PHYS 201: Fundamentals of Physics II

Lecture 20

 - Quantum Mechanics II


Lecture begins with a detailed review of the double slit experiment with electrons. The fate of an electron traversing the double slit is determined by a wave putting an end to Newtonian mechanics. The momentum and position of an electron cannot both be totally known simultaneously. The wave function is used to describe a probability density function for an electron. Heuristic arguments are given for the wave function describing a particle of definite momentum.

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Fundamentals of Physics II

PHYS 201 - Lecture 20 - Quantum Mechanics II

Chapter 1: Review of Double Slit Experiment Using Electrons [00:00:00]

Professor Ramamurti Shankar: Well, this is just informal discussion till everybody’s in here. So any questions on the subject? What?

Student: Every question.

Professor Ramamurti Shankar: Everything. Okay, well, you know what, you guys should stop and ask more things as you go along, because there is just no way you could get all of this. And it’s a little strange and only by talking about it, you will at least know what’s going on. There’s no way to make it reasonable. It’s not a reasonable world out there. I can only tell you what it is. I take that view; when I teach quantum mechanics, just tell the rules and say, “This is what happens. This is how we calculate things.” And whether you like the formulas or not, it’s not my concern. And the fact that it doesn’t look like daily life, also not my concern, because this is not daily life. Strange things happen. But you have to keep me informed on how much you’re following and what you are understanding, at any stage. Don’t wait for this to end, because it’s not something where you can go on the last day and figure everything out. And I will try to repeat at every stage what has gone up to that point, because the whole thing is only a few lectures, maybe six or so. I can afford to go back every time to the beginning.

But I know that it makes sense to me, because I’ve seen it, and I don’t know how it sounds to you. I have no clue. You know that and so you have to speak up. You can ask any question you want, and I will try to answer you, if it’s within the realm of possibility. Okay, so what have I said so far? So let me summarize. Even if you never came to last lecture, here is what you should know about the last lecture, okay? Here’s what I said. First thing I said is, everything is really particles, all things, electrons, photons, protons, neutrons. They are all particles, so let there be no doubt about that. By that, I mean if one of them hits your face, like an electron, you will feel it in only one tiny region, one spot. Electron dumps all its charge, all its momentum, all its energy to one little part of your face. So there’s nothing wavelike about that. It’s not like getting hit by a boxing glove, which can hit your whole face. An electron hits one dot, or if it’s an electron-detecting screen, only 1 pixel is hit by the electron. And into that pixel is given all the charge, all the momentum, all the energy of that electron.

That’s exactly what particles do. So when you encounter an electron, it is simply a particle. So where does the problem come in? Where does the quantum mechanics come in? It comes in when you do the famous double slit experiment. That’s the key. The entire quantum mystery is in the double slit. Part of the resolution is in the double slit, but the rest are a little more difficult, and I’ll try to tell you. First I want to tell you what goes wrong with Newtonian mechanics. After all, if everything is a particle, what’s the big deal, what’s the problem? The double slit experiment is a problem. That’s what puts the nail on the coffin for Newtonian physics, and here it is in the basic version. You’ve got two slits.

By the way, I’m going to call the particle the electron. They’re all doing the same thing, so what applies to one, applies to all of them. There is a source, like an electron gun, that emits electrons. In the old days, televisions had the electron gun. And the gun emits the electrons, they go and hit the screen, they make a little dot, and then the dot moves around, and you see your favorite show. Okay, this is the electron gun, and the electron gun has been engineered to send electrons off a definite momentum. That you can get by accelerating the electrons over a definite potential, and the gain of so many electron volts will turn into kinetic energy. As for direction, if this gun is really far away to the left, in principle 1 mile, then the only way electrons are going to go 1 mile and hit the screen is they’re all basically moving in the horizontal direction. Then you put a row of detectors in the back, which will detect electrons. Then this is slit 1 and this is slit 2. You block slit 1.

In fact, let me say the following thing: what do we really know when we do the experiment? Once in a while this gun will emit an electron, and we know it’s emitted the electron, because it will recoil one way, just like a gun, rifle. It will recoil. That’s when we know the electron left. Then we don’t know anything, and suddenly, one of these guys says click. That means electron’s arrived here. This is what we really know. Everything else you say about the electron is conjecture at this point. You know it was here, you know it was there. The question is, what was it doing in between? Now if you say, “Look, things cannot go from here to there, except by following some path, I don’t know what path it is.” Maybe if it’s an ordinary particle, like a Newtonian particle, it will take some straight line, hit that slit, or go through that slit and arrive here. So you might say, “I don’t know the trajectory, but it’s got to be some trajectory, maybe like that, or maybe like that.” So the electron takes some path and you can label the path as either through slit 1 or through slit 2.

Okay, now here is the problem. Suppose I do the experiment with slit 2 blocked, so you cannot even get through this one, and I sit at a certain location for a certain amount of time, maybe 1 hour and I see how many electrons come, and I get 5 electrons, with only 1 slit open. And if I move that observation point, I get some pattern, pretty dull, looking like that, and I’m going to call it I1. That’s the count, as a function of position up and down that wall of detectors. Then I repeat the experiment with this guy closed and that guy open, and I get another count, looks like that. Now I’m going to pick a location. These are not drawn to scale or anything, so I’m not responsible for any of that. Maybe I’ll at least show you one thing, which is pretty important. This graph will be big in front of the second slit, which is somewhere here. It will look like that. So I get I1 when 1 is open and I get I2 when 2 is open. This is I1, this is I2. Now I’m going to open the two slits and I’m going to pick a particular location. It doesn’t happen everywhere. I’m going to pick one location called x. Where I used to get 5 electrons per hour with one thing open, and 5 electrons per hour with the second thing open.

Now I want to open both and ask, what will I get? In Newtonian mechanics, there’s only one possible answer to that question, and that is 10, because we’ve got 5 this way and you’ve got 5 that way. And you open both, whoever is going this way will keep going that way; whoever is going this way will keep going this way. They will add up to give you 10. Now I told you, some people may say, “Well, maybe it’s not 10, because with both slits open, maybe someone from here can collide with someone from there. How do you know that will not happen?” So I’m saying, do the experiment with such a feeble beam of electrons, there’s only one electron at a time in the whole lab. It’s not going to collide with itself. Then you wait long enough, and you have to get 5 + 5 = 10. And what I’m telling you is that if you go to the location marked x, where you’ve got 5 with each one open, when you open both, you will get 0. You don’t get anything. That is a great mystery.

