Stanford University. All right, so we were in the midst of talking about symmetry. Symmetries are operations that you can do on a system which

don't change the description, which don't change the phenomena, which don't change the energy levels, which don't change the values of the energies of systems. That's not ours, I don't think.

In particular, they don't change the values of the energies of a system. Examples that we talked about were translation. If you take an atom with a given energy and you move it from one place to another, the energy doesn't change. Doesn't matter which atomic state you study.

It's not just true for the ground state. It's true for every atomic state. And so the Hamiltonian doesn't change when you move things. That's a symmetry. The Schrodinger equation doesn't change.

Same Schrodinger equation. Same thing is true for rotating a system. We're going to talk a lot tonight about rotating systems. Another example is just interchanging two particles, two identical particles.

You might call one electron, electron one, and the other electron, electron two. We might call them Harry and Fred. Changing the names of the electrons. Interchanging electron one with electron two doesn't change the properties of an atom.

So that's also a symmetry. We tend not to think of it as a symmetry just because we're so used to it. Identical particles are identical. So it doesn't matter what you name them. But it is a symmetry. And there are other symmetries of physics. For example, if you have a crystal lattice. A crystal lattice has a kind of translation symmetry.

It's not translation symmetry in space. It is translation symmetry in space, but a different kind of translation and symmetry of space. It's just translating everything by one lattice unit. If we had a very, very long, almost infinite crystal, then

it's almost a perfect symmetry to translate the whole system by one lattice spacing. Of course, there it's not exactly a symmetry if the system is not infinite.

If the system is not infinite, then there are boundaries to the crystal, well, then you can't really translate it. But if the crystal went on and on and on forever, then you could do two things. You could imagine taking the crystal and moving the entire

crystal a little bit. This one goes to here. This one goes to here. This one goes to here. This one goes to here. That is a symmetry of a crystal. That's true. It's a symmetry of anything, the ability to move it from one place to another.

But the crystal has a different symmetry, which is just translating from, let's call this one, two, three, four, and let's suppose it goes on and on and on forever and ever, just relabeling the crystal so

that this one is now called minus one. This one is called, no, no, not one, minus one, zero, one, two, three, imagining the crystal goes on and on. That's also a symmetry. Translation of crystals by the crystal lattice spacing. What else?

What other symmetries are there? There are lots. There are lots, but some of them are more abstract than others. The physically intuitive ones tend to have to do with properties of space.

The homogeneity of space, the fact that every point is like every other point, is translation symmetry. The isotropy of space, the fact that every direction is like every other direction, is rotational symmetry. Now, another concept that we'll see is related to

symmetry, not the same thing, is called degeneracy. Degeneracy, and then specifically degeneracy of energy levels. Degeneracy of energy levels is simply the statement, an energy level of a system is said to be degenerate if

there's more than one state with the same energy level. Then that energy level is called degenerate. When does it happen? I mean, for a random system with a generic Hamiltonian, assuming that the spectrum of it is discrete spectrum, some

sort of discrete spectrum of energy levels, it would be an accident, or it would be a rather odd accident, that two levels would have exactly the same energy. And when I say exactly, I mean exactly the same energy. They may have the same energy to one part in 10 to the

15th, but to say they have exactly the same energy, what kind of coincidence would that take? Well, as we'll see, it doesn't necessarily take a coincidence. It takes a symmetry. In fact, those are the only cases where we really believe

that there are true degeneracies, exact degeneracies of nature, when there are symmetries. Degeneracies sometimes, no, sometimes, not degeneracy, sometimes means symmetry, the other way.

Symmetries sometimes imply degeneracy, but not always. And I'm going to show you now, I'm going to begin with an example in which you can see sort of intuitively how symmetries and when symmetries imply degeneracy.

That, incidentally, is one of the main uses of symmetry to analyze the energy spectrums of systems and see that they have levels that exactly match. That can be very important in atoms. It's extremely important.

Let's start with rotation symmetry. Let's start with rotation symmetry. And I want to think about a very simple system, namely a particle moving on a circle. It could be a particle moving on a circular wire. I don't want it to get off the circle, so it's a little

particle which is kind of a bead moving on the circle. And what is the symmetry? The symmetry is rotational invariance. How do we describe the location of the particle? Well, if we embed it in two dimensional space, then we

could describe it by an x and a y coordinate. But as long as a particle can't get off the circle, if it's only moving in the angular space, then it's a little bit excessive to think about an x and a y coordinate. We can just think of a theta coordinate or an angle.

