, you get a little bit of the particle here? No. Max Born said, we've done this experiment. The particle chooses some way to go. And it goes one way, and when you measure, you get the full particle. The particle never disintegrates. So Schrodinger hated what Max Born said. Einstein hated it. But never mind. Max Born was right. Max Born said, it represents probabilities. And why did they hate it? Because suddenly you lose determinism. You can just talk about probability. So that was sort of funny. And in fact, neither Einstein nor Schrodinger ever reconciled themselves with the probabilistic interpretation. They never quite liked it. It's probably said that the whole Schrodinger cat experiment was a way of Schrodinger to try to say how ridiculous the probability interpretation was. Of course, it's not ridiculous. It's right. And the important thing is summarized, I think, with one sentence here. I'll write it. Psi of x and t does not tell how much of the particle-- is at x at time t. But rather-- what is the probability-- probability-- --bility-- to find it-- at x at time t. So in one sentence, the first clause is what Schrodinger said, and it's not that. It's not what fraction of the particle you get, how much of the particle you get. It's the probability of getting. But that requires-- a little more precision. Because if a particle can be anywhere, the probability of being at one point, typically, will be 0. It's a continuous probability distribution. So the way we think of this is we say, we have a point x. Around that point x, we construct a little cube. d cube x. And the probability-- probability dp, the little probability to find the particle at xt in the cube, within the cube-- the cube-- is equal to the value of the wave function at that point. Norm squared times the volume d cube x. So that's the probability to find the particle at that little cube. You must find the square of the wave function and multiply by the little element of volume. So that gives you the probability distribution. And that's, really, what the interpretation means. So it better be, if you have a single particle-- particle, it better be that the integral all over space-- all over space-- of psi squared of x and t squared must be equal to 1. Because that particle must be found somewhere. And the sum of the probabilities to be found everywhere must add up to 1. So it better be that this is true. And this poses a set of difficulties that we have to explore. Because you wrote the Schrodinger equation. And this Schrodinger equation tells you how psi evolve in time. Now, a point I want to emphasize is that the Schrodinger equation says, suppose you know the wave function all over space. You know it's here at some time t0. The Schrodinger equation implies that that determines the wave function for any time. Why? Because if you know the wave function throughout x, you can calculate the right hand side of this equation for any x. And then you know how psi changes in time. And therefore, you can integrate with your computer the differential equation and find the wave function at a later time all over space, and then at a later time. So knowing the wave function at one time determines the wave function at all times. So we could run into a big problem, which is-- suppose your wave function at some time t0 satisfies this at the initial time. Well, you cannot force the wave function to satisfy it at any time. Because the wave function now is determined by the Schrodinger equation. So you have the possibility that you normalize the wave function well. It makes sense at some time. But the Schrodinger equation later, by time evolution, gives you another thing that doesn't satisfy this for all times. So what we will have to understand next time is how the Schrodinger equation does the right thing and manages to make this consistent. If it's a probability at some time, at a later time it will still be a probability distribution.

MIT_804_Quantum_Physics_I_Spring_2016

Harmonic_oscillator_Differential_equation.txt

PROFESSOR: Simple harmonic oscillator. So what is there about this simple harmonic oscillator? Well, it's a classical system that you understand perfectly well. An oscillator, a spring with a mass oscillates and has an energy, which is the kinetic energy plus the potential energy, and that's p squared over 2m plus 1/2 m omega squared x squared. And this may be a tiny bit unfamiliar, this way of writing it. But you may recall that omega is equal to the square root of k over m, the so-called spring constant, in which the potential in terms of k would be 1/2 k x squared. And that's the potential energy stored in a spring that you stretch at distance x. That's the total energy of a harmonic oscillator. So when physicists starting with quantum mechanics in the '20s decided, let's do a harmonic oscillator, a quantum harmonic oscillator, they had to invent the Hamiltonian. And the Hamiltonian they invented was a simple one. They looked at that and said, h is going to be p hat squared over 2m plus 1/2 m omega squared x hat squared. And now the difference is going to be that x with p are operators, and this is our h bar, and that's going to be my quantum system. So this is a quantum system that is inspired by classical mechanics in the purest and simple way. Anyone could have invented this quantum system. It was very natural. Still, the result of the quantization is very surprising because while this mechanical oscillator can oscillate with any amplitude, the quantum oscillator has quantized amplitudes and quantized energies therefore. So all kinds of interesting things happen with this oscillator. Now the reason this is also very ubiquitous is that this potential is exactly a quadratic potential v of x. v of x is 1/2 m omega squared x squared. We have x without the hat. This is a good approximation to almost any system we consider in nature, any oscillating system. Because for any potential that has a minimum, there is some parabolic approximation at the bottom. At the bottom the derivative vanishes, so the Taylor series says that approximately at the bottom is a quadratic potential. And therefore this quadratic potential will govern the quantum oscillations of a diatomic molecule, the quantum oscillations of a periodic system, all kind of quantum oscillations will be approximately governed by a harmonic oscillator. Light has a harmonic oscillator description for its photons. This Hamiltonian is the most famous Hamiltonian there is. In fact, when you have electrons in a magnetic field, somehow this shows up. And this becomes some sort of problem that you solve very well, understand very well, and suddenly it pops up in all kinds of contexts. So we need to understand it. And let's go directly to the issue of solving this problem, because it has many important lessons. So there's two ways to solve this problem of finding the bound states of the energy eigenstates of the harmonic oscillator. This is a very interesting potential because all its energy eigenstates are bound states. That's not the case for the delta function potential. In the delta function potential we found one bound state, but they're all kind of unbound states with positive energy. But this potential grows forever, never stops growing. So whatever energy you have, it is a bound state. It will decay. It will be localized. So you just have bound states. It's marvelously nice because of that property. Much simpler than anything you can imagine. So what do we have to do? We want to find the energy eigenstates. So we'll write h, and I will write phis. People write sometimes phis. Phi n of x is equal to E phi n of x. And we don't know the energy eigenstates. You know it's a symmetric potential. It's a real potential. This we used to go psi n's, but I will write them as phi n's as many people do, because they are the harmonic oscillator ones that are very famous. And we don't know-- this could be En, energy of the n-th state. And we don't know what are the energies nor how the wave functions look. And we have to solve a differential equation. As I was saying, there's two ways of solving this differential equation. One, treating this as a differential equation and understanding why the energy is quantized from the differential equation. This actually gives you a lot of insight as to what's going on, and it will relate to the kind of things you're doing in the homework this week. The other way is to be very clever and invent what are called raising and lowering operators, and sort of solve this whole system without solving the general differential equation, by solving a first order very simple differential equation, and then doing everything else algebraically with creation and annihilation operators. We will also do that. But we will not develop that too far. That's The applications of that method are mostly for 805. So we'll introduce creation and annihilation operators, which are very nice and very useful. But we leave some of the applications to coherent states of harmonic oscillators, to squeezed states of harmonic oscillators, for later. So this is what we want to solve. So we have minus h squared over 2m d second phi n dx squared. And I will probably forget about the labels. The labels will come later as we solve the equation. Plus v of x 1/2 m omega squared x squared phi of x, is equal to E phi of x. OK. This is the question we want to solve. Now we're going to do one thing first with this differential equation. We don't like all these dimensionful constants. If you had to put it in the computer, what? Are you going to put 6 times 10 to the minus 23 over m? And you won't solve a differential equation. We have to clean this up. To clean this up, there is a procedure that is guaranteed to do the job for you. And the procedure is to change the x variable into a variable that has no units. This is guaranteed to lead you to the way to solve this differential equation. You will be using this throughout the semester. Cleaning differential equations is a nice skill. And the fact is that there's a method, and the method always proceeds by first writing x equals au, where this will be unit-free, this quantity u. So this will become a differential equation on a unit-free constant, which is ideal for your numerical solution and is much nicer. But then this a to have units of length. So the first thing you have to do is look up in your problem. What are your constants? And you have a mass m, a frequency omega, an h bar. And you need to find a constant with units of length. Wasn't that in your test? I think so. How do you find a constant with units of length here? Well, energy is equal to-- you can write it in two ways. The energy can be written as p squared over m, so it will be h squared over m times a length squared. But from the potential, it also has the units of m omega squared a squared. These are units, equations for units. The units of energy are these, and the units of energy from the second term in the Hamiltonian are those. From where you get that a squared is equal to h over m omega. So that's the constant that you need. If you have that constant, your differential equation becomes what? Well, it becomes the following. Let's write it out. It's very simple, because x is equal to au, and therefore you get minus h squared over 2m a squared d second phi du squared. x is equal to au, so it basically just shows up here. Plus 1/2 m omega squared a squared u squared phi, is equal to E phi. Now things have to work nicely. If you did the job well, they have to work nicely. Let's think of the units of this equation. Phi is here, phi is here, phi is here, so phi is not an issue. The units of phi are irrelevant. Here is units of energy, but u has no units. This has no units. So this must have units of energy. And since u has no units, this derivative has no units, and this must have units of energy. And these two numbers must be something nice if you substitute a squared. And indeed, if you substitute a squared here, this whole number becomes h bar omega. And this number becomes h bar omega as well, which is very nice. So this whole equation has become minus 1/2 times h bar omega d second phi du squared plus 1/2 h bar omega u squared phi equal E phi. The next step is to say, you know, I don't even want these energy units. Even though they don't look that bad, this equation looks much nicer than the original equation which had all kinds of strange units. So I will multiply this equation by 2 over h omega to cancel it. So I get minus d second phi du squared plus u squared phi is equal to 2E over h bar omega phi. So look at this. The equation is now almost in perfect form. And in order to make it perfect, I would say that the right hand side-- now, see again. Phi, whatever units it has, it doesn't matter. It's all over the place. But this has no units, this derivative, and this has no units, this multiplication. So this must have no units. And indeed, you know that h omega has units of energy, and that's energy. So it suggests that you define a unit-free energy, free energy, which is 2E over h omega. And calculating curly E is the same thing as calculating the energy, because if you know this number you know the energy. The energy is h bar omega over 2 times curly E. The advantage is that curly E will be either 1, 2, 1/2, a nice number, while E is some 0.87 Ev, or things like that. So all of this equation has been reduced to this very nice equation. Minus d second phi du squared plus u squared phi is equal to E phi. Or, d second phi du squared is equal to u squared minus E phi.

MIT_804_Quantum_Physics_I_Spring_2016

Threedimensional_Fourier_transforms.txt

PROFESSOR: We got here finally in terms of position and in terms of momentum. So this was not an accident that it worked for position and wave number. It works with position and for momentum. And remember, this phi of p now has interpretation of the weight that is associated with a plane wave of momentum p, and you're summing over P in here. So we'll do the natural thing that we did with x. We'll interpret phi of p squared-- phi of p squared-- dp is the probability. Find the particle with momentum in the range p, p plus dp. Just the same way as we would say that psi squared of x, dx is the probability to find the particle between x and x plus dx. So this is allowed now by the conservation of probability and this, therefore, makes sense. It's a postulate, though. It's not something that can derive. I can just argue that it's consistent to think in that way, and that's the way we finally promote this phi of p, which did encode psi of x. Phi of b has the same information as phi of x. phi of p is the weight of the superposition but, finally, it's given a probabilistic interpretation. It represents a probability to find the particle with some momentum. So this is what is going to allow us to do expectation values in a minute. But I want to close off this discussion by writing for you the three-dimensional versions of these equations. 3D version of Fourier transform. So this is what we want to rewrite. So what would it be? It's psi of the vector x. Since you're going to have three integrals-- because you're going into it over three components of momentum-- this factor appears three times. So actually, it's 2 pi h bar to the three halfs integral phi of vector p into the i vector p dot product vector x h bar d cube p, and phi of p vector the inverse theorem-- same factor, we keep the nice symmetry between x and p-- psi of x vector negative exponent same dot product but negative exponent d cube x. So these are the three dimensional versions of your x versus p. And there is a three-dimensional version of Parseval. So oh there's a three dimensional version of the delta function. Just like we had a delta function here-- a delta function in three dimensional space would be delta cubed x minus x prime would be one over 2 pi cubed integral d cube k e to the i k vector x minus x prime vector. It's all quite analogous. I think you should appreciate that you don't have to memorize them or anything like that. They won't be in any formula sheet, but they are very analogous expressions. Parseval also works in the same way. And you have-- just as you would imagine-- that the integral all over three dimensional space of psi of x squared is equal to the integral over three dimensional momentum space of phi of p squared. So the three results-- the Fourier theorem the delta function and Parseval hold equally well.

MIT_804_Quantum_Physics_I_Spring_2016

Galilean_transformation_of_ordinary_waves.txt

BARTON ZWIEBACH: Do normal wave analysis to demonstrate that indeed these things should not quite happen. So for that, so ordinary waves and Galilean transformations. So when you have a wave, as you've probably have seen many times before, the key object in the wave is something called the phaze of the wave. Phaze, the phaze. And it's controlled by this quantity kx minus omega t. k being the wave number, omega being the angular frequency and we spoke about. And the wave may be sine of that phaze or cosine of that phaze or a linear combination of sines and cosines, or E to this wave, any of those things could be your wave. And whenever you have such a wave, what we say is that the phaze of this wave is a Galilean invariant. Invariant. What it means is that two people looking at this wave, and they look at the point on this wave, both people will agree on the value of the phaze, because basically, the reality of the wave is based on the phaze, and if you have, for example, cosine of this phaze, the place where this cosine is 0 is some of the phaze, and if the cosine is 0, the wave is 0, and everybody should agree that the wave is 0 at that point. So if you have a place where the wave has a maximum or a place where the wave is 0, this is an ordinary wave, everybody would agree that at that place you have a maximum and in that place you have a 0. So observers should agree on the value of this phaze. It's going to be an invariant. And we can rewrite this phaze in a perhaps more familiar way by factoring the k, and then you have x minus omega over kt, and this is 2 pi over lambda, x minus-- this quantity is called the velocity of the wave, and we'll write it this way. And I'll write in one last way-- 2 pi x over lambda minus 2 pi V over lambda t. And this quantity is omega and this quantity is k. So this is our phaze. And we've said that it's a Galilean invariant, so I will say that S should see-- the observer S prime should see the same phaze-- phaze-- as S. So phi prime, the phaze that S prime sees, must be equal to phi when referring to the same point. When referring to the same point at the same time. Let's write this. So phi prime should be equal to phi. And phi, we've written there. 2 pi over lambda x minus Vt. And this is so far so good, but we want to write it in terms of quantities that S prime measures. So this x should be replaced by 2 pi over lambda x prime plus Vt minus Vt like this. And I could even do more if I wish. I could put t prime here, because the t and t primes are the same. So phi prime, by the condition that these phazes agree, it's given by this, which is by the relation between the coordinates and times of the two frames, just this quantity. So we can rewrite this as 2 pi over lambda x prime minus 2 pi over lambda V 1 minus little v over capital V t prime. I think I got the algebra right. 2 pi over lambda, the sine-- yes, I grouped those two terms and rewrote in that way. So that is the phaze. And therefore, we look at this phaze and see, oh, whenever we have a wave, we can read the wave number by looking at the factor multiplying x, and we can read the frequency by looking at the factor multiplying t. So you can do the same thing in this case and read, therefore, that omega prime, this whole quantity is this, omega prime. And this is k prime, because they can respond to the frame as prime. So omega prime is equal to this 2 pi V over lambda, which is omega, times 1 minus V over V. And k prime is equal to k or, what I wanted to show, that lambda prime for a normal wave is equal to lambda for ordinary wave moving in the medium. So at this moment, one wonders, so what happened? What have we learned? Is that this wave function is not like a sound wave. It's not like a water wave. We're doing everything non-relativistic. But still, we're seeing that you're not expected to have agreement. That is, if somebody looks at one wave function and you look at the same wave function, these two people will not agree on the value of the wave function necessarily. So the things that we conclude-- so the conclusions are that waves are surprising. So size are not directly measurable-- measurable-- because if you had a quantity for which you could measure, like a sound wave or a water wave, and you could measure aspects to it, they should agree between different observables. So this is going to be something that is not directly measurable-- not all of psi can be measured. Some of psi can be measured, and you're already heard the hints of that. Because we said any number that you multiply, you cannot measure, and in the phase that you multiply, you cannot measure. So complex numbers can't be measured, you measure real numbers. So at the end of the day, these are not directly measurable, per se. The second thing is that they're not Galilean invariant, and that sets the stage to that problem 6. You see, the fact that this phaze that controls these waves is Galilean invariant led you to the quality of the wavelengths, but these wavelengths don't do that. The de Broglie wavelengths don't transform as they would do for a Galilean invariant wave. Therefore, this thing is not Galilean invariant, and what does that mean? That if you have two people and you ask, what is the value of the wave function here at 103, the two observers might give you a different complex number for the wave function. They will just not agree. Not all is lost, because you will find how their measurements can be compared. That will be the task of the problem. How-- if you have a wave function, how does your friend, that is moving with some velocity, measure the wave function? What does this other person measure? So the end result, if you have a point here at some time t, the wave function psi of x and t is not going to be the same as the wave function measured by the prime observer at x prime t prime, so this point is the point x and t or x prime and t prime. These are two different labels for the same point. You're talking about the wave function at the same point at the same time. You still don't agree. These two people will not agree. If they agreed, this wave function would have a simpler transformation law with a wavelength that this can serve. So by simply discussing the Galilean properties of this wave, we're led to know that the de Broglie waves are not like normal matter waves that propagate in a medium or simple.

MIT_804_Quantum_Physics_I_Spring_2016

Associated_Legendre_functions_and_spherical_harmonics.txt

PROFESSOR: We're talking about angular momentum. We've motivated angular momentum as a set of operators that provided observables, things we can measure. Therefore, they are important. But they're particularly important for systems in which you have central potentials. Potentials that depend just on the magnitude of the radial variable. A v of r that depends just on the magnitude of the vector r relevant to cases where you have two bodies interacting through a potential that just depends on the distance between the particles. So what did we develop? Well, we discussed the definition of the angular momentum operator. You saw they were Hermitian. We found that they satisfy a series of commutators in which lx with ly gave ih bar lz, and cyclical versions of that equation, which ensure that actually you can measure simultaneously the three components of angular momentum. You can measure, in fact, just one. Happily we found there was another object we could measure, which was the square of the total angular momentum. Now, you should understand this symbol. It's not a vector. It is just a single operator. l squared is, by definition, lx times lx, plus ly times ly, plus lz times lz. This is this operator. And we showed that any component of angular momentum, be it lx, ly, or lz, commutes with l squared. Given that they commute, it's a general theorem that two permission operators that commute, you can find simultaneous eigenstates of those two operators. And therefore, we set up for the search of those wave functions that are simultaneous eigenstates of one of the three components of angular momentum. Everybody chooses lz and l squared. lz being proportional to angular momentum has an h bar m. We figured out by looking at this differential equation that if we wanted single valued wave functions-- wave functions would be the same at phi, and at phi plus 2 pi, which is the same point. You must choose m to be an integer. For the l squared operator we also explained that the eigenvalue of this operator should be positive. That is achieved when l, whatever it is, is greater than 0, greater or equal than 0. And the discussion that led to the quantization of l was a little longer, took a bit more work. Happily we have this operator, and operator we can diagonalize, or we can find eigenstates for it. Because the Laplacian, as was written in the previous lecture, Laplacian entering in the Schrodinger equation has a radial part and an angular part, where you have dd thetas, and sine thetas, and the second defies square. All these things were taken care of by l squared. And that's very useful. Well, the differential equation for l squared-- this can be though as a differential equation-- ended up being of this form, which is of an equation for the so-called Associate Legendre functions. For the case of m equals 0 it simplifies very much so that it becomes an equation for what were eventually called Legenre polynomials. We looked at that differential equation with m equals 0. We called it pl 0. So we don't write the zeros. Everybody writes pl for those polynomials. And looking at the differential equation one finds that they have divergences at theta equals 0, and a theta equal pi, north and south pole of this spherical coordinate system. There aren't divergences unless these differential equations has a polynomial solution that this is serious the recursion relations terminate. And that gave for us the quantization of l. And that's where we stopped. These are the Legendre polynomials. Solve this equation for m equals 0. Are there any questions? Anything about the definitions or? Yes? AUDIENCE: Why do we care about simultaneous eigenstates? PROFESSOR: Well, the question is why do we care about simultaneous eigenstates. The answer is that if you have a system you want to figure out what are the properties of the states. And you could begin by saying the only thing I can know about this state is its energy. OK, well, I know the energy at least. But maybe thinking harder you can figure out, oh, you can also know the momentum. That's progress. If you can also know the angular momentum you learn more about the physics of this state. So in general, you will be led in any physical problem to look for the maximal set of commuting operators. The most number of operators that you could possibly measure. You know you have success at the very least, if you can uniquely characterize that states of the system by observables. Let's assume you have a particle in a circle. Remember that the free particle in a circle has degenerate energy eigenstates. So you have two energy eigenstates for every allowed energy, except for 0 energy, but two energy eigenstates. And you would be baffled. You'd say, why do I have two? There must be some difference between these two states. If there are two states, there must be some property that distinguishes them. If there is no property that distinguishes them, they should be the same state. So you're left to search for another thing. And in that case the answer was simple. It was the momentum. You have a particle with some momentum in one direction, or in the reverse direction. So in general, it's a most important question to try to enlarge the set of commuting observables. Leading finally to what is initially called a complete set of commuting observables. So what do we have to do today? We want to complete this analysis. We'll work back to this equation. And then work back to the Schrodinger equation to finally obtain the relevant differential equation we have to solve if you have a spherical symmetric potential. So the equation will be there in a little while. Then we'll look at the hydrogen atom. We'll begin the hydrogen atom and this task why? Having a proton and an electron we can reduce this system to as if we had one particle in a central potential. So that will be also very important physically. So let's move ahead. And here there is a simple observation that one can make. Is that the differential equation for p l m depends on m squared. We expect to need values of m that are positive and negative. You have wave functions here, of this form. The complex conjugate ones should be thought as having m negative. So we expect positive and negative m's to be allowed. So how did people figure this out? They, in fact, figured out that if you have these polynomials you can create automatically the solutions for this equation. There's a rule, a simple rule that leads to solutions. You put p l m of x is equal to 1 minus x squared, to the absolute value of m over 2. So there are square roots here, possibly. An absolute value of m means that this is always in the numerator, whether m is positive or negative. And d, dx acting exactly absolute value of m times on p l x. The fact is that this definition solves the differential equation star. This takes a little work to check. I will not check it, nor the notes will check it. It's probably something you can find the calculation in some books. But it's not all that important. The important thing to note here is the following. That this provides solutions. Since this polynomial is like x to the l plus x to the l minus 2 plus coefficients like this. You can think that most m equal l derivatives-- if you take more than l derivatives you get 0. And there's no great honor in finding zero solution of this equation. These are no solutions. So this produces solutions for an, absolute value of m, less than l. So produces solutions for absolute value of m less or equal to l. And therefore m in between l and minus l. But that's not all that happens. There's a little more that takes mathematicians some skill to do. It's to show that there are no more solutions. You might seem that you were very clever and you found some solutions, but it's a theorem that there are no more solutions. No additional regular solutions. I mean solutions that don't diverge. So this is very important. It shows that there is one more constraint on your quantum numbers. This formula you may forget, but you should never forget this one. This one says that if you choose some l which corresponds to choosing the magnitude of the angular momentum, l is the eigenvalue that tells you about the magnitude of the angular momentum. You will have several possibilities for m. There will be several states that have the same l, but m different. So for example you'll have l equal 0, in which case m must be equal to 0. But if you choose state with l equals 1, or eigenfunctions with l equal 1, there is the possibility of having m equals minus 1, 0, or 1. So are three waves functions in that case. Psi 1, minus 1, psi 1, 0, and psi 1, 1. So in general when we choose a general l, if you choose an arbitrary l, then m goes from minus l, minus l plus 1 all the way up to l. These are all the values which are 2l plus 1 values. 2l and the 0 value in between. So it's 2l plus 1 values. The quantization in some sense is done now. And let me recap about these functions now. We mentioned up there that the y l m's are the objects. The spherical harmonicas are going to be those wave functions. And they have a normalization, n l m, an exponention, and all that. So let me write, just for the record, what a y l m looks like with all the constants. Well, the normalization constant is complicated. And it's kind of a thing you can never remember by heart. It would be pointless. OK. All of that. Then a minus 1 to the m seems useful. e to the i m phi, p l m of cosine theta. And this is all valid for 0 less 0 m positive m. When you have negative m you must do a little variation for m less than 0 y l m of theta and phi is minus 1 to the m y l minus m of theta and phi complex conjugated. Well, if m is negative, minus m is positive. So you know what that is. So you could plug this whole mess here. I don't advise it. It's just for the record. These polynomials are complicated, but they are normalized nicely. And we just need to understand what it means to be normalized nicely. That is important for us. The specific forms of these polynomials we can find them. The only one I really remember is that y 0 0 is a constant. It's 1 over 4 pi. That's simple enough. No dependents. l equals 0, m equals 0. Here is another one. y1 plus minus 1 is minus plus square root of 3 over 8 pi e to the plus minus i phi sine theta. And the last one, so we're giving all the spherical harmonics with l equals 1. So with l equals 1 remember we mentioned that you would have three values of m. Here they are. Plus or minus 1 and 0.

MIT_804_Quantum_Physics_I_Spring_2016

Scattered_wave_and_phase_shift.txt

PROFESSOR: Let's look at the magnitude squared of those waves that we've already defined here. We have two solutions, one for no potential and one for a real potential. Both are for some finite range potential. We have phi of x squared is equal to sine squared of kx. And psi of x squared is equal to sine squared of kx plus delta. Even from this information you can get something. Think of x plotted here. Here's x equals 0. And there's this sine squared. This wave for phi of x squared. Suppose you're looking at some feature-- a maximum, a minimum-- of this function. Suppose the feature happens when the argument, kx, is equal to some number, a0. Whatever feature-- this number a0 could be 0, in which case you're looking at a minimum, it could be pi over 2, in which case you're looking at a maximum-- some feature of sine squared. Well the same feature will appear in this case when the whole argument is equal to a0. So while this one happens at x equals a0 over k, here it happens at x still equals a0 over k minus delta over k. If this is the probability density associated to the solution for no potential and it has a maximum here, the maximum of the true solution-- say, here-- would appear at a distance equal to delta over k. Earlier-- so this is like the x, and this is like the x tilde-- that feature would appear, delta over k in that direction. So this is psi. This is psi squared. So we conclude, for example, that when delta is greater than 0, the wave is pulled. Delta equals 0, the two shapes are on top of each other. For delta different from 0, the wave function is pulled in. So delta greater than 0, psi is pulled in. What could we think of this? The potential is attractive. It's pulling in the wave function. Attractive. Delta less than 0, the wave is pushed out. It would be in the other direction, and the psi is pushed out. Potential is repulsive. So a little bit of information even from the signs of this thing. We want to define one last thing, and then we'll stop. It's the concept of the scattered wave. What should we call the scattered wave? We will define the scattered wave psi s as the extra piece in the solution-- the psi solution-- that would vanish without potential. So we say, you have a psi, but if you didn't have a potential, what part of this psi would survive? Think of writing the psi of x as the solution without the potential plus the extra part, the scattered wave, psi x. So this is the definition. The full scattering solution, the full solution when you have a potential, can be written in a solution without the potential and this scattered thing. Now, you may remember-- we just did it a second ago-- that this original solution and the psi solution have the same incoming wave. The incoming wave up there is the same for the psi solution as for this one. So the incoming waves are the same. So only the outgoing waves are different. And this represents how much more of an outgoing wave you get than from what you would have gotten with psi. So this must be an outgoing wave. We'll just plug in the formula here. psi s is equal to psi minus phi. And it's equal to 1 over 2i e to the ikx plus 2i delta minus e to the minus ikx minus 1 over 2i-- the phi-- into the ikx minus e to the minus ikx. So the incoming waves wee the same. Indeed, they cancelled. But the outgoing waves are not. You can factor an e to the ikx, and you get e to the 2i delta minus 1. Which is equal to e to the ikx e to the i delta times sine delta. There we go. We have the answer for the scattered wave. It's proportional to sine delta, which, again, makes sense. If delta is equal to 0, there is no scattering. It's an outgoing wave and all is good. So I'll write it like this. psi s is equal to As e to the ikx, with As equal to e to the i delta sine delta.

MIT_804_Quantum_Physics_I_Spring_2016

Quantization_of_the_energy.txt

PROFESSOR: We have to ask what happens here? This series for h of u doesn't seem to stop. You go a 0, a 2, a 4. Well, it could go on forever. And what would happen if it goes on forever? So if it goes on forever, let's calculate what this aj plus 2 over aj as j goes to infinity. Let's see how the coefficients vary as you go higher and higher up in the polynomial. That should be an interesting thing. So I pass the aj that is on the right side and divide it, and now on the right-hand side there's just this product of factors. And as j goes to infinity, it's much larger than 1 or e, whatever it is, and the 2 and the 1 in the denominator. So this goes like 2j over j squared. And this goes roughly like 2 over j. So as you go higher and higher up, by the time j is a billion, the next term is 2 divided by a billion. And they are decaying, which is good, but they're not decaying fast enough. That's a problem. So let's try to figure out if we know of a function that decays in a similar way. So you could do it some other way. I'll do it this way. e to the u squared-- let's look at this function-- is this sum from n equals 0 to infinity 1 over n u squared to the n. So it's u to the 2n-- 1 over n factorial, sorry. So now, since we have j's and they jump by twos, these exponents also here jump by two. So that's about right. So let's think of 2n being j, and therefore this becomes a sum where j equals 0, 2, 4, and all that of 1 over-- so even j's. 2n is equal to j-- so j over 2 factorial over 1, and then you have u to the j. So if I think of this as some coefficient c sub j times u to the j, we've learned that c sub j is equal to 1 over j divided by two factorial. In which case, if that is true, let's try to see what this cj plus 2 over cj-- the ratio of two consecutive coefficients in this series. Well, cj plus 2 would be j plus 2 over 2 factorial, like this. That's the numerator, because of that formula. And the denominator would have just j over 2 factorial. Now, these factorials make sense. You don't have to worry that they are factorials of halves, because j is even. And therefore, the numerators are even-- divided by 2. These are integers. These are ordinary factorials. There are factorials of fractional numbers. You've seen them probably in statistical physics and other fields, but we don't have those here. This is another thing. So this cancels. If you have a number and the number plus 1, which is here, you get j plus 2 over 2, which is 2 over j plus 2. And that's when j is largest, just 2 over j, which is exactly what we have here. So this supposedly nice, innocent function, polynomial here-- if it doesn't truncate, if this recursive relation keeps producing more and more and more terms forever-- will diverge. And it will diverge like so. If the series does not truncate, h of u will diverge like e to the u squared. Needless to say, that's a disaster. Because, first, it's kind of interesting to see that here, yes, you have a safety factor, e to the minus u squared over 2. But if h of u diverges like e to the u squared, you're still in trouble. e to the u squared minus u squared over 2 is e to the plus u squared over 2. And it actually coincides with what we learned before, that any solution goes like either plus or minus u squared over 2. So if h of u doesn't truncate and doesn't become a polynomial, it will diverge like e to the u squared, and this solution will diverge like e to the plus u squared over 2, which was a possibility. And it will not be normalizable. So that's basically the gist of the argument. This differential equation-- whenever you work with arbitrary energies, there's no reason why the series will stop. Because e there will have to be equal to 2j plus 1, which is an integer. So unless je is an integer, it will not stop, and then you'll have a divergent-- well, not divergent; unbounded-- far of u that is impossible to normalize. So the requirement that the solution be normalizable quantizes the energy. It's a very nice effect of a differential equation. It's very nice that you can see it without doing numerical experiments, that what's going on here is an absolute requirement that this series terminates. So here, phi of u would go like e to the u squared over 2, what we mentioned there, and it's not a solution. So if the series must terminate, the numerator on that box equation must be 0 for some value of j, and therefore there must exist a j such that 2j plus 1 is equal to the energy. So basically, what this means is that these unit-free energies must be an odd integer. So in this case, this can be true for j equals 0, 1, 2, 3. In each case, it will terminate the series. With j equals 0, 1, 2, or 3 there, you get some values of e that the series will terminate. And when this series terminates, aj plus 2 is equal to 0. Because look at your box equation. aj, you got her number, and then suddenly you get this 2j plus 1 minus e. And if that's 0, the next one is zero. So, yes, you get something interesting even for j equals 0. Because in that case, you can have a0, but you will have no a2, just the constant. So I will write it. So if aj plus 2 is equal to zero, h of u will be aj u to the j plus aj minus 2u to the j minus 2. And it goes down. The last coefficient that exists is aj, and then you go down by two's. So let's use the typical notation. We call j equals n, and then the energy is 2n plus 1. The h is an u to the n plus an minus 2. You do the n minus 2, and it goes on. If n is even, it's an even solution. If n is odd, it's an odd solution. And the energy e, remember, was h omega over 2 times e-- so 2n plus 1. So we'll move the 2 in, and e will be equal to h omega n plus 1/2. And n in all these solutions goes from 0, 1, 2, 3. We can call this the energy en. So here you see another well-known, famous fact that energy levels are all evenly spaced, h omega over 2, one by one by one-- except that there's even an offset for n equals 0, which is supposed to be the lowest energy state of the oscillator. You still have a 1/2 h bar omega. This is just saying that if you have the potential, the ground state is already a little bit up. You would expect that-- you know there's no solutions with energy below the lowest point of the potential. But the first solution has to be a little bit up. So it's here and then they're all evenly spaced. And this begins with E0; for n equals 0, e1. And there's a little bit of notational issues. We used to call the ground state energy sometimes e1, e2, e3, going up, but this time it is very natural to call it E0 because it corresponds to n equals 0. Sorry. Those things happen. No, it's not an approximation. It's really, in a sense, the following statement. Let me remind everybody of that statement. When you have even or odd solutions, you can produce a solution that you may say it's a superposition, but it will not be an energy eigenstate anymore. Because the even solution that stops, say, at u to the 6 has some energy, and the odd solution has a different energy. So these are different energy eigenstates. So the energy eigenstates, we prove for one-dimensional potentials, are not chosen to be even or odd for bound states. They are either even or odd. You see, a superposition-- how do we say like that? Here we have it. If this coefficient is even, the energy sum value-- if this coefficient is odd, the energy will be different. And two energy eigenstates with different energies, the sum is not an energy eigenstate. You can construct the general solution by superimposing, but that would be general solutions of the full time-dependent Schrodinger equation, not of the energy eigenstates. The equation we're aiming to solve there is a solution for energy eigenstates. And although this concept I can see now from the questions where you're getting, it's a subtle statement. Our statement was, from quantum mechanics, that when we would solve a symmetric potential, the bound states would turn out to be either even or odd. It's not an approximation. It's not a choice. It's something forced on you. Each time you find the bound state, it's either even or it's odd, and this turned out to be this case. You would have said the general solution is a superposition, but that's not true. Because if you put a superposition, the energy will truncate one of them but will not truncate the other series. So one will be bad. It will do nothing. So if this point is not completely clear, please insist later, insist in recitation. Come back to me office hours. This point should be eventually clear. Good. So what are the names of these things? These are called Hermite polynomials. And so back to the differential equation, let's look at the differential equations when e is equal to 2n plus 1. Go back to the differential equation, and we'll write d second du squared Hn of u. That will be called the Hermite polynomial, n minus 2udHn du plus e minus 1. But e is 2n plus 1 minus 1 is 2n Hn of u is equal to 0. This is the Hermite's differential equation. And the Hn's are Hermite polynomials, which, conventionally, for purposes of doing your algebra nicely, people figured out that Hn of u is convenient if-- it begins with u to the n and then it continues down u to the n minus 2 and all these ones here. But here people like it when it's 2 to the n, u to the n-- a normalization. So we know the leading term must be u to the n. If you truncate with j, you've got u to the j. You truncate with n, you get u to the n. Since this is a linear differential equation, the coefficient in front is your choice. And people's choice has been that one and has been followed. A few Hermite polynomials, just a list. H0 is just 1. H1 is 2u. H2 is 4u squared minus 2. H3 is our last one, 8u cubed minus 12u, I think. I have a little typo here. Maybe it's wrong. So you want to generate more Hermite polynomials, here is a neat way that is used sometimes. And these, too, are generating functional. It's very nice actually. You will have in some homework a little discussion. Look, you put the variable z over there. What is z having to do with anything? u we know, but z, why? Well, z is that formal variable for what is called the generating function. So it's equal to the sum from n equals 0 to infinity. And you expand it kind of like an exponential, zn over n factorial. But there will be functions of u all over there. If you expand this exponential, you have an infinite series, and then you have to collect terms by powers of z. And if you have a z to the 8, you might have gotten from this to the fourth, but you might have gotten it from this to the 3 and then two factors of this term squared or a cross-product. So after all here, there will be some function of u, and that function is called the Hermite polynomial. So if you expand this with Mathematica, say, and collect in terms of u, you will generate the Hermite polynomials. With this formula, it's kind of not that difficult to see that the Hermite polynomial begins in this way. And how do you check this is true? Well, you would have to show that such polynomials satisfy that differential equation, and that's easier than what it seems. It might seem difficult, but it's just a few lines. Now, I want you to feel comfortable enough with this, so let me wrap it up, the solutions, and remind you, well, you had always u but you cared about x. So u was x over a. So let's look at our wave functions. Our wave functions phi n of x will be the Hermite polynomial n of u, which is of x over a, times e to the minus u squared over 2, which is minus x squared over 2a squared. And you should remember that a squared is h bar over m omega. So all kinds of funny factors-- in particular, this exponential is e to the minus x squared m omega over h squared over 2. I think so-- m omega over 2h bar. Let me write it differently-- m omega over 2h bar x squared. That's that exponential, and those are the coefficients. And here there should be a normalization constant, which I will not write. It's a little messy. And those are the solutions. And the energies en were h bar omega over 2 n plus 1/2, so E0 is equal to h bar omega over 2. E1 is 3/2 of h bar omega, and it just goes on like that.

MIT_804_Quantum_Physics_I_Spring_2016

Halfwidth_and_time_delay.txt

PROFESSOR: What is gamma? Well, it seems to have a beta in there, alpha. It's a little unclear what gamma means, so let's try to make sense of it. Gamma has units of energy, because this term has units of energy, and therefore the other term must have units of energy as well. So gamma has units of energy. And from a unit of energy you could get a time. So tau, you can define it as h over gamma. That has units of time. h bar has units of energy times time, so if you divide by energy, you get units of time. And what is this number? So this is some time. And it's equal to m over 2 alpha beta h bar. We have this guy. So the natural thing, of course, is to compare to the time delay associated due to this process that we've been studying. So all the time delay, delta t, was 2 h bar dE-- no, d delta dE. And this is 2 h bar dk dE times d delta dk. Now, d delta dk, we calculated it up on the blackboard. So this was equal to-- this had been calculated as 1 over beta at resonance. At the resonance, this time delay is calculated above, and it's equal to 1 over beta. This thing is also easily calculated, because this is 1 over dE dk, which is h bar squared k over m. Therefore, what do we get? 2 h bar divided by h bar squared. k, you're at rest so that's alpha over m, m here. And d delta dk is a beta. So what do we get here? Delta t is 2m over h bar alpha beta. And compare with this one, it's 4 times tau. So h over-- so the end conclusion is that h bar over gamma, which is a time, is tau, which is equal to delta t over 4. So that's the intuition for the half width. So in the distribution of this scattering amplitude, there's a half width. It's an energy distribution. And there's a time associated with it, which is h bar over the half width. And, being a time, it must be related to some physical time that has a period, and there's nothing else than the delay. So if the delay is large, gamma is small, and the width is small. It's a narrow resonance. So a narrow resonance is one in which the width is very small. So this has enormous applications. It's used in nuclear physics all the time. It's used in particle physics as well. The Higgs-Boson was discovered over a year ago, it's a particle but it's unstable. It decays very fast. If you thought in terms of that, it just gets created by a reaction, stays in the well for a little while, and poof, disappears. So its mathematics in cross-sections is governed by resonances. So we call the Higgs particle a particle, but anything that is unstable it's more like, more precisely viewed as an object that represents a resonance. The Higgs particle doesn't live too long. Lives about 10 to the minus 22 seconds. Very little time. And then it decays and it goes into b-bbar, bottom bottom bar quark. As it goes into z's, it goes into tau lambda. So it goes into two photons. It can decay into all these things. So in the case of the Higgs-Boson, the center energy of the resonance is observed in scattering amplitudes at a center energy E alpha of 125 GEv. The width is very small. In fact, the width is about gamma, it's about 4 MEv plus minus 5%. 4 MEv, that's very little, because an MEv is 1/1000 of a GEv. So it's a very narrow resonance. And from this gamma, you can get a time. And the time is lifetime. It's about 10 to the minus 22 seconds. So this is the language that we use to describe any unstable particle. We think of it as resonance. It's sometimes called resonance. And the title of papers that appear at that time is the resonance observed in the data, the Higgs-Boson. And the answer was yes. This was observed as a resonance. It is a scattering amplitude that is uniform, and it has some energy, has a little peak. You can imagine people can't quite measure the width directly. It's too narrow. So the plots don't show the width, but the width is implied, and there's related calculations over here with the lambda. So it's all a nice story. OK. So look what we've done. We began by discussing that we could not get time delays that were negative. Times advance, then we get long time delays, and that's a resonance. We showed that they correspond to changes, rapid changes of delta in the positive direction. And we've modeled them so that in general, we have a general description of what's happening near a resonance.

