20. Multiparticle States and Tensor Products (continued) and Angular Momentum
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Zwiebach, Barton
In this lecture, the professor talked about EPR and Bell inequalities, orbital angular momentum and central potentials, etc.
 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. I'm going to get started. After a couple of minutes there will be an announcement. So today's lecture will begin by looking at the singlet state for a few minutes, a couple of its properties. And then we go into this Einstein-Pololski-Rosen argument. We'll talk about what they said, what they wanted to criticize quantum mechanics for. And then we'll go through the answer given by Bell that actually demonstrated that Einstein, Pololski, and Rosen were wrong. And after we'll do those Bell inequalities, we'll close this chapter on quantum states of spins and begin our treatment of angular momentum. So I want to remind you of a couple of things. You've been discussing this so-called singlet state plus minus, for the second particle, minus, minus, plus first and second particle. This state. A few things we've discovered about this state. You've been working with it. We calculated the z component of total angular momentum. The angular momentum of the first particle and the second particle, and it was 0. The x component was 0. This y component is 0. In fact, it doesn't have any total angular momentum. So it's a state that is rotational invariant, we say, because it doesn't have angular momentum. You did verify in the homework that the state is in fact rotational invariant. Now, this state is a very interesting state. It's one of those entangled states that we discussed last time when we were talking about teleportation and Bell states. And apart from that, it's a state that is not hard to realize physically. In fact, it typically takes place, for example, in reactions. In the case of certain particles, for example, pi 0 can decay into two photons, then the two photons can be in a state of total angular momentum 0. But more precisely, since we're talking spin 1 1 1 1 half particles, if you have a Mason called an eta 0, it's a interacting particle of strong interactions, a Mason, decays rather quickly into a mu plus plus a mu minus. Actually, it decays to other things as well. So it decays into two spin 1 1 1 half particles. And this particle has zero angular momentum. It's a scalar. It's not spinning. And therefore, if these two particles go into a state of zero orbital angular momentum, conservation of angular momentum implies that these two particles are in a state of this form that has zero spin angular momentum, total spin angular momentum. So the realization of an entangled state like that is fairly common and fairly easy. So you have a decay of this form, and you get particles that are entangled this way. We also, you showed that this state actually could be written as 1 over square root of 2 n plus n minus 1 2 minus n minus n plus 1 2, precisely because it's invariant under rotations. You could use instead of the plus minus basis, any basis n for any direction n, you have this state. Now, we'll talk about the probability that is of interest to us. We'll write this symbol, probability to get a plus, say, b plus. And this means the probability to get the first particle with spin along a, so that is first particle to be in the state a plus. And the second particle with spin along b, so in the state b plus. So in the state b plus. Now, that looks this calculation of this probability, so you're going to do some measurements. And you ask, what is the probability I find the first particle in this direction, second particle pointing in this direction? It may look like a somewhat nontrivial calculation, and it is. But if you use the fact that this state over there, because we're asking for this probability on this state, we're going to be talking about this state. So if you put the state and you write the state in the form, you pick one of the two vectors, say a. Well, the state is a plus, a minus, 1, 2, minus, a minus, a plus, because I could choose n to be anything. So might as well choose one of the two vectors to be a. And then you ask, what is the probability to find first particle in this state and second particle in this state? Well, when somebody tells you what is the probability to find a particle in a state, you put that state into your sandwich, your psi, with that state, and this overlap, which is a number, you square it. So we're going to do the same thing here. So what is this probability? Probability to find a plus b plus would be the absolute value. And then we put here a plus in the first state, tensor product, we could say, or b plus in the second state. And we should put, because this is what we want to find, the state on the tensor product that we're looking for. And we put the psi here. So we must put the 1 over square root of 2 times the a plus 1, a minus 2, minus a minus 1, a plus 2. So overlap time, well, we go 1 with 1, 2 with 2. So well, a plus with a plus will give me 1. b plus with a minus, I don't know. Second term, a plus with a minus gives me 0. So this term is irrelevant. I just need this. So the a plus with a plus gives me 1. I have the 1 over square root of 2 here. And I must close this and square. I forgot to write that. So what is this probability? a plus b plus is equal to the 1 over square root becomes 1 half. 2, the 1 over square root becomes 1 half. And then all we have left is b plus with a minus in the second state space, in the particle 2 state space. The label doesn't matter at the end of the day now that we disentangle the 1 and 2. So it's b plus a minus. I don't have to write that just 2, 2, or you could write it. And it's this square. So it simplified a lot, but not quite yet the answer. What we need here is the overlap between these two spin states. And I remind you that when you had any arbitrary two spin states with n and n prime, long ago, homework 3 or 4 or something like that, you calculated the overlap between these two spin states. And the answer was that you would take cosine of 1 half of the angle, if there's an angle gamma, the overlap between the spin states squared was cosine squared of 1 half the angle. So here we have the vector a, say, the vector b. Here is the vector a minus, a minus direction. And if we call this theta ab, this is pi. This is theta ab, this is pi minus theta ab. So this should be 1 half cos squared, 1 half of pi minus theta ab, 1 half cosine squared of half the angle between the two relevant vectors. So this is cosine of pi over 2 minus theta over 2, that's sine squared of theta ab over 2. So here it is. It's our calculation of this thing. It's a neat formula that we're going to need later today. So one more comment before a little stop. If b is equal to minus a, now in that case you should be in luck, because precisely what's happening here is that if one spin is along the plus direction, the other has to be along the minus direction, whichever you choose. So in order to have that a be in plus and b be in plus, well, this first term would do it. a is in plus, the first particle. Here, no. And b, which is minus a, would be in plus. So the probability should be 1 half. So if b is minus a, the angle theta ab is equal to pi. And the probability of ab is 1 half correctly. And it's 1 half, because half of the cases a is in plus. The other cases a is in minus. So the other case, for example, that could be interesting is if what is the probability that the first particle is in z plus and the second particle is in x plus? Well, these two vectors form 90 degrees. So you should have 1 half of the sine of half that. So that should give you 1 half of the sine squared of pi over 4. And it's 1 half times 1 over square root of 2. That's another 1 half. So it's 1 quarter, for example. OK, I went long enough for a moment. There's an announcement that Prashant wants to make. So please listen to him. Hi. I'm back for another announcement. So tonight, the MIT SPS is going to be holding its fall year off, like in lectures. So if you don't have a year off, it's a great opportunity for you to come and see what other research your classmates are doing in physics. If you do have a year off, I would highly encourage you to come as well, because you can actually come and share your stories about the technical content of what you're doing in your year off. It's 7.30 this evening in the PCR Room 8329. There will be free food, so please join us then. And see you all then. Thank you. All right. So OK, this was the introduction to what we really need to do today. So before we get started, are there any questions on what we've done so far on these properties of this entangled state? So this is a measurement on an entangled state. Two observers, these two particles could have flown away a big distance. Two observers, one tries to see what is the probability. But the first observer sees the spin pointing in some direction. The other observer sees the spin pointing in another direction. It's a natural question which can be done experimentally. And we've calculated that answer. OK, so let's begin with this EPR story. Now you've seen some of EPR last semester in 804. The only complication with that is that you really needed to have these mathematics to appreciate it completely. So this second look at EPR should be fairly complete, in that we won't leave almost anything out of the story. There are many ways of doing that. Sorry, there are many ways of doing EPR. And essentially, this bell inequality, which is the really non-trivial thing that comes after that. So some are in the homework, some elaborations, and probably in recitation or later in the course, we'll see a little more. But it all begins with a strange thing, that the kind of thing that you wouldn't expect people in physics to discuss. And it's the point of the so-called Einstein, Podolski, and Rosen wrote a paper. And they talked about local realism. Now that sounds like philosophy. And for a while, people thought, well, this is interesting, but undecidable, can't really do anything with this. So what is local realism? Now, again, not being exactly physics is not all that easy to say what it is. And people discuss that. But some notion of it is fairly clear. And the notion is that this reflects something. It's basically two assumptions about measurement results. So you measure something, and you obtain a number. And the first assumption, one, is that this measurement results correspond to some aspect of reality. Just say like that. It seems a little funny. But you measure something. And if you get some numbers, because that was something real about this object, you've had this property. And so measurement corresponds to some aspect of reality. So measurements, local assumptions about measurement results. So measurements, measurements, M, correspond to some aspect of reality. Two, the measurements that you do in your lab are not affected whatsoever by the measurements that somebody else is doing at the same time at the moon. There's no time for the information of what that result in the moon has given to reach you. So at that instant of time, what they are doing at the moon doesn't affect the result of your experiment. So its measurement is independent of actions performed at a distant location at the same time. Now to Einstein and Polovsky and Rosen, but Einstein was very