That particle has a wave function. Here's the particle over here. It has a wave function. The wave function could be thought of as a function of x and y, but as I said, that's excessive, too many variables.

We can just think of it as psi of theta. The wave function, it's square, or it times its absolute, excuse me, it times its complex conjugate defines the probability to find particle on the circle.

What does the operation of rotation do? Let's say rotation counterclockwise. Rotation counterclockwise takes psi to psi of theta minus epsilon, where epsilon is a small angle of

rotation. I constantly get confused about whether it's plus epsilon or minus epsilon. I hope I have it right, but if not, don't worry about it, it just means all of my signs will be wrong. OK.

So psi goes to psi of theta minus epsilon. Or another way to say the same thing is that the change in psi, epsilon of course is our small number which is so small that its square is zero. It's an infinitesimal number.

The change in psi is equal to the derivative of psi with respect to theta times epsilon. I can rewrite that as minus i epsilon times minus i

epsilon minus i derivative of psi with respect to theta. Now I've put those i's in there because I want you to remember, well, let's think about ordinary momentum.

Ordinary momentum is described by an operator which is minus i d by dx. Angular momentum is simply basically the same as momentum except the coordinate is angle instead of being

linear position. The operator minus i d by d theta, that's an operator. It's a linear operator. It's a Hermitian operator. The i is there to make it Hermitian. Without the i, it would not be Hermitian.

Is one of the i's what? Do I have a minus i minus i? Yeah, I think it may be plus. Let's see, did I get it wrong?

No, I actually had it right. Minus. I had it right. This side was wrong. Yeah, OK. It's correct now. OK, but minus i d by d theta is a Hermitian operator.

You may be surprised that there has to have an i in front of it for it to be Hermitian. After all, Hermitian is the analog among operators as being real. But I guarantee you that i has to be there. Go back in your notes to the study of momentum and

you'll find out why that i has to be there. And that's called the angular momentum. That's usually denoted L. Why L? Don't know why L. Why P for momentum? Don't know why P for momentum.

But L is the angular momentum. And so we can write that the change in psi under the operation of infinitesimal rotation is governed by minus i times the action of the angular momentum

operator on psi. That identifies, according to the definitions of last week, that identifies L as the generator of rotations. If you can't remember the precise definition, let's just

say that this is the definition of the generator of a symmetry operation. If the symmetry operation is such that it has a version of it for arbitrarily small parameter, in this case angle,

then the definition of the generator is just in terms of the small change in the wave function when you shift by an infinitesimal amount. And that will just give you the generator. I did lose an epsilon.

Right. I did lose an epsilon. And this defines L. And L is minus i d by d theta. OK. And L is known as the angular momentum. Now before I go on, I want to put in some h bars.

I'll surely drop them again. But just to remind you where they go, the precise definition, including h bars, when we don't set h bar to one, contains an h bar over here. That's the same h bar that occurs in the momentum, p equals minus i h bar d by dx.

Same h bar. And it's a little bit arbitrary where we put it. But putting it over here, I guess we would say that the change in psi epsilon over h bar times L.

I'll probably, here and there, just remind you. And to keep you awake, I will put in h bars. Now and then. Every so often. Randomly.

OK. What are the eigenvectors of the angular momentum? The eigenvectors and the eigenvalues. The eigenvalues are the values of the possible outcomes

of experiments. And the eigenvectors are the states of the system for which those outcomes are certain. All right. To answer that, we write the equation L on psi. That's the action of the angular momentum. Is equal to an eigenvalue.

I'm going to call that eigenvalue m. It's standard. M times psi. Now, if I were doing it, I would have put little L here. But somebody did this long in the past. M happens to stand for magnetic quantum number.

But that's a historical anachronism. Nevertheless, the eigenvalue of L. And let's add a little more to this picture. Let's put a z-axis in. The z-axis comes out of the blackboard. So we're talking about rotations about the z-axis.

We could call this the z-component of the angular momentum. It is the z-component of the angular momentum. For the moment, I'm going to suppress that. And m is the eigenvalue of that angular momentum. So strictly speaking, it's the eigenvalue of the z-component of angular momentum.

Does it matter which axis is z and which axis is x? Or the rotator? No, it doesn't matter. But it's convention. OK, how do we find the eigenvectors? In other words, how do we find the wave functions that correspond to eigenvectors? And for that, we just do the obvious thing.

