 Anglementum is enormously important in physics, for example, it's central to all kinds of scattering experiments and scattering experiments are at the core of high energy physics. They play a very big role in condensed matter physics. Anglementum plays a central role in the theory of the application of quantum mechanics to atoms to get atomic structure. So, from the very beginning of the subject, it played a very big role, and people write whole books, horrifically, they write whole books on Anglementum in quantum mechanics. So, we are going to have to spend a few lectures on it, even though there won't be quite as much physical content. We're building foundations for later work, is what I suppose I should be saying. But we will, on Wednesday, in the lecture, we will at least be able to do something interesting and useful with Anglementum, so the outlook is not entirely bleak. But I'm afraid today's lecture is a bit on the formal side. So, you will recall, I hope, at the end of last term, we talked about operators that generated translations. They turned out to be momentum operators, and we concluded that there must be operators that affect rotations. So, there must be a unitary operator, U of alpha, which generates a state like the state you've already got, that generates a state that's the same as the system you've already got, except rotated by an angle mod alpha around the unit vector. In the direction of alpha. There must be some unitary operator like this. This is a unitary operator, depending on a continuous parameter. You can shrink the angle of your rotation down to nothing continuously. It's in that class of continuous unitary operators. So, it's generated, we can write it as the exponential of something or other. By putting an i in there, this thing becomes the Hermitian operator, these j's. And because there are three components of this vector alpha, which describes the rotation that you're planning, there must be three of these operators that generate these rotations. And we're calling them, of course, GX, JY and JZ. I claimed, I said that they are the angular momentum operators, but we haven't really done a great deal. We didn't do a great deal of that time to justify this claim. Okay, so we have those three operators. They're the generators of rotations respectively around the X axis, the Y axis and the Z axis. Out of them, because they're Hermitian operators, we can construct another operator called J squared as the sum of the squares of the operators. And we have a set of four operators and we showed, by considering what happens when you make rotations around different axes, we demonstrated that these operators must have the commutation relations that J squared commutes with all of them, with all of JX, JY and JZ. But these operators do not strangely commute with each other. They have the commutation relations that JX, JY is IJZ and similar things, which can be encapsulated in this way where epsilon IJK is the object that keeps changing its sign. And it's zero if any two of its subscripts are identical. So what we want to do now, so that sort of show the existence of these things, what we have to do in the next section is find out more about these operators and the eigenstates of these operators. We need to justify the claim that these operators really are the anglimentum operators. We need to find crucially, well it will turn out that the orientation of something like an electron, well indeed the orientation of any quantum system, is encoded in the amplitudes to find the possible results, the possible eigenvalues when you make a measurement of JX, JY or JZ, there will be possible answers, you'll get a number which belongs to the spectrum of that and the amplitude for that event strangely encodes the orientation of the object. We need to understand about that. So what we want to know now really is what is the spectrum of these operators. You want to know what are the possible results of measuring J squared or JX squared or JZ squared or whatever, right? So this is what the next spectrum is about, it's about the spectrum of J squared at L, right, these operators. So, since the J, the JX, JY and JZ don't commute with each other, there isn't a complete set of mutual eigencets of JX, JY and JZ. But there is a complete mutual set of eigencets because of that commutation relation because J squared commutes with all of its subordinates. There is a complete set of mutual eigencets of J squared and any one of those and it's conventional to pick just at random that we choose to have mutual eigencets of J squared and JZ. So it's just a convention that we choose JZ out of the three things connected to the fact that Z is the singular axis, is the special axis in systems of spherical polar coordinates, right? So in spherical polar coordinates X and Y have pretty much the same role in life but Z axis is special and that's why we choose this one. So that's what we're going to do, so we're going to say, look there must be some eigencets, we're going to label them by beta M. This label is going to tell us how the thing responds to this operator, concretely it's going to be this, right? So obviously we're labeling this ket by its eigenvalue with respect to J squared and JZ on beta M is going to be M, beta M. So the second label in this thing tells you how it