 Okay, so we finished last time where just about pushed through the calculation of what L squared is as a differential operator, which we did, if you recall, by multiplying the ladder operators L minus and L plus together, and it was rather a tedious calculation, at the end of the day with luck we ended up with this. And we should recognize that this L squared is minus this combination of partial derivatives vector theta and phi is the, it's minus R squared times the angular part of, of del squared, the Laplacian when looked in, when put into spherical polar coordinates. So if you take, if you take this thing and put it here, this minus sign cancels that minus sign and we get a one over R squared sine theta d with theta sine theta, et cetera, which I hope you recognize as, as del squared. So, so you might ask yourself, so physically what's happening? We have that the kinetic energy operator, i.e., so we've got this, this h sub k, which means p squared over 2m, where this is px squared plus py squared plus pz squared, is also minus h bar squared over 2m times del squared. So in the position representation, this operator becomes this, right? Because p is minus ih bar times gradient. And classically, we have that L is equal to v tangential times the radius. So L squared is v tangential squared times radius squared, L squared over R squared is equal to v tangential squared. So we have, we have that hk is equal to, well, sorry, that's suggestive something, isn't it? That this, this nabla squared can be written in terms of some radial derivative plus, so we could say that, yeah, sorry, we could say that hk is equal to some radial minus h bar squared over 2m, 1 over R squared d by dr, R squared d by dr. And then we're going to have plus h bar squared L squared over R squared, I think, just by substituting into there over 2m, sorry, over 2m. And what's this going to be? We've defined the angular momentum operator L squared to be dimensionless. So putting an h bar in front of it, h bar L is the classical animal, right? So h bar operator is the analog, I shouldn't say it's equal to, it's the analog of classical angular momentum, total angular momentum. So this that you have here has the dimensions of total angular momentum squared, it's the classically understood thing, so this term here is, looks, is looking awfully like v tangential squared, sorry, I need a mass here, right? So the classical angular momentum is m v tangential R. So the square is m squared v tangential squared R squared. Move the R squared down here and this is the classical relationship that L squared over R squared is m squared v tangential squared. So this is looking like, this in the back here is looking like half m v tangential squared. That's what this suggests to us. This is a quantum mechanical formula which is correct, but it's suggesting to us that it's this, the sort of natural translation in classical physics is this, and this is clearly the tangential kinetic energy. So this is the k, the tangential kinetic energy associated with tangential motion which suggests that this here should be the kinetic energy associated with radial motion and that's what we want now to put on a firm intellectual footing. So we're going to show that this thing is minus h bar squared over 2m times pr squared where pr is the radial momentum. So the question we want to address now is what is the radial momentum operator? We found the tangential operator, we found it in some sense the tangential momentum operator in the sense L, now we want to find the radial one. So classically momentum is a vector and we can say that the radial momentum is simply r dot p over r. In other words the unit vector r dotted into p must surely be radial momentum, momentum in the radial direction. But there's a problem with this from the perspective of quantum mechanics because this operator doesn't commute with this operator. So it's, well what does that mean? This thing r, so in qm r dot p over r is not Hermitian. Let me prove that to you. It's easier to prove that in general than in particular. So let me have two Hermitian operators. a dagger is a and b dagger is b. Then let's look at a b dagger. If I multiply these together, what is, I get an operator a b, is this operator Hermitian? Find out. That's b dagger a dagger because the rule for taking Hermitian adjoint is you reverse the order and dagger the individual bits. But b is b dagger and a is a dagger so this is equal to ba. So is this equal to ab? Well clearly it is if and only if b and a commute. So this is only if, whoops, a comma b equals naught. In words the product of two Hermitian operators is itself Hermitian only if those two operators commute. If and only if those two operators commute. So this r dot p which is shorthand of course for x px plus y p y plus z p z would be Hermitian, could be Hermitian only if x and p x commuted, y and p y commuted. Well they don't. Therefore this is not a Hermitian therefore this is not an observable. So it can't be what we're looking for. We're looking for something, the momentum