We put in minus i h-bar d by d theta. That's L acting on psi of theta is equal to m psi of theta. That's a standard type of equation. The derivative of something is proportional to that

something else. We could bring the i h-bar over to the right-hand side. And that tells us that the eigenvector with eigenvalue m is equal to e to the i m theta.

Very similar to the case of ordinary momentum. Oops. What do I have missing here? I'm missing the h-bar. The h-bar would go downstairs here. Because if you divide by h-bar, you see the thing that comes in is m over h-bar.

Now, unlike the case of ordinary linear momentum, there's another constraint. You know the other constraint. I'm sure that most of you know it. But let's spell it out anyway. The other constraint is that when theta is equal to 2 pi, the wave function should be exactly what it was when

theta was equal to zero. Why? Because going around by 2 pi brings you back to the same point. And so if there's a sensible wave function, that sensible wave function should be unchanged if you go around the circle by 2 pi.

So that says that e to the i m over h-bar times 2 pi. Well, let's see. First of all, the wave function for theta equals zero is just one. If I set theta equal to zero, this is just one.

If I go around by 2 pi, the answer will be e to the i 2 pi m over h-bar. And that also has to be equal to one. What's the constraint? The constraint is that m over h-bar has to be an integer.

In order to make sure that the wave function comes back to itself, that it's single valued, that it has a definite value at each point, it's necessary that the eigenvalues m be integer multiples of Planck's constant.

So let's write that out. m is equal to an integer. No. Yeah. m is an integer times h-bar. But the usual notation, the usual definition, is to set m

equal to the integer. Just call m the integer. And then the angular momentum, the eigenvalue of the angular momentum, is an integer times h-bar. OK, that's the quantization of angular momentum in units

of h-bar. Hereafter, at least for a time being, I will drop h-bar and will just say that the angular momentum must be equal to an integer. Angular momentum must be equal to an integer, and that's a fundamental fact of quantum mechanics that we've

known since the time of Niels Bohr. OK, now I have the following question for you.

Must it be the case, let's suppose there's some energy. And let's suppose the energy depends on the angular momentum. Of course it does. The energy, the bigger the angular momentum, the bigger the energy. Particles whipping around with a large angular momentum that has more energy than a small angular momentum.

So we expect that the angular momentum, that the energy will depend on the angular momentum. But must it be that the energy of a state with angular

momentum m, that's the eigenvalue, must it be the same as the energy of a state with angular momentum minus m? What's the difference between m and minus m? Well, the only difference is that positive angular momentum

means the particle is going around the circle counterclockwise. And negative angular momentum means it's going around clockwise. Take a vote. How many people think that the energy must be the same for m and minus m?

Of course. I'll raise my hand too. How many think it does not have to be the same? I will also raise my hand. Yeah. Just when you start out, you said the side theta goes to the side theta minus epsilon. Should I think then, that implies we're looking at it as

going in a clockwise direction? Not telling you anything about the direction of motion of the particle. It's only telling you which way you counted the theta as positive. It's like telling you I've decided that the positive x axis points that way.

That tells me nothing about whether I'm walking that way or I'm walking this way. All right. It does set the definition of what you mean by positive and negative angular momentum. But it does not tell you whether the system has positive or negative angular momentum.

OK. Art, you raised your hand the second time. Do you know an example where? I was just thinking that you could have an asymmetrical field somewhere. OK. Right. Does anybody know an example of a field which would lead to an asymmetry between positive and negative angular momentum

or angular velocity? Magnetic field. If there were a magnetic field, that's exactly what magnetic fields do. Magnetic field, for example, magnetic field into the

blackboard will raise the energy a little bit. Maybe I may have backward. Raise the energy a little bit of positive angular momentum and lower the energy a little bit of negative angular momentum. And it will break the degeneracy. It'll break the degeneracy between m and minus m.

That's the language. If you didn't have the magnetic field, the energy levels would be degenerate in pairs. Angular momentum zero would not be degenerate. There's only one of them.

But angular momentum plus one and minus one would be degenerate. Angular momentum two and minus two would be degenerate. In the presence of the magnetic field, that degeneracy is broken. But that magnetic field does not violate in any way rotational invariance. The magnetic field is pointing into the blackboard.

It doesn't in any way violate the invariance with respect to rotation. And so we see that symmetry is itself not enough. Not this symmetry. Rotation symmetry by itself is not enough to tell you that the energy levels are degenerate.

