The Theoretical Minimum | Lecture 8
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Susskind, Leonard
(February 27, 2012) Leonard Susskind spends some time in the beginning of the lecture discussing some of the basic qualities of systems to lay a foundation for the rest of the lecture and the class.
 Stanford University was talking about what happens during or from beginning, from before a measurement to after a measurement. They were talking about what happens to the evolution of an isolated black hole left to itself between the time that it was prepared and the time it was measured. Okay. Nevertheless, I could answer the question, is entanglement inevitably irreversible? If two things which are not entangled come together, interact with each other, do whatever they do, and become entangled, is it irreversible? Not necessarily. Anything that can happen, the opposite can happen. And now we're not talking about an external apparatus measuring a system. We're simply talking, for example, about two spins. Two spins interact with each other, come together, go off, and they're entangled. What happens is it possible to disentangle them by, for example, reflecting them, letting them come together and interact again, can they become disentangled? Yes. Yes, there's no rule that says entanglement is irreversible. On the other hand, if an entanglement happens and you take one of the electrons off and throw it away, then the other one is stuck there, left being entangled with a thing that just is not going to come back. Not going to come back unless, of course, somebody out there reflects it back in. So yeah, entanglement is reversible. But as far as you're concerned, the measurements that you do are irreversible. That's a very peculiar situation, and you have to think about it to understand what it means. We're going to move off the question of these very simple systems, spins, sometimes called qubits. Tonight at some point, but I want to just go over just very briefly. I think I probably discussed this in class last time, but nevertheless, it's so important that I want to hammer on it. First of all, there was a notion of a density matrix. If we have two systems, Alice's system and Bob's system, and Alice and Bob's system are far away from each other so that Alice doesn't get to do any experiments on Bob's system, Bob doesn't get to do any experiments on Alice's system, but they're entangled, either they are or they are not entangled, they may be entangled, is there a description that Alice by herself uses to describe just her subsystem? The answer is yes, and the working description of her subsystem is called the density matrix. So let me just review that very, very quickly, and go back over some concepts. If the wave function of the Alice-Bob system, the combined system of Alice's subsystem and Bob's, if the wave function is called psi of A and B, B for Bob, A for Alice, then Alice's density matrix is defined to be psi star, the complex conjugate of the wave function. Now let's say it's rho, called the density matrix, it's indicated by rho. It has subscripts A and A prime because it's a matrix, but it doesn't have anything to do with Bob, so it's purely a matrix with respect to Alice's variables. Let's indicate that by calling this Alice's density matrix. And it's given by psi star of A, B, A prime B, I always get confused, yeah, A prime, psi of A, B summed over Bob's variables. It doesn't depend on Bob's variables because they've been summed over. When you sum over something, it no longer depends on that set of variables, so this is only a function of Alice's variables, and it contains all possible statistical information that Alice can ever know about her half of a system unless Bob comes and interacts with her system. Of course, if Bob comes from wherever he is and gives her a system a shot of some sort, that will change her density matrix, but as long as Bob stays away from her system, Bob's and Palo Alto, Alice's on Alpha Centauri, as long as Bob does not interact directly with her system, her density matrix doesn't change no matter what Bob does. Let me give a little proof of that. Let's take Bob's subsystem to not be so narrow as just to be the little spin that he's carrying around, and also Alice's. Let's imagine that Alice's subsystem consists of everything in Alice's neighborhood, everything that she can have any contact with, and everything in Bob's neighborhood, maybe even including Bob, and now Bob is going to do something. He's going to do something, interact, of course he can only interact, interact may mean he may turn on a magnetic field, he may do all sorts of things to change the evolution of his subsystem. He's going to make a subsystem evolve in some way. One of the ways which Bob can cause his subsystem to evolve, many, many ways, but whatever they are, they are always governed by unitary matrices. Let's come over here and forget Alice for a minute and ask what Bob can do to his subsystem. He has a wave function, let's leave Alice out of the picture altogether, just let's suppose that Bob has a wave function, and he's going to manipulate the system somehow, he's going to manipulate the system somehow, and send it to a new wave function. The rule is the new wave function of his subsystem must be related to the old wave function by a unitary operator. Unitary operators are the things which govern time evolution. Go back, remember about time evolution, unitary operators, and that means that psi of B becomes replaced by some over B, some unitary matrix, BB prime, psi of B prime. In terms of wave functions and matrices, this is the description of what happens to Bob's matrix, sorry, Bob's column vector, this is basically Bob's column vector, describing his half of the world or his system, and to find out what happens to it no matter what he does, even if he doesn't do anything, but let's suppose he does something, it generates a unitary evolution of his wave function. Now, how do we describe the same thing in the composite language? In the composite language, if the wave function is psi of A and B, Alice's and Bob's variable, what happens to it when Bob manipulates his half of the world? Exactly the same thing with Alice's variables being just spectators, just being passive spectators. So psi of A and B just becomes psi of, whoops, let's put in there, B prime, this is sum of B prime. This is a function of B and you sum over B prime. Okay, that's again a sum over B prime of U of B and B prime, psi of A and B prime. A just passed through. Alice knew nothing whatever about this whole process. Her subsystem was not in any way influenced, her index, or her part of the wave function, or half of the dependence of the wave function, was uninfluenced by Bob's manipulation. So let's go back and take the manipulated wave function, here's the manipulated wave function, let's call it psi manipulated, the manipulated wave function of B and A, of A and B, and that's this. It's been modified by the action of Bob's unitary operator. Okay, now let's recalculate Alice's density matrix. After all, Alice's density matrix contains all information that she can ever know about her half of the world. And so let's see what happened to it. Well, we first of all have to take the manipulated psi star, we still have sum over B, and now we take the manipulated psi star, oh, we haven't written psi star, sorry about that, we have to write psi star, manipulated of, and we want to know about it with index A prime and B, A prime and B, and that we get just by complex conjugating. This tells us how psi changes, let's just work out the complex conjugate. The complex conjugate of A prime. Where is it A and B prime? This is A prime and B, just A prime and B. But here, since B prime is a summation variable, we do not want to write B prime here. B prime was the summation variable used over here, let's just invent another one called B double prime. That's going to be a summation variable in a moment. And U complex conjugate of B and B double prime. Yeah, U complex conjugate, and of course there's a sum over B double prime here. And the only reason I call it B double prime instead of B prime, was so as not to confuse it with B prime. Here's one sum and here's a separate sum. Now, here's something, what do we know about unitary operators? Unitary operators are operators that if we interchange B, if we interchange the two indices, and complex conjugate, what happens? That's called the inverse operator, or the Hermitian conjugate. In general, it's called the Hermitian conjugate. Specifically for a unitary, it becomes the inverse operator. Now, let's just call it the Hermitian conjugate. So, if we interchange the two indices, the Hermitian conjugate, and the B double prime, B, then this becomes Hermitian conjugate. So, this is the usual story. If operators operate to the right on ket vectors, the corresponding action on bra vectors is given by the Hermitian conjugate. So, now what we have to do is plug this in. We plug this in for the manipulated psi star, that's summation over B double prime, psi star of A prime and B double prime times U dagger of B double prime and B. That's a no-sense. Too many indices for me to be comfortable on the blackboard, but I think this is worth doing. Now, what about psi? Psi is the other half that comes into the density matrix, and that's U BB prime psi of what? Psi of A and B prime. Complicated little mess. Oh, what are we summing over? We're summing over all the B's basically. We're summing over this B over here, let's put a B summation here. We're summing over B double prime to get the manipulated wave form of this complex conjugate, but we also have to sum over B prime here. Sum over all the B's in sight. Each B is a summation variable by the time we're finished. All right, but now look at what we have. It looks complicated, but what is this thing over here? Well, not quite, yeah, one is essentially correct, but of course what you mean is the unit matrix. This is the product matrix of U dagger. This is a unitary matrix, so U dagger is U's inverse. B double prime B, BB prime, this is just the chronicle delta. B double prime, B prime. Let's instruct you to set B double prime equal to B prime when you do the sum, when you do one of the two sums, let's say the B double prime sum. So all of this then becomes sum over B prime, well, let's see, what have we got? We've set B double prime, yeah, we've done the sum over B, that's done. No sum over B, we just did it. And now we set B double prime equal to B prime. Psi star of A prime, B prime, psi of A and B prime. That's it. Now it doesn't matter whether we call B prime, B prime, or whether we call it B, because it's a summation variable, so let's do that, let's just call it B. Now what do we have? We have exactly, I hope, I pray, the original density matrix. So Bob, no matter what he does, as long as what he does is described by a unitary matrix, Bob will have absolutely no effect on Alice's density matrix. It does not matter whether they're entangled or not. Now who wears that I specify that this was a product state? That is what physicists call locality, the inability of Bob to influence any statistical prediction of Alice's by doing a manipulation on his own degrees of freedom. Locality is not violated, this form of locality, this form of locality is not violated in any way by the phenomenon of entanglement. That's the technical argument. Notice that it does depend on unitarity. So what did unitarity do with? Unitarity had to do with conservation of distinctions. What I call the minus first law, at least in its quantum mechanical version. So there's a funny interesting interplay, funny interesting interplay between the conservation of distinctions on the one hand and locality. If Bob somehow had access to changes which were not described by unitary operators, this would not have become the unit matrix over here and the final density matrix would, Alice's final density matrix indeed would have been modified. This is interesting the way the whole thing holds together. This is why it's so hard to change quantum mechanics. You want to play a little game? Let's let Bob, let's let Bob's evolution not quite be unitary, only approximately unitary. Well, then immediately you start sending signals faster than the speed of light. How? By instantly modifying Alice's density matrix. So things hold together tightly and it's hard to change them. Good, so yes? Is it correct to say that Bob's matrix is unitary? The evolution matrix is a vector? Yeah. You can't do it, it's not really an infectious system. He's not able to change Alice's system. He's changed his own system by doing something and that doing something was described by this unitary matrix. Okay, that's the part that confuses me. If he's doing something, isn't he going to be violating unitary? No, for example, he might be turning a magnet. He might be turning a magnet in such a way that it has an influence on his spin. What is that influence? It will change the state of the spin. For example, yeah. If you have a spin in a given direction and it's coupled to a magnetic field and you turn the magnetic field in some way, it will typically change the evolution of that spin. So Bob can do that. He can change a magnetic field that's acting on a certain spin. The result will be to change the way the spin evolves. That will change his own measurements of the spin, of his spin. If he does act by rotating a magnetic field, which is mathematically described by some sort of evolution, doesn't matter what, some sort of evolution that's described by a unitary operator, then he does change the results of his own measurements. For example, he rotates the magnetic field but he doesn't rotate his apparatus. The result is that he's going to change the probability distributions of his own spin, but he will not change any of the probabilities of Alice's spin. Okay. So are you saying quantum mechanics is always a basal calving? Or is the probability defined by the inability of Bob to influence Alice's measurements? Yeah, now the question is, so what's the big deal? Is that what you're going to ask? Yeah. I was guessing what you were going to ask. Why don't you ask it? Single particle wave function, put a photon through a beam splitter. It goes off in two different directions. You catch it at one detector faster than the speed of light, the wave function collapses. Whoa, whoa, whoa, whoa, what? You do what? You detect a photon at one detector. The wave function at the other detector collapses to zero, instantaneously. I showed you what happens. The density matrix of the other system does not change, period. It did average measurements, still changed. What happened? I know, I know, I know, people, look, I'm doing my absolute best to tell it to you as it really is and to cut through the crap that's out there. To cut through the crap that's out there, because there is a huge amount of, there is a pile a mile high. And we somehow want to cut through that. This is the mathematics of cutting through it. Now, nevertheless, there was some pretty damn smart people who are uncomfortable about something, something that they call nonlocality. I mean, you know, Einstein was no jerk, he was pretty smart. And they didn't call him Einstein for nothing. That's my favorite joke. John Bell was pretty smart. They didn't call him Einstein, though. Yeah. So there's something going on. There's something going on. Yeah, go ahead. I want to get, I'm reviewing last week, so make it quick. To me, this is real clear, I think, and that is your idea of locality means that the states don't affect each other. In other words, there's two things. The density. Yeah. And it seems that in the confusing part of quantum mechanics is that those two different things get mixed up somehow. Is that accurate or? No, it's fuzzy. But it's true. So I will try to tell you as I understand it what the concrete operational facts are. So I told you last time, I was a little bit tired last time, and I'm not sure I was completely clear. So I won't go over it again because it's very, very interesting. And it really is at the heart of all of the confusions about confusions. I mean, I must say, including my own internal confusions about quantum mechanics. When I first learned about, er, Einstein, Rosen, Podolski, that was the discovery. Well, I don't know if it was a discovery. I think maybe Schrodinger said some things about entanglement. But it was Einstein, Rosen, and Podolski who really called attention to how peculiar it was. Basically what they said is we have a very funny situation here where although we can know everything that can possibly be known about a composite system, we know nothing at all about its constituents. That's crazy. That violates my sense of reality. That violates my sense of, er, of what it means to know what a system is doing. Violates all sorts of things. I remember when I first heard about it, it was actually before John Bell sometime in the 60s when I was a student. No, no, I wasn't a student. I was a young assistant professor. And I started to think about it and I thought about it a lot and tried to figure out what it means. I was sort of inclined to try to think about it in terms of simulating quantum mechanics. And I still am. What are the limits? If, if, if quantum mechanics is so very strange and so very different than classical physics, does it mean that you can't fool somebody into thinking a system is quantum mechanical if all you have, your resources are entirely classical? And the answer is yes and no. So I know I talked about this last time, but it really is at the center of things. So let's go back over it quick. Quick as I can. We're going to build ourselves a quantum simulator. I think somebody built one on the, on the site. I came across it about an hour ago. Who, who's that? Yeah. Does it work? Well, it doesn't actually simulate yet. Not yet. But is that picture, is that picture your picture? It was a picture of Einstein and Heisenberg and Einstein says I don't like this and Heisenberg said what? Einstein says I think I'm not going to like this. Yeah. Right. Oh. Yes. Right. I think that captures it. Why go on? But okay, let's, let's talk about the nature of quantum simulators. First of all, I just want to simulate a single spin. I think I told you how to do it, but let's write down the steps. We have a computer screen. Here's our computer screen. On the screen, we have a picture of a detector, not a detector, an apparatus. And the apparatus can be oriented in any direction in three-dimensional space. We need to do some three-dimensional graphics here. And it has a window, and the window shows one of three possible states. One is called blank B. That's the detector or the apparatus before it's interacted with anything. So it can be B or it can be plus or minus one. B is not important. It's just to say before the detector interacts with anything. There's a button over here called M. And the operator of the system, of the computer game, basically all you ever get to do is press M, and then look at the result of the M stands for measure. It performs a measurement. Now, over here connected to the computer by wires, by cables, connected to the screen by cables, is a computer. And the computer stores information in its memory, and the basic information that it stores is two complex numbers, alpha up and alpha down. Those are supposed to be the components of the wave function of the spin. I didn't say yet that there was a spin here, but that's what we're talking about. Alpha up and alpha down. The computer only can generate alphas which satisfy alpha star, alpha up, plus alpha star, alpha