Yeah. Okay, so let's just go. So we were, 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. 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 one. And the next item on the agenda is to check that P is a commission operator and therefore an observable.
And the proof of that is that we we take two states, find cy and we evaluate this complex conjugate. Let's make sure that is what I exactly what I plan to do. Yes. So. All right. Yeah. So 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 find the P so we right this is d cubed of x phi x x p ci. Then we can and we need to star the whole thing because I decided to start the thing on the left.
Right. That star is that star X of course is real interesting is real. And then we can use that to replace this. So that becomes d cubed x phi x minus x oops minus x minus x sign. And then we need to do the starring operation. So that's the integral de. Q Dex, take the complex conjugate of that and it becomes by excuse me, minus X and then here we're going to have X and I could write just five but just for fun went into write p squared phi because p squared is the identity operator.
So that's safe enough except I regard p squared is p times p and then I can take I can say, look, this p this out of p can be got rid of by replacing this by a minus x because I'm using x, p, some newfangled state phi primed which is PFI.
So this can be written as the integral d cubed of x sine minus x minus x. A P that's this inner p that out p has been dealt with by changing that sign on fi and then I can change my variable of integration from x to x prime, which is minus x, and that's going to produce that. He's going to be. But this take away the identity operator story that's next. Primed, which is minus X, take away the identity operator. And we're looking at. Which says that p since fine and CYA arbitrary.
That tells me comparing with the initial thing that P dagger is equal to P, which we already know is equal to P minus one. So first of all, this says that it's her mission. So it's an observable and this P dagger equals minus one says it's also at the same time unitary. So it leaves the length of all states the same. So since it's a mission. So. So what are its eigenvalues? We have if p upside is equal to M upside, this is its eigenvalue.
Or maybe we should call it lambda more traditional. So if PFC is equal to emit size. So it's a nice sine eigen ket. Then we can apply again and get that piece squared of ASI, which is actually equal to upside because p squared is one is also equal to Lambda PSI, which is lambda squared CI. So so we have the size lambda squared CI. And that implies that lambda squared is one, which implies that lambda must be plus and minus one. It has two eigenvalues plus and minus one. And we say that.
So if P upside is equal to CI, we say that size and even parity state. Correspondingly, of course, if PSI is minus of PSI. So that's the nagaon cat with the one with a plus one eigenvalue. This is an I can cat with the minus one eigenvalue. We say it's an it's an old parity state. What does that mean? From what we have up there, it means that. It means that when you from the top there the question is so so. So let's let's have a look at this. Sorry. From the wave function point of view, right?
For even parity, we can say that s p psi which is equal to minus x of psi by the operation of the p thing. But since size equal to up psi is also equal to x abassi. In other words, the wave function. Isn't even a function of x and similarly odd parity states have way functions which are odd functions of x, etc. etc. And when we did the harmonic oscillator, we found that, for example, we found that. 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 eigen functions of this parity operator. By no means all states are eigen functions of the parity operator. However. Okay. Now we do something considerably more interesting, which is transformations of operators.
So we introduced this. We introduced the displacement operator loss 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 so ci primed being U of a times that CI is a new state of the system, the state that it would have if 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 if so. So we can make the following. We can make the following statement that the expectation value in the state of SCI primed of x. Has to equal the expectation value in the state website plus a. Because we have 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 that is logical necessity.
But this but this we can write in a different way. This we can write using that expression as a ci times you dagger x you times of ci. All 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 one. I could say that this is this is the following. This is X plus a times the identity operator on its side.
Right. Because it's clear that it's 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 sign whatsoever. And it it's shown in a box. It's in the book.
It's a it's a little box which leads to the conclusion that not surprising conclusion, that if that's true, that this expectation is equal to this expectation of 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 that's relegated to a box in the book that you dagger X you is equal to X plus AA where this eye 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 here right now. Let's let's now make we'll make a small right. If a small we know we can expand this in terms of its of its generator. Right. So we can write this is this thing here can be written as the identity operator minus I a dot p over h bar plus order of h squared.