in the radial direction which is, which is observable-ish. All right, well there's a fix to this problem. There's a general fix and we're going to use it but if we do a half of ab plus ba, this is Hermitian, right? Because if you take the dagger of this, this one, we've just proved the dagger of this one is that and the dagger of that one is this so this thing, the dagger of this bracket is itself. So when you've got two non-Hermitian operators, sorry, you've got two Hermitian operators that don't commute and you want to make the product the way to go is to, is to take the average of them, you know, it's a really naive thing to do. So let's do that. So let's, so we try, let's have a look at the Hermitian operator pr which we're going to define to be a half of r dot p over r where it's important that this r here is in front of the pr plus p dot r over r. So this thing here will be Hermitian and I'm going to show that it is what we require. So in the position representation, so we, we, you can do this calculation in the abstract, not in the position representation, but it's easiest in position representation, so that's how we'll do it. So pr is equal to, so this p gets replaced by minus i h bar grad, right? So this is going to be minus i h bar over two common factor, we're going to have r dot grad over r, that's, this of course is the scalar r, plus the divergence of r over r. Well, now the issue is this, when, what, this isn't quite the divergence of this, what this means is, is, this is remember an operator, it's waiting for a wave function to come and stand in front and get operated on, right? So this differential operator operates on everything to its right, it operates on this and it operates on these two. This here, this differential operator operates on everything to its right, which is only the psi. So we have to, we, we, when we, when we expand this out, we're going to get three terms because we're going to get this thing operating on this, that, that's standing idly by, this thing operating on this with this and this standing idly by, and this thing operating on this with these two standing idly by, which is the same as that. So this is going to be minus i h bar r dot grad over r, right? So I'm taking this one that I've got and the one that I'm promised at the end of all this reduction here, so that's where the two went to. So that's, that's those two and now I've got these two bits minus i h bar brackets, we're going to have over two, sorry, that's this factor here, then I'm going to have this thing operating on this, the divergence of r is three and then I've got this thing operating on that, it's going to be r dotted into the gradient of one over r. So that's going to be minus r dotted into the gradient of one over r, the gradient of one over r has to be, oh well, it is the vector r over r squared. This minus sign comes from the differentiating of the one over r, right? Because I'm reminding you of prelims maths now, the gradient of r itself is the, is the vector r divided by r, this is a dimensionless animal because that has dimensions of one over length, that has dimensions of length, so it's the vector, it's the unit vector r and that's what we've been using. Oops, excuse me, so I made a mistake, this should be r cubed, maybe we need to do this, sorry. So what, what, just to fill in here, so the gradient of one over r is equal by the ordinary rules of differentiation minus one over r squared times the gradient of r. But the gradient of r, I've just said, is the vector r divided by r, hence the r cubed. These two dot together make an r squared, which cancel most of those, so, so this minus sign, these two can be combined to a two over r, the twos go away and guess what we end up with is minus i h bar r dot grad over r plus one over r. So that's what this, this stuff reduces to. What we next want to know is, so what is r dot gradient? Well, I want to know what this is in spherical polar coordinates. Well, the thing to do is just to write down, let's write down r d by dr. It's easy to see that that is going to be x, d, well, let's, let's do it, r, then we're going to have, by the chain rule, dx by dr d by dx plus dy by dr d by dy plus dz by dr d by dz. That's just the chain rule. But, but what is dx by dr? You know, x is equal to r sine theta cos phi, so dx by dr is equal to x over r, and for the same reason, dy by dr is equal to y over r and so on and so forth. So this is equal to x d by dx plus y d by dy plus z d by dz, because this is x over r, but this r cancels the r on the bottom, y over r, r cancels what's on the bottom. And what's this, this is a vector product of x comma y comma z with nabla, d by dx, d by dy, d by dz. In other words, this is, this is the animal that we're interested in, r dot grad. So I have now that pr is equal to minus i h bar, r dot grad, we've just agreed is r d by dr, so those r's go away, and we have d by dr plus 1 over r. So we have an interesting result, we have the momentum associated with the radius is not simply d by d radius, like the momentum associated