down is equal to one. Whenever it generates them. And it can do something else, it can do two other things. It can update the alphas by solving the Schrodinger equation for them. There's some Hamiltonian, for example, it could be that the spin is in an imaginary magnetic field, in which case the alphas would change with time. The computer knows how to do that. It knows how to solve Schrodinger's equation. Computers can solve Schrodinger's equation beautifully. And it updates these as a function of time. Updates them according to the rules of quantum mechanics. One other thing, the cables are instantaneous. They convey information instantly. So we'll have one more element. The other element will be a random number generator. Let me just say some words about random number generators. Random number generators, sounds like we're introducing quantum mechanics. No, we don't need quantum mechanics. You can simulate to a very, very high degree of confidence a sequence of random numbers or a random number generator purely out of classical considerations. You know how to do it. You take the digits of pi or whatever and you make something that looks, would fool anybody, including a sophisticated mathematician, into believing that you were generating a set of random numbers. So there's a random number generator inside the computer. All right, the operator starts by initializing the system. And initializing the system simply means starting out some orientation for the detector, for the apparatus, doesn't matter what, and setting the alphas to some initial values. You can plug in what he wants the initial values to be, or the computer just sets them according to whatever rule it likes. Then the system runs for a period of time, an indeterminate period of time, updating the alphas until the operator decides to press M. A certain amount of time is elapsed, the operator decides how much time and presses M. At that point, the random number generator generates a plus one or a minus one, but with probability alpha star alpha up, that it's plus one, sorry, alpha up, alpha up star, that's plus one, or else alpha star alpha down, that it's minus one. Remember, the sum of these two alpha star alphas add up to one. So armed with the random number generator, the computer sends a signal with a probability of the appropriate probability to the detector, and the detector detects or registers what the computer sent it. That's one measurement. So as soon as that happens, the computer re-initializes the alphas. How does it re-initialize the alphas? If the detector detected plus one, then it sends back a message, it doesn't really need to send back a message, after all, this message got from here to here, so the computer over here knows whether it got plus one or minus one. If the detector saw plus one, then it sets alpha u back to one and sends alpha d to zero. In other words, it does what we call collapsing the wave function. According to the result, or according to the outcome of the measurement, it re-initializes the alphas. So whatever was measured, you start with probability one for the outcome, and probability zero for the, what shall we call it, the anti-outcome. And then it runs again. Solves a Schrodinger equation, does it again. The operator makes a ramp, makes whatever decision he likes. I mean, what makes him decide to press M? He's an experimental physicist. What makes experimental physicists do what they do? I don't really know, but he does what experimental physicists do. Now and then he presses M and does the whole operation over again. Oh, sorry. Before he does, he may decide to rotate the apparatus. Now, the computer in here knows what the orientation of the apparatus is. So when it calculates, it not only calculates the updated alphas, but it also calculates the probabilities for the rotated detector to detect whatever it's going to detect. That's all algebra and a little bit of, that's purely algebra. The computer makes the kind of calculations that you did, presumably did, makes the kind of calculations to calculate what the probability would be that if the detector was rotated in some arbitrary direction, then what's the probability that it will get up or down? Now, who is it that decides to rotate the computer? Well, it's the operator again, the experimentalist. The experimentalist not only can press M, but he can also reorient the detector any time he wants. This is about as much as an experimental physicist can do with that spin. Rotate the apparatus, measure. Rotate the apparatus again, measure. And what's the result of the outcome of this? The outcome of this is indistinguishable from what the experimental physicist examining the spin would see. So we've simulated quantum mechanics. Now we want to know whether we can simulate quantum mechanics of two spins, one on Alice's side of the world over here, and one on Bob's side of the world. Yeah? I don't understand the purpose of the button M in ours. In other words, when he decides to put it, some amount of time is involved. Has that affected anything going on in the computer? Yeah, the computer has been updating all this time. Right, but when we talked about our base model of how spin works, and how it was funded, and I come into it at all. We did talk about Hamiltonians. We talked about Hamiltonians and how Hamiltonians update the state of the system. You're right, we didn't discuss how... We didn't discuss much about Hamiltonians for spins. Yes, we did. Yes, we did. I believe I described a spin and a magnetic field and how the average spin precesses. We didn't have the magnetic field, because that's how we started out talking about all this stuff, right? Then time should not be a factor. Okay, I understand that. Right, now there's no real magnetic field. There's just an imaginary magnetic field that the computer knows about. And whereas I come into it, it comes into the nature of the exact Schrodinger equation. Okay. Now, we do this over here and precisely the same thing over here, except instead of alphas, we have betas. Beta stands for Bob. Exactly the same setup, replicated twice. Bob does his measurement, Alice does her measurements. They're perfectly happy. Each one says, I have a spin. It's quantum mechanical. They don't talk to each other. And each one says, my spin is in a state. The language that we would describe this by quantum mechanically is they have simulated a product state. A product state in which Bob's... Sorry, Alice's wave function is this, and Bob's wave function is this, and the combined system is in a product state. Supposing now we want to simulate any state, a more complete description that includes entangled states. Well, we can do it this way. Let's start with a way that's sort of guaranteed to work. We take the computer and we put it in the middle, a single computer. We can think of the single computer if we like as a pair of computers connected by wires, but just by the fact that they're communicating with each other and interacting with each other really makes them a single computer. And in that single computer we now store four complex numbers. The four complex numbers that we store in there are alpha up, up, alpha up, down, alpha down, up, alpha down, down. These complex numbers here are not necessarily products. In fact, let's assume they're not products. In other words, this is not a product state. Let's assume it's a highly entangled state. Even a highly entangled state is described by four complex numbers like this. That's what's in the computer. And what does a computer do with it? It evolves the Schrodinger equation. First of all, we initialize. We start with some alphas. The alphas are allowed to run according to whatever Schrodinger equation describes them. And at the end, either Bob or Alice or both, without knowing what the other one is doing, press or as the case may be, don't press their measurement buttons. Once the measurement button has been pressed, an instantaneous signal goes back to the computer. The computer grinds out its random number generator, and whichever side was measured sends a signal to the detector with appropriate probability, either plus one or minus one, and then reinitializes itself, reinitializes itself in a way that doesn't touch, for example, if Alice does the experiment, then it doesn't touch Bob's end of things. If Bob does the experiment, the computer reinitializes his half of the wave function, which is another way of saying it throws away the piece that you didn't measure, throws away the pieces you didn't measure and renormalizes the wave function so that the total probability is warning. Exactly the same thing. There's no real difference with a single system you can simulate it, and you can simulate it in a way which would describe all the possible things that Alice and Bob could measure, whether or not the system was entangled. Where do we get in trouble? We get in trouble if after initializing, we try to separate the computers and break the wires. For example, here's what we might do. Given these alphas, we might put them into both computers. Take the set of alphas that we generated, put this one into this computer, this one into this computer. Now, since the computers are not going to talk to each other, they have to have separate random number generators. The random number generators will not talk to each other. Each one contains the full entangled state, and then we go through the same operation. What's going to happen? The computer is simply going to forget, for example, for a highly entangled state, let's say the entangled state is the singlet state, then the right answer would be that if Bob and Alice both measure, they should get opposite values for any spin. If they measure the x component of spin, they should get opposite values. If they measure y components of spin, they should get opposite values. But if when we separate them, they have separate random number generators, and the random number generators don't talk to each other, then guess what's going to happen? There's going to be no correlation of any kind between what Bob measures and what Alice measures. No, Bob may measure plus, and Alice may also measure plus, because their random number generators are not connected to each other. And so even though both computers may know about the full state of the system, if the random number generators don't talk to each other, then the results of the experiments will not contain the correlation. The correlation that Bob and Alice could later come together and report, Bob would say, my first experiments are a plus. If things were working right, then Alice should say, my first experiments are a minus. And so forth and so on, that correlation will not be there once you try to separate the computers. Are the random numbers correlated if you don't separate the computers? You can have a single random number generator that sends out signals to both computers. If it sends out a signal plus one to this computer, it sends out a signal minus one to that computer. So all you need is a single random number generator, but it has to be able to talk to both computers. So there's got to be wires in the world. There's got to be wires in the world connecting this side and this side for no other reason than they can both connect to the same random number generator. You've got four outputs there, but it sounds like you just described that the outputs are independent. Now you've got to think about it. Yeah, yeah, yeah, yeah. Yeah. Yeah. Well, that's sort of right. Once you've separated them, then what happens next after some measurements are made, doesn't preserve the relationship between them. They become disconnected from each other. Yeah, so think about it a little bit. I mean, even better. Try to simulate it. This is this, I've never tried, but I don't think it should be that hard if you can do a little bit of computer programming. I've simulated much harder things than that when my son was a little kid. We had this thing called BASIC. Anybody remember BASIC? And he and I designed the basketball game, and it had friction, and it had, it was much harder than this. So that was my last experience of programming anything. I think you'll learn a lot by trying to simulate this. You have a detector called a cell phone, and I get a signal from Alice or Bob about some random event. It could be a spin, let's suppose, but we're, will you please speak concretely? We get a signal from Bob of some random event. What does it mean? Can you please be exact? Very exact. I'm getting two text messages, one from A, one from B, okay, and they're going to have a value of one or zero, each of them. Okay? And I get a million of these things, and I noticed the distribution that the A signal had a 50% chance of being one or zero, same with the B signal, okay? It could be from a spin. Yes, yes, okay, go ahead. Now, I look at the correlation that turns out that whether it's A or B, the other one is the same. In other words, it's correlated, a correlation of one. That's in the real world of entangled states. This could be, we don't know where it's coming from. I'm just getting information. I get information to A comes through, I don't look at it, B comes through. I know it's a 50-50 chance, I don't look at it, but as soon as I look at A, I know what B is, so the wave function completely changes. Now, the example I'm using is not a state, I'm just a Bob and Alice or sexting twins, one's a boy, one's a girl. You've got to hurry up because I've got to get on to particles. So there is nothing magical about that. There is nothing weird about that. If it's correlated like that, it's very probable they're identical twins instead of eternal twins. So what, how am I missing? What is weird about that? If you don't find it weird, good. What's weird is that you can't simulate it classically without having a central processor connected by wires. So what's weird about it? So don't those wires that connect the two distant computers represent non-locality? Oh, okay, now we ask. All right, those wires are there. For goodness sakes, surely Bob can send a message through the wires to Alice. It takes no time at all because they're instantaneous. Yeah, true enough if this were a real computer setup, but if Bob is restricted to only do the things quantum mechanics allows him to do, if he is restricted by the rules of quantum mechanics, then we find out amazingly that Bob can never send a message through the wires that violates locality. Why? Because all Bob can do is manipulate his own system with unitary operators and so that's what's curious here. Does that mean that the world is filled up with wires that we can't see? I don't think that's the message. I think the right message, but of course here you can hear with this room for debate what the right message of all of this is. The message I take away from it is basically quantum mechanics cannot be simulated by a classical system. The logic is too different. The two theories are too different. I don't take away the message that the world is full of wires that if only we could access them, we could send messages back and forth. Other people do, but I think at least if we know how to do quantum mechanics, we'll agree on the facts. We may disagree on basically whether there are or are not wires there. Are there really wires there or are there really not wires there? Well, perhaps the culprit is the word really. But this is, oh, there is one other element to it, which is quite interesting. We've made an assumption that Bob and Alice get to kind of randomly make decisions about when to press the button, about what orientations, about what to decide to measure. Bob sits here and says, hmm, okay, Northwest, hmm, Southeast, or whatever, makes these random decisions according to the principle of free will, so to speak. That's fine as long as Alice and Bob are outside the system and not part of it. We don't need to ask the question of consciousness and the question of free will. We'll just admit that Alice and Bob can make random decisions about when to measure various things. It's obviously a pretty good approximation. But once we take Alice and Bob and put them into the system so that they are governed by the laws of physics, then things get a little bit touchy. What do we mean by saying that Alice can decide to do this or Bob can decide to do that? And that, I think, is where the possible loopholes in all of this can come from. Now, I don't want to get too deep into heavy philosophical discussions of all sorts of psychological questions of consciousness. I don't know the answers. That's it. I don't know the answers beyond what I've told you. I am certain of two things about it. I am certain that there are a lot of people who think they do know the answers, and I'm also certain that they don't. That's it. A question. Actually, not a question. If you Google for Quantum Randy Challenge, you'll find a blog entry that has the setup for this simulation. Really? For the entangled case? It doesn't solve it. It says, if you can, here's a program. Your challenge is to fill in this blank place. I see. If you can, he'll give you a million dollars. Actually, he won't give you a million dollars. Somebody in Sweden will give you a million dollars. Yeah. OK, that's very interesting. I'm not the first one to think about it this way, but I've been telling this to people for 50 years. No. I don't know. I probably wasn't. I think maybe Bohr thought about it this way. Who knows? It doesn't matter. We could spend the rest of eternity talking about spins. Why? Because they really are interesting. But we want to get to something else. We want to get to particles. We want to get to more complex systems. The most complex system we've discussed is two spins, four-dimensional Hilbert space. We've gotten about to the point where in classical physics, we got through the first lecture. The first lecture was about these simple discrete systems. And then we rather immediately moved on to continuous motion of things like particles moving continuously with an infinite, not just an infinite, but a continuously infinite range of possibilities. That's where we have to go now. Now fortunately, we've set up so much of the apparatus of quantum mechanics that we can do it fairly quickly. There's a little bit of subtlety about replacing sums by integrals. That's about it. That's about the nature of the subtlety. Summs and integrals, we've expressed the general principles of quantum mechanics and then exhibited it in a variety of simple spin system examples. Now we'll plunge right into quantum gravity. No. We'll plunge right into the problem of a single particle moving on a one-dimensional axis. We'll take the axis to be infinite, although that's mostly because they don't have time to be more careful. If we want to get anywhere, we're going to have to be slightly less mathematically rigorous. So we have, and in order to be rigorous, you really have to contain a system in a finite volume and so forth, but we don't need to do that. So we have a line, infinite line. Along that infinite line, one particle can exist. We won't do two particles yet. One particle can exist in one dimension, and that line is labeled by a coordinate x. I mean, not right down the coordinate x, but also the states of the particle are also labeled by the position of a particle. Now here we are departing from classical logic. In classical logic, the states of a particle are labeled by a position and a momentum. If we were talking about the states of a spin, we would say the state of a spin is described by all of the, or maybe at least two of the components of spin. But we can't do that in quantum mechanics. We