vocal, physics must satisfy that. It's kind of sad, I think, actually. A person that managed to see through and discover how nature works at so many deep levels, the photoelectric effect, special relativity, general relativity, somehow became convinced that this had to be true. And unfortunately, he was wrong. Or fortunately, I guess it's not worth trying to qualify that. But these two things that seem just so reasonable are just not true. This one, measurements correspond to some aspect of reality. You see, you have a stern girl, a apparatus, and you throw a spin, it goes up. You say it ended up with spin up. Einstein would say, you know, it always had spin up. It was a reality about that object, that it had spin up. You just didn't know. You did the experiment, and you discovered it. So the thing people try to do in order to understand this concept, that it corresponds to something having to do with reality, is that you admit that half of the particles go up and half go down. But you say, actually, there's something about these particles you don't know. And if you knew that, you would just be able to tell. This is a particle that has spin up, and it will go up. But in quantum mechanics, we have abandoned that. We've said, these particles, they are superposition of a state up and a state down, and there's nothing definitely up about this particle or definitely down. So the way people do this to correspond to this aspect of reality that you don't know is by introducing what is called hidden variables. Some things that allow you, there's something hidden about this spin particle that you don't know. But if you knew it, you would know exactly how it's going to come out through the Stern-Gerlach experiment. And you say, well, that sounds fairly untestable, but the fact is that it's not. Now, so this is implemented, this assumption is when people try to modify quantum mechanics, they use what is called hidden variables. Some things that you don't know about the particle, but if you knew, you would see that, in fact, this particle has spin up. The second is in some ways even more disturbing, because we got accustomed to the idea that locally simultaneous things that cannot be reached, events that cannot talk to each other via the exchange of light cannot affect each other. So simultaneous things that happen far away can't affect each other. So this also sounds very reasonable, but that's also wrong. And there's the obvious question. So if this is wrong, can you send information faster than the speed of light? And the answer is people have looked at it in many, many ways. And it's very interesting, and we could discuss that, but it would take us long. So I will leave it to maybe recitations, maybe other forms. But here, actually, there's no contradiction, no way of finding information, real information, going faster than the speed of light, even though things far away at the same time can affect you. So two very interesting things that seem very dangerous to discard, but turn out to be wrong. So this is what EPR did. And they made some thought experiments that we're going to review to some degree and see if we can discard these assumptions. So that's what we're going to do now. I'm going to try to understand that. Now, if you're interested in what hidden variables are, my discussion will not use hidden variables. Although they are kind of implicit, you will see as I state some things, I'm basically going to be explicit on the fact that things, they're real facts, as Einstein would like you to think. And we try to see if those real qualities about particles get us in trouble. So let's begin and try to discuss this first experiment. You see observer 1, Alice and Bob again, if you want. Alice is going to measure about the z-axis. Bob is going to measure about the z-axis. And therefore, if Alice finds spin up, Bob finds spin down. If Alice finds spin down, Bob finds spin up. And that's a correlation. And it's very interesting. But Einstein, Bell, and Poloski would say, look, it's not all that interesting. There's nothing all that mysterious happening here. EPR would say, says, the pairs that you've built, so-called entangled pairs, are pairs of particles with definite spins, spin directions, spin vectors. So what you've built, EPR would say, if you have here particle 1 and particle 2, the proper, I'm going to list the properties. Suppose you have a particle 1. You've created, Einstein would say, some particles, particle 1s with spin up in z, and particle 2 with spin down in z. Or you've created particles 1 with spin down in z, and particle 2 with spin up in z. And you've created, in fact, 50% of the particles are of this type, of the pairs. And 50% of the pairs are of this kind. So don't tell me all this superposition, Hocus Pocus. Half of your pairs, one particle has spin up, one particle has spin down. The other 50% of your pairs, one particle 1 is down, the other particle is up. No wonder they're correlated. You get plus, gets minus, get minus, get plus. What is the probability that you get plus? 50%. What is the probability that you get minus? 50%. Everything is reproduced. Mystery over, no quantum superpositions. There's no mistake here, no mathematical mistake. You're saying that the particle has a definite spin. You maybe not know it, but we say, look, your particles, in fact, had a definite spin, z plus and z minus, and those things. And this is where hidden variables would come along. You would say, well, what is, if you have the particle 1, its spin is a function of some hidden variable that you don't know. But if you knew it, you would know what the spin is, because it has a definite spin. You don't know what is the hidden variable, but as a function of the hidden variable, the spin is known, definite, not a superposition, nothing like that. It's a very aggressive attack on quantum mechanics and something that troubled people and, in fact, fascinates people even up to now, because the idea that these kind of things are really absolutely wrong, it's very shocking, perhaps the second, even more shocking, because, OK, by now maybe you're accustomed to all kinds of variables that are a little more detached from reality. You have electromagnetic fields, and they force you to learn about potentials that seem to be a little more abstract, and similarly here. So no problem. So people say, OK, this is a simple situation, but it may be that we're going to do more measurements, and we're going to consider two directions that are different. Maybe Alice has two Stern-Gerlach machines, one that measures about z and one that measures about x, and Bob has also two Stern-Gerlach machines, one that measures about z and one that measures about x. And they are going to ask different questions, because we know the spin, how it transforms, so getting those results right with two directions is going to be a little more interesting. So we're going to try measuring in two possible directions, both A and B, Alice and Bob, can measure in two directions z and x. And Einstein would say, look, yeah, you can measure z and in x. Let's for to avoid confusion, let's not talk about one measurement after another. Particle comes, you measure in z or you measure in x. And realism says that you don't know what you will get, because maybe they're hidden variables, but each particle has a definite answer if you ask what is the z direction of the spin and has a definite answer if you ask what is the x component of the spin. Definite answers, real answers, realism again. So I'm going to label the particles. For example, this is a label for a particle z plus x minus. This particle, labeled like that, would be such that if you measure its spin in the z direction, its plus, if you measure the spin in the x direction, its minus. So measured in z, it is up, measured in x, it is down. These are the kind of particles that EPR would say exist. It's not that you got a particle out and it's in some strange entangled state. These particles are flying away. They're not talking to each other anymore. And this particle, since you can measure in two directions, there's some reality and the measurements correspond to reality. So they're attributes and this particle is classified by having these attributes. If you measure z plus, if you measure x minus. OK, so how about this situation? Well, let me make a list now, particles 1 and particle 2. And this would be the list that EPR would do for you. EPR comes along and says, look, here's what you're doing. Particle 1, suppose it's a z plus x plus. Well, in your beams, actually, these particles, when particle 1 is that, your other particles are z minus x minus. So if particle 1 is of this type, particle 2 is of this type, that way EPR protect themselves because they say, look, if you measure z plus and you measure z minus, you get correlation. If you get plus, plus, you get correlation as well. Now, there can be a particles in z plus x minus. This would go with a z minus x plus. There could be a particle z minus x plus and this as z plus x minus. And there could be particles z minus x minus z plus x plus. So there are four cases, four types of particles, they say, have been produced. And 25% of pairs are of this form. 25% of pairs are of this, 25 of this, and 25 of this. So you could ask some questions to EPR. What is the probability that you get z plus for 1 and z minus on 2? Well, z plus on 1, it's these two cases. And z minus on 2, well, those two cases. So this is 50% of the times. So it's 1 half that probability. It's correct. That's what we would predict from the entangled state viewpoint. Oh, but let's ask something mixed. Let's see if we get in trouble. P of z plus on 1 and let's say x plus on 2. Well, z plus on 1 and x plus on 2, that's not it. This case, z plus on 1, x plus on 2, 25% of the times, 1 fourth. This was the probability we calculated. So you tried with one direction. You don't need quantum mechanics to produce a result. You try with two directions. You don't need quantum mechanics to get the result. So people got stuck and they said, well, maybe it's undecidable. Maybe this is philosophy. Maybe this is something. And people had to wait for Bell that he said, I'm going to try three directions. Now, three directions makes a big difference. This is the first time it really goes wrong. So it's kind of surprising, perhaps, at some level. But this is subtle stuff. This is subtle stuff, so it takes a while before you find something wrong. You're talking about showing Einstein this wrong, so that's not so easy. So three directions. So particles are going to be of types. EPR would say, look, I'm going to use the same strategy. I'm going to say that particles have three attributes now. They're all physical. They correspond to reality. Because if you measure in either of three directions, they have to have an answer for that. So here is a label for a particle. And for example, A plus B minus C plus. So if you measure, it would give spin in the A direction plus h bar over 2, spin in the B direction minus, well, in the B direction, it would give you minus h bar over 2, spin in the C direction, it would give you plus h bar over 2. So we're not measuring simultaneously. We're just asking, well, you take a particle, do a measurement, and see what you get. And we're always going to be asking for probabilities of this kind, probability that the first particle is doing this and the second is doing