MIT_804_Quantum_Physics_I_Spring_2016

Finite_square_well_Setting_up_the_problem.txt

BARTON ZWIEBACH: Finite square well. So this brings us also to a little common aside. So far, we could find every solution. Now we're going to write the equations for the finite square well, and we're not going to be able to find the solution. But we're going to understand the solution. So you're going to enjoy a little-- mathematicians usually say it's the most important thing, understanding the solution. Finding it, it's no big deal. But we're physicists as well. So we sometimes have to find the solutions. Even if we don't understand them very well, it's nice to find them. And then you're going to use numerical methods, and this is the part of the course where you're going to be using numerical methods a lot. Here is the finite square well, and now we draw it symmetrically. Here it is. Here is x. We're drawing the potential V of x. It extends from a to minus a. It's 0. This is the 0 of the potential. It's here. And it goes down. The potential is negative here. Its value is minus V0 with V0 positive. And then a possible energy for a bound state-- we're going to look for bound states. Bound states are normalizable states, normalizable solutions of the Schrodinger equation. So we're going to look for them. Look for bound states. And they have energy less than 0. So something that has energy less than 0 is bound. You would have to give it some energy to put it at 0 energy, and the particle could escape. The other thing about the bound state is that it will be some probability to find it here. Very little probability to find it in the forbidden region. And this energy, which is negative, it's somewhere here. We don't know what are the possible energies, but we will assume there's some energy that is negative, that is there. And that's the energy of the solution. Let me write the equation again. The equation is psi double prime is minus 2m over h bar squared, E minus V of x times psi. This is the Schrodinger equation. You recognize it if you multiply h squared here, 2m there. And we've been solving already two examples where there was no potential, but finally, there is a potential. So this is the energy of the particle. The energy of the particle can be interpreted as the potential energy plus the kinetic energy. Think intuitively here. If you have some energy over here-- this is the potential energy. Well, all this much is kinetic energy. On the other hand, in this region, the energy is smaller than the potential energy. So it has negative kinetic energy, which is classically not understandable. And in quantum mechanics, it just will have some probability of being here, but that probability will go down and eventually go to 0 in such a way that the wave function is normalizable. So this is a very mysterious thing that happens here, that the wave function will not be just over here, but it will leak. And it leaks because there's a finite discontinuity. If you take the barrier to be infinitely high, it would leak so little that eventually it would not leak. The wave function would vanish there. And it will be the end of the story, and you're back to the infinite square well. So the infinite square well is a limit of this as V0 goes to minus infinity. OK, so a few numbers we can put in this graph. E, and this-- what is the energy difference here? It's E minus minus V0 , which is V0 plus E. And many times because E is negative, we'll write it as V0 minus absolute value of E. A negative number is equal to minus its absolute value. So how about this whole constant here? Well, my tongue slipped, and I said this constant, but it's V of x. So what do you mean a constant? Well, the potential is piecewise constant. So actually, we are not in such difficult situation because in this region, the potential is a constant. In this region, the potential is a constant. So this is the constant here, and we wish to understand what it is and what are the sines of this constant. This constant, that we call alpha-- see-- is going to be-- let's see what it is, the different circumstances. Suppose you are here. The energy is bigger than the potential. So energy minus the potential is positive, and the constant is negative. So alpha is negative for x less than a. If it's negative and it's a constant, you're going to have trigonometric solutions for x, absolute value less than a, so trig. On the other region, alpha will be positive for x greater than a, and you will have real exponentials of-- I'll just write exponentials. So E to the minus 3x, E to the 5x, things like that. This is the difference between trigonometric and real exponential solutions depend on the signs. So we're going to have to impose boundary conditions as well because we're going to solve the equation here inside with one value of alpha and then outside with another value of alpha. And then we're going to match them. So that's how this will go. Now in this process, somehow the energy will be fixed to some value, some allowed values. There's a counting we could do to understand that, and we'll probably do it next time. At this moment, we'll just proceed. But we imagine there must be a quantization because in the limit as V0 goes to infinity and the potential well becomes infinitely deep, you're back to an infinite square well that has quantized energies that we calculated. So that should be quantized, and it should be no problem OK, ready to do some work? Let's-- Yes? AUDIENCE: So when alpha is 0 and you get polynomial solutions, does that case not matter? BARTON ZWIEBACH: When alpha is 0-- well, alpha is going to be never 0. You see, either E is less than the potential or is better than the potential. So you're not going to get alpha equals 0. You could say, do we have some energy maybe here or here? Well, you could have those energies, in which case you will see what happens. It's not quite polynomial solutions. There's one property about potentials that actually is in the homework due this week, which is that there cannot be solutions of the Schrodinger equation with less energy than the minimum of the potential. You have to have more energy than the minimum of the potential. So alpha equal to 0 is not going to show up in our analysis. OK, so let's begin. Moreover, we have to use this result. The potential is symmetric, completely symmetric, so there are going to be even solutions and odd solutions. So let's consider the even solutions. Solutions. So those are solutions. Psi of minus x is equal to psi of x. Now, you will see how I solve this thing now, and the lesson of all of this is the relevance of unit-free numbers. Unit-free numbers are going to be your best friends in solving these equations. So it will look like I'm not solving anything except making more and more definitions, and all of the sudden, the solution will be there. And it's a power of notation as well. This problem can be very messy if you don't have the right notation. If you were solving it alone, if we would stop the lecture now and you would go home, you would probably find something very messy for quite a bit. And then maybe you clean it up little by little, and eventually, something nice shows up. So here's what we're going to do. We're solving for x in between a and minus a, even solutions. So what does the equation look like? d second psi d x squared is minus 2m over h squared E minus V of x-- in that region, the potential is minus V0. So this is minus 2m over h squared, V0 plus E, which I write as minus absolute value of E, psi. And I forgot a psi here. OK. V0 is positive. Minus V0 is going down, and V0 minus the absolute value of E is positive. So this whole quantity over here is positive. And here it comes, first definition. I will define k squared to be that quantity-- 2m over h squared V0 minus absolute value of E. And that's greater than 0 in this region, as we're trying to solve, and I'll call this equation one. And the differential equation has become psi double prime is minus k psi. So the solutions are simple. It's trigonometrics, as we said, and the only solution-- k squared, I'm sorry. k squared-- and the only solution that is possible, because it's a symmetric thing, is cosine of kx. So psi of x is going to be cosine of kx, and that will hold from a to minus a. End of story, actually, for that part of the potential. You could ask-- one second. You could say, you're going to normalize these things, aren't you? Well, the fact that I won't normalize them, it's just a lot of work, and it's a little messy. But that's no problem. You see, you're finding energy eigenstates, and by definition, solving the differential equation is not going to give you a normalization. So this is a good solution of the differential equation, and let's leave it at that. This we'll call solution two. How about the region x greater than a? Well, what does the differential equation look like? Well, it looks like psi double prime is equal to minus 2, m over h squared, E psi because the potential is 0. V0 outside. x equals a. And here again, you want to look at the equation and know the sign. So maybe better to put the absolute value here. So this is 2m absolute value of E over h squared psi. And one more definition-- kappa squared is going to be 2mE over h squared. That's equation number three. The equation has become psi double prime equals kappa squared psi, and the solutions for that are exponentials. Solutions are psi-- goes like E to the plus minus kappa x. You see the solution we're constructing is symmetric. It's even. So let's worry just about one side. If one side works, the other side will work as well. So I will just write that for psi of x is equal to the minus kappa x. That's a solution for x greater than a. If I just did that, I would be making a mistake. And the reason is that yes, I don't care about the normalization of the wave function, but by not putting a number here, I'm selecting some particular normalization. And The wave function must be continuous and satisfy all these nice things. So yes, here I can maybe not put a constant, but here, already, I must put a constant. It may be needed to match the boundary conditions. I cannot ignore it here. So I must put the constant a that I don't know, and that's going to be equation number four. Now, look at your definitions. k squared for a trigonometric, kappa-- many people use kappa for things that go with exponentials. But look, k squared and kappa squared satisfy a very nice relation. If you add them up-- k squared plus kappa squared-- the energy part cancels, and you get 2mV0, which is positive, over h squared. Well, that's not so bad, but we want to keep defining things. How can I make this really nice? If it didn't have units, it would be much nicer. This is full of units. k times a length has no units, and kappa times a length has no unit. So multiply by a squared, and you have k squared a squared plus kappa squared a squared is equal to 2mV0 a squared over h squared. Now, say, well yes, this looks nice. So let's make the new definitions. So don't lose track of what we are doing. We need to find the energy. That's basically what we want. What are the possible energies? And we already included two constants-- k and kappa- and they have these properties here. So let me define psi to be kappa a, and it will be defined to be positive. Kappa is defined to be positive, and k is defined to be positive. eta, you will define it to be ka. a It's unit-free. No units. And from that equation, now we have eta squared plus psi squared is equal to the right-hand side, which I will call z0 squared. So z0 squared-- that's another definition-- is 2mV0 a squared over h squared. This is the list of your definitions. OK. What did we do? We traded kappa and k that control the behavior of the wave function-- k inside the well, kappa outside the well-- we traded them for eta and psi, unit-free. And a new constant came up, z0. What is z0? Well, z0 is a very interesting constant. It's a number that you can construct out of the parameters of your potential. It involves V0 and the width. If V0 is very large, z0 is large. If the width is very big, z0 is big. If the potential is shallow or very narrow, z0 is small. But the most important thing about z0 is that it will give you the number of bound states. If z0 is very big, it's a very deep potential, we'll have lots of bound states. If z0 is very shallow, there will be one bound state, no more. You will see it today. But z0 controls the number of bound states. And this is its role, and it will be very important-- these dimensionless quantities and number. If somebody says, I have a potential with z0 equals 5, you can tell immediately three bound states or some number of bound states. That's the nice thing about z0. And this is a very nice-looking equation, this equation of a circle in the eta-xi plane.

MIT_804_Quantum_Physics_I_Spring_2016

Is_probability_conserved_Hermiticity_of_the_Hamiltonian.txt

PROFESSOR: Let's do a work check. So main check. If integral psi star x t0, psi x t0 dx is equal to 1 at t equal to t0, as we say there, then it must hold for later times, t greater than t0. This is what we want to check, or verify, or prove. Now, to do it, we're going to take our time. So it's not going to happen in five minutes, not 10 minutes, maybe not even half an hour. Not because it's so difficult. It's because there's so many things that one can say in between that teach you a lot about quantum mechanics. So we're going to take our time here. So we're going to first rewrite it with better notation. So we'll define rho of x and t, which is going to be called the probability density. And it's nothing else than what you would expect, psi star of x and t, psi of x and t. It's a probability density. You know that has the right interpretation, it's psi squared. And that's the kind of thing that integrated over space gives you the total probability. So this is a positive number given by this quantity is called the probability density. Fine. What do we know about this probability density that we're trying to find about its integral? So define next N of t to be the integral of rho of x and t dx. Integrate this probability density throughout space, and that's going to give you N of t. Now, what do we know? We know that N of t, or let's assume that N of t0 is equal to 1. N is that normalization. It's that total integral of the probability what had to be equal to 1. Well, let's assume N at t0 is equal to 1. That's good. The question is, will the Schrodinger equation guarantee that-- and here's the claim-- dN dt is equal to 0? Will the Schrodinger equation guarantee this? If the Schrodinger equation guarantees that this derivative is, indeed, zero, then we're in good business. Because the derivative is zero, the value's 1, will remain 1 forever. Yes? AUDIENCE: May I ask why you specified for t greater than t0? Well, I don't have to specify for t greater than t naught. I could do it for all t different than t naught. But if I say this way, as imagining that somebody prepares the system at some time, t naught, and maybe the system didn't exist for other times below. Now, if a system existed for long time and you look at it at t naught, then certainly the Schrodinger equation should imply that it works later and it works before. So it's not really necessary, but no loss of generality. OK, so that's it. Will it guarantee that? Well, that's our thing to do. So let's begin the work by doing a little bit of a calculation. And so what do we need to do? We need to find the derivative of this quantity. So what is this derivative of N dN dt will be the integral d dt of rho of x and t dx. So I went here and brought in the d dt, which became a partial derivative. Because this is just a function of t, but inside here, there's a function of t and a function of x. So I must make clear that I'm just differentiating t. So is d dt of rho. And now we can write it as integral dx. What this rho? Psi star psi. So we would have d dt of psi star times psi plus psi star d dt of psi. OK. And here you see, if you were waiting for that, that the Schrodinger equation has to be necessary. Because we have the psi dt. And that information is there with Schrodinger's equation. So let's do that. So what do we have? ih bar d psi dt equal h psi. We'll write it like that for the time being without copying all what h is. That would take a lot of time. And from this equation, you can find immediately that d psi dt is minus i over h bar h hat psi. Now we need to complex conjugate this equation, and that is always a little more scary. Actually, the way to do this in a way that you never get into scary or strange things. So let me take the complex conjugate of this equation. Here I would have i goes to minus i h bar, and now I would have-- we can go very slow-- d psi dt star equals, and then I'll be simple minded here. I think it's the best. I'll just start the right hand side. I start the left hand side and start the right hand side. Now here, the complex conjugate of a derivative, in this case I want to clarify what it is. It's just the derivative of the complex conjugate. So this is minus ih bar d/dt of psi star equals h hat psi star, that's fine. And from here, if I multiply again by i divided by h bar, we get d psi star dt is equal to i over h star h hat psi star. We obtain this useful formula and this useful formula, and both go into our calculation of dN dt. So what do we have here? dN dt equals integral dx, and I will put an i over h bar, I think, here. Yes. i over h bar. Look at this term first. We have i over h bar, h psi star psi. And the second term involves a d psi dt that comes with an opposite sign. Same factor of i over h bar, so minus psi star h psi. So the virtue of what we've done so far is that it doesn't look so bad yet. And looks relatively clean, and it's very suggestive, actually. So what's happening? We want to show that dN dt is equal to 0. Now, are we going to be able to show that simply that to do a lot of algebra and say, oh, it's 0? Well, it's kind of going to work that way, but we're going to do the work and we're going to get to dN dt being an integral of something. And it's just not going to look like 0, but it will be manipulated in such a way that you can argue it's 0 using the boundary condition. So it's kind of interesting how it's going to work. But here structurally, you see what must happen for this calculation to succeed. So we need for this to be 0. We need the following thing to happen. The integral of h hat psi star psi be equal to the integral of psi star h psi. And I should write the dx's. They are there. So this would guarantee that dN dt is equal to 0. So that's a very nice statement, and it's kind of nice is that you have one function starred, one function non-starred. The h is where the function needs to be starred, but on the other side of the equation, the h is on the other side. So you've kind of moved the h from the complex conjugated function to the non-complex conjugated function. From the first function to this second function. And that's a very nice thing to demand of the Hamiltonian. So actually what seems to be happening is that this conservation of probability will work if your Hamiltonian is good enough to do something like this. And this is a nice formula, it's a famous formula. This is true if H is a Hermitian operator. It's a very interesting new name that shows up that an operator being Hermitian. So this is what I was promising you, that we're going to do this, and we're going to be learning all kinds of funny things as it happens. So what is it for a Hermitian operator? Well, a Hermitian operator, H, would actually satisfy the following. That the integral, H psi 1 star psi 2 is equal to the integral of psi 1 star H psi 2. So an operator is said to be Hermitian if you can move it from the first part to the second part in this sense, and with two different functions. So this should be possible to do if an operator is to be called Hermitian. Now, of course, if it holds for two arbitrary functions, it holds when the two functions are the same, in this case. So what we need is a particular case of the condition of hermiticity. Hermiticity simply means that the operator does this thing. Any two functions that you put here, this equality is true. Now if you ask yourself, how do I even understand that? What allows me to move the H from one side to the other? We'll see it very soon. But it's the fact that H has second derivatives, and maybe you can integrate them by parts and move the derivatives from the psi 1 to the psi 2, and do all kinds of things. But you should try to think at this moment structurally, what kind of objects you have, what kind of properties you have. And the objects are this operator that controls the time evolution, called the Hamiltonian. And if I want probability interpretation to make sense, we need this equality, which is a consequence of hermiticity. Now, I'll maybe use a little of this blackboard. I haven't used it much before. In terms of Hermitian operators, I'm almost there with a definition of a Hermitian operator. I haven't quite given it to you, but let's let state it, given that we're already in this discussion of hermiticity. So this is what is called the Hermitian operator, does that. But in general, rho, given an operator T, one defines its hermitian conjugate P dagger as follows. So you have the integral of psi 1 star T psi 2, and that must be rearranged until it looks like T dagger psi 1 star psi 2. Now, these things are the beginning of a whole set of ideas that are terribly important in quantum mechanics. Hermitian operators, or eigenvalues and eigenvectors. So it's going to take a little time for you to get accustomed to them. But this is the beginning. You will explore a little bit of these things in future homework, and start getting familiar. For now, it looks very strange and unmotivated. Maybe you will see that that will change soon, even throughout today's lecture. So this is the Hermitian conjugate. So if you want to calculate the Hermitian conjugate, you must start with this thing, and start doing manipulations to clean up the psi 2, have nothing at the psi 2, everything acting on psi 1, and that thing is called the dagger. And then finally, T is Hermitian if T dagger is equal to T. So its Hermitian conjugate is itself. It's almost like people say a real number is a number whose complex conjugate is equal to itself. So a Hermitian operator is one whose Hermitian conjugate is equal to itself, and you see if T is Hermitian, well then it's back to T and T in both places, which is what we've been saying here. This is a Hermitian operator.

MIT_804_Quantum_Physics_I_Spring_2016

Nondegeneracy_of_bound_states_in_1D_Real_solutions.txt

PROFESSOR: You first are facing the calculation of the energy eigenstate with some arbitrary potential. You probably want to know some of the key features of the wave functions you're going to calculate. So in fact, all of today's lecture is going to be devoted to this intuitive, qualitative insights into the nature of the wave function. So we will discuss a few properties that help us think clearly. And these are two of those properties. I want to begin with them. Then we'll do a third one that we have already used, and we will prove it completely. And then turn to the classical and semi-classical intuition that lets us figure out how the wave function will look. And that's a great help for you. Even if you're solving for your wave function numerically, you always need to know what the answer should look like. And it's ideal if before you calculate, you think about it. And you realize, well, it should have this t properties. And if you find out that those are not true, well, you will learn something about your intuition and see what was wrong with it. So we're talking about one dimensional potentials, time independent potentials. And a first statement that is very important, and you will prove in an exercise after spring break, and that is the fact that one dimensional potentials, when you look at what are called bound states, you never find degeneracies, energy degeneracies. And this is when x extends from minus infinity to infinity. You've seen already, in the case of a particle in a circle, there are degenerate energy eigenstates. But if the potential extends to infinity, there is no such things. Now what is a bound state? A bound state sounds like a complicated concept. But it is not. It really means an energy eigenstate that can be normalized. Now if an energy eigenstate can be normalized and you live in the full real line, that the wave function must go to 0 at infinity. Otherwise you would never be able to normalize it. And if the wave function goes 0 at infinity, the bound state is some sort of bump in the middle region or something like that. And it eventually decays. So this is bound by the potential in some way. And that's basically what we use to define a bound state. We'll take it to be that generally. So this is something, this property, which is very important, is something you will prove. But now we go to another property. We've emphasized forever that the Schrodinger equation is an equation with complex numbers. And the solutions have complex numbers. And suddenly, I wrote a few lectures ago a wave function was real. And I was asked, well, how can it be real? Well, we've discussed stationary states in which the full wave function, capital PSI, is equal to a little psi of x times the exponential of e to the minus i et over h bar. And there in that exponential, there is complex numbers on this little psi of x in front of that exponential, which is what we called basically those energy eigenstates. The e to the minus i et over h bar, it's understood that little psi of x is the thing we've been looking for. And this psi of x solves the time independent Schrodinger equation h psi equal e psi. And that equation has no complex number in it. So little psi of x can be real. And there's no contradiction. Because the full solution to the time dependent Schrodinger equation is complex. But here is a statement. With v of x real, the energy eigenstates can be chosen to be real. And the words can be chosen are very important here. It means that you may find a solution that is complex, but you need not stick to that solution. There is always a possibility to work with real solutions. And what is the way you prove this? This I will put this in the notes. You don't have to worry about the proof. You consider the Schrodinger equation for psi. And you show that psi star, the complex conjugate of psi, solves the same equation that psi solves. And therefore, if psi is a solution, psi star is a solution with the same energy. That part is very important. Therefore, if you have two energy eigenstates with the same energy, you can form the sum. That's still an energy eigenstate with the same energy. Even formed in difference, that's still an energy eigenstate with the same energy. And the sum of psi plus psi star is real. And the difference psi minus psi star, if you divide by 2i, is real as well. Therefore you can construct two solutions, the real part of psi and the imaginary part of psi. And both are solutions to the Schrodinger equation. So I've said in words what is the proof of the first line. It's that if you have a psi, psi star is also a solution. Therefore, psi plus psi star and psi minus psi star are solutions. So given a complex psi, then psi psi of x. Then psi real of x that we define to be psi of x plus psi star of x over 2. And the imaginary part of the wave function 1 over 2i psi of x minus psi star of x are both solutions with the same energy as this one has. So these are the two solutions. So far so good. You don't like to work with complex psi? No need to work with complex psi. Work with real psi. But here comes the second part of the argument, the second sentence. I want you to be alert that the second sentence is very powerful. It says that if you have a bound state of a one dimensional potential, more is true. There are no genuinely complex solutions in this case. Any solution that you will find, it's not that it's complex and then you can find the real and imaginary part. No, any solution that you will find will be basically real. And how can it fail to be real? It just has a complex number in front of it that you can ignore. So it is a very strong statement. That the wave function, it's not that you can choose to work it. You're forced to do it up to a phase. So how is that possible? How is that true? And here is the argument for the second line. If we're talking bound states, then these two are real solutions with the same energy. So now suppose these are bound states. There is a problem if there are two real solutions with the same energy. They would be degenerate. And property number 1 says there's no such thing as degenerate energy bound states. So they cannot be degenerate. So if you start with a complex psi, and you build these two, they must be the same solution. Because since there are no degenerate bound states, then psi, I will write it as psi imaginary, of x must be proportional to psi real of x. And both are real, so the only possibility is that they are equal up to a constant, where the constant is a real constant. You see there cannot be degenerate bound states. So the two tentative solutions must be the same. But that means that the original solution, psi, which is by definition the real part plus i times the imaginary part, is now equal to psi r plus i times c times psi r again, which is 1 plus ic times psi r. And that is basically the content of the theorem. Any solution is up to a number, just the real solution. So you're not going to find the real solution has non-trivial different real imaginary parts here. No, just the real solution and a complex number. Now if you want, you can just write this as e to the i argument of 1 plus ic times square root of 1 plus c squared psi r. And then it's literally the way it's said here. The wave function is proportional to a real wave function up to a phase. So that's a very neat situation. And therefore, you should not be worried that we are going to have to assume many times in our analysis that the bound states were trying to look for are real. And we plot real bound states. And we don't have to worry about, what are you plotting? The real part? The imaginary part? Many times we can just work with real things.

MIT_804_Quantum_Physics_I_Spring_2016

Levinsons_theorem_part_2.txt

PROFESSOR: We have two equations now relating this number of states. And now you can say, oh, OK, so I look at the k line. And I look at the little piece of the k and say, oh, how many states were there with 0 potential, some number, first blackboard. How many states are there now with some potential, some other number? It has changed. For every-- because these two equations, the n for equal dk, the n is not equal to the dn prime. In one case, the energy levels or the momentum levels are more compressed or more separated, but whatever it is, whatever the sign is, there is a little discrepancy. So both of them are giving me the total number of positive energy states in the little dk. Case So if I take the difference, I will get some information. So I would say the following, if I want to calculate the number of positive energy solutions and now I think the following, I take the potential V equals 0 and slowly but surely deepen it, push it, and do things and create the potential V of x slowly, slow the formation. In this process, I can look at a little interval dk and tell how many states are positive energy states I lost. So if, for example, dn is bigger than dn prime, dn equal 5 and dn prime is equal to 3, I started with 5 positive energy states in this little interval and by the time I change the potential I ended up with 3. So I lost 2. So let me write here the number of positive energy solutions lost in the interval dk as the potential is turned on is dn the original number minus the dn prime. If that's positive, I've lost state. If it's negative, I gained state, positive energy states. In this number, we can calculate the difference. This is minus 1 over pi d delta dk dk. I'll put it here. We'll we're not far. We'll this is what you lost. The number of positive energy eigenstates that you lost in little dk. To see how many positive energy states you lost over all, you must integrate over all the dk's and see how much you lost in every little piece. So the number of positive energy solutions lost, not in the dk, but lost as the potential is turned on is equal to the integral over k from 0 to infinity of minus 1 over pi d delta dk is in the way of that expression of that right coincide. But this is a total derivative. So this is minus 1 over pi delta of k evaluated between infinity and 0. And therefore, the number of states lost is 1 over pi, because of the sign down to 0 minus delta of infinity. So we're almost there. This is the number of positive energy solutions lost. Now I want to emphasize that the situation is quite interesting. Let me make a little drawing here. So suppose here is the case where you have the potential equal to 0 and here is energy equal to 0. Then you have all these states. Now even though we've put the wall, the wall allows us to count the states, but there are still going to be an infinite number of states. The infinite square wall has an infinite number of states. So that thing really continues, but what happens by the time v is deferred from 0? Here is that the E equals 0 line and here is the E equals 0 line. As we've discussed, as you change the potential slowly, this are going to shift a little and some are going to go down here, are going to become bound states. They're going to be a number of bound states, N bound states, number of bound states equal N. And then there's going to be still sub states here that's also go to infinity. So you cannot quite say so easily, well, the number of states here minus the number of states here is the number lost. That's not true, because that's infinite, that's infinite, and subtracting infinity is bad. But you know that you've lost a number of finite number of positive energy solutions. So as you track here, the number of states must-- the states must go into each other. And therefore, if these four states are now here, before they were here, and those were the positive energy solutions that were lost, in going from here to here, you lost positive energy solutions. You lost a finite number of positive energy solutions. Even though there's infinite here and infinite here, you lost some. And you did that by keeping track at any place how much you lost. And therefore the states lost are never really lost. They are the ones that became the band states here. So the positive energy states that got lost are the bound states. So the number bound states is equal to the number of positive energy solutions, because there are no lost states. So this is equal to a number of bound states, because there are overall no lost states.

MIT_804_Quantum_Physics_I_Spring_2016

Modelling_a_resonance.txt

PROFESSOR: That's how it looks, a resonance. You can see it basically in the phase shift. And great increase of the phase of almost minutely pi over a very small change of energy. And it should [INAUDIBLE] with a very big [INAUDIBLE].. So this is how it looks. And I want to now proceed, after if there are some questions, of how do we search for residences a little more mathematically rather than plotting them. How could I write an equation for a resonance. Cannot say, oh, the phase changes fast. Well, that's not a very nice way of saying it. It's good. It's intuitive. But we should be able to do better. So how do I find resonances? So let's model resonances a little bit. How do we find resonances? So let's model this behavior. By that is writing a formula that is simple enough that seems to capture what's happening. And that formula's going to inspire us to think of resonances perhaps a little more clearly. So suppose you have a resonance near k equal alpha. I claim the following formula would be a good way to represent the resonance? We would say that tan delta is equal to beta over alpha minus k. Or-- yeah, we would say that. [INAUDIBLE] Or if you wish, delta is tan minus 1 of beta over alpha minus k. Why is that reasonable? It's a little surprising, but not that surprising. You see that-- delta is equal to minus pi over 2. The tangent of delta goes to infinity. So there's something going on here in which you have this property. So let's plot this. So let's plot beta over alpha minus k. You need a clock to understand this. So this is k, and we're plotting this quantity. Well, it's going to go crazy at k equal alpha. That we know. When k is less than alpha, I'm going to assume that alpha and beta are positive. They both have units of k. And when k is less than alpha-- we begin here-- then this denominator is positive, the numerator is positive, ratio is positive. It's small, maybe. And then suddenly, when k reaches alpha it goes to infinity. So it's going to be like that. Now, it actually is true that when k differs by alpha by beta, it reaches value 1. So here is alpha minus beta. That point it reaches value 1. So if I want this thing to be very sharp, I need beta to be small so that it's little until it reaches beta within-- distance beta within alpha, and then it shoots up. So I want beta to be small for sharp behavior. On the other hand here, it goes the other way. It goes from minus infinity back to 0, and has value minus 1 at alpha plus beta. So within minus beta, and beta off of the center alpha, most of the things happen. If we plot now the tangent of this, or the arctangent of this, tan minus 1 of beta, alpha minus k, well, if the tangent of an angle is very little, the angle can be taken to be very little. At this point, it will reach pi over 2, so the angle is little, will go to pi over 2, and then quickly becomes larger than pi over 2, you're thinking tangents. So the tangent is going up, is blowing up at pi over 2. Then continuously, it goes to minus pi over 2, and then continuously goes to 0 so it reaches pi. So this is the behavior of delta. Delta is this tan minus 1 of beta over that. And delta is doing the right thing. It's doing this kind of behavior. There is a shift. I could add a constant here to produce this shift, but it's not important at this moment. The resonance is doing this thing, up to a total shift of pi that doesn't change the tangent of an angle. So this is one way of modeling what's happening to the phase shift near our resonance. So let's explore it a little more. I can do a couple of calculations. For example, I can compute what is d delta dk at k equals alpha. That's should be a nice quantity. Is a derivative of the face. At the resonance, at the position alpha of the resonance. So here's delta, here is k, and there's the derivative of this k. And how should it be? Well, basically, the phase changes by amount pi over a distance beta or 2 beta. So this must be a number divided by beta. You can calculate this derivative from this equation. It's a nice exercise. It's actually just 1 over beta. That's a result. 1 over beta. The other quantity that is nice to understand is how does this scattering amplitude behave near the resonance. So what is the value of as squared? Oh, that's the absolute value of psi s squared, which is sine squared delta. That's the same thing as As squared. Well, you know what is the tangent of delta? A little trigonometric play should be able to do it, and can give you the sin squared delta. And here is the answer. It's beta squared over beta squared plus alpha minus k squared. Kind of a nice, almost bell shape. Of course, it's polynomial, but it looks a little like just a nice symmetric shape around alpha equal k. Now, this division is so famous it has been given a name. It's called the Breit-Wigner distribution. Breit-Wigner. But it's described as the Breit-Wigner distribution, and it's usually referring to terms of energy. Of energy, not momentum. So-- and it's-- what should happen to scattering amplitude in general when you have a resonance. So the way to do this calculation now is to say, well, what is alpha minus k? Let's try to relate it to the energy minus the energy at k equal alpha. Well, this is h squared k squared over 2 m minus h squared alpha squared over 2 m, which is h squared over 2 m, k squared minus alpha squared. On the other hand, I have here alpha minus k squared. I don't have k squared minus alpha squared. So-- approximations. if the resonance is narrow enough, if beta is small, let's do an approximation. We do h squared over-- everybody knows this approximation, shouldn't be afraid of doing it. It's alpha-- how could I write it-- k minus alpha times k plus alpha. And the approximation is that all the interesting thing comes from the difference between k and alpha, how close k is to alpha. So when k is close to alpha, all the dependence is going to be here. This is going to be about 2 alpha when k is near alpha. And if it's a little more than that, it doesn't matter, because it's [INAUDIBLE].. So this could be approximated to 2 alpha, and therefore this becomes h squared alpha over m times k minus alpha. So that's a little help. Then size of s squared, doing a little more of algebra with the constants there. Probably you want to do it with two lines. It's tricky. It's really simple. It's always written in this form-- 1 over 4 gamma squared over e minus e alpha squared plus 1 over four gamma squared, and that's the so-called Breit-Wigner distribution. And gamma is a funny constant here. We'll try to understand it better. 2 alpha beta h squared over m. It has to be something that depends on alpha and beta, because after all, we weren't modeling the resonance with alpha beta. So this curve is very famous. That's the distribution of the scattering amplitude over energies whenever you have a resonance. So we should plot it. You have an e alpha. You have an e alpha plus gamma over 2 and e alpha minus gamma over 2. But the energy minus e alpha is equal to the gamma over 2, you get the gamma squared over 4 so the total amplitude goes down to 1/2 of the usual amplitudes. When the energy is equal to e alpha, you get 1. 1 for psi s squared. But when the energy differs from e alpha by gamma over 2, you get half. So-- actually, I'm not sure of the deflection point, where it is. Probably not there. Or is it there? I don't know. I drew it as if it is there. So that's the distribution, and the width over here is gamma. So gamma is called the width at half power, or at half intensity. Yeah. Width-- the half width of the distribution.

MIT_804_Quantum_Physics_I_Spring_2016

Three_dimensional_current_and_conservation.txt

Three-dimensional case. Now, in the future homework, you will be doing the equivalent of this calculation here with the Laplacians-- it's not complicated-- so that you will derive with the current is. And the current must be a very similar formula as this one. And indeed, I'll just write it here. The current is h bar over m, the imaginary part of psi star. And instead of ddx, you expect the gradient of psi. That is the current for the probability in three dimensions. And the analog of this equation, d rho dt plus dj dx equals 0, is d rho dt plus divergence of j is equal to 0. That is current conservation. Perhaps you do remember that from your study of electromagnetism. That's how Maxwell discovered the displacement current when he tried to figure out how everything was compatible with current conservation. Anyway, that argument I'll do in a second so that it will become clearer. So one last thing here-- it's something also-- you can check the units here of j is 1 over l squared times 1 over t, so probability per unit area and unit time. So what did we have? We were doing the integral of the derivative of the integral given by n. It was over here, dn dt. We worked hard on it. And dn dt was the integral of d rho dt. So it was the integral of d rho dt dx. But we showed now that d rho dt is minus dj dx. So here you have integral from minus infinity to infinity dx of dj dx. And therefore, this is-- I should have a minus sign, because it was minus dj dx. This is minus the current of x equals infinity times p minus the current at x equals minus infinity nt. And as we more or less hinted before, since the current is equal to h over 2im psi star [INAUDIBLE] psi dx minus psi [INAUDIBLE] psi star dx, as you go to plus infinity or minus infinity, these things go to 0 given the boundary conditions that we put. Because psi or psi star go to 0 to infinity, and the derivates are bound at the infinity. So this is 0, dn dt 0. All is good. And two things happened. In the way of doing this, we realized that the computation we have done pretty much established that this is equal to that, because dn dt is the difference of these two integrals, and we showed it's 0. So this is true. And therefore, we suspect h is a Hermitian operator. And the thing that we should do in order to make sure it is is put two different functions here, not two equal functions. This worked for two equal function, but for two different functions, and check it as well. And we'll leave it as an exercise. It's a good exercise. So this shows the consistency. But we discover two important ideas-- one, the existence of a current for probability, and two, h is a Hermitian operator. So last thing is to explain the analogy with current conservation. I think this should help as well. So the interpretation that we'll have is the same as we have in electromagnetism. And there's a complete analogy for everything here. So not for the wave function, but for all these charge densities and current densities. So we have electromagnetism and quantum mechanics. We have rho. Here is the charge density. And here is the probability density. If you have a total charge q in a volume, here is the probability to find the particle in a volume. There is a j in Maxwell's equations as well, and that's a current density. Amber's law has that current. It generates the curl of b. And here is a probability current density. So that's the table. So what I want to make sure is that you understand why these equations, like this or that, are more powerful than just showing that dn dt is 0. They imply a local conservation of probability. You see, there has to be physics of this thing. So the total probability must be 1. But suppose you have the probability distributed over space. There must be some relation between the way the probabilities are varying at one point and varying in other points so that everything is consistent. And those are these differential relations that say that whenever you see a probability density change anywhere, it's because there is some current. And that makes sense. If you see the charge density in some point in space changing, it's because there must be a current. So thanks to the current, you can learn how to interpret the probability much more physically. Because if you ask what is the probability that the particle is from this distance to that distance, you can look at what the currents are at the edges and see whether that probability is increasing or decreasing. So let's see that. Suppose you have a volume, and define the charge inside the volume. Then you say OK, does this charge change in time? Sure, it could. So dq dt is equal to integral d rho dt d cube x over the volume. But d rho dt, by the current conservation equation-- that's the equation we're trying to make sure your intuition is clear about-- this is equal to minus the integral of j-- no, of divergence of j d cube x over the volume. OK. But then Gauss's law. Gauss's Lot tells you that you can relate this divergence to a surface integral. dq dt is therefore minus the surface integral, the area of the current times that. So I'll write it as minus jda, the flux of the current, over the surface that bounds-- this is the volume, and there's the surface bounding it. So by the divergence theorem, it becomes this. And this is how you understand current conservation. You say, look, charge is never created or destroyed. So if you see the charge inside the volume changing, it's because there's some current escaping through the surface. So that's the physical interpretation of that differential equation, of that d rho dt plus divergence of j is equal to 0. This is current conservation. And many people look at this equation and say, what? Current conservation? I don't see anything. But when you look at this equation, you say, oh, yes. The charge changes only because it escapes the volume, not created nor destroyed. So the same thing happens for the probability. Now, let me close up with this statement in one dimension, which is the one you care, at this moment. And on the line, you would have points a and b. And you would say the probability to be within a and b is the integral from a to b dx of rho. That's your probability. That's the integral of psi squared from a to b. Now, what is the time derivative of it? dp ab dt would be integral from a to b dx of d rho dt. But again, for that case, d rho dt is minus dj dx. So this is minus dx dj dx between b and a. And what is that? Well, you get the j at the boundary. So this is minus j at x equals bt minus j at x equal a, t. So simplifying it, you get dp ab dt is equal to minus j at x equals bt plus j of a, t. Let's see if that makes sense. You have been looking for the particle and decided to look at this range from a to b. That's the probability to find it there. We learned already that the total probability to find it anywhere is going to be 1, and that's going to be conserved, and it's going to be no problems. But now let's just ask given what happens to this probability in time. Well, it could change, because the wave function could go up and down. Maybe the wave function was big here and a little later is small so there's less probability to find it here. But now you have another physical variable to help you understand it, and that's the current. That formula we found there for j of x and t in the upper blackboard box formula is a current that can be computed. And here you see if the probability to find the particle in this region changes, it's because some current must be escaping from the edges. And let's see if the formula gives it right. Well, we're assuming quantities are positive if they have plus components in the direction of x. So this current is the current component in the x direction. And it should not be lost-- maybe I didn't quite say it-- that if you are dealing with a divergence of j, this is dj x dx plus dj y dy plus dj z dz. And in the case of one dimension, you will have those, and you get this equation. So it's certainly the reduction. But here you see indeed, if the currents are positive-- if the current at b is positive, there is a current going out. So that tends to reduce the probability. That's why the sign came out with a minus. On the other hand, if there is a current in a, that tends to send in probability, and that's why it increases it here. So the difference between these two currents determines whether the probability here increases or decreases.