That is the end of Newtonian physics. And I told you that something like that never happens in your daily life. I gave an example with machine guns. This is a machine gun. This is a concrete wall with 2 holes in it, and there’s some target here, you. And then you see how many are coming through this and how many are coming through that. Then you go there and you wait. And both are open. Somehow, nothing comes. With the second hole in the wall, you are safe. With one hole in the wall, you’re not safe. That can never happen with bullets. So these electrons are not following any path, because the minute you commit yourself to saying it follows one path, either through this one or that, you cannot avoid the fact that with both of them open, the intensity with 1 + 2 has to = I1 + I2. That’s the Newtonian prediction. And I1 + I2 gives you 5 + 5 = 10 here and you get 0. In other places, instead of 10, you will get 20. Some places you get more, but more dramatic thing is where you get less, where you get nothing.

Therefore you abandon the notion that electrons have any trajectories. You don’t want to abandon it, but you have to, because that assumption, which is very reasonable, just doesn’t agree with experiment. Then you say, “Okay, what should I do? Newtonian mechanics is wrong. What’s going to take its place?” To find that, you have to move away from this x and move up and down this row here and see what you get, and I think I told you what you get. You get a pattern that looks like this. So the real I1 + 2 looks like this: it’s got ups and downs and ups and downs. And let’s say these downs really correspond to 0. That means nobody comes here, a lot of them come here, no one comes here, a few come here and so on. That’s what you find. I’m just telling you what happens when you do the experiment. So you put yourself in the place of a person who did the experiment. You thought of moving away from that point x and you plot it, it makes no sense in the language of particles.

But this is such a familiar pattern. If you’re a trained physicist, which you guys are, you will say, “Hey, this reminds me of this wave interference, with water waves or sound waves or any waves. ” Obviously there’s some wavelength. The minute you give me wavelength and a slit separation, I can calculate this pattern. dsinθ = l and whatnot, and from that sinθ, where you get a minimum or a maximum, and if there’s a certain separation to the screen, you can find the precise location of these maxima or minima. Or given the maxima and minima, you can work back and find the wavelength. And the wavelength happens to be some number called ℏ, which is 10-34 joule seconds, divided by the momentum. p is the momentum. In other words, you find that if you send more energetic electrons, accelerate them through bigger voltage, increase the p, l goes down, the pattern gets squeezed. You slow down the electrons, p reduces, l increases, the pattern spreads out, and the dependence on momentum is inversely proportional to momentum. And you fool around and find out the coefficient of proportionality, which people used to call h in the old days is now written as 2pxℏ, but it doesn’t matter, it’s some constant. And the number is 10-34.

So you can successfully reproduce this pattern, but what does it tell you about what’s going on? What good is that pattern? The pattern tells you that if you repeat the experiment with this electron gun a million times or a billion times and you plotted the histogram patiently, the histogram will eventually fill out and take this shape. So this wave is not the wave associated with a huge stream of electrons. A single electron in the lab is controlled by this wave. You need a whole wave for 1 electron, so it’s obviously not a wave of electrons. It’s not a wave of charge, like the wave of water. It’s a mathematical function and you are drawn to it, because the only way you know how to get this wiggly graph is to take something with definite wavelength and let it interfere. So you’re forced to think about this wave. And the intensity of the wave, the brightness if you like, the square of its height, gives you what? Gives you the graph you will get if you repeat the experiment many times. And what does it mean for the individual trial? What does it mean for the millionth + one electron? For the millionth + one electron, it gives you the odds of where it will land on that screen, okay?

You can never tell exactly where it will be. You can tell what the odds are, and the only way to test the statistical theory is to do the experiment many times. And if you do it, it works, and it seems to work for everything, for electrons, for protons, for photons, whatever it is, the wavelength and momentum are connected in this fashion. So this wave is forced upon us, and it gives you the odds of finding the electron somewhere. And we say that the probability — I’ll be a little more precise in a minute on what I mean by the probability to find it at a location x, but let’s just say, if you draw the graph of Y2, wherever it is big, the probability is larger; wherever it is small, probability is small; wherever it is 0, probability is 0. So there seems to be a function whose amplitude or whose square gives you the probability. That function is called the wave function, and we know it exists, because it’s the only way to calculate the result of this experiment.

Once you tell me that the fate of a particle is controlled by a wave, you’re immediately led to some other conclusions, so I’m going to tell you what they are.

First conclusion is this: if I make a single slit, let’s call this the x direction, let’s call this the y direction, and I’m sending a bunch of particles in the x direction with some momentum p0. In Newtonian mechanics, I can manufacture for you an electron of known momentum and known position, or known to arbitrary accuracy. If Dx is the uncertainty in my position, and Dp is the uncertainty in momentum, I can make each of them as small as I like. So here’s an actual practical way to prepare such a state. If I say, “Give me an electron of known position, known momentum,” here’s what I will do. I will take a slit with a very tiny hole in it. The width of that hole is d, and whoever comes out on the other side, what can I say about that particle? Its position has an uncertainty of order d, because if it was not — I’m sorry, Dy now, because this is the y direction. Right? Anything who came out of hole right after it had to have a y position, whose known to within d. It’s got to come anywhere within the slit, but that’s it. So that is how I have prepared for you, that’s how we filter electrons of definite position. And you can make Dy as small as you like by making the slit as thin as you like. What about its momentum? If you had a momentum p0 in the x direction, no momentum in the y direction, therefore py was strictly 0, no uncertainty. Dpy is 0 and the fellow I catch here is moving horizontally with that momentum p0 whose y position is within d, and I can make the d as small as I like. This is Newtonian physics. But we have now learned that the fate of the particle is not in its own hands. It’s contained in this wave. So what should I do in this context to find out what it will do? Any idea what I should do to find out what will happen in this experiment, given what we learned? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Right, but how will I calculate what will happen in this experiment? What will decide? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Yes, this is one hole, and a light beam is coming from the left — I mean, a wave is coming from the left, and if the particle has momentum p0, it’s got some wavelength, which is 2pℏ/p0. But if you want to know what will happen on the other side of this slit, I have to find the fate of that wave. Yes, you can put a screen, but what will I see on a screen? Will the wave just hit this region? You know it will spread out from diffraction. I’ve told you, the light will spread out. There are tiny wiggles we ignore, and this point, where you get most of the action, that angle, θ, satisfies dsinθ= l. This is just wave theory. That’s when you can pair up the points in the slit, in the hole, to things shifting, differing by half a wavelength, so for every one I can find a partner that cancels it, you will get 0 here. Beyond that, you may get a few more rises, but it’s pretty much dead outside this cone. That means you will observe this particle anywhere in this angular width.

