 Let's go. We just began work on the parity introduced, the parity operator P yesterday. So what does it do? It makes out of your left hand, your right hand, if you orient it correctly, by reflecting, by producing a state, which is the same as the state you first had, but with everything reflected through the origin. So what you used to find the amplitude to be at minus x is now the amplitude to be at x. at x. We showed, not surprisingly, that the square of p is the identity operator because if you reflect something twice, you have it back where it was, which formally implies that p is p minus 1. The next item on the agenda is to check that p is a Hermitian operator and therefore an observable. The proof of that is that we take two states, find psi, and we evaluate this complex conjugate. Let's make sure that is exactly what I plan to do, yes. We need to do what we want to do because we know what p does with an x here on the left of p. We slide an identity operator in between the phi and the p. We write this as d cubed of x phi x p psi. We need to star the whole thing because I decided to star the thing on the left. That star is that star. x is real, integral sign is real. We can use that to replace this. That becomes d cubed x phi x minus x, oops, minus x, minus x of psi. Then we need to do the starring operation. That's the integral d cubed x. Take the complex conjugate of that and it becomes psi minus x. Here we're going to have x. I could write just phi, but just for fun, why don't I write p squared phi because p squared is the identity operator. That's safe enough. Except I regard p squared as p times p. Then I can say look this p, this outer p, can be got rid off by replacing this by a minus x because I'm using x p some newfangled state phi primed, which is p at phi. This can be written as the integral d cubed of x psi minus x minus x p. That's this inner p that outer p has been dealt with by changing that sign on phi. Then I can change my variable of integration from x to x primed, which is minus x. That's going to produce, that is going to be, but this, take away the identity operator, sorry, that's x primed, which is minus x. Take away the identity operator and we're looking at, which says that p, since phi and psi are arbitrary, that tells me, comparing with the initial thing, that p dagger is equal to p, which we've already know is equal to p minus 1. First of all, this says that it's her mission. It's an observable. This p dagger equals minus 1, says it's also at the same time unitary. It leaves the lengths of all states the same. Since it's her mission, so what are its eigenvalues? We have, if p of psi is equal to m of psi, well this is its eigenvalue, or maybe we should call it lambda, more traditional, so if p of psi is equal to m of psi, so it's psi and eigen ket, then we can apply p again and get that p squared of psi, which is actually equal to psi, because p squared is 1, is also equal to lambda p of psi, which is lambda squared of psi, so we have that psi is lambda squared of psi, and that applies that lambda squared is 1, which implies that lambda must be plus and minus 1. It has two eigenvalues, plus and minus 1, and we say that, so if p of psi is equal to psi, we say that psi is an even parity state correspondingly, of course, if p of psi is minus of psi, so that's an eigen ket with a plus 1 eigenvalue, this is an eigen ket with a minus 1 eigenvalue, we say it's an odd parity state. What does that mean? From what we have up there, it means that when you, from the top there, the question is, so let's have a look at this, sorry, let's look at the wave function point of view, for even parity, we can say that xp of psi, which is equal to minus x of psi by the operation of the p thing, but since p of psi is equal to psi, it's also equal to x of psi, in other words, the wave function is an even function of x, and similarly, odd parity states have wave functions which are odd functions of x, et cetera, et cetera, and when we did the harmonic oscillator, we found that, for example, we found that n is even parity for n an even number, and correspondingly it's odd parity for an odd number, just as a concrete example. So we very often classify our states, it's very useful to know whether our states are even parity or odd parity, we like to work with ones that have well-defined parity, that is to say, our eigenfunctions of this parity operator, by no means all states are eigenfunctions of the parity operator, however. Now we do something considerably more interesting, which is transformations of operators. So we introduced this, we introduced a displacement operator yesterday, so it was called u of a, and it was e to the minus i a dot p over h bar, where p is the momentum operator, and we understood that what it did was, it made out of a state, so psi primed being u of a times of psi is a new state of the system, the state that it would have, it was identical in all respects, except it was shoved along by the vector a. If you shove your system on by the vector a, the expectation value of the position obviously has to increment by a, right? So we can make the