MIT_804_Quantum_Physics_I_Spring_2016

Node_Theorem.txt

BARTON ZWIEBACH: The next thing I want to talk about for a few minutes is about the node theorem. Theorem. And it's something we've seen before. We've heard that if you have a one-dimensional potential and you have bound states, the ground state has no nodes. The first excited state has 1 node. Second, 2, 3, 4. All I want to do is give you a little intuition as to why this happens. So this will be an argument that is not mathematically very rigorous, but it's fairly intuitive and it captures the physics of the problem. So it begins by making two observations. So in the node theorem, if you have psi 1, psi 2, psi 3, all energy-- energy-- eigenstates of a one-dimensional potential-- bound states. Bound states. With energy E1 less than E2, less than E3 and E4, psi n has n minus 1 nodes. Those are points where the wave function vanishes inside the range of x. So for this square well, you've proven this by calculating all the energy eigenstates. The first state is the ground state. It has no nodes. The next state is the first excited state. It has one node. And you can write all of them, and we saw that each one has one more node than the next. Now I want to argue that in an arbitrary potential that has bound states, this is also true. So why would that be true for an arbitrary potential? The argument we're going to make is based on continuity. Suppose you have a potential like this-- V of x-- and I want to argue that this potential will have bound states and will have no node, 1 node, 2 nodes, 3 nodes. How could I argue that? Well, I would do the following. Here is the argument. Identify the minimum here. Call this x equals 0. Oh, I want to say one more thing and remind you of another fact that I'm going to use. So this is the first thing, that the square well realizes this theorem, and the second is that psi of x0 being equal to psi prime at x0 being equal to 0 is not possible. The wave function and its derivative cannot vanish at the same point. Please see the notes about this. There is an explanation in last lecture's notes. It is fact that for a second order differential equation, psi and psi prime tell you how to start the solution, and if both psi and psi prime are equal to 0, the general solution of the differential equation is always 0 everywhere. So this kind of thing doesn't happen to a wave function-- the point where it's 0 and the derivative is 0. That never happens. This happens-- 0 wave function with the derivative. But this, no. Never happens. So those two facts. And now let's do the following. Let's invent a new potential. Not this potential, but a new one that I'll mark the point minus a here and the point a here and invent a new potential that is infinite here, infinite there, and has this part I'll write there. So this will be called the screened potential. Screened potential. Va of x in which Va of x is equal to V of x for x less than a, and it's infinity for x greater than a. So that's a potential in which you turn your potential into an infinite square well whose bottom follows the potential. It's not flat. And now, we intuitively argue that as I take a to infinity, the bound states of the screened potentials become the bound states of your original potential. Because when the screen is very, very, very far away, up to infinity, you've got all your potential, and by the time you have bound states that are decaying, so the screen is not going to do much at infinity. And anyway, you can move it even further away. If you move it one light year away or two light years away, shouldn't matter. So the idea is that the bound states-- bound states-- of Va of x as a goes to infinity are the bound states of V of x. And moreover, as you slowly increase the width of the screen, the bound states evolve, but they evolve continuously. At no point a bound state blows up and reappears or does something like that. It just goes continuously. These are physically reasonable, but a mathematician would demand a better explanation. But that's OK. We'll stick to this. So let's continue there. So here is the idea, simply stated. If a is going to 0, if the width of the screen is extremely narrow, you're sitting at the bottom of the potential at x equals 0. And the screened potential is basically a very, very narrow thing, and here, there's the bottom of the potential. And for sufficiently small a-- since you picked the bottom of the potential there-- it's basically flat. And then I can use the states of the infinite square well potential. As a goes to 0, yes, you have a ground state with no nodes, a first excited state with one node, and all the states have the right number of nodes because they are the states of the infinite square well, however narrow it is. So the only thing we have to now show is that if you have a wave function-- say, let's begin with one with no nodes-- as you increase the width of the screen, you cannot get more nodes. It's impossible to change the number of nodes continuously. So here it is. I'm going to do a little diagram. So for example, let's assume the screen is this big at this moment, that you have some ground state like this. You've been growing this, and then as the screen grows bigger, you somehow have maybe a node. Could this have happened? As you increase this screen, you get a node. Now I made it on this point. I didn't intend to do that, so let me do it again somewhere. Do you get a node? Well, here was the original screen, and here the derivative psi prime is negative. Psi prime is negative. On the other hand, psi prime here is already positive. So as you grew this screen, this PSI prime that was here must have turned from negative to positive, the way it looks here. But for that, there must have been a point somewhere here when it was horizontal if it's continuous. And therefore, there must have been some point at which psi and psi prime were both 0 at the endpoint x equals a, whatever the value of a was, because psi prime here is positive, and here is negative. So at some point it was 0, but since it's at the point where you have the infinite square well, psi is also 0. And you would have both psi and psi prime equal 0, which is impossible. So basically, you can't quite flip this and produce a node because you would have to flip here, and you can't do it. One could try to make a very precise, rigorous argument, but if you have another possibility that you might think, well, you have this wave function maybe. And then suddenly it starts doing this, and at some stage, it's going to try to do this. But before it does that, at some point, it will have to be just like this and cross, but at this point, psi and psi prime would be 0. So you can intuitively convince yourself that this thing doesn't allow you to produce a node. So if you start with whatever wave function that has no nodes, as you increase the screen, you just can't produce a node. So the ground state of the whole big potential will have no nodes. And if you start with the first excited state that has one node, as you increase the screen, you still keep one node. So the next state of the full potential will have one node as well. And that way, you argue that all your bound states of the complete potential will just have the right number of nodes, which is 0, 1, 2, 3, 4. And it all came, essentially, from the infinite square well and continued.

MIT_804_Quantum_Physics_I_Spring_2016

MachZehnder_interferometers_and_beam_splitters.txt

PROFESSOR: Mach-Zehnder-- interferometers. And we have a beam splitter. And the beam coming in, it splits into 2. A mirror-- another mirror. The beams are recombined into another beam splitter. And then, 2 beams come out. One to a detector d0-- and a detector d1. We could put here any kind of devices in between. We could put a little piece of glass, which is a phase shifter. We'll discuss it later. But our story is a story of a photon coming in and somehow leaving through the interferometer. And we want to describe this photon in quantum mechanics. And we know that the way to describe it is through a wave function. But this photon can live in either of 2 beams. If a photon was in 1 beam, I could have a number that tells me the probability to be in that beam. But now, it can be in either of 2 beams. Therefore, I will use two numbers. And it seems reasonable to put them in a column vector. Two complex numbers that give me the probability amplitudes-- for this photon to be somewhere. So you could say, oh, look here. What is the probability that we'll find this photon over here? Well, it may depend on the time. I mean, when the photon is gone, it's gone. But when it's crossing here, what is the probability? And I have 2 numbers. What is the probability here, here, here, here? And in fact, you could even have 1 photon that is coming in through 2 different channels, as well. So I have 2 numbers. And I want, now, to do things in a normalized way. So this will be the probability amplitude to be here. This is the probability amplitude to be down. And therefore, the probability to be in the upper one-- you do norm squared. The probability to be in the bottom one, you do norm squared. And you get 1. Must get 1. So if you write 2 numbers, they better satisfy that thing. Otherwise, you are not describing probabilities. On the other hand, I may have a state that is like this. Alpha-- oh, I'll mention other states. State 1-0 is a photon in the upper beam. No probability to be in the lower beam. And state 0-1 is a photon in the lower beam. So these are states. And indeed-- think of superposition. And we have that the state, alpha beta-- you know how to manipulate vectors-- can be written as alpha 1-0. Because the number goes in and becomes alpha 0. Plus beta 0-1. So the state, alpha beta, is a superposition with coefficient alpha of the state in the upper beam plus the superposition with coefficient beta of the state in the lower beam. We also had this little device, which is called the beam shifter of face delta. If the probability amplitude completing is alpha to the left of it, it's alpha e to the i delta to the right of it, with delta a real number. So this is a pure phase. And notice that alpha is equal to e to the i delta. The norm of a complex number doesn't change when you multiply it by a phase. The norm of a complex number times a phase is the norm of the complex number times the norm of the phase. And the norm of any phase is 1. So actually, this doesn't absorb the photon, doesn't generate more photons. It preserves the probability of having a photon there, but it changes the phase. How does the beam splitter work, however? This is the first thing we have to model here. So here is the beam splitter. And you could have a beam coming-- A 1-0 beam hitting it. So nothing coming from below. And something coming from above. And then, it would reflect some and transmit some. And here is a 1-- is the 1 of the 1-0. And here's an s and a t. Which is to mean that this beam splitter takes the 1-0 photon and makes it into an st photon. Because it produces a beam with s up and t down. On the other hand, that same beam splitter-- now, we don't know what those numbers s and t are. That's part of designing a beam splitter. You can ask the engineer what are s and t for the beam splitter. But we are going to figure out what are the constraints. Because no engineer would be able to make a beam splitter with arbitrary s and t. In particular, you already see that if 1-0-- if a photon comes in, probability conservation, there must still be a photon. You need that s squared plus t squared is equal to 1 because that's a photon state. Now, you may also have a photon coming from below and giving you uv. So this would be a 0-1 photon, giving you uv. And therefore, we would say that 0-1-- gives you uv. And you would have u plus v norm squared is equal to 1. So we need, apparently, 4 numbers to characterize the beam splitter. And let's see how we can do that. Well, why do we need, really, 4 numbers? Because of linearity. So let's explore that a little more clearly. And suppose that I ask you, what happens to an alpha beta state-- alpha beta state if it enters a beam splitter? What comes out? Well, the alpha beta state, as you know, is alpha 1-0 plus beta 0-1. And now, we can use our rules. Well, this state, the beam splitter is a linear device. So it will give you alpha times what it makes out of the 1-0. But out of the 1-0 gives you st. And the beta times 0-1 will give you beta uv. So this is alpha s plus beta u times alpha t plus beta v. And I can write this, actually, as alpha beta times the matrix, s u t v. And you get a very nice thing, that the effect of the beam splitter on any photon state, alpha beta, is to multiply it by this matrix, s u t v. So this is the beam splitter. The beam splitter acts on any photon state. And out comes the matrix times the photon state. This is matrix action, something that is going to be pretty important for us. How do we get those numbers? After all, the beam splitter is now determined by these 4 numbers and we don't have enough information. So the manufacturer can tell you that maybe you've got-- you bought a balanced beam splitter. Which means that if you have a beam, half of the intensity goes through, half of the intensity gets reflected. That's a balanced beam splitter. That simplifies things because the intensity here, the probability, [INAUDIBLE] must be the same as that. So each norm squared must be equal to 1/2, if you have a balanced-- beam splitter. And you have s squared equal t squared equal u squared equal v squared equal 1/2. But that's still far from enough to determine s, t, u, and v. So rather than determining, them at this moment, might as well do a guess. So can it be that the beam splitter matrix-- Could it be that the beam splitter matrix is 1 over square root of 2, 1 over square root of 2, 1 over the square root of 2, and 1 over square root of 2. That certainly satisfies all of the properties we've written before. Now, why could it be wrong? Because it could be pluses or minuses or it could be i's or anything there. But maybe this is right. Well, if it is right, the condition that it be right is that, if you take a photon state, 1 photon-- after the beam splitter, you still have 1 photon. So conservation of probability. So if you act on a normalized photon state that satisfies this alpha squared plus beta squared equal 1, it should still give you a normalized photon state. And it should do it for any state. And presumably, if you get any numbers that satisfy that, some engineer will be able to build that beam splitter for you because it doesn't contradict any physical principle. So let's try acting on this with on this state-- 1 over square root of 2, 1 over square root of 2. Let's see. This is normalized-- 1/2 plus 1/2 is 1. So I multiply. I get 1/2 plus 1/2 is 1, and 1. Sorry, this is not normalized. 1 squared plus 1 squared is 2, not 1. So this can't be a beam splitter. No way. We try minus 1 over square root of 2. Actually, if you try this for a few examples, it will work. So how about if we tried in general. So if I try it in general, acting on alpha beta, I would get 1 over square root of 2 alpha plus beta and alpha minus beta. Then, I would check the normalization. So I must do norm of this 1 squared. So it's 1/2 alpha plus beta squared plus 1/2 alpha minus beta norm squared. Well, what is this? Let me go a little slow for a second. [INAUDIBLE] plus beta star. Plus alpha minus beta. Alpha star minus beta star. Well, the cross terms vanish. And alpha alpha star, alpha alpha star, beta beta star, beta beta star add. So you do get alpha squared plus beta squared. And that's 1 by assumption because you started with a photon. So this works. This is a good beam splitter matrix. It does the job. So actually-- Consider this beam splitters. Actually, it's not the unique solution by all means. But we can have 2 beam splitter that differ a little bit. So I'll call beam splitter 1 and beam splitter 2. Beam splitter or 1 will have this matrix. And beam splitter 2 will have the matrix were found here, which is a 1 1 1 minus 1. So both of them work, actually. And both of them are good beam splitters. I call this-- beam splitter 1. And this, beam splitter 2. And we'll keep that. And so we're ready, now, to think about our experiments with the beam splitter.

MIT_804_Quantum_Physics_I_Spring_2016

Consistency_condition_Particle_on_a_circle.txt

PROFESSOR: Let me do a little exercise using still this manipulation. And I'll confirm the way we think about expectations values. So, suppose exercise. Suppose you have indeed that psi is equal to alpha i psi i. Compute the expectation value of Q in the state of psi. Precisely, the expectation value of this operator we've been talking about on the state. So this is equal to the integral dx psi star Q psi. And now I have to put two sums before. And go a little fast here. dx sum over i alpha i psi i star Q sum over j alpha j psi j. No star. This is equal to sum over i sum over j alpha i star alpha j integral dx psi i star Q psi j. But Q psi j is equal to qj psi j. Therefore, this whole thing is equal to qj times the integral dx of psi i star psi j, which is qj delta ij. So here we go. It's equal to sum over i, sum over j, alpha i star alpha j, qj delta ij, which is equal to the sum over i. The j's disappear. And this is alpha i squared qi. That's it. OK. Now you're supposed to look at this and say, yay. Now why is that? Look. How did we define expectation values? We defined it as the sum of the value times the probability that this value have. It's for a random variable. So here our random variable is the result of the measurement. And what are the possible values? qi's. And what are the probabilities that they have Pi? OK. So the expectation value of q should be that, should be the sum of the possible values times their probabilities, and that's what the system gives. This is how we defined expectation value of x. Even though it's expectation value of P. And it all comes from the measurement postulate and the definition. Now, this definition and the measurement postulate just shows that this is what we expect. This is the result of the expectation value. OK. I think I have a nice example. I don't know if I want to go into all the detail of these things, but they illustrate things in a nice way. So let's try to do it. So here it is. It's a physical example. This is a nice concrete example because things work out. So I think we'll actually illustrate some physical points. Example. Particle on a circle. x 0 to L. Maybe you haven't seen a circle described by that, but you take the x-axis, and you say yes, the circle is 0 to L. L and 0. And the way you think of it is that this point is identified with this point. If you have a line and you identify the two endpoints, that's called a circle. It's in the sense of topology. A circle as the set of points equidistant to a center is a geometric description of a round circle. But this, topologically speaking, anything that is closed is topologically a circle. We think of a circle as this, physically, or it could be a curved line that makes it into a circle. But it's not important. Let's consider a free particle on a circle, and suppose the circle has an end L. So x belongs here. And here is the wave function, psi equals 2 over L, 1 over square root of 3 sine of 2 pi x over L, plus 2 over square root of 3 cosine 6 pi x over L. This is the wave function of your particle on a circle. At some time, time equals 0, it's a free particle. No potential. And it lives in the circle, and these functions are kind of interesting. You see, if you live on the circle you would want to emphasize the fact that this point 0 is the same as the point L, so you should have that psi and L must be equal to psi at 0. It's a circle, after all, it's the same point. And therefore for 0 or for L, the difference here is 0 or 2 pi, and the sine is the same thing. And 0, when x equals 0, and 6 pi, so that's also periodic, and it's fine. It's a good wave function result. The question is, for this problem, what are, if you measure momentum, measure momentum, what are the possible values and their probabilities? Probabilities. So you decide to measure momentum of this particle. What can you get? OK. It looks a little nontrivial, and it is a little nontrivial. Momentum. So I must sort of find the momentum eigenstates. Momentum eigenstates, they are those infinite plane waves, e to the ikx, that we could never normalize. Because you square it, it's 1, and the integral over all space is infinite. So are we heading for disaster here? No. Because it lives in a finite space. Yes, you have a question? STUDENT: Should it be a wave function [INAUDIBLE] complex? Because right now, it just looks like it's a real value. And we can't [INAUDIBLE] real wave functions, can we? PROFESSOR: Well, it is the wave function at time equals 0. So the time derivative would have to bring in complex things. So you can have a wave function that is 0, that is real at some particular time. Like, any wave function psi of x e to the minus iEt over h bar is a typical wave function. And then at time equal 0 it may be real. It cannot be real forever. So you cannot assume it's real. But at some particular times it could be real. Very good question. The other thing you might say, look, this is too real to have momentum. Momentum has to do with waves. That's probably not a reliable argument. OK, so, where do we go from here? Well, let's try to find the momentum eigenstates. They should be things like that, exponentials. So how could they look? Well, e to the 2 pi i, maybe. What else? x, there should be an x for a momentum thing. Now there should be no units here, so there better be an L here. And now I could put, maybe, well the 2 maybe was-- why did I think of the 2 or the pi? Well, for convenience. But let's see what. Suppose you have a number m here. Then the good thing about this is that when x is equal to 0, there is some number here, but when x is equal to L, it's a multiple of e to the 2 pi i, so that's periodic. So this does satisfy, I claim, it's the only way if m is any integer. So it goes from minus infinity to infinity. Those things are periodic. They satisfy psi. Actually they satisfy psi of x plus L is equal to psi of x. OK. That seems to be something that could be a momentum eigenstate. And then I have to normalize it. Well, if I square it and integrate it. If I square it then the phase cancels, so you get 1. If you integrate it you get L. If you put 1 over the square root of L, when you square it and integrate, you will get 1. So here it is. Psi m's of x are going to be defined to be this thing. And I claim these things are momentum eigenstates. In fact, what is the value of the momentum? Well, you calculate h bar over i d dx on psi m. And you get what? You get 2 pi m over L times h bar times psi. The h bar is there, the i cancels, and everything then multiplies, the x falls down. So this is the state with momentum P equals to h bar 2 pi m over L. OK. Actually, doing that, we've done the most difficult part of the problem. You've found the momentum eigenfunctions. So now the rest of the thing is to rewrite this in terms of this kind of objects. I'll do it in a second. Maybe I'll leave a little space there and you can check the algebra, and you can see it in the notes. But you know what you're supposed to do. A sine of x is e to the ix minus is e to the minus ix over 2i. So you'd get these things converted to exponentials. The cosine of x is equal to e to the ix plus e to the minus ix over 2. So if you do that with those things, look. What the sine of 2 pi x going to give you? It's going to give you some exponentials of 2 pi ix over L. So suppose that m equals 1. And m Equals minus 1. And this will give you m equals 3, 3 times 2 is 6. And m equal minus 3. So I claim, after some work, and you could try to do it. I think it would be a nice exercise. Psi is equal square root of 2 over 3, 1 over 2 i psi 1 minus square root of 2 over 3, 1 over 2i psi minus 1 plus 1 over square root of 3 psi 3, plus 1 over square root of 3 psi minus 3. And it should give you some satisfaction to see something like that. You're now seeing the wave function written as a superposition of momentum eigenstates. This theorem came through. In this case, as a particle in the circle, the statement is that the eigenfunctions are the exponentials, and it's Fourier's theorem. Again, for a series. So finally, here is the answer. So psi 1, we can measure psi 1. What is the momentum of psi 1? So here are p values. And probabilities. The first value, psi 1, the momentum is 2 pi h bar over L. So 2 pi h bar over L. And what is its probability? It's this whole number squared. So square root of 2/3, 1 over 2i squared. So how much is that? It's 2/3 times 1/4. 2/3 times 1/4, which is 1/6. And the other value that you can get is minus this one, so minus 2 pi h bar over L. This minus doesn't matter, probability also 1/6. The next one is with 3. So you can get 2, 6 pi, 6 pi h bar over L, with probability square of this, 1/3. And minus 6 pi h bar over L with probability 1/3. Happily our probabilities add up. So there you go. That's the theorem expressed in a very clear example. We had a wave function. You wrote it as a sum of four momentum eigenstates. And now you know, if you do a measurement, what are the possible values of the momentum. This should have been probably 1/6. You can do anything you want.

MIT_804_Quantum_Physics_I_Spring_2016

Algebraic_solution_of_the_harmonic_oscillator.txt

PROFESSOR: SHO algebraically. And we go back to the Hamiltonian, p squared over 2m plus 1/2 m omega squared x hat squared. And what we do is observe that this some sort of sum of squares plus p squared over m-- p squared over m squared omega squared. So the sum of two things squared. Now, the idea that we have now is to try to vectorize the Hamiltonian. And what we call vectorizing is when you write your Hamiltonian as the product of two vectors, V times W. Well actually, that's not quite the vectorization. You want kind of the same vector, and not even that. You sort of want this to be the Hermitian conjugate of that. And if there is a number here, that's OK. Adding numbers to a Hamiltonian doesn't change the problem at all. The energies are all shifted, and it's just how you're defining the zero of your potential, is doing nothing but that. So vectorizing the Hamiltonian is writing it in this way, as V dagger V. And you would say, why V dagger V? Why not VV dagger or VV or V dagger V dagger? Well, you want the Hamiltonian to be Hermitian. And this thing is Hermitian. You may recall that AB dagger. The Hermitian conjugate of AB dagger is B dagger A dagger. So the Hermitian conjugate of this product is V dagger times the dagger of V dagger. A dagger of a dagger is the same operator, when you dagger it twice, you get the same. So this is Hermitian. V dagger times V is a Hermitian operator, and that's a very good thing. And there will be great simplifications. If you ever succeed in writing a Hamiltonian this way, you've gone 90% of the way to solving the whole problem. It has become infinitely easier, as you will see in a second, if you could just write this vectorization. So if you had x minus-- x squared minus this, you would say, oh, clearly that's-- A squared minus B squared is A minus B time A plus B, but there's no such thing here. It's almost like A squared plus B squared. And how do you sort of factorize it? Well, actually, since we have complex numbers, this could be A minus IB times A plus IB. That is correctly A squared plus B squared, and complex numbers are supposed to be friends in quantum mechanics, so having Is, there's probably no complication there. So let's try that. I'll write it. So here we have x squared plus p squared over m squared omega squared. And I will try to write it as x minus i p hat over m omega times x plus I p hat over m omega. Let's put the question mark before we are so sure that this works. Well, some things work. The only danger here is that these are operators and they don't commute. And when we do this, in one case, in the cross-terms, the A is to the left of B, but the other problem the B is to the left of A. So we may run into some trouble. This may not be exactly true. So what is this? This x with x, fine. x squared. This term, p with ps, correct. Plus p squared over m squared omega squared. But then we get plus i over m omega, x with p minus p with x, so that x, p commutator. So vectorization of operators in quantum mechanics can miss a few concepts because things don't commute. So the cross-terms give you that, and this x, p is I h bar, so this whole term will give us the following statement. What we've learned is that what we wanted, x squared plus p squared over m squared omega squared is equal to-- so I'm equating this line to the top line-- is equal to x hat minus i p hat over m omega times x hat plus i p hat over m omega. And then, from this whole term, i with i is minus, so it's h bar over m omega. So I'll put it in-- it's a minus h bar over m, [INAUDIBLE]. So here is plus h bar over m omega times a unit vector, if you wish. OK. So this is very good. In fact, we can call this V dagger and this V. Better call this V first and then ask, what is the dagger of this operator? Now, you may remember that, how did we define daggers? If you have phi with psi and the inner product-- with an integral of five star psi-- if you have an A psi here, that's equal to A dagger phi psi. So an operator is acting on the second wave function, moves as A dagger into the first wave function. And you know that x moves without any problem. x is Hermitian. We've discussed that p is Hermitian as well, moves to the other side. So the Hermitian conjugate of this operator is x, the p remains means p, but the i becomes minus i. So this is correct. If this second operator is called V, the first operator should be called V dagger. That is a correct statement. One is the dagger of the other one. So the Hamiltonian is 1/2 m omega squared times this sum of squares, which is now equal to V dagger V plus h bar over m omega. So h hat is now 1/2 m omega squared V dagger V plus a sum, which is plus 1/2 h bar omega. So we did it. We vectorized the Hamiltonian V dagger V, and this is quite useful. So the Vs, however, have units. And you probably are aware that we like things without units, so that we can see the units better. This curve is perfectly nice. It's a number added to the Hamiltonian. It's h omega, it has units of energy, but this is still a little messy. So let's try to clean up those Vs, and the way I'll do it is by computing their commutator, to begin with. So let's compute the commutator of V and V dagger and see how much is that commutator. It's a simple commutator, because it involves vectors of x and V. So it's the commutator of x plus ip over m omega, that's V, with x minus ip over m omega. So the first x talks only to the second piece, so it's minus i over m omega x, p. And for the second case, you have plus i over am omega p with x. This is i h bar, and this is minus i h bar. Each term will contribute the same, i times minus i is plus, so h bar over and, omega times the 2. That is V dagger V. VV dagger, I'm sorry. 2 h bar over m omega. So time to change names a little bit. Let's do the following. Let's put square root of m omega over 2 h bar V. Have a square root of m omega over 2 h bar V dagger, commute to give you 1. That's a nice commutator. It's one number-- or an operator is the same thing. So I brought the square root into each one. And we'll call the first term-- because of reasons we'll see very soon-- the destruction operator, A square root of m omega over 2 h bar V. It's called the destruction operator. And the dagger is going to be A dagger. Some people put hats on them. I sometimes do too, unless I'm too tired. 2h bar V dagger. And those A and A daggers are now unit-free-- and you can check That-- Because they have the same units. And A with A dagger is the nicest commutator, 1. Is A a Hermitian operator? Is it? No. A is not Hermitian. A dagger is different from A. A is basically this thing, A dagger is this thing. So not Hermitian. So we're going to work with these operators. They're non-Hermitian. I need to write the following equations. It's very-- takes a little bit of writing, but they should be recorded, they will always make it to the formula sheet. And it's the basic relation between A, A dagger, and x and p. A is this, A dagger, as you know, is x minus ip hat over m omega. Since I'm copying, I'd better copy them right. x, on other hand, is the square root of h bar over 2m omega A plus A dagger, and p is equal to i square root of m omega h bar over 2 A dagger minus A. So these four equations, A and A dagger in terms of x and p and vice versa, are important. They will show up all the time. Here are the things to notice. A and A dagger is visibly clear that on is the Hermitian conjugate of the other. Here, x is Hermitian. And indeed, A plus A dagger is Hermitian. When you do the Hermitian conjugate of A plus A dagger, the first A becomes an A dagger. The second A, with another Hermitian conjugation, becomes A. So this is Hermitian. But p is Hermitian, and here we have A dagger minus A. This is not Hermitian, it changes sign. Well, the i is there for that reason, and makes it Hermition. So there they are, they're Hermitian, they're good.

MIT_804_Quantum_Physics_I_Spring_2016

Expectation_value_of_Hermitian_operators.txt

PROFESSOR: Today we'll talk about observables and Hermitian operators. So we've said that an operator, Q, is Hermitian in the language that we've been working so far, if you find that the integral, dx psi 1 Q psi 2, is actually equal to the integral dx of Q, acting this time of Psi 1 all star psi 2. So as you've learned already, this requires some properties about the way functions far away, at infinity, some integration by parts, some things to manage, but this is the general statement for a large class of functions, this should be true. Now we want to, sometimes, use a briefer notation for all of this. And I will sometimes use it, sometimes not, and you do whatever you feel. If you like to use this notation, us it. So here's the definition. If you put up Psi 1, Psi 2 and a parentheses, this denotes a number, and in fact denotes the integral of psi 1 star of x, psi 2 of x dx. So whatever you put in the first input ends up complex motivated. When you put in the second input, it's like that, it's all integrated. This has a couple of obvious properties. If you put a number times psi 1 times psi 2 like this, the number will appear, together with psi 1, and will complex conjugated. So it can go out as a star psi 1 psi 2. And if you put the number on the second input, it comes out as is. Because the second input is not complex conjugated in the definition. With this definition, a Hermitian operator, Q is Hermitian, has a nice look to it. It becomes kind of natural and simple. It's the statement that if you have psi 1, Q psi 2, you can put the Q in the first input. Q psi 1 psi 2. This second term in the right hand side is exactly this integral here. And the first tern in the left hand side is the left hand side of that condition. So it's just maybe a briefer way to write it. So when you get tired of writing integral dx of the first, the second, you can use this. Now with distance last time, the expectation values of operators. So what's the expectation value of Q in some state psi of x? And that is denoted as these braces here and of psi is equal to the integral of psi. The expectation value depends on the state you live in and it's psi Q psi. Or if you wish, dx in written notation psi Q. I should put the hats everywhere. This is the expectation value of Q. I'm sorry, I missed here a star. So so far, so good. We've reviewed what a Hermitian operator is, what an expectation value is, so let's begin with some claims. Claim number one. The expectation value of Q, with Q Hermitian. So everywhere here, Q will be Hermitian. The expectation value of Q is real. A real number, it belongs to the real numbers. So that's an important thing. You want to figure out the expectation value of Q, you have a psi star, you have a psi. Well, it'd better be real if we're going to think, and that's the goal of this discussion, that Hermitian operators are the things you can measure in quantum mechanics, so this better be real. So let's see what this is. Well, Q psi, that's the expectation value. If I complex conjugate it, I must complex conjugate this whole thing. Now if you want to complex conjugate an integral, you can complex conjugate the integrand. Here it is. I took this right hand side here, the integrand. I copied it, and now I complex conjugated it. That's what you mean by complex conjugating an integral. But this is equal, integral dx. Now I have a product of two functions here. Psi star and Q that has acted on psi. So that's how I think. I never think of conjugating Q. Q is a set of operations that have acted on psi and I'm just going to conjugate it. And the nice thing is that you never have to think of what is Q star, there's no meaning for it. So what happens here? Priority of two functions, the complex conjugate of the first-- now if you [INAUDIBLE] normally something twice, you get the function back. And here you've got Q psi star. But that, these are functions. You can move around. So this Q hat psi star Q psi. And so far so good. You know, I've done everything I could have done. They told to come to complex conjugate this, so I complex conjugated it and I'm still not there. But I haven't used that this operator is Hermitian. So because the operator is Hermitian, now you can move the Q from this first input to the second one. So it's equal to integral dx psi star Q psi. And oh, that was the expectation value of Q on psi, so the star of this number is equal to the number itself, and that proves the claim, Q is real. So this is our first claim. The second claim that is equally important, claim two. The eigenvalues of the operator Q are real. So what are the eigenvalues of Q? Well you've learned, with the momentum operator, eigenvalues or eigenfunctions of an operator are those special functions that the operator acts on them and gives you a number called the eigenvalue times that function. So Q, say, times, psi 1, if psi 1 is a particularly nice choice, then it will be equal to some number. Let me quote Q1 times psi1. And there, I will say that Q1 is the eigenvalue. That's the definition. And psi1 is the eigenvector, or the eigenfunction. And the claim is that that number is going to real. So why would that be the case? Well, we can prove it in many ways, but we can prove it kind of easily with claim number one. And actually gain a little insight, cold calculate the expectation value of Q on that precise state, psi 1. Let's see how much is it. You see, psi 1 is a particular state. We've called it an eigenstate of the operator. Now you can ask, suppose you live in psi 1? That's who you are, that's your state. What is the expectation value of this operator? So we'll learn more about this question later, but we can just do it, it's the integral of dx psi 1 Q psi 1. And I keep forgetting these stars, but I remember them after a little while. So at this moment, we can use the eigenvalue condition, this condition here, that this is equal to dx psi 1 star Q1 psi 1. And the Q1 can go out, hence Q 1 integral dx of psi 1 star psi 1. But now, we've proven, in claim number one, that the expectation value of Q is always real, whatever state you take. So it must be real if you take it on the state psi 1. And if the expectation value of psi 1 is real, then this quantity, which is equal to that expectation value, must be real. This quantity is the product of two factors. A real factor here-- that integral is not only real, it's even positive-- times Q1. So if this is real, then because this part is real, the other number must be real. Therefore, Q1 is real. Now it's an interesting observation that if your eigenstate, eigenfunction is a normalized eigenfunction, look at the eigenfunction equation. It doesn't depend on what precise psi 1 you have, because if you put psi 1 or you put twice psi 1, this equation still holds. So if it hold for psi 1, if psi 1 is called an ideal function, 3 psi 1, 5 psi 1, minus psi 1 are all eigenfunctions. Properly speaking in mathematics, one says that the eigenfunction is the subspace generated by this thing, by multiplication. Because everything is accepted. But when we talk about the particle maybe being in the state of psi 1, we would want to normalize it, to make psi 1 integral squared equal to 1. In that case, you would obtain that the expectation value of the operator on that state is precisely the eigenvalue. When you keep measuring this operator, this state, you keep getting the eigenvalue. So I'll think about the common for a normalized psi 1 as a true state that you use for expectation values. In fact, whenever we compute expectation values, here is probably a very important thing. Whenever you compute an expectation value, you'd better normalize the state, because otherwise, think of the expectation value. If you don't normalize the state, you the calculation and you get some answer, but your friend uses a wave function three times yours and your friend gets now nine times your answer. So for this to be a well-defined calculation, the state must be normalized. So here, we should really say that the state is normalized. Say one is the ideal function normalized. And this integral would be equal to Q1 belonging to the reals. And Q1 is real. So for a normalized psi 1 or how it should be, the expectation value of Q on that eigenstate is precisely equal to the eigenvalue.

MIT_804_Quantum_Physics_I_Spring_2016

Momentum_operator_energy_operator_and_a_differential_equation.txt

PROFESSOR: Last time, we talked about the Broglie wavelength. And our conclusion was, at the end of the day, that we could write the plane wave that corresponded to a matter particle, with some momentum, p, and some energy, E. So that was our main result last time, the final form for the wave. So we had psi of x and t that was e to the i k x minus i omega t. And that was the matter wave with the relations that p is equal to h bar k. So this represents a particle with momentum, p, where p is h bar times this number that appears here, the wave number, and with energy, E, equal to h bar omega, where omega is that number that appears in the [? term ?] exponential. Nevertheless, we were talking, or we could talk, about non-relativistic particles. And this is our focus of attention. And in this case, E is equal to p squared over 2m. That formula that expresses the kinetic energy in terms of the momentum, mv. So this is the wave function for a free particle. And the task that we have today is to try to use this insight, this wave function, to figure out what is the equation that governs general wave functions. So, you see, we've been led to this wave function by postulates of the Broglie and experiments of Davisson, and Germer, and others, that prove that particles like electrons have wave properties. But to put this on a solid footing you need to obtain this from some equation, that will say, OK, if you have a free particle, what are the solutions. And you should find this solution. Perhaps you will find more solutions. And you will understand the problem better. And finally, if you understand the problem of free particle, there is a good chance you can generalize this and write the equation for a particle that moves under the influence of potentials. So basically, what I'm going to do by trying to figure out how this wave emerges from an equation, is motivate and eventually give you, by the middle of this lecture, the Schrodinger equation. So that's what we're going to try to do. And the first thing is to try to understand what kind of equation this wave function satisfies. So you want to think of differential equations like wave equations. Maybe it's some kind of wave equation. We'll see it's kind of a variant of that. But one thing we could say, is that you have this wave function here. And you wish to know, for example, what is the momentum. Well you should look at k, the number that multiplies the x here, and multiply by h bar. And that would give you the momentum. But another way of doing it would be to do the following. To say, well, h bar over i d dx of psi of x and t, calculate this thing. Now, if I differentiate with respect to x, I get here, i times k going down. The i cancels this i, and I get h bar k. So, I get h bar k times the exponential. And that is equal to the value of the momentum times the wave. So here is this wave actually satisfies a funny equation, not quite the differential equation we're looking for yet, but you can act with a differential operator. A derivative is something of a differential operator. It operates in functions, and takes the derivative. And when it acts on this wave function, it gives you the momentum times the wave function. And this momentum here is a number. Here you have an operator. An operator just means something that acts on functions, and gives you functions. So taking a derivative of a function is still a function. So that's an operator. So we are left here to think of this operator as the operator that reveals for you the momentum of the free particle, because acting on the wave function, it gives you the momentum times the wave function. Now it couldn't be that acting on the wave function just gives you the momentum, because the exponential doesn't disappear after the differential operator acts. So it's actually the operator acting on the wave function gives you a number times the wave function. And that number is the momentum. So we will call this operator, given that it gives us the momentum, the momentum operator, so momentum operator. And to distinguish it from p, we'll put a hat, is defined to be h bar over i d dx. And therefore, for our free particle, you can write what we've just derived in a brief way, writing p hat acting on psi, where this means the operator acting on psi, gives you the momentum of this state times psi of x and t. And that's a number. So this is an operator state, number state. So we say a few things, this language that we're going to be using all the time. We call this wave function, this psi, if this is true, this holds, then we say the psi of x and t is an eigenstate of the momentum operator. And that language comes from matrix algebra, linear algebra, in which you have a matrix and a vector. And when the matrix on a vector gives you a number times the same vector, we say that that vector is an eigenvector of the matrix. Here, we call it an eigenstate. Probably, nobody would complain if you called it an eigenvector, but eigenstate would be more appropriate. So it's an eigenstate of p. So, in general, if you have an operator, A, under a function, phi, such that A acting on phi is alpha phi, we say that phi is an eigenstate of the operator, and in fact eigenvalue alpha. So, here is an eigenstate of p with eigenvalue of p, the number p, because acting on the wave function gives you the number p times that wave function. Not every wave function will be an eigenstate. Just like, when you have a matrix acting on most vectors, a matrix will rotate the vector and move it into something else. But sometimes, a matrix acting in a vector will give you the same vector up to a constant, and then you've got an eigenvector. And here, we have an eigenstate. So another way of expressing this, is we say that psi of x and t, this psi of x and t, is a state of definite momentum. It's important terminology, definite momentum means that if you measured it, you would find the momentum p. And the momentum-- there would be no uncertainty on this measurement. You measure, and you always get p. And that's what, intuitively, we have, because we decided that this was the wave function for a free particle with momentum, p. So as long as we just have that, we have that psi is a state of definite momentum. This is an interesting statement that will apply for many things as we go in the course. But now let's consider another aspect of this equation. So we succeeded with that. And we can ask if there is a similar thing that we can do to figure out the energy of the particle. And indeed we can do the following. We can do i h bar d dt of psi. And if we have that, we'll take the derivative. Now, this time, we'll have i h bar. And when we differentiate that wave function with respect to time, we get minus i omega times the wave function. So i times minus i is 1. And you get h bar omega psi. Success, that was the energy of the particle times psi. And this looks quite interesting already. This is a number, again. And this is a time derivative of the wave function. But we can put more physics into this, because in a sense, well, this differential equation tells you how a wave function with energy, E, what the time dependence of that wave function is. But that wave function already, in our case, is a wave function of definite momentum. So somehow, the information that is missing there, is that the energy is p squared over 2m. So we have that the energy is p squared over 2m. So let's try to think of the energy as an operator. And look, you could say the energy, well, this is the energy operator acting on the function gives you the energy. That this true, but it's too general, not interesting enough at this point. What is really interesting is that the energy has a formula. And that's the physics of the particle, the formula for the energy depends on the momentum. So we want to capture that. So let's look what we're going to do. We're going to do a relatively simple thing, which we are going to walk back this. So I'm going to start with E psi. And I'm going to invent an operator acting on psi that gives you this energy. So I'm going to invent an O. So how do we do that? Well, E is equal to p squared over 2m times psi. It's a number times psi. But then you say, oh, p, but I remember p. I could write it as an operator. So if I have p times psi, I could write it as p over 2m h bar over i d dx of psi. Now please, listen with lots of attention. I'm going to do a simple thing, but it's very easy to get confused with the notation. If I make a little typo in what I'm writing it can confuse you for a long time. So, so far these are numbers. Number, this is a number times psi. But this p times psi is p hat psi which is that operator, there. So I wrote it this way. I want to make one more-- yes? AUDIENCE: Should that say E psi? PROFESSOR: Oh yes, thank you very much. Thank you. Now, the question is, can I move this p close to the psi. Opinions? Yes? AUDIENCE: Are you asking if it's just a constant? PROFESSOR: Correct, p is a constant. p hat is not a constant. Derivatives are not. But p at this moment is a number. So it doesn't care about the derivatives. And it goes in. So I'll write it as 1 over 2m h/i d dx, and here, output p psi, where is that number. But now, p psi, I can write it as whatever it is, which is h/i d dx, and p psi is again, h/i d dx psi. So here we go. We have obtained, and let me write the equation in slightly reversed form. Minus, because of the two i's, 1 over 2m, two partials derivatives is a second order partial derivative on psi, h bar squared over 2m d second dx psi. That's the whole right-hand side, is equal to E psi. So the number E times psi is this. So we could call this thing the energy operator. And this is the energy operator. And it has the property that the energy operator acting on this wave function is, in fact, equal to the energy times the wave function. So this state again is an energy eigenstate. Energy operator on the state is the energy times the same state. So psi is an energy eigenstate, or a state of definite energy, or an energy eigenstate with energy, E. I can make it clear for you that, in fact, this energy operator, as you've noticed, the only thing that it is is minus h squared over 2m d second dx squared. But where it came from, it's clear that it's nothing else but 1 over 2m p hat squared, because p hat is indeed h/i d dx. So if you do this computation. How much is this? This is A p hat times p hat, that's p hat squared. And that's h/i d dx h/i d dx. X And that gives you the answer. So the energy operator is p hat squared over 2m. All right, so actually, at this moment, we do have a Schrodinger equation, for the first time. If we combine the top line over there. I h bar d dt of psi is equal to E psi, but E psi I will write it as minus h squared over 2m d second dx squared psi.