can't specify two components simultaneously. Same in classical mechanics, specifying, as I'm sure you know, specifying the position and the momentum is too much. And we'll see how that works. But for the moment, let's take it as a postulate that a complete description of the state of a particle, and you can't get any more out, is, for example, to know that it's located at a position x, and if it's located at a position x, we'll describe that by a ket vector. The ket vector x. That's the state of a particle when it's located at position x. Now, of course, we can consider more complicated, more complicated states. Linear superpositions of these in quantum mechanics, we are always allowed to make linear superpositions. And so the space of states is bigger than just the space of particles located at definite positions. Let's call the general state psi. This is the general state of a particle. It doesn't mean anything now. It's just an abstract symbol. And it's a vector in some vector space. The vector space now has a continuously infinite number of basis vectors. Now, strictly speaking, that's a mathematically bad thing, but we're going to play that game. If you want to understand more deeply the mathematics of quantum mechanics, you go and you learn about Hilbert space. But we're not going to do that tonight. So the general state of a particle is called psi. Remember what the wave function of a system is. The wave function of a system, in a particular basis, is the projection of a state onto the basis vectors. And we call that psi of A. Now we come to a particle. The basis of states are the position states. And what is the wave function? The wave function is the inner product of the state vector with the eigenvectors of position. And that's called psi of X. That's the wave function. That's the classical, not classical, but that's the standard quantum mechanical wave function of a particle. It plays the same role as in the simple systems we've discussed, the inner product of the state vector with A. Now we can say a little more. In this context, the probability that the system, whatever it is, exhibits property A is given by this thing squared. In fact, it's given by psi star of A, psi of A. That's the probability that the system is, but the outcome of an experiment to determine whether the system had property A, that's the probability. Likewise, in exact parallel, the probability to find the particle at X is equal to psi star of X times psi of X. I really haven't done anything that we haven't done before. Now, in quantum mechanics, like in any other sensible statistical theory, the sum of all probabilities should add up to one. So the sum of psi star psi should add up to one. But when probabilities refer to a continuous variable, we don't talk about the probability that the particle is located exactly at point X. That's too sharp. The probability that a thing is located exactly at any given X is zero. We talk about the probability that it's located within a range of X. Or better yet, we talk about the density of probability. We talk about the probability density, and we integrate this over a little range of X to find the probability that you're in that range of X. We do something like that here. If we want to know not whether the system is definitely specified at A, but we want to know whether it's at a range of A, we sum over them. But here we sort of have to do it. Total probability equal to one? That's the condition that the integral of psi star of X, psi of X, dX is equal to one. Total probability of the system has to add up to one. And the only, as I said, the only new thing here is that we have to replace sums by integrals. And that does make mathematical subtlety that we have to worry about, but it's sort of in the mathematical details. Okay, what about the inner product between two state vectors, one at X and one at X prime? Well, here we go back for inspiration, right back to, right back to this case. Zero if A is not equal to A prime and one if A is equal to A prime. And we're not going to be able to quite get away with that here. We're not going to quite be able to get away with a statement that X, X prime is equal to zero or one. We are going to say that if X is not equal to X prime, then it's zero. But we're not going to say that when it's equal to X prime, that it's one, that would not leave us anywhere. What we're going to say is that it's a function which vanishes unless equals X, X equals X prime, but has area one under it. We'll see how that works out. We'll see what that does for us in a moment. What it really comes down to is a different normalization here. But again, the basic idea that states are orthogonal, if X is not equal to X prime, that we will retain. But the numerical value of what we get when X is equal to X prime, we're going to take to be slightly different. And in fact, we're going to invent a function that we are, derac did, the derac delta function, X minus X prime. The derac delta function is not a real mathematical function, but by now, I think mathematicians have defined it to death so that it is a real mathematical entity. But for our purposes, we can think of it in a very simple way. It's a function which is zero, well, let's take delta of X, not delta of X minus X prime, this is basically delta of X minus zero, X prime zero, the function delta of X. The function which is zero unless X is equal to zero, that's the same as saying this thing is zero unless X is equal to X prime. So if this is X, it's zero everywhere except X equals X prime, and X equals X prime, it's a very high narrow spike. How high and narrow? High enough and narrow enough that the area under it is one unit. That means if it's approximated by a narrow interval of size epsilon, that it has height one over epsilon. And then you imagine a limit where you take epsilon smaller and smaller, one over epsilon higher and higher, always keeping the area under it equal to one. That's the derac delta function, derac delta function, zero everywhere except when X equals X prime, infinite in some sense when X is equal to X prime, but in such a way that the area under it is equal to one. All right, now let's use this to further elaborate on what the meaning of psi of X is. I'm going to guess and then we're going to check. I'm going to guess that the meaning of psi of X is the following, that if I want to write a general state vector, psi, that I can think of it as a sum over the basis vectors, but not a sum, an integral. Of course an integral because the basis vectors are continuous. The X of some coefficients, some coefficients, well let's know, let's write it this way, psi is equal to an integral of the X of some set of coefficients. I'm going to guess what they are. I'm going to guess that the psi of X times the, the state, the quantum state where the particle is located at X, but then we're going to add them up for different X's. It's exactly the same thing that we did for ordinary quantum systems except this was replaced by a sum. All right, now I've given you two distinct definitions of psi of X. One is that it's the inner product of the state vector with the eigenvector X, I'll call it the eigenvector of X. The other is that it's the expansion coefficients in writing psi as a sum over vectors. Now if you're a little bit uncomfortable with the idea of a continuous basis of vectors that you integrate over, you're in good company. The mathematics of this is, as I said, by now it has been made rigorous. But it's, it's, it's, it took pages and pages and pages and pages and books and books and books on something called distribution theory to make sense out of this with precision, but it's pretty intuitive. It's pretty intuitive and we'll stick with that. All right, so let's calculate, oh let's put a prime here. This is a sum, an integral. It doesn't matter, it doesn't matter what you call the integration variable. Let's call it X prime. And now let's calculate the inner product of X with psi. As I said, there were two different definitions of psi of X. Let's check that they're the same. So here we have it. Integral the X prime, psi of X prime, inner product of X with X prime. What's the inner product of X with X prime? I think I wrote it someplace, where is it? I think I erased it. That's equal to the inner product, to the integral of the X prime, psi of X prime, delta of X minus X prime. Okay, what is that? All right, first of all, delta is a function which is zero everywhere except that X equals X prime. So that means that in doing this integral here, you'll only get a contribution when X prime is equal to X. That means the answer is going to come out to be proportional to psi of X. Furthermore, the area under this delta function is exactly one. If we imagine that this delta function is a pinched, narrow, or narrow, or high spike, so that the area under it is one, then this integral would tend to psi of X. The rule is, if you have an integral over a delta function like this, it simply picks out the value of the integration variable where the argument of the delta function vanishes, where X minus X prime vanishes. That's where the function is not zero. And the integral of a function with a delta function just gives you the value of the function at the point X. That's a rule. We might as well write that down. Let's write that down as a rule. Integral delta of X minus X prime, F of X prime is always equal to F of X. It picks out the value X prime equal to X. That's a part of the definition of the delta function. Okay, so it's just equal to psi of X, and we find consistency that the thing that we called psi of X over here is the same as the thing that we called psi of X over here. All of this has its exact parallel for the ordinary systems. Top integral on that bottom board, yeah. I'm trying to figure out which are the base inspectors. These are the base inspectors. Let's compare that with psi is equal to sum over a prime. Let's call it a prime. Psi of a prime, a prime. Okay? If we use the fact that a and a prime are orthogonal and do exactly the same thing except replacing sums by integrals, we will also find that the psi that appears here is the same as the inner product of psi with the basis vectors. So I've really done nothing that we haven't done before. I appreciate it's a little, possibly a little unfamiliar, but all it is, is the same thing we've done before. So the psi of X prime sort of plays the same role as the alpha's and some other. The X prime plays the same role as the A prime here. Yeah. The psi plays the same role as the psi. The sum is replaced by the integral. Sum over A prime is replaced by integral, the X prime, and otherwise it's exactly the same. Now let's calculate the inner product. One quick question. It may be nitpicking or something, but on the left-hand side where you have the Dirac's formula, lower right-hand corner, yeah, you've got a vertical line between the axis. Over here in the middle formula, there's no vertical line. Thank you. Okay. Great. Because it could have been a line or a problem. Got you. Got you. Got you. Right? Nope. Absolutely. How about the inner product of two vectors? Well, the inner product of two vectors is some inner product, but one way of calculating the inner product, let's go back to ordinary quantum mechanics, ordinary means for the discrete systems. We don't need this right now. How do we calculate the inner product? Let's take it high. Inner product of psi with phi. We first write psi and phi as sums over the basis vectors. So this is equal to sum over two sets of basis vectors, a prime and a. Psi star of a prime times a prime, that's the bra vector. We complex conjugate, of course, when we deal with bra vectors. That's the psi half of it. And then the phi half of it is the sum over a of phi of a times a. Phi of a being the coefficients of the phi vector, psi star being the coefficients of the psi bra vector. All right. Now, phi of a is just a number. And the inner product of a prime with a is just the chronic a delta. So all of this adds up to sum of psi star of a prime, phi of a. And then a chronic a delta of a and a prime, which is exactly the instruction to set a equal to a prime. Now, I think you've seen that before. This is simply the rule about row vectors. If psi star of a is represented by a row vector, psi star of a1, blah, blah, blah, blah, blah, and phi is represented by a column vector, then this is just the rule to take this times this, plus this times this, plus this times this, plus this times that. So this is the common garden variety quantum mechanical inner product. I'm not going to, we don't need to go through the same exercise here. We can immediately write down what the result is going to be just by analogy. It's almost not even an analogy. It's just replacing wherever we see a's put x's and wherever we see sum put integral. The inner product of two state vectors for a particle on a line is integral the x psi star of x phi of x. It satisfies, incidentally, one of the basic rules of inner products. What happens to an inner product if you want to change the two vectors? What's the relation between the inner product psi phi and the inner product phi psi? Complex conjugation. Now what would happen in here if you want to change psi with phi? You'd wind up writing phi star here and psi here. That again would be the inner product, the complex conjugate. And for exactly the same reasons. All right, so this is equal to this for both discrete systems and for these continuous systems. Inner products, the inner product of a vector with itself is the norm of the vector. And just as in previous examples, you normalize the vector to one to express the fact that total probability adds up to one. So not much difference. Just a difference of notation. I think what gets to be confusing about it is you know both of these things. You know about these simple row vectors and column vectors that you know about functions, but you're not used to thinking of functions as vectors. Or maybe you are if you are good. But if you're not used to using final thinking of functions as vectors, f of x's, or function psi of x's, then you'll have to get used to it. But functions are vectors in a vector space. You can add them, you can multiply them by complex constants, and you can add them, you can multiply them by numbers. And so they satisfy the axioms of a vector space. Space of functions is a vector space. Now sometimes you may want, yeah. In more classical mathematics, would you put as the limits of integration minus and plus infinity? You would. Which incidentally says something. It says that we shouldn't be thinking about functions which grow at infinity, for example. We should be only thinking about functions which are square integrable. Now we're going to have to modify that. Square integrable means you square them, this is called squaring them, that the integral exists. Functions which grow at infinity won't do that. Now those are crazy because they would say that all of the, if you had functions which grew at infinity, where would the probability be? The probability would be all located off at infinity, and that's not what we want. We want to be talking about probability distributions which have a total finite probability under them. So that means we should take perhaps with a grain of salt the idea that we can write down any function. But for the moment we're going to be a little bit loose about that. Okay. I think I'll slow down here for a few minutes and would you digest that and ask for questions. Yeah. Can I go back to the end of this? No, no, no, no, no. Yeah, go ahead. No, I didn't want, I meant about this, but okay. Maybe this is quick. It seems like just about everything must be entangled with something almost all the time. Oh yeah. Yeah. You're entangled with all sorts of things in the environment. Now we haven't talked about the importance of a complex environment. It's not so important in the very foundations of the subject, but it is important to why we see the world the way we are, that we're constantly being bombarded and getting entangled with things. As a theory of distribution has been useful for physicists, or is it just... Oh boy. Certainly. Okay, so it's not just some, it didn't just clean up the direct delta function, I appreciate it. Oh, you mean, has it, well... I see what you mean. You mean has it done anything else besides clean up the... I don't know. I don't know. I don't know. I think it's made rigorous a set of concepts that physicists use all the time. Has it really changed? No, no. The theory of distributions... Yeah, I'll tell you what it did for physicists. It made them have to argue less with mathematicians. It was one of the many examples where mathematicians got dragged screaming and kicking by physicists. Of course, the opposite also happens, but it was an example where mathematicians at first looked at this thing and said, it's crazy, that doesn't make any sense. It's non-rigorous, and in time they came to peace with it through defining it in their own way. But no, I don't think the mathematical structure of it has influenced physics as much as you might have expected. Unfortunately, because there are big problems of confusion related to it in quantum field theory, and it hasn't resolved those problems. Okay. Okay, next place we want to go is observables. We want to talk about observables and give some examples of observables. Observables are linear operators. What are linear operators acting on a space of functions? They're operations on a function that gives another function. If you have a linear operator, let's call it M. For some reason, I've saved L to stand for observables. I've used previously M for a general linear operator, so let's continue to do so. Whatever M is, it's a machine, and thought of as an abstract operator in a space, it takes a state vector into another state vector, a side prime, let's say, or a vector in the vector space. But it also can be thought of as a thing which acts on wave functions. Every state has a wave function that describes it. And the machine, oh, that's why we called it M. Yes, I remember now. We called it M for machine. The machine can also be thought of as acting on a wave function to give another wave function. So let's think of it that way. Let's think of it in concrete terms as an action that we do on a wave function to produce another wave function. Now, it's linear. Linear means that when it acts on the sum of two functions, it gives back the sum of the results. It means that if it acts on a function which is a numerical product, a numerical multiple of some function, it gives back the same numerical multiple. All the ideas of what a linear operator are continue to be exactly the same. Let's talk about some linear op- oh, okay. That's the notion of a linear operator. Now, what about a Hermitian linear operator? I'll tell you the definition of Hermitian. We've talked about it before, but I'm going to give it to you in a condensed form. The condensed form is also true for all of the vector spaces we've studied. And you may recognize it if not go back and prove that it follows from the previous definitions. A Hermitian operator, Hermitian for operators is roughly speaking the analog of real as opposed to complex. Okay, here's what's true about Hermitian operators. If you have a Hermitian operator which I will now call L, then it has the property that if you take, if you act on a wave function psi, or a state vector psi, and take its inner product of phi, that that is equal to psi L phi, but not quite, what do I have to do with it? Complex conjugate it. Now, if L were not, let's go back a step. Supposing L was