Now a particle cannot go from this slit to there unless it had a momentum, which had a component in the y direction. You cannot get there from here, unless you have y momentum. It’s the y momentum is uncertain to within that cone. So what’s the uncertainty in the y momentum? For a vector of length p0 when it gets shifted by an angle Dθ or by an angle θ, it is just p0. Or, if you like, precisely, p0 times sinθ. But p0sinθ, sinθ is controlled by dsinθ = l, so this is l/d. But l is 2pℏ/p0 x d. You cancel that, you find D py X d = 2p. That means D py D y is roughly — forget the 2p’s — of our . So you should understand this much completely without any doubt: if the future of the particle, the fate of the particle, is controlled by the wave, you try to narrow the location of the particle by making the hole smaller and smaller, the wave fans out more and more.

Chapter 2: Heisenberg’s Uncertainty Principle [00:20:28]

That’s just wave theory. People knew this about waves hundreds of years back. What is novel is that this wave is going to tell you where the particle will end up. This wave is going to control the odds of where the particle will end up, and the odds are pretty much concentrated in this cone, not of 0 opening angle, but an angle θ, so that dsinθ = l. l in turn is connected to the momentum of the particle. This is where the uncertainty principle comes in. What Heisenberg said is, “You had in your mind the classical notion that you can have a particle of known position and known momentum. Let me see you produce that particle. So you try to do it by putting a slit and catching guys with a very narrow range in y, but now you find out that the momentum gets broader and broader, and that’s the result of the wave associated with the project. ” It was not in the Newtonian picture. Yes?

Student: Why with the single slit, like you’ve drawn there, do you get the little wiggle at the outside — ?

Professor Ramamurti Shankar: You mean, why would it have those wiggles? Okay, so if I look at the single slit, I could think of many little point sources. In the forward direction, if you go very far, so you treat them as roughly parallel, they are all in step. You’ve got a big maximum. In another direction where this difference is l, dsinθ = l, this guy and this guy are differing by l/2.

Student: So you’re assuming the slit is large enough for them to do that ________.

Professor Ramamurti Shankar: These are little mathematical dots inside the slit. You know, when you make a slit, every point on the slit looks like a source of light, a point source of light. They are not really light bulbs, but if you make any hole in the wall, a light comes through that hole, looks like it’s a source of light. I’m taking every point on the hole to be a source of light. And what I’m saying is, there is a direction in which it will cancel, but if you go a little further out, it won’t cancel completely. This is the direction for perfect cancelation, where I can pair them, this one with this one, that one to that one, and so on. They pair to give 0. But if you move further up, you no longer cancel completely, but you don’t add perfectly either, so things will get better, then they’ll get worse again, and better and worse and so on. So that’s the origin of that pattern. Yes?

Student: So you’re saying this is the first minimum. After that, it would increase _______.

Professor Ramamurti Shankar: Yes. For a single slit diffraction, the big thing in the middle is pretty much all you have. It’s not like double slit experiment with two holes, where you get many times the pattern. That’s because — let’s understand why that is true. Here, if these two differ by 1 wavelength, they don’t differ at all. I can find another angle where they differ by 2 wavelengths. They don’t differ at all. There’s only 2 sources, so you can engineer them to differ by either 1 wavelength or 2 wavelength or 3 wavelengths. Here in a single slit, each point is like a source of light. You got them all to agree. You can get them to agree only in the forward direction. In any other direction, you can get them to neutralize each other, but never for perfect reinforcement. So you cannot get it more than — this is the only real maximum here. Everything else is tiny. Yes?

Student: Why can’t you do the same trick with the double slit, where each of the one slits has its _______?

Professor Ramamurti Shankar: No, in double slit, what’s happening is, we take — in double slit, the number d I used in double slit was not the size of a slit, but the space in between the slit, okay? So there I took the size of every slit to be vanishingly small. That means a light coming out of this slit spreads out completely in all directions, okay? It’s as if this packet became that broad. Likewise a light from this one, we are still in the first maximum of that slit. So you’ve got to understand, in a single slit experiment of diffraction, the slit size is what you are varying. In a double slit experiment, the slit is taken to be mathematically point like, so it fans out completely. It’s the interference between those two point sources that you’re adding. They can add and cancel, add and cancel, many, many times as you move along this line.

Okay, so what we learn is it’s when you combine waves and particles and go back and forth that you run into the situation. So you cannot make a state of perfect momentum. By the way, I said one thing, I thought about it, which is incorrect, which is, in the microscope, I said if you want to locate the position of an electron in a microscope, take a microscope with an opening, and electron is somewhere on this line. I said you’re shining light down here. It hits the electron, but it goes in through the slit by spreading out. So the photon that came in goes into the eyepiece with a certain uncertainty in its final angle. That means we know the incoming momentum, but we don’t know the outgoing momentum of the photon. The lens picks up everything inside that cone. That means we don’t know how much momentum it gave to the electron. It gives an indefinite amount of momentum to the electron. Therefore the x momentum of the electron is uncertain by that little shape, the conical shape of the momentum. And if you do the uncertainty principle argument, you’ll again find DxDp is h, h.

Now what I don’t like about my experimental setup is that I had the incoming light also coming from inside the microscope, but that means incoming light, when it comes through this hole here, will itself spread. Then it will hit this guy, and that will go to the aperture. That will spread some more. This uncertainty in incoming momentum is unnecessary. We can do better than that, because in other words, when the light is picked up, it is picked up by this tiny hole. There’s no reason it should also come from the tiny hole. It can come from a source far away, say on the other side, so that it is a well defined direction, it’s not diffracting at all. So I want it to come in through a very broad hole, so it’s got well defined direction, so the light here has known momentum. It hits the electron and goes into the microscope. It is the final momentum of the photon I don’t know. And I cannot make it better. If I make it better, I’ve got to open this eyepiece a lot. If I open the eyepiece a lot, I don’t know where I caught this guy. So again the problem between taking a very tiny eyepiece, so that if I see a flicker, I know the electron was in front of it, but the light coming from the reflected electron fans out more and more.