responds to JZ and this is by definition a member of the complete set of mutual eigencets of this operator and this operator which the mathematicians have promised us exists. Now we introduce some ladder operators. We're going to follow a line of reasoning that's very similar to how we got the eigenvalues of the Hamiltonian of a simple harmonic oscillator. So we're going to introduce J plus as JX plus IJY. So this is a little bit analogous to when we introduced in the simple harmonic oscillator the destruction operator we said that A was equal to X plus IP. So because of this I, this is not Hermitian, it's not an observable, it's a tool of the trade and correspondingly needless to say we have J minus which is equal to JX minus IJY and we also have that J plus dagger is equal to J minus. So this thing here is the Hermitian adjoint of that thing there because if you take the dagger of this equation this dagger goes into this because it's an observable, that goes to minus I and this goes into this. So these are tools of the trade. Now we ask ourselves what are the commutation relations. We have that J squared on J plus is nothing but J squared comma JX plus IJ squared comma JY is nothing because this vanishes and this vanishes, right? So J squared commutes with J plus and of course it commutes with J minus as well, right? So this is plus or minus, it vanishes. What does that tell us? That tells us that if you take J plus of beta M you use this nontrivial operator on this state, you get some other state. What can we say about this other state? Well one thing we can say is that J squared applied to it because you can swap these two over is the same as J plus beta beta M. So you swap these two over then J squared looks at this and says ah ha that's my eigenket out pops a beta. This is a mere number, can be popped over here is equal to beta J plus beta M. So when you use J plus on this eigenstate of J squared you get another eigenstate of J squared for the same eigenvalue, right? Encouraged by that the next thing to do is to have a look at JZ on J plus beta M. Now when we swap these two over, we want to swap the two over but of course we can't so we do the usual business. J plus JZ I've swapped them over and then add in what we should have and take away what we're not entitled to. JZ comma J plus commutator brackets brackets beta M. Now this we found what this thing was. We found that oops sorry we didn't. I'm getting ahead of myself. Apologies, right. So that's what we need to. Okay well we're going to find out what this is. We're going to find out what this is. That's the next thing we have to do. Alright so what is JZ comma J plus? Well it's JZ comma JX plus IJZ comma JY. This is minus IJY from the rule given way up there. And this is minus I, so this is going to be I, this I minus coming up another IJX again from the rule above. So this is going to be minus, sorry this is going to be J plus because these two I's are going to make a minus sign cancel list. So we're going to have JX, it has to be plus. So what the hell's gone wrong here is I've goofed presumably in that X, Y. Yes I've goofed in that, sorry. I'm always bad at this cyclical ordering. So it is equal to J plus. So we take this important result. Stuff it in there and we have that JZ on J plus beta M is equal to, so this is going to be J plus and JZ working on that is going to produce an M times that. So we're going to have an M plus one times this. So what does that show? That shows that when you apply J plus to this object you get a new Eigen ket of this operator, one which has this for an Eigen value. So let's write that down. J plus on beta M is equal to M plus one, sorry, is equal to some amount of, which we will call alpha plus, the state beta and M plus one. So the point is that what goes in here is the Eigen value of this thing with respect to JZ. So this thing here turns out to be, this shows that it is an Eigen ket of this operator with this Eigen value, M plus one. So that's what should go in there. And this is some constant, this is some normalising constant. So what have we achieved? When we applied J plus to this state JM we made a new state with the same total amount of angular momentum, the same response to J squared. But the amount of this parallel to the Z axis has increased. So we have reoriented our system, right, we have here a spinning top, well, okay, some angular momentum along here, and we've moved it a bit towards the Z axis. That's what J plus does. It realigns the angular momentum that you've got. Strictly speaking, it makes you a new state, and this new state has the same angular momentum as the old state, but more of its parallel to the Z axis. Okay, we could repeat all this stuff. I recommend that after the lecture you do repeat all this stuff using J minus, and you will find that J minus on beta M is going to equal some amount, not to be determined, not known yet, of beta M minus 1. It does the reverse trick. It moves it away from the Z axis, or if you like, towards the minus Z axis. So showing this is precise repeat of what was done up here except every plus sign gets turned into a minus sign. Okay, now we have that. The expectation value of, for example, Jx squared is equal to Jx of psi, sorry, for any state of psi. So take any state of psi and work out this expectation