following statement that the expectation value in the state of psi primed of x has to equal the expectation value in the state of psi plus a, because we have displaced our system. This system is the same as this system except its location has been incremented, it's been moved by the vector a, and that is logical necessity. But this we can write in a different way, this we can write using that expression as psi times u dagger x u times psi, right? That's just a rewrite of that using this operator here, and this I can rewrite in a different way because I can say this vector a, which is just an ordinary boring vector, I could multiply by, well I could say that this is the following, this is x plus a times the identity operator on a psi, right? Because it's clear that the expectation value of a times the identity operator is the vector a. So these are equivalent expressions. So I found that the expectation value of this operator is equal to the expectation value of this operator for any psi whatsoever. And it's shown in a box, it's in the book, it's a little box which leads to the conclusion and not surprising conclusion that if that's true that this expectation is equal to this expectation. If two operators have the same expectation for every state whatsoever the two operators have to be equal. So this implies with a little bit of footwork that's relegated to a box in the book that u dagger x u is equal to x plus a, where this I is perhaps understood. Well let's put it in because I want that to be an operator, right? This is an operator on the right hand side. Right, now let's now make, we'll make a small, right? If a is small we know we can expand this in terms of its generator, right? So we can write this as, this thing here can be written as the identity operator minus I a dot p over h bar plus order of a squared. So we've done that before and we're going to do it again. So that becomes the identity operator plus I a dot p over h bar plus dot dot dot, which we will ignore, times x bracket 1 minus I a dot p over h bar dot dot dot and that is equal to x plus a, the I can be understood. So we multiply this all up to find what the terms on the order of a proportional to a are. So a is small but still arbitrary. I mean you can still fiddle around with it but it's small. So this is going to give me, this is going to give me I x I is going to equal that, right? We're going to get an x on the left hand side which will cancel with that and in the next order we're going to get a dot p which is small because a is small on x times I and then we'll have an I times x times minus a dot p. So what we're going to left within the order of a is I over h bar a dot p comma x, the commutator because we're going to have this times this and also with a plus sign and we're going to have this times this with the minus sign and once that's equal to, that's equal to a because that's the terms on the order of a on the right hand side. The higher order terms must all cancel that we leave that to the magic of mathematics and not interested in it. So we have this relationship here and let's look what this looks like in terms of components. If I look at the, so this is a set of three equations, one for the x component of this, one for the y component of this and one for the z component of this. So what does it look like? It looks like let's multiply by I over h bar and swap the order here. Then this is going to be x comma a dot p commutator is equal to I h bar a. Now let's use, write this out in its components. This is a set of three equations so I can say that x, j comma the sum over k of a, well, it's a, j, sorry, it's sum over k a, k p, k, but I can take the a, k outside the commutator because it's a mere number is equal to I h bar. I have to write now a, j because this is the, this, this component to this vector here matches this here, right? This is a dot product which is the sum p k a, k. And now I can identify, okay, so, so this has to be true for all small a, k. So I can write this right hand side as the sum over k if I want to, sorry, I h bar the sum over k of delta, j, k, a, k. Posh a way of writing it. And now I can say because a, k is arbitrary, the coefficient of a, k on the right side has to equal the coefficient of a, k on the left side. So that leads to the conclusion that x, j comma p, k is equal to I h bar delta j, k. So we've recovered the canonical commutation relation between x and p as a consequence of p being the operator which generates translations. So we've, we've come at this in rather a roundabout way. I, just to review how this has happened. I wrote down a rather arbitrary rule. I introduced p by an arbitrary rule. I said that, I said that x, p of psi is minus I h bar d by dx x of psi. Using Ehrenfest theorem, I tried to persuade you that this wasn't completely crazy. But really it wasn't a very satisfactory job to start in that way. Then we showed that because p has this d by dx structure, it is the generator of translations. And as a consequence of its being the generator of translations, it must have this commutation rule. And what we should have done really is we should have