not Hermitian, would this be right? Not quite. What would we have to do? We'd have to put a dagger here. In other words, we'd have to roughly speaking complex conjugate L. Okay. If L is Hermitian, then the rule for that is that it's matrix element, it's when sandwiched. When sandwiched in opposite order, you don't have to complex conjugate L because it's already the analog of real. This is the condition that an operator is Hermitian. That's important because Hermitian operators are the class of observables. So let's work out one or two Hermitian operators while we're at it. I guess we may need this, so let's work out a couple. First of all, one of the simplest things that I can do to a wave function, I don't know what's there. There are many simple things I can do to it. But a very simple thing is to multiply it by x. An even simpler thing would be to multiply it by a simple number. But let's take slightly more complicated and multiply it by x. Multiplying by x gives back a function. This is at minimum an operation on a function. Is it linear? Sure it is. If x multiplies psi of x, let's say, plus phi of x, it gives x psi of x plus x phi of x. In other words, when it acts on a sum, it gives the sum of the results. If it acts on a numerical multiple of a psi, it gives back the numerical multiple. If it acts on a linear operator, question, is it a Hermitian? Is it a Hermitian operator? Okay, so let's check. Here's what we have to take. We have to take psi in a product of x times phi. That's this one. Oh, sorry. I think I want to do it the other way, phi x psi, the top, phi x psi. And ask whether this is true or not. We'll check in a minute. No, x is not a number. x is a function x. We've multiplied one function by another function. That takes functions to functions. x is not a number. It's a variable. Any given value of it is a number. But it's a machine that takes the function psi of x and replaces it by x psi of x. All right, so let's calculate. The wave function of the ket vector is x psi of x. The wave function of the bra vector is phi star of x. The inner product is, let's put some brackets around it, bracket. This is a ket vector inside the red bracket, and this is a bra vector. And the rule for taking the inner product, whoops, I think I interchanged star and not star, but the rule for taking the inner product is to integrate. So let's get rid of a bracket here. And just integrate dx. Oops, sorry, dx. All right, that's in this order. What about this order? This order over here, the inner product without the star, without the star, the thing we're going to compare it with, is the same thing except with psi star x phi of x. Well, look at it. Is this integral the complex conjugate of this? Yes, it is. The complex conjugate of phi star is phi. The complex conjugate of psi is psi star. And what about x? x is its own complex conjugate. So yes, indeed. x is an operator which satisfies the basic rule of Hermitian operator that if you interchange the psi and the phi, it complex conjugates. All right? If, yeah. What observable does x correspond to? Position. It is the position. Is the delta function a eigenvector of position? Well, maybe we'll come to that next time. I think I won't. Yes, it is. But we'll come back to that. x is an observable. What does it correspond to? It corresponds to the position of a particle. We will prove that by showing that the eigenvectors x, that the vectors x are eigenvectors of that operator, but not right now. We'll do that the next time. It's easy. It's trivial. Just a big question. So my correctness in assuming that when you write x up there, it's implied that that is really f of x. The x is a function of x. x is an n f of x. Yes, it's implied that that x is a function of x. It has a whole bunch of values. It has a whole bunch of values? That's right. It itself is a function which is multiplying another function. All right? So it's an operation on a function which gives another function. Yeah. That's exactly right. Let's take another operation or operator, differentiation. If I differentiate a function, I get a function. That's the first test of whether it's an operator. If I differentiate the sum of two functions, I get the sum of derivatives. If I differentiate a numerical multiple of a function, I get the numerical multiple of the derivative. It's a linear function. Is it a Hermitian operator? So here we have to get ourselves a fresh blackboard and check once and for all whether it's, well, check whether it's Hermitian. Okay. What happens? Let's call a derivative operation. Let's call it capital D. D on psi gives us some psi prime. What about wave functions? D when it acts on psi of x by definition now. This is the definition of how D acts. It simply gives d psi by dx. I may sometimes lapse into calling this partial derivative with respect to x. That's fine. Doesn't matter. It's just a function of x. So it just differentiates. It creates a new function. It's a linear operator. Is it Hermitian? Okay. So let's write down the two sides of the Hermitian condition here and see if they're the same. That's all there is to it. The left side is integral phi star of x times the operation D on psi, which is just the psi by dx. What's the other way of doing it? Or the right-hand side? The other way of doing it is integral psi star. Not doing the same thing, doing something different. Psi star of x, the phi by dx. And the question is whether this is the complex conjugate of this. So let's write equals with a question mark. We don't know. Complex conjugate. So let's do this integral over here. First of all, the thing in the bracket here is just equal to integral psi of x d phi star by dx. Just complex conjugate. The complex conjugate of a product is just a product of complex conjugates. So the thing in the bracket is psi of x d phi star by dx. And we're asking whether this is equal to this. They look similar. This has a psi. This has a psi. This has a phi star. This has a phi star. But unfortunately, this one has psi differentiated. And this one has psi, you know what I mean, psi and phi. How do you go from this to this? Integration by parts. And integrate like this, assuming, what are we assuming when we say we can integrate by parts? We're assuming that far away the functions go to zero. That the endpoint contributions from integration by parts don't give us anything. Well, of course, the functions must go to zero far away because the integral of the total thing must be one. So we can integrate by parts. Everybody know what integrating by parts is? If not, find out fast. It's simply a rule. I'll tell you what the rule is. You can shift the derivative from one, if you have a product, dx. There's a dx here. If you have a product, psi of x times the derivative of phi star of x, then you can simply shift the derivative from one function to the other, but it's at a cost. A cost? A minus sign, right? Good. So this is equal to minus the integral phi star of x d psi by dx. Whoops. We're in bad shape. This is integral phi star d psi dx. This is minus the integral of phi star d psi dx. They are not equal. In fact, worse than that, they're absolutely the opposite of each other. Opposite in the sense of the negative of each other. Can we fix it? Well, no, we can't fix this, but we can redefine an operator which is her mission. The redefined operator which is her mission is not derivative. It is minus i. The minus sign doesn't matter. We can put i or minus i d by dx. So let's go through it again and check whether now it's her mission. This is the new thing that we're going to call, and we're going to give it a different name. I'm not going to call it d because it's not derivative. That's minus i times the derivative. Anybody have a good suggestion for what I might call this? P, brilliant. P. P4? Momentum. Momentum. All right? And if I really wanted it to be momentum with units of momentum, what else would I put there? Where? Right or left? Right. Well, you can put it on the left, but then you'd better put it downstairs. All right, but we're not going to worry about each bar. We're going to use our usual convention of ignoring each bar for the moment. P is minus i d by dx. Let's see what happens. Over here, we have to replace this by an i over here. Minus i, thank you. Minus i. What about this one? If we have a complex conjugated everything, we'd better change the sign because the complex conjugate of minus i is plus i. And so, let's do it this way. Minus i, and we should get plus i here if it's her mission. Why? Because in the process of taking the complex conjugate, minus i gets converted to a minus i, and every minus i gets converted to a plus i. Is it true now? Yeah. Why? Because this is equal to minus this. That's what we proved over here. Okay? So if we stick an i here, because i changes sign on the complex conjugate, this little operation here tells us that minus i d by dx is a her mission operator. It's a little odd, isn't it, that minus i d by dx should somehow be a real thing, should represent a real thing, and that d by dx should not. But let's check something out. Let's check something out right now. Let's just, we're going to finish up in a moment. We haven't talked about expectation values yet tonight, but I think you can guess how to think about them. Expectation values in the state psi of an observable L are just the sandwich composed of psi on both sides and L in the middle. So let's take the expectation value of momentum. It's integral psi star of x, psi of x, and then according to the rule here, p is minus i d by dx. Is this real or imaginary? You're wrong. It's real. So let's prove that it's real. I hope it's real. God, I hope it's real. All right. So let's integrate by parts. All right. This is equal, first of all, let's take out the minus i. Minus i integral psi star of x d psi by dx. Now let's integrate that by parts. This is equal to minus i, and then we can switch the derivative. We can switch the derivative to the star variable, integral d psi star by dx times psi of x. But then in doing the integration by parts, we have to change the sign. So here we have a little theorem. The integral of this thing here with a minus i is the integral of this thing here with a plus i. What's the complex conjugate of this whole thing? The complex conjugate it, you replace psi star by psi. Well, psi is over here. You replace the psi by the x by the psi star by the x, and you replace minus i by plus i. This over here is the complex conjugate of this. Can everybody see that, or should I explain it further? Let's put a minus i out here and write that this is equal to d psi by the x. Minus i d psi by the x. This is clearly the complex, the integral, are complex conjugates of each other. The psi by the x is the complex conjugate of this. That one's a complex conjugate of this, and i and minus i are complex conjugates of each other. So, given that this is the complex conjugate of this and they're equal to each other, what do you say about a thing which is equal to its own complex conjugate? It's real. So we prove a rather startling fact that the integral of psi star of x minus i d psi by the x is real. No, yes, I did. And again, it's because when you integrate by parts, you pick up an extra minus sign. So instead of this thing being minus its complex conjugate, it's plus its complex conjugate. Okay, so what did we learn? We learned that this is a Hermitian operator. It's a Hermitian operator and therefore an observable in quantum mechanics. The next time we'll explore a little more. We'll discuss what this has to do with Fourier transforms. We'll discuss questions about what the probability for finding the particle at different locations are. But then we'll also talk about what the probability for having different momentum are. And then we'll move to the Schrodinger equation. We'll move to the evolution of the wave function. This, after all, now is really in the classic sense of the word a wave function. And we'll think about how it evolves with time. Oh, one other thing we could do with tonight. I think we're probably at the point of diminishing returns or you exhausted. We could do one more thing. Let me do one more thing for you. Let's calculate an interesting quantity and why it's an interesting quantity we'll come through in a moment. So let's take a quantity which seems very dull, but which is interesting and calculate it. It's the commutator of x with p. Take any wave function, whatever it is, psi of x, anyone. And we're going to do two things with it. We're first going to multiply it by x and then we're going to multiply it by minus i d by, or operate with minus i d by dx. This can be called px operating on psi. This is p, this is x. In the order first x then p. Then we're going to calculate it the other way. We're going to write xp on psi. What is that? On the other half of the blackboard over here, we're going to write x times minus i d by dx psi of x. Let's see what we get. In fact, what we're going to do is we're going to compute the commutator, which means we're going to calculate this one with a plus sign, we're going to calculate this one with a plus sign and this one with a minus sign. I don't know the minus sign, but before we subtract them, let's just calculate them. All right. This is just minus i. And now the derivative of a product. The derivative of a product. This derivative here now has to act on the whole result of x acting on psi of x. x acting on psi of x gives a new wave function and then p acts on it. So what do you get when you hit the product? You get two terms. The first term is the derivative of x times psi. The derivative of x is just 1. So the first term is just psi of x. And the second term is plus x d psi by dx. Everybody agree with that? The derivative hit x and left psi and then the derivative hit psi and left x. That's the side over here. Now what about this side over here? Here, the d by dx doesn't act on x, it just acts this way. It just acts on psi of x and gives us minus i d by dx psi of x. X, sorry, x minus i x d psi by dx. Now let's subtract them. Let's take this one with minus and this one with plus. This corresponds to taking the commutator of x with p acting on psi of x. The commutator of x with p is another operator and it acts on psi of x. What do we get when we subtract? Let's see, this one now has to come with a plus sign since we're subtracting. Since we're subtracting, we have to change sign here. And this term cancels. Yeah. This term cancels, goes away, and all we're left with is psi of x. That's all we're left with. This term cancels and we're just left with what is it? i psi of x. i psi of x. All right, so here's what we found. For any psi of x whatever, for any psi of x, the commutator of x with p acting on psi of x just gives, is it plus i? I think it's plus i psi of x. If an operator acting on any wave function gives another operator, if the operator xp acting on any wave function is the same as i acting on that same wave function, for any wave function whatever, it follows that this operator is equal to this operator. Another way of saying it is if xp minus i, put it all in a big bracket, when it hits any wave function at all, gives zero, then it means that this is zero. There's only one operator which gives zero when it acts on every state, on every vector. There's no vector which escapes when this thing hits it. The implication is that this thing is zero. There's only one operator which annihilates, which kills every vector and it's zero. xp minus, well, commutator xp is equal to i. If I put back the h bars, there would be an h bar here, i h bar. We've seen things like this before. You may have to go back to your classical mechanics. We've seen things before like this. We've seen Poisson brackets which had similar relationships. The Poisson bracket of x with p is equal to one. Earlier in this course, we discussed the potential relationship. We had reason to discuss why commutators. It had to do with Schrodinger's equation. Go back, but we discussed the relationship that there might be a relationship between commutators and Poisson brackets, and in particular, the relation involved this i h bar. This is actually where Dirac began his discussion of quantum mechanics. He said there must be something analogous to a Poisson bracket in quantum mechanics. The only thing he could think of which had the right algebraic structure, was anti-symmetric, it had a certain mathematical structure, was the commutator. He postulated that commutators and Poisson brackets were related with each other to an i h bar, and then proceeded to go exactly backwards from the way that we went. He said if this is the Poisson bracket, then this must be the commutator, and so there must be an operator p, whose commutator with x is i, and he went back through the operation and discovered that p was minus i d by dx. So he went in the opposite direction. We discovered p by just playing a game. He said how about this operated d by dx? What does it do? What we found is that it was not her mission, we had to multiply it by i to make it her mission. We explored it a little bit, found out some interesting properties of it, and the most interesting property is that its commutator is equal to i h bar, which suggests that it's connected with the Poisson bracket, the classical Poisson bracket being one. Of course the upshot of it is, and of course we need to do a lot more to convince ourselves that the operator minus i d by dx really behaves like quantum mechanical momentum. We can't just say, well, it looks like a Poisson bracket, it must be the quantum mechanical momentum. We have to show that wave packets move with velocities that are governed by this, that it's conserved, that it has all the properties of momentum, but we will. There will be some connection between this abstract operator minus i derivative of mass. And what? And the godney of mass. Well, we would like to show that in some way that it's the mass times the velocity. So what we'll eventually show is that if you have a wave packet, in other words, a psi of x, which is a concentrated distribution like this, that that wave packet moves along the x-axis with a velocity which is given by the expectation value of this momentum operated divided by the mass. That's what we would like to show, that the velocity with which this wave packet propagates along the axis, that velocity, you can call it the group velocity, it's usually called the group velocity, the velocity that the group of waves moves along times the mass of the particle is equal to the expectation value of the momentum. Once we do that, then I think we're, and that it's conserved, that'll take us a long way to calling it momentum. At some point, we have to say quantum mechanics is more fundamental in classical mechanics, and the definition of momentum should be through quantum mechanics, and then we should show it behaves in a certain way when particles are heavy and so forth. Okay. For more, please visit us at stanford.edu.