value of Jx squared. It's Jx of psi mod squared, right? Because if you take the mod square of this, what you're doing is taking the emission adjoint of this, which is that, the emission adjoint of Jx, which is Jx itself, and multiplying it into this, so you end up with this, and this is clearly, this is the length squared of a vector, so it's greater than or equal to nothing for all of psi. So let's ask ourselves about Jm J squared, Jm. That's clearly equal to beta, because J squared onto, sorry, M of beta M, beta M. So J squared on this produces beta times this. This is correctly normalised, so we get beta, but this can also be looked at as beta M Jx squared, beta M. So it's equal to this, plus beta M Jy squared, plus beta M Jz squared. But this, this last one here is clearly M squared, because J, one of these Js looks at this and produces an M times beta M, and the other one then looks at that and reduces another M times beta M, and we end up with just M squared. So what have we got? We've got that beta is equal to, what should we call this? We'll call this A, and we'll call this B is equal to A plus B plus M squared, where these numbers are greater than or equal to naught. In other words, we concluded that beta is greater than equal to M squared. So there's a problem. We've got an operator J plus, which can make us a new state with M increased by one, which has, but this new state has the same value of beta. So apparently we can make states with bigger and bigger M for the same beta. And that we just shown mathematically that that's absurd, physically it's absurd, because I'm saying that I've got a fixed amount of angular momentum and J plus just moves it towards the z axis. Well eventually it'll have it parallel to the z axis and it won't be able to increase M any more. So what truncates this, something has to give, and it's just like the harmonic oscillator, what gives is that eventually, so series of states of bigger M truncated at beta M max, for maximum value of M such that, how does this happen? It happens because when we use J plus on this state, we get exactly nothing. So what does this mean? This implies that alpha plus equals naught in this particular case. That's the only way we can be stopped from making states of bigger and bigger M and it's clear we have to be stopped, so we are stopped in this way. So what we have to do now is look at the mod square of this state and show that it's zero. So we have that naught is equal to mod. So the mod square of this is going to be this emission adjointed times J plus dagger, which is J minus times J plus times beta M max. So this thing here, this is J plus dagger, which is appearing here, and I pointed out earlier on that J plus dagger is J minus. So let's have a look, see what we've got here by staring inside. So this is going to be, I don't need the mod square, that's already taken care of. So this is beta M max J x minus i J y, J x plus i J y, close brackets, beta M max. So we multiply this stuff out and we get J x squared plus J y squared and then we get, we have a minus i J y x, the plus i J x y, so we have plus i commutator J x comma J y. Well, when we've got this much of J squared, you might as well have a whole of J squared, so we write this as beta M max J squared minus J z squared. So we add a J z squared and take it away again. And this of course is i J z. So along with that i, we get minus J z beta M max. And now we're in clover because we know what every single one of these operators produces when it bangs into that so we can evaluate this. This of course produces a beta. So this is going to be, this is going to produce a beta times this thing. Then this thing will meet this thing and produce one so I only need to write down beta. This J z will produce an M max times this thing which will then bang into this thing and produce a one. So I have a minus M max and this one is going to produce M max squared also with a minus sign. So in fact let me write this as, oh never mind, M max squared. So this is more conveniently, so what do we have? We have that nothing is equal to this stuff from which it follows that beta. We've discovered now what beta is in terms of M max. It's equal to M max brackets M max plus one. So if we apply J minus to beta M I claimed that this was alpha minus beta M minus one. So M will start, let's imagine M starts off positive as we take units from it. It's going to get smaller and if we keep going presumably it will become negative and M will start to be a negative number of growing magnitude. But we still have this condition that M squared has got to be less than beta. So this series of operations has got to terminate as well. So series of kets with ever smaller M has to stop. So there must be a minimum value of M which we imagine will be negative. So we're going to have that beta M min times J, well I should write it differently. I should say that nothing has to equal the mod square of J minus applied to beta M min. And when we expand that out there will be other things happen here. So in other words nothing is going to be beta M min J plus J minus beta M min. That's awfully similar to what we had here. Well we had a J minus J plus. So you can see that it's going to produce the same stuff except that the sign of the commutator is going to be different. Otherwise everything