with x is d by dx, there's also this additional term in here. But just to convince you this really is the momentum associated with radius, let's for fun calculate r comma pr. So what is that? That is minus i h bar of r d by dr plus 1, which is r times pr minus d by dr of r minus 1. Yeah, right? Yes, sorry, I mean danger of getting zero on time, that's the trouble. I want to get minus, I want to get i h bar out of this, what the hell did I do wrong? d by dr pr minus d by dr plus 1 over r working on r. Yes, yes, yes, yes, sorry, sorry, yes, yes, yes, yes, yes, it's perfectly correct. Okay, so the ones go away, this sort of thing is confusing. Right, now as I say, what does this mean? This means d by dr of everything to its right, and there's a phantom wave function here waiting to be operated on. So this is the derivative of r of psi. When we take the derivative of the r, we get one times of psi, and then the r stands idly by and we do the gradient of psi. The second term cancels on this because r times the gradient of psi is occurring here with a plus sign, and there it will be occurring with a minus sign. So what we're left with is the d by dr times of psi, the d by dr which makes one. So this is equal to plus because there's a minus sign coming here, i h bar, so that it's these two operators satisfy the canonical commutation relations, right, canonical commutation relation. So pr really is the momentum associated with r. Okay, so what what are we really trying to do here, we're trying to show that that one over r squared d by dr of r squared d by dr is essentially pr squared. So let's calculate pr squared. Pr squared is going to be minus h bar squared because there'll be two, there'll be two minus i h bars, and then it's d by dr plus one over r brackets d by dr plus one over r, which is equal to minus h bar squared. Obviously this on this is d two by dr squared. We will get this differential operator, we'll differentiate that and reduce a minus one over r squared. We will otherwise get a one over r d by dr and also a one over r d by dr. So we'll get two of one over r d by dr, and sorry and I haven't finished, and I also get this thing on this thing is a plus one over r squared. So these two terms, these two terms cancel and we're left staring at this. I should have had two of these terms. I think I said I was going to get two terms because I have a one over r d by dr, and I have a one over r d by dr. After this operator, when this operator works on this, it reduces that, but also it works on the phantom wave function sitting over here without standing idly by, so we get two of these. I think I said that but I didn't write it, I'm not sure. So we have a minus h bar squared d two by dr squared plus two over r d by dr, which can also be written as minus h bar squared over r squared d by dr of r squared d by dr. Because if you differentiate out this product, you get r squared on r squared times d two by dr squared, which is this term, and you also get a two r over r squared two over r times d by dr. So yeah, this term here, we've now shown that hk, the momentum operator, which is minus h bar squared over two m del squared is also minus h bar squared over two m of, sorry, of, well, let's leave that outside. Let's take the h bar squared into the bracket. We're going to have a one over r squared d by dr all that stuff, which we've just shown is pr squared. And then, whoops, there was a minus sign, so that soaks up this minus sign, a pr squared. And then similarly, there's plus h bar squared l squared over r squared. This is a very important formula that we will need when doing hydrogen, and therefore fundamental to, so it's expressing your kinetic energy in terms of your radial kinetic energy and your tangential kinetic energy. And that's one of the reasons why the total orbital angular momentum operator is important because it encodes your sort of energy of going around and around. So with that, we are now finished with, we can mercifully finish with orbital angular momentum and we can get on to spin. And this is somewhat more interesting in the sense that it's quantum mechanics says more remarkable things to say and it's less tedious because all that stuff with those partial differential operators, d by d theta and stuff is not much fun, it has to be said. All right, so we have identified two types of generators of rotations. The total angular momentum operators and they generate, we introduce them in order to generate complete rotations. So u of alpha being e to the minus i alpha dot j rotates system as on a turntable. So it moves, it moves your system around the origin, if you put your system, it's as if you put your system on a turntable, centered at the origin with its axis at the origin and you turn the turntable around, your system moves through space and it rotates simultaneously, whatever internal structure it has. But we also have shown that Li, the orbital angular momentum, moves system on circles. So it moves it around, physically it translates it around the origin, but it does not rotate