So we we've done that before and we're going to do it again. So that becomes the identity operator, plus a dot p overage plus dot dot dot which we will ignore times x bracket one minus i a dot p over par dot dot dot 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 also a small but still arbitrary.
I mean, you can still fiddle around with it, but it's small. So this is going to give me the this is going to give me r, i x is going to equal that.
So I'm 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 I is small on x times I and then we'll have an I times x times minus eight dot P. So what we're going to left with, left with in the order of a is I have for a dot p, comma x the commentator 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 what's that 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 this so this is a set of three equations, one for the X component of this,
one for the Y components of this and one for the Z component of this. So what does it look like? It it looks like this multiply by I over each bar and swap the order here and this is going to be x comma. A dot p commentator is equal to i. H bar a. Now let's use write this out in its components. This is to say a set of three equations so I can say that x j comma the sum over k of a. Well, it's AJ sorry.
It's sum of a k a.k.a k, but I can take the a.k.a outside the commentator because it's a man number is equal to i h far. I have to right now a j because this is the this this component to this vector here matches this here, here, right. This is a dot product which is the sum PKK. 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 of a k if I want to. Sorry, I h bar the sum of a k of delta j k a k posh way of writing it.
And now I can say because a k is arbitrary, the coefficient of k on the right side has to equal the coefficient of K on the left side. So that leads to the conclusion that x j comma k is equal to 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 just to review how this has happened. I wrote down a rather arbitrary rule.
I introduced P by arbitrary rule. I said that I said that x p psi is minus h bar d by x x ci. 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 DVD 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 is represented like this. And for the angular momentum operations, 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 torturous way. This is the main job, the 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 you dagger x you is equal to x plus a.
Even when A is big. Right. So we've this this stuff has all been for an infinitesimal a. And it's good to check that the other thing works. It sorts us even for a big. So now is just talk about for any a including a big one any big displacement. So we're going to be looking at you. Dagger. Sorry. Excu. You just checked my. Yeah. Which I can. Right. Of course. As you dagger you x plus you dagger.
Excuse me of you. So I've just swapped the order of those two and put it in the commentator that compensates. This of course, is going to be ex because you dagger you 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 commentator 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 that the commentator is 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 d by well, DVT, P of this is a minus. I. Now. But. 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 in 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 you because I get the exponential back and then I have times minus I over bar, right? That's the derivative of the argument of the exponential with respect to a P. And now I have to write down the commentator of X with respect to a p.
So this, of course, produces one. And what does this produce? This can be. Let's let's write that down. That's X plus groups 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 of a k of x. Comma. Peak times AK. Right. It doesn't matter what order I put on because it's a number and this. It may be that we ought now to introduce an index on X, otherwise we're going to get into 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 h bar of delta i k the I and the I make a minus one which cancels this, the bar kills that. So this is equal to x and then this is nothing. This, this, this is nothing is k is summed except when k equals I. So that becomes an I. So this book just becomes an I.
And yes, it does sort us. That thing is equal to x plus a as advertised at the top that. So now let's think about rotations. So we have we introduced these operators, we had J, X, j y and Jay Z. So that alpha dot j generated. Rotation. Through Model Alpha about the unit vector in the direction of alpha. That's that's what we establish. We use that notation. We said there had to be such a thing. And what we want to do now is talk about is apply is is is adapt that argument to this case.
So the thing is the expectation value. It so sorry we've let us let it slip primed be the state that you get when you use the ulfa on its side. 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 too.
If you've rotated the system, you've rotated the expectation value of X. So we can say that upside primed X upside 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 instead 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 this course operating on a sci x sign. So this was the old expectation, the expectation value in our around rotated system. If you rotate that expectation value, you must get the expectation value in the rotated system. So that's the analogue for the rotational case of. Of that statement. You 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 a. You dagger of Alpha. X you of alpha upside.
All right. So so I'm replacing this with that. And now I'm saying that for any state upside, this expectation value equals this expectation value. Ergo, since this is a set of boring numbers, it can go inside the 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 combinations, combination of the X, Y and Z position operators.