Okay, so anyway, this is the uncertainty principle and the uncertainty principle told us something very interesting. I asked you, what can be the function here that produces this interference pattern in the double slit? We know the wavelength. Wavelength was 2p/p. And you know from basic physics that a function like cosine 2px/l has got wavelength l. So let’s put in the formula for l here. You get cosine 2px. l is 2p/p. Cancel the 2 p’s, you get A cosine px/ℏ. That function, when you throw it at a double slit, will form two little wavelets, and they will interfere, that produce an interference pattern of the type you want.

Do you understand that the experiment only showed you there’s a wavelength. It did not tell you what the actual function is? That’s very, very important. When Young did the experiment with the double slit, he found the oscillations and he could read off the wavelength. It’s just geometry. But he didn’t know what was oscillating. He didn’t know there’s an electric and magnetic field underneath all of that. But you can always read out the wavelength without knowing what’s going on. Likewise, we have the wavelength. We know it comes from a function with a well defined wavelength, so I make my first guess to be this function. But I told you what was wrong with this choice. You guys remember that? I said this function violates the uncertainty principle. The uncertainty principle says if you know the position to an accuracy Dx, and if you know the momentum to accuracy Dp, the product must be at least as big as hx some number of order 1. We have taken the particles to have well defined momentum. If they have well defined momentum, Dp is 0. Dp is 0, Dx is infinity. Now I told the square of the wave function is the probability to find it somewhere, and you have no idea where it is. In other words, a particle of perfectly known momentum has totally unknown position. So the probability should look flat, Y2 should look flat. But the Y2, due to cosine, of course does this. It prefers some locations to another, but you’re not supposed to have any preference for any x, so we have a problem. How do I put in a wavelength into a function whose square is flat? That’s the problem we have. When you think about it, you realize your trigonometric functions will not do the trick. If they have a wavelength, their square is not flat. The square is also oscillating.

But then what came to the rescue is the following function, not a cosine but Aeipx/, rather than cosine xp/ℏ. Look at this function. This function, I’ve told you many, many times, if you don’t know your complex numbers, you’re definitely going to have trouble. It looks like a vector of length A and angle θ, which is px/ℏ. As you vary x, this x changes and this will rotate round, but as it rotates the amplitude of this complex number, absolute value of Y2, which is Y times Y*, which is A — I’m taking A to be real here, so A* is this, times eipx/ℏ, times e −ipx/ℏ.

That cancels out, you just get A2. In other words, the complex number describing the wave function changes in phase but not amplitude. It’s the amplitude that gives the probability. Now there is no problem with this guy having a wavelength, because this oscillates in x. Its real part and imaginary part both oscillate, but the square of the real + the imaginary square is 1 [A^2]. That’s why the amplitude doesn’t change.

So we are driven now to the very interesting result that the wave function for a particle of definite momentum p is this. So this is a very important lesson. Let me label this function by label p to tell you, “Hey, I’m not talking about any old wave function.” This guy has a definite momentum p. Its wave function looks like e ipx/ times any number A you want in front of it. That’s a very important thing to know. This is called a plane wave, and a plane wave with a p right where it is describes a particle of momentum p in particle mechanics. And I told you particles of momentum p are everywhere. Every machine produces them, every accelerator produces them, and if you want to describe them in quantum mechanics, you have to know complex numbers. There’s just no way you can get a real answer to our predicament. It’s complex.

So that’s roughly where I left you, and I want to remind you of a few other things, this further discussion of the result we have, okay? The discussion is, if the world is really this messed up at the microscopic level, why do I think it’s the world I see in the macroscopic level? Where are all these oscillations? Why is it that when there’s a concrete wall, making another hole is bad for me and not good? Why do all these things happen? Why do I think particles have definite momentum and position? Why do I think that if I make a hole in the wall and I send a beam, the beam will go on the other side of the wall with a shape precisely like the shape of the hole, no spreading out? It all has to do with the size of the object. The laws of physics are always quantum mechanical laws, but when you apply it to an elephant, you get one kind of answer; when you apply it to an electron, you get another kind of answer. You don’t have separate laws for big and small things. The real question is, how do these very same laws, when applied to big things — by big things, I mean things you see in daily life — give the impression that the world is Newtonian? So let’s look at the double slit experiment.

Here’s a double slit and we are told, “Send something. See what happens on the other side.” And the prediction is that you get these oscillations, with the peculiar property that with two holes open, you don’t get anything somewhere. We don’t seem to see that in daily life, and you can ask, “Why is that so?” Well, you remember that the condition for the next minimum is like dsinθ, is l/2. So if θ is very small, it’s like θ xd is l/2, or θ = l/d. That’s the angle you’ve got to go through from the central maximum. That’s the central maximum to the first minimum here. That tiny angle is given by l/d. The reason you don’t see the oscillations is when you put in the values for l and d. Let’s pick a reasonable value for this angle θ. Do you understand what θ is? In that maximum, there are some oscillations. I want to go to the first minimum near that. The distance between these two is roughly the spacing between maxima and minima, maxima and minima. That angle is l/2. l is 2p over the momentum and that is d. In a microscopic world, p is mv, and let’s take an object of mass 1 kilogram moving at 1 meter per second and a slit hole is 1 meter. The size of the slit is 1 meter. Put everything equal to 1 for a typical estimate.

You find this number is 10-34 radians. That means that the angular difference between the maximum and the minimum and the maximum and the minimum is 10-34 radians. What that means is, if you put a screen 1 meter away, the distance between one maximum and the next maximum or one minimum to the next minimum and so on, that spacing will be 10-34 meters, because that’s how a radian is defined. If that angle is θ, that distance is just the distance to the screen times θ. That will be 10-34 meters. That means the wavelength of the oscillation on your screen is 10-34 meters. Can you see it? Well, make the world’s smallest detector. It’s as big as 1 proton, okay. Nothing can be smaller. That’s your whole detector, all the parts, everything, 1 proton. Size of a proton is 10-15 meters. That means you will have 1019 oscillations inside your tiny detector. So don’t be fooled by the 19. Let’s take a minute to savor this.