it at the same time, it leaves its orientation fixed. And we have some, well okay, so we found the commutation relations here, we found that J i comma J j is equal to i sum over k epsilon i J k J k and we found that it was also true that Li Lj was equal to i summed over k epsilon i J k Lk. They had the same commutation relations amongst themselves, these operators, which is why we could use the work we did demonstrating what the eigenvalues of these could be. Also down here, this implied that M, that J squared has i values J J plus one, for J is nothing a half, one, three halves, etc. And from these commutation relations, we inferred that those are possible values for the eigenvalues of these operators, but we also had the principle that if we wrote, if we translated something completely around the origin, we proved that that was the identity transformation. So we concluded that L, L plus one had to be L equals, L equals naught one two integers only allowed in this case. What we're now going to do is introduce S i is by definition J i minus L i. It's the difference between these two. What does that mean physically? It means that S i is going to be the generator of rotations of a thing about its own axis. So we're not going to be, this rotates it on a turntable, so it rotates it and moves it. This simply moves it around a circle. So this is going to only rotate it on its own axis. It's not going to move it. It's only going to rotate it. That's what we expect to happen, but we'll have to be guided to some extent by the mathematics. Well, it is what is going to happen. Right, so what about the, so what we want to, having introduced these newfangled operators, it's important to figure out what the commutation relations are going to be. Now S i comma S j is going to be J i minus L i comma J i minus, sorry, J j minus L j. We're going to get this, commuting with this, so we're going to get I summed over K epsilon i J k. This commuting with this will produce a J k. This commuting with this will produce an L k. This commuting with this, now we didn't write that down, but J i commuting with L k, this is a vector operator and therefore this thing, when you, when it commutes with this always produces the missing component of this vector. So this is going to be minus L k and similarly this thing on this thing is going to produce, swap them over and you're going to, well there are several signs here that we could hound down, but this thing is, we're looking fundamentally at the same thing as the commutator of this on this. We're looking at minus L i comma J j is equal to, is equal to obviously J j L i is equal to i epsilon J i k L k is equal to minus i epsilon, that's summed over K, this is summed over K of epsilon i J k L k. So, all right, J i, J i k and, but I would like to have this in the order i J k, so I swap those two over and introduce the minus sign to compensate and then at the end and then you can see that this thing including that minus sign is the same thing as this thing including that minus sign, so we have a minus another L k, so this is the justification for that last term there. So what do we end up with at the end of the day, these three L k's collapse into just one L k, it's going to be J k minus L k, in other words this is going to be i summed over K epsilon i J k S k. So these spin operators, sorry did I say that they're going to be the, we call them the spin operators, they have exactly the same commutation relations as the J, therefore we know what their eigenvalues are, so this implies that the eigenvalues of S squared, which is of course S, which is S x squared plus S y squared plus S z squared are S, S plus 1 where S is equal to a half, sorry nothing a half, 1, 3 halves blah blah blah, okay, because these results follow nearly from the fact of having the commutation relations S i comma S j is i epsilon i J k S k. Are the half integer values allowed? Answer, you will have a half integer when J does because L does not, all right, why is that? That's because S z is equal to J z minus L z and J z comma S z equals nothing, which is also the same as L z comma J z, etc. All these three operators commute with each other, so there's a complete set of mutual eigenstates, and if, so we can now see that if this has half integer, this is using half integer eigenvalues, so the eigenvalues of this are going to be, are going to be the difference between the eigenvalues of this and the eigenvalues of this, so if this has half integer eigenvalues, therefore this will have to have half integer eigenvalues because this one has integer eigenvalues, so if J has half integer eigenvalues then S does, correspondingly if J doesn't, S doesn't, it just tags along behind J, and indeed that's how we tend to think about it, we tend to think that the integer amounts of angular momentum come from orbital motion L z and the half integer values, if present, come from S z and that's why J has half integer values, that's how we tend to think about it. And I think I've claimed a few times that spin is something to do with, spin