that. Well, EPR would start now with the particles again. Particle 1, particle 2, and population. So let's list quickly the particles. A plus B plus C plus. Then you would go A plus B plus C minus. Then you've done the C plus C minus. Here you go for two more. A plus B minus C plus. A plus B minus C minus. Four, I'm supposed to fit four more there. Can I? Well, I'll try A minus. Now you've done all the four A pluses, so you need four A minuses. Then you have B plus, B plus, B minus, B minus, C plus, C minus, C plus, C minus. OK, I got all I think. And particle 2, of course, is correlated. You would say, well, I don't need to write it, but it helps seeing what's going on. So I'll write it A minus B minus C minus. A minus B minus C plus. A minus, let's use bars here, A minus B plus C minus. Plus C minus A minus B plus C plus. A plus B minus C minus. A plus B minus C plus. A plus B plus C minus. A plus B minus C. This was minus, so it's plus here. OK, we're done. Lots of labels. And you could say, well, maybe you want to put 1, 8 in each of them. But actually, the argument is more interesting. It doesn't need you to try to put fractions. So let's consider that there's a total number of particles, N, which is N1 plus up to N8. And here are N1 of this, N2 of this, N3 of this, N4 of this, N5, N6, N7, and N8. OK, let's see. All right, so we have that. And then, well, it takes imagination to see how you're going to run into some contradiction. So what is the basis for the contradiction? Somehow, this formula, which is really quantum mechanical, must eventually go wrong with all these attempts to deny that the world is quantum mechanical. So we could split, again, those particles into equal fractions, but there's no need to do that. And it's clearer if you don't. So you try to combine the three directions into one equation. So one way to do that would be to say, OK, what is the probability that you get A plus and B plus? So this is for the particle number 1, and this is particle number 2. So then you must look at the table and say, which ones do that? Well, you need A plus in the first, so it's one of the first four rows, and B plus in the second. So actually, it's these two cases, N3 and N4. 4 over N. We want to involve three directions, so let's go for another one, P of A plus with C plus. Let's see how much is that. Again, this is for the first particle. This is for the second particle. So I must look at the first four rows and see that you have an A plus. All of the first particles in A plus for the first four rows, but the second should be in C plus, and that is cases N2 and N4. So we get N2 plus N4 here over N. Well, we've involved this A with B, A with C. How about involving B with C? So I'll put C plus, B plus, for example. OK. C plus, B plus. I must look at C pluses and B pluses. No. C plus and B pluses. Yes. N3 is there, which is good because it already was there. And which else? C plus and B plus. C plus, no. C plus here, B plus. Yes. N7. OK. Now, N3 plus N4 is less than N3 plus N7 plus N4. It's less than N3 plus N7 plus N4 plus N2. You have N3 and N4. And now I add whatever N7 and N2 are. And that's then an inequality because it's going to be more cases. Now I divide by N. So you obtain inequality that P of A plus, B plus, is less than or equal than N3 plus N7 and N4 plus N2. I'll write the second first. P of A plus C plus plus P of C plus B plus. So I didn't put specific populations 1, 3, 1, 4. But in general, whatever populations you choose, this inequality must hold. Now that's a more clever strategy because suppose you choose some populations and you don't get in trouble. Well, maybe with some other populations you would get in trouble. Maybe it's not so easy to guess the relative factors. So here is something that must be true whatever populations you choose. And now that is Bell's inequality. So the achievement of Bell is to somehow translate this assumption of realism into an inequality. And now quantum mechanics has a formula for these things, for this probability. So we can test whether this is true. So let's do that. This is the so-called Bell inequality. So if quantum mechanics is true, the following should hold. Let's see that. If Qm is true, well, there should be a problem with this inequality. So let's see what it happens. Let's see if it's true, this inequality. Does it really true? Well, the left-hand side would be, given the formula that we had, 1 half of the sine squared of theta AB over 2. So let me just emphasize. This was derived using local realism. Local realism gives that. So you do the experiment, get these probabilities. And if realism is true, this should hold. Let's see what quantum mechanics has to say. Let's plug in the values that you get from quantum mechanics. Now we can see that the value of the quantum mechanics we calculated this probability. We put the first term. Here is sine squared of theta AC over 2 and 1 half of sine squared theta BC over 2. So does that work? Does that always work? Can I orient this axis in such a way to disprove EPR? And in fact, it turns out to be quite easy to do that. So you choose three vectors like this, AC BC. So C here, A here, and B here, I believe. Yep. So put an angle theta here, an angle theta here. And then what do you have? Theta AB would be 2 theta. Theta AC would be equal to theta BC equal to theta. That's a pretty nice simple choice of angles. If you choose these angles now, let's see what happens with our inequality. So you get 1 half sine squared theta AB over 2 would be theta. Is it less than or equal to this? Well, these two become the same. So you get sine squared of theta over 2. Violated or not violated? Is it true or false this for all theta? What do you say? Yes? Give theta is less than pi. That's not true. Close. It's not true for small theta. So if you're this and you're desperate to know, the thing you have to do is assume theta is very, very small. See if you get in trouble. How much is this? 1 half theta squared. Let's see. How about this one? Yeah. And how much is this? This is theta squared over 4. I was not seeing it. Sine theta for small theta is roughly theta. So here is theta over 2 squared. But here's 1 half theta squared. And it's false. The 1 half theta squared is not smaller than 1 quarter theta squared. And in fact, for theta equal pi over 2, I think this is an equality. Because theta equal pi over 2, you get on the left-hand side 1 half less than or equal than sine of 45 degrees, which is correct. So it's an equality at pi over 2 fails below. So it was a shock that if you could do an experiment in quantum mechanics and an experiment with correlated entangled particles, that you could measure these probabilities, these correlations. And you would obtain a result that actually contradicts for certain alignments of your experiments the assumptions of local realism. So it was a great result of Bell to show that quantum mechanics is in contradiction with local realism. There's no way to keep the ideas of quantum mechanics and put hidden variables and assume that there's real values for things and that there's no effect at the distance. It would all be contradicted by an experiment that was done later. And the definite version of the experiments around 1980 or 82 by Allen Aspect and others, very clever experiments, worth reading and understanding. But confirm that the quantum mechanical result is really true by measuring correlated pairs, and these inequalities are violated. Yes? So what about David Baum's theory of hidden variables quantum mechanics? David Baum's theory of quantum mechanics theory. I know that allegedly it reproduces the same results as quantum mechanics, but it's still a hidden variable theory. I don't think there's any hidden variable theory that works. David Baum, I think, actually was created by rewriting EPR, who essentially talked about position and momenta to talk about spins. And he might have been the first one that began to try to do hidden variable theory, but no hidden variable theory works at this moment. And this shows it. So people say that actually this assumes that there's local hidden variable theories and there's non-local hidden variable theories and all kinds of strange things, but it's more and more unnatural. So it doesn't seem to do something very interesting. Yes, there are many questions. Aaron, maybe you want to say something? I think Baum has a non-local hidden variable theory. It's kind of awful looking, but I guess violates the second principle rather than the first one. But also, it doesn't extend to, you know, it doesn't really work for everything. Like, we don't know how to make it work for a spin one-half particle or finite dimensional. So I think it's really believed to be a true theory of rules. Thank you. One more question. Steve. In the case with EPR, is there a problem that we can have scenarios where the spin is greater than spin one-half particle would have? Like, if we had a state a plus b plus c plus, we would have 3h forever, too? No, the statement that is done here is not that, well, this is a label for a particle. EPR just assumed that if you measured a, you would be able to get this. If you measure along b, you would get this. And if you measured along c, you would get this. So this is a single particle. Any measurement gives some results. These are the lists. There's no sense in which these are added. But even when you measure one, the other value still exists with those measures? Well, there's no, I believe there's no need to discuss that. So they don't talk about the statement of doing subsequent measurements within this statement. You just take this particle and you decide. You measure a or measure b or measure c. You don't try to measure simultaneously. You don't try to measure one after another. You just do one measurement. And that already, which is the minimum you can do, gets you in trouble. So I'm not sure how EPR would phrase subsequent measurements after they've done the first measurement or things like that. But they're not necessary for this stage. OK, so look, it's a very interesting thing. There's lots to discuss here. But it's best if you read it. James had a question. Just let's take one quick question. I was just wondering if there was any other extension beyond three directions or belts and qualities. Are there like a bay and direction belts and qualities or some type of form of it? Or is there really a need for it? There are other forms of belly inequalities. I believe, not sure if it's popular with four directions or other things, but certainly Weinberg, for example, discusses other ways. There are alternative ways to phrase it. I've talked about here probabilities to observe results. There's a more perhaps common way talking about expectation values or correlation functions. This is something you'll do in the homework. This sort of game that is done in the homework but was a suggestion by Aram to put it in the homework, this game in which with a quantum strategy and entangled pair you beat the system is yet another formulation of the belly inequalities as well. So lot to do, but I think it's better now that we stop and talk about angular momentum from now until the end of the semester. This was about spins, but we now have to put together angular momentum, which is orbital and spin and all the various kinds. So we'll begin with angular momentum. So there are notes on the web on that that I wrote and modified a little