Why am I even talking about it? Well, you need to add, if you add one more symmetry, one more symmetry is enough to tell you that the energy levels have to be degenerate. And the symmetry is very simple. It's not a continuous symmetry. It's a discrete symmetry.

It's simply the discrete symmetry of reflection about one of the axes. Reflection about an axis is basically mirror reflection. Imagine you had a horizontal mirror here, and any system above the mirror has a mirror image below the mirror.

And if there's mirror symmetry, it's not called mirror symmetry, it's called reflection symmetry, I'm going to call it mirror symmetry and use the notation M to describe that symmetry. The reason is I don't want to use R for reflection

because I don't want you to confuse it with rotation. So M, big M. Big M stands for reflection about a mirror. And in particular, we could imagine the mirror is oriented along the x-axis.

Of course, a mirror is two-dimensional, but in this situation, along the x-z plane. But we don't have to think about z, we just have a mirror. And the mirror image of, I don't know if I can draw the

mirror image of that. I don't think I can. Did I get that right? Maybe not, I don't know. The mirror image of this is this.

OK. All right. If there is not mirror symmetry in a problem, then a configuration like this, whatever it happens to be, I don't know what it is, it's a couple of building blocks or something, the energy of this does not have

to be the same as the energy of that if there is no mirror symmetry in the system. But if there is mirror symmetry, it means that the description of this and the description of this are indistinguishable and that they must have the same energy.

So let's add mirror symmetry to rotation symmetry and see what we buy from it. Well, we will certainly find that the energy levels e of m must be the same as e of minus m.

Why is that? Let's think about in the upper half plane here, let's think of a circular motion moving counterclockwise. What's the mirror image of that? That's a circular motion moving clockwise.

So the mirror image of a particle moving with positive angular momentum is a particle moving with negative angular momentum. And if the world or our system has such a mirror symmetry, such a reflection symmetry, then it means that

the energies of the two states must be exactly the same. There we have an example where rotation symmetry, which tells you that the energy levels depend on the angular momentum and that the angular momentum is conserved, the

added ingredient is the mirror symmetry, which tells you that the energy levels have to be degenerate. So that's a case in which two symmetries combined tells you that they have to be degenerate energy levels. If we is determined by the probability, i.e. by the

square of psi, we look at psi, and we'll think of the Moivre expression of the cosine, and we change m to minus m, and you do the algebra, at least, it seems that it made.

But it seems that the sign goes away, and you're left with the same number. So in a way, replacing m with minus m algebraically will show that it is the same probability, meaning they're equal, the energies are equal. Is that not a good way to look at it? It's just wrong.

A function of theta is not the same as a function of minus theta. For example, here's a function of theta, the probability is concentrated over here. Here's a corresponding function of minus theta, the

kind that's over here. They're not the same thing. Also, e to the im theta is not the same as e to the minus im theta. But the square, this is the equation for the eigenvectors

of psi, or the eigenvectors of angular momentum, it involves psi. It doesn't involve psi star psi. Yeah.

That's the whole point of quantum mechanics. We have to work with wave functions, not probabilities. Of mirror symmetry? No. No. No.

It doesn't. It doesn't in itself. What's necessary is to have two symmetries and one more condition, that the two symmetries don't commute with each other. That the symmetry operations for the two symmetries don't commute with each other.

I'm going to let you prove that, that if symmetries don't commute with each other, that it implies degeneracy. It's not hard to prove, but I'm going to show you how it works in a couple of examples. Any two symmetries, if they don't commute with each other. So let's ask whether rotation symmetry and reflection

symmetry, incidentally, the term mirror symmetry happens to have another meaning in physics. It's a very fancy term, very mathematical term, that applies to certain manifolds in string theory. Don't get confused by it. It's a completely different concept.

I just use mirror to represent the reflection in space. Yes. Very interesting. Good. Good point. Yeah. The mirror symmetry of a magnetic field, the mirror inversion of a magnetic field, is a magnetic field

pointing in the other direction. If you take a magnetic field pointing into the blackboard, and you reflect the configuration about a plane like this, what you get is a configuration with the

magnetic field coming out of the blackboard. This is because magnetic field is an axial vector. You can see it another way. Magnetic fields are made by currents. Let's not imagine that the magnetic field is there.

Let's instead imagine that the magnetic field is being made by a current going this way. A current going around in a loop like this. It could be going around a solenoid. We could be talking about a solenoid into the blackboard with the current going clockwise. That creates a magnetic field going into the blackboard.