 Stanford University  was talking about what happens during or from beginning, from before a measurement to after a measurement.  They were talking about what happens to the evolution of an isolated black hole  left to itself between the time that it was prepared and the time it was measured.  Okay. Nevertheless, I could answer the question, is entanglement inevitably irreversible?  If two things which are not entangled come together, interact with each other, do whatever they do,  and become entangled, is it irreversible? Not necessarily.  Anything that can happen, the opposite can happen.  And now we're not talking about an external apparatus measuring a system.  We're simply talking, for example, about two spins.  Two spins interact with each other, come together, go off, and they're entangled.  What happens is it possible to disentangle them by, for example, reflecting them,  letting them come together and interact again, can they become disentangled?  Yes. Yes, there's no rule that says entanglement is irreversible.  On the other hand, if an entanglement happens and you take one of the electrons off and throw it away,  then the other one is stuck there, left being entangled with a thing that just is not going to come back.  Not going to come back unless, of course, somebody out there reflects it back in.  So yeah, entanglement is reversible.  But as far as you're concerned, the measurements that you do are irreversible.  That's a very peculiar situation, and you have to think about it to understand what it means.  We're going to move off the question of these very simple systems, spins, sometimes called qubits.  Tonight at some point, but I want to just go over just very briefly.  I think I probably discussed this in class last time, but nevertheless, it's so important that I want to hammer on it.  First of all, there was a notion of a density matrix.  If we have two systems, Alice's system and Bob's system,  and Alice and Bob's system are far away from each other so that Alice doesn't get to do any experiments on Bob's system,  Bob doesn't get to do any experiments on Alice's system, but they're entangled, either they are or they are not entangled,  they may be entangled, is there a description that Alice by herself uses to describe just her subsystem?  The answer is yes, and the working description of her subsystem is called the density matrix.  So let me just review that very, very quickly, and go back over some concepts.  If the wave function of the Alice-Bob system, the combined system of Alice's subsystem and Bob's,  if the wave function is called psi of A and B, B for Bob, A for Alice,  then Alice's density matrix is defined to be psi star, the complex conjugate of the wave function.  Now let's say it's rho, called the density matrix, it's indicated by rho.  It has subscripts A and A prime because it's a matrix, but it doesn't have anything to do with Bob,  so it's purely a matrix with respect to Alice's variables.  Let's indicate that by calling this Alice's density matrix.  And it's given by psi star of A, B, A prime B, I always get confused, yeah, A prime, psi of A, B summed over Bob's variables.  It doesn't depend on Bob's variables because they've been summed over.  When you sum over something, it no longer depends on that set of variables,  so this is only a function of Alice's variables, and it contains all possible statistical information  that Alice can ever know about her half of a system unless Bob comes and interacts with her system.  Of course, if Bob comes from wherever he is and gives her a system a shot of some sort,  that will change her density matrix, but as long as Bob stays away from her system,  Bob's and Palo Alto, Alice's on Alpha Centauri, as long as Bob does not interact directly with her system,  her density matrix doesn't change no matter what Bob does.  Let me give a little proof of that.  Let's take Bob's subsystem to not be so narrow as just to be the little spin that he's carrying around,  and also Alice's. Let's imagine that Alice's subsystem consists of everything in Alice's neighborhood,  everything that she can have any contact with, and everything in Bob's neighborhood, maybe even including Bob,  and now Bob is going to do something.  He's going to do something, interact, of course he can only interact, interact may mean he may turn on a magnetic field,  he may do all sorts of things to change the evolution of his subsystem.  He's going to make a subsystem evolve in some way.  One of the ways which Bob can cause his subsystem to evolve, many, many ways,  but whatever they are, they are always governed by unitary matrices.  Let's come over here and forget Alice for a minute and ask what Bob can do to his subsystem.  He has a wave function, let's leave Alice out of the picture altogether, just let's suppose that Bob has a wave function,  and he's going to manipulate the system somehow, he's going to manipulate the system somehow,  and send it to a new wave function.  The rule is the new wave function of his subsystem must be related to the old wave function by a unitary operator.  Unitary operators are the things which govern time evolution.  Go back, remember about time evolution, unitary operators, and that means that psi of B becomes replaced by some over B,  some unitary matrix, BB prime, psi of B prime.  In terms of wave functions and matrices, this is the description of what happens to Bob's matrix,  sorry, Bob's column vector, this is basically Bob's column vector,  describing his half of the world or his system, and to find out what happens to it no matter what he does,  even if he doesn't do anything, but let's suppose he does something,  it generates a unitary evolution of his wave function.  Now, how do we describe the same thing in the composite language?  In the composite language, if the wave function is psi of A and B, Alice's and Bob's variable,  what happens to it when Bob manipulates his half of the world?  Exactly the same thing with Alice's variables being just spectators, just being passive spectators.  So psi of A and B just becomes psi of, whoops, let's put in there, B prime, this is sum of B prime.  This is a function of B and you sum over B prime.  Okay, that's again a sum over B prime of U of B and B prime, psi of A and B prime.  A just passed through. Alice knew nothing whatever about this whole process.  Her subsystem was not in any way influenced, her index, or her part of the wave function,  or half of the dependence of the wave function, was uninfluenced by Bob's manipulation.  So let's go back and take the manipulated wave function, here's the manipulated wave function,  let's call it psi manipulated, the manipulated wave function of B and A, of A and B, and that's this.  It's been modified by the action of Bob's unitary operator.  Okay, now let's recalculate Alice's density matrix.  After all, Alice's density matrix contains all information that she can ever know about her half of the world.  And so let's see what happened to it.  Well, we first of all have to take the manipulated psi star, we still have sum over B,  and now we take the manipulated psi star, oh, we haven't written psi star, sorry about that,  we have to write psi star, manipulated of, and we want to know about it with index A prime and B, A prime and B,  and that we get just by complex conjugating.  This tells us how psi changes, let's just work out the complex conjugate.  The complex conjugate of A prime.  Where is it A and B prime?  This is A prime and B, just A prime and B.  But here, since B prime is a summation variable, we do not want to write B prime here.  B prime was the summation variable used over here, let's just invent another one called B double prime.  That's going to be a summation variable in a moment.  And U complex conjugate of B and B double prime.  Yeah, U complex conjugate, and of course there's a sum over B double prime here.  And the only reason I call it B double prime instead of B prime,  was so as not to confuse it with B prime. Here's one sum and here's a separate sum.  Now, here's something, what do we know about unitary operators?  Unitary operators are operators that if we interchange B, if we interchange the two indices,  and complex conjugate, what happens? That's called the inverse operator, or the Hermitian conjugate.  In general, it's called the Hermitian conjugate.  Specifically for a unitary, it becomes the inverse operator.  Now, let's just call it the Hermitian conjugate.  So, if we interchange the two indices, the Hermitian conjugate,  and the B double prime, B, then this becomes Hermitian conjugate.  So, this is the usual story.  If operators operate to the right on ket vectors,  the corresponding action on bra vectors is given by the Hermitian conjugate.  So, now what we have to do is plug this in.  We plug this in for the manipulated psi star, that's summation over B double prime,  psi star of A prime and B double prime times U dagger of B double prime and B.  That's a no-sense.  Too many indices for me to be comfortable on the blackboard,  but I think this is worth doing.  Now, what about psi?  Psi is the other half that comes into the density matrix,  and that's U BB prime psi of what?  Psi of A and B prime.  Complicated little mess.  Oh, what are we summing over?  We're summing over all the B's basically.  We're summing over this B over here, let's put a B summation here.  We're summing over B double prime to get the manipulated wave form of this complex conjugate,  but we also have to sum over B prime here.  Sum over all the B's in sight.  Each B is a summation variable by the time we're finished.  All right, but now look at what we have.  It looks complicated, but what is this thing over here?  Well, not quite, yeah, one is essentially correct,  but of course what you mean is the unit matrix.  This is the product matrix of U dagger.  This is a unitary matrix, so U dagger is U's inverse.  B double prime B, BB prime, this is just the chronicle delta.  B double prime, B prime.  Let's instruct you to set B double prime equal to B prime when you do the sum,  when you do one of the two sums, let's say the B double prime sum.  So all of this then becomes sum over B prime, well, let's see, what have we got?  We've set B double prime, yeah, we've done the sum over B, that's done.  No sum over B, we just did it.  And now we set B double prime equal to B prime.  Psi star of A prime, B prime, psi of A and B prime.  That's it.  Now it doesn't matter whether we call B prime, B prime, or whether we call it B,  because it's a summation variable, so let's do that, let's just call it B.  Now what do we have? We have exactly, I hope, I pray, the original density matrix.  So Bob, no matter what he does, as long as what he does is described by a unitary matrix,  Bob will have absolutely no effect on Alice's density matrix.  It does not matter whether they're entangled or not.  Now who wears that I specify that this was a product state?  That is what physicists call locality, the inability of Bob to influence any statistical prediction of Alice's  by doing a manipulation on his own degrees of freedom.  Locality is not violated, this form of locality, this form of locality is not violated in any way by the phenomenon of entanglement.  That's the technical argument.  Notice that it does depend on unitarity.  So what did unitarity do with?  Unitarity had to do with conservation of distinctions.  What I call the minus first law, at least in its quantum mechanical version.  So there's a funny interesting interplay, funny interesting interplay between the conservation of distinctions on the one hand and locality.  If Bob somehow had access to changes which were not described by unitary operators,  this would not have become the unit matrix over here and the final density matrix would, Alice's final density matrix indeed would have been modified.  This is interesting the way the whole thing holds together.  This is why it's so hard to change quantum mechanics.  You want to play a little game?  Let's let Bob, let's let Bob's evolution not quite be unitary, only approximately unitary.  Well, then immediately you start sending signals faster than the speed of light.  How?  By instantly modifying Alice's density matrix.  So things hold together tightly and it's hard to change them.  Good, so yes?  Is it correct to say that Bob's matrix is unitary?  The evolution matrix is a vector?  Yeah.  You can't do it, it's not really an infectious system.  He's not able to change Alice's system.  He's changed his own system by doing something and that doing something was described by this unitary matrix.  Okay, that's the part that confuses me.  If he's doing something, isn't he going to be violating unitary?  No, for example, he might be turning a magnet.  He might be turning a magnet in such a way that it has an influence on his spin.  What is that influence?  It will change the state of the spin.  For example, yeah.  If you have a spin in a given direction and it's coupled to a magnetic field and you turn the magnetic field in some way,  it will typically change the evolution of that spin.  So Bob can do that.  He can change a magnetic field that's acting on a certain spin.  The result will be to change the way the spin evolves.  That will change his own measurements of the spin, of his spin.  If he does act by rotating a magnetic field, which is mathematically described by some sort of evolution,  doesn't matter what, some sort of evolution that's described by a unitary operator,  then he does change the results of his own measurements.  For example, he rotates the magnetic field but he doesn't rotate his apparatus.  The result is that he's going to change the probability distributions of his own spin,  but he will not change any of the probabilities of Alice's spin.  Okay.  So are you saying quantum mechanics is always a basal calving?  Or is the probability defined by the inability of Bob to influence Alice's measurements?  Yeah, now the question is, so what's the big deal?  Is that what you're going to ask?  Yeah.  I was guessing what you were going to ask.  Why don't you ask it?  Single particle wave function, put a photon through a beam splitter.  It goes off in two different directions.  You catch it at one detector faster than the speed of light, the wave function collapses.  Whoa, whoa, whoa, whoa, what?  You do what?  You detect a photon at one detector.  The wave function at the other detector collapses to zero, instantaneously.  I showed you what happens.  The density matrix of the other system does not change, period.  It did average measurements, still changed.  What happened?  I know, I know, I know, people, look, I'm doing my absolute best to tell it to you as it really is  and to cut through the crap that's out there.  To cut through the crap that's out there, because there is a huge amount of, there is a pile a mile high.  And we somehow want to cut through that.  This is the mathematics of cutting through it.  Now, nevertheless, there was some pretty damn smart people who are uncomfortable about something,  something that they call nonlocality.  I mean, you know, Einstein was no jerk, he was pretty smart.  And they didn't call him Einstein for nothing.  That's my favorite joke.  John Bell was pretty smart.  They didn't call him Einstein, though.  Yeah.  So there's something going on.  There's something going on.  Yeah, go ahead.  I want to get, I'm reviewing last week, so make it quick.  To me, this is real clear, I think, and that is your idea of locality means that the states don't affect each other.  In other words, there's two things.  The density.  Yeah.  And it seems that in the confusing part of quantum mechanics is that those two different things get mixed up somehow.  Is that accurate or?  No, it's fuzzy.  But it's true.  So I will try to tell you as I understand it what the concrete operational facts are.  So I told you last time, I was a little bit tired last time, and I'm not sure I was completely clear.  So I won't go over it again because it's very, very interesting.  And it really is at the heart of all of the confusions about confusions.  I mean, I must say, including my own internal confusions about quantum mechanics.  When I first learned about, er, Einstein, Rosen, Podolski, that was the discovery.  Well, I don't know if it was a discovery.  I think maybe Schrodinger said some things about entanglement.  But it was Einstein, Rosen, and Podolski who really called attention to how peculiar it was.  Basically what they said is we have a very funny situation here where although we can know everything that can possibly be known about a composite system,  we know nothing at all about its constituents.  That's crazy.  That violates my sense of reality.  That violates my sense of, er, of what it means to know what a system is doing.  Violates all sorts of things.  I remember when I first heard about it, it was actually before John Bell sometime in the 60s when I was a student.  No, no, I wasn't a student. I was a young assistant professor.  And I started to think about it and I thought about it a lot and tried to figure out what it means.  I was sort of inclined to try to think about it in terms of simulating quantum mechanics.  And I still am.  What are the limits?  If, if, if quantum mechanics is so very strange and so very different than classical physics,  does it mean that you can't fool somebody into thinking a system is quantum mechanical if all you have, your resources are entirely classical?  And the answer is yes and no.  So I know I talked about this last time, but it really is at the center of things.  So let's go back over it quick.  Quick as I can.  We're going to build ourselves a quantum simulator.  I think somebody built one on the, on the site.  I came across it about an hour ago.  Who, who's that?  Yeah. Does it work?  Well, it doesn't actually simulate yet.  Not yet.  But is that picture, is that picture your picture?  It was a picture of Einstein and Heisenberg and Einstein says I don't like this and Heisenberg said what?  Einstein says I think I'm not going to like this.  Yeah.  Right.  Oh.  Yes.  Right.  I think that captures it.  Why go on?  But okay, let's, let's talk about the nature of quantum simulators.  First of all, I just want to simulate a single spin.  I think I told you how to do it, but let's write down the steps.  We have a computer screen.  Here's our computer screen.  On the screen, we have a picture of a detector, not a detector, an apparatus.  And the apparatus can be oriented in any direction in three-dimensional space.  We need to do some three-dimensional graphics here.  And it has a window, and the window shows one of three possible states.  One is called blank B.  That's the detector or the apparatus before it's interacted with anything.  So it can be B or it can be plus or minus one.  B is not important.  It's just to say before the detector interacts with anything.  There's a button over here called M.  And the operator of the system, of the computer game,  basically all you ever get to do is press M,  and then look at the result of the M stands for measure.  It performs a measurement.  Now, over here connected to the computer by wires, by cables,  connected to the screen by cables, is a computer.  And the computer stores information in its memory,  and the basic information that it stores is two complex numbers,  alpha up and alpha down.  Those are supposed to be the components of the wave function of the spin.  I didn't say yet that there was a spin here, but that's what we're talking about.  Alpha up and alpha down.  The computer only can generate alphas which satisfy alpha star, alpha up,  plus alpha star, alpha down is equal to one.  Whenever it generates them.  And it can do something else, it can do two other things.  It can update the alphas by solving the Schrodinger equation for them.  There's some Hamiltonian, for example,  it could be that the spin is in an imaginary magnetic field,  in which case the alphas would change with time.  The computer knows how to do that.  It knows how to solve Schrodinger's equation.  Computers can solve Schrodinger's equation beautifully.  And it updates these as a function of time.  Updates them according to the rules of quantum mechanics.  One other thing, the cables are instantaneous.  They convey information instantly.  