this time and lots of little exercises. So what I want to do now is guide you through the things that happen there so that you get a view of what we're going to do. We're going to work with this thing in an elegant way. Using vector notations for operators is going to help us understand things better. So we've seen angular momentum before, and let me summarize simple things that we know about it. If you have angular momentum, and let's begin with orbital, this is what we have when we take Lx to be ypz minus zpy Ly equals zpx minus xpz and Lz to be equal to xpy minus ypz. These are the angular momentum, orbital angular momentum operators. Now, it's better for many things to use labels like x, y, and z. Those operators call them x1 hat, x2 hat, x3 hat. And px, py, pz, p1, p2, p3. In that way, you can write commutation relations like xi, pj equal ih bar delta ij xi with xj equal pi with pj equals 0. So it allows you to write things more quickly. In fact, the angular momentum operators become also a little simpler. It's sort of x2p3 minus x3p2. And you don't have the x, y, z labels and all that. So we want to use vector notation. Now, vector notation, you can do it in two ways. You can talk about triplets. Those are vectors. Or you can form the vectors themselves. Now, if you form the vectors, you get objects that sometimes are a little disconcerting. But we try that they not be so. So here is the r vector operator. You could think of it as the triplet of x, y, and z. But let's call it like this. x times the first basis vector plus y operator times the second basis vector plus z operator times the third basis vector. And we've done things like that. And we understand that these basis vectors really don't talk with these operators. They can be moved across each other. The basis vectors are sort of things that help you write expressions rather than talking about triplets. Same thing for momentum. You would put, let's do it this way, p1 e1 plus p2 e2 plus p3 e3. And finally, well, angular momentum, the vector operator. You've done a lot of the angular momentum vector operator for spin. So here you would put lx or l1 e1 plus l2 e2 plus l3 e3. So those are ways to write equations that carry all the operators and treat them as vectors, even though they're operators. So they're unusual vectors. They're vectors whose components are operators. They're not numbers. So the obvious question is what changes them? So we're going to define dot product and cross products as we had before. But we have to be a little aware that when we write these things, we could make a mistake unless we're not careful. So here are two vector operators. Operators. What is the dot product of these two vector operators? Well, you know it's supposed to be the first component of this and the first component of that, second, second, third, third. So it should be a i, b i summed. Repeated indices are summed. Now, I should not write b i, a i with a reverse order because this thing, the components, are now operators. And maybe they don't commute. So I've defined this once and for all to be this. And A cross B, the i-th component of this thing, is going to be defined once and for all to be epsilon i j k a j b k. A definition, the A to the left of the B. And with this, we can check our usual rules of manipulation of operators. So one more definition. A squared is going to be A a and it's going to be a i a i. OK. Simplest calculation is A dot B equals to B dot A. Yes or no? No, they're operators. So let's calculate the difference. Let's get a little practice, calculating differences. So I write A dot B is equal to A i B i. It's summed. So then I say, oh, that's A i B i plus B i A i. You see, the commutator is A i B i minus B i A i. And I add it back. But this thing, B i A i, is B dot A. So here I got a formula. A dot B is equal actually to B dot A plus this commutator A i B i. And now you've got a new formula for operator vector analysis. A dot B and B dot A are not the same, but they differ by this thing. And a very important corollary, very famous corollary, what is r dot p? Is it the same as p dot r? If you're working in quantum mechanics, you may be tempted. Oh, r dot p and p dot r are the same, but r and p don't commute. So what is the difference? r dot p is equal to p dot r plus xi pi. And how much is that? Sorry? I h bar. I h bar? I h bar. 3 I h bar. Yes, don't forget the sum. This is supposed to be sum. So this says x1 commutator p1, x2 commutator 3 I h bar. So here is a famous formula. rp differs from pr plus 3 h bar i. I h bar, people write. OK. Another formula that you would be curious to know, well, the dot product was supposed to be symmetric. It's not. The dot product, the cross product, is supposed to be anti-symmetric. Is it or not? A cross B sub i is equal to epsilon ijk ajbk. OK. What do I do next? I want to move the a and b's around. So I'm going to replace this by a commutator plus the other ordering, ijk. And I put here, well, this should be a parenthesis, ajbk commutator plus bk aj. Now, what do we get? Well, you have to look at the second term. Let's put the first term here, because that's pretty much we're not going to be able to do too much with it, ajbk. And the first term, I would write it like this, minus epsilon. Flip this to ikj bk aj. If you do it like that, then it sort of fits nicely with the definition of the cross product. Because in the cross product, the first label goes here. The second label goes with the last label of the epsilon. So this thing is minus b cross a. And that was a cross b plus epsilon ijk ajbk. So that's your formula for how the cross product now fails to be antisymmetric. It's not necessarily antisymmetric unless you're lucky. So here is a property that you should try to think about. How about r cross r? Is it 0 or not 0? R cross r. You could say, oh, what can I do here? Well, r cross r minus r cross r, I would have 2r cross r should be equal to this. But both components are x's. So r cross r is 0. p cross p. That's also 0. And therefore, say, l cross l, is it 0? Maybe, is that right? No. l cross l, we'll see it a little later. But it's not 0 because one l with another l don't commute. Lights. OK. So l cross l is actually not 0 because this thing is not 0. We'll talk about it a little later. But actually, this is a very famous one. l cross l is proportional to l with an h bar. It's a lovely formula. So another interesting thing, well, what is r cross p? r cross p, for this formula, with minus p cross r. And how about the other term? Is it 0 or not 0? Is r cross p equal to minus p cross r, or does that fail? Well, r and p don't commute. So actually, this formula, somebody should have complained. They should have been the i-th component, the i-th component. And here is the i-th component. So the i-th component, the i-th component, let's look at here. Epsilon ijk xj pk. But xj and pk is i h bar delta jk. Now, this delta jk is symmetric. This is anti-symmetric. You should get accustomed to the idea that that is 0. If this is anti-symmetric and this is symmetric, this is 0. Let me do it one more time. Maybe this is not completely obvious, but epsilon ijk delta jk. The intuition is also obvious. Epsilon must have the three numbers different, and this forces them to be the same. But it's more general. If this is anti-symmetric and this is symmetric, this should be 0. And the way you do that is you say relabel j and k. Whatever was called j, call it k. Whatever was called k, call it j. So this is epsilon ikj delta kj. And then use the symmetry property. So when you exchange this back, you get a minus sign. When you exchange this back, you don't get a sign. So minus ijk delta jk. So you've shown that this thing is equal to minus itself, and therefore it's 0. Something that's equal to minus itself is 0. So this is 0. And you got that r cross p is really minus p cross r. And that's what we call the angular momentum, r cross p. But now you know that it's also equal to minus p cross r. OK. Let's see. One other one, for example, that is classically true. Let's see if it's quantum mechanically true. r dot l. You know, the angular momentum, l, is supposed to be perpendicular to r and perpendicular to p, because it's the cross product. Is that true quantum mechanically or not true? Well, maybe I didn't say, well, l is r cross p. So this is r dot r cross p. So this would be ri times epsilon ijk r, no, x. xj pk. OK. So what is this? It's epsilon ijk xi xj pk. Well, this is 0 because these two x's are operators, but they commute. Therefore, this object is symmetric in i and j. This is anti-symmetric. This is 0. So r dot l is actually 0. How about p dot l? Well, there's two ways of doing that. One that looks pretty obvious, one that looks a little harder. Let's do the one that looks a little harder. This would be pi epsilon ijk r, xj pk. OK. So here is the temptation. Must be 0 because it's a pi pk symmetric in i and k. And here is anti-symmetric in i and k. But it's wrong. These are not obviously symmetric unless you can move them across. And there's an x. In the middle, laughing at you and saying, beware. This could go wrong. So you have to be a little more careful. So let's do be careful ijk. And you put pi xj pk. All are operators. All are operators. But pi with xj, the commutator, would be a delta ij. And that vanishes. So this p actually can be moved across the x. And therefore, show that that thing is on the other side because the epsilon is there. So it's a lucky thing. So pi xj pk is not symmetric in i and k. But if there's an epsilon, you're in better shape. So here, the p can be moved. ijk xj pi pk. And now, nobody can stop you from saying, this is symmetric. They're next to each other. They can be moved across each other. It's really symmetric in i and k. And therefore, i and k are anti-symmetric. This is 0. p dot l is equal to 0 as well. So this was a little bit hard way. If you had used that l is actually equal to p cross r, then the p would have been next to the p from the beginning. And you would have saved a little time, the same time that it took you to realize that this equality is true. So those are true equalities. p dot l is 0. And it's also equal to l dot p. It's not obvious that l and p commute. But in this case, l with p dot p is 0. If you use this formula for l, it's obvious. r dot l is 0, it's also equal to l dot r. Doesn't make a difference. It's also true as well. So that's roughly what goes on. Look, I would like you to just read those pages. It's a continuation of this. It's about eight pages. I leave exercises to be done. That's in the homework of this type playing with the indices. It's a good thing that you get accustomed to it. And the material that you need just for the last problem of the homework will be covered on Monday. And we can talk about it in recitation tomorrow. So that's it for today.