In the lower half plane, the mirror image of that is a solenoid in which the current is going in the opposite direction. And because the current is going in the opposite direction, it makes a magnetic field pointing out of the blackboard. So yes, the mirror inversion of a magnetic field is not

the same magnetic field. So I think you foresaw that that had to be the case. Let's ask the mathematical, let's go through the mathematics now, of showing that mirror reflection does

not commute with rotation. And we'll do that by showing that the generator of rotation doesn't commute with the action of the reflection operator. We need to figure out what the reflection operator does.

The reflection operator we'll call M for mirror. It takes any wave function and replaces it. I shouldn't write arrow, I should write equals. And replaces it by a wave function of minus theta.

Reflection about the horizontal axis just takes theta to minus theta. So any wave function, when you reflect it, goes to psi of minus theta. In particular, if you apply it to ether, well all right,

we'll do it in a minute. Now let's consider the commutator of the mirror symmetry transformation with the angular momentum. The angular momentum is just a version of a rotation operator for very small rotation.

Let's consider the commutator. What the commutator is telling us about is what happens if we do the two operations in opposite order. A rotation about an axis and a reflection versus a reflection and a rotation about the same axis.

Are they the same operations? Rotating about an axis and then reflecting, is that the same as reflecting and then rotating? OK, the answer is obviously no. Well, if it's not obvious to you, think about it. But we'll work it out by doing the mathematical commutator and seeing what we get. Let's look at the commutator of M with L. We

know what L does. Here it is. It differentiates with respect to theta. What does M do? It reflects theta. So we can figure out what the commutator does. We can figure out what it does when it acts on a wave function. I'm going to act on a particular wave function. I'm going to act on the wave function e to the i M

theta, an eigenvector of the angular momentum. I'm leaving h bar out of it. We don't need it. And don't confuse big M with little m. Little m and big M have nothing to do with each other.

This stands for mirror. This stands for magnetic quantum number. OK, let's see if we can work it out. Let's first do M L times e to the i M theta. What does L do when it hits e to the i M theta? It brings an M down, right?

It just brings an M down. So this gives us M times little m e to the i M theta. And what does capital M do when it hits e to the i M theta?

It just changes the sign of theta. Incidentally, once this is just a number, it doesn't matter which order big M and little m appear. And so this is just equal to little m times e to the minus i M theta.

The big M just reflected theta. OK, let's do it in the opposite order. L M e to the i M theta. Now in the opposite order, first M acts.

And that gives us L. When M acts, it changes the sign of theta and gives us e to the minus i M theta. But what happens when L hits e to the minus i M theta? Anybody got a good idea? It brings down a minus M. By reflecting, we've changed

the sign of the angular momentum. So the angular momentum operator gives us a minus of what it gave before. It gives us a minus M e to the minus i M theta. So the answer is that M and L, angular momentum and the

reflection operator, don't commute with each other. That's the key to whether or not symmetries create

degeneracies of energy levels. We've seen in this case that two symmetries, rotational symmetry together with reflection symmetry, tell us that there are degeneracies of the energy levels. That M and minus M have to have the same energy.

That's a pattern. It's also a theorem, but I don't think we'll do the theorem right now. Does this mean that any Heisenberg pair generates degeneracies? A Heisenberg pair like x and p?

Well, the problem is that x and p are not conserved. Right, and they're not symmetries in themselves. In the case of a particle not having a magnetic field, then angular momentum, rotation, and reflection are

both symmetries. They both commute with the Hamiltonian. They both are conserved. x and p are typically and generally not conserved. If they were, then the answer would be yes. Then the answer would be yes.

We're going to study an example in some detail tonight. I think we'll get through it. Oh, let's just write down just some formal observations of what I mean. Supposing you have two symmetries.

Now, what does it mean to have a symmetry? Let's call the two symmetries a and b. The a and b stand for operators. Operators that act on the state vectors. They could stand for infinitesimal generators, or they could stand for the full operation of the symmetry, a and b.

Let's say they refer to the generators. Let's take the case of where there are small versions of the transformations. Then, to say they're symmetries means that both of them commute with the Hamiltonian.

It does not in itself say that a commutes with b. In general, here's an example where they didn't. Commutator of a with b may be non-zero.

And there are two possibilities. Both possibilities will lead to degeneracies, but let's think of two possibilities. In one possibility, the commutator of a and b may be a itself, it may be b itself, or it may be some linear combination of a and b.