So I'm going to be able to say that u dagger, uh, x u is equal to oh alpha of x. That's a matrix. And now, of course, I'm going to express this as one plus alpha. Dot J. Dot, dot, dot. So this is now. Now we're now making small alpha. So we're rotating through a small angle and this can be written. Plus, here is an x, here is one minus I alpha dot j plus dot dot dot. And that is equal to the. To this vector. It is a vector of operators, but still it is a vector rotated by a small angle.
Now we know what that is. At least I hope we do. Come on. Come on. Oh, no. Go to sleep. Do I have to draw? We're. Yeah. System seems to. Just. No, it's gone to sleep. So this requires a bit of this is this is a piece of just standard geometry. What I want to do is write the action of a rotation matrix. When for a small angle, if I rotate something through a small angle. And I hate drawing these diagrams. It's going to be something as it come alive.
Oh, right. Yeah. Okay. It just went to sleep. It needs warming up. So we're looking at this second diagram. Can you see it? Because I'm sure it 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 the, the displacement that you have up there. This is the rotated vectors on the right, the unrooted vectors on the left.
The displacement is, is this thing here, which is the, the vector delta alpha or the vector. So the small, the small rotation vector crossed with the original V so that we can say that V primed the rotated vector is equal to the original vector. Plus this, this, this, this small rotation vector crossed into into V. So the right side. So I'm not going to draw this horrible diagram. It's the right side. 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 it's just alpha. We've made it small to get rid of symbols. So we do the same old stuff. We multiply this out on the left to the to two, up toward Alpha. We notice that the eye, the x of the eye produces an x, which cancels with the eye on the right. And we find that what we're left with toward alpha is alpha dot g times x minus from the eye, the x and the alpha dot j the thing the other way around.
So we find it eye alpha whoops alpha dot j comma x is equal to alpha cross x. Now we need to write this in, uh, 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 of the k of alpha k which will come out times j k. All right, that's alpha k k, comma x j. Now this is sorry. Let's change that. Let's just change that to something we some of it to be.
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 case component of this vector on the right. Oh, sorry. We should call that I. Right. This is going to be the IV component of the vector on the right. Now, across 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 sine 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 intensive note in Cartesian tends to notation nothing to do with quantum mechanics, it's just standard vector algebra. And we have arranged it so that we have the eyes components of the left side here and the eyes 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 xy comma JJ is equal to I times that's this I brought across the sum of a k. The sum of a j will go away because we're awaiting 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 Jth component of angular momentum computes 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 v i j j equals i epsilon some of a k of epsilon i j k vk for any vector operator. So a vector operator is just a set of three operators, if you like. Whose components? 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 VII is p the momentum. And we can also apply it to VII is equal to G. The angular momentum. Why is that? And it would be the other way round, which is right. Oh, sorry. In which one? This one. And this one just before the act. Okay. I lost a sign somewhere. No, I think. I think this is right. But I. Yeah, I multiply through by I and I swap the order of these two. Today is the boat from. No, I don't think so. No.
But this order is the same as this order. Surely to goodness. I don't think you invoke the other way around. Okay. Let me let me take advice on that. I'm. Yeah, I can't help being sceptical, but I suppose I should look here. I suppose I should look. Yeah. True. You know the thing I'm thinking of. Okay, so maybe I have, dear. I think we. Have I drifted a sign somewhere? I feel that that's what they. Well, this is definitely the ice 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, uh. This is 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 sine problems on blackboards. 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 and write up a for tomorrow's lecture the way it should be. We need to be persuaded that this will. The next thing I want to do is be persuaded that this is. So can we apply this to the Anglo momentum operations is going to be very important that we can. And what's the argument? The argument is that. Alpha J has to be a scalar.
Why? Because the operator ulfa which is e to the minus eye alpha dot j. This thing is the rotation around a certain vector. What this operator is this is a physically meaningful thing, and it is defined not by the three numbers that we happen to use to define the vector, 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.