 The following content is provided under a Creative  Commons license.  Your support will help MIT OpenCourseWare continue to  offer high quality educational resources for free.  To make a donation or to view additional materials from  hundreds of MIT courses, visit MIT OpenCourseWare at  ocw.mit.edu.  I'm going to get started.  After a couple of minutes there will be an announcement.  So today's lecture will begin by looking at the singlet  state for a few minutes, a couple of its properties.  And then we go into this Einstein-Pololski-Rosen  argument.  We'll talk about what they said, what they wanted to  criticize quantum mechanics for.  And then we'll go through the answer given by Bell that  actually demonstrated that Einstein,  Pololski, and Rosen were wrong.  And after we'll do those Bell inequalities, we'll close  this chapter on quantum states of spins and begin our  treatment of angular momentum.  So I want to remind you of a couple of things.  You've been discussing this so-called singlet state plus  minus, for the second particle, minus, minus, plus  first and second particle.  This state.  A few things we've discovered about this state.  You've been working with it.  We calculated the z component of total angular momentum.  The angular momentum of the first particle and the second  particle, and it was 0.  The x component was 0.  This y component is 0.  In fact, it doesn't have any total angular momentum.  So it's a state that is rotational invariant, we  say, because it doesn't have angular momentum.  You did verify in the homework that the state is in fact  rotational invariant.  Now, this state is a very interesting state.  It's one of those entangled states that we discussed last  time when we were talking about teleportation and Bell  states.  And apart from that, it's a state that is not hard to  realize physically.  In fact, it typically takes place, for example, in  reactions.  In the case of certain particles, for example, pi 0  can decay into two photons, then the two photons can be in a  state of total angular momentum 0.  But more precisely, since we're talking spin 1 1 1 1  half particles, if you have a Mason called an eta 0, it's  a interacting particle of strong interactions, a Mason,  decays rather quickly into a mu plus plus a mu minus.  Actually, it decays to other things as well.  So it decays into two spin 1 1 1 half particles.  And this particle has zero angular momentum.  It's a scalar.  It's not spinning.  And therefore, if these two particles go into a state of  zero orbital angular momentum, conservation of angular  momentum implies that these two particles are in a state of  this form that has zero spin angular momentum, total spin  angular momentum.  So the realization of an entangled state like that is  fairly common and fairly easy.  So you have a decay of this form, and you get particles  that are entangled this way.  We also, you showed that this state actually could be  written as 1 over square root of 2 n plus n minus 1 2  minus n minus n plus 1 2, precisely because it's  invariant under rotations.  You could use instead of the plus minus basis, any basis  n for any direction n, you have this state.  Now, we'll talk about the probability that is of  interest to us.  We'll write this symbol, probability to get a plus,  say, b plus.  And this means the probability to get the first  particle with spin along a, so that is first particle to be  in the state a plus.  And the second particle with spin along b, so in the state  b plus.  So in the state b plus.  Now, that looks this calculation of this  probability, so you're going to do some measurements.  And you ask, what is the probability I find the first  particle in this direction, second particle pointing in  this direction?  It may look like a somewhat nontrivial  calculation, and it is.  But if you use the fact that this state over there, because  we're asking for this probability on this state,  we're going to be talking about this state.  So if you put the state and you write the state in the form,  you pick one of the two vectors, say a.  Well, the state is a plus, a minus, 1, 2, minus, a minus,  a plus, because I could choose n to be anything.  So might as well choose one of the two vectors to be a.  And then you ask, what is the probability to find first  particle in this state and second particle in this state?  Well, when somebody tells you what is the probability to  find a particle in a state, you put that state into your  sandwich, your psi, with that state, and this overlap,  which is a number, you square it.  So we're going to do the same thing here.  So what is this probability?  Probability to find a plus b plus would be the absolute value.  And then we put here a plus in the first state, tensor  product, we could say, or b plus in the second state.  And we should put, because this is what we want to find,  the state on the tensor product that we're looking for.  And we put the psi here.  So we must put the 1 over square root of 2 times the a  plus 1, a minus 2, minus a minus 1, a plus 2.  So overlap time, well, we go 1 with 1, 2 with 2.  So well, a plus with a plus will give me 1.  b plus with a minus, I don't know.  Second term, a plus with a minus gives me 0.  So this term is irrelevant.  I just need this.  So the a plus with a plus gives me 1.  I have the 1 over square root of 2 here.  And I must close this and square.  I forgot to write that.  So what is this probability?  a plus b plus is equal to the 1 over square root becomes 1 half.  2, the 1 over square root becomes 1 half.  And then all we have left is b plus with a minus in the  second state space, in the particle 2 state space.  The label doesn't matter at the end of the day now that we  disentangle the 1 and 2.  So it's b plus a minus.  I don't have to write that just 2, 2, or you could write it.  And it's this square.  So it simplified a lot, but not quite yet the answer.  What we need here is the overlap between these two spin  states.  And I remind you that when you had any arbitrary two spin  states with n and n prime, long ago, homework 3 or 4 or  something like that, you calculated the overlap between  these two spin states.  And the answer was that you would take cosine of 1 half of  the angle, if there's an angle gamma, the overlap between  the spin states squared was cosine squared of 1  half the angle.  So here we have the vector a, say, the vector b.  Here is the vector a minus, a minus direction.  And if we call this theta ab, this is pi.  This is theta ab, this is pi minus theta ab.  So this should be 1 half cos squared, 1 half of pi minus  theta ab, 1 half cosine squared of half the angle between  the two relevant vectors.  So this is cosine of pi over 2 minus theta over 2, that's  sine squared of theta ab over 2.  So here it is.  It's our calculation of this thing.  It's a neat formula that we're going to need later today.  So one more comment before a little stop.  If b is equal to minus a, now in that case you should be in  luck, because precisely what's happening here is that if one  spin is along the plus direction, the other has to  be along the minus direction, whichever you choose.  So in order to have that a be in plus and b be in plus, well,  this first term would do it.  a is in plus, the first particle.  Here, no.  And b, which is minus a, would be in plus.  So the probability should be 1 half.  So if b is minus a, the angle theta ab is equal to pi.  And the probability of ab is 1 half correctly.  And it's 1 half, because half of the cases a is in plus.  The other cases a is in minus.  So the other case, for example, that could be interesting is  if what is the probability that the first particle is in z  plus and the second particle is in x plus?  Well, these two vectors form 90 degrees.  So you should have 1 half of the sine of half that.  So that should give you 1 half of the sine squared of pi over 4.  And it's 1 half times 1 over square root of 2.  That's another 1 half.  So it's 1 quarter, for example.  OK, I went long enough for a moment.  There's an announcement that Prashant wants to make.  So please listen to him.  Hi.  I'm back for another announcement.  So tonight, the MIT SPS is going to be holding its fall  year off, like in lectures.  So if you don't have a year off, it's a great opportunity  for you to come and see what other research your classmates  are doing in physics.  If you do have a year off, I would highly encourage you  to come as well, because you can actually come and share  your stories about the technical content of what you're doing  in your year off.  It's 7.30 this evening in the PCR Room 8329.  There will be free food, so please join us then.  And see you all then.  Thank you.  All right.  So OK, this was the introduction to what we really  need to do today.  So before we get started, are there  any questions on what we've done so far  on these properties of this entangled state?  So this is a measurement on an entangled state.  Two observers, these two particles  could have flown away a big distance.  Two observers, one tries to see what is the probability.  But the first observer sees the spin pointing in some direction.  The other observer sees the spin pointing in another direction.  It's a natural question which can be done experimentally.  And we've calculated that answer.  OK, so let's begin with this EPR story.  Now you've seen some of EPR last semester in 804.  The only complication with that is  that you really needed to have these mathematics  to appreciate it completely.  So this second look at EPR should be fairly complete,  in that we won't leave almost anything out of the story.  There are many ways of doing that.  Sorry, there are many ways of doing EPR.  And essentially, this bell inequality,  which is the really non-trivial thing that comes after that.  So some are in the homework, some elaborations,  and probably in recitation or later in the course,  we'll see a little more.  But it all begins with a strange thing,  that the kind of thing that you wouldn't expect people  in physics to discuss.  And it's the point of the so-called Einstein, Podolski,  and Rosen wrote a paper.  And they talked about local realism.  Now that sounds like philosophy.  And for a while, people thought, well,  this is interesting, but undecidable,  can't really do anything with this.  So what is local realism?  Now, again, not being exactly physics  is not all that easy to say what it is.  And people discuss that.  But some notion of it is fairly clear.  And the notion is that this reflects something.  It's basically two assumptions about measurement results.  So you measure something, and you obtain a number.  And the first assumption, one, is that this measurement results  correspond to some aspect of reality.  Just say like that.  It seems a little funny.  But you measure something.  And if you get some numbers, because that was something real  about this object, you've had this property.  And so measurement corresponds to some aspect of reality.  So measurements, local assumptions  about measurement results.  So measurements, measurements, M, correspond to some aspect  of reality.  Two, the measurements that you do in your lab  are not affected whatsoever by the measurements  that somebody else is doing at the same time at the moon.  There's no time for the information  of what that result in the moon has given to reach you.  So at that instant of time, what they are doing at the moon  doesn't affect the result of your experiment.  So its measurement is independent of actions performed  at a distant location at the same time.  Now to Einstein and Polovsky and Rosen,  but Einstein was very vocal, physics must satisfy that.  It's kind of sad, I think, actually.  A person that managed to see through and discover  how nature works at so many deep levels,  the photoelectric effect, special relativity,  general relativity, somehow became  convinced that this had to be true.  And unfortunately, he was wrong.  