MIT_804_Quantum_Physics_I_Spring_2016

Resonant_transmission.txt

PROFESSOR: Here is the answer, answer. It's easier apparently to write 1 over T. And 1 over T is equal to 1 plus 1 over 4 V0 squared over E times E plus V0 times sine squared of 2k2a. So the one thing to notice in this formula, it's a little complicated, is that the second term is positive. Because V0 squared is positive, the energy is positive, and sine squared is positive. So if this is positive, the right-hand side is greater than 1. And therefore, the T is less than 1. So this implies T less than or equal to 1. And there seems to be a possibility of T being equal to 1 exactly if the sine squared of this quantity, or the sine of this quantity, vanishes. So there is a possibility of very interesting saturation, in which the transmission is really equal to 1. So we'll see it. The other thing you can notice is that, as E goes to 0, this is infinite. And therefore, T is going to 0. No transmission as the energy goes to 0. As the energy goes to infinity, well, this term goes to 0. And you get transmission, T equals to one. So these are interesting limits. Now, to appreciate this better, we can write it with unit-free language. So for that, I'll do the following. It's a little rewriting, but it helps a bit. So think of 2k2 times a, this factor, as the argument of the sine function. Well, it's 2. k2 was defined up there, so it's 2m a squared E plus V0 over h squared. And I put the a inside the square root. So what do we have here? 2 times the square root of 2m a squared. Let's factor a V0, so that you have 1 plus E over V0. And you have h squared here. So this is OK. There's clearly two things you can do. First, define a unit-free energy. So the energy is now described by this little E. Without units, that compares the energy of your energy eigenstate to the depth of the potential. So it should be over V0. So this is nice. You don't have to talk about EVs or some quantity. Just a pure number. And here, there is another number that is famous. This is the number Z0 squared of a potential well. This is the unit-free number that tells you how deep or profound is your potential, and controls the number of zeros. So at this moment, this is simply 2 Z0, because the square root is there and takes the Z0 squared out as Z0. Square root of 1 plus e, which is nice. So here, you can divide by V0 squared, numerator and denominator. So you have an E over V0, and a 1 plus an E over V0. So the end result is that 1 over T is now 1 plus 1 over 4e 1 plus e sine squared of 2 Z0 square root of 1 plus e. So it's ready for numerical calculation, for plotting, and doing all kinds of things with it. But what we want to understand is this phenomenon that you would expect, in general, some reflection and some transmission. But there is a possibility when T is equal to 1, and in particular, when this sine squared function is equal to 0, and that will make T equal to 1, then you have a perfect transmission. So let's see why it is happening, or under what circumstances it happens. So for what the energies will we have? For what energies? Energies is T equal to 1. It's perfect transmission. No reflection whatsoever. So we need, then, that the argument of this sine function be equal to multiples of pi, 2 Z0 square root of 1 plus e is equal to a multiple of pi. Now, we would say what the multiple of pi? Well, it could be 0, 1, 2, 3. Not obvious, because the only thing you have here to adjust is the energy. The energy is positive. And that's that little e in here. So this number n must exceed some number, because this left-hand side never becomes very small. The smallest it can be is 2 Z0. So n must be greater than or equal to 2 Z0 over pi. This is because e, since e is greater than 0. So the left hand side is a number that is greater than 2 Z0, and the right-hand side must therefore be that way. All right, so this is a possibility. But then, let's calculate those values of the energies. Calculate those en's. So what do we have? We squared the left hand side for Z0 squared times and 1 plus en is equal to pi squared n squared. And en is equal to minus 1 plus n squared pi squared over 4 Z0 squared. OK, this is quantitatively nice. But probably still doesn't give us much intuition about what's going on. So let me go back to the total energy. en, remember, was energy divided by V0. So multiply all terms by V0. E equals minus V0 plus n squared pi squared V0 over 4. Z0 squared, I'm going to go all the way back to conventional language. And, too, 4 times Z0 squared, which is 2ma squared V0 over h bar squared. So E is minus V0 plus n squared pi squared. The V0s cancel. h squared over 2n times 2a squared. I think I got every term right. So what does this say? Well, think of the potential. In this region, there's an e here. And there's minus V0 there. So it says E is minus V0 plus this quantity. So minus V0 plus this quantity, which is n squared pi squared h squared over 2m times 2a squared. So the resonance happens if the energy is a distance above the bottom of the potential, which is equal to this quantity. And now, you see something that we could have seen maybe some other way. That what's happening here is a little strange at first sight. These are the energy levels of an infinite square well of width, 2a. If you remember, the energy levels of an infinite square well are n squared pi squared h squared over 2m times the width squared. And those are exactly it. So the energies at which you find the transmission, and the name is going to become obvious in a second, it's called the resonant transition, are those in which the energy coincides with some hypothetical energy of the infinite square well that you would put here. If it is as if you would have put an infinite square well in the middle and look at where are the energies of bound states that are bigger than 0, that might be bouncing the energies here, but those are not relevant, because you only consider energies positive. So if you find an energy that is positive, that corresponds to a would-be of infinite square well, that's it. That's an energy for which you will have transmission. And in fact, if we think about this from the viewpoint of the wave function, this factor over here, look at this property over here. So what do we have? The condition was that k2 time 2a, the argument of the sine function would be a multiple of pi. But k2 is 2 pi over the wavelength of the wave that you have in this range, over 2a. It's equal to n pi. So we can cancel the pis and the 2s so that you get 2a over lambda is equal to n over 2. And what does that say? It says that the de Broglie wavelength that you have in this region is such that it fits into 2a. Let me write it yet in another way. Let me try this a as-- I won't write it like that. Leave it like that. The wavelength lambda fits into 2a a half-integer number of times. And that's exactly what you have in an infinite square well. If you have a width, well, you could have half a wavelength there for n equals 1, a full thing for n equals 2, 3 halves for n equals 3. You always get half and halves and halves increasing and increasing all the time. Yeah. So the way I think I wanted to do it, this equation can be written as n is equal to 2a over lambda over 2. That's the same equation. So in this way, you see an integer a number of times is 2a divided by lambda over 2, which is precisely the condition for infinite square well energy eigenstate. So there is no infinite square well anywhere in this problem. But somehow, when the wavelength of the de Broglie representation of the particle in this region is an exact number of half-waves, there's resonance. And this resonance is such that it allows a wave to go completely through. It's a pretty remarkable phenomenon. So the infinite square well appears just as a way to think of what are the energies at which you will observe the resonances. But the resonance is simply due to having an exact number of half-waves in this region. So we can do on a little numerical example to show how that works.

MIT_804_Quantum_Physics_I_Spring_2016

The_general_Schrödinger_equation_x_p_commutator.txt

PROFESSOR: ih bar d psi dt equal E psi where E hat is equal to p squared over 2m, the operator. That is the Schrodinger equation. The free particle Schrodinger equation-- you should realize it's the same thing as this. Because p is h bar over i ddx. And now Schrodinger did the kind of obvious thing to do. He said, well, suppose I have a particle moving in a potential, a potential V of x and t-- potential. Then the total energy is kinetic energy plus potential energy. So how about if we think of the total energy operator. And here is a guess. We'll put the just p squared over 2m, what we had before. That's the kinetic energy of a particle. But now add plus V of x and t, the potential. That is reasonable from your classical intuition. The total energy is the sum of them. But it's going to change the Schrodinger equation quite substantially. Now, most people, instead of calling this the energy operator, which is a good name, have decided to call this the Hamiltonian. So that's the most popular name for this thing. This is called the Hamiltonian H. And in classical mechanics, the Hamiltonian represents the energy expressed in terms of position and momenta. That's what the Hamiltonian is, and that's roughly what we have here. The energy is [? in ?] [? terms ?] [? of ?] momenta and position. And we're going to soon be getting to the position operator, therefore. So this is going to be the Hamiltonian. And we'll put the hat as well. So Schrodinger's inspiration is to say, well, this is going to be H hat. And I'm going to say that ih bar d psi dt is equal to H hat psi. Or equivalently, ih bar ddt of psi of x and t is equal to minus h squared over 2m, [? v ?] [? second ?] dx squared-- that's the p squared over 2m-- plus V of x and t, all multiplying psi. This is it. This is the full Schrodinger equation. So it's a very simple departure. You see, when you discover the show that the equation for a free particle, adding the energy was not that difficult. Adding the potential energy was OK. We just have to interpret this. And maybe it sounds to you a little surprising that you multiply this by psi. But that's the only way it could be to be a linear equation. It cannot be that psi is acted by this derivative, but then you add v. It would not be a linear equation. And we've realize that the structure of the Schrodinger equation is d psi dt is equal to an energy operator times psi. The whole game of quantum mechanics is inventing energy operators, and then solving these equations, then see what they are. So in particular, you could invent a potential and find the equations. And, you see, it looks funny. You've made a very simple generalization. And now you have an equation. And now you can put the potential for the hydrogen atom and calculate, and see if it works. And it does. So it's rather unbelievable how very simple generalizations suddenly produce an equation that has the full spectrum of the hydrogen atom. It has square wells, barrier penetration, everything. All kinds of dynamics is in that equation. So we're going to say a few more things about this equation now. And I want you to understand that the V, at this moment, can be thought as an operator. This is an operator, acts on a wave function to give you a function. This is a simpler operator. It's a function of x and t. And multiplying by a function of x and t gives you a function of x and t. So it is an operator. Multiplying by a given function is an operator. It changes all the functions. But it's a very simple one. And that's OK, but V of x and t should be thought as an operator. So, in fact, numbers can be operator. Multiplication by a number is an operator. It adds on every function and multiplies it by a number, so it's also an operator. But x has showed up. So it's a good time to try to figure out what x has to do with these things. So that's what we're going to do now. Let's see what's x have to do with things. OK, so functions of x, V of x and t multiplied by wave functions, and you think of it as an operator. So let's make this formal. Introduce an operator, X hat, which, acting on functions of x, multiplies them by x. So the idea is that if you have the operator X hat acting on the function f of x, it gives you another function, which is the function x times f of x-- multiplies by x. And you say, wow, well, why do you have to be so careful in writing something so obvious? Well, it's a good idea to do that, because otherwise you may not quite realize there's something very interesting happening with momentum and position at the same time, as we will discover now. So we have already found some operators. We have operators P, x, Hamiltonian, which is p squared over 2m. And now you could put V of x hat t. You know, if here you put V of x hat, anyway, whatever x hat does is multiplied by x. So putting V of x hat here-- you may want to do it, but it's optional. I think we all know what we mean by this. We're just multiplying by a function of x. Now when you have operators, operators act on wave functions and give you things. And we mentioned that operators are associated or analogs of matrices. And there's one fundamental property of matrices. The order in which you multiply them makes a difference. So we've introduced two operators, p and x. And we could ask whether the order of multiplication matters or not. And this is the way Heisenberg was lead to quantum mechanics. Schrodinger wrote the wave equation. Heisenberg looked at operators and commutation relations between them. And it's another way of thinking of quantum mechanics that we'll use. So I want to ask the question, that if you have p and x and you have two operators acting on a wave function, does the order matter, or it doesn't matter? We need to know that. This is the basic relation between p and x. So what is the question? The question is, if I have-- I'll show it like that-- x and p acting on a wave function, phi, minus px acting on a wave function, do I get 0? Do I get the same result or not? This is our question. We need to understand these two operators and see how they are related. So this is a very good question. So let's do that computation. It's, again, one of those computation that is straightforward. But you have to be careful, because at every stage, you have to know very well what you're doing. So if you have two operators like a and b acting on a function, the meaning of this is that you have a acting on what b acting on phi gives you. That's what it means to have two things acting. Your first act with the thing on the right. You then act on the other one. So let's look at this thing-- xp phi minus px phi. So for the first one, you would have x times p hat on phi minus p hat times x on phi, phi of x and t maybe-- phi of x and t. OK, now what do we have? We have x hat acting on this. And this thing, we already know what it is-- h over i ddx of phi of x and t-- minus p hat and x, acting on phi, is little x phi of x and t. Now this is already a function of x and t. So an x on it will multiply it by x. So this will be h over i x ddx of phi. It just multiplies it by x at this moment-- minus here we have h bar over i ddx of x phi. And now you see that when this derivative acts on phi, you get a term that cancels this. But when it acts on x, it gives you an extra term. ddx of x is minus h over i phi-- or ih phi. So the derivative acts on x or an phi. When it acts on phi, gives you this term. When it acts on x, gives you the thing that is left over. So actually, let me write this in a more clear way. If you have an operator, a linear operator A plus B acting on a function phi, that's A phi plus B phi. You have linear operators like that. And we have these things here. So this is actually equal to x hat p hat minus p hat x hat on phi. That's what it means when you have operators here. So look what we got, a very surprising thing. xp minus px is an operator. It wants to act on function. So we put a function here to evaluate it. And that was good. And when we evaluate it, we got a number times this function. So I could say-- I could forget about the phi. I'm simply right that xp minus px is equal to ih bar. And although it looks a little funny, it's perfectly correct. This is an equality between operators-- equality between operators. On the left-hand side, it's clear that it's an operator. On the right-hand side, it's also an operator, because a number acts as an operator on any function it multiplies by it. So look what you've discovered, this commutator. And that's a notation that we're going to use throughout this semester, the notation of the commutator. Let's introduce it here. So if you have two operators, linear operators, we define the commutator to be the product in the first direction minus the product in the other direction. This is called the commutator of A and B. So it's an operator, again, but it shows you how they are non-trivial, one with respect to the other. This is the basis, eventually, of the uncertainty principle. x and p having a commutator of this type leads to the uncertainty principle. So what did we learn? We learned this rather famous result, that the commutator of x and p in quantum mechanics is ih bar.

MIT_804_Quantum_Physics_I_Spring_2016

Commutators_matrices_and_3dimensional_Schrödinger_equation.txt

PROFESSOR: This is very important. This is the beginning of the uncertainty principle, the matrix formulation of quantum mechanics, and all those things. I want to just tabulate the information of matrices. We have an analog, so we have operators. And we think of them as matrices. Then in addition to operators, we have wave functions. And we think of them as vectors. The operators act on the wave functions or functions, and matrices act on vectors. We have eigenstate sometimes and eigenvectors. So matrices do the same thing. They don't necessarily commute. There are very many examples of that. I might as well give you a little example that is famous in the theory of spin, spin 1/2. There is the Pauli matrices. Sigma 1 is equal to 1, 1, 0, 0. Sigma 2 is 0 minus i, i 0, and sigma 3 is 1 minus 1, 0, 0. And a preview of things to come-- the spin operator is actually h bar over 2 sigma. And you have to think of sigma as having three components. That's where it is. Spins will be like that. We won't have to deal with spins this semester. But there it is, that spin 1/2. Somehow these matrices encode spin 1/2. And you can do simple things, like sigma 1 times sigma 2. 0, 1, 1, 0 times 0 minus i, i, 0. Let's see if I can get this right. i, 0, 0 minus i. And you can do sigma 2 sigma 1 0 minus i, i 0, 0, 1, 1, 0 equals minus i, 0, 0, i, i. So I can go ahead here. And therefore, sigma 1 commutator with sigma 2 is equal to sigma 1, sigma 2 minus sigma 2, sigma 1. And you can see that they're actually the same up to a sign, so you get twice. So you get 2 times i 0, 0 minus i. And this is 2i times 1 minus 1, 0, 0. And that happens to be the sigma 3 matrix. So sigma 1 and sigma 2 is equal to 2i sigma 3. These matrices talk to each other. And you would say, OK, these matrices commute to give you this matrix. This thing commutes to give you a number so that surely it's a lot easier. You couldn't be more wrong. This is complicated, extraordinarily complicated to understand what this means. This is very easy. This is 2 by 2 matrices that you check. In fact, you can write matrices for x and p. This correspondence is not just an analogy. It's a concrete fact. You will learn-- not too much in this course, but in 805-- how to write matrices for any operator. They're called matrix representations. And therefore, you could ask how does the matrix for x look. How does the matrix for p look? And the problem is these matrices have to be infinite dimensional. It's impossible to find two matrices whose commutator gives you a number. Something you can prove in math is actually not difficult. You will all prove it through thinking a little bit. There's no two matrices that commute to give you a number. On the other hand, very easy to have matrices that commute to give you another matrix. So this is very strange and profound and interesting, and this is much simpler. Spin 1/2 is much simpler. That's why people do quantum computations. They're working with matrices and simple stuff, and they go very far. This is very difficult. x and p is really complicated. But that's OK. The purpose of this course is getting familiar with those things. So I want to now generalize this a little bit more to just give you the complete Schrodinger equation in three dimensions. So how do we work in three dimensions, three-dimensional physics? There's two ways of teaching 804-- it's to just do everything in one dimension, and then one day, 2/3 of the way through the course-- well, we live in three dimensions, and we're going to add these things. But I don't want to do that. I want to, from the beginning, show you the three-dimensional thing and have you play with three-dimensional things and with one-dimensional things so that you don't get focused on just one dimension. The emphasis will be in one dimension for a while, but I don't want you to get too focused on that. So what did we have with this thing? Well, we had p equal h bar over i d dx. But in three dimensions, that should be the momentum along the x direction. We wrote waves like that with momentum along the x direction. And py should be h bar over i d dy, and pz should be h bar over i d dz-- momentum in the x, y, and z direction. And this corresponds to the idea that if you have a wave, a de Broglie wave in three dimensions, you would write this-- e to the i kx minus omega t, i omega t. And the momentum would be equal to h bar k vector, because that's how the plane wave works. That's what de Broglie really said. He didn't say it in one dimension. Now, it may be easier to write this as p1 equal h bar over i d dx1, p2 h bar over i d dx2, and p3 h bar over i d dx3 so that you can say that all these three things are Pi equals h bar over i d dxi-- and maybe I should put pk, because the i and the i could get you confused-- with k running from 1 to 3. So that's the momentum. They're three momenta, they're three coordinates. In vector notation, the momentum operator will be h bar over i times the gradient. You know that the gradient is a vector operator because d dx, d dy, d dz. So there you go. The x component of the momentum operators, h bar over i d dx, or d dx1, d dx2, d dx3. So this is the momentum operator. And if you act on this wave with the momentum operator, you take the gradient, you get this-- so p hat vector. Now here's a problem. Where do you put the arrow? Before or after the hat? I don't know. It just doesn't look very nice either way. The type of notes I think we'll use for vectors is bold symbols so there will be no proliferation of vectors there. So anyway, if you have this thing being the gradient acting on this wave function, e to the i kx minus i omega t, that would be h over i, the gradient, acting on a to the i kx vector minus i omega t. And the gradient acting on this-- this is a vector-- actually gives you a vector. So you can do component by component, but this gives you i k vector times the same wave function. So you get hk, which is the vector momentum times the wave function. So the momentum operator has become the gradient. This is all nice. So what about the Schrodinger equation and the rest of these things? Well, it's not too complicated. We'll say one more thing. So the energy operator, or the Hamiltonian, will be equal to p vector hat squared over 2m plus a potential that depends on all the coordinates x and t, the three coordinates. Even the potential is radial, like the hydrogen atom, is much simpler. There are conservation laws. Angular momentum works nice. All kinds of beautiful things happen. If not, you just leave it as x and p. And now what is p hat squared? Well, p vector hat squared would be h bar over-- well, I'll write this-- p vector hat dotted with p vector hat. And this is h over i gradient dotted with h over i gradient, which is minus h squared Laplacian. So your Schrodinger equation will be ih bar d psi dt is equal to the whole Hamiltonian, which will be h squared over 2m. Now Laplacian plus v of x and t multiplied by psi of x vector and t. And this is the full three-dimensional Schrodinger equation. So it's not a new invention. If you invented the one-dimensional one, you could have invented the three-dimensional one as well. The only issue was recognizing that the second dx squared now turns into the full Laplacian, which is a very sensible thing to happen. Now, the commutation relations that we had here before-- we had x with p is equal to ih bar. Now, px and x failed to commute, because d dx and x, they interact. But px will commute with y. y doesn't care about x derivative. So the p's failed to commute. They give you a number with a corresponding coordinate. So you have the i-th component of the x operator and the j-th component of the p operator-- these are the components-- give you ih bar delta ij, where delta ij is a symbol that gives you 1 if i is equal to j and gives you 0 if i is different from j. So here you go. X and px is 1 and 1. Delta 1, 1 is 1. So you get ih bar. But if you have x with py or p2, you would have delta 1, 2, and that's 0, because the two indices are not the same. So this is a neat way of writing nine equations. Because in principle, I should give you the commutator of x with px and py and pc, y with px, py, pc, and z with px, py, pc. You're seeing that, in fact, x just talks to px, y talks to py, z talks to pz. So that's it for the Schrodinger equation. Our goal is going to be to understand this equation. So our next step is to try to figure out the interpretation of this psi. We've done very nicely by following these things. We had a de Broglie wave. We found an equation. Which invented a free Schrodinger equation. We invented an interacting Schrodinger equation. But we still don't know what the wave function means.

MIT_804_Quantum_Physics_I_Spring_2016

Fourier_transforms_and_delta_functions.txt

BARTON ZWIEBACH: Today's subject is momentum space. We're going to kind of discover the relevance of momentum space. We've been working with wave functions that tell you the probabilities for finding a particle in a given position and that's sometimes called coordinate space or position space representations of quantum mechanics, and we just talked about wave functions that tell you about probabilities to find a particle in a given position. But as we've been seeing with momentum, there's a very intimate relation between momentum and position, and today we're going to develop the ideas that lead you to think about momentum space in a way that is quite complimentary to coordinate space. Then we will be able to talk about expectation values of operators and we're going to be moved some steps into what is called interpretation of quantum mechanics. So operators have expectations values in quantum mechanism-- they are defined in a particular way and that will be something we're going to be doing in the second part of the lecture. In the final part of the lecture, we will consider the time dependence of those expectation values, which is again, the idea of dynamics-- if you want to understand how your system evolves in time, the expectation value-- the things that you measure-- may change in time and that's part of the physics of the problem, so this will tie into the Schrodinger equation. So we're going to begin with momentum space. So we'll call this uncovering momentum space. And for much of what I will be talking about in the first half of the lecture, time will not be relevant-- so time will become relevant later. So I will be writing wave functions that don't show the time, but the time could be put there everywhere and it would make no difference whatsoever. So you remember these Fourier transform statements that we had that a wave function of x we could put time, but they said let's suppress time. It's a superposition of plane waves. So there is a superposition of plane waves and we used it last time to evolve wave packets and things like that. And the other side of Fourier's theorem is that phi of k can be written by a pretty similar integral, in which you put psi of x, you change the sign in the exponential, and of course, integrate now over x. Now this is the wave function you've always learned about first, and then this wave function, as you can see, is encoded by phi of k as well. If you know phi of k, you know psi of x. And so this phi of k has the same information in principle as psi of x-- it tells you everything that you need to know. We think of it as saying, well, phi of k has the same information as psi of x. And the other thing we've said about phi of k is that it's the weight with which you're superposing plane waves to reconstruct psi of x. The Fourier transform theorem is a representation of the wave function in terms of a superposition of plane waves, and here, it's the coefficient of the wave that accompanies each exponential. So phi of k is the weight of the plane waves in the superposition. So one thing we want to do is to understand even deeper what phi of k can mean. And so in order to do that, we need a technical tool. Based on these equations, one can derive a way of representing this object that we call the delta function. Delta functions are pretty useful for manipulating objects and Fourier transforms, so we need them. So let's try to obtain what is called the delta function statement. And this is done by trying to apply these two equations simultaneously. Like, you start with psi of x, it's written in terms of phi of k, but then what would happen if you would substitute the value of phi of k in here? What kind of equation you get? What you get is an equation for a delta function. So you begin with psi of x over square root of 2 pi, and I'll write it dk e to the ikx, and now, I want to write phi of k. So I put phi of k-- 1 over square root of 2 pi integral psi, and now I have to be a little careful. In here in the second integral, x is a variable of integration, it's a dummy variable. It doesn't have any physical meaning, per se-- it disappears after the integration. Here, x represents a point where we're evaluating the wave function, so I just cannot copy the same formula here because it would be confusing, it should be written with a different x and x-prime. Because that x certainly has nothing to do with the x we're writing here. So here it is, we've written now this integral. And let's rewrite it still differently. We'll write as integral dx-prime. We'll change the orders of integration with impunity. If you're trying to be very rigorous mathematically, this is something you worry about. In physics problems that we deal with, it doesn't make a difference. So here we have dx-prime, psi of x-prime, then a 1 over 2 pi integral dk e to the ik x minus x-prime. So this is what the integral became. And you look at it and you say, well, here is psi of x, and it's equal to the integral over x-prime of psi of x-prime times some other function of x and x-prime. This is a function of x minus x-prime, or x and x-prime if you wish. It doesn't depend on k, k is integrated. And this function, if you recognize it, it's what we call a delta function. It's a function that, multiplied with an integral, evaluates the integrand at a particular point. So this is a delta function, delta of x-prime minus x. That is a way these integrals then would work. That is, when you integrate over x-prime-- when you have x-prime minus something, then the whole integral is the integrand evaluated at the point where we say the delta function fires. So the consistency of these two equations means that for all intents and purposes, this strange integral is a representation of a delta function. So we will write it down. I can do one thing here, but-- it's kind of here you see the psi of x-prime minus x, but here you see x minus x-prime. But that's sign, in fact, doesn't matter. It's not there. You can get rid of it. Because if in this integral, you can let K goes to minus k, and then the dk changes sign, the order of integration changes sign, and they cancel each other. And the effect is that you change the sign in the exponents, so if you let k goes to minus k, the integral just becomes 1 over 2 pi integral dk e to the ik x-prime minus x. So we'll say that this delta function is equal to this thing, exploiting this sign ambiguity that you can always have. So I'll write it again in the way most people write the formula, which is at this moment, switching x and x-prime. So you will write this delta of x minus x-prime is 1 over 2 pi integral from infinity to minus infinity dk e to the ik x minus x-prime. So this is a pretty useful formula and we need it all the time that we do Fourier transforms as you will see very soon. It's a strange integral, though. If you have x equal to x-prime, this is 0 and you get infinity. So it's a function, it's sort of 0-- when x minus x-prime is different from 0, somehow all these waves superimpose to 0. But when x is equal to x-prime, it blows up. So it's does the right thing, it morally does the right thing, but it's a singular kind of expression and we therefore manipulate it with care and typically, we use it inside of integrals. So it's a very nice formula, we're going to need it, and it brings here for the first time in our course, I guess, the delta functions. And this is something if should-- if you have not ever play without the functions, this may be something interesting to ask in recitation or you can try to prove, for example, just like we show that delta of minus x is the same as delta of x, the delta of a being a number times x is 1 over absolute value of a times delta of x. Those are two simple properties of delta functions and you could practice by just trying to prove them-- for example, use this integral representation to show them.

MIT_804_Quantum_Physics_I_Spring_2016

Defining_uncertainty.txt

PROFESSOR: Uncertainty. When you talk about random variables, random variable Q, we've said that it has values Q1 up to, say, Qn, and probabilities P1 up to Pn, we speak of a standard deviation, delta Q, as the uncertainty, the standard deviation. And how is that standard deviation defined? Well you begin by making sure you know what is the expectation value of the-- or the average value of this random variable, which was defined, last time, I think I put braces, but bar is kind of nice sometimes too, at least for random variables, and it's the sum of the Pi times the Qi. The uncertainty is also some expectation value. And expectation value of deviation. So the uncertainty squared is the expectation value, sum over i, of deviations of the random variable from the mean. So you calculate the expected value of the difference of your random variable and the mean squared, and that is the square of the standard deviation. Now this is the definition. And it's a very nice definition because it makes a few things clear. For example, the left hand side is delta Q squared, which means it's a positive number. And the right hand side is also a positive number, because you have probabilities times differences of quantities squared. So this is all greater and equal to zero. And moreover, you can actually say the following. If the uncertainty, or the standard deviation, is zero, the random variable is not that random. Because if this whole thing is 0, this delta squared, delta Q squared must be 0 and this must be 0. But each term here is positive. So each term must be 0, because of any one of them was not equal to zero, you would get a non-zero contribution. So any possible Qi that must have a Pi different from 0 must be equal to Qbar. So if delta cubed is equal to 0, Qi is equal to Q as not random anymore. OK, now we can simplify this expression. Do the following. By simplifying, I mean expand the right-hand side. So sum over i, Pi Qi squared, minus 2 sum over i, Pi Qi Q bar plus sum over i, Pi Q bar squared. This kind of thing shows up all the time, shows up in quantum mechanic as well, as we'll see in a second. And you need to be able to see what's happenening. Here, you're having the expectation value of Qi squared. That's the definition of a bar of some variable, you'd multiply with variable by the exponent of [INAUDIBLE].. What is this? This a little more funny. First, you should know that Q bar is a number, so it can go out. So it's minus 2 Q bar. And then all that is left is this, but that's another Q bar. So it's another Q bar. And here, you take this one out because it's a number, and the sum of the probabilities is 1, so it's Q bar squared as well. And it always comes out that way, this minus 2 Q bar squared plus Q bar squared. So at the end, Delta Q, it's another famous property, is the mean of the square minus the square of the mean. And from this, since this is greater or equal than 0, you always conclude that the mean of the square is always bigger than the-- maybe I shouldn't have the i here, I think it's a random variable Q squared. So the mean, the square of this is greater or equal than Q bar squared. OK. Well, what happens in quantum mechanics, let give you the definition and a couple of ways of writing it. So here comes the definition. It's inspired by this thing. So in quantum mechanics, permission operator Q will define the uncertainty of Q in the state, Psi O squared as the expectation value of Q squared minus the expectation value of Q squared. Those are things that you know in quantum mechanics, how you're supposed to compute. Because you know what an expectation value is in any state Psi. You so Psi star, the operator, Psi. And here you do this thing, so it's all clear. So it's a perfectly good definition. Maybe it doesn't give you too much insight yet, but let me say two things, and we'll leave them to complete for next time. Which is claim one, one, that Delta Q squared Psi can be written as the expectation value of Q minus absolute expectation value of Q squared. Like that. Look. It looks funny, and we'll elaborate this, but the first claim is that this is a possible re-writing. You can write this uncertainty as a single expectation value. This is the analog of this equation in quantum mechanics. Claim two is another re-writing. Delta Q squared on Psi can be re-written as this. That's an integral. Q minus Q and Psi. Look at that. You act on Psi with the operator, Q, and multiplication by the expectation value of Q. This is an operator, this is a number multiplied by Psi. You can add to this on the [? wave ?] function, you can square it, and then integrate. And that is also the uncertainty. We'll show these two things next time and show one more thing that the uncertainty vanishes if and only if the state is an eigenstate of Q. So If the state that you are looking for is an eigenstate of Q, you have no uncertainty. And if you have no uncertainty, the state must be an eigenstate of Q. So those all things will come from this planes, that we'll elaborate on next time.

MIT_804_Quantum_Physics_I_Spring_2016

Delta_function_potential_I_Solving_for_the_bound_state.txt

PROFESSOR: So, what is the wave function that we have? We must have a wave function now that is symmetric, and built with e to the k x, kappa x, and into the minus kappa x. This is the only possibility. E to the minus of kappa absolute value of x. This is psi of x for x different from 0. This is-- as you can quickly see-- this is e to get minus kappa x, when x is positive, A e to the kappa x, when x is negative. And, both of them decay. The first exponential negative is the standard decaying exponential to the right. The one with positive-- well, here x is negative as you go all the way to the left. This one decays case as well. And, this thing plotted is a decaying exponential with amplitude A, like that. And, a decaying exponential with amplitude A, and a singularity there, which is what you would have expected. So, this seems to be on the right track-- it's a continuous wave function. The wave function cannot fail to be continuous, that's a complete disaster to show that an equation could not be satisfied. So, this is our discontinuous wave function. So, at this moment you really haven't yet used the delta function-- the delta function with intensity alpha down. I should have made a comment that it's very nice that alpha appeared here in the numerator. If it would have appeared in the denominator, I would be telling you that I think this problem is not going to have a solution. Why? Because if it appears in the numerator, it means that as the delta function potential is becoming stronger and stronger, the bound state is getting deeper and deeper-- which is what you would expect. But, if it would be in the numerator-- in the denominator-- as the potential gets deeper and deeper, the boundary is going up. That makes no sense whatsoever. So, it's good that it appeared there, it's a sign that things are in reasonable conditions. So, now we really have to face the delta function. And, this is a procedure you are going to do many times in this course. So, look at it, and do it again and again until you're very comfortable with it. It's the issue of discovering what kind of discontinuity you can have with the delta function. And, it's a discontinuity in the derivative, so let's quantify it. So, here it is-- we begin with the Schrodinger equation, again. But, I will write now the potential term as well. The potential is plus v of x psi of x equals E psi of x. And the idea is to integrate this equation from minus epsilon to epsilon. And, epsilon is supposed to be a small positive number. So, you integrate from minus epsilon to epsilon the differential equation, and see what it does to you in the limit as epsilon goes to 0. That's what we're going to try to do. So, what do we get? If you integrate this, you get minus h squared over 2 m. And now, you have to integrate the second derivative with respect to x, which is the first derivative, and therefore this is the first derivative at x equal epsilon minus the first derivative at x equals minus epsilon. This is from the first term, because you integrate d x d second d x squared psi is the same thing as d x d d x of d psi d x between A and B. And, the integral of a total derivative is d psi d x at B A-- I think people write it like this-- A to B. Evaluate it at the top, minus the evaluation at the bottom. Now, the next term is the integral of psi times v of x. So, I'll write it plus the integral from minus epsilon to epsilon d x minus alpha delta of x psi of x-- that's the potential. Now, we use the delta function. And, on the right hand side this will be E times the integral from minus epsilon to epsilon of psi of x d x. So, that's the differential equation integrated. And now, we're going to do two things. We're going to do some of these integrals, and take the limit as epsilon goes to 0. So, I'll write this minus h squared over 2 m limit as epsilon goes to 0 of d psi d x at epsilon minus d psi d x at minus epsilon plus. Let's think of this integral. We can do this integral, it's a delta function. So, it picks the value of the wave function at 0, because 0 is inside the interval of integration. That's why we integrate it from minus epsilon to epsilon, to have the delta function inside. So, you get an alpha out, a psi of 0, and that's what this integral is. It's independent of the value of epsilon as long as epsilon is different from 0. So, this gives you minus alpha psi of 0. And now, the last term is an integral from minus epsilon to epsilon of the wave function. Now, the wave function is continuous-- it should be continuous-- that means it's finite. And, this integral, as of any function that is not divergent from minus epsilon to epsilon as epsilon goes to 0, is 0. Any integral of a function that doesn't diverge as the limits of integration go to 0, the area under the function is 0. So, this is 0-- the limit. And this thing goes to 0, so we put a 0 here. So, at this moment we got really what we wanted. I'll write it this way. I'll go here, and I'll say minus h squared over 2 m, and what is this? This expression says, calculate the derivative of the function a little bit to the right of 0, and subtract the derivative of the function a little bit to the left of 0. This is nothing but the discontinuity in psi prime. You're evaluating for any epsilon greater than 0-- the psi prime a little to right, a little too the left, and taking the difference. So, this is what we should call the discontinuity delta at 0-- at x equals 0. And, this and this is for discontinuity of psi prime at x equals 0 minus alpha psi of 0 equals 0. And from here, we discover that delta zero psi prime is equal to minus 2 m alpha over h squared psi of 0. This is the discontinuity condition produced by the delta function. This whole quantity is what we call delta 0 of psi prime. And, what it says is that yes, the wave function can have a discontinuous first derivative if the wave function doesn't vanish there. Once the wave function doesn't vanish at that point, the discontinuity is in fact even proportional to the value of the wave function at that point. And, here are the constants of proportionality. Now, I don't think it's worth to memorize this equation or anything like that, because it basically can be derived in a few lines. This may have looked like an interesting or somewhat intricate derivation, but after you've done is a couple of times-- this is something you'll do in a minute or so. And, you just integrate and find the discontinuity in the derivative-- that's a formula there. And, that's a formula for a potential, minus alpha delta of x. So, if somebody gives you a different potential, well, you have to change the alpha accordingly. So, let's wrap this up. So, we go to our case. Here is our situation. So, let's apply this. So, what is the value? Apply this equation to our wave function. So, what is the derivative at epsilon? It's minus kappa A E to the minus kappa epsilon. That's the derivative of psi on the positive side. I differentiated the top line of this equation minus the derivative on the left side-- this one, the derivative. So, this is kappa A E to the kappa epsilon-- no, kappa minus epsilon again. So, that's the left hand side. The right hand side would be minus 2 m alpha h squared psi at 0. Psi at 0 is A, so that's what it gives us. And we should take the limit as epsilon goes to 0. So, this is going to 1, both of them. So, the left hand side is minus 2 kappa A, and the right hand side is 2 m alpha over h squared A. So, the 2-- it's also minus, I'm sorry-- so the 2s cancel, the A cancels-- you never should have expected to determine A unless you tried to normalize the wave function. Solving for energy eigenstates states will never determine A. The Schrodinger equation is linear, so A drops out, the minus 2 drops out, and kappa is equal to m alpha over h squared. So, that said that's great because kappa is just another name for the energy. So, I have kappa m alpha over h squared, so that's another name for the energy. So, let's go to the energy. The energy is h bar squared kappa minus h bar squared kappa squared over 2 m. So, it's minus h bar squared. Kappa squared would be m squared alpha squared h to the fourth, and there's a two m. All these constants. So, final answer. E, the bound state energy is minus m alpha squared minus m alpha squared. The m cancels it over h squared minus one half. So, back here the units worked out, everything is good, and the number was determined as minus one half. That's your bound state energy for this problem. So, this problem is instructive because you basically learn that in delta functions, with one delta function you get a bound state. If you have two delta functions, you may get more bound states-- three, four-- people study those problems, and you will investigate the two delta function cases.