That’s how many oscillations you have, okay? You’ve got enough now? 3, 6, 9, 12, well, I don’t have enough time. That’s a lot of oscillations. You should check the numbers though, okay? I’m saying they typical angle will be 10-34 radians and if you put the screen 1 meter away, the spacing will be 10-34meters, and you look at it with an object, a detector, whose size is 10-15 meters, which looks very small, size of a proton. But look, 1019 fit into that length, so your proton detector looks huge. In fact, I cannot even show it here. So you don’t see the oscillations; you see the average only. If you see only the average, you can show that with 2 slits open, the sum is the sum of the two averages, so you don’t see the oscillation. That’s the first reason. Now you can say, “You took a kilogram. Let me take a gram.” I said, “Go ahead. Take a gram, take a milligram and take a slit which is not 1 meter wide but — 1 meter apart, but 1 millimeter apart.” It doesn’t matter. You’re playing around with factors like 10 and 100 and 1,000. I got 1019 here. So nothing you do will make any dent on that. So in the macroscopic world, you will not see this interference.

Another reason you won’t see it is that the particle should have a definite momentum. It’s got an indefinite momentum, it’s coming in with different momenta, then each will have its own interference pattern and they’ll get washed out. Finally, I told you, if you ever try to see which slit the particle took by putting a light beam here, the minute you catch the electron going through one slit or the other, this pattern is gone. It will do this “I’m not here and I’m not there” routine only if you never catch it being anywhere. That’s very interesting. The electron behaves like it does not go through any one particular slit, as long as you don’t catch it going through one slit. You put enough light to catch every electron, then you can add the numbers and you must get the sum of the two numbers. Now for the atomic world, it’s possible for the electron to go for a long time without encountering anything, and the interference effects come into play. In a macroscopic world, there is no way a macroscopic object can travel for any length of time without running into something. It will run into other air molecules. It will run into cosmic ray radiation. It can collide with black body radiation from the big bang, anything. The minute you have any contact with it, this funny thing will disappear. So that’s one reason you don’t see it.

Now we can go on and on and give other numbers. I’ve given examples in my notes, which I will post later on. One of them is the uncertainty principle. Why does it look like the uncertainty principle is not important? Take again, this is 10-34. Everything is in MKS units. So take an object of mass 1 kilogram whose location is known to the accuracy of 1 proton. Okay? So this number is 10-15 meters, do you understand? You take an object made of 1023 protons and you know its location to the width of 1 proton.

That’s all you don’t know about its location. That’s your Dx. What’s the Dp? Well, Dp is now 10 to the, what, 19? 10-19. Now Dp is m times Dv. That’s 10-19. If this is 1 kilogram, Dv is 10-19meters per second. You don’t know its velocity to 1 part in 1019. Now how bad is that? Well, suppose I start a particle off exactly known velocity, I know where it will be forever. But suppose I don’t know the velocity to this accuracy, and I let it run for 1 year. So I don’t know precisely where it is, but how bad is it? How badly do I not know? Well, 1 year is 107 seconds, so if it runs for 1 year, it will be unknown to 10-12 meters. 10-12meters is what, let’s see? It’s 1/100 of an atom size, 1/100 the size of an atom.

So you see, these uncertainties are not important in real life. So everything that you think has a definite position and momentum actually has a slight uncertainty, but the uncertainties don’t lead to any measurable consequences over any distances that you can actually have. So what I’m trying to tell you is, there are these waves. They do all kinds of things, they do interference and everything, but the condition for them is really the microscopic world. The minute the masses become comparable to gram or kilogram and distances and slits and so on, or like a meter or a centimeter, these effects get washed out. But in the atomic scale, they are seen.

Chapter 3: The Probability Density Function of an Electron [00:42:34]

Now the final thing I want to mention before moving on to a completely new topic is the role of probability in quantum mechanics. We have seen that quantum mechanics makes probabilistic predictions. It says if you do the double slit experiment, I don’t know where this guy will land, but I’ll give you the odds. Okay, now that looks like something we have done many times in classical mechanics. For example, if you have a coin and you throw the coin, you flip it and you say, “Which way will it land?” well, it’s a very difficult calculation to do, but it can be done in principle, because a coin, once released from your hand, can only land in one way. That’s the determinism of Newtonian mechanics. If you knew exactly how you released it with what angle or momentum, what’s the viscosity of air, whatever you want, if you give me all the numbers, I’ll tell you it’s head or tails. There’s no need to guess. In practice, no one can do the calculation. What you do in practice is, you throw the same coin 5,000 times and you find out the odds for head or tails and you say, “I predict that when you throw it next time, it will be .51 chance that it will be heads.” That’s how you give statistical predictions. Now you did not have to use statistics. You use it, because you cannot really in practice do the hard calculation. In principle, you can. Secondly, if you toss a coin and I hide it in my hand, I don’t show it to you, it’s either head or tails, and I say, “What do you think it is?” you’ll say, “It’s .1 chance that it’s heads.” And I look at it, it may be head or it may be tail. Suppose I got head. It means that it was head even before I opened my hand, right? The correct answer’s already inside my hand.

I just didn’t know it. I’m using odds, but when I look at it, I get an answer. That’s the answer it had even before I looked.

So I’ll give another analogy here. So this is a probability of locating me somewhere. This is my home town, Cheshire, this is Yale, and this is the infamous Route 10. So somebody has studied me for a long time and said, “If you look for this guy, here are the odds.” Either he’s working at home or he’s working at Yale, and sometimes he’s driving, okay? This is the probability. First thing to understand is the spread out probability does not mean I am myself spread out, okay? Unless I got into a terrible accident on Route 10, I’m in only one place. So probability’s being extended doesn’t mean the thing you’re looking at is extended. I am in some sense a particle which can be somewhere. These are the odds.

Well, suppose you catch me here on one of your many trials. If you catch me only once, you don’t know if the prediction’s even good, so you repeat it. You study me over many times and you agree the person had got the right picture, because after observing me many, many days you in fact get the histogram that looks like this. The important thing is, every time you catch me somewhere, I was already there; you just happened to catch me there. I had a definite location. It was not known to you, but I had it. I had a definite location because in the macroscopic world I’m moving in, my location is being constantly measured. You didn’t ask or you didn’t find out, but I’m running through air molecules. I’ve slammed into them. They know that. I ran over this ant. That was the last thing the ant knew, okay? So I’m leaving behind a trail of destruction and they all keep track of where I am. My location is well known. You just happened to find out.