Or fortunately, I guess it's not worth trying to qualify that.  But these two things that seem just so reasonable  are just not true.  This one, measurements correspond to some aspect of reality.  You see, you have a stern girl, a apparatus,  and you throw a spin, it goes up.  You say it ended up with spin up.  Einstein would say, you know, it always had spin up.  It was a reality about that object, that it had spin up.  You just didn't know.  You did the experiment, and you discovered it.  So the thing people try to do in order  to understand this concept, that it  corresponds to something having to do with reality,  is that you admit that half of the particles go up  and half go down.  But you say, actually, there's something  about these particles you don't know.  And if you knew that, you would just be able to tell.  This is a particle that has spin up, and it will go up.  But in quantum mechanics, we have abandoned that.  We've said, these particles, they  are superposition of a state up and a state down,  and there's nothing definitely up about this particle  or definitely down.  So the way people do this to correspond  to this aspect of reality that you don't know  is by introducing what is called hidden variables.  Some things that allow you, there's  something hidden about this spin particle that you don't know.  But if you knew it, you would know exactly how it's  going to come out through the Stern-Gerlach experiment.  And you say, well, that sounds fairly untestable,  but the fact is that it's not.  Now, so this is implemented, this assumption  is when people try to modify quantum mechanics,  they use what is called hidden variables.  Some things that you don't know about the particle,  but if you knew, you would see that, in fact, this particle  has spin up.  The second is in some ways even more disturbing,  because we got accustomed to the idea  that locally simultaneous things that cannot be reached,  events that cannot talk to each other via the exchange of light  cannot affect each other.  So simultaneous things that happen far away  can't affect each other.  So this also sounds very reasonable, but that's also wrong.  And there's the obvious question.  So if this is wrong, can you send information faster  than the speed of light?  And the answer is people have looked at it in many, many ways.  And it's very interesting, and we could discuss that,  but it would take us long.  So I will leave it to maybe recitations, maybe other forms.  But here, actually, there's no contradiction,  no way of finding information, real information,  going faster than the speed of light,  even though things far away at the same time can affect you.  So two very interesting things that  seem very dangerous to discard, but turn out to be wrong.  So this is what EPR did.  And they made some thought experiments  that we're going to review to some degree  and see if we can discard these assumptions.  So that's what we're going to do now.  I'm going to try to understand that.  Now, if you're interested in what hidden variables are,  my discussion will not use hidden variables.  Although they are kind of implicit,  you will see as I state some things,  I'm basically going to be explicit on the fact  that things, they're real facts, as Einstein would  like you to think.  And we try to see if those real qualities about particles  get us in trouble.  So let's begin and try to discuss this first experiment.  You see observer 1, Alice and Bob again, if you want.  Alice is going to measure about the z-axis.  Bob is going to measure about the z-axis.  And therefore, if Alice finds spin up, Bob finds spin down.  If Alice finds spin down, Bob finds spin up.  And that's a correlation.  And it's very interesting.  But Einstein, Bell, and Poloski would say, look,  it's not all that interesting.  There's nothing all that mysterious happening here.  EPR would say, says, the pairs that you've built,  so-called entangled pairs, are pairs of particles  with definite spins, spin directions, spin vectors.  So what you've built, EPR would say,  if you have here particle 1 and particle 2,  the proper, I'm going to list the properties.  Suppose you have a particle 1.  You've created, Einstein would say, some particles,  particle 1s with spin up in z, and particle 2  with spin down in z.  Or you've created particles 1 with spin down in z,  and particle 2 with spin up in z.  And you've created, in fact, 50% of the particles  are of this type, of the pairs.  And 50% of the pairs are of this kind.  So don't tell me all this superposition, Hocus Pocus.  Half of your pairs, one particle has spin up,  one particle has spin down.  The other 50% of your pairs, one particle 1 is down,  the other particle is up.  No wonder they're correlated.  You get plus, gets minus, get minus, get plus.  What is the probability that you get plus?  50%.  What is the probability that you get minus?  50%.  Everything is reproduced.  Mystery over, no quantum superpositions.  There's no mistake here, no mathematical mistake.  You're saying that the particle has a definite spin.  You maybe not know it, but we say, look,  your particles, in fact, had a definite spin, z plus and z  minus, and those things.  And this is where hidden variables would come along.  You would say, well, what is, if you have the particle 1,  its spin is a function of some hidden variable  that you don't know.  But if you knew it, you would know what the spin is,  because it has a definite spin.  You don't know what is the hidden variable,  but as a function of the hidden variable,  the spin is known, definite, not a superposition,  nothing like that.  It's a very aggressive attack on quantum mechanics  and something that troubled people and, in fact, fascinates  people even up to now, because the idea that these kind of things  are really absolutely wrong, it's very shocking,  perhaps the second, even more shocking,  because, OK, by now maybe you're accustomed to all kinds  of variables that are a little more detached from reality.  You have electromagnetic fields, and they  force you to learn about potentials that  seem to be a little more abstract, and similarly here.  So no problem.  So people say, OK, this is a simple situation,  but it may be that we're going to do more measurements,  and we're going to consider two directions that are different.  Maybe Alice has two Stern-Gerlach machines,  one that measures about z and one that measures about x,  and Bob has also two Stern-Gerlach machines, one  that measures about z and one that measures about x.  And they are going to ask different questions,  because we know the spin, how it transforms,  so getting those results right with two directions  is going to be a little more interesting.  So we're going to try measuring in two possible directions,  both A and B, Alice and Bob, can measure in two directions  z and x.  And Einstein would say, look, yeah, you can measure z and in x.  Let's for to avoid confusion, let's not  talk about one measurement after another.  Particle comes, you measure in z or you measure in x.  And realism says that you don't know what you will get,  because maybe they're hidden variables,  but each particle has a definite answer  if you ask what is the z direction of the spin  and has a definite answer if you ask  what is the x component of the spin.  Definite answers, real answers, realism again.  So I'm going to label the particles.  For example, this is a label for a particle z plus x minus.  This particle, labeled like that,  would be such that if you measure its spin in the z  direction, its plus, if you measure the spin in the x  direction, its minus.  So measured in z, it is up, measured in x,  it is down.  These are the kind of particles that EPR would say exist.  It's not that you got a particle out  and it's in some strange entangled state.  These particles are flying away.  They're not talking to each other anymore.  And this particle, since you can measure in two directions,  there's some reality and the measurements  correspond to reality.  So they're attributes and this particle  is classified by having these attributes.  If you measure z plus, if you measure x minus.  OK, so how about this situation?  Well, let me make a list now, particles 1 and particle 2.  And this would be the list that EPR would do for you.  EPR comes along and says, look, here's what you're doing.  Particle 1, suppose it's a z plus x plus.  Well, in your beams, actually, these particles,  when particle 1 is that, your other particles are z minus x  minus.  So if particle 1 is of this type, particle 2 is of this type,  that way EPR protect themselves because they say,  look, if you measure z plus and you measure z minus,  you get correlation.  If you get plus, plus, you get correlation as well.  Now, there can be a particles in z plus x minus.  This would go with a z minus x plus.  There could be a particle z minus x plus and this  as z plus x minus.  And there could be particles z minus x minus z plus x plus.  So there are four cases, four types of particles,  they say, have been produced.  And 25% of pairs are of this form.  25% of pairs are of this, 25 of this, and 25 of this.  So you could ask some questions to EPR.  What is the probability that you get z plus for 1 and z  minus on 2?  Well, z plus on 1, it's these two cases.  And z minus on 2, well, those two cases.  So this is 50% of the times.  So it's 1 half that probability.  It's correct.  That's what we would predict from the entangled state  viewpoint.  Oh, but let's ask something mixed.  Let's see if we get in trouble.  P of z plus on 1 and let's say x plus on 2.  Well, z plus on 1 and x plus on 2, that's not it.  This case, z plus on 1, x plus on 2, 25% of the times, 1 fourth.  This was the probability we calculated.  So you tried with one direction.  You don't need quantum mechanics to produce a result.  You try with two directions.  You don't need quantum mechanics to get the result.  So people got stuck and they said, well, maybe it's undecidable.  Maybe this is philosophy.  Maybe this is something.  And people had to wait for Bell that he said,  I'm going to try three directions.  Now, three directions makes a big difference.  This is the first time it really goes wrong.  So it's kind of surprising, perhaps, at some level.  But this is subtle stuff.  This is subtle stuff, so it takes a while  before you find something wrong.  You're talking about showing Einstein this wrong,  so that's not so easy.  So three directions.  So particles are going to be of types.  EPR would say, look, I'm going to use the same strategy.  I'm going to say that particles have three attributes now.  They're all physical.  They correspond to reality.  Because if you measure in either of three directions,  they have to have an answer for that.  So here is a label for a particle.  And for example, A plus B minus C plus.  So if you measure, it would give spin in the A direction  plus h bar over 2, spin in the B direction minus,  well, in the B direction, it would give you minus h bar  over 2, spin in the C direction, it  would give you plus h bar over 2.  So we're not measuring simultaneously.  We're just asking, well, you take a particle,  do a measurement, and see what you get.  And we're always going to be asking  for probabilities of this kind, probability  that the first particle is doing this  and the second is doing that.  Well, EPR would start now with the particles again.  Particle 1, particle 2, and population.  