That does not lead to a new symmetry. It doesn't lead to a new operator. It just leads to a linear combination of the old ones. But let's suppose that it leads to a genuinely new operator that cannot be written as a linear combination of a and b. Let's call it c.

OK, first theorem, theorem, it's not a theorem. It's hardly a theorem. c also commutes with h. How do I know? Let's put an i in here. Why am I putting an i in there?

The commutator of two Hermitian operators is not Hermitian, it's anti-Hermitian. It's like the commutator of x and p is i. If you want a Hermitian operator, you have to put an i here. So let's suppose that c is a commutator of a and b, except with an extra i, then c is a Hermitian operator.

And let me show you why it's conserved. It's conserved for a trivial reason. Let's think about the commutator of c with h. That's abh minus hab, right? ab is equal to c, apart from a factor of i.

So the commutator of c with h is proportional to abh minus hab. But now all we have to do is say, look, we know that b commutes with h. That's an assumption. That means that we can move b to the other side.

Once we've moved b to the other side, we know that a commutes with h. That's also an assumption. So we can move a to the other side. And that, of course, is equal to 0. So if a and b both commute with the Hamiltonian, then the commutator of a and b also commute with the

Hamiltonian. In fact, the product a and b commute with the Hamiltonian. But the reason we don't consider the commutator is because the product of a and b commute with the Hamiltonian. Certainly, the commutator does.

I'm not so interested in the product because it's not generally Hermitian. It's the commutator which is Hermitian. So there's an example where, given two symmetries, if they don't commute, you generate a third symmetry.

And it might be an independent one. It might be a new thing. Once you have a and b and c, you can play the game again. Commute c with a and b. See what you get. You may get something that just reproduces or that's just a linear combination of the three that you started with.

Or you may get something new. You may get d. You keep doing this and doing this and doing this until you don't get something new. You might never end. You might get an infinite number of different symmetries. That's rare and that's unusual and it's bizarre.

Or it will close eventually and you'll have some finite number of symmetries which are all independent of each other and which form a kind of closed system when you commute them. When you commute the generators.

Question? Yeah. Just to make sure I understand your argument. What you showed was that if you have a and b that commute with a, then the product a, b commutes with a. It does. And then since the commutator of a, b is just a, b minus b, a, then you'll get the full commutation.

Right. So eventually, if you have a few symmetries, you can start commuting them and commuting them among themselves, being guaranteed that what you get is another symmetry, which may be just a trivial superposition of the original ones or it may be something new.

If it's something new, good. Put it into your list and commute it some more. Eventually, you'll either run out of them or you won't. But I'm particularly interested in the situation where you run out and then you'll have what is called a commutator algebra. The commutator algebra includes the commutators of all

the symmetries and it includes the commutation relations with the Hamiltonian, which just say zero. When you're in that situation, whenever that happens, you have a great deal of power over the system to be able to say a lot about the nature of the energy

levels and the degeneracies among them. OK. So that's the general. Oh, what is such a structure called the collection of symmetry operations? A group. This is a symmetry group.

This is a symmetry group. Why do we care that these two things are degenerate? Why do we care that they're degenerate? It would certainly be important if, for example, we wanted to know if from one state you could emit a photon and then go down to the other state.

Can't do it if the energies are the same. We want to classify the energy levels of systems. Why? Because we're spectroscopists and we want to study the spectroscopy of atoms. Oh, we're spectroscopists and we want to study the way hadrons, excited states of hadrons emit

pions or whatever. Same rules. Well, these things themselves, these I imagined were generators.

The generators form an algebra, a commutator algebra, a Lie algebra. The symmetries themselves, which are built up from the generators by applying small transformations repeatedly, those are the group elements. But we're going to mainly work with the generators.

The generators have a lot of power. The algebra of them. The C generator is not in the right combination. They're all generators. Yep. Right. They're all. Now, you can have a situation where you may have many

symmetries, but they commute with each other. That can happen. Then you don't get anything terribly interesting. You just have separate symmetries that commute with each other. Symmetries of that kind are called abelian. Abelian means everything commutes with everything else. When there are non-zero commutators between symmetry

generators, it's called non-abelian. Is there any fundamental definition of a generator? A generator of a symmetry is, if you take the symmetry operations, unitary symmetry operations, and you

consider the case of a symmetry operation very close to the identity. In the case of rotations, it means a rotation by a very small angle. And you write it in the form 1, I can never remember. I think it's minus i g times epsilon, when g is called the generator.