MIT_804_Quantum_Physics_I_Spring_2016

Scattering_states_and_the_step_potential.txt

PROFESSOR: Scattering states are energy eigenstates that cannot be normalized. And when you say this cannot be normalized, so what's the use of them? They don't represent particles. Well, it's like they e to the ipx over h bar, those infinite plane waves. Each one by itself cannot be normalized, but you can conserve wave packets that are normalized. So the whole intuition that you get with scattering states is based on the idea that we're going to construct energy eigenstates. This time we cannot think of them as states of a particle. Bound states, yes. We can think of them, they're normalizable. But this energy eigenstates and bounded scattering states are not states of one particle. So we definitely have to go back and produce wave packets. But the intuition from those energy eigenstates is very valuable. So scattering states. And we call them sometimes scattering states because they look like the process of scattering. This will be non-normalizable energy eigenstates. And you've played a little with some of them. And we'll now study one case in detail. We'll try a couple of cases between today and next lecture. So the step potential. And the step potential is a potential that is 0 up to x equals 0. Here's the x-axis, and then suddenly there's a step at v0. And here is the potential. But then the wave, this is here, goes up. It's a step. And any energy eigenstates here has to be bigger. The energy has to be bigger than the lowest point of the potential. You know that, you kind of have an energy that is like that less because this would have decayed exponentially for infinite distance. It just, all over it would have to decay exponentially. It's impossible. So all the energy states, eigenstates here, must have positive energy. So we have actually qualitatively two possibilities. The energy may be less than v0, might be greater than v0. It would look like you have to solve the problem two times. Happily, we'll solve one, then let the other happen by analytic continuation. So here is the energy. I'll take the energy greater than v0. But whatever is the energy, even if it's less than v0, the solution over here is going to be an exponential or a cosine and a sine, a non-decaying function, and therefore can't be normalized because it's non-decaying forever and ever. So it cannot be normalized. So how do we write the solution for the energy eigenstate? It's a psi of x. Well, I should write two formulas: a formula for what's happening on the left side, and the formula for what's happening on the right side. Now I have a choice actually here. There's two ways of visualizing this. I can visualize it as a wave that is coming from the left, moving here. Or a wave that is coming from the right. So let's visualize this solution as a wave that's coming from the left. It will be a little easier. So I will write it. A e to the ikx. OK. Why is it coming from the left? Because if you put the energy-- that I will not put it, it's in stationary state, presumably this is a state with some fixed energy. You will have a factor e to the minus iEt over h bar. And when you see kx minus Et, you know that that's a wave that is moving to the right. So this A e to the ikx is moving to the right. And then what will happen? Now it's a matter of finding a solution of Schrodinger's equation. So you can try to find the solution of Schrodinger equation, but you have to write some answers for what's happening on the right. I will write an answer here, that we'll put C e to the i k bar x. And another k. Well, we'll see now what those k's are. I say the following. Here, the energy is bigger than the potential so it has to be a wave. But here they energy is still bigger than the potential so it also must be a wave. But a wave with different kinetic energy, different momentum, therefore different de Broglie wavelength and different k. But we know from Schrodinger's equation what that should be. This wave is also moving to the right, because probably if I have a wave moving to the right here, it produces some transmitted wave to the right. But then, you could try solving the Schrodinger equation with this. It won't be enough because physically you would expect the wave bouncing back as well from here. So I will put a B e to that minus ikx. That's a wave moving towards the left with an unknown coefficient. And, now let's get those constants. I'll finish in two minutes. What is k? Well if you have energy E, you know that the energy is h squared k squared over 2 m. You can look at the Schrodinger equation with 0 potential over there. And therefore, k squared is also 2mE over h squared. It's a combination you've been seeing quite a bit. The intuition for k bar should be that k bar squared is 2m times the kinetic energy, so it should be e minus v0 over h squared. So these are k and k bar. And the wave function must be continuous at x equals 0. That gives you A plus B equal to C. At 0, all the exponentials vanish. And the derivative must be continuous at x equals 0. And the derivative being continuous because there's no delta function anywhere here. So you have ikA, that's the derivative of the first term, minus ikB, the same k in that region of course, is equal to i k bar C. So from this you get A minus B is equal to k bar over kC. Two equations and two unknowns. And that's OK, even though there are three coefficients, because the way to think of this is that you're sending in some wave and you're going to get some reflection and some transmission. So in some sense, A is the input. You could want to call it 1 or whatever. So what we're looking for is what is B over A? And what is C over A? And these two equations, it's a one line computation. I'll write the answer. B over A is k minus k bar over k plus k bar. Do it for fun. And C over A is 2k over k plus k bar. B gives you a sense of how much is reflected. C, how much is transmitted. But this is the beginning. Because this is not a particle coming in. So we'll have to build the packet and send it in and see how this relations tell you what's going to happen. So this is a nice story that we will develop next time.

MIT_804_Quantum_Physics_I_Spring_2016

Time_evolution_of_a_free_particle_wavepacket.txt

PROFESSOR: Time evolution of a free particle wave packet. So, suppose you know psi of x and 0. Suppose you know psi of x and 0. So what do you do next, if you want to calculate psi of x and t? Well, the first step, step one, is calculate phi of k. So you have phi of k is equal 1 over square root of 2 pi integral dx psi of x, 0 e to the minus ikx. So you must do this integral. Step two-- step two-- with this, now rewrite and say that psi of x, 0 is 1 over square root of 2 pi dk e to the-- no, I'm sorry-- phi of k, e to the ikx. So that has achieved our rewriting of psi of x and 0, which was an arbitrary function as a superposition of plane waves. Step three is the most fun step of all. Step three-- you look at this, and then you say, well, I know now what psi of x and t is. Evolving this is as easy as doing nothing. What I must do here is 1 over square root of 2 pi-- just copy this-- dk, phi of k, e to the ikx. And I put here minus omega of k, t. And I remind you that h bar omega of k is the energy, and it's equal to h squared k squared over 2m. This is our free particle. And I claim that, just by writing this, I've solved the Schrodinger equation and I've time-evolved everything. The answer is there-- I didn't have to solve the differential equation, or-- that's it. That's the answer. Claim this is the answer. And the reason is important. If you come equipped with a Schrodinger equation, what should you check, that ih bar d psi dt is equal to h psi-- which is minus h-- squared over 2m, d second, dx squared psi. Well, you can add with ih d dt on this thing. And you remember all that happens is that they all concentrate on this thing. And it solves this, because it's a plane wave. So this thing, this psi of x and t, solves the Schrodinger equation. It's a superposition of plane waves, each of which solves the free Schrodinger equation. So, we also mention that since the Schrodinger equation is first ordered in time, if you know the wave function at one time, and you solve it, you get the wave function at any time. So here is a solution that is a solution of the Schrodinger equation. But at time equals 0-- this is 0-- and we reduce this to psi of x and 0. So it has the right condition. Not only solve the Schrodinger equation, but it reduces to the right thing. So it is the answer. And we could say-- we could say that there is a step four, which is-- step four would be do the k integral. And sometimes it's possible. You see, in here, once you have this phi of k, maybe you can just look at it and say, oh, yeah, I can do this k integral and get psi of x and 0, recover what I know. I know how to do-- this integral is a little harder, because k appears a little more complicated. But it has the whole answer to the problem. I think one should definitely focus on this and appreciate that, with zero effort and Fourier's theorem, you're managing to solve the propagation of any initial wave function for all times. So there will be an exercise in the homework, which is called evolving the free Gaussian-- Gaussian. So you take a psi a of x and time equals 0 to be e to the minus x squared over 4a squared over 2 pi to the 1/4-- that's for normalization-- square root of a. And so what is this? This is a psi-- this is a Gaussian-- and the uncertainty's roughly a-- is that right? Delta x is about a, because that controls the width of the Gaussian. And now, you have a Gaussian that you have to evolve. And what's going to happen with it? This Gaussian, as written, doesn't represent a moving Gaussian. To be a moving Gaussian, you would like to see maybe things of [? the ?] from e to the ipx that represent waves with momentum. So I don't see anything like that in this wave function. So this must be a Gaussian that is just sitting here. And what is it going to do in time? Well, it's presumably going to spread out. So the width is going to change in time. There's going to be a time in which the shape changes. Will it be similar to what you have here? Yes. The time will be related. So time for changes. So there will be some relevant time in this problem for which the width starts to change. And it will be related to ma squared over h bar. In fact, you will find that with a 2, the formulas look very, very neat. And that's the relevant time for the formation of the Gaussian. So you will do those four steps. They're all doable for Gaussians. And you'll find the Fourier transform, which is another Gaussian. Then you will put the right things and then try to do the integral back. The answer is a bit messy for psi, but not messy for psi squared, which is what we typically ask you to find.

MIT_804_Quantum_Physics_I_Spring_2016

Time_delay_and_resonances.txt

PROFESSOR: Let me begin by introducing the subject. The subject is resonances. And we have seen, actually, a little bit of this in the resonant transmission of the Ramsauer-Townsend effect. Because of a resonance phenomenon within the square well obstacle, somehow, for some particular frequencies, for some particular energies, the particles were able to zoom by without experiencing any reflection, whatsoever. So let's begin the subject of resonances by asking a question. If you have the usual potential, the short range potential, which means, that for some distance, R, greater than 0, the potential is 0. Here we put a barrier, and over there could be anything, some potential. We've computed some-- this concept of time delay, there's a formula for the time delay. In fact, it was given by 2 h bar d delta dE, the time delay, 2 h bar d delta dE. And we discussed that this time delay can be positive or it can be negative. If it's positive, it really means a time delay. You send in a wave packet. And it takes time to come back, more time than it would have taken if there had been no potential. You see, the time delay, you have a packet coming in from time minus infinity. And then it bounces back a time equal infinity. But nevertheless, you compare that with a situation in which there's no potential. And you see that there is some time delay. If you time the wave packet to reach at time equals 0, here, it will not reach back to where you were by time-- by whatever time-- suppose you have the wave packet here at t equal minus 10, then it goes here, and it delays, and at t equal 10, the packet has not reached, there is a time delay, a positive time delay. A negative time delay is the opposite. The packet arrives a little earlier. And the question I want to ask you, if you have a negative time delay, can it be arbitrarily large. Well, if you send in a wave packet, it may find an infinite wall here, and then may bounce, and then yes, it comes back earlier than you expected, because the free packet would have gone here and back. But you wouldn't expect it to be able to come earlier than if there was an infinite wall here, because there is no infinite wall here, nor an infinite wall here. So it's just not going to bounce before it reaches here. The best it can do is bounce when it reaches here. So you should not expect, and this, sometimes, will [INAUDIBLE], there is nothing that can make it bounce until you reach here. So you cannot expect that the time advance is larger as if it would have bounced before reaching the obstacle, whatsoever. So this is important. We cannot have a negative time delay that this infinitely large. So, in fact, the time delay as, we're right in here, should be greater than the total travel distance that you may save. If you bounce here, you would save 2R over v. And you must be greater than that negative number, which is the total travel time that it would take to go back and forth, here. So we can do a little arithmetic, here. This is equal to 2 h bar d delta dk, and here, dE dk. This is still greater than or equal than minus 2R over v. And I want to put a sim, because our argument is not completely rigorous as to what's happening when it reaches here. It seems very plausible classically, but there's a bit of a correction if you do it exactly. So it's not an exact inequality we're deriving. And what is the E dk is h bar times the velocity. Remember, dE dk, you are differentiating h squared k squared over 2m. And you get h bar times hk over m. So therefore, this is h bar and the velocity. And the h bars cancel. The velocities cancel. Between these two sides, the 2s cancel. And you'll get that d delta dk must be greater than or equal, approximately, to R. And that's sometimes called Wigner's condition on scattering. And it basically is the idea that the time delay, the time advance cannot be too large. OK, so now we can ask the second question. How about the time delay, a true time delay, can it be very large? Can it be arbitrarily large? Suppose we have a barrier of this form. And now you send a particle with a little bit higher energy here. Now, this particle is going to have very little kinetic energy. So it's going to travel quite slowly here, and go back. And this time, it's going to delay quite a bit, probably. But the problem is, if you create-- there's nothing very peculiar about this, if you go a little lower, than you're advanced, and then suddenly, it gets delayed. It's not that evident, but the phenomenon of resonance is precisely what we get when we, sort of, trap the particle. And then we make it be, as far as it seems, arbitrarily large, if you design a well properly. But the thing that we have to design, the example of what we're going to design, is different from all the things I've drawn so far. It's the following way, this is just an example. I have this zero line of the energy. This is v of x. This is x. And then I put an attractive potential here. And here is minus v0. And then I put a barrier here with a v1. So what I'm going to aim at is, you see, if v1 will be extremely large, there will be-- well, if v0 is extremely large, then begin there would be bound states here, but these are not scattering states. On the other hand, if v1 will also be infinite, you would have bound states here, but they could not escape. So certainly, if I combine these two, I put a v0 and maybe a larger v1, I can almost create bound states here. But they're not really bound states, because they can leak out and produce scattering states. But these are going to be resonances. This part and this, this being a attractive, trying to keep the particle in, and this being a barrier, can combine to produce a state that gets trapped here, and stays a very long, time, will have a very long time delay. And that's the phenomenon of resonances. We need to trap that particle, somehow. And we're going to see now the details of how this works, and what the properties are. Now, it's very interesting that actually, these resonances occur at some particular energies. And they have different properties. But we can identify energies of resonances. And these are not bound states. They're just resonances. They eventually escape. And they're not normalizable, really, but in some ways they behave as bound states for awhile. They stay there for a while and do nice things. So let's set this off. Now we're going to spare you a little bit of these calculations, because the important thing is that you know how to set it up, and if you get an answer, you know how to plot it, how to get the units out, how to try to understand it. So that's what we're going to do. I'm going to put an energy here, an energy, E. And I'm going to receive E to be less than v1 and greater than 0. I don't expect true resonances beyond, because the particle just bounces out. It doesn't get trapped. The phenomenon of resonance is a little more intricate than just having a long time delay. There's more that has to happen. Another thing that will happen, is if the particle spends a lot of time here, you would find, in this spirit of resonance, that the amplitude of the wave function here is going to be very big. So you will scan the energy and the amplitude. It will be normal, normal, normal. And suddenly for some energy it becomes very big. And we're going to do that. The way I'm going to develop that, we're going to calculate this, plot these things. And then we are going to ask whether there is a mathematical condition that picks resonances. Well, how do I, if I want to explain to somebody in 30 seconds where are the resonances, how do you calculate them, you cannot tell that somebody, OK, calculate it for all energies, do all the plots, and see some peak in some thing, and this is a resonance. This is what we're going to do to begin with, but then we'll get more sophisticated. So let's put k in. So let's call this k prime, the wave number in this area. Kappa here, because it's a forbidden region, and k over here, as usual. So k squared is 2mE over h bar squared. K prime squared is equal to 2m, the total kinetic energy is E plus v0, over h squared. And kappa squared is again, similar formula, but this time is the energy differential between v1 and E, so 2m v1 minus E over h squared. All of these three numbers are positive. And they are the relevant constants to write wave functions. So we have to write a wave function. And I'm going to write a wave function because it takes a little tinkering to do it in an efficient way. There is one that you don't have to think, you just have to remember. It's the one outside. It's the universal formula, e to the i delta sine kx plus delta is valid for for x greater than R. This one we derived at the beginning of our analysis of scattering. How about the other region. Oops, I should have put letters here. These are a and 2a they are positions. And therefore, it's not R in here. Well, it's R, it's the range of the potential, but here is 2a. How about the other one? In this region, it's kind of simple again. The wave function has to vanish here, has to be sines or cosines of k prime. So it has to be a sine function of k prime. And since we don't put an extra constant in here, we kind of put an extra constant in here, there must be a constant here, A sine of k prime x. And that must be for x between 0 and a. We used k prime, the wave from over there. And there is A. And what we were saying about resonances, is that, well, A may depend on k. And when you have a resonance, A is going to [INAUDIBLE], presumably because the particle spends a long time inside the well. And now I have to write this one in here. And this is the one that, you can do it, do a little bit more work, or do it kind of, efficiently. In that region we have exponentials, like we have e to the kappa x and e to the minus kappa x. Or I may want to have sinh of kappa x and cosh of kappa x to write my solutions. But I actually don't want either of them too much, because I would like to write an answer that almost imposes continuity in a nice way. So I could use sinh of kappa x minus a and cosh of kappa x minus a. These are all solutions. You can choose whichever pair you want. So for example, if I want to implement continuity with this thing, this wave function, I want to write something that I don't have to write another equation for continuity. So I will write A sine of k prime a-- so far, this wave function, if x equal a, coincides with this one. But this is no wave function yet, not with an x dependence, so I have to put more. But then, I know that cosh is 1 for x equals zero. So I put here a cosh kappa x minus a. And now this is a solution that matches that one at x equals a. At x equal a, the cosh becomes 1 and matches. But this kind of need a complete solution. It's not general enough. So you have to put a B sinh of kappa x minus a. And this won't ruin the matching, because at x equal a, that second term vanishes. So we're still matching well there. And Well matching here is non-trivial when I impose some conditions. So you still have to match derivatives and do a little bit of work but not too much work.

MIT_804_Quantum_Physics_I_Spring_2016

Units_of_h_and_Compton_wavelength_of_particles.txt

BARTON ZWIEBACH: Now that we've introduce h, h is a very important quantity in quantum mechanics. So let's talk a little more about h, its units, and we already put one number that I really wish you will remember. Now let's talk about the units of h and some other things you can do with h. So units of h. So if you have a quantity that appears for the first time and as it appears here, E equal h nu, this is a good place to understand the units of h because the units of h would be units of energy divided by units of frequency. And I put this square brackets to denote units. Now what are the units of energy? We're going to work with units that are characterized by M, L and T-- mass, length, and time. So energy, you think kinetic energy and you say, MV squared, so that's a mass and velocity squared is L squared over T squared. So that's units of energy. Frequency is cycles per unit time. Cycles have a number of units, so it's 1 over time here. So then you say that it's ML squared over T. So that's the first answer and that's a nice answer, although it's never quite that useful in this way, so we try to rearrange it. And I will rearrange it the following way to think-- you see, it's nice to think of what physical quantity that we are familiar, hats units of h-bar. We know these units of h-bar are energy over frequency, but that's not a single physical quantity, so let's look at it and separate this as L times MLT. That's the same thing. And then I see an interesting thing-- this is the units of position, or length. Length or radius. Distance. And this has the units of momentum p. Momentum. So this product has the units of angular momentum. And perhaps that's the most important quantity that has the units of h-bar. It's something that you should remember. Of h-bar has units of angular momentum, that's why when people talk about a particle of spin 1/2, they say the angular momentum is 1/2 of h-bar, and that has the right units. So spin 1/2 particle-- 1/2 particle-- means that the magnitude of the intrinsic angular momentum is 1/2 of h-bar. h or h-bar have the same units, they just differ by a 2 pi that-- unfortunately, we have to be careful about that 2 pi, it affects numbers and some formulas are nicer without the bar, some formulas are less nice. So OK. So another thing that you could say is that this h allows you to construct all kinds of new quantities. And that's a nice thing to do. Whenever you have a new constant of nature that comes up, and we have the speed of light, Planck's constant, Newton's constant-- seem to be the three fundamental units of nature-- you can do some things. And you can look at this quantity-- h is proportional to rp and get an inspiration. So you can think h has units of r times p. And you can say, look-- if I have any particle with mass M, I can now associate a length to it. I can invent a length associated to that particle. And how do I do it? Well, this has units of length, so all I have to do is take h and divide by p. Well, that will be one way to get the length where p is the momentum, and it will be called the de Broglie wavelength. But there is another way. Suppose this particle is just not moving and you have the momentum and you say, wow, momentum is not moving, so what's going on here? So think of it at rest and then you say, well, you still can construct a length. You can put h and divide by the mass times the velocity of light, why not? That's a velocity, it is a constant of nature. So that way, you associate a length to any particle of a given mass. You don't have to tell me what is the momentum. You can just know the mass and it has a length associated to it. So it's called the Compton-- Compton-- wavelength of a particle. And I want to make sure you don't confuse, it's not the same as de Broglie wavelength that we will see later. It's not the same as the de Broglie wavelength. This is the Compton wavelength of the particle. And you can say, all right, good, you give me a particle of some mass, I can tell you what a length associated to it-- why would it be important? It will be important in two different ways-- through an experiment and through a thought experiment, which I want to do right now. You see, I could ask the following question-- I have this particle, has a mass M. I use the speed of light, so with that mass M, I could associate out to this particle a rest energy. MC squared. That's the rest energy. And then I could ask, what is the wavelength of a photon that has the same energy as the rest energy of this particle? So you translate the question into a question of a length. Once you have some energy, there is a natural length, which is the wavelength of a photon with that energy. So let's ask this question independently of what we did. So what is the wavelength-- wavelength-- of a photon whose energy is the rest mass-- rest mass-- of a particle? So the rest mass is MC squared, and that's the energy of this photon. And we know that energy of a photon h nu or hC over lambda, and there, we can calculate the lambda. Lambda is hC over MC squared, and no surprise, it gives us h over MC and that thing is the Compton wavelength. So it's sometimes called l-Compton of the particle of mass M. So this is a way that you can think of this particle. You think of a particle, you have a Compton wavelength, and that Compton wavelength is the wavelength of light that has that rest energy. And that actually has experimental implications in high energy particle physics. Because if you have an electron and it has a Compton wavelength, and you shine a photon that has that size, that photon is carrying as much energy as the rest energy of the electron. And in particle theory and quantum field theory, particles can be created and destroyed, so this photon maybe can do some things and create more particles out of this electron, particle equation could start. happening. So it will be difficult to isolate a particle to a size smaller than its Compton wavelength, because the photons could do such damage to the particle by creating new particles or doing other things to it. So for an electron, let's calculate the Compton wavelength. So l-Compton of an electron would be h over MeC, and you would do h-- you would do 2 pi h-bar C over and MeC squared. And you've got 2 pi 197.33 MeV fermi, and here you would have 0.511 MeV. So this gives you 2,426 fermi, or about 2.426 picometers. Picometers is kind of a natural length. Picometer is 10 to the minus 12 meters. The Bohr radius is about 50 picometers, so that's how big this thing is. Is it still much bigger than the size of the nucleus? The nucleus is a few fermis. A single proton is about a fermi big. And nucleus grow slowly, so you can have a big nucleus with 200 particles maybe of a radius of 7 or 8 fermi. So this is still a lot bigger and it's a very interesting quantity that will show up very soon.

MIT_804_Quantum_Physics_I_Spring_2016

Interferometer_and_interference.txt

PROFESSOR: And let me I assume, for example, that I'll put the state alpha beta in. Alpha and beta. What do I get out? So you have this state, alpha beta. What do you get out? Well, state comes in and is acted by beam splitter 1. So you must put the beam splitter, 1 matrix. And then it comes the mirrors. And lets assume mirrors do nothing. In fact, mirrors-- the two mirrors would multiply by minus 1, which will have no effect. So lets ignore mirrors. And then you get to beam splitter 2 and you must multiply by the matrix of beam squared 2. And that's the output. And that output is a two-component vector. That gives you the amplitude up and the amplitude down. So I should put BS2 here, BS1 over here, alpha beta. The numbers move away, 1 over square root of 2 and 1 over square root of 2. Commute in matrix multiplication. Then you multiply these two matrices. You get 0, 2, minus 2, and 0, alpha beta. And you put the 2 in so you get beta minus alpha. So here is the rule. If you have alpha and beta, you get, here, beta and minus alpha here, or a beta minus alpha photon at the end. Good. So let's do our first kind of experiment. Our first experiment is to have the beam splitters here. D0-- detector D0 and detector D1 over there. And let's send in a photon over here only-- 1 or input 0, 1. Well this photon, 0, 1, splits here. You act with BS1-- the matrix BS1. You get two things. You act with the matrix BS2, and it gives you this. But we have the rule already. If you have an alpha beta, out comes a beta minus alpha. So it should have as 1, here, and minus 0, here, which is 0. So you get a 1, 0. So what is really happening? What's really happening is that your photon that came in divided in two, recombined, and, actually, there was a very interesting interference here. From the top beam came some amplitude and gave some reflected and some transmitted. From the bottom beam, there was some transmission and some reflection. The transmission from the top and reflection from the bottom interfered, to give 0. And this, too, the reflection from the top and transmission from the bottom, were coherent and added up to 1. And every single photon ends up in D0. If you would put the beam-- well, Mach and Zehnder were working in the late 1800s, 1890s. And they would shine light. They had no ability to manipulate photons. But they could put those beam splitters and they could get this interference effect, where everything goes to D0. So far, so good. Now let me do a slightly different experiment. I will now put the same thing, a BS1 and a beam going in, mirror, mirror, BS2 here. But now, I will put a block of concrete here on the way. I'll put it like this. So that if there is any photon that wants to come in this direction, it will be absorbed. Photon could still go like this, but nothing would go through here. And here, of course, there might be D0 and D1. And here are the mirrors, M and M. Now the bottom mirror is of no use anymore because there is a big block of concrete that will stop any photon from getting there. And we are asked, again, what happens? What do the detectors see? And this time, we still have a 01. Now I would be tempted to use this formula, but this formula was right under the wrong assumption-- that there was no block here. So I cannot use that formula. And certainly, things are going to be different. So I have to calculate things. And we're doing a quantum mechanical calculation. Well, up to here, before it reaches here, I can you do my usual calculation. Certainly, we have BS1 acting on the state, 01, and this is 1 over square root of 2, I think, minus 1, 1, 1, 1. Yup, that B is 1, acting on 01. And that gives me 1 over square root of 2, 1 over square root of 2. So, yes, here I have one over square root of 2 amplitude. And here I also have 1 over square root of 2 amplitude. OK. Now that's the end of this amplitude. It doesn't follow. But on the other hand, in this branch, the mirror doesn't change the amplitude, doesn't absorb. So you still have 1 over square root of 2 here. And now you're reaching BS2. Now what is the input for BS2? The input is a one over square root 2 in the top beam, and nothing in the lower beam because nothing is reaching BS2 from below. This is blocked. So yes, there was some times when something reached from below, but nothing here. So to figure out the amplitudes, here, I must do BS2 acting on 1 over the square root of 2, 0. Because 1 over square root of 2 is coming in, but nothing is coming in from below. And, therefore, I get 1 over the square root of 2, 1, 1, 1, minus 1, 1 over square root of 2, 0. This time, I get 1/2 and 1/2. OK, we must trust the math. 1/2 here and 1/2 there, so 1/2 a column vector, 1/2, 1/2. OK, let me maybe tabulate this result, which is somewhat strange, really. So what is strange about it is the following. In the first case, where the interferometer was totally clear, nothing in the middle, everything went to D2. And nothing went into D1. But now, you do something that should block some photons. You block some photons in the lower path, and yet, now you seem to be able to get something into D1. There is an amplitude to get into the D1. So by blocking a source, you're getting more somewhere. It's somewhat counterintuitive. You will see by the end of the lecture in 10 minutes, that it's not just somewhat counterintuitive, it's tremendously counterintuitive. Let's summarize the result here-- the outcome in the blocked lower branch case and the probability for those events. So photon at the block-- the photon can end in three places. It can end on the block. It can end on the D0. Or it can end on D1. So photon at the block-- well, the amplitude to be here is one over square root of 2. The probability should be 1/2. Photon at D0, probability amplitude, 1/2, probability, 1/4-- photon at D1, probability, 1/4. You could put another table here-- outcome all open, probability. And in this case, there's just photon at D0. That's 1. And photon at D1 was 0.

MIT_804_Quantum_Physics_I_Spring_2016

RamsauerTownsend_phenomenology.txt

PROFESSOR: If you have potential transmission coefficient for a potential where z0 is equal to 13 pi over 4. That's a square well of certain depth, and we represent it in this way. Remember n must be greater than or equal than 2z0 over pi. So this will be 13/2. And 13/2 means that we can start with n equals 7, 8, 9, and all those. Remember, this n counts which bound state of the infinite square well you're talking about. And the energy that you must use are your integers, are positive energy. So positive energies mean that you have sufficiently large n, and the n that this sufficiently large is 7 in this case. So you can then determine from this formula what is the value of e n over v0. So for example E7 over v0 turns out to be 0.15976. E8 over v0 turns out to be 0.514793. And E9 over v0 is 0.91716. So if you plot it, you have here E, or capital energy, over v0. And you want to plug the transmission probability. And it begins with 0. That was the question a second ago. And then it may reach 1. And it will reach 1 at each one of those values. So if, here is 1, 0.15. There will be 0.15, 0.51, and 0.92. So you get this, and here another one, and here another one. Probability like that. So that's a typical graph for the transmission probability. It oscillates, and it reaches 1 at several points forever and ever. And the amplitude become smaller, so it's really overall tending to 1. So these two people we're talking about, Ramsauer and Townsend. They lived from 1860s to 1940s and '50s. And they did their famous experiment in 1921. So their experiment was elastic scattering of low energy electrons off of rare gas atoms. So Ramsauer and Townsend, in 1921, they scattered elastically. That means the particles didn't change their identities. They didn't create more particles. It was just electrons came in and electrons went out. Electrons. And these are low energy electrons, off of rare gas atoms. So these are noble gases. Their shells are completely filled. And they're rather inert, very unreactive, high ionization energies, no low energy states you can scatter these atoms into. So basically very unreactive atom. And you can imagine it as a very beautiful spherical cloud. We can draw some electrons, there's some protons, a nucleus here, and an electron cloud. So how does this look to an electron? Well, you know from electrostatics that if you have total charge 0 and it's totally spherically symmetric, no electric field outside. So the electron comes in, feels nothing. And as soon as you penetrate this, at any point here, the electric field points in. Or, well, it actually points out, but the electron will feel a force in. Because the charge in the outside shell doesn't produce any field. But now, the protons in the nucleus beat the effect of the electrons. So there's a force in, a force in, that goes in. So basically this is like a deep square well, or spherical well, representing the atom. The atom can be some sort of spherical well that attracts the electrons. So what these people did were throwing these electrons. And they considered that this electron scattered a lot when they bounced back. On the other hand, if they continued, if the electrons pass by, they said nothing has happened. So the reflection coefficient for them, the reflection coefficient. Reflection coefficient is a proxy, a good representation for the scattering cross-section. So the reflection coefficient, what they found experimentally was a reflection coefficient, R, that as a function of energy was very high. And people thought at this moment, OK, these are like particles colliding with particles. Their energies shouldn't make much difference, you know. You either collide or you don't collide, and you bounce back or you don't bounce back. So they thought that this would be flat. But nevertheless, it actually went down enormously, and then it went up again. So they found that for electrons with about 1 Ev, that's very low energy electrons, but they were going pretty fast. And E1, Ev electron is going like at 600 kilometers per second. So the reflection was going like this. And they had no explanation why it was so sensitive with energy, and why there would be a funny effect going on, that the reflection would suddenly go down, and just basically the particles would get transmitted. But if you think of reflection here as a continuous line and transmission as a dotted line, the transmission that must alter the reflection to be 1 would be going up here and would have reached near 1 at this value of the energy. So the explanation eventually was this effect, that you should do well and there is a resonant effect in the well, and for some energies the resonance is such that it allows the particles to just go through and not scatter. So this had to wait some time, because the experiment was done in 1921, and Schrodinger and everybody started doing good work in 1925, and of wave mechanics took a while. But eventually it was recognized that basically it's resonant transmission, what is happening there. Well, if you want to get the numbers right, if you want to get that Ev better, you have to do a spherical model of the square, finite square well, you have this spherical well, and do it a little more precisely. But then the agreement is pretty reasonable.

MIT_804_Quantum_Physics_I_Spring_2016

Orthonormality_of_spherical_harmonics.txt

PROFESSOR: How do we state the issue of normalization? See, the spherical harmonics are functions of theta and phi. So it makes sense that you would integrate over theta and phi-- solid angle. The solid angle is the natural integration. And it's a helpful integration, because if you have solid angle integrals and then radial integrals, you will have integrated over all volume. So for this spherical harmonic, solid angle is the right variable. And you may remember, if you have solid angle, you have to integrate over theta and phi. Solid angle, you think of it as a radius of one. Here is sine theta. So what is solid angle? It's really the area on a sphere of radius one. The definition of solid angle is area over radius squared, the part of the solid angle that you have. If you're working with a sphere of radius one, it's the area element is the solid angle. So the area element in here would be, or the integral over solid angle-- this is solid angle, d omega-- you would integrate from theta equals 0 to theta equals pi of sine theta, d theta. And then you would integrate from 0 to 2 pi of d phi. Now, you've seen that this equation that began as a differential equation for functions of theta ending up being the differential equation for a function of cosine theta. Cosine theta was the right variable. Well, here it is as well, and you should always recognize that. This is minus d of cosine theta. And here you would be degrading from cosine theta equals 1 to minus 1. But the minus and the order of integration can be reversed, so you have the integral from minus 1 to 1 of d cos theta and then the integral from 0 to 2 pi of alpha. So this is the solid angle integral. You integrate d of cosine theta from minus 1 to 1 and 0 to 2 pi of d phi. And we will many times use this notation, the omega, to represent that integral so that we don't have to write it. But when we have to write it, we technically prefer to write it this way, so that the integrals should be doable in terms of cosine theta. So what does this all mean for our spherical harmonics? Well, our spherical harmonics turned out to be eigenfunctions or Hermitian operators. And if they have different l's and m's, they are having different eigenvalues. So eigenfunctions of Hermitian operators with different eigenvalues have to be orthogonal. So we'll write the main property, which is the integral of 2l, say l prime, m prime, of theta and phi. And he put the star here. I'll put the star just at the Y, and he's complex conjugated the whole thing. Remember that in our problem, one wave function was complex conjugated. The other wave function is not complex conjugated. They're different ones because l and m and l prime and m prime could be different. So orthogonality is warranted. Two different ones with different l's and different m's should give you different values of this integral. So at this moment, you should get a delta l prime l delta m prime m. And if l and m are the same as l prime and m prime, you have the same spherical harmonic. And all this tremendous formula over there, with 2l plus 1 and all these figures, you're guaranteed that in that case you get 1 here. So this formula is correct as written. That is the orthonormality of this solution. Probably, this stage might be a little vague for you. We saw this a long time ago. We may want to review why eigenfunctions of Hermitian operators with different eigenvalues are orthogonal and see if you could prove. And you do it, or is it kind of a little fuzzy already? We saw it over a month ago. So time to go back to the Schrodinger equation. So for that, we remember what we have. We have minus h squared over 2m Laplacian of psi plus V of r psi equals E psi. And the Laplacian has this form, so that we can write it the following way-- minus h squared over 2m 1 over r d second dr squared r psi-- I won't close the brackets here-- minus this term. So I'll write it minus one over h squared r squared l squared psi plus V of r psi is equal to E psi. Now, you could be a little concerned doing operators and say, well, am I sure this l squared is to the right of the r squared? r and l-- l has momentum. Momentum [INAUDIBLE] with r. Maybe there's a problem there. But rest assured, there is no problem whatsoever. You realize that l squared was all these things with dd phetas and dd phis. There was no r in there. It commutes with it. There is no ambiguity. We can prove directly that l squared commutes with r, and it takes a little more work. But you've seen what l squared is. It's the dd thetas and dd phis. It just doesn't have anything to do with it. So now for the great simplification. You don't want any of your variables in this equation. You want to elicit to a radial equation. So we try a factorized solution. Psi is going to be-- of all the correlates-- is going to be a product a purely radial wave function of some energy E times a Ylm of theta and phi. And we can declare success if we can get from this differential equation now a radial differential equation, just for r. Forget thetas and phi. All that must have been taken care of by the angular momentum operators. And we have hoped for that. In fact, if you look at it, you realize that we've succeeded. Why? The right-hand side will have a factor of y and m untouched. V of r times psi will have a factor of Ylm untouched. This term, having just r derivatives will have some things acting on this capital R and Ylm untouched. The only problem is this one. But l squared on Ylm is a number times Ylm. It's one of our eigenstates. Therefore, the Y's and m's drop out completely from this equation. And what do we get? Well, you get minus h squared over 2m 1 over r, d second, dr squared, r capital RE minus l squared on psi lm-- or Ylm now-- is h squared times l times l plus 1. So the h squared cancels. You get l times l plus 1 r squared, and then we get the RE of r times the psi lm that has already-- I started to cancel it from the whole equation. So I use here that L squared from the top blackboard over there, has that eigenvalue, and the psi lm has dropped out. Then I have the V of r RE equals E time RE of r. So this is great. We have a simplified equation, all the angular dependencies gone. I now have to solve this equation for the radial wave function and then multiply it by a spherical harmonic. And I got a solution that represents a state of the system with angular momentum l and with z component of angular momentum m. The only thing I have to do, however, is to clean up this equation a little bit. And the way to clean it up is to admit that, probably, it's better as an equation for this product. So let's clean it up by multiplying everything by r. dr squared of little r RE. If I multiplied by r here, I will have plus h squared over 2m l times l plus 1 over r squared rRE plus V of r times r times RE equals E times r times RE of r. So we'll call u of r r times RE of r, and look what we've got. We've got something that has been adjusted, but things worked out to look just right. Minus h squared over 2m d second, dr squared u of r plus-- let me open a parentheses V of r plus h squared over 2m r squared, l times l plus one u of r is equal to E times u of r. Here it is. It's just a nice form of a one-dimensional Schrodinger equation. The radial equation for the wave function dependents, a long r, has become a radial one-dimensional particle in that potential, in which you should remember two things. That this u is not quite the full radial dependent. The radial dependent is RE, which is u over r. But this equation is just very nice. And what you see is another important thing. If you look at the given particle in a potential, you have many options. You can look first for the states that have 0 angular momentum-- l equals zero-- and you must solve this equation. Then you must look at l equals 1. There can be states with l equals one. And then you must solve it again. And then you must solve for l equals 2 and for l equals 3 and for all values of l. So actually, yes, the three-dimensional problem is more complicated than the one-dimensional problem, but only because, in fact, solving a problem means learning how to solve it for all values of l. Now, you will imagine that if you learn how to solve for one value of l, solving for another is not that different. And that's roughly true, but there's still differences. l equals 0 is the easiest thing. So if the particle is in three dimensions but has no angular momentum-- and remember, l equals 0 means no angular momentum-- it's this case. l equals 0 means m equals 0. l squared is 0. lz is 0. This is 0 angular momentum.