But now let’s change this picture and say this is not me. This is an electron and it’s got two nuclei. This is nucleus 1 and this is nucleus 2. It can be either near this nucleus or that nucleus, and this is the Y2 for the electron. That means it’s the probability you’ll catch it here and you’ll catch it there. Once again, if you catch the electron, you will catch all of it in one place. It is wrong to think the electronic charge is somehow spread out around the atom. It’s not true. The charge is in one place. The odds are spread out. That looks just like my case. But the difference is, if you catch the electron — let’s in fact simplify life and say there are only 2 possibilities. Either it is near atom 1 — nucleus 1 or nucleus 2, only 2 discrete choices. If you catch it near 2, it is wrong to think that it was there before you got it. So where was it? It was not in any one place. It had no location till you found its location. That’s very strange.

We think of measurement as revealing a pre-existing property of the object. But in quantum theory, it’s not that you don’t know where the electron is. It does not know. It is not anywhere. It’s the act of measurement that confers a location or position on the electron. That state of being, where you can be either here or there, or simultaneously here or there, has no analog in the classical world. If anybody tries to give you an example, don’t believe it, because there are no examples in the macroscopic world that look like this. No analogies should satisfy you, because this has no analog in the real world, okay? So this is the interesting thing in quantum mechanics. If this is a possible wave function Y, electron near nucleus 1, and that’s a possible wave function Y, electron near nucleus 2, you can add the two functions. That’s another possible function. But what does that describe?

It describes an electron which upon measurement could be found here and could be found there. It’s not like finding me in Cheshire or finding me in New Haven, because in those cases, on a given day on a given measurement, you can only get one answer, depending on where I am. Right now if they look for me, they can only find me here. They cannot find me anywhere else. But in the case of the electron, the one and the same electron, on a given trial, at a given instant, is fully capable of being here or there. It’s like tossing a coin and it’s in my hand. We all know that when I reveal it to you, you can only get one answer, now that the toss has been done, it’s got one answer. If it’s a quantum mechanical coin, you don’t know, and it doesn’t have a value till you look. When you look, it has a definite value. Before you look, it doesn’t have a definite value. That’s exactly like saying, when you looked, it goes through a definite slit. When you don’t look, it’s wrong to assume it went through a definite slit. Yes?

Student: Say you had a double slit experiment but instead of having a screen that went all the way in both directions, you just sort of had _____ screen. So then you would only be looking at the final location of a _______ electron. The other half you would know. How would that work, because location for some of them has to be ____________.

Professor Ramamurti Shankar: The minute you find the location to the accuracy of knowing which slit it went through, you’ve got 2 slits or only 1 in the experiment?

Student: You have 2 slits but only a half screen, and nothing on the other one.

Professor Ramamurti Shankar: Oh, you’ve got a screen that only comes to here, you mean?

Student: Yes.

Professor Ramamurti Shankar: Yes, the real problem of location that I’m talking about here is not when it hits the screen, but here, when you try to see which hole it went through, by putting a light source here. I was referring to the fact that if you have the right kind of light to know which hole it went through, if you give it enough momentum to wash out the pattern. As far as the screen goes, once it comes out, it’s the sum of these 2 waves coming from the 2 holes, and it also doesn’t have a well defined position. The probability for finding it may look like this. In fact, the probability will look like this. Forget your screen. This is the probability. The minute you catch it, it is found there, that’s the location after that measurement. Prior to the measurement, it can really be anywhere where the function is not 0. There are many wave functions. There’s the 1 to the left of the slit and there is 1 — then it becomes 2 wave functions coming from the 2 slits. They form the interference pattern and that gives you the odds that if you looked for it, you will find it.

Now till you find it, it’s not anywhere. It can be anywhere on this line at that instant. It’s only the act of measurement, or hitting a detector that tells you that’s my location. So you will have to get used to that. You’ll have to get used to the fact that things don’t have position, momentum, angular momentum, energy or anything, until you measure it. Okay, so I’ve got to tell you a little more now about just position. So let’s take — by the way, any questions so far? Yes?

Student: Can you explain again how you can tell which hole the electron goes through with the light?

Professor Ramamurti Shankar: Well, you just see it. You see a flicker and if it’s near this hole, you know this guy went through that. And if it’s near that one, you know it went through that. But to have such good resolution, the wavelength should be much smaller than the space in between the slits. Otherwise you’ll get a big blur and you won’t know which one it went through. That’s a soft measurement that doesn’t do you any good. It won’t destroy the pattern. That’s because you don’t know which hole it went through. If you make it fine enough to know which hole it went through, you will disrupt the electron’s momentum enough to wipe out the pattern. Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Oh, here? You mean what happened to the 2? Oh yeah, forget the 2s. There are 2p’s I forgot, right? There’s a lot of 2p’s too you’ve got to put in. 10-34 is not exactly the answer, because I got 2p there. I’m just saying, look, if it’s 10-34, suppose you’re talking about how much money Bill Gates has. It’s 109. Now is it 2 x 109, 3 x 109? I don’t know, but I’m not worried about his financial wellbeing, because it’s up there in the 109s. So whenever I do these arguments, you should get used to this notion, it’s very common for physicist when they argue in quantum mechanics, will use the symbol that’s not quite an =. It looks like a wiggle and =. Basically it means, I’m not quite sure, but the number is in this ballpark. And if this ballpark is 1,000 miles from that ballpark, you just have to know it’s in the ballpark. You don’t have to know where it is. Some things are definitely macroscopic; some things are definitely microscopic. But something very interesting is happening at Yale right now. People are asking the following question: how small an object has to be before I can see its quantum mechanical fluctuations? We know that if it’s like an electron, it’s completely fluctuating. You don’t know anything. If it’s like a bowling ball, it seems to have a well defined position. Make the objects smaller and smaller and smaller. How small can it be before it’s first beginning to show quantum effects, like quantization of energy or quantization of momentum, depending on the problem, or quantum fluctuations in position? So those measurements are now being carried out at Yale. It’s a very exciting time and it’s so many years after the discovery of quantum mechanics. Because we knew the really big world and we knew the really small world, but now we’re trying to go, because of nanotechnology, continuously from big to that small, and how small is small?

That’s the question? Can a little macroscopic object simultaneously be in two places? Most of them seem to have a well defined location. Can you create a situation when it’s capable of being found here and found there? It’s very hard, because you have to isolate the particle from the outside world. That’s the first condition. That’s what ruins everything.