You can build up, if you know the generator, that allows you to make a small rotation. If you keep doing small rotations repeatedly about a given axis, you'll eventually build up a whole rotation, a finite rotation. So the generators contain basically all the interesting

information about the group. Well, almost. Is there any fundamental difference between ab and then c or e? No. We could have started with b and c and then b and a. Yeah. Well, maybe yes, maybe no. You might take b and c, and you might not get a.

You might get d. You might. Or you might get a plus b. If you got a plus b, then it wouldn't be anything new. You might get a plus b plus c back again, and it still wouldn't be anything new.

But if you got something new, d, which was not among your previous list, then you have a genuinely new symmetry. Yeah. Could you please repeat what you said about what's an algebra? That's not a question. That's a Dirac story.

What did you ask me? No, what did you tell me? What did you request of me? You said something was an algebra, and first of all, I want to know what it is that's the algebra, and what is the most? The algebra is the collection of generators. You can add them. They are still generators.

You can subtract them. You can even multiply them by ordinary numbers, and you can commute them. So commute product in that algebra. That's correct. Commuting is a product in that algebra. An algebra is a collection of elements that you can add,

you can multiply by numbers, and you can multiply by each other. In this algebra, multiplication is commutation. That's correct. That's correct. OK, so we'd like an example. Not only would we like an example, we would like the

preeminent, most important example, and that is angular momentum and rotation about different axes. As an example of what I'm talking about here, we can consider rotation about the x-axis. Which way is the x-axis? That way, I guess. Rotation about the x-axis, rotation about the y-axis.

And if you think about it, and you probably know this, I could never do this. I'm not going to try it. But if you do rotate about the x-axis, then rotate about the y-axis by the right amount, what you get is a rotation about the z-axis. What's more, if you rotate a little bit about the x-axis,

and then a little bit about the y-axis, no, a little bit about the x-axis, and then a little bit about the y-axis, it's not the same as rotating a little bit about the y-axis first, and then about the x-axis. They don't commute with each other. We're going to show that. We're going to show that by studying the commutation relations of the angular momentum.

You look puzzled. I'm puzzled, because I know that that's true for finite rotations, but I thought somehow that infinitesimal rotations. OK. They don't commute the second order in the small numbers.

That's enough. That's enough. Remember, these generators have the small numbers factored out of them. And they don't commute. They really don't commute. OK, we're going to work out an example in detail. We'll do it right now. The example is the three components of angular momentum.

The three components of angular momentum form a, the three rotations about any axis form a group. But for our purposes now, we're not concentrating on groups. The angular momentum generators form a Lie algebra, a commutator algebra. That simply means that the commutators of components of

the angular momentum give you other components of the angular momentum. And we'll work that out. But first, let's talk a little bit about classical angular momentum for a minute, just to understand the corresponding classical concept.

OK, we can do this. OK, let's imagine a planetary orbit. For my purpose now, it can be a circular orbit. It's orbiting about an axis, and it has an angular

momentum which is perpendicular to the orbit. That could be Lz. It has a certain energy, e. Question, at a classical level, is that energy level degenerate? The meaning of that is, is there another orbit with

exactly the same energy? Of course, there are many other orbits with exactly the same energy. But in particular, there's some that you can easily prove have the same energy because of rotational invariance. All you have to do is rotate this about an axis.

Let's rotate it about the y-axis. Take this configuration, rotate it about the y-axis. It's a new configuration. And the angular momentum has changed.

Yeah, I guess that's OK. I guess, what did I do? I don't like my picture. My picture, this is in the wrong place. It should be coming from here. There's Lz. And now there's a change in L. It's

picked up an x component. Yes, in fact, it's picked up an x component. And so because of rotation about the y-axis, rotation configuration with the same energy, but it's going to be

given as a configuration with a shift of the angular momentum, and in particular, a shift of the x component of the angular momentum. Classically, that's an indication that quantum mechanically, the components of angular momentum don't commute. A rotation of a component of an angular momentum about a

different axis changes the angular momentum about the third axis. That has a great deal to do with the commutation relations of rotations. But we'll work it out in detail. So here's an example. Classical physics, just at the classical level, the

combination of rotation and variance, the combination of the different rotation and variances about different axes tells us unequivocally that there must be degeneracy of the orbits. That there must be orbits with the same energy with rotated angular momentum.