MIT_804_Quantum_Physics_I_Spring_2016

Energy_below_the_barrier_and_phase_shift.txt

PROFESSOR: Let's do E less than V not. So we're back here. And now of the energy e here is v not is x equal 0. X-axis. And that's the situation. Now you could solve this again. And do your calculations once more. But we can do this in an easier way by trusting the principle of analytic continuation. In this case, it's very clear and very unambiguous. So the big words, analytic continuation, don't carry all the mathematical depth. But it's a nice, simple thing. We first say that the solution is the same for x less than 0. So for x less than 0, we write the same solution. Because the energy is greater than 0, or all what we said here, the value of k squared, a into the ikhd e to the minus i k x. It's all good. And k squared is still 2 m e over h squared. The problem is the region where x is greater than 0. Because there you have an exponential. But now you must have a decaying exponential. But we know how that should work. It should really be an e to the minus some kappa x. So how could I achieve that? If I let k bar replace-- everywhere you see k bar, replace it by i kappa. Park then one thing that happens is that-- you learn from here, from this k bar squared, would be minus kappa squared equal to that. So kappa squared would be 2 m, v not minus e over h squared. The sign is just the opposite from this equation. That's what that equation becomes upon that substitution. Now that substitution would not make any difference, how they put a plus i or minus i, I wouldn't have gotten my sine change, and this. But if I look at the solution there, the solution psi becomes-- on the region x greater than 0, turns into c, e to the i. kappa bar is a i kappa x. Therefore, it's equal to c, e to the minus kappa x, which is the right thing. And that sine that I chose, of letting k equal i kappa proves necessary to get the right thing. So it's clear that to get the right thing, you have that. And now you know that, of course, if you would have written the equation from the beginning, you would have said, yes, in this region, there is a decaying thing. And looking at the Schrodinger equation, you have concluded that kappa is given by [? that. ?] But the place where you now save the time is that, since I just must do this change in the equations, I can do that change in the solutions as well. And I don't have to write the continuity equations again, nor solve them. I can take the solutions and let everywhere that was a kappa bar replaced by i, that was a k bar, replace it by i kappa bar. So what do we get? It should go here and believe those circuits. OK, so b over a, that used to be. Top blackboard there. Middle, k minus k bar becomes k plus i-- no, minus I k bar, minus i kappa. And k plus i kappa. so it has changed. Suddenly this ratio has become complex. It's kind of interesting. Well, let's make it clearer by factoring a minus i here. So this becomes kappa. And you need plus k, so this must be plus i k over i. This would be kappa minus i k. So this is just minus kappa plus i k over kappa minus i k. But when you see that ratio, you're seeing the ratio of two complex numbers of equal length. And therefore, that ratio is just a phase. It's not any magnitude. So this is just a phase, and it deserves a new name. There is a phase shift between the b coefficient and the a coefficient. And we'll write it as e minus-- the minus I'll keep. e to the 2 i delta. That depends on the energy. I'll put delta of the energy, because after all, kappa, k, everybody depends on the energy. So let's call it 2 i delta of e. And what is delta of e? Well, think of the number kappa plus i k. This is the k, i k here. The angle-- this complex number has an angle that, in fact, is delta. e to i delta is that phase. And delta is the arc tangent, k over kappa. So I'll write it like that. Delta. Now you get a delta from the numerator, a minus delta, as you can imagine, from the denominator. And that's why you get a total 2 i delta here. So delta of e is 10 minus 1, k over kappa. And if you look at what k and kappa were, k over kappa is like the ratio of the square root of the energy over v not minus the energy. So delta of e is equal to 10 minus 1 square root of energy over e v not minus and energy. Now I got the question about current conservation. What happens to current conservation this time? Well, you have all these waves here. But on the region x greater than 0, the solution is real. If the solution is real, there is no probability current on the right. There's really no probability that you get this thing, and you get current flowing there. And you get this pulse, or whatever you send to keep moving and moving and moving to the right. Indeed, the solution decays. And it looks like the one of the bound state in this region. So eventually there's no current here, because there's no current here. No current there. No current there. Because the solution drops down. But it's a real solution anyway. So there's no current there. So Ja-- Jc is equal to 0. 0. Solution is real for x greater than 0, and any way goes to 0 at infinity. So the fact that it's real is a mathematical nicety that help us realize that it must be 0. But the fact that there's no current far away essentially telling you better be 0. So if the current Jc is 0, Ja must be equal to Jb. And therefore that means a squared is equal to b squared. And happily that's what happened because b over a is a complex number of magnitude one. So the fact that b and a differ by just the phase was required by current conservation. A over b is a number that has norm equal to 1. So that's a consistent picture. This phase is very important. So what happens for resolution for x less than 0? Well, psi of x would be a into the i k x, plus b all the way to the left there. Your solution is a plus b equal minus i k x. Of course, we now know what the b is, so this is minus a from the ratio over there. e to the 2 i delta of e. E to the minus i k x. For x less than 0. And for x greater than 0, psi of x is going to c e to the minus kappa x. And I'm not bothering to write the coefficient c in Terms of a. Now this expression for x less than 0 can be simplified a little. You can factor an a. But it's very nice, and you should have an eye for those kind of simplifications. It's very nice to factor more than an a and to factor one phase like an i delta. I delta of e, because in that way, you get e to the i k x minus delta of e from the first term, where the two delta appearances cancel each other. Because the first term didn't have a delta. But then the second term will have the same argument here, of the exponential but with a minus sign. Minus i k x. And I claim also a minus delta of e. And this time minus and minus gives you plus. And the other e to the i delta gives you back the 2. But now you've created the trigonometric function, which is simpler to work with. So psi effects is equal to 2 i a e to i delta of e, sine of k x minus delta of e. And if you wish psi squared, the probability density is proportional to 4 a squared, sine squared of k x minus delta of e. So this can be plotted. Sine squared is like that. And where is x equal 0. OK, I'll say x equal 0, say is here. So this is not really true anymore. But this point here is x 0. It vanishes. Would be the point at which k x 0 is equal to delta of e. And the sine squared vanishes. So this is not a solution either. Solution is like that. And then this is for psi squared. And then it must couple to the k exponential on this side. So that the true solution must somehow be like this and well, whatever. I don't know how it looks. That is the e to the minus 2 kappa x decay and exponential. It must decay. There's continuity of the derivative and continuity of the wave function. So that's how this should look. A couple more things we can say about this solution that will play a role later. I want to get just a little intuition about this phase, delta of e, this phase shift. So we have it there. Delta of e, I'll write it here, so you won't have to-- 10 minus 1, square root of e over v not minus e. So this is interesting. This phase shift just applies for energies up to v not. And that corresponds to the fact that we've been solving for energies under the barrier. And if we solve for energies under the barrier, well, the solutions as we're writing with these complex numbers, apply up to energies equal to the barrier, but no more. So we shouldn't plot beyond this place. And here is delta of e. The phase shift. And when the energy is 0, when your particle you're sending in, or the packet eventually is very low energy here. Then the phase shift is the arctangent of 0, which is 0. As the energy goes to the value v not, then the denominator goes to 0. The ratio goes to infinity. And the arctangent is pi over 2. So it's a curve that goes from here to here. And it's not quite like a straight line. But because of the square roots, it sort of begins kind of vertical, then goes like this. It's not flat either, in the middle. So maybe my curve doesn't look too good. Those more vertical here. Wow, I'm having a hard time with this. Something like this. In fact, it's kind of interesting to plug the derivative, d delta, d energy. A little calculation will give you this expression. You can do this with mathematica or v not minus e. And shows, in fact, that here is v not, and here is v delta, v e. We could call it delta prime of e, because we wrote the phase shift as a function of the energy. So that the delta, d e is really delta prime of e. And it sort of infinite-- goes to a minimum and infinite again, in that direction. That's how it behaves.

MIT_804_Quantum_Physics_I_Spring_2016

de_Broglies_proposal.txt

PROFESSOR: This is Louis-- L-O-U-I-S d-e Broglie. And this is not hyphenated nor together. They are separate. And the d is not capitalized apparently too. And it's 1924, the photon as a particle is clear, and the photon is also a wave. And Louis de Broglie basically had a great insight in which he said that if this is supposed to be a universal or a real basic physical property that photons are waves and particles, we knew them as waves and now we know they're particles. But if they are dualed with respect to each other, both descriptions are in different regimes and in a sense, a particle at the end of the day has wave attributes and particle attributes. Wave attributes because it interferes and is described by waves. And particle attributes is because it has a definite amount of energy, it comes in packets, they cannot be broken into other things-- this could be a more general property. And in a sense, you could say that de Broglie did a fundamental step almost as important as Schrodinger when he claimed that all matter particles behave as waves as well. Not just the photon, that's one example, but everybody does. So associated to every modern particle, there is a wave. But that is quite interesting because in quantum mechanics, you have the photon and it's a particle, but it's associated to a wave and if you are a little quick, you say, oh sure, the electromagnetic wave, but no, in quantum mechanics, it's the probability amplitude to be some work. Those are the numbers we tracked in the mass and the interferometer, the probability to be sampled. We didn't track the waves or a single photon, the wave was a wave of probability amplitude, something they didn't know at all about yet at that time. So de Broglie's says just like the photons have properties of particles and properties of waves, every particle has properties of waves as well and every wave has a property of particles. But what is left unsaid here is yes, you have a wave, but a wave of what? And we've already told you a little bit, the answer has to do with probability waves. So it's very strange that the fundamental equation for a wave that represents a particle is not an electric field or a sound wave or this, it's for all of them is a probability wave. Very, very surprising. But that's what de Broglie's ideas led to. So if you had a photon, you would say it's a particle, and when you think of it as a particle, you would say it's a bundle of some energy and some momentum. And if you think of it as a wave, you would say it has a frequency. And that's a particle wave duality or in some sense, a particle wave description of this object-- you have a particle and a wave at the same time. When we have this, we have a particle wave duality. And de Broglie said that this is universal for all particles. Universal. And it appeared the name of matter waves. These are the matter waves that we're going to try to discuss. These are the waves of something that are probability amplitudes we're going to try to discuss. So you could say wave of what? What. And that comes later, but the answer is probability amplitudes, those complex numbers whose squares are probabilities. So just like we had for a photon, de Broglie's idea was that we would associate to a particle a wave that depends on the momentum. So remember, the Compton wavelength was a universal-- for any particle, the Compton wavelength is just one number, but just for photons, the wavelength depends on the momentum, so in general, it should be dependent on the momentum. So we say that for a particle of momentum p, we associate a wave-- a plane wave, in fact-- a plane wave, so we're getting a little more technical, with of lambda equals h over p, which is the de Broglie wavelength-- de Broglie wavelength. So it's a pretty daring statement. It was his PhD thesis and there was no experimental evidence for it. It was a very natural conjecture-- we'll discuss it a lot more next lecture-- but there are very little evidence for it. So experiments can a few years later, and people saw that you could interfere or diffract electrons. They would behave, colliding into lattices like waves, and those are rather famous experiments of Davisson and Germer. So particles, just like you do as an interference effect-- a two slit interference effect in which you have a screen, a slit and a screen, and you shine photons and then you get an interference effect over here because of the wave nature of photons, or in quantum mechanics, you would say, because there are probability amplitudes, that are complex numbers that have to be interfered between the possibilities of the two paths, because every photon goes through both paths at the same time, these experiments of interference, or two slit interference, were done for electrons. And then, eventually, they've been done for bigger and bigger particles, so that it's not just something that you do with elementary particles now. There's experiments done about three years ago-- I will put on the web site or on the notes some of these things so that you can see them, but now you can throw in molecules here, molecules that have a weight of now 10,000 atomic mass units, like 10,000 protons, like hundreds of-- 430 atom molecules, and you can get an interference pattern, so it's pretty ridiculous. It's almost like, you one day so you throw a baseball and you're going to see an interference pattern, but, you know, we've got to things with about 10,000 hydrogen atoms and de Broglie wavelengths of 1 picometer, which are pretty unbelievable. So the experiments are done with those particles and in fact with electrons. People do those experiments and they're in very beautiful movies in which you see those electrons hitting on the screen and then-- I'll give you some links so you can find them as well-- and you see one electron falls here and it gets detected and two electrons, three electrons, four electrons, five electrons, six electrons-- by the time you get 10,000 electrons, you see lots of electrons here, very well here, lots of electrons here, and the whole interference pattern is created by sending one electron at that time in an experiment that takes several hours and it's reduced to a movie of about one minute. So particles, big particles interfere, not just photons interfere. So those particles have some waves, some matter waves discovered by de Broglie, and next lecture, we're going to track the story from de Broglie to the Schrodinger equation where the nature of the wave suddenly becomes clear.

MIT_804_Quantum_Physics_I_Spring_2016

Number_operator_and_commutators.txt

PROFESSOR: Last time we discussed the differential equation. I'll be posting notes very soon. Probably this afternoon, at some time. And last time, we solved the differential equation, we found the energy eigenstates, and then turned into an algebraic analysis in which we factorized the Hamiltonian. Which meant, essentially, that you could write the Hamiltonian-- up to an overall constant that doesn't complicate matters-- as the product of an a dagger a. And that was very useful to show, for example, that any energy eigenstate would have to have energy greater than h omega over 2. We call this a dagger a the number operator n, which is a Hermitian operator. Recall that the dagger of a product of operators is the reverse order product of the daggered operators. So the dagger of a dagger a is itself. And then a was related to x and p, and so was a dagger. Recall that x and p are Hermitian. And there are overall constants here that were wrote last time, but now they're not that urgent. And a and a dagger, the commutator is equal to one. That was very useful. Finally, we also show that while the energy of any state would have to be greater than h omega over 2, if you had a state that is killed by a hat, it would have the lowest allowed energy-- which is h omega over 2. And, therefore, that is the ground state. And we looked at this differential equation, and we found this Gaussian wave function. And it's a first order differential equation. And, therefore, it has just one solution. And, therefore, there is just one ground state, and it's a bound state. And, of course, you wouldn't expect more than one ground state, because there's no degeneracies in the bound state spectrum of a one-dimensional potential. So we found one ground state was phi 0, and it's killed by a-- which means that it's killed by n-hat, because a is to the right in n-hat. So the a finds phi 0 and just kills it. Now, the other thing to note is that the Hamiltonian is really, pretty much, the same thing as the number operator multiplied by something with units of energy. The number operator has no units, because a and a dagger have no units. And that's very useful. So it's like a dimensionless version of the energy. And, certainly, if you have an eigenstate of h it must be an eigenstate of n. And the eigenvalue of n-- if we call it capital n. Therefore, you can imagine this equation acting on an eigenstate-- which happens to be an eigenstate of n or of h. On the left-hand side you would read the energy, and on the right-hand side you would read the eigenvalue of the n operator. So that gives you a very nice simple expression. You see that the energy is the number plus a 1/2 multiplied by h-bar [INAUDIBLE]. So that's pretty much the content of what we reached last time. And now we have to complete the solution. And the plan for today is to complete the solution, familiarize ourselves with these operators, learn how to work with a harmonic oscillator with them. And then we'll leave the harmonic oscillator for the time being-- let you do some exercises with it-- but turn to scattering states. So the second part of today's lecture we'll be talking about scattering states. OK. So when we look at this thing and you have a number operator-- which encodes the Hamiltonian-- it's a good idea to try to understand how it interacts with the other operators that you have here. And a good question, whenever you have operators, is the commutator. So you can ask, what is the commutator of n with a? And this commutator is going to show up. But it's basically that kind of thing. If you have a and a dagger, you ask, what is the commutator? If you have n, you ask, what is the commutator with the other thing? So n with a would be the commutator of a dagger with a-- a like that. And sometimes I will not write the hats to write things more quickly. Now, in this commutator, you can move the a out, and you have a dagger a a. And a dagger a is minus 1. Because aa dagger is 1. So this is minus a. So that's pretty nice. It's simple. How about n with a dagger? Well, this would be a dagger a with a dagger. A dagger with a dagger commute. So this a dagger can go out, and you get aa dagger. And that's 1, so you get a dagger. So it's a nice kind of computation relation. You would have commutation reserve x with p given a constant. Now n with a gives a number times a. N commutated with a dagger gives a number times a dagger. And those numbers are pretty significant, so I'll write this again. N with a is minus a. And n with a dagger is plus a dagger. This is part of the reason-- as we will see soon-- that the name of a-- which we call destruction operator, because it destroys the vacuum-- it's sometimes called lowering operator, because it comes with a negative sign here. And we'll see a better reason for that name. A dagger is sometimes called the creation operator or the raising operator, because it increases some number, as you will see. And here it's reflected by these plots. But we need a little more than that. We need a little more commutators than this. So for example, if I would have the commutator of a with a dagger to the k-- you can imagine this. You have to become very used and very comfortable with these commutation relations. And sometimes the only way to do that is to just do examples. So I'm doing this with a k here. Maybe-- when you review this lecture-- you should do it with k equals 2 or with k equals 3, and do it a few times. Until you're comfortable with these things, and you know what identities you've been using. If this was a little quick, then go more slowly and make absolutely sure you know how to do those commutators. In here, I'm going to say what happens. You have an a and you have to move it across a string of a-hats. Now, moving an a cross an a-hat-- because of the commutator-- gives you a factor of 1, but it destroys the a and the a-hat. As you move the a across the a-hats-- because this is a with all the a-hats, here, minus the a-hats times the a there. So if you could just move the a all the way across-- then you cancel with this. What you get is what happens when you're moving it to across. And you're moving across a string of those. So each time you try to move on a across an a-hat, you get this factor of 1 and you kill the a and you kill one a dagger. So this answer will not have an a, and it will have one less a dagger. So a dagger to the k, minus 1. And then I would argue-- and you should do it more slowly-- that you have to go across k of those. And each time you get a factor of 1, and you lose the a and the a dagger. So at the end you get a k. You should realize that this is not all that different from the kind of commutators you had. Like, with x to the n. This was very similar-- it might be a good time to review how that was done-- in which that pretty much gives you an x to the n minus 1, times a factor of n, because p is a derivative. You could almost think of a as the derivative with respect to a dagger. And then this commutator would be 1. So this is true. And there is also-- if you want-- an a dagger with a to the n or a to the k. This would give you-- if you had just one of them you would get a minus sign, because a dagger with a is that. But the same thing holds, you're going to get one less a-hat. So a-hat to the k minus 1. A factor of k-- because k times you're going to move an a dagger across an a. And a minus because you're getting a dagger commutator with a, as opposed to a commutator with a dagger-- which is 1. So these are two very nice and useful equations that you should be comfortable with. Now, this implies that you can do more with an n operator. So n with a-hat to the k, this time will be minus k a-hat to the k. It doesn't change the number of a-hats, because you're now making commutators with a dagger a. So each time you have this commuted with one a-hat, the a dagger and the a give you 1, but you have another a back. So the power is the same. The sign comes from this sign. AUDIENCE: Shouldn't the n there have a hat? PROFESSOR: Yes, it should have a hat. I'm sorry. Yes. And, similarly, n a-hat dagger to the k. This is k a-hat dagger to the k. So what happened before, that n-hat operator leaves the a same but puts a number-- leaves the a dagger the same and puts a number. Here, you see it happening again. N with a collection-- with a string of a-hats-- gives you the same string, but the number. And with a collection of a daggers gives you the same collection of a daggers with a number. And the number happens to be the number of a's or the number of a daggers. So that's the reason it's called the number operator, because the eigenvalues are the number of creation operators or the number of destruction operators. I was a little glib by calling it the eigenvalue. But it almost looks like an eigenvalue equation, which have an operator, another operator, and a number times the second operator. It is not exactly an eigenvalue equation, though, because with eigenvalues you would just have this acting on the second one. But the fact that this case appear here are the reason these are number operators. So it was a little quick for many of you. Some of you may have seen this before. It was a little slow, but the important thing is after a couple of days from now, or by Friday, you find all this very straightforward.

MIT_804_Quantum_Physics_I_Spring_2016

Eigenfunctions_of_a_Hermitian_operator.txt

PROFESSOR: So here comes the point that this quite fabulous about Hermitian operators. Here is the thing that it really should impress you. It's the fact that any, all Hermitian operators have as many eigenfunctions and eigenvalues as you can possibly need, whatever that means. But they're rich. It's a lot of those states. What it really means is that the set of eigenfunctions for any Hermitian operator-- whatever Hermitian operator, it's not just for some especially nice ones-- for all of them you get eigenfunctions. And these eigenfunctions, because it has vectors, they are enough to span the space of states. That is any state can be written as a superposition of those eigenvectors. There's enough. If you're thinking finite dimensional vector spaces, if you're looking at the Hermitian matrix, the eigenvectors will provide you a basis for the vector space. You can understand anything in terms of eigenvectors. It is such an important theorem. It's called the spectral theorem in mathematics. And it's discussed in lots of detail in 805. Because there's a minor subtlety. We can get the whole idea about it here. But there are a couple of complications that mathematicians have to iron out. So basically let's state we really need, which is the following. Consider the collection of eigenfunctions and eigenvalues of the Hermitian operator q. And then I go and say, well, q psi 1 equal q 1 psi 1 q psi 2 equal q2 psi 2. And I actually don't specify if it's a finite set or an infinite set. The infinite set, of course, is a tiny bit more complicated. But the result is true as well. And we can work with it. So that is the set up. And here comes the claim. Claim 3, the eigenfunctions can be organized to satisfy the following relation, integral dx psi i of x psi j of x is equal to delta ij. And this is called orthonormality. Let's see what this all means. We have a collection of eigenfunctions. And here it says something quite nice. These functions are like orthonormal functions, which is to say each function has unit norm. You see, if you take i equal to j, suppose you take psi 1 psi 1, you get delta 1 1, which is 1. Remember the [INAUDIBLE] for delta is 1 from the [INAUDIBLE] are the same. And it's 0 otherwise. psi 1 the norm of psi 1 is 1 and [INAUDIBLE] squared [INAUDIBLE] psi 1, psi 2, psi3, all of them are well normalized. So they satisfied this thing we wanted them to satisfy. Those are good states. psi 1, psi 2, psi 3, those are good states. They are all normalized. But even more, any two different ones are orthonormal. This is like the 3 basis vectors of r3. The x basic unit vector, the y unit vector, the z unit vector, each one has length 1, and they're all orthonormal. And when are two functions orthonormal? You say, well, when vectors are orthonormal I know what I mean. But orthonormality for functions means doing this integral. This measures how different one function is from another one. Because if you have the same function, this integral and this positive, and this all adds up. But for different functions, this is a measure of the inner product between two functions. You see, you have the dot product between two vectors. The dot product of two functions is an integral like that. It's the only thing that makes sense So I want to prove one part of this, which is a part that is doable with elementary methods. And the other part is a little more complicated. So let's do this. And consider the case if qi is different from qj, I claim i can prove this property. We can prove this orthonormality. So start with the integral dx of psi i star q psi j. Well, q out here at psi j is qj. So this is integral dx psi i star qj psi j. And therefore, it's equal to qj times integral psi i star psi j. I simplified this by just enervating it. Because psi i and psi j are eigenstates of q. Now, the other thing I can do is use the property that q is Hermitant and move the q to act on this function. So this is equal to integral dx q i psi i star psi j. And now I can keep simplifying as well. And I have dx. And then I have the complex conjugate of qi psi i psi i, like this, psi j. And now, remember q is an eigenvalue for Hermitian operator. We already know it's real. So q goes out of the integral as a number. Because it's real, and it's not changed. Integral dx psi i star psi j. The end result is that we've shown that this quantity is equal to this second quantity. And therefore moving this-- since the integral is the same in both quantities, this shows that q i minus qj, subtracting these two equations, or just moving one to one side, integral psi i star psi j dx is equal to 0. So look what you've proven by using Hermiticity, that the difference between the eigenvalues times the overlap between psi i and psi j must be 0. But we started with the assumption that the eigenvalues are different. And if the eigenvalues are different, this is non-zero. And the only possibility is that this integral is 0. So this implies since we've assumed that qi is different than qj. We've proven that psi i star psi j dx is equal to 0. And that's part of this little theorem. That the eigenfunctions can be organized to have orthonormality and orthonormality between the different points. My proof is good. But it's not perfect. Because it ignores one possible complication, which is that here we wrote the list of all the eigenfunctions. But sometimes something very interesting happens in quantum mechanics. It's called degeneracy. And degeneracy means that there may be several eigenfunctions that are different but have the same eigenvalue. We're going to find that soon-- we're going to find, for example, states of a particle that move in a circle that are different and have the same energy. For example, a particle moving in a circle with this velocity and a particle moving in a circle with the same magnitude of the velocity in the other direction are two states that are different but have the same energy eigenvalue. So it's possible that this list not all are different. So suppose you have like three or four degenerate states, say three degenerate states. They all have the same eigenvalue. But they are different. Are they orthonormal or not? The answer is-- actually the clue is there. The eigenfunctions can be organized to satisfy. It would be wrong if you say the eigenfunctions satisfy. They can be organized to satisfy. It means that, yes, those ones that have different eigenvalues are automatically orthonormal. But those that have the same eigenvalues, you may have three of them maybe, they may not necessarily be orthonormal. But you can do linear transformations of them and form linear combinations such that they are orthonormal. So the interesting part of this theorem, which is the more difficult part mathematically, is to show that when you have degeneracies this still can be done. And there's still enough eigenvectors to span the space.

MIT_804_Quantum_Physics_I_Spring_2016

Wavepackets.txt

PROFESSOR: In order to learn more about this subject, we must do the wave packets. So this is the place where you really connect this need solution of Schrodinger's equation, the energy eigenstates, to a physical problem. So we'll do our wave packets. So we've been dealing with packets for a while, so I think it's not going to be that difficult. We've also been talking about stationary phase and you've practiced that, so you have the math ready. We should not have a great difficulty. So let's new wave packets with-- I'm going to use A equals 1 in the solution. Now, I've erased every solution, and we'll work with E greater than v naught to begin with. The reason I want to work with E greater than v naught is because there is a transmitted wave. So that's kind of nice. So what am I going to do? I'm going to write it this following way. So here is a solution with A equals to 1. e to the i kx plus k minus k bar over k plus k bar e to the minus i kx. And this was for x less than 0. And indeed, when there was an A, there was a B, if A is equal to 1, remember we solve for the ratio of B to A and it was this number so I put it there. I'm just writing it a little differently, but this is a solution. And for x greater than 0, the solution is C, which was 2k over k plus k bar e to the i k bar x. Now we have to superimpose things. But I will do it very slowly. First, is this a solution of the full Schrodinger equation? No. Is this a solution a time-independent Schrodinger equation? What do I need to make it the solution of the full Schrodinger equation? I need e to the minus i Et over h. And I need it here as well. And this is a psi now of x and t. There's two options for x, less than 0 and x less than 0, and those are solutions. So far so good. Now I'm going to multiply each solution. So this is a solution of the full Schrodinger equation, not just the time-independent one, all of it. It has two expressions because there's a little discontinuity in the middle, but as a whole, it is a solution. That is the solution. Some mathematicians would put a theta function here, theta of minus x and add this with a theta of x so that this one exists for less than x, x less than 0, and this would exist for less greater than 0. I would not do that. I will write cases, but the philosophy is the same. So let's multiply by a number, f of k. Still a solution. k is fixed, so this is just the number. Now superposition. That will still be a solution if I do the same superposition in the two formulas, integral dk and integral dk. That's still a solution of a Schrodinger equation. Now I want to ask you what limits I should use for that integral. And if anybody has an opinion on that, it might naively be minus infinity to infinity, and that might be good, but maybe it's not so good. It's not so good. Why is not so good, minus infinity to infinity? Would I have to do from minus infinity on my force. Does anybody force me? No. You're superimposing solutions. For different values of k, you're superimposing. You had a solution, and another solution for another, another solution. What goes wrong if I go from minus infinity to infinity? Yes. AUDIENCE: [INAUDIBLE] the wave packet's going to be in one direction, so it [? shouldn't ?] be [? applied. ?] PROFESSOR: That's right. You see here, this wave packet is going to be going in the positive x direction, positive direction, as long as k is positive. It's just that the direction is determined by the relative sign within those quantities. E is positive in this case. k is positive. This moves to the right. If I start putting things where k is negative, I'm going to start producing things and move to the left and to the right in a terrible confusion. So yes, it should go 0 to infinity, 0 to infinity. And f of k, what is it? Well, in our usual picture, k f of k is some function that is peaked around some k naught. And this whole thing is psi of x and t, the full solution of a wave packet. So now you see how the A, B, and C coefficients enter into the construction of a wave packet. I look back at the textbook in which I learned quantum mechanics, and it's a book by Schiff. It's a very good book. It's an old book. I think was probably written in the '60s. And it goes through some discussion of wave packets and then presents a jewel, says with a supercomputer, we've been able to evaluate numerically these things, something you can do now with three seconds in your laptop, and it was the only way to do this. So you produce an f of k. You fix their energy and send in a wave packet and see what happens. You can do numerical experiments with wave packets and see how the packet gets distorted at the obstacle and how it eventually bounces back or reflects, so it's very nice. So there is our solution. Now we're going to say a few things about it. I want to split it a little bit. So lets go here. So how do we split it? I say the solution is this whole thing, so let's call the incident wave that is going to be defined for x less than 0 and t, this is x less than 0. And the incident wave packet is dk 0 to infinity f of k e to i kx e to the minus i et over h bar. And this is just defined for x less than 0, and that's so important that write it here. For x less than 0, you have an incident wave packet. And then you also have a reflected wave packet, x less than 0 t is the second part dk f of k k minus k bar over k plus k bar e to the minus i kx e to the minus i et over h bar. It's also for x less than 0, and we have a psi transmitted for x greater than 0 and t, and that would be 0 to infinity dk f of k 2k over k plus k bar e to the i k bar x e to the minus i et over h bar. Lots of writing, but that's important. And notice given our definitions, the total psi of x and t is equal to psi incident plus psi reflected for x less than 0, and the total wave function of x and t is equal to psi transmitted for x greater than 0. Lots of equations. I'll give you a second to copy them if you are copying them. So now comes if we really want to understand this, we have to push it a little further. And perhaps in exercises we will do some numerics to play with this thing as well. So I want to do stationary phase approximation here. Otherwise, we don't see what these packets, how they're moving. So you have some practice already with this. You're supposed to have a phase whose derivative is 0, and it's very, very slowly at that place where there could be a contribution. Now every integral has the f of k. So that still dominates everything, of course. You see, if f of k is very narrow, you pretty much could evaluate these functions at the value of k naught and get a rather accurate interpretation of the answer. The main difficulty would be to do the leftover part of the integral. But again, here we can identify phases. We're going to take f of k to be localized and to be real. So there is no phase associated with it, and there is no phases associated with these quantities either, so the phases are up there. So let's take, for example, psi incident. What is this stationary phase condition? Would be that the derivative with respect to k that we are integrating of the phase, kx minus Et over h bar must be evaluated at k naught and must be equal to 0. So that's our stationary phase approximation for the top interval. Now remember that E is equal to h squared k squared over 2m. So what does this give you? That the peak of the pulse of the wave packet is localized at the place where the following condition holds. x minus de dk, with an h bar will give an h k t over m evaluated at k naught equals 0. So this will be x equals h bar k naught over m t. That's where the incident wave is propagating. Now, look at that incident wave. What does it do for negative time? As time is infinite and negative, x is negative, and it's far away. Yes, the packet is very much to the left of the barrier at time equals minus infinity. And that's consistent because psi incident is only defined for x less than 0. It's only defined there. So as long as t is negative, yes, the center of the packet is moving in. I'll maybe draw it here. The center of the packet is moving in from minus infinity into the wall, and that is the picture. The packet is here, and it's moving like that, and that's t negative. The psi incident is coming from the left into the barrier, and that's OK. But then what happens with psi incident as t is positive? As t is positive, psi incident, well, it's just another integral. You might do it and see what you get, but we can see what we will get, roughly. When t is positive, the answer would be you get something if you have positive x. But psi incident is only for negative x. So for negative x, you cannot satisfy the stationarity condition, and therefore, for negative x and positive time, t positive, psi incident is nothing. It's a little wiggle. There's probably something, a little bit-- look at it with Mathematica-- there will be something. But for positive t, since you only look at negative x, you don't satisfy stationarity, so you're not going to get much. So that's interesting. Somehow automatically psi incident just exists for negative time. For time near 0 is very interesting because somehow stationary [INAUDIBLE], when you assess, you still get something, but you're going to see what the packet does as it hits the thing. Let's do the second one of psi reflected. d dk this time would be kx with a different sign, minus kx minus E t over h bar at k naught equals 0. For the reflected wave, the phase is really the same. Yeah, this factor is a little more complicated, but it doesn't have any phase in it. It's real, so [INAUDIBLE]. So I just change a sign, so this time I'm going to get the change of sign. x is equal to minus h bar k naught over m t. And this says that for t positive, you get things. And in fact, as t is positive your are at x negative. And remember psi reflected is only defined for x negative, so you can satisfy stationary, and you're going to get something. So for t positive, you're going to get as t increases, a thing that goes more and more to the left as you would expect. So you will get psi reflected going to the left. I will leave for you to do the psi transmitted. It's a little different because you have now k bar, and you have to take the derivative of k bar with respect to k. It's going to be a little more interesting example. But the answer is that this one moves as x equals h bar k bar over m t. k bar is really the momentum on the right, and since psi transmitted exists only for positive x, this relation can be satisfied for positive t. For positive t, there will be a psi transmitted. The psi transmitted certainly exists for negative t, but for negative t, stationarity would want x to be negative, but that's not defined. So for negative t on the right, yes, psi transmitted maybe it's a little bit of something especially for times that are not too negative. But the picture is that stationary phase tells you that these packets, psi incident, pretty much exist just for negative t and psi reflected and psi transmitted exist for positive t. And these are consequences of the fact that psi incident and psi reflected exist for negative x. The other exists for positive x, and that coupled with stationarity produces the physical picture that you expect intuitively, that the incident wave is just something, part of the solution that exists just at the beginning. And somehow it whistles away. Some of it becomes transmitted, some of it becomes reflected.

MIT_804_Quantum_Physics_I_Spring_2016

Step_potential_probability_current.txt

PROFESSOR: I've put on the blackboard here the things we were doing last time. We began our study of stationary states that are not normalizable. These are scattering states. Momentum eigenstates were not normalizable, but now we have more interesting states that represent the solutions of the Schrodinger equation, that are stationary states with some energy e. Because they are not normalizable, but we cannot directly interpret any of these solutions as the behavior of a particle. I kind of tell you a story, OK. So this is-- a particle is coming, colliding, doing something. These are not normalizable states. So part of what we're going to be trying to do today is connect to the picture of wave packets and see how this is used to really calculate what would happen if you send in a particle of potential. Nevertheless, we wrote a solution that has roughly that interpretation, at least morally speaking. We think of a wave that is coming from the left, that's ae to the ikx. Now, in order to have a wave, you would have to have time dependents, and this is a stationary solution. And there is some time dependents. There is exponential of e to the minus iet over h bar, where e is the energy. So this could be added here to produce the full stationary state psi of x and t. But we'll leave it understood-- it's a common phase factor for both the solutions of x less than 0, and x greater than 0, because the whole solution represents a single solution. It's not like a solution on the left and a solution on the right. It's a single solution over all of x 4 a psi that has some definite energy. So we looked at the conditions of continuity of the wave function and continuity of the derivative of the wave function, at x equals 0. Those were two conditions. And they gave you this expression for the ratios of c over a and b over a. We could even imagine, since we can't normalize this, setting a equal to 1, And. Then calculating b and c from those numbers. We have two case a k in a k bar. The k is relevant to the wave function for x less than 0. The k bar is relevant for x greater than 0. And they have to be different because this represents a DeBroglie wavelength, and the DeBroglie wavelength encodes the momentum of the particle, and the momentum of the particle that we imagine here classically is different here where this has this much kinetic energy, and in the region on the right, where the particle only has a much smaller kinetic energy. So that's represented by k bar. And k bar, being the energy proportional to the energy minus v 0, while k squared has just the energy, it's smaller than k. So this is what we did. We essentially solved the problem, and this qualifies as a solution, but we still haven't learned anything very interesting from it. We have to understand more what's going on. And one thing we can do is think of a particular limit. The limit, or the case, when e is equal to v 0, exactly equal to b 0, what happens? Well, k bar would be equal to 0. And if k bar is equal to 0, you're going to have just the constant c in here. But if k bar is equal to 0, first b is equal to a, because then it's k over k, so b is equal to a. And c is equal to 2a. And the solution would become psi of x equals-- well, a is equal to b. So this is twice a cosine of kx, when a equal to b is a common factor, call it a. And this thing is the sum of two exponentials with opposite signs. That gives you the cosine. And for c, you have 2a. And since k bar is equal to 0, well, it's just 2a. It's a number. This is unnormalizable, but even the original solution is unnormalizable, so we wouldn't worry too much about it. So how does that look for x over here? You have a cosine of kx, so it's going to be doing this to the left. That's for x less than 0. And at this point, it goes like that. Just flat side effects, and here's 2a. so there's nothing wrong with the solution in this case. It's kind of a little strange that it becomes a constant, but perfectly OK. What really helps you here is to find some conditions that express the conservation of probability. You see, you have a stationary state solution. Now, stationary states are funny states. They're not static states, completely static states. For example, if you have a loop, and you have a current that never changes in time. This is a stationary condition, even in electromagnetism. So what we imagine here is that we're going to have some current, probability current, that is coming from the left, and some of it maybe bounces back, and some of it goes forward. But essentially, if you think of the barrier, whatever-- you look a little to the left of the barrier and a little to the right of the barrier. Whatever is coming in, say, must be going out there, because probability cannot increase in this region. It would be like saying that the particle suddenly requires larger and larger probability to be in this portion of the graph. And that can't happen. So probabilistic current gives you a way to quantify some of the things that are happening. So probability current plus j effects, which was h bar over m, imaginary part of psi star d psi dx. So let's compute the probability current. Let's compute for x less than 0. What is the probability current j of x? We then call it the probability current on the left side. I would have to substitute the value of the wave function for x less than 0, which is the top line there, into this formula. And see what is the current. In fact, I believe you've done that in an exercise some time ago. And as you can imagine, the current is proportional to the modulus of a squared, the length of a squared, the length of b squared also enters. And the funny thing is, between those two waves, one that is going to the right and one is going to the left, that that's very visible in the current. This is h bar k over m. A squared minus b squared. It's a short computation, and it might be an OK and a good idea to do it again. It was done in the homework. And x greater than 0. J right of x would be equal to h bar k bar. Now that, you can almost do it in your head. This c into the ik bar x. Look what's happening. From psi star you get a c star. From psi you get the c, so that's going to be a c squared. The face is going in a cancel between the one here and the one on psi, but the derivative will bring down an ik bar, and the imaginary part of that is just k bar. So the answer is this. And these are the two currents. Now, if we are doing things correctly, the two currents should be the same. Whatever current exists to the left, say a positive current that is coming in, to the right must be the same. Another pleasant thing is that this current doesn't depend on the value of x, as x is less than 0. Nor of the value of x when x is greater than 0. And that's good for conservation. It would be pretty bad as well, if you look at two places for x less than 0, and you find that the current is not the same. So where is it accumulating? What's going on? So the independence of these things-- from x, this is constant, and a constant is very important, because this constant should be the same. Now, whether or not about current conservation, it's encoded in Schrodinger's equation. And we solved Schrodinger's equation. That's how we got these relations between b, a and c So it better be that these two things are the same. So Jay elsewhere-- jl, for example. I won't put off x, because it's just clear. It doesn't depend on x. 1 minus b over a squared a squared. I'm starting to manipulate the left one, and see if indeed the currents are the same. And now, I get h bar k over m. 1 minus b over a squared. I put the modulus squared, but so far everything is real in this-- k, k bar, b, a, c-- all real. Is that right? No complex numbers there. So I don't have to be that careful. B over a squared is k minus k bar over k plus k bar squared a squared. And I gfit that here, maybe. H bar k over m. Now what? This is squared, so the si, squared passes here to the numerator minus the difference squared, that's going to be a 4k k bar over k plus k bar squared a squared. And yes, this seems to be working quite well. Flip these k's-- these two k's, flip them around. So the answer for jl so far is h bar, now k bar, because I flipped that, m, and now I would have 4 k squared over this thing, which is 2k over k plus k bar squared of a squared. And this quantity, if I remember right, is just c squared, as you can see from there. So indeed, this is j right, and it worked out. This is j right. So j left is equal to j right by current conservation. So that's nice. That's another way of getting insight into these coefficients. And that kind of thing that makes you feel that there's no chance you got this wrong. It all works well.