A quantum computer, you must know, has got these bits called qbits and unlike the bits in your laptop, which are either 0 or 1, a qbit can also give you 0 or 1, but it can also be in a state where it can give either 0 or 1 on a given trial. The bits in your computer, the particular bit right now is either a 0 or a 1. Maybe you don’t know it, but it can only give you that answer because that’s what it is. Because that bit is in contact with the world and the world is constantly measuring its value. A qbit is a quantum system which can do one of two things, but it’s isolated and it’s neither this nor that. It’s like the electron going through both slits. So a quantum bit can explore many possibilities. If you build a computer with 10 qbits it can be doing 210 things at the same time. And if it’s got a million bits, it’s 2million operations, things it can be exploring at the same time.

That’s why, as you know, one of the ways to securely send your credit card information is to use very large numbers, on the assumption that no one can factorize them. You can always multiply a 100 digit number by a 100 digit number on your computer instantaneously. But if I gave you the 200 digit number and told you to find the two factors you won’t find it. You won’t find it in the age of the universe. It’s amazing, but that simple problem of factorization cannot be done if the numbers are 100 digits long, and that’s the reason why people openly broadcast the product, they may broadcast one of the numbers, and only the other person knows the second number. Now there is something somebody called Shore, Peter Shore, showed that if you have a quantum computer, made up of these qbits, it can actually factor the number exponentially faster, namely, instead of taking 1010 seconds, it will take 10 seconds. So if you build a quantum computer, you have two options. Either you can become famous, or get tenure at Yale, maybe, or you can go on the biggest shopping spree of your life, because you can get anybody’s credit card number. So that’s the choice. When you come to that fork, you decide which way you want to go. Maybe you can go through both choices, I don’t know. That’s something in your future.

So why is it so hard to build a quantum computer? There are many, many quantum systems which can do one or two things, and can be the state, but they are both this and that. The problem is, they cannot be in contact with the outside world, because single contact with them is like a dream. Think about it, it’s gone. Same thing. Any measurement destroys it. Any unintended measurement also destroys it, so you’ve got to keep your system fully isolated. A system that is not talking to the outside world, unfortunately, is also not talking to you. So you cannot ask it any questions, and if it knows the answer, it cannot tell you. So sometimes you want it to talk. It’s like relationships. Sometimes you don’t want it to talk. So what do you do? You’ve got to build a system where sometimes, in a controlled way, you can make contact with your system, namely give it the problem. Then it does its quantum thing, then you’ve got to make a measurement to find out what the answer is. Then you want to be able to get into it again. So the challenge for quantum computers is how to keep them isolated long enough to do the calculation. That’s a challenge, how to keep it from — how to keep it in what’s called a quantum coherent state. A coherent state is really when it’s doing many things at the same time.

All right, so I want to tell you now more formally how to do more quantum mechanics. So let’s take a simple example, a particle living on a line. That’s the function Y(x). Let’s ask the following question: how do we do business in Newtonian mechanics? We say, “Here’s a particle. That’s its x. Here’s the momentum. That’s its p.” Given that, I know everything I need to know right now. Angular momentum, r x p, Kinetic energy, p2/2m. Everything is given in terms of the coordinates and momentum.

In quantum theory, you don’t even tell me where it is. For every possible x, there is a function whose square, if you now square this guy, everything will be now positive, definite. I don’t know, it’s something like this. This is Y2. We say the height is proportional to finding it everywhere. What is the condition on the function side?

The answer is, whatever you like. Anything I can draw, with no special effort, is a possible function for an electron. There are no restrictions. It’s like saying, what should the position of the particle be? x. Any x you want is a possible x. Likewise, any Y you draw is a possible Y. That’s a set of all possible ampli — it’s called wave functions — whose square is the set of all possible probabilities. So I said Y2, Y at the point x2, is connected to the probability of finding it at x. That has to be in fact to be refined. That’s not precisely the story. I’ll tell you why. If a statistical event has got 6 possible answers, like you throw the die, you want to get any number from 1 to 6, you can give the probability for 1, probability for 2, 3, 4, 5 and 6. These are all the odds for getting any number from 1 to 6. Since there are only a finite number of things that can happen, I call them I = 1 to 6, there’s a probability for each I, and if you add all the probabilities, you should get 1. But if the set of things that can happen is a continuous variable like x, in other words, the location of the electron is not a discrete set of numbers. It’s any real number is a possible location. Then you cannot give a finite probability for any one x. If that was finite, since the number of points is infinite, you cannot make the total probability 1.

So what we really mean by Y2 is called the probability density. That means draw the Y2, let this be the height of Y2. Take a little sliver of width dx. The area under the graph, P(x) dx, that is the probability of finding the electron, or whatever particle, between x and x + dx. You understand? It’s called a density. So you don’t give a finite probability to each point. For an infinitesimal region, you give it infinitesimal probability, which is the function P(x) dx. And the statement that the particle has to be somewhere, namely, all the probabilities add up to 0, is the statement that when you integrate this probability density from - to + infinity, namely, Y2dx, from - to + infinity, that should be 1. This is called normalization. It’s a mathematical term. Norm is connected to length in some way, and these can be viewed as length squared. Anyway, this is called the normalization. Now if somebody gave you a Y, which did not obey this condition, here is a Y. This already tells you a nice story, right? It tells you the odds are pretty big here, pretty small there, 0 here. Now take a function that’s twice as tall. That gives you the same relative odds, you understand.

So when you multiply Y by any number, you don’t change the basic predictive power of the theory. It is just that if your original Y had a square integral = 1, the new one may not have, but the information is the same. It’s really the relative height of the function. That’s another shocking thing in quantum mechanics. If Y stood for a string vibrating, 2 × Y (this is Y, this is 2 × Y) is a totally different configuration of the string. But in quantum mechanics, Y and any multiple of Y are physically equivalent, because what we extract from Y is the relative probability of finding it here and there and there and there. So scaling the whole thing by a factor, 2 or 4 or any number, it doesn’t matter. That’s a very new thing. That’s why the Y is not very physical. If you took a string and you pulled it by twice as much, it’s a totally different situation. If you took the electric field and made it twice as big, that’s a different situation. Forces on electrons are doubled now. In the quantum mechanical Y, when you double it, it stands for the same physical condition of the electron, because the odds of being here versus being there are not altered. The only job of Y is to give you the odds.

Therefore it’s like saying in 2 dimensions, that’s a vector, that’s a different vector. But suppose you only care about the direction of the vector. For some reason, you don’t care how long it is, you just want to know which way it points, then of course all of these are considered equal. And that’s really how it is for quantum mechanics. Every Y and every multiple of it stands for one situation only. So what one normally does is from all these vectors in that direction, you may pick one whose length is 1 and say, “Let me use that member of the family to stand for the situation.” That’s like saying of all the Y’s obtained by scaling up and down, I’ll pick one whose square integral is 1.