said, look, there must be some operator which generates translations. This operator is going to have this commutation relation and we should have worked our way down to finding out that in the position representation it's represented like this. And for the angular momentum operators, this is the line of argument we're pursuing. We are using, we are introducing them as the generators of rotations and then we're going to find out what they look like in the position rotation and the position representation later on. So we've come at this in a slightly tortuous way. This is the main job that momentum, the momentum operator has. It's interesting, so it's probably worthwhile just checking that this commutation relation guarantees that rule up there that u dagger x u is equal to x plus a even when a is big. So this stuff has all been for an infinitesimal a and it's good to check that the other thing works, that it sorts us even for a big. So now let's just talk about for any a, including a big one, any big displacement. So we're going to be looking at u dagger, sorry, x u, let me just check my, yep, which I can write of course as u dagger u x plus u dagger x comma u. So I've just swapped the order of those two and put in the commutator that compensates. This of course is going to be x because u dagger u is one and what's this going to be? So it's going to be x plus u dagger of x comma e to the i minus i p, excuse me, a dot p over h bar, close brackets. So this is a, this is a classic example, we studied this problem before, we're doing the commutator of x and a function of p, this is the function of p we're doing it. And do you recall that the answer to that problem was that we could write the commutator as the rate of change of this function with respect to t p, sorry, times x comma p. So this can be written as x plus u dagger, the rate of change of this with respect to the derivative of this with respect to p, which is going to be of course, is going to be u because the derivative of an exponential is the exponential times the derivative of what's up here with respect to p. So it's going to be u, db, well dbdp of this is minus i, no, sorry, let us do this as the derivative with respect to a dot p, right? We regard this as a function of a dot p, I'm worried about components and the way I can get out of that is considering this to be a function a dot p, which is just one thing. So if I take the derivative of this with respect to a dot p, I get u because I get the exponential back and then I have times minus i over h bar, right? That's the derivative of the argument of the exponential with respect to a dot p and now I have to write down the commutator of x with respect to a dot p. So this of course produces one and what does this produce? This can be, let's write that down, that's x plus, whoops, minus because of this, i over h bar, this is producing a one and now I need the derivative of this which is the sum over k of x comma pk times ak, right? Doesn't matter what order I put down a because it's mere number and this, it may be that we ought now to introduce an index on x, otherwise we're going to get into a confusion. So let's make that i. So I'm making, this was a vector x, an arbitrary component, let me call that component i, then this becomes i, then this becomes, this becomes delta i h bar of delta ik, the i and the i make a minus one which cancels this, the h bar kills that, so this is equal to xi and then this is nothing, this is nothing as k is summed except when k equals i so that becomes an i, so this just becomes an ai and yes it does sort us. That thing is equal to x plus a as advertised at the top there. So now let's think about rotations. So we have, we introduced these operators, we had jx, jy and jz so that alpha dot j generated rotation mod alpha about the unit vector in the direction of alpha, right? That's, that's what we established. Well we used that notation, we said there had to be such a thing and what we want to do now is talk about, is apply, is adapt that argument to this case. So the thing is the expectation value, so sorry, let's let up psi primed be the state that you get when you use u alpha on psi. So this is the state of the system which is identical to this state except it's been turned around through an angle around the axis as advertised up there, right? So we can say something about the expectation value of x of this system must be the same as the expectation value of that system but rotated to. If you've rotated the system you've rotated the expectation value of x. So we can say that psi primed x psi primed which is now this thing is a boring vector, right? It's the expectation value of a vector operator so it is a boring vector. It's a set of three numbers and a set of three numbers we can use a boring rotation matrix on which I'm going to call r alpha. So this is a three by three rotation matrix, an ordinary boring rotation matrix such as I think you must have studied in Professor Esler's course operating on psi x psi. So this was the old expectation, the expectation value in our unrotated system. If you rotate that expectation value you