MIT_804_Quantum_Physics_I_Spring_2016

Group_velocity_and_stationary_phase_approximation.txt

PROFESSOR: Velocity. So we assume that we have an omega of k. That's the assumption. There are waves in which, if you give me k, the wavelength, I can tell you what is omega. And it may be as simple as omega equal to kc, but it may be more complicated. In fact, the different waves have different relations. In mechanics, omega would be proportional to k squared. As you've seen, the energy is proportional to p squared. So omega will be proportional to k squared. So in general, you have an omega of k. And the group velocity is the velocity of a wave packet-- packet-- constructed-- by superposition of waves. Of waves. All right. So let's do this here. Let's write a wave packet. Psi of x and t is going to be done by superposing waves. And superposition means integrals, summing over waves of different values of k. Each wave, I will construct it in a simple way with exponentials. ikx minus army of kt. And this whole thing, I will call the phase of the wave. Phi of k. So that's one wave. It could be sines or cosines, but exponentials are nicer. And we'll do with exponentials, in this case. But you superimpose them. And each one may be superimposed with a different amplitude. So what does it mean? It means that there is a function, phi of k, here. And for different k's, this function may have different values. Indeed, the whole assumption of this construction is based on the statement that phi of k peaks. So phi of k-- as a function of k is 0 almost everywhere, except a little bump around some frequency that we'll call k0. Narrow peak. That is our wave. And depending on how this phi of k looks, then we'll get a different wave. We're going to try to identify how this packet moves in time. Now-- There is a quick way to see how it moves. And there is a way to prove how it moves. So let me do, first, the quick way to see how it moves. It's based on something called the principle of stationary phase. I doubt it was said to you in [? A03 ?] in that way. But it's the most powerful wave to see this. And in many ways, the quickest and nicest way to see. Takes a little bit of mathematical intuition, but it's simple. And intuition is something that. I think, you have. If you're integrating-- a function-- multiplied-- a function, f of x, multiplied by maybe sine of x. Well, you have f of x, then sine of x. Sine is 1/2 the times positive, 1/2 the times negative. If you multiply these two functions, you're going to get the function that is 1/2 time positive and 1/2 the time negative. And in fact, the integral will contribute almost nothing if this function is slowly varying. Because if it's slowly varying, the up peak and the down peak hasn't changed much the function. And they will cancel each other. So the principle of stationary phase says that if you're integrating a function times a wave, you get almost no contribution, except in those places where the wave suddenly becomes of long wavelength and the phase is stationary. Only when the wave doesn't change much for a while, and then it changes again. In those regions, the function will give you some integral. So that's the principle of stationary phase. And I'll say it here, I'll write it here. Principal-- of stationary phase. We're going to use that throughout the course. Phase. And I'll say the following. Since-- phi anyway only peaks around k0-- This is the principle of stationary phase applied to this integral. Since-- since only for k roughly equal to k0. The integral-- has a chance-- To be non-zero. So here is what I'm saying. Look. The only place where this integral contributes-- it might as well integrate from k0 minus a little delta to k0 plus a little delta. Because this whole thing vanishes outside. And if we're going to integrate here, over this thing, it better be that this wave is not oscillating like crazy. Because it's going to cancel it out. So it better stop oscillating there in order to get that contribution, or send in another way. Only when the phase stops, you get a large contribution. So on the phase stops varying fast with respect to k. So you need-- need-- that the phase becomes stationary-- with respect to k, which is the variable of integration-- at k0. So around k0, better be that the phase doesn't change quickly. And the slower it changes, the better for your integral. You may get something. So if you want to figure out where you get the most contribution, you get it for k around k0, of course. But only if this thing it's roughly stationary. So being roughly stationary will give the following result. The result is that the main contribution comes when the phase, phi of k, which is kx minus omega of kt, satisfies the condition that it just doesn't vary. You have 0 derivative at k0. So the relative with respect to k is x minus d omega of k dk at k0 t must be 0. Stationary phase. Function phase has a stationary point. Look what you get. It says there that you only get a contribution if this is the case. So the value of x, where you get a big bump in the integral, and the time, t, are related by this relation. The hump in this packet will behave obeying this relation. So x is equal to d omega dk at k0 t. And it identifies the packet as moving with this velocity. x equal velocity times times. This is the group velocity. End of the answer by stationary phase. Very, extremely simple.

MIT_804_Quantum_Physics_I_Spring_2016

Shape_changes_in_a_wave.txt

PROFESSOR: Next is this phenomenon that when you have a wave packet and it moves it can change shape and get distorted. And that is a very nice phenomenon that takes place in general and causes technological complications. And it's conceptually interesting. So let's discuss it. So it's still wave packets. But now we have to go back and add some time to it. So shape changes. So we had a psi of x and t is equal to 1 over square root of 2 pi phi of k e to the ikx e to the minus i omega of kt. And what did we do with this to analyze how it propagates? We expanded omega of k as omega of k0, which, again, this quantity is centered and peaks around k0, plus k minus k0 times d omega dk at k0 plus 1/2 k minus k0 squared, the second omega, dk squared at k0. And it might seem that this goes on forever. And what did we do before? We looked at this thing and we did the integral with this term and ignored the next. And with this term, we discovered that the profile moves with this velocity, the group velocity. Now we want to go back and at least get an idea of how this term could change the result. And it would change the result by deforming the shape of the packet. So it is of interest to know, for example, how long you have to wait before your packet gets totally deformed, or how do you evolve a packet. So we need to recall these derivatives. So the omega vk is the same as de dp by multiplying by h bar. And this you'll remember, was p over m. The edp is p over m and is equal to h bar k over m. So the second omega, dk squared. I must differentiate the first derivative with respect to k. So I differentiate the first derivative with respect k. And now I get just h bar over m, which is quite nice. And the third derivative, the 3 omega, dk cubed, is 0. And therefore, I didn't have to worry about these terms. The series terminates. The Taylor series terminates for this stuff. Yes? AUDIENCE: The reason this happens is because we're [INAUDIBLE]. PROFESSOR: That's right. So of what is it that we get? Well, this term is roughly then 1/2 k minus k0 squared times h bar over m. And we can go back to the integral that we're trying to do. We don't do it again or not by any means. But just observe what's going on there. And we have an e to the minus i omega of kt that we did take into account. But the term that we're dropping now is a term that is minus i omega of k, well, whatever we have here, 1/2 k minus k0 squared h bar over mt. That's the phase that we ignored before. But now we'll just say, that we expect, therefore, that the shape doesn't change as long as we can ignore this phase. And this phase would start changing shape of the object. So our statement is going to be that we have no shapes. So let's imagine you started with a packet that sometime t equals 0. And then you let time go by. Well, there's some numbers here and time is increasing. At some point, this phase is going to become unignorable. And it's going to start affecting everything. But we have no shape change, or no appreciable shape change, as long as this quantity is much less than 1. So as long as say, k minus k0 squared h bar over m absolute value of t is much less than 1, no shape change. Now it's convenient to write it in terms of things that are more familiar. So we should estimate this thing. Now we're doing estimates in a very direct and rough way here. But look, your integrals are around k0. And as you remember, they just extend a little bit because it has some width. So k minus k0, as you do the integral over k, you're basically saying this thing is about the size of the uncertainty in k. So I'll put here delta k squared. Then you'll have h bar t over m much less than 1. Now h bar times delta k is delta p. So this equation is also of the form delta p squared t over h bar m much less than 1. There's several forms of this equation that is nice. So this is a particularly nice form. So if you know the uncertainty and momentum of your packet, or wave packet, up to what time, you can wait and there's no big deformation of this wave packet. Another thing you can do is involve the uncertainty in x. Because, well, delta p delta x is equal to h bar. So we can do that. And so with delta p times delta x equal to about h bar, you can write t less than h bar over m over a delta p squared, which would be h squared delta x squared. I think I'm getting it right. Yep, so t much less than m over h bar delta x squared. That's another way you could write this inequality. There is one way to write the inequality that you can intuitively feel you understand what's happening. And take this form a from a. Write it as delta p t over m is less than h bar over delta p. And h bar over delta p is delta x. So you go delta p over mt much less than delta x. I think this is understandable. Why does the packet change shape? The reason it changes shape is because the group velocity is not the same for all the frequencies. The packet mostly moves with k0. And we haven't rated the group velocity in k0. But if it would have a definite velocity, we would have a definite momentum. But that's not possible. These things have uncertainty in momentum. And they have uncertainty in k that we use it to write it. So different parts of the wave can move with different velocities, different group velocities. The group velocity you evaluated at k0. But some part of the packet is propagating with group velocities that are near k0 but not exactly there. So you have a dispersion in the velocity, which is an uncertainty in the velocity or an uncertainty in the momentum. Think, the momentum divided by mass is velocity. So here it is, an uncertainty in the velocity. And if you multiply the uncertainty in the velocity times this time that you can wait, then the change in shape is not much if this product, which is the difference of how one part moves with respect to the other, the difference of relative term, is still smaller than the uncertainty that controls the shape of the packet. So the packet has a delta x. And as long as this part, the left part of the packet, then the top of the packet, the difference of velocities times the time, it just still compared to delta x is small, then the thing doesn't change much. So I think this is one neat way of seeing what an equation that you sometimes use in this form, sometimes use in in this form-- it's just things that you can use in different ways. So for example, I can do this a little exercise. If you have delta x equals 10 to the minus 10 meters, that's atomic size for an electron. How long does it remain localized? So you have an electron. And you produce a packet. You localize it to the size of an atom. How long can you wait before this electron is just all over the room? Well, when we say this t, and we say this time, we're basically saying that it's roughly still there. Maybe it grew 20%, 30%. But what's the rough time that you can expect that it stays there? So in this case, we can use just this formula. And we say the time could be approximately m over h bar delta x squared. It's fun to see the numbers. You would calculate it with mc squared over h bar c times delta x over c, this squared. The answer is about 10 to the minus 16 seconds, not much. This is a practical issue in accelerators as well. Particle physics accelerators, they concern bunches, a little bunch of protons in the LHC. It's a little cylinder in which the wave functions of the protons are all collimated very thin, short, a couple of centimeters short. And after going around many times around the accelerator, they always have to be compressed and kept back, sent back to shape. Because just of diffusion, these things just propagate. And so it's a rather important thing.

MIT_804_Quantum_Physics_I_Spring_2016

Linearity_and_nonlinear_theories_Schrödingers_equation.txt

PROFESSOR: So here is of something funny. You might say, OK, what is simpler? A theory that is linear or a theory that is not linear? And the answer, of course, a linear theory is much simpler. General-- Maxwell's equations are linear. Einstein's theory of relativity is very nonlinear, very complicated. How about classical mechanics? Is classical mechanics linear or nonlinear? What do we think? Can't hear anyone. Linear, OK. You may think it's linear because it's supposed to be simple, but it's not. It's actually is very nonlinear. Newton could solve the two body problem but he couldn't solve the three body problem. Already with three bodies, you cannot superpose solutions that you get with two bodies. It's extraordinarily complicated, classical mechanics. Let me show you. If you have motion in one dimension, in 1D, you have the equation of motion, motion in one dimension, and there are potential V of x, that this time independent-- a particle moving in one dimension x with under the influence of a potential, V of x. The second-- the dynamical variable is x of t. The dynamical variable. And the equation of motion is-- so let me explain this. This is force equal mass times acceleration. This is mass, this is acceleration, the second derivative of the position, and V force is minus the derivative of the potential evaluated at the position. You know, derivatives of potentials-- if you think of a potential, the derivative of the potential is here positive, and you know if you have a mass here, it tends to go to the left, so the force is on the left, so it's minus. So V prime is the derivative of V with respect to its argument. And the problem is that while this, taking derivatives, is a linear operation. If you take two derivatives of a sum of things, you take two derivatives of the first plus two derivatives of the second. But yes, its-- this side is linear, but this side may not be linear. Because a potential can be arbitrary. And that the reverse-- so suppose the potential is cubic in x. V of x goes like x cubed. Then the derivative of V goes like x squared, and x squared is not a linear function. So this, Newton's equation, is not a linear equation. And therefore, it's complicated to solve. Very complicated to solve. So finally, we can get to our case, quantum mechanics. So in quantum mechanics, what do we have? Quantum mechanics is linear. First, you need an equation, and whose equation is it? Schrodinger's equation, 1925. He writes an equation for the dynamical variable, and the dynamical variable is something called the wave function. This wave function can depend on t-- depends on time-- and it may depend on other things as well. And he describes the dynamics of the quantum system as it evolved in time. There is the wave function, and you have an equation for this wave function. And what is the equation for this wave function? It's a universal equation-- i hbar partial derivative with respect to time of psi is equal to H hat of psi, where H hat is called the Hamiltonian and it's a linear operator. That's why I had to explain a little bit what the linear operator is. This is the general structure of the Schrodinger equation-- time derivative and the linear operator. So if you wish to write the Schrodinger equation as an L psi equals 0, then L psi would be defined i hbar del/del t of psi minus H hat psi. Then this is the Schrodinger equation. This equation here is Schrodinger's equation. And as you can see, it's a linear equation. You can check it, check that L is a linear operator. Therefore, it is naturally linear, you can see, because you do it differently, because the derivative with respect to time is a linear operation. If you have the ddt of a number of times a function, the number goes out, you differentiate the function. ddt of the sum of two functions, you differentiate the first, you differentiate. So this is linear and H we said is linear, so L is going to be linear and the Schrodinger equation is going to be a linear equation, and therefore, you're going to have the great advantage that any time you find solutions, you can scale them, you can add them, you can put them together, combine them in superpositions, and find new solutions. So in that sense, it's remarkable that quantum mechanics is simpler than classical mechanics. And in fact, you will see throughout this semester how the mathematics and the things that we do are simpler in quantum mechanics, or more elegant, more beautiful, more coherent, it's simpler and very nice. OK, i is the square root of minus 1, is the imaginary unit, and that's what we're going to talk next on the necessity of complex numbers. hbar, yes, it's a number. It shows up in quantum mechanics early on. It it's called Planck's constant and it began when Planck tried to fit the black value spectrum and he found the need to put a constant in there, and then later, Einstein figured out that it was very relevant, so yes, it's a number. For any physical system that you have, you will have a wave function and you will have a Hamiltonian, and the Hamiltonian is for you to invent or for you to discover. So if you have a particle moving on a line, the wave function will depend on time and on x. If you have a particle moving in three dimensions, it will depend on x vector. It may depend on other things as well or it maybe, like, one particle has several wave functions and that happens when you have a particle with spin. So in general, always time, sometimes position, there may be cases where it doesn't depend on position. You think of an electron at some point in space and it's fixed-- you lock it there and you want understand the physics of that electron locked into place, and then position is not relevant. So what it does with its spin is relevant and then you may need more than one wave function-- what is one describing the spin up and one describing the spin down? So it was funny that Schrodinger wrote this equation and when asked, so what is the wave function? He said, I don't know. No physical interpretation for the wave function was obvious for the people that invented quantum mechanics. It took a few months until Max Born said it has to do with probabilities, and that's what we're going to get next. So our next point is the necessity of complex numbers in quantum mechanics.

MIT_804_Quantum_Physics_I_Spring_2016

Reflection_and_transmission_coefficients.txt

PROFESSOR: We can write, however, J left as J of A, minus J of B. Where J of A would be h bar k over m A squared. And J of B would be h bar k over m B squared. You see the current that exists to the left of the barrier has two components and-- it's very intuitive. It's the current that would have been brought alone by the incoming A-wave. Minus the current that would have existed alone from the reflected B-wave. B is the reflected wave. So that's very nice. There's no interference between these two terms. We can really think of a current that is associated to the incoming wave, and a current that is associated with the reflected wave. So this suggests how you should define a reflection coefficient. Reflection coefficient R would give me the amount of current I get reflected, compared to the amount of current that there is incident. You see, the incident current is going to be partially reflected and partially transmitted. So an idea of a reflection is the value of the reflected current divided by the incident current. It's a definition, but it's a reasonable definition. And then, if it is this ratio-- because of these expressions-- it happens to be B over A squared. And that's an interesting number. Now, there's some physics in it. It tells me how much of the probability gets reflected as a function of the probability that is incident. So that's a good measure. If you get a reflection coefficient of 1/10, then you would expect 1/10 of the particles to be reflected. Now we don't have particles yet. This is non-normalizable solution. But, still, this will be the intuition very soon. Now we could have a transmission coefficient, as well. And here is something where we sometimes make a mistake. T is going to be the transmission coefficient. Transmission coefficient. And how should we define it? There is a temptation to define it-- well, coefficient B over A gives me this. Then maybe it should be C over A. But actually-- while c over a gives you some idea of how big the wave to the right is compared to the wave of the left-- that's not what we should call a reflection coefficient. And the reason is that-- I will call this current J C. And that's the amount of probability-- because it's a current associated to the wave C. And that's the amount of probability that is being carried by the transmitted wave. That is the probability. Not necessarily C over A. So the transmission coefficient will be defined to be J C divided by J A. And then J C divided by J A-- J C has an h-bar, k-bar. And J A has a k. So this is not equal to this ratio, but is actually k-bar over k, C over A. So it's not just this number. The reflection coefficient-- and transmission coefficients-- really originate from probabilities. And the probabilities for this current. And, therefore, there is no-- it would have been very hand-wavy, and actually wrong to think it's C over A. These definitions-- because after all, this is a definition-- makes some nice sense. Because you have J L-- we said is equal to J right, but J L is J A minus J B, is equal to J C. And therefore R plus T. The reflection coefficient plus the transmission coefficient-- which is J B over J A plus J C over J A is equal to J B plus J C over J A. And you see here that J A plus J C is indeed equal to J B. I'm sorry. I got this wrong. Yes. Where is the eraser. I should have passed the B to the other side. This force implies J A equals J B plus J C. And this ratio is equal to 1, which is something you usually want when you define reflection and transmission coefficients. They should add up to 1. So now we've got an idea. Yes, with this solution I can understand the reflection and transmission coefficients. But do these apply to particles? Well, the good news is that it roughly applies to particles-- as we will see with the wave packets soon. If you send the wave packet, it's going to have some uncertainty and momentum. It's going to have some uncertainty and energy. But for some energy-- suppose the wave packet doesn't have that much uncertainty-- basically, the probability that the wave packet bounces is the reflection at that energy, that is the main energy that the wave packet sends. If your wave packet is very broad over energies, then it's a more complicated thing. But as long as the wave packet is such that it basically has one narrow band of energy, the reflection coefficient associated to this calculation is the reflection coefficient or reflection probability for the wave packet that you're sending in.

MIT_804_Quantum_Physics_I_Spring_2016

Normalizable_wavefunctions_and_the_question_of_time_evolution.txt

BARTON ZWIEBACH: We were faced last time with a question of interpretation of the Schrodinger wave function. And so to recap the main ideas that we were looking at, we derive this Schrodinger equation, basically derived it from simple ideas-- having operators, energy operator, momentum operator, and exploring how the de Broglie wavelength associated to a particle would be a wave that would solve the equation. And the equation was the Schrodinger equation, a free Schrodinger equation, and then we added the potential to make it interacting and that way, we motivated the Schrodinger equation and took this form of x and t psi of x and t. And this is a dynamical equation that governs the wave function. But the interpretation that we've had for the wave function, we discussed what Born said, was that it's related to probabilities and psi squared multiplied by a little dx would give you the probability to find the particle in that little dx at some particular time. So psi of x and t squared dx would be the probability to find the particle at that interval dx around x. And if you're describing the physics of your Schrodinger equation is that of a single particle, which is the case here-- one coordinate, the coordinate of the particle, this integral, if you integrate this all over space, must be 1 for the probability to make sense. So the total probability of finding the particle must be 1, must be somewhere. If it's in one part of another part or another part, this probabilities-- for this to be a probability distribution, it has to be well-normalized, which means 1. And we said that this equation was interesting but somewhat worrisome, because if the normalization of the wave function satisfies, if this holds for t equal t-nought then the Schrodinger equation, if you know the wave function all over space for t equal t-nought, which is what you would need to know in order to check that this is working, a t equal t-nought, you take the psi of x and t-nought, integrate it. But if you know psi of x and t-nought for all x, then the Schrodinger equation tells you what the wave function is at a later time. Because it gives you the time derivative of the wave function in terms of data about the wave function all over space. So automatically, the Schrodinger equation must make it true that this will hold at later times. You cannot force the wave function to satisfy this at all times. You can force it maybe to satisfy at one time, but once it satisfies it at this time, then it will evolve, and it better be that at every time later, it still satisfies this equation. So this is a very important constraint. So we'll basically develop this throughout the lecture today. We're going to make a big point of this trying to explain why the conditions that we're going to impose on the wave function are necessary; what it teaches you about the Hamiltonian, we'll teach you that it's a Hermitian operator; what do you learn about probability-- you will learn that there is a probability current; and all kinds of things will come out of taking seriously the interpretation of this probability, the main point being that we can be sure it behaves as a probability at one time, but then for later times, the behaviors and probability the Schrodinger equation must help-- must somehow be part of the reason this works out. So that's what we're going to try to do. Now, when we write an equation like this, and more explicitly, this means integral of psi star of x and t, psi of x and t dx equal 1. You can imagine that not all kind of functions will satisfy it. In particular, any wave function, for example, that at infinity approaches a constant will never satisfy this, because if infinity, you approach a constant, then the integral is going to be infinite. And it's just not going to work out. So the wave function cannot approach a finite number, a finite constant as x goes to infinity. So in order for this to hold-- order to guarantee this can even hold, can conceivably hold, it will require a bit of boundary conditions. And we'll say that the limit as x goes to infinity or minus infinity-- plus/minus infinity of psi of x and t will be equal to 0. It better be true. And we'll ask a little more. Now, you could say, look, certainly the limit of this function could not be in number, because it would be non-zero number, the interval will diverge. But maybe there is no limit. The wave function is so crazy that it can be integrated, but suddenly, it has a little spike and it just doesn't have a normal limit. That could conceivably be the case. Nevertheless, it doesn't seem to happen in any example that is of relevance. So we will assume that the situations are not that crazy that this happened. So we'll take wave functions that necessarily go to 0 at infinity. And that certainly is good. You cannot prove it's a necessary condition, but if it holds, it simplifies many, many things, and essentially, if the wave function is good enough to have a limit, then the limit must be 0. The other thing that we will want is that d psi/dx, the limit as x goes to plus/minus infinity is bounded. That is, yes, the limit may exist and it may be a number, but it's not infinite. And In every example that I know of-- in fact, when this goes to 0, this goes to 0 as well-- but this is basically all you will ever need in order to make sense of the wave functions and their integrals that we're going to be doing. Now you shouldn't be too surprised that you need to say something about this wave function in the analysis that will follow, because the derivative-- you have the function and its derivative, because certainly, there are two derivatives here. So when we manipulate these quantities inside the integrals, you will see very soon-- single derivatives will show up and we'll have to control them. So the only thing that I'm saying is that when you see a wave function that satisfies this property, you know that unless the function is extremely crazy, it's a function that goes to 0 at plus/minus infinity. And it's the relative pursuant it also goes to 0, but it will be enough to say that it maybe goes to a number. Now there's another possibility thing for confusion here with things that we've been saying before. We've said before that the physics of a wave function is not altered by multiplying the wave function by a number. We said that psi added to psi is the same state; psi is the same state as square root of 2 psi-- all this is the same physics, but here it looks a little surprising if you wish, because if I have a psi and I got this already working out, if I multiply psi by square root of 2, it will not hold. So there seems to be a little maybe something with the words that we're been using. It's not exactly right and I want to make sure there is no room for confusion here, and it's the following fact. Here, this wave function has been normalized. So there's two kinds of wave functions that you can have-- wave functions that can be normalized and wave functions that cannot be normalized. Suppose somebody comes to you and gives you a psi of x and t. Or let's assume that-- I'll put x and t. No problem. Now suppose you go and start doing this integral-- integral of psi squared dx. And then you find that it's not equal to 1 but is equal to some value N, which is different from 1 maybe. If this happens, we say that psi is normalizable, which means it can be normalized. And using this idea that changing the value-- the coefficient of the function-- doesn't change too much, we simply say, use instead psi prime, which is equal to psi over square root of N. And look what a nice property this psi prime has. If you integrate psi prime squared, it would be equal-- because you have psi prime here is squared, it would be equal to the integral of PSI squared divided by the number N-- because there's two of them-- dx, and the number goes out and you have the integral of psi squared dx, but that integral was exactly N, so that's 1. So if your wave function has a finite integral in this sense, a number that is less than infinity, then psi can be normalized. And if you're going to work with probabilities, you should use instead this wave function, which is the original wave function divided by a number. So they realize that, in some sense, you can delay all of this and you can always work with wave functions that are normalizable, but only when you're going to calculate your probabilities. You can take the trouble to actually normalize them and those are the ones you use in these formulas. So the idea remains that we work flexibly with wave functions and multiply them by numbers and nothing changes as long as you realize that you cannot change the fact that the wave function is normalizable by multiplying it by any finite number, it will still be normalized. And if it's normalizable, it's equivalent to a normalized wave function. So those two words sound very similar, but they're a little different. One is normalizable, which means it has an integral of psi squared finite, and normalize is one that already has been adjusted to do this and can be used to define a probability distribution. OK. So that, in a way of introduction to the problem that we have to do, our serious problem is indeed justifying that the time evolution doesn't mess up the normalization and how does it do that?

MIT_804_Quantum_Physics_I_Spring_2016

Correspondence_principle_amplitude_as_a_function_of_position.txt

PROFESSOR: There's one more property of this thing that is important, and it's something called the correspondence principle, which is another classical intuition. And it says that the wave function, and it addresses the question of what happens to the amplitude of the wave function. It says that the wave function should be larger in the regions where the particle spends more time. So in this problem, you have the particle going here. It's bouncing and it's going slowly here, it's going very fast here. So it spends more time here, spends a lot of time here, spends a lot of time here. So it should be better in these regions and smaller in the regions that spends little time. So this was called the correspondence principle, which is a big name for a somewhat vague idea. But nevertheless, it's an interesting thing and it's true as well. So let me explain this a little more and get the key point about this. So we say, if you have a potential, you have x and x plus dx, so this is dx, the probability to be found in the x is equal to psi squared dx, and it's proportional to the time spent there. So we'll say that it's-- we'll write it in the following way. It's proportional to the fraction of time spent in dx. And that, we'll call little t over the period of the motion in this oscillation. The classical particle is doing, the period there. That's the fraction of time it spends there. Up two factors of 2, maybe, because it spends going there and there for the whole period, it doesn't matter, it's anyway approximate. It's a classical intuition expressed as the correspondence principle. So this is equal to dx over v, over the velocity that positioned the [INAUDIBLE] velocity T. And this is there for dx. And the velocity is p over m, so the mass over period and the momentum. So here we go. Here's the interesting thing. We found that the magnitude of the wave function should be proportional to 1 over p of x, or lambda over h bar of x. So then the key result is that the magnitude of the wave function goes like the square root of the position the [INAUDIBLE] de Broglie wavelength. So if here the de Broglie wavelength is becoming bigger because the momentum is becoming smaller, the logic here says that yes indeed, in here, the particle is spending more time here, so actually, I should be drawing it a little bigger. So when I try to sketch a wave function in a potential, this is my best guess of how it would be. And you will be doing a lot of numerical experimentation with Mathematica and get that kind of insight. They position the [INAUDIBLE] de Broglie wavelength as you have, it is a function of the local kinetic energy. And that's what it gives for you. OK so that is one key insight into the plot of the wave function. Without solving anything, you can estimate how the wave length goes, and probably to what degree the amplitude goes. What else do you know? There's the node theorem that we mentioned, again, in the case of the square well. The ground state, the bounce state, the ground state bounce state is a state without the node. The first excited state has one node, the next excited state has two nodes, the next, three nodes, and the number of nodes increase. With that information, it already becomes kind of plausible that you can sketch a general wave function.

MIT_804_Quantum_Physics_I_Spring_2016

Scattering_in_1D_Incoming_and_outgoing_waves.txt

BARTON ZWIEBACH: Scattering in dimension. And we will consider a world that is just one dimensional, x. And, in fact, there's an infinite barrier at x equals 0. Infinite barrier, nothing goes beyond there. On the other hand, in here, up to some distance R, there could be a potential V of x. So we will have a potential V of x, it will have the following properties-- it will be identically 0 for x greater than R, which is the range of the potential; it will be some function V of x for x in between R and 0; and it will be infinity for x less than or equal than 0. This is your potential, it's a potential of range R-- range of the potential. And the experiment that we think about is somebody at x equal plus infinity throwing waves into this potential. And this observer can only get back a reflected wave, and from that reflected wave, the observer wants to deduce the type of the potential that you have there. And that's absolutely the way physics goes in particle physics. In LHC, you throw protons or electrons together and you just catch what flies out of the collision with all the detectors and read that they then deduce what happened to the collision, what potential was there, what forces were there, was there a new particle? It's all found by looking at what comes out and flies away. So there's enormous amount of information on the potential from the data that comes out of you throwing in some particles in and waiting to see what comes back to you. So we will always have this infinite wall. And this infinite wall at x equals 0 means that x less than 0 is never relevant. And this is analogous to R, the variable R in radial coordinates for which the radial distance is never negative. So in fact, what we'll do here has immediate applications when we will consider-- not in this course-- scattering in three dimensions. So to begin this, we'll solve the simplest case where you have no potential whatsoever. Now no potential means still the barrier at x equals 0, the infinite barrier, but in between 0 and R, nothing is happening. So you have just a case of no potential. Is the case where you have the barrier here and x is over there and up to R, nothing's happening, it's just the wall. That's all there is, just one wall. So this is V-- no potential is V is equal to V of x is equal 0 for x greater than 0. And it's infinity for x less or equals than 0. So in this case, let's assume we have an incident wave. An incident wave must be propagating in this way, so an incident wave is an e to the minus ikx. And if you have an outgoing wave, it would be some sort of e to the ikx. These are the only things that can be there. They correspond to energy eigenstates, this is the de Broglie wave function of a particle with momentum, in one direction or in the other direction. But let's combine them in a way to produce a simple solution. So this solution, phi of x, will be the solution. Will be a combination that's similar-- e to the ikx and e to the minus ikx, and I should make the wave function vanish at x equals 0. At x equals 0, both exponentials are equal to 1, so if I want them to cancel, I should put a minus. So in order to simplify this the best possible way, we can put a 1 over to 2i's over there so that we have a sine function, and the sine function is particularly nice. So we'll have e to the-- output like this-- e to the minus ikx plus e to the ikx over 2i, and this is just the sine function side of kx, which you would admit, it's a V solution over here, a sine of kx. On the other hand, I can think of this solution as having an incoming wave, which is minus e to the minus ikx over 2i, and an outgoing wave of e to the ikx over 2i. So this is the representation of the solution when nothing is happening, and the good thing about this solution is that it tells us what we should write-- gives us an idea of what we should write when something really is happening. So now let's consider how we would write the general experiment in which you send in a wave but this time, there is really a potential. So let's consider now, if no potential was there and now yes, potential-- so no potential here, so what does it mean yes potential? Well, it means you have this and you have some potential there up to some distance R, and then it flattens out, and something happens. So in order to compare, we'll take an incoming wave, the same as the one where there was no potential. So let's take an incoming wave, which is of the form e to the minus ikx times minus over 2i. But I must say here, I must write something more-- I must say that x is greater than R, otherwise this is not the solution. You see, in the region where the potential really exists, where the-- goes up and down, you don't know the solution. It would take solving the Schrodinger equation. You know the solution where the potential is 0, so yes, this incoming wave is the solution of the Schrodinger equation in this potential when x is greater than R. And how about the outgoing wave? Well, we would like to write it like that. So we'll say 1 over 2i e to the ikx is also an outgoing wave, and we have no hope of solving it here, finding what's happening here unless we solve a complicated equation, but then let's look outside-- we're still looking outside. But that cannot be the outgoing wave. This is the same as the other one and there is a potential, so something must be different. On the other hand, if you think about it, very little can be different because you must have a solution with 0 potential and-- you know these plane waves going out are the only things that exist. And now you decide, oh, if that's the case, I cannot put another function of x in there because that's not a solution. The best I can do is multiply by a number, because maybe there's very little outgoing wave or there is not, but then I think of another thing-- remember if you had e to the A, e to the minus ikx plus B e to the ikx, well, the probability current was proportional to A squared minus B squared. And this time, however, you have-- you're sending in a wave and you're getting back a wave and this is a stationary state-- we're trying to get energy eigenstate, solutions of some energy just like this energy. And the only way it can happen is if they carry the same amount of probability-- probability cannot be accumulating here, nor it can be depleted there as well, so the currents associated to the two waves must be the same. And the currents are proportional to those numbers that multiply these things squared, so in fact, A squared must be equal to B squared, and therefore we cannot have like a 1/3 here, it would just ruin everything. So the only thing I can have is a phase. It's only thing-- cannot depend on x, because that was an unsolved equation. Cannot be a number that is less than 1 or bigger than 1. The only thing we can put here is a phase. So we'll put an e to the 2i delta. And this delta will depend on k or will depend on the energy, and it will depend on what your potential is, but all the information of this thing is in this delta that depends on k. And you say, well, that's very little, you just have one phase, one number that you could calculate and see, but remember, if you have a delta of k, you could measure it for all values of k by sending particles of different energies and get now a whole function. And with a whole function delta of k, you have some probability of getting important information about the potential. So we'll have a phase there, e to the i delta of k. And let's summarize here, it's due to current conservation-- the current of this wave and the current of the outgoing waves should be the same. And also note that no extra x dependents is allowed. So this will produce the J incident will be equal to J reflected. Now you could say, OK, very good, so there's delta, there's a phase-- should I define it from 0 to 2 pi? From minus pi to pi? It's kind of natural to define it from minus pi to pi, and you could look at what it is, but as we will see from another theorem, Levinson's theorem, it will be convenient to just simply say, OK, you fixed the phase delta at k equals 0. At 0 energy scattering, you read what is your delta-- unless you increase the energy, the phase will change. So if you have a phase, for example, on a circle, and the phase starts to grow and to grow and to grow and to grow and to go here, well, should you call this pi and this minus pi? No. You probably should just-- if it keeps growing with energy-- and it might happen, that the phase keeps growing with energy-- well, pi, 2 pi, 3 pi, 4 pi, just keep the phase continuous. So keeping the phase continuous is probably the best way to think about the phase. You start at some value of the phase and then track it continuously. There is always a problem with phases and angles, they can be pi or minus pi's the same angle, but try for continuity in defining the phase when we'll face that problem. So let's write the solution. We have this, so the total solution. We call the solution with no potential phi, this one we'll call psi of x will be 1 over 2i, the first term-- e to the ikx plus 2i delta minus e to the minus ikx. It's convenient to pull out of i delta to make the two terms have opposite arguments, so e to the ikx plus i delta, and-- or e to the ikx plus delta parenthesis minus e to the minus ikx plus delta. So this is e to the i delta times sine of kx plus delta. So that's this full scattered wave, not the full reflected-- well, that word again. This is the full wave that you have for x greater than R. So let's write it here-- psi of x. It's not the reflected wave nor that it covers everything. We include-- it's for x greater than R, but we include the incoming and outgoing things, because both are defined for x greater than R, so the total wave is this one. And you notice that if the phase shift is 0, you are nicely back to the wave function phi that we found before.

MIT_804_Quantum_Physics_I_Spring_2016

Energy_eigenstates_on_a_generic_symmetric_potential_Shooting_method.txt

PROFESSOR: Here is your potential. It's going to be a smooth, nice potential like that. V of x. x, x. And now, suppose you don't know anything about the energy eigenstates. Now, this potential will be assumed to be symmetric. So here is one thing you can do. You can exploit some things that you know about this potential. And here's the wave function that we're going to try to plot. And we could say the following. Let's see. Whatever energy are here, for bound states, I'm going to eventually be in the forbidden region. So far on the right here, I will be in the forbidden region. And I must meet the wave function that looks like the forbidden region wave function. And the only possibility is something like that. You could say it's the [INAUDIBLE] of 1, but actually, if it's a [INAUDIBLE] then I could multiply by minus 1 and use this one conventionally. That wave function is always going to be like that over there. On the other hand, very important-- the on the very left, how will the wave function look? Well it also has to decay, so it can decay like that, or it might be decaying like this. And in fact, you don't know until you figure out what's happening in the middle. It may be decaying like this or like that. I fix the sign here, so whatever this does it should either end up like that or end up like that. So these are the guidelines that you have to solve this. Should begin like that, and we'll see. And it should be either symmetric or anti-symmetric. This would be the case anti-symmetric, this would be the case symmetric. OK. So let's draw one, two, three, four lines there. One, two, three, and four. And I don't know where the energies lie. I don't know what is the ground state energy. And I want to give you an insight into how you can figure out why you get this energy one decision when this happens. So let's plot the wave function for the first case. I don't know if I have a label, but let's assume this is E0, E1, E2, E3. Three energies [INAUDIBLE], and here's the one for e0. OK. So I begin here, that's how it goes. And then I go through my Schrodinger equation, integrate it. You see? Numerical, you can always integrate the Schrodinger equation. And this should be always in this region, let me-- like this. And it's growing. And, oops, there should be turning points here. There should be turning points-- suppose this is-- I'm not going to get this so well, but it goes like this. And now this should be a turning point. So I should change to the other type of curvature, curvature down. But what probably will happen with E0 is that it will switch and it will go like and start to curve, maybe. Well, if it looks like that, it must match to the development of the odd piece or the even piece. Now, it's never going to match with the odd one, so it might be with the even. And yes, it would match turning point here, but look what has happened here. You got the corner there. You know, this was turning slowly, and this is starting to turn slowly, but here there is a discontinuity in the derivative. So this is not the solution. You try, but you fail. But that's-- right. Not every energy gives a solution. So they should have matched continuously and derivative continuously at that point, but it didn't have enough time to do that. On the other hand, if we try the next one, maybe. The turning points will be here. Let's see what happens. Well, now the forbidden energies are over here, and now you have a turning point here that-- in here, the curvature is negative, the second derivative's curvature. And it's larger than it was here. Here it was small, here it's larger. So it's going to curve faster. Maybe if you get the E1 right, it will curve enough so this flat here, in which case the other side will match nicely and you've got the solution. So you probably have to go little by little until this becomes flattened and, boom, you've got the solution. Energy eigenstate. Let's go a little further. This graph continues there. Now I want to go to E2. How am I going to do that? I'm going to do it this way. So this was here, this was there. There is the vertical line here. And for E2, the turning points are even further out. And here is the wave function. And let's look at this thing that I have. Now, the turning point in this one corresponds to the E2 turning point. This is E1. And now this will go in here, we'll turn, and we will go curve and maybe do something like that. Because it's curving more and more, and faster. So by the time you reach here, this is no good, because this one will be symmetric. You know, you would have an anti-symmetric one that is no use. But now you don't have a solution, again. So as you increase the energy, this is starting to do this, and that is not quite so good. And then when you go to E3, you have a turning point over here. So maybe in this case it will go up here and it will start turning, and it will turn enough to just-- this dip go to the origin. OK. You're saying no good either, because this is terrible, this continuous. But, ah, you were supposed to draw the other one as well. The old one is actually perfect for it. So this is dash, it doesn't exist, and this one matches here. So by the time the dip-- this is not a solution, but the dip goes down and down, and eventually goes to zero, it matches with this one. That's why I said sometimes you don't know whether this matches with the one that comes from here or the one that comes from the bottom. So there you go. This is an energy eigenstate again. It's odd and it has one node. And that gives you the intuition how, as you sort of come from the end and you reach the middle, you sometimes match things or sometimes don't match. And explains why you get energy quantization. The other way in which you're going to gain intuition is with the so-called shooting method, which is the last thing I want to discuss for a minute. So the shooting method in differential equations is quite nice. Shooting method. Suppose you have a potential that this symmetric may be something like this-- it doesn't look very symmetric. It looks a little better now. And you want to find energy eigenstates. You do the following. You say, well, the normalization of the energy eigenstates is not so important. Let's look for even states. Now, you can look for even or odd states if the potential is symmetric. Sometimes the potential will have a wall, in which case you have to require a symmetric potential. It's easy to solve, as well. But let's consider the case when the potential is symmetric and you look for even states. So what you do is just, say, you pick an energy. Pick some energy E0. And then you put some boundary condition. You say that the wave function at x equals 0 is 1. And then you say that the derivative of the wave function at x equals zero is how much? Any suggestion? How much should it be? You see, you have a second order differential equation. The second psi is equal to E minus V. That's the Schrodinger equation. You need-- the boundary conditions are the value of psi and the derivative at one point, and then the computer will integrate for you. Mathematica will do it. But you have to give me the derivative, so what should I put? A number there, 1, 2, 3? Is that an unknown? What should I pick? We must put in the 0, because if you had a wave function whose derivative is not 0, and it's an even wave function, it would look like this. And there would be a discontinuity in psi prime-- discontinuous. And that's not possible unless you have a hard wall or you have a delta function. So you must put this. And then you integrate numerically. Numerically. And what will happen? Well, if you integrate numerically, the computer is just going to integrate and see no problem. Basically, it's just going to do the interview. Ask the computer to calculate the wave function out to x equal 5, it will calculate it. So the problem is that-- you can see visually, if you pick some energy, the wave function will do like something, and then will start blowing up. And then you say, oh, that energy is no good because the wave function won't be normalizable. And then you go back to the computer and change the energy a little bit, and then you will find, well, maybe this. Now it blows up in the other direction. No good either. But in some energy in between, as you change, there must be one in which it does the right thing, which is BK. Somewhere in between. And numerically, you change the value of the energy, you go-- in the shooting method, when you shoot it either goes up or down. And you start working within those two numbers to restrict it until you get here. If you have a solution with five-digit accuracy, it will do this, this, this, and then blow up. If you have a solution with 10 digits after it, it will do this and go up to here and blow up again. You need 500-digits accuracy to get that wall. But it's a fun thing that you can do numerically and play with it. You can calculate five digits accuracy, ten digits accuracy within a matter of minutes. It's very practical, and it's very nice, but one thing you have to do is clean up your equation before you start. You cannot have an equation in Mathematica with h-bar and m and all that. So you have to clean up the units, is the first step, and write it as an equation question without units. Your And this plots very nicely in Mathematica, and you will have lots of practice.