So let me do a concrete example, so you know what I’m talking about. So let’s take a function that looks like this. It is 0 everywhere, and it has a height A between +a and -a. That’s my Y. So Y(x) = A for absolute value(x) less than a and = 0 outside. That’s a possible wave function Y. Now what does it tell you in words? If it’s a word problem, what does it tell you about the electron? Can anybody tell me? What can you say about the electron given by this function? What do you know about it? Yes?

Student: It must be found within + or -a.

Professor Ramamurti Shankar: It must be found within + or -a and more than that.

Student: [inaudible]

Professor Ramamurti Shankar: That is correct.

Student: With the same probability.

Professor Ramamurti Shankar: With the same probability, okay? The probability is it restricted to -a to +a, and it’s the same throughout the interval. After all, if you just set it restricted to -a to +a, it’s true for this function too, but that’s not the same everywhere. I’ve got a guy who’s same everywhere. Do you agree that this function has exactly the same property, restricted to -a to +a, and the probability’s constant? So there are many, many functions you can draw, all with the same statement that this object has got equal likelihood to be in this interval and 0 outside. Of this family, we are going to pick one guy whose square integral is 1. So I’m going to keep this number A, the height of the function, as a free parameter, and I’m going to choose is so that A2 — so that the Y2dx from - to + infinity, I want it to be 1. And I’ll pick A so that that is true. Well, we can do this integral in our head. What is this integral? This is just the square of A times the width of this region. That’s got to be 1.

That tells me that A must be chosen to be 1/(2a)1/2, where this little a, 2a, is now the width of this region. Therefore from this whole family, the normalized Y will look like 1/(2a)1/2 for mod x < or = a and 0 outside. And we can all see at a glance that if you squared this normalized Y and integrated from -a to +a you will get 1. This is normalization. So sometimes, people will give you a wave function and they will say as a first step, “Normalize this wave function.” What you have to do is, you’ve got to square the wave function and then put a number in front of it, and choose it so that the number makes the square integral 1.

Let me give you another example. There’s a very famous function, called the Gaussian function. It looks like this. The function e-α(x (squared))dx from - to + infinity happens to have an area which is square root of p/a. That is just one of those tabulated integrals. So here’s a bell shaped function with this property. Now I want to make a quantum mechanical wave function that looks like the Y(x) = A e- x squared over 2 D squared. That’s a possible wave function, right? Nothing funny about it, but what do you know about the wave function? It’s biggest at x = 0. It’s symmetric between + and -x. And it dies off very quickly, but how far should you go? You can easily guess that when x is much bigger than D, this function is gone, because x/D2 is going on the exponent. So if that number’s big, it’s e to the - big, which is very small. So roughly speaking the width of this graph is of order D or 2D. I’m just going to call it D, just to give you an order of magnitude. So that’s an electron whose location is roughly known to an amount D. But this is not normalized, because if I take the square of this, I won’t get 1.

So I will choose A so that 1 = Y2xdx. This being real, I don’t need the absolute value of Y. That gives me A2e -x(squared)/D2dx. The 2 went away, because I squared the function Y, so don’t forget that. Now I look at the table of integrals and what is a? When I compare these 2, a is just 1/D2. So it’s square root of pD2. This is an easy thing, because I’m already giving you the integral you need to do, but I want you to get used to it. So this whole thing should be 1. That means A is 1/pD2 to the fourth root, the power ¼. Therefore the normalized wave function, Y normalized, looks like 1/pD2 to the ¼, e −x(squared)/2D2.

Normalization is just a discipline. You discipline yourself to take all functions and normalize them, because why do you normalize them? If you normalize them to 1, then Y2 is directly the absolute probability density. That means when you add it all up, you’ll get 1. If you don’t normalize it to 1, Y2 is the relative probability density. It will still tell you the relative odds of this and that, but you cannot say this interval from here to here, the chances are 30 percent for catching it. You must take the region that you’re integrating, divide by the whole thing. But you don’t have to divide by the whole thing if you’ve normalized it to 1.

Okay, this is just practice in normalization. So I’m going to give you a little hint on what is going to happen next, but I won’t do it now, so you guys don’t have to take down anything. Just ask the following question and we’ll come back to it on Wednesday. I’ve told you that in Newtonian mechanics, every particle has an x and it has a p. In quantum theory, instead we traded for a function Y(x) and we learned the meaning of the Y(x) is that absolute value of Y2 is the probability density, meaning P(x) dx is the probability of finding it between x and x + dx.

Now we can say, “Okay, that’s enough about position. What about momentum?” I can measure the momentum of a particle. You talked about momentum on and off in the lecture. If I measure momentum, what answer will I get?” What are the odds for getting this or that answer? So given Y(x) that looks like this, you square it, you get Y(P(x)). The question is, x is not the only thing we’re interested in. Even in Newtonian mechanics, x and p were equally important. What do you think will happen now? How do I find out what happens if, instead of being interested in where I find it, I ask, with what momentum will I find it? Can you imagine a guess on what the answer might be or in what form the answer will be given to you? This is a wild guess. Nobody expects you to invent quantum mechanics in 30 seconds, so make a wild guess. Yes? Anybody there want to make a wild guess? No? Go ahead, yes, you’re smiling. Make a guess. I want the odds for different values for momentum. How do you think that information will be contained in this theory?

Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

Student: [inaudible]

Professor Ramamurti Shankar: Maybe, based on the uncertainty principle, but I want for every value of momentum a probability, right? I want the odds of getting this p or that p or that. So what do you think we need to get the odds for every momentum? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

Student: If you have a more defined location _________ calculate the probabilities of momentum, same way we did it for location?

Professor Ramamurti Shankar: Right. So what you will need, it seems reasonable to think, that this guy contained all the information on where you will find it, maybe there’s a different function of momentum, whose square will give you the probability density that you get — if that function looks like this and you square that, that’s the odds for getting one momentum versus another momentum. After all, every variable in classical mechanics you can measure in the quantum theory and you can give the odds. And for every variable, it looks like you need a function. What I will show you is that you don’t need that. Y(x) itself contains information on what happens when you measure momentum, what happens when you measure energy, what happens when you measure anything — and how do you extract it is what we’ll talk about.

[end of transcript]

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