So we'll have one more element.  The other element will be a random number generator.  Let me just say some words about random number generators.  Random number generators,  sounds like we're introducing quantum mechanics.  No, we don't need quantum mechanics.  You can simulate to a very, very high degree of confidence  a sequence of random numbers or a random number generator  purely out of classical considerations.  You know how to do it.  You take the digits of pi or whatever  and you make something that looks, would fool anybody,  including a sophisticated mathematician,  into believing that you were generating a set of random numbers.  So there's a random number generator inside the computer.  All right, the operator starts by initializing the system.  And initializing the system simply means starting out  some orientation for the detector, for the apparatus,  doesn't matter what, and setting the alphas  to some initial values.  You can plug in what he wants the initial values to be,  or the computer just sets them according to whatever rule it likes.  Then the system runs for a period of time,  an indeterminate period of time, updating the alphas  until the operator decides to press M.  A certain amount of time is elapsed,  the operator decides how much time and presses M.  At that point, the random number generator  generates a plus one or a minus one,  but with probability alpha star alpha up,  that it's plus one, sorry, alpha up,  alpha up star, that's plus one,  or else alpha star alpha down, that it's minus one.  Remember, the sum of these two alpha star alphas add up to one.  So armed with the random number generator,  the computer sends a signal with a probability  of the appropriate probability to the detector,  and the detector detects or registers  what the computer sent it.  That's one measurement.  So as soon as that happens, the computer re-initializes the alphas.  How does it re-initialize the alphas?  If the detector detected plus one,  then it sends back a message,  it doesn't really need to send back a message,  after all, this message got from here to here,  so the computer over here knows  whether it got plus one or minus one.  If the detector saw plus one,  then it sets alpha u back to one  and sends alpha d to zero.  In other words, it does what we call collapsing the wave function.  According to the result,  or according to the outcome of the measurement,  it re-initializes the alphas.  So whatever was measured,  you start with probability one for the outcome,  and probability zero for the, what shall we call it,  the anti-outcome.  And then it runs again.  Solves a Schrodinger equation, does it again.  The operator makes a ramp,  makes whatever decision he likes.  I mean, what makes him decide to press M?  He's an experimental physicist.  What makes experimental physicists do what they do?  I don't really know,  but he does what experimental physicists do.  Now and then he presses M  and does the whole operation over again.  Oh, sorry.  Before he does,  he may decide to rotate the apparatus.  Now, the computer in here  knows what the orientation of the apparatus is.  So when it calculates,  it not only calculates the updated alphas,  but it also calculates the probabilities  for the rotated detector  to detect whatever it's going to detect.  That's all algebra and a little bit of,  that's purely algebra.  The computer makes the kind of calculations that you did,  presumably did,  makes the kind of calculations  to calculate what the probability would be  that if the detector was rotated  in some arbitrary direction,  then what's the probability that it will get up or down?  Now, who is it that decides to rotate the computer?  Well, it's the operator again,  the experimentalist.  The experimentalist not only can press M,  but he can also reorient the detector  any time he wants.  This is about as much as an experimental physicist  can do with that spin.  Rotate the apparatus, measure.  Rotate the apparatus again, measure.  And what's the result of the outcome of this?  The outcome of this is indistinguishable  from what the experimental physicist  examining the spin would see.  So we've simulated quantum mechanics.  Now we want to know whether we can simulate  quantum mechanics of two spins,  one on Alice's side of the world over here,  and one on Bob's side of the world.  Yeah?  I don't understand the purpose of the button M in ours.  In other words, when he decides to put it,  some amount of time is involved.  Has that affected anything going on in the computer?  Yeah, the computer has been updating all this time.  Right, but when we talked about our base model  of how spin works,  and how it was funded, and I come into it at all.  We did talk about Hamiltonians.  We talked about Hamiltonians  and how Hamiltonians update the state of the system.  You're right, we didn't discuss how...  We didn't discuss much about Hamiltonians for spins.  Yes, we did.  Yes, we did.  I believe I described a spin and a magnetic field  and how the average spin precesses.  We didn't have the magnetic field,  because that's how we started out talking about all this stuff, right?  Then time should not be a factor.  Okay, I understand that.  Right, now there's no real magnetic field.  There's just an imaginary magnetic field  that the computer knows about.  And whereas I come into it,  it comes into the nature of the exact Schrodinger equation.  Okay.  Now, we do this over here  and precisely the same thing over here,  except instead of alphas, we have betas.  Beta stands for Bob.  Exactly the same setup, replicated twice.  Bob does his measurement, Alice does her measurements.  They're perfectly happy.  Each one says, I have a spin.  It's quantum mechanical.  They don't talk to each other.  And each one says, my spin is in a state.  The language that we would describe this by quantum mechanically  is they have simulated a product state.  A product state in which Bob's...  Sorry, Alice's wave function is this,  and Bob's wave function is this,  and the combined system is in a product state.  Supposing now we want to simulate any state,  a more complete description that includes entangled states.  Well, we can do it this way.  Let's start with a way that's sort of guaranteed to work.  We take the computer and we put it in the middle,  a single computer.  We can think of the single computer if we like  as a pair of computers connected by wires,  but just by the fact that they're communicating with each other  and interacting with each other really makes them a single computer.  And in that single computer we now store four complex numbers.  The four complex numbers that we store in there are  alpha up, up, alpha up, down, alpha down, up, alpha down, down.  These complex numbers here are not necessarily products.  In fact, let's assume they're not products.  In other words, this is not a product state.  Let's assume it's a highly entangled state.  Even a highly entangled state is described by four complex numbers like this.  That's what's in the computer.  And what does a computer do with it?  It evolves the Schrodinger equation.  First of all, we initialize.  We start with some alphas.  The alphas are allowed to run according to whatever  Schrodinger equation describes them.  And at the end, either Bob or Alice or both,  without knowing what the other one is doing,  press or as the case may be, don't press their measurement buttons.  Once the measurement button has been pressed,  an instantaneous signal goes back to the computer.  The computer grinds out its random number generator,  and whichever side was measured sends a signal to the detector  with appropriate probability, either plus one or minus one,  and then reinitializes itself,  reinitializes itself in a way that doesn't touch, for example,  if Alice does the experiment,  then it doesn't touch Bob's end of things.  If Bob does the experiment,  the computer reinitializes his half of the wave function,  which is another way of saying it throws away the piece  that you didn't measure,  throws away the pieces you didn't measure  and renormalizes the wave function  so that the total probability is warning.  Exactly the same thing.  There's no real difference with a single system  you can simulate it, and you can simulate it in a way  which would describe all the possible things  that Alice and Bob could measure,  whether or not the system was entangled.  Where do we get in trouble?  We get in trouble if after initializing,  we try to separate the computers and break the wires.  For example, here's what we might do.  Given these alphas, we might put them into both computers.  Take the set of alphas that we generated,  put this one into this computer,  this one into this computer.  Now, since the computers are not going to talk to each other,  they have to have separate random number generators.  The random number generators will not talk to each other.  Each one contains the full entangled state,  and then we go through the same operation.  What's going to happen?  The computer is simply going to forget,  for example, for a highly entangled state,  let's say the entangled state is the singlet state,  then the right answer would be that if Bob and Alice both measure,  they should get opposite values for any spin.  If they measure the x component of spin,  they should get opposite values.  If they measure y components of spin, they should get opposite values.  But if when we separate them,  they have separate random number generators,  and the random number generators don't talk to each other,  then guess what's going to happen?  There's going to be no correlation of any kind  between what Bob measures and what Alice measures.  No, Bob may measure plus,  and Alice may also measure plus,  because their random number generators are not connected to each other.  And so even though both computers may know about the full state of the system,  if the random number generators don't talk to each other,  then the results of the experiments will not contain the correlation.  The correlation that Bob and Alice could later come together and report,  Bob would say,  my first experiments are a plus.  If things were working right,  then Alice should say,  my first experiments are a minus.  And so forth and so on,  that correlation will not be there once you try to separate the computers.  Are the random numbers correlated if you don't separate the computers?  You can have a single random number generator  that sends out signals to both computers.  If it sends out a signal plus one to this computer,  it sends out a signal minus one to that computer.  So all you need is a single random number generator,  but it has to be able to talk to both computers.  So there's got to be wires in the world.  There's got to be wires in the world connecting this side and this side  for no other reason than they can both connect to the same random number generator.  You've got four outputs there,  but it sounds like you just described that the outputs are independent.  Now you've got to think about it.  Yeah, yeah, yeah, yeah.  Yeah.  Yeah.  Well, that's sort of right.  Once you've separated them,  then what happens next after some measurements are made,  doesn't preserve the relationship between them.  They become disconnected from each other.  Yeah, so think about it a little bit.  I mean, even better.  Try to simulate it.  This is this, I've never tried,  but I don't think it should be that hard  if you can do a little bit of computer programming.  I've simulated much harder things than that  when my son was a little kid.  We had this thing called BASIC.  Anybody remember BASIC?  And he and I designed the basketball game,  and it had friction, and it had,  it was much harder than this.  So that was my last experience of programming anything.  I think you'll learn a lot by trying to simulate this.  You have a detector called a cell phone,  and I get a signal from Alice or Bob about some random event.  It could be a spin, let's suppose,  but we're, will you please speak concretely?  We get a signal from Bob of some random event.  What does it mean?  Can you please be exact?  Very exact.  I'm getting two text messages,  one from A, one from B, okay,  and they're going to have a value of one or zero, each of them.  Okay?  And I get a million of these things,  and I noticed the distribution that the A signal  had a 50% chance of being one or zero,  same with the B signal, okay?  It could be from a spin.  Yes, yes, okay, go ahead.  Now, I look at the correlation that turns out  that whether it's A or B,  the other one is the same.  In other words, it's correlated, a correlation of one.  That's in the real world of entangled states.  This could be, we don't know where it's coming from.  I'm just getting information.  I get information to A comes through,  I don't look at it, B comes through.  I know it's a 50-50 chance, I don't look at it,  but as soon as I look at A, I know what B is,  so the wave function completely changes.  Now, the example I'm using is not a state,  I'm just a Bob and Alice or sexting twins,  one's a boy, one's a girl.  You've got to hurry up  because I've got to get on to particles.  So there is nothing magical about that.  There is nothing weird about that.  If it's correlated like that, it's very probable  they're identical twins instead of eternal twins.  So what, how am I missing?  What is weird about that?  If you don't find it weird, good.  What's weird is that you can't simulate it classically  without having a central processor connected by wires.  So what's weird about it?  So don't those wires that connect the two distant computers  represent non-locality?  Oh, okay, now we ask.  All right, those wires are there.  For goodness sakes, surely Bob can send a message  through the wires to Alice.  It takes no time at all because they're instantaneous.  Yeah, true enough if this were a real computer setup,  but if Bob is restricted to only do the things  quantum mechanics allows him to do,  if he is restricted by the rules of quantum mechanics,  then we find out amazingly that Bob can never send  a message through the wires that violates locality.  Why?  Because all Bob can do is manipulate his own system  with unitary operators  and so that's what's curious here.  Does that mean that the world is filled up with wires  that we can't see?  I don't think that's the message.  I think the right message, but of course here you can hear  with this room for debate what the right message  of all of this is.  The message I take away from it is basically quantum mechanics  cannot be simulated by a classical system.  The logic is too different.  The two theories are too different.  I don't take away the message that the world is full of wires  that if only we could access them,  we could send messages back and forth.  Other people do, but I think at least if we know how  to do quantum mechanics, we'll agree on the facts.  We may disagree on basically whether there are or are not wires there.  Are there really wires there or are there really not wires there?  Well, perhaps the culprit is the word really.  But this is, oh, there is one other element to it,  which is quite interesting.  We've made an assumption that Bob and Alice get to kind of  randomly make decisions about when to press the button,  about what orientations, about what to decide to measure.  Bob sits here and says, hmm, okay, Northwest, hmm, Southeast,  or whatever, makes these random decisions according to the principle  of free will, so to speak.  That's fine as long as Alice and Bob are outside the system  and not part of it.  We don't need to ask the question of consciousness  and the question of free will.  We'll just admit that Alice and Bob can make random decisions  about when to measure various things.  It's obviously a pretty good approximation.  But once we take Alice and Bob and put them into the system  so that they are governed by the laws of physics,  then things get a little bit touchy.  What do we mean by saying that Alice can decide to do this  or Bob can decide to do that?  And that, I think, is where the possible loopholes  in all of this can come from.  Now, I don't want to get too deep into heavy philosophical  discussions of all sorts of psychological questions of  consciousness.  I don't know the answers.  That's it.  I don't know the answers beyond what I've told you.  I am certain of two things about it.  I am certain that there are a lot of people who think  they do know the answers, and I'm also certain that they don't.  That's it.  A question.  Actually, not a question.  If you Google for Quantum Randy Challenge,  you'll find a blog entry that has the setup for this simulation.  Really?  For the entangled case?  It doesn't solve it.  It says, if you can, here's a program.  Your challenge is to fill in this blank place.  I see.  If you can, he'll give you a million dollars.  Actually, he won't give you a million dollars.  Somebody in Sweden will give you a million dollars.  Yeah.  OK, that's very interesting.  I'm not the first one to think about it this way,  but I've been telling this to people for 50 years.  No.  I don't know.  I probably wasn't.  I think maybe Bohr thought about it this way.  Who knows?  It doesn't matter.  We could spend the rest of eternity talking about spins.  Why?  Because they really are interesting.  But we want to get to something else.  We want to get to particles.  We want to get to more complex systems.  The most complex system we've discussed is two spins,  four-dimensional Hilbert space.  We've gotten about to the point where in classical physics,  we got through the first lecture.  The first lecture was about these simple discrete systems.  And then we rather immediately moved on to continuous motion  of things like particles moving continuously with an infinite,  not just an infinite, but a continuously infinite range  of possibilities.  That's where we have to go now.  Now fortunately, we've set up so much of the apparatus  of quantum mechanics that we can do it fairly quickly.  There's a little bit of subtlety about replacing sums  by integrals.  That's about it.  That's about the nature of the subtlety.  Summs and integrals, we've expressed the general principles  of quantum mechanics and then exhibited it in a variety  of simple spin system examples.  Now we'll plunge right into quantum gravity.  No.  We'll plunge right into the problem of a single particle  moving on a one-dimensional axis.  We'll take the axis to be infinite, although that's mostly  because they don't have time to be more careful.  If we want to get anywhere, we're going to have to be slightly  less mathematically rigorous.  So we have, and in order to be rigorous, you really have to  contain a system in a finite volume and so forth,  but we don't need to do that.  So we have a line, infinite line.  Along that infinite line, one particle can exist.  We won't do two particles yet.  One particle can exist in one dimension,  and that line is labeled by a coordinate x.  I mean, not right down the coordinate x,  but also the states of the particle are also labeled  by the position of a particle.  Now here we are departing from classical logic.  In classical logic, the states of a particle are labeled  by a position and a momentum.  If we were talking about the states of a spin,  we would say the state of a spin is described by all  of the, or maybe at least two of the components of spin.  But we can't do that in quantum mechanics.  We can't specify two components simultaneously.  Same in classical mechanics, specifying, as I'm sure you know,  specifying the position and the momentum is too much.  And we'll see how that works.  