MIT_804_Quantum_Physics_I_Spring_2016

Local_picture_of_the_wavefunction.txt

PROFESSOR: OK, so, local picture. It's all about getting insight into how the way function looks. That's what we'll need to get. These comments now will be pretty useful. For this equation you have one over psi, d second psi, d x squared is minus 2 m over h squared, E minus v of x. Look how I wrote it, I put the psi back here, and that's useful. Now, there's a whole lot of discussion-- many textbooks-- about how the way function looks, and they say concave or convex, but it depends. Let's try to make it very clear how the wave function looks. For this we need two regions. So, the first case, A, is when the energy minus v of x is less than 0. The energy is less than v of x, that's a forbidden region-- as you can see there-- so it's a classically forbidden. Not quantum mechanically forbidden, but classically forbidden. What is the main thing about this classically forbidden region is that the right hand side of this equation is positive. Now, this gives you two possibilities. It may be that psi at some point is positive, in which case the second psi must also be positive, because psi and the second psi appear here. If both are positive, this is positive. Or, it may be case two, that psi is negative, and the second psi-- the x squared-- it's also negative. Well, how do we plot this? Well, you're at some point x, and here it is, a positive wave function seems to be one type of convexity, another type of convexity for a negative, that's why people get a little confused about this. There's a way to see in a way that there's is no confusion. Look at this, it's positive, second derivative positive. When you think of a second derivative positive, I think personally of a parabola going up. So, that's how it could look. The wave function is positive, up, it's all real. We're using the thing we proved at the beginning of this lecture: you can work with real things, all real. So, the wave from here is x, and here negative. And the negative opening parabola, that's something they got. So nice. So, the wave function at any point could look like this if it's positive, or, it could look like this if it's negative. So, it doesn't look like both, it's not double value. So, either one or the other. But, this is easy to say in words, it is a shape that is convex towards the axis. From the axis it's convex here and convex there. So, convex towards the axis. Now, there's another possibility I want to just make sure you visualize this. Sometimes this looks funny-- doesn't mean actually the way function can look like that-- but, it's funny because of the following reason. It's funny because if you imagine it going forever, it doesn't make sense because you're in a classically the forbidden region. And the way function's becoming bigger and bigger is going to blow up. So, eventually something has to happen. But, it can look like this. So, actually what happens is that when you're going to minus infinity-- here is x and we use minus infinity-- it can look like this. This is an example of this piece that is asymptotic, and it's positive, and the second derivative is positive. Or, negative and the second derivative is negative. So that's a left asymptote. Or, you could have a right asymptote, and it looks like this. Again, second derivative positive, positive wave function. Second derivative negative, negative wave function. So, you may find this at the middle of the potential, but then eventually something has to take over. Or, you may find this behavior, or this behavior, at plus minus infinity. But, in any case you are in a classically forbidden, you're convex towards the axis. That's the thing you should remember. On the other hand, we can be on the classically allowed region. So, let's think of that. Any questions about the classically forbidden? Classically allowed, B. E minus v of x greater than 0, classically allowed. On the right hand side of the equation is negative. So, you can have, one, a psi that is positive, and a second derivative that is negative. Or, two, a psi that is negative, and a second derivative that this positive. So, how does that look? Well, positive and second derivative negative, I think of some wave function as positive, and negative is parabolic like that. And then, negative and second derivative positive, it's possible to have this. The wave function there it's negative, but the second derivative is positive. These things are not very good-- they're not very usable asymptotically, because eventually if you are like this, you will cross these points. And then, if you're still in the allowed region you have to shift. But, this is done nicely in a sense if you put it together you can have this. Suppose all of this is classically allowed. Then you can have the wave function being positive, the second derivative being negative, matching nicely with the other half. The second derivative positive, the wave function negative, and that's what the psi function is. It just goes one after another. So, that's what typically things look in the classically allowed region. So, in this case, we say that it's concave towards the axis. That's probably worth remembering. So, one more case. The case C, when E is equal to E minus v of x not is equal to 0. So, we have the negative, the positive, 0. How about when you have the situation where the potential at some point is equal to the energy? Well, that's the turning points there-- those were our turning points. So, this is how x 0 is a turning point. And, something else happens, see, the right hand side is 0. We have that one over psi, the second psi, the x squared is equal to 0. And, if psi is different from 0, then you have the second derivative must be 0 at x not. And, the second derivative being 0 is an inflection point. So, if you have a wave function that has an inflection point, you have a sign that you've reached a turning point. An inflection point in a wave function could be anything like that. Second derivative is positive here-- I'm sorry-- is negative here, second derivative is positive, this is an inflection point. It's a point where the second derivative vanishes. So, that's an inflection point. And, it should be remarked that from that differential equation, you also get that the second psi, the x squared, is equal to E minus v times psi, which is constant. And, therefore, when psi vanishes, you also get inflection points automatically because the second derivative vanishes. So, inflection points also at the nodes. Turning point is an inflection point where you have this situation. Look here, you have negative second derivative, positive second derivative, the point where the wave function vanishes and joins them is an inflection point as well. Is not the turning point-- turning point are more interesting-- but inflection points are more generic.

MIT_804_Quantum_Physics_I_Spring_2016

Excited_states_of_the_harmonic_oscillator.txt

PROFESSOR: Suppose you define now, one state called phi 1 as a dagger acting on phi 0. You could not define any interesting state with a acting on phi 0 because a kills phi 0, so you try phi 0 like this. Now you could ask, OK, what energy does it have? Is it an energy eigenstate? Well it is an energy eigenstate if it's a number eigenstate. And we can see if it's a number eigenstate by acting with the number operator. So N phi 1 is equal to N a dagger phi 0. OK. Here comes trick. Maybe it's too much to even call it a trick, number one. This thing you look at it and you say, I want to sort of simplify this, learn something about it. If this is supposed to be an eigenstate of N hat, I have to make it happen somehow. Now n hat kills phi 0. So if I would have a term a dagger times N hat near phi 0, it would be 0. So I claim, and this is a step that I want you to be able to do also quickly, that I can replace this by the commutator of these two operators. The product is replaced by the commutators. Why? Aren't products simpler than commutators? No. We have formulas for commutators. And products are, in general, more complicated. And why is this correct? And you say, well, it is correct because this has two terms. The term I want minus a dagger N hat. But the term a dagger N hat is 0 because N hat kills phi 0. So I can do that because this is N dagger a hat, which is what I had, minus a hat dagger N on phi 0. And this term is 0. So you would have put a 2 here or a 3 here, or any number even. But the right one to put is the commutators. So that's this. And now this commutator is already known. That's why we computed it. It's just a dagger. So this is a dagger phi 0, and that's what we call phi 1. So N hat on phi 1 is phi 1. N hat has eigenvalue 1 on phi 1. So N is equal to 1. That's the eigenvalue. It is an eigenstate. It is an energy eigenstate. In fact how much energy, E, is h bar omega times N, which is 1, plus 1/2, which is 3/2 h bar omega? And look what this is. This is the reason this is called a creation operator. Because by acting on the ground state, what people sometimes call the vacuum, the lowest energy state, the vacuum is called the lowest energy state, by acting on the vacuum you get a state. I mean, you've created a state, therefore. How is this concretely done? Remember you had phi 0 of x, what it is, and a dagger over there is x minus ip over m omega. So this is x minus-- or minus h bar over m omega d dx. So you can act on it. It may be a little messy. But that's it. It's a very closed form expression. Now, phi 0 was defined, the ground state such that it's a normalized state. This means the integral of phi 0 multiplied with phi 0 over x is 1. That's how we had the ground state. You could ask, if I've defined phi 1 this way, is simply normalized? So I'll try it. And now you could say, oh, this is going to be a nightmare. Normalizing phi 0 is difficult. Now I have to act with a dagger, which means act with x, take derivatives. It's going to grow twice as big. Then I'm going to have to square it and integrate it. It looks very bad. The good thing is those with these a's and a daggers, you have to compute anything, pretty much. See how we do it. I want to know how much is phi 1 with phi 1. Is it 1? And it's normalized or not? Then I say, look, phi 1 is a dagger phi 0, a dagger phi 0. So far so good. But I just know things about phi 0. So let's clear up one phi 0. At least I can move the a dagger as an a. So this is phi 0 a a dagger phi 0. Can I finish the computation in this line? Yes, I think we can. Phi 0. a with a dagger, same story as before. a would kill phi 0. So you can replace that by a commutator. Commutator of a with a dagger phi 0. But the commutator of a with a dagger is 1, so this is phi 0 phi 0 and it's equal to 1. Yes, it is properly normalized. So that's the nice thing about these a's and a daggers. Just start moving them around. You have to get practice. Where should you move it? Where should you put it? When you replay something by a commutator, when you don't. It's a matter of practice. There's no other way. You have to do a lot of these commutators to get a feeling of how they work and what you're supposed to do. Let's do another state. Let's try to do phi 2. I'll put a prime because I'm not sure this is going to work out exactly right. And this time, I'll put an a dagger a dagger on the vacuum. Two a daggers, two creation operators on the vacuum. And now I want to see if this is an energy eigenstate. Well, this is a dagger squared on the vacuum. So let's ask, is N hat-- is phi 2 prime an eigenstate of N hat? Well I would have N hat on a dagger squared on phi 0. Again, by now you know, I should replace this by a commutator because N hat kills the phi 0, so N hat with a dagger squared phi 0. And that commutator has been done. It's two times a hat dagger squared, two times a dagger squared on phi 0, which is 2 phi 2. That's what we call the state phi 2 prime. I'm sorry. So again, it is an energy eigenstate. Is it normalized? Well, let's try it. Phi 2 prime phi 2 prime is equal to a dagger a dagger. Let me not put the hats. I'm getting tired of them. a dagger a dagger phi 0. Now I move all of them. This a dagger becomes an a, the next a dagger becomes an a here. So this is phi 0 a a a dagger a dagger phi 0. Wow, this looks a little more complicated. Because we don't want to calculate that thing, really. We definitely don't want to start writing x and p's. But, you know, you decide. Take it one at a time. This a is here and wants to act on this thing. And then this other a will, but let's just concentrate on the first a that wants to act. a would kill phi 0, so we can replace this whole thing by a commutator. So this is phi 0. The first a is still there, but the second, we'll replace it by the commutator, this commutator. I've replaced this product, the product of a times this thing, by the commutator of those two operators. And then I say, oh look, you've done that. a with a dagger to the k is k a dagger k minus 1. So I'll write it here. This will be a factor of 2 phi 0 a. And this is supposed to be now a dagger to one power less, so it's just a dagger phi 0. So this is supposed to be 2a dagger. So that's what I did. And again, this a wants to act on phi 0 and it's just blocked by a dagger, but you can replace it by a commutator. a with a dagger phi 0. And this is therefore a 1, so this whole result is a 2. So this phi 2 prime, yes, it is the next excited state. Two creation operators on the ground state. Energy and eigenvalues too. You had N equal zero eigenvalue for the ground state 1 for phi 1, 2 for phi 2 prime. But it's not properly normalized. Well, if the normalization gives you 2, then you should define phi 2 as 1 over the square root of 2 a dagger a dagger on phi 0. And that's proper. So it's time to go general. The n-th excited state, we claim is given by an a dagger a dagger, n of them, acting on phi 0 with a coefficient 1 over square root of-- we might think it's n, but it's actually, you can't tell at this far-- this one is n factorial. That's what you need. That is the state. And what is the number of this state? What is the number eigenvalue on phi n? Well, it is 1 over square root of n factorial. The number acting on the a daggers, the n of them, phi 0. You can replace by the commutator, which then is 2 times already. So it's N commutator with a dagger to the little n phi 0 times 1 over square root of n. And how much is this commutator? Over there. This is N times a dagger to the n phi 0. So between these three factors, you're still getting n phi to the n. So the number for this state is little n. It is an energy eigenstate. The N eigenvalue is little n. And the energy is h bar omega. The eigenvalue of N hat, which is little n plus 1/2. So it is the energy eigenstate of number little n. This is the definition. And the last thing you may want to check is the normalization. Let me almost check it here. No, I will check it. Let's say I think this is a full derivation. Phi n with phi n would be two factors of those, so I would have 1 over n factorial a dagger a dagger, n of them on phi 0, a dagger a dagger, n of them again on phi 0. So then that's equal to 1 over n factorial phi 0 a a, lots of a's, n of them, n a daggers, phi 0, like that. That's what it is. We had to move all the a daggers that were acting on the left input of the integral, or the inner product, all the way to the right. And that's it. So now comes this step. And I think you can see why it's working. Think of moving the first a all the way here. Well, you can replace the first a with a commutator. But that a with lots of a daggers, with n a daggers, would give you a factor of n, with n a daggers will give you a factor of n times one a dagger less. So to move the first a, there are n a daggers and you get one factor of n from this a. But for the next a, there's now n minus 1 a daggers, so this time you get a factor of n minus 1 when you move it. From the next one, there's going to be n minus 2 a daggers, so n minus 2. All of them all the way up to one, cancels this n factorial, and that's equal to 1.

MIT_804_Quantum_Physics_I_Spring_2016

Infinite_square_well_energy_eigenstates.txt

PROFESSOR: Square well. So what is this problem? This is the problem of having a particle that can actually just move on a segment, like it can move on this eraser, just from the left to the right. It cannot escape here. So the way we represent it is the interval 0 to a on the x-axis. And there's going to be two walls, one wall to the left and one wall to the right, and no potential in between. That is, I write the potential V of x as 0, for x in between a and 0, and infinity for x less than or equal to 0, and x greater than or equal to a. So basically the particle can move from 0 to a, and nowhere else. The potential is infinity. Now, this problem, meaning that the wave function-- the particle cannot be outside the interval, means that the wave function must vanish outside the interval. And you could say, how do you know? Well, if the potential is close to infinite amount of energy to be there, so the particle cannot really be there if it's really infinite energy that you need. You will see in the finite square well that the particle has probability to be in regions where it classically cannot be. But that probability will go to 0 if the potential is infinite. So we can think of it as a limit and we will reconfirm that. But in fact, if the potential is infinity, we will take it to mean that psi of x is equal to 0 for x less than 0 and for x greater than a. I am putting this equals or-- there are many ways of doing this. If this function, as this continues, you have a wall at a, is the potential 0 at a or is it infinity? Well it doesn't quite matter. The issue is that the wave function is 0 here, is 0 there, and we've said that the wave function must be continuous. So it should be 0 by that time you're at 0 or at a. So therefore we will take psi of 0 to be 0, and psi of a to be 0 by continuity. So we discuss why the wave function has to be continuous. If the wave function is not continuous, the second derivative of the wave function is terribly singular. It's like a derivative of a delta function, which is an impossible situation. So the wave function, we will take it to vanish at these two places, and this is what is called a hard wall. So what is the Schrodinger equation? The Schrodinger equation is, again, a free Schrodinger equation. Nothing, no potential here, so it's the same Schrodinger equation we had there, psi double prime equals minus 2mE over h squared, psi of x. Or, again, minus k squared psi of x. Let's solve this. So how do we do it? Well it's, again, a very simple equation, but this time it's conveniences-- we don't have a circle or periodicity to use sines and cosines. So I'll take psi of x to be c1 cosine of kx plus c2 sine of kx. But the wave function must vanish at 0. And at 0, the cosine is 1, so you get c1. And the sine is 0, so this must be 0, so c1 is gone. There's no c1 contribution to the solution. So psi of x is c2 sine of kx. But we're not done. We need this function to vanish at the other side. So psi of x equals a must be 0, and that c2 sine of ka must be 0. And therefore we realize that ka must be equal to a multiple of pi because sine vanishes for 0, pi, 2 pi, 3 pi, minus pi, minus 2 pi, minus 3 pi, all the multiples of pi. And therefore we will write kn equals 2 pi n-- not 2 pi n. Pi n over a. OK, well, let me ask you, what should we take for n? All integers? Should we skip some? We took all integers for the circle, but should we take all integers here? So what happens here, n equals 0. What's the problem with n equals 0? n equals 0, k equals 0, the wave function vanishes. Well, wave function vanishing is really bad because there is no particle then. There is nowhere in the probability to find the particle. So n equals 0 is not allowed, for sure. n equals 0, no. So why did we allow it, n equals 0, in the circle? In the circle for n equals 0, exponential doesn't vanish. It's a constant and that constant is a fine wave function. 0 is not fine, but the constant is good. But n equals 0 is not. So how about positive ends or negative ends. And here comes the problem, see we're getting to it. For n equals minus 2 or for n equals 2. So in one case, k is a number. And in the other case, k is the opposite sign number. And sine of a number, or minus a number, that number goes out. So if you have a sine of minus kx, that's minus sine of kx. And two wave functions that differ by a sign are the same wave function, physically. There's nothing different. They could differ by an i and other things. So when you pick negative n minus 1, or pick n equals plus 1, you get the same wave function, but just different by a sign. So it's not new. So in this case, it's very interesting that we must restrict ourselves. We can correct all this and just say n equals 1, 2, 3, all the way to infinity. The wave function, then, is psi n of x, is proportional to sine of n pi x over a. And you look at it and you say, yes, that looks nice. For x equals 0, it vanishes. For x equals a, it vanishes. n and minus n would give me the same wave function up to a sine. So this is good. I just have to normalize it. And normalizing it would be done by putting an n here. And then the integral psi n squared dx from 0 to a only would be n squared integral from 0 to a dx of sine squared n pi x over a. Now, you can do this integral by calculation. And our sine squared is written in terms of a double angle cosine of double angle plus a 1/2. The intuition with these things are that if you're integrating over the right interval that contains an integer number of cycles of the sine squared, then the sine squared has average 1/2. Because sine squared plus cosine squared is equal to 1. So you don't have to do the interval in general. This is n squared times 1/2 times the length of the interval, which is a. And therefore n squared, this is equal to 1, and therefore n is equal to square root of 2/a and we can write now our solutions. Our solutions are m psi n of x equals the square root of 2/a sine n pi x over a. And n equals 1, 2, up to infinity. And En is equal to h bar squared, k squared, so pi squared, n squared, a squared, to m. That's it for the solutions of-- are there degeneracies? No. Every energy state is different because there's any single 1, 2, 3, infinity, each one has more energy than the next. No, I'm sorry, the energy increases as you increase n. The energy levels actually become more and more spaced out. And the last thing I want to do with this box is to look at the states and see how they look and gather some important properties that are going to be very relevant soon.

MIT_804_Quantum_Physics_I_Spring_2016

Simultaneous_eigenstates_and_quantization_of_angular_momentum.txt

PROFESSOR: Simultaneous eigenstates. So let's begin with that. We decided that we could pick 1 l and l squared, and they would commute. And we could try to find functions that are eigenstates of both. So if we have functions that are eigenstates of those, we'll try to expand in terms of those functions. And all this operator will become a number acting on those functions. And that's why the Laplacian simplifies, and that's why we'll be able to reduce the Schrodinger equation to a radial equation. This is the goal. Schrodinger equation has r theta and phi. But theta and phi will deal with all the angular dependents. We'll find functions for which that operator gives a number acting on them. And therefore, the whole differential equation will simplify. So simultaneous eigenstates, and given the simplicity of l z, everybody chooses l z. So we should find simultaneous eigenstates of this two things. And let's call them psi l m of theta and phi. Where l and m are numbers that, at this moment, are totally arbitrary, but are related to the eigenvalues of this equations. So we wish that l z acting on psi l m is going to be a number times psi l m. That is to be an eigenstate. The number must have the right units, must be an H bar. And then we'll use m. We don't say what m is yet. M. Where m belongs to the real numbers. Because the eigenvalues of a Hermitian operator are always real. So this could be what we would demand from l z. From l squared on psi l m, I can demand that this be equal because of units and h squared. And then a number, lambda psi l m. Now this lambda-- do I know anything about this lambda? Well, I could argue that this lambda has to be positive. And the reason is that this begins as some sort of positive operator, is L. Squared. Now that intuition may not be completely precise. But if you followed it a little more with an inner product. Suppose we would have an inner product, and we can put psi l m here. And l squared, psi l m from this equation. This would be equal to h squared lambda, psi l m, psi I m. An inner product if you have it there. And then if your wave functions are suitably normalized, this would be a 1. But this thing is l x-- l x plus l y, l y plus l z, l z. And l x, l x-- you could bring one l x here, and you would have l x, psi l m, l x psi l m. Plus the same thing for y and for z. And each of these things is positive. Because when you have the same wavefunction on the left and on the right, you integrate the norm squared. It's positive. This is positive. This is positive. So the sum must be positive, and lambda must be positive. So lambda must be positive. This is our expectation. And it's a reasonable expectation. And that's why, in fact, anticipating a little the answer, people write this as l times l plus 1 psi l m. And where l is a real number, at this moment. And you say, well, that's a little strange. Why do you put it as l times l plus 1. What's the reason? The reason is-- comes when we look at the differential equation. But the reason you don't get in trouble by doing this is that as you span all the real numbers, the function l times l plus 1 is like this. l times l plus 1. And therefore, whatever lambda you have that is positive, there is some l for which l times l plus 1 is a positive number. So there's nothing wrong. I'm trying to argue there's nothing wrong with writing that the eigenvalue is of the form l times l plus 1. Because we know the eigenvalue's positive, and therefore, whatever lambda you give me that is positive, I can always find, in fact, two values of l, for which l times l plus 1 is equal to lambda. We can choose the positive one, and that's what we will do. So these are the equations we want to deal with. Are there questions in the setting up of these equations? This is the conceptual part. Now begins a little bit of play with the differential equations. And we'll have to do a little bit of work. But this is what the physical intuition-- the commutators, everything led us to believe. That we should be able to solve this much. We should be able to find functions that do all this. All right, let's do the first one. So the first equation-- The first equation is-- let me call it equation 1 and 2. The first equation is h bar over i d d 5. That's l z, psi l m, equal h bar m psi l m. So canceling the h bars, you'll get dd phi of psi l m is equal to i m, psi l m. So psi l m is equal to e to the i m phi times some function of theta. Arbitrary function of theta this moment. So this is my solution. This is up psi l m of theta and phi. With the term in the phi dependants, and it's not that complicated. So at this moment, you say, well, I'm going to use this for wavefunctions. I want them to behave normally. So if somebody gives me a value of phi, I can tell them what the wavefunction is. And since phi increases by 2 pi and is periodic with 2 pi, I may demand that psi l m of theta, and 5 plus 2 pi be the same as psi l m theta and phi. You could say, well, what if you could put the minus sign there? Well, you could try. The attempt would fail eventually. There's nothing, obviously, wrong with trying to put the sine there. But it doesn't work. It would lead to rather inconsistent things soon enough. So this condition here requires that this function be periodic. And therefore when phi changes by 2 pi, it should be a multiple of 2 pi. So m belong to the integers. So we found the first quantization. The eigenvalues of l z are quantized. They have to be integers. That was easy enough. Let's look at the second equation. That takes a bit more work. So what is the second equation? Well, it is most slightly complicated differential operator. And let's see what it does. So l squared. Well, we had it there. So it's minus h squared 1 over sine theta, dd theta, sine theta, dd theta, plus 1 over sine squared theta, d second d phi squared psi l m equal h squared l times l plus 1 psi l m. One thing we can do here is let the dd phi squared act on this. Because we know what dd phi does. Dd phi brings an i n factor, because you know already the phi dependents of psi l m. So things we can do. So we'll do the second d 5 squared gives minus-- gives you i m squared, which is minus m squared, multiplying the same function. You can cancel the h bar squared. Cancel h bar squared. And multiply by minus sine squared theta. To clean up things. So few things. So here is what we have. We have sine theta, dd theta. This is the minus sine squared that you are multiplying. The h squared went away. Sine theta, d p l m d theta. Already I substituted that psi was into the i m phi times the p. So I have that. And maybe I should put the parentheses here to make it all look nicer. Then I have in here two more terms. I'll bring the right-hand side to the left. It will end up with l l plus 1, sine squared theta, minus m squared, p l m equals 0. There we go. That's our differential equation. It's a major, somewhat complicated, differential equation. But it's a famous one, because it comes from [? Laplatians. ?] You know, people had to study this equation to do anything with Laplatians, and so many problems. So everything is known about this. And the first thing that is known is that theta really appears as cosine theta everywhere. And that makes sense. You see, theta and cosine theta is sort of the same thing, even though it doesn't look like it. You need angles that go from 0 to pi. And that's nice. But [? close ?] and theta, in that interval goes from 1 to minus 1. So it's a good parameter. People use 0 to 180 degrees of latitude. But you could use from 1 to minus 1, the cosine. That would be perfectly good. So theta or cosine theta is a different variable. And this equation is simpler for cosine theta as a variable. So let me write that, do that simplification. So I have it here. If x is closer in theta, d d x is minus 1 over sine theta, d d theta. Please check that. And you can also show that sine theta, d d theta is equal to minus 1 minus x squared d d x. The claim is that this differential equation just involves cosine theta. And this operator you see in the first term of the differential equation, sine theta, dd theta is this, where x is cosine theta. And then there is a sine squared theta, but sine squared theta is 1 minus cosine squared theta. So this differential equation becomes d d x-- well, should I write the whole thing? No. I'll write the simplified version. It's not-- it's only one slight-- m of the x plus l times l plus 1 minus m squared over 1 minus x squared p l m of x equals 0. The only thing that you may wonder is what happened to the 1 minus x squared that arises from this first term. Well, there's a 1 minus x squared here. And we divided by all of it. So it disappeared from the first term, disappeared from here. But the m squared ended up divided by 1 minus x squared. So this is our equation. And so far, our solutions are psi l m's. Are going to be some coefficients, m l m's, into the i m phi p l m of cosine theta. Now I want to do a little more before finishing today's lecture. So this equation is somewhat complicated. So the way physicists analyze it is by considering first the case when m is equal to 0. And when m is equal to 0, the differential equation-- m equals 0 first. The differential equation becomes d d x 1 minus x squared d p l 0. But p l 0, people write as p l. The x plus l times l plus 1, p l equals 0. So this we solve by a serious solution. So we write p l of x equals some sort of a k-- sum over k, a k, x k. And we substitute in there. Now if you substituted it and pick the coefficient of x to the k, you get a recursion relation, like we did for the case of the harmonic oscillator. And this is a simple recursion relation. It reads k plus 1-- this is a two-line exercise-- k plus 2, a k plus 2, plus l times l plus 1, minus k times k plus 1, a k. So actually, this recursive relation can be put as a [? ratio ?] form. The [? ratio ?] form we're accustomed, in which we divide a k plus 2 by a k. And that gives you a k plus 2 over a k. I'm sorry, all this coefficient must be equal to 0. And a k plus 2 over a k, therefore is minus l times l plus 1 minus k times k plus 1 over k plus 1 times k plus 2. OK, good. We're almost done. So what has happened? We had a general equation for phi. The first equation, one, we solved. The second became an [? integrated ?] differential equation. We still don't know how to solve it. M must be an integer so far. L we have no idea. Nevertheless we now solve this for the case m equal to 0, and find this recursive relation. And this same story that happened for the harmonic oscillator happens here. If this recursion doesn't terminate, you get singular functions that diverge at x equals 1 or minus 1. And therefore this must terminate. Must terminate. And if it terminates, the only way to achieve termination on this series is if l is an integer equal to k. So you can choose some case-- you choose l equals to k. And then you get that p l of x is of the form of an x to the l coefficient. Because l is equal to k, and a k is the last one that exists. And now a l plus 2, k plus 2 would be equal to 0. So you match this, the last efficient is the value of l. And the polynomial is an elf polynomial, up to some number at the end. and you got a quantization. L now can be any positive integer or 0. So l can be 0, 1, 2, 3, 4. And it's the quantization of the magnitude of the angular momentum. This is a little surprising. L squared is an operator that reflects the magnitude of the angular momentum. And suddenly, it is quantized. The eigenvalues of that operator, where l times l plus 1, that I had in some blackboard must be quantized. So what you get here are the Legendre polynomials. The p l's of x that satisfy this differential equation are legendre polynomials. And next time, when we return to this equation, we'll find that m cannot exceed l. Otherwise you can't solve this equation. So we'll find the complete set of constraints on the eigenvalues of the operator.

MIT_804_Quantum_Physics_I_Spring_2016

Compton_Scattering.txt

PROFESSOR: So we're building this story. We had the photoelectric effect. But at this moment, Einstein, in the same year that he was talking about general relativity, he came back to the photon. And there there's actually a quote of Einstein's saying, his greatest discoveries, for sure, were special relativity and general relativity. The photo-- he got the Nobel Prize for the photoelectric effect, and he certainly helped invent the quantum theory and many important things in this subject, but in retrospect, his greatest successes were that. But he may have not quite seen it exactly that way. He wrote, that some stage of his whole life had been a difficult struggle against the quantum, pulsed by some small happy interludes of some other discoveries. But the quantum theory certainly made him very-- well, he was very suspicious about the truth, the deep truth of the quantum theory. So 1916, he is busy with general relativity, but then he's more ready to admit that the photon is a particle, because he adds that the photon now has momentum as well. So it's a-- these photons that were not called photons yet are quanta for energy. But now he adds it's also for momentum. So this already characterizes particles. You see, there is the relativistic relation, well known by then, that e squared minus p squared c squared is equal to m squared c to the fourth. You might say, well, this is a little surprising. And If you don't remember too much special relativity, this may not quite be your favorite formula. Your favorite formulas might be that the energy is mc squared divided by 1 minus v squared over c squared. And that the momentum is the ordinary momentum again multiplied by this denominator like this. But these two equations, with a little bit of algebra, yield this equation, which summarizes something about a particle, that basically if you know the energy and the momentum of a particle, you know its mass. And it comes out from this. This is the relativistic version of similar equations in which you have energy one half mv squared, momentum, mv, and then energy equal p squared over 2m, a very important relation that you can check. For out of these two comes this one. And this is nonrelativistic. So for photons, we will have particles of zero mass. Photons have zero mass, m of the photon equals zero, and therefore e is equal to pc for a photon. So we can look at what the photon momentum is, for example photon momentum. We can treat it as some particle and the photon momentum would be e of the photon divided by c, or h Nu of the photon divided by c. And it's h over lambda of the photon of gamma. So this is a very interesting relation between the momentum of the photon and the wavelength of photons. So the idea that the photon is really a particle is starting to gather evidence, but people were not convinced about it until Compton did his work. So the same Compton that we used this length over there, he works on this problem and does the following. So is Compton scattering, the name of the work. Compton Scattering. So what is Compton scattering? It is x-rays shining on atoms again, but this time, these are very energetic photons, energetic x-rays. X-rays can have anything from a hundred EV to 100kEV, 100,000 electron volts. And what are the energies, of binding energies of electrons in atoms? 10 EV, 13 EV for hydrogen. So you're talking about 100,000 EV coming in, so it's easily going to shake electrons and release them very easily. So you're going to have almost-- even though you're shining on electrons that are bound to atoms, it's almost like shining light on free electrons if it's x-rays. So a few things happening. So this is photons scattering on electrons. Scattering on electrons that are virtually free. And the first thing that happens is that there is a violation of what was called the classical Thomson scattering, that you may have started in 802. So the reason Compton scattering did the job and physicists finally admitted the photon was a particle is that it made it look like particle collision of a photon with an electron, it could calculate and measure and treat the photon as a particle, just like another particle like the electron, and out came the right result. So the classical Thomson scattering was a photon as a wave. And what does that do? Well, you have a free electron, and here comes an electromagnetic wave, E and B. And if it's low frequency wave, low energy electron, this electric-- the magnetic field does very little, because this election doesn't move too fast and the velocity is being small, the Lorentz force is very small. But the electric field shakes the electron. And as the electron is being shaken, it's accelerating, and therefore it radiates itself. And it radiates in a pattern, so you get photons out. And the pattern is the following. I'll write the formula with this cross-section. We'll maybe not explain too much about what this cross-section means, it could be a nice thing for recitation. This is the formula for the Thomson cross-section as a function of the angle between the incident direction of the photon and the photon that emerges. So you detect photons out and this is the cross-section. What does it mean, cross-section? Well this has units. I will say, very briefly, units of area. Area per solid angle, but solid angle has no units. So if you imagine a little solid angle here and you multiply by this cross-section, it gives you some area that represents the solid angle you're looking at to see how many photons you get. And the solid angle that you have multiplied here to give you an area, the area should be thought of as the area that captures from the incoming beam the energy that is being sent into this solid angle. So it represents an area, and an area represents an energy, because if you have a beam coming in from a magnetic wave through a little area, some energy goes in. So that area that you get is that area that extracts from the incident beam the energy that you need to go in this solid angle direction. So basically, this is a plot of intensity of the radiation as a function of angle. But the most important thing, not only-- this is not quite accurate when the photon is of high energy. The thing that is pretty wrong about this is that the outgoing photon or wave has the same frequency as the original wave. So that's a property of this scattering. The electron is being moved at the frequency of the electric field, and therefore the frequency of the radiation is the same. And this is all classical. But out comes, when you have a high energy, this thing is not accurate, and you have a different result. So what did Compton find? Well, the first thing is a couple of observations. . Treat the photon as a particle. OK, so it has some energy and some momentum. The electron has some energy and momentum. You should analyze the collision using energy and momentum conservation. So before the collision, you have an incoming photon that has some energy and some momentum, and you have an electron, maybe here. And then after a while, the electron flies away in some direction, E minus. And the photon also, a photon prime of different frequency flies away. It's like a collision. You can do this calculation and maybe it could even be done in recitation. It's a relativistic calculation. You were asked in first homework to show that the photon can not be absorbed by the electron, and that uses the relativistic relations if you want to show you just can't absorb it. It's not consistent with energy and momentum conservation. So it's something you can try to figure out. The other thing that should become obvious is that the photon is going to lose some energy, because as it hits the electron, it gives the electron a kick. The electron now has kinetic energy. Think of this in the lab. The electron was static, the photon was coming. After a while, the electron has moved, it's moving now with some velocity. The photon must have lost some energy. So photon loses energy. And therefore, the final lambda must be bigger than the initial lambda. Remember the shorter the wavelength, the more energetic the photon is. So what is the difference? That's the result of a calculation. It's a nice calculation, all of you should do it. It's probably in some book, in many books. And it's a nice exercise also for recitation. Lambda final minus lambda initial. Or I'll write it differently. Lambda final is equal to lambda initial plus something that depends on the angle theta, in fact, has a one minus cosine theta dependence. But here has to be something with units of length And the only party you have here is the electron. And this electron has some length, which is the quantum wavelength, which is very natural for a Compton scattering problem, of course. And it's here, h over mec. So the l Compton of the electron. And that's the correct formula for the loss of energy, or change in frequency. So the most you can get is if you don't interact when theta is equal to zero, the photon keeps going, doesn't even kick the electron. And then you get zero, the initial lambda is equal to the final lambda. But this can be as large as two, for totally backwords photon emitted. So theta equals pi, cosine pi is minus 1, you get 2. At 90 degrees, you get the Compton shift. So it's a very nice thing, you even know already what's happening here. So let's describe the experiment itself, of how it was done. So he used, Compton, the experiment have a source of molybdenum x-rays that have lambda equals 0.0709 nanometers. So smaller than nanometers, it's 70 picometers. And that corresponds to e photon-- that's pretty small, so it must be high energy-- and it's 17.49 kEV. That's very big energy. And there was a carbon foil here. And you send the photons in this direction. And they were observed at several degrees, but in particular, I'll show you a plot of how it looked for theta equal 90 degrees, so detector. So source comes in, carbon is there, and what do you get? You get the following plot. Intensity-- so you plot the intensity of the photons that you detect, as a function of the wavelength of the photons that you get, because there's supposed to be a wavelength, a shift of wavelength. So it's actually quite revealing, because you get something like this. A bump and a bigger bump here. Something like that. Pretty surprising, I think, to first approximation. And here is about-- the first bump happens to have about the same wavelength as the incoming radiation, 0.0709. I should write it a little more to the left. 0.0709 nanometers over here. And then there is another peak at lambda f, about 0.0731 nanometers. And the question is, what is the interpretation? Why are there two peaks and what's going on? Anybody has any idea? Let me ask a simpler question. Which is the lambda that corresponds to the prediction of the fact that the wavelength must change, the smaller one or the bigger one? The bigger one. You certainly should loose energy, so the lambda f, the thing we were expecting to see, presumably that thing, because we were expecting to see that at 90 degrees, the photons have this thing. So we seem to observe this one. And let's look at it in a little more detail. You have this 0.0731 nanometers and you have the original light was at 0.0709 nanometers. So the difference is 0.0022 nanometers, which is 10 to the minus 9, but it's exactly, or pretty close, to this thing, because a picometer is 10 to the minus 3 nanometers. So this is 0.0024 nanometers. So this is pretty nice. Look, at 90 degrees, cosine theta is zero. So the difference between initial and final wavelengths should be equal to the Compton wavelength with about 0.0024 nanometers. And that's about it pretty close. So this peak is all right. Should've been there. The other peak, why is it there?