So let's list quickly the particles.  A plus B plus C plus.  Then you would go A plus B plus C minus.  Then you've done the C plus C minus.  Here you go for two more.  A plus B minus C plus.  A plus B minus C minus.  Four, I'm supposed to fit four more there.  Can I?  Well, I'll try A minus.  Now you've done all the four A pluses,  so you need four A minuses.  Then you have B plus, B plus, B minus, B minus, C plus, C  minus, C plus, C minus.  OK, I got all I think.  And particle 2, of course, is correlated.  You would say, well, I don't need to write it,  but it helps seeing what's going on.  So I'll write it A minus B minus C minus.  A minus B minus C plus.  A minus, let's use bars here, A minus B plus C minus.  Plus C minus A minus B plus C plus.  A plus B minus C minus.  A plus B minus C plus.  A plus B plus C minus.  A plus B minus C. This was minus, so it's plus here.  OK, we're done.  Lots of labels.  And you could say, well, maybe you  want to put 1, 8 in each of them.  But actually, the argument is more interesting.  It doesn't need you to try to put fractions.  So let's consider that there's a total number of particles, N,  which is N1 plus up to N8.  And here are N1 of this, N2 of this, N3 of this,  N4 of this, N5, N6, N7, and N8.  OK, let's see.  All right, so we have that.  And then, well, it takes imagination  to see how you're going to run into some contradiction.  So what is the basis for the contradiction?  Somehow, this formula, which is really quantum mechanical,  must eventually go wrong with all these attempts  to deny that the world is quantum mechanical.  So we could split, again, those particles into equal fractions,  but there's no need to do that.  And it's clearer if you don't.  So you try to combine the three directions into one equation.  So one way to do that would be to say, OK,  what is the probability that you get A plus and B plus?  So this is for the particle number 1,  and this is particle number 2.  So then you must look at the table  and say, which ones do that?  Well, you need A plus in the first,  so it's one of the first four rows,  and B plus in the second.  So actually, it's these two cases, N3 and N4.  4 over N.  We want to involve three directions,  so let's go for another one, P of A plus with C plus.  Let's see how much is that.  Again, this is for the first particle.  This is for the second particle.  So I must look at the first four rows  and see that you have an A plus.  All of the first particles in A plus for the first four rows,  but the second should be in C plus,  and that is cases N2 and N4.  So we get N2 plus N4 here over N.  Well, we've involved this A with B, A with C.  How about involving B with C?  So I'll put C plus, B plus, for example.  OK.  C plus, B plus.  I must look at C pluses and B pluses.  No.  C plus and B pluses.  Yes.  N3 is there, which is good because it already was there.  And which else?  C plus and B plus.  C plus, no.  C plus here, B plus.  Yes.  N7.  OK.  Now, N3 plus N4 is less than N3 plus N7 plus N4.  It's less than N3 plus N7 plus N4 plus N2.  You have N3 and N4.  And now I add whatever N7 and N2 are.  And that's then an inequality because it's  going to be more cases.  Now I divide by N. So you obtain inequality  that P of A plus, B plus, is less than or equal than N3 plus N7  and N4 plus N2.  I'll write the second first.  P of A plus C plus plus P of C plus B plus.  So I didn't put specific populations 1, 3, 1, 4.  But in general, whatever populations you choose,  this inequality must hold.  Now that's a more clever strategy  because suppose you choose some populations  and you don't get in trouble.  Well, maybe with some other populations  you would get in trouble.  Maybe it's not so easy to guess the relative factors.  So here is something that must be true whatever populations  you choose.  And now that is Bell's inequality.  So the achievement of Bell is to somehow translate  this assumption of realism into an inequality.  And now quantum mechanics has a formula for these things,  for this probability.  So we can test whether this is true.  So let's do that.  This is the so-called Bell inequality.  So if quantum mechanics is true, the following should hold.  Let's see that.  If Qm is true, well, there should  be a problem with this inequality.  So let's see what it happens.  Let's see if it's true, this inequality.  Does it really true?  Well, the left-hand side would be, given the formula that we  had, 1 half of the sine squared of theta AB over 2.  So let me just emphasize.  This was derived using local realism.  Local realism gives that.  So you do the experiment, get these probabilities.  And if realism is true, this should hold.  Let's see what quantum mechanics has to say.  Let's plug in the values that you get from quantum mechanics.  Now we can see that the value of the quantum mechanics  we calculated this probability.  We put the first term.  Here is sine squared of theta AC over 2  and 1 half of sine squared theta BC over 2.  So does that work?  Does that always work?  Can I orient this axis in such a way to disprove EPR?  And in fact, it turns out to be quite easy to do that.  So you choose three vectors like this, AC BC.  So C here, A here, and B here, I believe.  Yep.  So put an angle theta here, an angle theta here.  And then what do you have?  Theta AB would be 2 theta.  Theta AC would be equal to theta BC equal to theta.  That's a pretty nice simple choice of angles.  If you choose these angles now, let's  see what happens with our inequality.  So you get 1 half sine squared theta AB over 2 would be theta.  Is it less than or equal to this?  Well, these two become the same.  So you get sine squared of theta over 2.  Violated or not violated?  Is it true or false this for all theta?  What do you say?  Yes?  Give theta is less than pi.  That's not true.  Close.  It's not true for small theta.  So if you're this and you're desperate to know,  the thing you have to do is assume theta is very, very small.  See if you get in trouble.  How much is this?  1 half theta squared.  Let's see.  How about this one?  Yeah.  And how much is this?  This is theta squared over 4.  I was not seeing it.  Sine theta for small theta is roughly theta.  So here is theta over 2 squared.  But here's 1 half theta squared.  And it's false.  The 1 half theta squared is not smaller than 1 quarter  theta squared.  And in fact, for theta equal pi over 2,  I think this is an equality.  Because theta equal pi over 2, you get on the left-hand side  1 half less than or equal than sine of 45 degrees, which  is correct.  So it's an equality at pi over 2 fails below.  So it was a shock that if you could  do an experiment in quantum mechanics  and an experiment with correlated entangled particles,  that you could measure these probabilities, these correlations.  And you would obtain a result that actually contradicts  for certain alignments of your experiments  the assumptions of local realism.  So it was a great result of Bell to show  that quantum mechanics is in contradiction  with local realism.  There's no way to keep the ideas of quantum mechanics  and put hidden variables and assume  that there's real values for things  and that there's no effect at the distance.  It would all be contradicted by an experiment that was done  later.  And the definite version of the experiments around 1980  or 82 by Allen Aspect and others, very clever experiments,  worth reading and understanding.  But confirm that the quantum mechanical result is really  true by measuring correlated pairs,  and these inequalities are violated.  Yes?  So what about David Baum's theory of hidden variables  quantum mechanics?  David Baum's theory of quantum mechanics theory.  I know that allegedly it reproduces the same results  as quantum mechanics, but it's still  a hidden variable theory.  I don't think there's any hidden variable theory that works.  David Baum, I think, actually was  created by rewriting EPR, who essentially  talked about position and momenta to talk about spins.  And he might have been the first one that  began to try to do hidden variable theory,  but no hidden variable theory works at this moment.  And this shows it.  So people say that actually this assumes  that there's local hidden variable theories  and there's non-local hidden variable theories  and all kinds of strange things, but it's  more and more unnatural.  So it doesn't seem to do something very interesting.  Yes, there are many questions.  Aaron, maybe you want to say something?  I think Baum has a non-local hidden variable theory.  It's kind of awful looking, but I guess  violates the second principle rather than the first one.  But also, it doesn't extend to, you know,  it doesn't really work for everything.  Like, we don't know how to make it work for a spin  one-half particle or finite dimensional.  So I think it's really believed to be a true theory of rules.  Thank you.  One more question.  Steve.  In the case with EPR, is there a problem  that we can have scenarios where the spin is greater  than spin one-half particle would have?  Like, if we had a state a plus b plus c plus,  we would have 3h forever, too?  No, the statement that is done here is not that, well,  this is a label for a particle.  EPR just assumed that if you measured a,  you would be able to get this.  If you measure along b, you would get this.  And if you measured along c, you would get this.  So this is a single particle.  Any measurement gives some results.  These are the lists.  There's no sense in which these are added.  But even when you measure one, the other value still  exists with those measures?  Well, there's no, I believe there's  no need to discuss that.  So they don't talk about the statement  of doing subsequent measurements within this statement.  You just take this particle and you decide.  You measure a or measure b or measure c.  You don't try to measure simultaneously.  You don't try to measure one after another.  You just do one measurement.  And that already, which is the minimum you can do,  gets you in trouble.  So I'm not sure how EPR would phrase subsequent measurements  after they've done the first measurement or things like that.  But they're not necessary for this stage.  OK, so look, it's a very interesting thing.  There's lots to discuss here.  But it's best if you read it.  James had a question.  Just let's take one quick question.  I was just wondering if there was any other extension  beyond three directions or belts and qualities.  Are there like a bay and direction  belts and qualities or some type of form of it?  Or is there really a need for it?  There are other forms of belly inequalities.  I believe, not sure if it's popular with four directions  or other things, but certainly Weinberg, for example,  discusses other ways.  There are alternative ways to phrase it.  I've talked about here probabilities to observe results.  There's a more perhaps common way talking  about expectation values or correlation functions.  This is something you'll do in the homework.  This sort of game that is done in the homework  but was a suggestion by Aram to put it in the homework,  this game in which with a quantum strategy and entangled  pair you beat the system is yet another formulation  of the belly inequalities as well.  So lot to do, but I think it's better now that we stop  and talk about angular momentum from now  until the end of the semester.  