But for the moment, let's take it as a postulate that a  complete description of the state of a particle,  and you can't get any more out, is, for example,  to know that it's located at a position x,  and if it's located at a position x,  we'll describe that by a ket vector.  The ket vector x.  That's the state of a particle when it's located at position x.  Now, of course, we can consider more complicated,  more complicated states.  Linear superpositions of these in quantum mechanics,  we are always allowed to make linear superpositions.  And so the space of states is bigger than just the space of  particles located at definite positions.  Let's call the general state psi.  This is the general state of a particle.  It doesn't mean anything now.  It's just an abstract symbol.  And it's a vector in some vector space.  The vector space now has a continuously infinite number  of basis vectors.  Now, strictly speaking, that's a mathematically bad thing,  but we're going to play that game.  If you want to understand more deeply the mathematics of quantum  mechanics, you go and you learn about Hilbert space.  But we're not going to do that tonight.  So the general state of a particle is called psi.  Remember what the wave function of a system is.  The wave function of a system, in a particular basis,  is the projection of a state onto the basis vectors.  And we call that psi of A. Now we come to a particle.  The basis of states are the position states.  And what is the wave function?  The wave function is the inner product of the state vector with  the eigenvectors of position.  And that's called psi of X.  That's the wave function.  That's the classical, not classical,  but that's the standard quantum mechanical wave function of a  particle.  It plays the same role as in the simple systems we've discussed,  the inner product of the state vector with A.  Now we can say a little more.  In this context, the probability that the system,  whatever it is, exhibits property A is given by this thing  squared.  In fact, it's given by psi star of A, psi of A.  That's the probability that the system is,  but the outcome of an experiment to determine whether the system  had property A, that's the probability.  Likewise, in exact parallel, the probability to find the particle  at X is equal to psi star of X times psi of X.  I really haven't done anything that we haven't done before.  Now, in quantum mechanics, like in any other sensible statistical  theory, the sum of all probabilities should add up to one.  So the sum of psi star psi should add up to one.  But when probabilities refer to a continuous variable,  we don't talk about the probability that the particle is located  exactly at point X.  That's too sharp.  The probability that a thing is located exactly at any given X  is zero.  We talk about the probability that it's located within a range  of X.  Or better yet, we talk about the density of probability.  We talk about the probability density,  and we integrate this over a little range of X to find the  probability that you're in that range of X.  We do something like that here.  If we want to know not whether the system is definitely specified  at A, but we want to know whether it's at a range of A,  we sum over them.  But here we sort of have to do it.  Total probability equal to one?  That's the condition that the integral of psi star of X,  psi of X, dX is equal to one.  Total probability of the system has to add up to one.  And the only, as I said, the only new thing here is that we  have to replace sums by integrals.  And that does make mathematical subtlety that we have to worry  about, but it's sort of in the mathematical details.  Okay, what about the inner product between two state vectors,  one at X and one at X prime?  Well, here we go back for inspiration, right back to,  right back to this case.  Zero if A is not equal to A prime and one if A is equal to A  prime.  And we're not going to be able to quite get away with that here.  We're not going to quite be able to get away with a statement  that X, X prime is equal to zero or one.  We are going to say that if X is not equal to X prime,  then it's zero.  But we're not going to say that when it's equal to X prime,  that it's one, that would not leave us anywhere.  What we're going to say is that it's a function which vanishes  unless equals X, X equals X prime, but has area one under it.  We'll see how that works out.  We'll see what that does for us in a moment.  What it really comes down to is a different normalization here.  But again, the basic idea that states are orthogonal,  if X is not equal to X prime, that we will retain.  But the numerical value of what we get when X is equal to X prime,  we're going to take to be slightly different.  And in fact, we're going to invent a function that we are,  derac did, the derac delta function, X minus X prime.  The derac delta function is not a real mathematical function,  but by now, I think mathematicians have defined it to death  so that it is a real mathematical entity.  But for our purposes, we can think of it in a very simple way.  It's a function which is zero, well, let's take delta of X,  not delta of X minus X prime, this is basically delta of X minus zero,  X prime zero, the function delta of X.  The function which is zero unless X is equal to zero,  that's the same as saying this thing is zero unless X is equal to X prime.  So if this is X, it's zero everywhere except X equals X prime,  and X equals X prime, it's a very high narrow spike.  How high and narrow?  High enough and narrow enough that the area under it is one unit.  That means if it's approximated by a narrow interval of size epsilon,  that it has height one over epsilon.  And then you imagine a limit where you take epsilon smaller and smaller,  one over epsilon higher and higher, always keeping the area under it equal to one.  That's the derac delta function, derac delta function,  zero everywhere except when X equals X prime,  infinite in some sense when X is equal to X prime,  but in such a way that the area under it is equal to one.  All right, now let's use this to further elaborate on what the meaning of psi of X is.  I'm going to guess and then we're going to check.  I'm going to guess that the meaning of psi of X is the following,  that if I want to write a general state vector, psi,  that I can think of it as a sum over the basis vectors,  but not a sum, an integral.  Of course an integral because the basis vectors are continuous.  The X of some coefficients, some coefficients, well let's know,  let's write it this way, psi is equal to an integral  of the X of some set of coefficients.  I'm going to guess what they are.  I'm going to guess that the psi of X times the,  the state, the quantum state where the particle is located at X,  but then we're going to add them up for different X's.  It's exactly the same thing that we did for ordinary quantum systems  except this was replaced by a sum.  All right, now I've given you two distinct definitions of psi of X.  One is that it's the inner product of the state vector with the eigenvector X,  I'll call it the eigenvector of X.  The other is that it's the expansion coefficients in writing psi  as a sum over vectors.  Now if you're a little bit uncomfortable with the idea of a continuous basis  of vectors that you integrate over, you're in good company.  The mathematics of this is, as I said, by now it has been made rigorous.  But it's, it's, it's, it took pages and pages and pages and pages  and books and books and books on something called distribution theory  to make sense out of this with precision, but it's pretty intuitive.  It's pretty intuitive and we'll stick with that.  All right, so let's calculate, oh let's put a prime here.  This is a sum, an integral.  It doesn't matter, it doesn't matter what you call the integration variable.  Let's call it X prime.  And now let's calculate the inner product of X with psi.  As I said, there were two different definitions of psi of X.  Let's check that they're the same.  So here we have it.  Integral the X prime, psi of X prime, inner product of X with X prime.  What's the inner product of X with X prime?  I think I wrote it someplace, where is it?  I think I erased it.  That's equal to the inner product, to the integral of the X prime,  psi of X prime, delta of X minus X prime.  Okay, what is that?  All right, first of all, delta is a function which is zero everywhere  except that X equals X prime.  So that means that in doing this integral here,  you'll only get a contribution when X prime is equal to X.  That means the answer is going to come out to be proportional to psi of X.  Furthermore, the area under this delta function is exactly one.  If we imagine that this delta function is a pinched, narrow, or narrow, or high spike,  so that the area under it is one, then this integral would tend to psi of X.  The rule is, if you have an integral over a delta function like this,  it simply picks out the value of the integration variable  where the argument of the delta function vanishes, where X minus X prime vanishes.  That's where the function is not zero.  And the integral of a function with a delta function just gives you the value of the function at the point X.  That's a rule.  We might as well write that down.  Let's write that down as a rule.  Integral delta of X minus X prime, F of X prime is always equal to F of X.  It picks out the value X prime equal to X.  That's a part of the definition of the delta function.  Okay, so it's just equal to psi of X, and we find consistency that the thing that we called psi of X over here  is the same as the thing that we called psi of X over here.  All of this has its exact parallel for the ordinary systems.  Top integral on that bottom board, yeah.  I'm trying to figure out which are the base inspectors.  These are the base inspectors.  Let's compare that with psi is equal to sum over a prime.  Let's call it a prime.  Psi of a prime, a prime.  Okay?  If we use the fact that a and a prime are orthogonal and do exactly the same thing except replacing sums by integrals,  we will also find that the psi that appears here is the same as the inner product of psi with the basis vectors.  So I've really done nothing that we haven't done before.  I appreciate it's a little, possibly a little unfamiliar, but all it is, is the same thing we've done before.  So the psi of X prime sort of plays the same role as the alpha's and some other.  The X prime plays the same role as the A prime here.  Yeah.  The psi plays the same role as the psi.  The sum is replaced by the integral.  Sum over A prime is replaced by integral, the X prime, and otherwise it's exactly the same.  Now let's calculate the inner product.  One quick question.  It may be nitpicking or something, but on the left-hand side where you have the Dirac's formula,  lower right-hand corner, yeah, you've got a vertical line between the axis.  Over here in the middle formula, there's no vertical line.  Thank you.  Okay.  Great.  Because it could have been a line or a problem.  Got you.  Got you.  Got you.  Right?  Nope.  Absolutely.  How about the inner product of two vectors?  Well, the inner product of two vectors is some inner product, but one way of calculating the inner product,  let's go back to ordinary quantum mechanics, ordinary means for the discrete systems.  We don't need this right now.  How do we calculate the inner product?  Let's take it high.  Inner product of psi with phi.  We first write psi and phi as sums over the basis vectors.  So this is equal to sum over two sets of basis vectors, a prime and a.  Psi star of a prime times a prime, that's the bra vector.  We complex conjugate, of course, when we deal with bra vectors.  That's the psi half of it.  And then the phi half of it is the sum over a of phi of a times a.  Phi of a being the coefficients of the phi vector,  psi star being the coefficients of the psi bra vector.  All right.  Now, phi of a is just a number.  And the inner product of a prime with a is just the chronic a delta.  So all of this adds up to sum of psi star of a prime, phi of a.  And then a chronic a delta of a and a prime,  which is exactly the instruction to set a equal to a prime.  Now, I think you've seen that before.  This is simply the rule about row vectors.  If psi star of a is represented by a row vector,  psi star of a1, blah, blah, blah, blah, blah,  and phi is represented by a column vector,  then this is just the rule to take this times this,  plus this times this, plus this times this,  plus this times that.  So this is the common garden variety quantum mechanical inner product.  I'm not going to, we don't need to go through the same exercise here.  We can immediately write down what the result is going to be  just by analogy.  It's almost not even an analogy.  It's just replacing wherever we see a's put x's  and wherever we see sum put integral.  The inner product of two state vectors for a particle on a line  is integral the x psi star of x phi of x.  It satisfies, incidentally, one of the basic rules of inner products.  What happens to an inner product if you want to change the two vectors?  What's the relation between the inner product psi phi and the inner product phi psi?  Complex conjugation.  Now what would happen in here if you want to change psi with phi?  You'd wind up writing phi star here and psi here.  That again would be the inner product, the complex conjugate.  And for exactly the same reasons.  All right, so this is equal to this for both discrete systems  and for these continuous systems.  Inner products, the inner product of a vector with itself  is the norm of the vector.  And just as in previous examples, you normalize the vector to one  to express the fact that total probability adds up to one.  So not much difference.  Just a difference of notation.  I think what gets to be confusing about it is you know both of these things.  You know about these simple row vectors and column vectors  that you know about functions, but you're not used to thinking of functions as vectors.  Or maybe you are if you are good.  But if you're not used to using final thinking of functions as vectors, f of x's,  or function psi of x's, then you'll have to get used to it.  But functions are vectors in a vector space.  You can add them, you can multiply them by complex constants,  and you can add them, you can multiply them by numbers.  And so they satisfy the axioms of a vector space.  Space of functions is a vector space.  Now sometimes you may want, yeah.  In more classical mathematics, would you put as the limits of integration  minus and plus infinity?  You would.  Which incidentally says something.  It says that we shouldn't be thinking about functions which grow at infinity, for example.  We should be only thinking about functions which are square integrable.  Now we're going to have to modify that.  Square integrable means you square them, this is called squaring them,  that the integral exists.  Functions which grow at infinity won't do that.  Now those are crazy because they would say that all of the, if you had functions which grew at infinity,  where would the probability be?  The probability would be all located off at infinity, and that's not what we want.  We want to be talking about probability distributions which have a total finite probability under them.  So that means we should take perhaps with a grain of salt the idea that we can write down any function.  But for the moment we're going to be a little bit loose about that.  Okay.  I think I'll slow down here for a few minutes and would you digest that and ask for questions.  Yeah.  Can I go back to the end of this?  No, no, no, no, no.  Yeah, go ahead.  No, I didn't want, I meant about this, but okay.  Maybe this is quick.  It seems like just about everything must be entangled with something almost all the time.  Oh yeah.  Yeah.  You're entangled with all sorts of things in the environment.  Now we haven't talked about the importance of a complex environment.  It's not so important in the very foundations of the subject, but it is important to why we see the world the way we are, that we're constantly being bombarded and getting entangled with things.  As a theory of distribution has been useful for physicists, or is it just...  Oh boy.  Certainly.  Okay, so it's not just some, it didn't just clean up the direct delta function, I appreciate it.  Oh, you mean, has it, well...  I see what you mean.  You mean has it done anything else besides clean up the...  I don't know.  I don't know.  I don't know.  I think it's made rigorous a set of concepts that physicists use all the time.  Has it really changed?  No, no.  The theory of distributions...  Yeah, I'll tell you what it did for physicists.  It made them have to argue less with mathematicians.  It was one of the many examples where mathematicians got dragged screaming and kicking by physicists.  Of course, the opposite also happens, but it was an example where mathematicians at first looked at this thing and said,  it's crazy, that doesn't make any sense.  It's non-rigorous, and in time they came to peace with it through defining it in their own way.  But no, I don't think the mathematical structure of it has influenced physics as much as you might have expected.  Unfortunately, because there are big problems of confusion related to it in quantum field theory,  and it hasn't resolved those problems.  Okay.  Okay, next place we want to go is observables.  We want to talk about observables and give some examples of observables.  Observables are linear operators.  What are linear operators acting on a space of functions?  They're operations on a function that gives another function.  If you have a linear operator, let's call it M.  For some reason, I've saved L to stand for observables.  I've used previously M for a general linear operator, so let's continue to do so.  Whatever M is, it's a machine, and thought of as an abstract operator in a space,  it takes a state vector into another state vector,  a side prime, let's say, or a vector in the vector space.  But it also can be thought of as a thing which acts on wave functions.  Every state has a wave function that describes it.  And the machine, oh, that's why we called it M.  Yes, I remember now.  We called it M for machine.  The machine can also be thought of as acting on a wave function to give another wave function.  So let's think of it that way.  Let's think of it in concrete terms as an action that we do on a wave function  to produce another wave function.  Now, it's linear.  Linear means that when it acts on the sum of two functions,  it gives back the sum of the results.  It means that if it acts on a function which is a numerical product,  a numerical multiple of some function, it gives back the same numerical multiple.  All the ideas of what a linear operator are continue to be exactly the same.  Let's talk about some linear op- oh, okay.  That's the notion of a linear operator.  Now, what about a Hermitian linear operator?  I'll tell you the definition of Hermitian.  We've talked about it before, but I'm going to give it to you in a condensed form.  The condensed form is also true for all of the vector spaces we've studied.  And you may recognize it if not go back and prove that it follows from the previous definitions.  A Hermitian operator, Hermitian for operators is roughly speaking the analog of real  as opposed to complex.  Okay, here's what's true about Hermitian operators.  If you have a Hermitian operator which I will now call L,  then it has the property that if you take, if you act on a wave function psi,  or a state vector psi, and take its inner product of phi,  that that is equal to psi L phi, but not quite, what do I have to do with it?  