is something to do with the orientation of our system and now is time to make good this claim that the eigenvalues of the spin operator or your response to the spin operators encodes how a particle is oriented, and this is a strange area, a very quantum mechanical area. Okay, so in general the internal configuration of a system could be written, we could write psi is equal to the sum of S m psi S m, we got a complete set of mutual eigenstates of S squared and S z, right, so this is the, so we're saying that S squared on S m is equal to S S plus one, whoops, one of S m, and we're saying that S z on S m is equal to m S m, and there should be a complete set of eigenstates of this, of these, mutual eigenstates of these operators, so I should be able to expand any state as a linear combination of these states, and what are we going to have to sum over, we're going to have to sum over S is equal to naught, a half, you know, and we're going to have to sum over m is in modulus less than or equal to S. This should be a generally valid statement, but for the wave functions or the states of real occurring systems in the laboratory, the good news is, you don't, you only have to, there's only one value of S will occur in this sum, so this is, this would be generally the case, but for systems in the lab, for microscopic systems, this wouldn't be true, so if we were talking about a cricket ball, we would need to do a sum over all of these things, up to 10 to the 40 or something, but if we're dealing with an electron or an atom or whatever, we don't have to sum over all these things, we only, only one value of S has non-vanishing amplitudes, so this is going to vanish for real systems except when S takes one particular value, so for real, for microscopic, microscopic, only one value of S occurs, non-vanishing S, M, psi. In other words, this, we can say at the outset that the amplitude to find a value of S other than a value that's peculiar to the system is zero, so we just don't need to consider it, so we're able to, so we're able to write that psi is equal to the sum from M equals minus S to S of S M psi S M. That means, what does that mean? We're encoding, well the state of the system is described by these numbers here, there will be two S plus one complex numbers, these numbers here, and they're telling us they're encoding somehow the way the system is oriented, so this is two S plus one complex numbers encoding, encoding orientation, and what we want to do now is get some feel for how this encoding works in simple cases. How would you encode the orientation of a macroscopic body? Well the traditional procedure is you use Euler angles, you write down three angles, we should describe how you get some axis in the body, which way it points with respect to the z axis of some fixed coordinates and then how it rotates around that, so you usually encode the orientation of a body in three Euler angles for a classical system. For a quantum mechanical system you encode it in a certain number of complex amplitudes, the amplitudes defined it in various orientations, but it's carrying the same information, right, in a very funny way. Okay, so let's consider the various cases, the case S equals naught, in other words if the only occurring thing here is naught, there's basically no information, there's only one amplitude, naught, the state of your system, and also if you rotate your system, you try and rotate system with u alpha, which is e to the minus i alpha dot s, this is the, so s is the generator of rotations of the system around its own axis, right, so I've written down the unitary rotation that makes me a new system, which is rotated around the axis alpha by the magnitude of alpha, then if I use this u alpha on psi, what do I get? I get, using that expansion, I get nothing, nothing psi times u alpha on nothing, nothing, well let me replace this by e to the minus i alpha dot s, when this operator sees this ket it thinks, you know, nothing doing zero, right, you simply get all of these things, that becomes nothing, so we have e to the nothing, which is one, so this goes back to itself, which is the same thing as psi, in other words when s is one, you can't tell the difference between the system before you rotate it and after you rotate it, it's like an absolutely immaculate and perfect sphere, if you rotate it it stays the same, so a particle that does this, that has s equals naught, so no spin implies same after rotation, so if you like, for a classical animal we would say this is spherically symmetric and there's only one amplitude we need to bother with, so we say it's a scalar, so the simplest particles are spin zero particles, there's just one amplitude which is used to say what the energy is or to say what the location is, because there's no issue of how is it oriented, it's a silly question, how is it oriented, it doesn't have an orientation, you can't tell it, well it doesn't have an orientation, that's the best thing to say, nothing changes if you try and reorient it, and because of the one amplitude business