will be the same. So this is going to be nothing is going to be beta M min J squared minus JZ squared plus JZ. Which is going to lead to the conclusion that beta nothing is going to be beta minus M min squared plus M min. In other words beta is also equal to M min M min minus one. So we have a relationship here between beta and the largest value that M can take. And between beta and the smallest value that M can take. And we could, between these two equations we can eliminate beta and learn that M min squared minus M min. Which is this is equal to beta or minus beta equals naught. But minus beta is the same as minus M max M max plus one. So we have this equation and this can be thought of as a quadratic equation for M min in terms of M max. So this is a quadratic equation and it tells me that M min is equal to minus B. Well B is minus one so this is equal to one plus or minus the square root of B squared minus four AC. A is one, C is minus this stuff. So plus four M max brackets M max plus one all over two. Looks ugly but actually it's very beautiful because this is going to be a half of one plus or minus the square root of. This is, well let me write down what it is and you can tell me whether you agree with it. It's M max plus one squared. If you square this stuff up you get four M max squared. You also get four M max from the cross terms. Two times two makes four so that's that and that and you also get a one that's that. So we can extract the square root because we've got the square root of a square so we have plus or minus this. M min is obviously smaller than M max so the plus root can be ignored because that would tell me that M min was bigger than M max. So only the minus root is wanted and you soon find that that is equal to minus M max. So there's a biggest value that M can take and there's a smallest value that M can take and we've shown that that's minus the biggest value. In other words we've got a picture like this. We have a biggest value here then we have a next value, then we have a next value, then we have a next value and suppose that this is the end then zero lies, so this is a plot with M going up here. So here would be zero say and in this case this would be a half, this would be three halves, this would be minus a half and this would be minus three halves. Or it might work out like this that we'd start but the key thing is we can start slightly higher up and then we would have this one and this one and this one and this one. So if we started at two we could have one, nothing, minus one, minus two, these are the possibilities. But the key thing is that I know that in an integer number of steps, here three steps, I can go from the biggest value to the smallest value. Here there are four steps, one, two, three, four. So the key thing is that twice M max is an integer. Now we could carry on talking about beta and M max but it's extremely boring and nobody does that. What they do is they use a new notation. They say that the new notation is that J is what you mean by M max. The biggest value of M is called J, little J. And what have we got? We've got that beta is equal to M max, M max plus one, that's on the board just here. It's therefore equal to J, J plus one. And we know that two J is an integer, in other words J is a half integer. It may be an even number of half integers in which case it's an integer itself or it may be an odd number of half integers. So in this left-hand column, J is a half integer, all the values of J, sorry, J is a half integer, consequently all the values of M are half integers. In the right column, J happens to be an integer and therefore all the values of M are integers. Therefore this beta number is sometimes an integer so if J is an integer, this is an integer. For example if J, a possibility is that J comes out being nought, in which case beta is nought, or J might come out being one, in which case beta would be two, or J might come out being two, in which case beta would come out being six. We have a sort of funny selection of integers. But worse than that, when beta is, sorry, when J is a half integer, the values of beta are really quite weird. So we don't use beta as a label. So we relabel beta M to J M. Instead of using as the label in here that tells you how this state responds to J squared, instead of using the actual eigenvalue, you use this number which is either an integer or a half integer from which you can work out this eigenvalue because this eigenvalue is J, J plus one. That is to say, we have that J squared on J M is equal to J, J plus one, J M. And we have that J Z on J M is equal to M of J M. This is the new notation universally used. So we've changed notation only because we've discovered that the numbers beta are themselves rather unpleasant and don't make for handy labels, but they are related through this equation to something that's very simple, which will be an integer or a half integer, and moreover tells us immediately what the largest value of M is that you were allowed to have. So we have. If J equals two, there are five states. There is two comma two, two comma one, two comma nothing, two comma minus one, and two comma minus two. Now, what does that mean? What statement has been made physically? It's being said that if my pen has two units of angular momentum, well, it has J