Complex conjugate it.  Now, if L were not, let's go back a step.  Supposing L was not Hermitian, would this be right?  Not quite.  What would we have to do?  We'd have to put a dagger here.  In other words, we'd have to roughly speaking complex conjugate L.  Okay.  If L is Hermitian, then the rule for that is that it's matrix element,  it's when sandwiched.  When sandwiched in opposite order, you don't have to complex conjugate L  because it's already the analog of real.  This is the condition that an operator is Hermitian.  That's important because Hermitian operators are the class of observables.  So let's work out one or two Hermitian operators while we're at it.  I guess we may need this, so let's work out a couple.  First of all, one of the simplest things that I can do to a wave function,  I don't know what's there.  There are many simple things I can do to it.  But a very simple thing is to multiply it by x.  An even simpler thing would be to multiply it by a simple number.  But let's take slightly more complicated and multiply it by x.  Multiplying by x gives back a function.  This is at minimum an operation on a function.  Is it linear?  Sure it is.  If x multiplies psi of x, let's say, plus phi of x,  it gives x psi of x plus x phi of x.  In other words, when it acts on a sum, it gives the sum of the results.  If it acts on a numerical multiple of a psi,  it gives back the numerical multiple.  If it acts on a linear operator, question, is it a Hermitian?  Is it a Hermitian operator?  Okay, so let's check.  Here's what we have to take.  We have to take psi in a product of x times phi.  That's this one.  Oh, sorry.  I think I want to do it the other way, phi x psi, the top, phi x psi.  And ask whether this is true or not.  We'll check in a minute.  No, x is not a number.  x is a function x.  We've multiplied one function by another function.  That takes functions to functions.  x is not a number.  It's a variable.  Any given value of it is a number.  But it's a machine that takes the function psi of x  and replaces it by x psi of x.  All right, so let's calculate.  The wave function of the ket vector is x psi of x.  The wave function of the bra vector is phi star of x.  The inner product is, let's put some brackets around it, bracket.  This is a ket vector inside the red bracket,  and this is a bra vector.  And the rule for taking the inner product,  whoops, I think I interchanged star and not star,  but the rule for taking the inner product is to integrate.  So let's get rid of a bracket here.  And just integrate dx.  Oops, sorry, dx.  All right, that's in this order.  What about this order?  This order over here, the inner product without the star,  without the star, the thing we're going to compare it with,  is the same thing except with psi star x phi of x.  Well, look at it.  Is this integral the complex conjugate of this?  Yes, it is.  The complex conjugate of phi star is phi.  The complex conjugate of psi is psi star.  And what about x?  x is its own complex conjugate.  So yes, indeed.  x is an operator which satisfies the basic rule of Hermitian operator  that if you interchange the psi and the phi,  it complex conjugates.  All right?  If, yeah.  What observable does x correspond to?  Position.  It is the position.  Is the delta function a eigenvector of position?  Well, maybe we'll come to that next time.  I think I won't.  Yes, it is.  But we'll come back to that.  x is an observable.  What does it correspond to?  It corresponds to the position of a particle.  We will prove that by showing that the eigenvectors x,  that the vectors x are eigenvectors of that operator,  but not right now.  We'll do that the next time.  It's easy.  It's trivial.  Just a big question.  So my correctness in assuming that when you write x up there,  it's implied that that is really f of x.  The x is a function of x.  x is an n f of x.  Yes, it's implied that that x is a function of x.  It has a whole bunch of values.  It has a whole bunch of values?  That's right.  It itself is a function which is multiplying another function.  All right?  So it's an operation on a function which gives another function.  Yeah.  That's exactly right.  Let's take another operation or operator, differentiation.  If I differentiate a function, I get a function.  That's the first test of whether it's an operator.  If I differentiate the sum of two functions,  I get the sum of derivatives.  If I differentiate a numerical multiple of a function,  I get the numerical multiple of the derivative.  It's a linear function.  Is it a Hermitian operator?  So here we have to get ourselves a fresh blackboard  and check once and for all whether it's,  well, check whether it's Hermitian.  Okay.  What happens?  Let's call a derivative operation.  Let's call it capital D.  D on psi gives us some psi prime.  What about wave functions?  D when it acts on psi of x by definition now.  This is the definition of how D acts.  It simply gives d psi by dx.  I may sometimes lapse into calling this partial derivative  with respect to x.  That's fine.  Doesn't matter.  It's just a function of x.  So it just differentiates.  It creates a new function.  It's a linear operator.  Is it Hermitian?  Okay.  So let's write down the two sides of the Hermitian condition here  and see if they're the same.  That's all there is to it.  The left side is integral phi star of x  times the operation D on psi,  which is just the psi by dx.  What's the other way of doing it?  Or the right-hand side?  The other way of doing it is integral psi star.  Not doing the same thing, doing something different.  Psi star of x, the phi by dx.  And the question is whether this is the complex conjugate of this.  So let's write equals with a question mark.  We don't know.  Complex conjugate.  So let's do this integral over here.  First of all, the thing in the bracket here  is just equal to integral psi of x  d phi star by dx.  Just complex conjugate.  The complex conjugate of a product  is just a product of complex conjugates.  So the thing in the bracket is psi of x d phi star by dx.  And we're asking whether this is equal to this.  They look similar.  This has a psi.  This has a psi.  This has a phi star.  This has a phi star.  But unfortunately, this one has psi differentiated.  And this one has psi, you know what I mean,  psi and phi.  How do you go from this to this?  Integration by parts.  And integrate like this, assuming,  what are we assuming when we say we can integrate by parts?  We're assuming that far away the functions go to zero.  That the endpoint contributions from integration by parts  don't give us anything.  Well, of course, the functions must go to zero far away  because the integral of the total thing must be one.  So we can integrate by parts.  Everybody know what integrating by parts is?  If not, find out fast.  It's simply a rule.  I'll tell you what the rule is.  You can shift the derivative from one,  if you have a product, dx.  There's a dx here.  If you have a product, psi of x times the derivative of phi star of x,  then you can simply shift the derivative from one function to the other,  but it's at a cost.  A cost?  A minus sign, right?  Good.  So this is equal to minus the integral phi star of x d psi by dx.  Whoops.  We're in bad shape.  This is integral phi star d psi dx.  This is minus the integral of phi star d psi dx.  They are not equal.  In fact, worse than that, they're absolutely the opposite of each other.  Opposite in the sense of the negative of each other.  Can we fix it?  Well, no, we can't fix this,  but we can redefine an operator which is her mission.  The redefined operator which is her mission is not derivative.  It is minus i.  The minus sign doesn't matter.  We can put i or minus i d by dx.  So let's go through it again and check whether now it's her mission.  This is the new thing that we're going to call,  and we're going to give it a different name.  I'm not going to call it d because it's not derivative.  That's minus i times the derivative.  Anybody have a good suggestion for what I might call this?  P, brilliant.  P.  P4?  Momentum.  Momentum.  All right?  And if I really wanted it to be momentum with units of momentum,  what else would I put there?  Where?  Right or left?  Right.  Well, you can put it on the left, but then you'd better put it downstairs.  All right, but we're not going to worry about each bar.  We're going to use our usual convention of ignoring each bar for the moment.  P is minus i d by dx.  Let's see what happens.  Over here, we have to replace this by an i over here.  Minus i, thank you.  Minus i.  What about this one?  If we have a complex conjugated everything,  we'd better change the sign because the complex conjugate of minus i is plus i.  And so, let's do it this way.  Minus i, and we should get plus i here if it's her mission.  Why?  Because in the process of taking the complex conjugate,  minus i gets converted to a minus i, and every minus i gets converted to a plus i.  Is it true now?  Yeah.  Why?  Because this is equal to minus this.  That's what we proved over here.  Okay?  So if we stick an i here, because i changes sign on the complex conjugate,  this little operation here tells us that minus i d by dx is a her mission operator.  It's a little odd, isn't it, that minus i d by dx should somehow be a real thing,  should represent a real thing, and that d by dx should not.  But let's check something out.  Let's check something out right now.  Let's just, we're going to finish up in a moment.  We haven't talked about expectation values yet tonight,  but I think you can guess how to think about them.  Expectation values in the state psi of an observable L are just the sandwich  composed of psi on both sides and L in the middle.  So let's take the expectation value of momentum.  It's integral psi star of x, psi of x,  and then according to the rule here, p is minus i d by dx.  Is this real or imaginary?  You're wrong.  It's real.  So let's prove that it's real.  I hope it's real.  God, I hope it's real.  All right.  So let's integrate by parts.  All right.  This is equal, first of all, let's take out the minus i.  Minus i integral psi star of x d psi by dx.  Now let's integrate that by parts.  This is equal to minus i, and then we can switch the derivative.  We can switch the derivative to the star variable, integral d psi star  by dx times psi of x.  But then in doing the integration by parts, we have to change the sign.  So here we have a little theorem.  The integral of this thing here with a minus i is the integral of this thing here with a plus i.  What's the complex conjugate of this whole thing?  The complex conjugate it, you replace psi star by psi.  Well, psi is over here.  You replace the psi by the x by the psi star by the x, and you replace minus i by plus i.  This over here is the complex conjugate of this.  Can everybody see that, or should I explain it further?  Let's put a minus i out here and write that this is equal to d psi by the x.  Minus i d psi by the x.  This is clearly the complex, the integral, are complex conjugates of each other.  The psi by the x is the complex conjugate of this.  That one's a complex conjugate of this, and i and minus i are complex conjugates of each other.  So, given that this is the complex conjugate of this and they're equal to each other,  what do you say about a thing which is equal to its own complex conjugate?  It's real.  So we prove a rather startling fact that the integral of psi star of x minus i d psi by the x is real.  No, yes, I did.  And again, it's because when you integrate by parts, you pick up an extra minus sign.  So instead of this thing being minus its complex conjugate, it's plus its complex conjugate.  Okay, so what did we learn?  We learned that this is a Hermitian operator.  It's a Hermitian operator and therefore an observable in quantum mechanics.  The next time we'll explore a little more.  We'll discuss what this has to do with Fourier transforms.  We'll discuss questions about what the probability for finding the particle at different locations are.  But then we'll also talk about what the probability for having different momentum are.  And then we'll move to the Schrodinger equation.  We'll move to the evolution of the wave function.  This, after all, now is really in the classic sense of the word a wave function.  And we'll think about how it evolves with time.  Oh, one other thing we could do with tonight.  I think we're probably at the point of diminishing returns or you exhausted.  We could do one more thing.  Let me do one more thing for you.  Let's calculate an interesting quantity and why it's an interesting quantity we'll come through in a moment.  So let's take a quantity which seems very dull, but which is interesting and calculate it.  It's the commutator of x with p.  Take any wave function, whatever it is, psi of x, anyone.  And we're going to do two things with it.  We're first going to multiply it by x and then we're going to multiply it by minus i d by,  or operate with minus i d by dx.  This can be called px operating on psi.  This is p, this is x.  In the order first x then p.  Then we're going to calculate it the other way.  We're going to write xp on psi.  What is that?  On the other half of the blackboard over here, we're going to write x times minus i d by dx psi of x.  Let's see what we get.  In fact, what we're going to do is we're going to compute the commutator,  which means we're going to calculate this one with a plus sign,  we're going to calculate this one with a plus sign and this one with a minus sign.  I don't know the minus sign, but before we subtract them,  let's just calculate them.  All right.  This is just minus i.  And now the derivative of a product.  The derivative of a product.  This derivative here now has to act on the whole result of x acting on psi of x.  x acting on psi of x gives a new wave function and then p acts on it.  So what do you get when you hit the product?  You get two terms.  The first term is the derivative of x times psi.  The derivative of x is just 1.  So the first term is just psi of x.  And the second term is plus x d psi by dx.  Everybody agree with that?  The derivative hit x and left psi and then the derivative hit psi and left x.  That's the side over here.  Now what about this side over here?  Here, the d by dx doesn't act on x, it just acts this way.  It just acts on psi of x and gives us minus i d by dx psi of x.  X, sorry, x minus i x d psi by dx.  Now let's subtract them.  Let's take this one with minus and this one with plus.  This corresponds to taking the commutator of x with p acting on psi of x.  The commutator of x with p is another operator and it acts on psi of x.  What do we get when we subtract?  Let's see, this one now has to come with a plus sign since we're subtracting.  Since we're subtracting, we have to change sign here.  And this term cancels.  Yeah.  This term cancels, goes away, and all we're left with is psi of x.  That's all we're left with.  This term cancels and we're just left with what is it?  i psi of x.  i psi of x.  All right, so here's what we found.  For any psi of x whatever, for any psi of x, the commutator of x with p acting on psi of x just gives, is it plus i?  I think it's plus i psi of x.  If an operator acting on any wave function gives another operator,  if the operator xp acting on any wave function is the same as i acting on that same wave function,  for any wave function whatever, it follows that this operator is equal to this operator.  Another way of saying it is if xp minus i, put it all in a big bracket,  when it hits any wave function at all, gives zero,  then it means that this is zero.  There's only one operator which gives zero when it acts on every state, on every vector.  There's no vector which escapes when this thing hits it.  The implication is that this thing is zero.  There's only one operator which annihilates, which kills every vector and it's zero.  xp minus, well, commutator xp is equal to i.  If I put back the h bars, there would be an h bar here, i h bar.  We've seen things like this before.  You may have to go back to your classical mechanics.  We've seen things before like this.  We've seen Poisson brackets which had similar relationships.  The Poisson bracket of x with p is equal to one.  Earlier in this course, we discussed the potential relationship.  We had reason to discuss why commutators.  It had to do with Schrodinger's equation.  Go back, but we discussed the relationship that there might be a relationship between  commutators and Poisson brackets, and in particular, the relation involved this i h bar.  This is actually where Dirac began his discussion of quantum mechanics.  He said there must be something analogous to a Poisson bracket in quantum mechanics.  The only thing he could think of which had the right algebraic structure,  was anti-symmetric, it had a certain mathematical structure, was the commutator.  He postulated that commutators and Poisson brackets were related with each other  to an i h bar, and then proceeded to go exactly backwards from the way that we went.  He said if this is the Poisson bracket, then this must be the commutator,  and so there must be an operator p, whose commutator with x is i,  and he went back through the operation and discovered that p was minus i d by dx.  So he went in the opposite direction.  We discovered p by just playing a game.  He said how about this operated d by dx?  What does it do?  What we found is that it was not her mission, we had to multiply it by i to make it her mission.  We explored it a little bit, found out some interesting properties of it,  and the most interesting property is that its commutator is equal to i h bar,  which suggests that it's connected with the Poisson bracket,  the classical Poisson bracket being one.  Of course the upshot of it is, and of course we need to do a lot more to convince ourselves  that the operator minus i d by dx really behaves like quantum mechanical momentum.  We can't just say, well, it looks like a Poisson bracket,  it must be the quantum mechanical momentum.  We have to show that wave packets move with velocities that are governed by this,  that it's conserved, that it has all the properties of momentum, but we will.  There will be some connection between this abstract operator minus i derivative of mass.  And what?  And the godney of mass.  Well, we would like to show that in some way that it's the mass times the velocity.  So what we'll eventually show is that if you have a wave packet,  in other words, a psi of x, which is a concentrated distribution like this,  that that wave packet moves along the x-axis with a velocity which is given by the expectation value  of this momentum operated divided by the mass.  That's what we would like to show, that the velocity with which this wave packet propagates along the axis,  that velocity, you can call it the group velocity,  it's usually called the group velocity, the velocity that the group of waves moves along  times the mass of the particle is equal to the expectation value of the momentum.  Once we do that, then I think we're, and that it's conserved,  that'll take us a long way to calling it momentum.  At some point, we have to say quantum mechanics is more fundamental in classical mechanics,  and the definition of momentum should be through quantum mechanics,  and then we should show it behaves in a certain way when particles are heavy and so forth.  Okay.  For more, please visit us at stanford.edu.
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