must get the expectation value in the rotated system. So that's the analog for the rotational case of of that statement u dagger x u is x plus a. Well except I haven't yet written what this is so I'm going to write this as psi u dagger of alpha x u of alpha of psi. So I'm replacing this with that and now I'm saying that for any state of psi this expectation value equals this expectation value. Ergo, since this is a set of boring numbers it can go inside the expectation and just rotate the operator. So it's just taking a linear, this thing, this is a boring three by three rotation matrix so if I allow it to go inside there it's simply going to be taking a linear combination of the x, y and z position operators. So I'm going to be able to say that u dagger x u is equal to r alpha of x. That's a matrix. And now of course I'm going to express this as 1 plus i alpha dot j dot dot dot. So this is now, we're now making small alpha so we're rotating it through a small angle then this can be written thus here is our x here is our 1 minus i alpha dot j plus dot dot dot and that is equal to the to this vector, it's a vector of operators but still it's a vector rotated by a small angle. Now we know what that is, well at least I hope we do. Come on, come on, no it's gone to sleep, do I have to draw? Yeah sorry system seems to have, no it's gone to sleep. So this requires a bit of, this is a piece of just standard geometry. What I want to do is write the action of a rotation matrix for a small angle. If I rotate something through a small angle and I hate drawing these diagrams it's going to be something, has it come alive? Oh right yeah okay it's just it went to sleep and it needs it warming up. So we're looking at this second diagram, can you see it because I sure will count. It's too faint. Anyway so the point is that this is the vector v, here is the rotation axis alpha, we're rotating through a small angle therefore this distance there is small, the displacement that you have up there, this is the rotated vectors on the right, the unrotated vectors on the left, the displacement is this thing here which is the vector delta alpha or the vector, so the small rotation vector crossed with the original v, so that we can say that v prime the rotated vector is equal to the original vector plus this small rotation vector crossed into v. So the right side, so I'm not going to draw this horrible diagram, so this is going to become x, that's the v up there plus alpha cross x. So our alpha is small so we don't need the delta alpha it's just alpha, we made it small to get rid of symbols. So we do the same old stuff, we multiply this out on the left up toward alpha, we notice that the i, the x of the i produces an x which cancels with the i on the right and we find that what we're left with to order alpha is alpha dot j times x minus from the i, the x and the alpha dot j, the thing the other way around, so we find that i alpha, whoops, alpha dot j comma x is equal to alpha cross x. Now we need to write this in we need to introduce indices in order to disentangle what's going on around here, so this this is going to be the sum over k i times the sum over k of alpha k which would come out times j k, all right, that's alpha k j k comma x j, no this is sorry let's change that let's let's change that to the sum thing we sum over to be j, it doesn't matter what we call it but let's call it that and call that k, all right, what's that going to be that is going to be the the the kth component of this vector on the right, sorry we should call that i, all right, this is going to be the i-th component of the vector on the right. Now a cross product can be written as the sum over j and k of epsilon i j k, this is the thing which keeps changing its sign, if you swap any two indices it changes its sign and epsilon one two three is one, which I hope you've met in Professor Restless' course, so this is just writing a cross product in Cartesian tensor notation, nothing to do with quantum mechanics it's just standard vector algebra and we have arranged it so that we have the i-th component of the left side here and the i-th component on the right side there, now we play our trick of saying that look alpha is arbitrary, it needs to be small but otherwise it's arbitrary, you can choose this direction any which way you like and its magnitude in detail you can choose any which way you like, so we can compare we can equate the coefficients on the two sides multiply by through through both sides by i to get rid of this, you'll get a minus sign swap the order of these in order to clean it up and we will have x i comma j j is equal to i times that's this i brought across the sum over k, the sum over j will go away because we're equating the coefficient of j on the two sides of epsilon i j k x k, this is a terribly important relation it tells us how j commutes how the j-th component of angular momentum commutes with any component of the position operator but crucially in this argument here we have used nothing about the position operators except that the components form that the three the x y and z operators are the components