you say it's a scalar particle, right so that's s equals naught, next in the hierarchy and for an example a pion is an elementary particle which is a scalar particle, right let's do s equals a half, so unfortunately there are not many particles, there's not much in physics that's a scalar particle or a scalar field, this is a rare case, the really important cases are s equals one and s equals a half and s equals one and we're all built out of s equals a half particles, right, so this is electrons, protons, quarks, therefore neutrons, a wide range of things, anything made out of an odd number of quarks will be an s equals a half particle, we'll prove that later on, okay so this is the really big class, we have that the state of the system can now be written as, there's going to be a half, a half of psi plus a half minus a half, oh sorry I need to times a half plus psi times a half minus a half, so any state of the system of this particle can be written as a linear combination of this state which is to say that m is, you know, you're guaranteed to get the answer half if you measure sz and a linear combination of this, so we now have a non-trivial linear combination with two possibilities, this is a very cumbersome notation so people don't use it, they either write that this is plus psi plus plus minus psi minus, that's a handy notation, right, there's really no point in writing down this half because the first half we know for certain will always be there and instead of writing a half we write plus a half and minus short n for minus a half or since this is only a boring complex number we often write a plus plus b minus, so any state is a linear combination of these two basis states where we're guaranteed to get plus a half for s or minus a half for s and there are amplitudes so this is the amplitude, it's mod square gives you the probability if you would measure the z components of spin of this particle you would get a half and b is obviously the amplitude mod square of that would be the probability that you've got, sz was minus a half, so let's do some stuff with spin of half particles, so if I take an arbitrary spin operator, well so let that be the, this is the spin component along the unit vector, what is it mathematically it's n dot s in other words it's nx sx plus ny sy plus nz sz this wouldn't this this that I've just written down would apply for any spin not just spin a half and I would like to to do calculations like if I do sn on a psi what do I get I get some new state phi say all right this is an operator I use it on a state I get a new state I want to be able to to do calculations like this we will we'll find this is crucial now we know that this everything every one of these states can be written as a linear combination of plus and minus so this can be written as c plus plus d minus here's sn and this can be written as a linear combination of a plus plus b minus and the name of the game is given the numbers a and b which characterize that to calculate the numbers c and d which characterize that we need an apparatus that does that for us and that's easily obtained what we do to find c of course is we bra through with plus and then on the left we get c plus nothing so we find that c is equal to plus sn plus times a plus uh plus sn minus times b and brawing through by minus we get an equation which tells us the value of d which is equal to minus sn plus a plus minus sn minus b and there's a handier way of writing this we write this as c d is equal to a matrix right this is just a boring complex number plus sn plus that's a boring complex number minus sn plus plus sn minus minus sn minus and that's operating on the column vector ab so here we have a concrete apparatus this is a matrix a two by two matrix of complex numbers working on the given complex numbers that characterize psi which gives us the two complex numbers that characterize phi so for example uh suppose we take just got time to do this suppose we take uh n is equal to is equal to nothing comma nothing comma one then this becomes sz and this becomes s i mean all these sns become sz and then sz on plus is a half of plus so this number evaluates to a half then c d is equal to uh half sz on minus is minus a half times minus but minus is orthogonal to this so we get a nothing nothing and sz on this is minus minus so this one evaluates minus a half of minus so this thing evaluates to minus a half so in this particular case it evaluates to this which people write as a half of sigma z ab a column b where sigma z is the power matrix one minus one nothing nothing a power matrix so these matrices we really want these matrices but it's handy to take the half outside of the matrix uh and write them in terms of these well this is anic this is the first of them and if we had time which we haven't now we'll have to do it tomorrow we will derive what sigma x is and sigma y are the three power matrices that enable us then to write the matrix belonging to any one of these spin operators and then do calculations uh on these so it's it's time to stop