equals two, which means, as I've said, it has, strictly speaking, J squared has an eigenvalue of six. But if we call that two units of angular momentum, it has five possible orientations. This one, this one, this one, this one, and this one. Only five. This is what they call space quantisation when Stern and Gerlach discovered this experimentally. I think it's a terrible, terrible term, right? I wouldn't call it, I think it's, no, I think it's a very bad idea to call it space quantisation, but I just tell you historically that's what they called it. This is the bizarre conclusion that we have a discrete set of orientations anyway, being possible for a pen with that amount of angular momentum. If J is a half, then what do we have? We have a half and a half and a half and minus a half, and that's it, only two states. So that's why we've been talking about electrons and things, objects with angular momentum, with spin, a half, half of units of spin angular momentum like electrons, protons, positrons, et cetera, as the archetypal two-state system because there are only two possible orientations. Now this is very misleading, right? But I've already given health warnings on this, but the naive interpretation is that your spin-a-half particle, your spin-a-half gyro, has two orientations, this one and this one, and nothing in between allowed. So that is a grossly oversimplified picture which leads to misunderstandings, but it gives us a bit of orientation, and people often do think in those terms. In the three-halves case, we would have three-halves, three-halves, three-halves, one-half, three-halves minus one-half, and three-halves minus three-halves. We would have four possible orientations. We'd have this, this, this, and this never pointing horizontally, et cetera, et cetera, et cetera. Almost done. Let's have a look at the effect of rotation around the z-axis. Okay, so psi goes to psi-primed, which is u of alpha of psi. These angular momentum operators came in as the things you put in an exponential in order to generate a rotation, the unitri matrix that makes you a new system which is the old system rotated. So we want to see what we get now. So let's see what happens when we rotate a state of well-defined, one of these eigen states here, right? So let's do e to the minus, so if we go about the z-axis, then alpha only has a component in the z-direction, and this becomes, and it has a magnitude phi, so this becomes e to the minus i phi is the rotation angle jz, and let's use that on one of these jm states. Well, this is a function of an operator used on... It's a function of an operator, so by the definition of a function of the operator, it has the same eigen states as the operator whose function it is. So this thing is an eigen state of this operator, and the eigen value is the function on the eigen value, so this is e to the minus i phi m jm. So one of these eigen states here, when you rotate it using this rotation operator, produces you the same state multiplied by this phase factor. OK, so if we rotate through 2 pi, if we rotate the thing completely around, so if we put phi to 2 pi, we are looking at... What are we going to call this? We're going to call this upside-primed, say, right? Upside-primed is going to be e to the minus 2 pi i m. Well, maybe we should say 2 m pi i times what we first thought of. If m is an integer, then this is going to be a number one, so this is equal to jm if m is an integer, but it's equal to minus jm if m is a half integer, because we know it can be. So we have the surprising result that if you rotate a system with half integer angular momentum completely around, complete through an entire rotation, its state doesn't return to its original state, it returns to minus its original state. And this seems strange to us because we don't have any concrete experience. We have no experience of this kind of thing for the following reason, that particles which have half integer j... Well, particles are described by fields. Particles that have half integer j are described by fields whose value never becomes... This is a result of quantum field theory, whose value never becomes large compared to the quantum fluctuations in the field, the quantum uncertainty in the field. So the values of these fields never become significant, and we have... These fields never enter classical physics. So the Dirac field, whose excitations are electrons and positrons, is not something that's part of classical physics. It's part of the vacuum, just the same as the electromagnetic field or the gravitational field, located at a macroscopic level, so it doesn't enter classical physics. So we have no experience as classical beings within classical physics of the fields associated with these half integer values of M, and therefore we're unaware of this fact that if you turn the thing completely around, it changes its sign. The fields we do have experience of, the electromagnetic field and the gravitational field, belong to integer values of M. The electromagnetic field has J rather than equal to 1, and the gravitational field has J equal to 2, and therefore these fields don't manifest this strange behaviour. Well, I think that is the right place to stop even though it's a little early, and we will look at the rotating molecules as a physical application on Wednesday.