of a vector so all of this argument could be repeated for the three component operators of any other vector for example for p so it follows immediately we've only used only property of x used if the operator x used is that it's a vector so we've really shown that for any vector this relationship holds so we've shown that vi jj equals i epsilon sum over k of epsilon i j k v k for any vector operator so a vector operator is just a set of three operators if you like whose components are the whose expectation values will be the components of some classical vector okay so we can apply what we can immediately apply this as well as to x to vi is pi the momentum and we can also apply it to vi is equal to ji the angular momentum why is that oh sorry in in in which one this one and this one i lost a sign somewhere no i think i think this is right yeah i multiplied through by i and i swapped the order of these two did i no i don't think so uh no but this order is the same as this order surely to goodness i don't think okay let me let me take advice on that uh i'm yeah i can't up being skeptical but i suppose i should look here um i suppose i should look yeah true yeah the thing i'm thinking of okay so maybe i have dear i think we have i drifted a sign somewhere well no this is definitely the ith component of that cross product this is definitely so do we agree about that that should we just check whether that is whether this ordering is as advertised uh it's the closest thing i can't see it at the moment i think i think it's probably it's incredibly hard to do these sign problems on blackboards uh let me leave that and i will confirm tomorrow what the case is i imagine the book is right and i'm wrong but i do not see where i have made the mistake as things stand everything looks respectable no i can't see i cannot see an errant sign oh that's nasty and it does as you say matter yeah yeah i'll try and sort that uh and write up before tomorrow's lecture that the way it should be we need to be persuaded that this well the next thing i want to do is be persuaded that this is so can we apply this to the angular momentum operators it's going to be very important that we can and what's the argument the argument is that alpha dot j has to be a scalar y because the operator u alpha which is e to the minus i alpha dot j this thing is the rotation around a certain vector what this operator is the this is a physically in you know meaningful thing and it is defined not by the three numbers that we happen to use to define the the direction of the vector but by exactly what that vector is if you change your coordinate systems you use a new coordinate system you'll be using a new set of three numbers to define this right um but you must get the same the same direction in space and that will be the case if the if the these three operators also transform so the new operators that the operator associated with j x primed where x primed is your new x axis will be a linear combination of the old operators the operators associated with the old axes using the proper rule for a rotation then this product will stay the same and this operator will stay the same as we require so this operator will be independent of your coordinate system only if these three things transform amongst themselves uh as for a vector so j must be a vector and that means we can use it in here that is to say we can say that j i comma j j commutator is equal to i epsilon sum sum of a k oops i sum of a k epsilon i j k uh j k so this is a crucial relationship and from that we will find out what the eigen values can be of these operators j i and j j and then we'll be able to find what they what the states of well-defined angular metamarr and everything else this expression is is right right because it's independent of of of any swap can it be that both expressions are right oh i can't i mustn't take time to think about it okay let's consider oh what's a scalar let's consider a scalar scalar operators so what is a scalar operator it's uh it's an operator which uh well scalar sorry a scalar in ordinary physics is a number whose value is unaffected by a rotation of your coordinates right like a dot product it's unaffected by a rotation of the coordinates so what can we say is that if s is a scalar operator and we then the expectation value the expectation value of a scalar operator between rotated states must be the same as the expectation value between the unrotated states because because this is a boring number and uh it's evidently by definition a scalar something's unaffected by rotation so the fact that you've rotated your system shouldn't have any effect so when we ask ourselves what is that what what implications of that have it's that u dagger su is equal to s uh we can multiply on the left by u which is the inverse of u dagger because u is a unitary operator so we have then that su is equal to us which means that s comma u equals nought where this of course is u of alpha the rotation operator throughout so a scalar operator commutes with this rotation operator and it's easy to see by expanding this as one plus so if we write u is