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 operator is the operator associated with J x primed. Where X prime does 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 rotation. Then this dot 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. 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 commentator is equal to I epsilon sum. Some of it i some have a k epsilon i. J k. J k.
So this is a crucial relationship. And from that, we will find out what the eigenvalues can be of these operators J, R and JJ, and then we'll be able to find what they what the states are well defined, angular momentum and everything else. This expression is right. Right. Because it's independence of of of any swap. Can it be that both expressions are right? Oh, I can't. I must not take time to think about it. Okay. Let's consider. Oh, what's a scale?
Let's consider scalar scalar operators. So what is a scalar operator? It's a it's an operator. Which. 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 rotation of the coordinates.
So what can we say is that if X 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 and rotated states. But because, because this is a boring number and it's evidently by definition a scalar, something is unaffected by rotation. So the fact that you've rotated your system shouldn't have any effect.
So when we ask ourselves, so what does that what implications that have is that you'd like to ask you is equal to X? We can multiply on the left by you, which is the inverse of you dagger because you is a unitary operator. So we have then that S2 is equal to us, which means that s comma u equals nought. But this of course is U of alpha, the rotation operator throughout.
So scalar operator commutes with this rotation operator and it's easy to see by expanding this is one plus so if we write you is the identity minus I, alpha dot j, etc. that immediately goes to the statement that s comma j i equals nought. So scalar operators compute 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 jc j k a.k.a also known as J.
J. Right. That's a scalar operator. Every product is a scalar operator. So we have statements like J squared, comma, GI is nought. We have statements like x squared, comma, GI equals nought. We have statements like p squared, comma, gi equals nought. These are all important results that we will use. Many times we have statements like x dot P that's a scalar operator, comma, gi equals nought and so on and so forth.
So there are many operators we can make out of the operators already on the table which compute 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 some have a k excellent i j k j k. So the individual components is a somewhat strange state of affairs. The individual components of angular momentum do not compute with each other.
So you can't expect to know simultaneously the angle we're on the x axis in the engagement. Mm. Around the z-axis. I mean, in individual cases you can, but as a general rule, you can't expect to know that. And so there isn't a complete set of states which are simultaneously in states of checks. And Jay Z, for example. But we do have j squared comma j ii equals zero, and therefore there is a complete set of mutual eigen states of the total angular momentum and the Anglo mentum along any axis.
And that's what one or what we we have to work with that we have to consider states which we have to work with, states which are mutually agreed. States have j squared and usually the axis we choose. We have to choose one because of this business. And the access we usually choose is the z axis. An important result about about Paris's.
Go back to the Paris operation 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 power of operations, I make a system which is like my existing system but reflect it 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 it in the original system, the reflective system will have minus that value. So this can be written as a CI p dagger xp up ci. Right. Because of ci prime just by definition p c p dagger here. But we know that p dagger is p, so we have the expectation value for any state whatsoever of minus x is equal to the expectation value of p xp. 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 xp is not is nothing very much. So you can say now that the the the parity or p anti commutes this condition with M plus sign there right with the minus sign it will be a commentator with a plus sign. 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 is this stuff important? The practical importance of this is as follows. Suppose we have a state of well-defined parity. Okay. So let's let Pepsi equal either plus or minus upside. Don't care which, but it's going to be so size a state of well-defined parity. And we've seen that the Asian states of the harmonic oscillator Hamiltonian are actually.
States are well defined parity. And now let's consider Asi x asi. Well, that we've just seen is equal to minus ACI. P x p of sizes is a pure rewrite of a line higher up there. Well, except I've taken the dagger of 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 up CI is equal to either plus or minus up ci mean C plus up ci and p on this upside is equal to plus upside so these two P's can be got rid of if we put in a couple of plus signs. Or if size minus, we get that we have we have an extract, we take a minus sign out, but then we get another minus sign from there. So either way, we're taking out two, some sign, and therefore this is definitely one. So we have but it's inevitably the case. 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 for all states and 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 done 2 minutes in hand, but I think that is the moment to stop because the next section is on symmetries and conservation laws.