This was about spins, but we now have  to put together angular momentum, which  is orbital and spin and all the various kinds.  So we'll begin with angular momentum.  So there are notes on the web on that  that I wrote and modified a little this time  and lots of little exercises.  So what I want to do now is guide you  through the things that happen there  so that you get a view of what we're going to do.  We're going to work with this thing in an elegant way.  Using vector notations for operators  is going to help us understand things better.  So we've seen angular momentum before,  and let me summarize simple things that we know about it.  If you have angular momentum, and let's begin with orbital,  this is what we have when we take Lx to be ypz minus zpy Ly  equals zpx minus xpz and Lz to be equal to xpy minus ypz.  These are the angular momentum, orbital angular momentum  operators.  Now, it's better for many things to use labels like x, y,  and z.  Those operators call them x1 hat, x2 hat, x3 hat.  And px, py, pz, p1, p2, p3.  In that way, you can write commutation relations like xi, pj  equal ih bar delta ij xi with xj equal pi with pj equals 0.  So it allows you to write things more quickly.  In fact, the angular momentum operators  become also a little simpler.  It's sort of x2p3 minus x3p2.  And you don't have the x, y, z labels and all that.  So we want to use vector notation.  Now, vector notation, you can do it in two ways.  You can talk about triplets.  Those are vectors.  Or you can form the vectors themselves.  Now, if you form the vectors, you get objects  that sometimes are a little disconcerting.  But we try that they not be so.  So here is the r vector operator.  You could think of it as the triplet of x, y, and z.  But let's call it like this.  x times the first basis vector plus y operator  times the second basis vector plus z operator  times the third basis vector.  And we've done things like that.  And we understand that these basis vectors really  don't talk with these operators.  They can be moved across each other.  The basis vectors are sort of things  that help you write expressions rather  than talking about triplets.  Same thing for momentum.  You would put, let's do it this way, p1 e1 plus p2 e2 plus p3 e3.  And finally, well, angular momentum, the vector operator.  You've done a lot of the angular momentum vector operator  for spin.  So here you would put lx or l1 e1 plus l2 e2 plus l3 e3.  So those are ways to write equations  that carry all the operators and treat them as vectors,  even though they're operators.  So they're unusual vectors.  They're vectors whose components are operators.  They're not numbers.  So the obvious question is what changes them?  So we're going to define dot product and cross products  as we had before.  But we have to be a little aware that when we write these things,  we could make a mistake unless we're not careful.  So here are two vector operators.  Operators.  What is the dot product of these two vector operators?  Well, you know it's supposed to be the first component of this  and the first component of that, second, second, third, third.  So it should be a i, b i summed.  Repeated indices are summed.  Now, I should not write b i, a i with a reverse order  because this thing, the components, are now operators.  And maybe they don't commute.  So I've defined this once and for all to be this.  And A cross B, the i-th component of this thing,  is going to be defined once and for all  to be epsilon i j k a j b k.  A definition, the A to the left of the B. And with this,  we can check our usual rules of manipulation of operators.  So one more definition.  A squared is going to be A a and it's going to be a i a i.  OK.  Simplest calculation is A dot B equals to B dot A.  Yes or no?  No, they're operators.  So let's calculate the difference.  Let's get a little practice, calculating differences.  So I write A dot B is equal to A i B i.  It's summed.  So then I say, oh, that's A i B i plus B i A i.  You see, the commutator is A i B i minus B i A i.  And I add it back.  But this thing, B i A i, is B dot A. So here I got a formula.  A dot B is equal actually to B dot A plus this commutator A i  B i.  And now you've got a new formula for operator vector  analysis.  A dot B and B dot A are not the same,  but they differ by this thing.  And a very important corollary, very famous corollary,  what is r dot p?  Is it the same as p dot r?  If you're working in quantum mechanics,  you may be tempted.  Oh, r dot p and p dot r are the same,  but r and p don't commute.  So what is the difference?  r dot p is equal to p dot r plus xi pi.  And how much is that?  Sorry?  I h bar.  I h bar?  I h bar.  3 I h bar.  Yes, don't forget the sum.  This is supposed to be sum.  So this says x1 commutator p1, x2 commutator 3 I h bar.  So here is a famous formula.  rp differs from pr plus 3 h bar i.  I h bar, people write.  OK.  Another formula that you would be curious to know,  well, the dot product was supposed to be symmetric.  It's not.  The dot product, the cross product,  is supposed to be anti-symmetric.  Is it or not?  A cross B sub i is equal to epsilon ijk ajbk.  OK.  What do I do next?  I want to move the a and b's around.  So I'm going to replace this by a commutator  plus the other ordering, ijk.  And I put here, well, this should be a parenthesis,  ajbk commutator plus bk aj.  Now, what do we get?  Well, you have to look at the second term.  Let's put the first term here, because that's pretty much  we're not going to be able to do too much with it, ajbk.  And the first term, I would write it like this, minus epsilon.  Flip this to ikj bk aj.  If you do it like that, then it sort of fits nicely  with the definition of the cross product.  Because in the cross product, the first label goes here.  The second label goes with the last label of the epsilon.  So this thing is minus b cross a.  And that was a cross b plus epsilon ijk ajbk.  So that's your formula for how the cross product now  fails to be antisymmetric.  It's not necessarily antisymmetric unless you're lucky.  So here is a property that you should try to think about.  How about r cross r?  Is it 0 or not 0?  R cross r.  You could say, oh, what can I do here?  Well, r cross r minus r cross r, I would have 2r cross r  should be equal to this.  But both components are x's.  So r cross r is 0.  p cross p.  That's also 0.  And therefore, say, l cross l, is it 0?  Maybe, is that right?  No.  l cross l, we'll see it a little later.  But it's not 0 because one l with another l don't commute.  Lights.  OK.  So l cross l is actually not 0 because this thing is not 0.  We'll talk about it a little later.  But actually, this is a very famous one.  l cross l is proportional to l with an h bar.  It's a lovely formula.  So another interesting thing, well, what is r cross p?  r cross p, for this formula, with minus p cross r.  And how about the other term?  Is it 0 or not 0?  Is r cross p equal to minus p cross r, or does that fail?  Well, r and p don't commute.  So actually, this formula, somebody  should have complained.  They should have been the i-th component, the i-th component.  And here is the i-th component.  So the i-th component, the i-th component,  let's look at here.  Epsilon ijk xj pk.  But xj and pk is i h bar delta jk.  Now, this delta jk is symmetric.  This is anti-symmetric.  You should get accustomed to the idea that that is 0.  If this is anti-symmetric and this is symmetric, this is 0.  Let me do it one more time.  Maybe this is not completely obvious, but epsilon ijk delta  jk.  The intuition is also obvious.  Epsilon must have the three numbers different,  and this forces them to be the same.  But it's more general.  If this is anti-symmetric and this is symmetric,  this should be 0.  And the way you do that is you say relabel j and k.  Whatever was called j, call it k.  Whatever was called k, call it j.  So this is epsilon ikj delta kj.  And then use the symmetry property.  So when you exchange this back, you get a minus sign.  When you exchange this back, you don't get a sign.  So minus ijk delta jk.  So you've shown that this thing is equal to minus itself,  and therefore it's 0.  Something that's equal to minus itself is 0.  So this is 0.  And you got that r cross p is really minus p cross r.  And that's what we call the angular momentum, r cross p.  But now you know that it's also equal to minus p cross r.  OK.  Let's see.  One other one, for example, that is classically true.  Let's see if it's quantum mechanically true.  r dot l.  You know, the angular momentum, l,  is supposed to be perpendicular to r and perpendicular to p,  because it's the cross product.  Is that true quantum mechanically or not true?  Well, maybe I didn't say, well, l is r cross p.  So this is r dot r cross p.  So this would be ri times epsilon ijk r, no, x.  xj pk.  OK.  So what is this?  It's epsilon ijk xi xj pk.  Well, this is 0 because these two x's are operators,  but they commute.  Therefore, this object is symmetric in i and j.  This is anti-symmetric.  This is 0.  So r dot l is actually 0.  How about p dot l?  Well, there's two ways of doing that.  One that looks pretty obvious, one that looks a little harder.  Let's do the one that looks a little harder.  This would be pi epsilon ijk r, xj pk.  OK.  So here is the temptation.  Must be 0 because it's a pi pk symmetric in i and k.  And here is anti-symmetric in i and k.  But it's wrong.  These are not obviously symmetric  unless you can move them across.  And there's an x.  In the middle, laughing at you and saying, beware.  This could go wrong.  So you have to be a little more careful.  So let's do be careful ijk.  And you put pi xj pk.  All are operators.  All are operators.  But pi with xj, the commutator, would be a delta ij.  And that vanishes.  So this p actually can be moved across the x.  And therefore, show that that thing is on the other side  because the epsilon is there.  So it's a lucky thing.  So pi xj pk is not symmetric in i and k.  But if there's an epsilon, you're in better shape.  So here, the p can be moved.  ijk xj pi pk.  And now, nobody can stop you from saying, this is symmetric.  They're next to each other.  They can be moved across each other.  It's really symmetric in i and k.  And therefore, i and k are anti-symmetric.  This is 0.  p dot l is equal to 0 as well.  So this was a little bit hard way.  If you had used that l is actually equal to p cross r,  then the p would have been next to the p from the beginning.  And you would have saved a little time, the same time  that it took you to realize that this equality is true.  So those are true equalities.  p dot l is 0.  And it's also equal to l dot p.  It's not obvious that l and p commute.  But in this case, l with p dot p is 0.  If you use this formula for l, it's obvious.  r dot l is 0, it's also equal to l dot r.  Doesn't make a difference.  It's also true as well.  So that's roughly what goes on.  Look, I would like you to just read those pages.  It's a continuation of this.  It's about eight pages.  I leave exercises to be done.  That's in the homework of this type playing with the indices.  It's a good thing that you get accustomed to it.  And the material that you need just for the last problem  of the homework will be covered on Monday.  And we can talk about it in recitation tomorrow.  So that's it for today.
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