the identity minus i alpha dot j et cetera that immediately goes to the statement but s comma j i equals nought so scalar operators commute with all the angular momentum operators there's a very important and interesting scalar operator and that's j squared which means the sum of a k j k j k a k also known as j dot j right that's a scalar operator every dot product is a scalar operator so we have statements like j squared comma j i is nought we have statements like x squared comma j i equals nought we have statements like p squared comma j i equals nought these are all important results that we'll use many times we have statements like x dot p that's a scalar operator comma j i equals nought and so on and so forth so there are many operators we could make out of the operators already on the table which commute with the all the angular momentum operators so just just a summary now of the angular momentum commutation relations we've got that j i comma j j is equal to i sum over k epsilon i j k j k so the individual components this is somewhat strange state of affairs the individual components of angular momentum do not commute with each other so the you can't expect to know simultaneously the angular momentum around the x axis and the angular momentum around the z axis in individual cases you can but as a general rule you can't expect to know that and that so there isn't a complete set of states which are simultaneous eigen states of j x and j z for example but we do have that j squared comma j i equals zero and therefore there is a complete set of mutual eigen states of the total angular momentum and the angular momentum along any axis and that's what one always what we have to work with that we have to consider states which we have to work with states which are mutual eigen states of j squared and usually the axis we choose we have to choose one because of this business and the axis we usually choose is the z axis an important result about about parities go back to the parity operator now so in the same spirit if i consider so the expectation value of x if i reflect my system through the origin right by using the parity operators i make a system which is like my existing system but reflected through the origin it's obvious that the reflected system is going to have an expectation value which of x which is minus the original expectation value right because you've reflected everything and therefore the if there was an average value of x of in the original system the reflected system will have minus that value so this can be written as a psi p dagger x p a psi right because psi primed is by definition p a psi p dagger here but we know that p dagger is p so we have that the expectation value for any state whatsoever of minus x is equal to the expectation value of p x p so it follows that minus x is equal to p x p multiply through by p and use the fact that p squared is equal to is equal to one and we conclude that p x plus x p is not is nothing very much so you can say now that the the um that parity or p anti commutes this condition with a plus sign there right with a minus sign it will be accommodated with a plus sign it it's anti commutation anti commutes with x and in fact with any vector right because this argument here really only exploited the fact that we were talking about a vector not necessarily the position vector now why why is this stuff important why the practical importance of this is as follows um suppose we have a state of well-defined parity okay so let let p of psi equal either plus or minus of psi don't care which but it's going to be so psi is a state of well-defined parity and we've seen that the eigen states of the harmonic oscillator Hamiltonian are actually states of well-defined parity and now let's consider psi x of psi well that we've just seen is equal to minus uh psi p x p of psi this is a pure rewrite of a line higher up there well except i've taken the dagger off the p but as we know p dagger is p so who cares so but we've acquired a minus sign that's that minus sign but p on psi is equal to either plus or minus of psi i mean say plus of psi and p on this of psi is equal to plus of psi so these two p's can be got rid of if we put in a couple of plus signs or if p of psi is minus that we get we have we have an extra we take a minus sign out but then we get another minus sign from there so either way we're taking out two of some sign and therefore this is definitely one so we have that it's inevitably the case that this expectation value is equal to minus itself the only number equal to minus itself is zero so that implies that the expectation value of x vanishes but all states any all states of well-defined parity this is a result we use very often and it doesn't just apply to x it applies to any vector operator right i could have made x any vector operator and repeated the argument so when you're in a state of well-defined priority the expectation values of all parity operators are nothing and i think that's that we've won two minutes in hand but i think that is the moment to stop because the next section is on symmetries and conservation laws