That's the basic idea classically. So let's go through it quantum mechanically. The first thing we have to do is understand the angular momentum generators better. So let's come back to the angular momentum generators that form. Previously, I just said minus i d by d theta.

But let's work them out, not assuming that a particle is moving in a circle, but that a particle could be moving in three dimensions. In fact, it doesn't even really have to be a particle. But let it be a particle. A particle moving in three dimensions.

First, before we do three dimensions, let's do two dimensions. Particle moving in the xy plane, not restricted to a circle. And let's ask, what's the generator of rotation now? Well, it may be i d by d theta. But we can represent it in terms of the xy coordinates.

So here's the position of the particle, let's say. Let's suppose we rotate the configuration. Rotating the configuration, whatever it happens to be, means whatever the particle is, or whatever the wave function of it is, shift it. So if the particle was at point xy.

It moves to a new point. What's the new point? The new point is given, if the old point was given by xy, then the new point is shifted by an amount delta x and delta y. I'm assuming that the angle is small.

And let's call the angle epsilon. Let's call the angle epsilon. Then what is delta x and delta y? I'll leave it to you to prove this is a very simple exercise. It's just simple geometry. For a small epsilon, can anybody tell me what delta x is?

It's of course proportional to epsilon minus y. If y is positive and you rotate, the change in x is negative. So it's got to be minus.

And it's proportional to y. What about my delta y? That's epsilon times x. OK, now let's consider a wave function, psi of x and y. What is the change in the wave function if you

reconfigure the system by rotation? It's equal to the derivative of psi with respect to x times the change in x plus the derivative of psi with respect to y times the change in y if I shift the wave function.

But delta x is minus epsilon times y, minus epsilon times

y. And this is plus epsilon times x. That's the change in the wave function. And of course, this must be, by definition, I guess it's

L sub z times psi. So I epsilon times Lz. Now it's much too late in the evening for me to follow the signs, so I won't.

I'll leave the signs for you to figure out yourself. But it has this form. But what is psi by dx in quantum mechanics? i times px. So this is another factor of i.

And this is y times px. Did it matter which order I put down y and px here? Did they commute?

Yeah, y commutes with px. x doesn't commute with px. y doesn't commute with py, but x commutes with py. What about this one? This one is plus i times x py.

All times. All right, apart from the sign, which I said I wouldn't try to track in any detail, it's quite clear

what the correspondence between the z component of the generator of rotations and the coordinates and momenta of the particle are. Lz is to be identified with x py minus y px.

And now I do have, this is the right sign. I lost track of the sign, but this is the correct sign. Have you seen that before? That's the component of r cross p, the standard

three-dimensional angular momentum, r cross p. So now we see a new interpretation of r cross p. Not just the mechanical angular momentum from classical physics, but it now has the role of the thing which

generates small rotations by acting on the wave function. What about the other components of angular momentum? You can read them off just by cycling through the xyz. x goes to y, y goes to z, z goes to x.

So going through it, z goes to x, x goes to y, y goes to z, minus z py, and Lx, xyz, py, pz, px, minus x py.

These, of course, you could confirm by imagining rotations about the other axes, but just by the parallel reasoning, this is what you would get.

Questions? Oh, you're right, pz. I have two different notations for z, don't I? Wait a minute.

Which one, which one? Ly, good, thank you. Ly, good, good, good, good. OK, these are, are they symmetries? Well, they're symmetries if the system is rotationally invariant.

Why? Well, because rotating a system shouldn't change the Hamiltonian. So we expect, we fully expect that if we have a system with rotational invariance, then these should commute with the Hamiltonian. And typically they do, they do.

In fact, they always do if the system has invariance. What kind of systems have rotational invariance? Well, a very particular system is a particle moving in a central force field. A particle moving in a central force field will have rotational invariance, and therefore for that kind of

system, the commutator of Lx with h equals zero, and in fact, all three of them. L sub i, one, two, and three, or x, y, and z, they all commute with the Hamiltonian, and they're all conserved. And this is just good old angular momentum conservation,

except in a quantum guise. Next question. Do they commute with each other? Suppose all we had was Lx and Ly. That's all we knew about, Lx and Ly. Why we only knew about Lx and Ly?

I don't know, our mothers only told us about X and Y. Would we discover Lz? That Lz is another symmetry. Well, our mothers only told us about X and Y, but they also told us about commuting the generators of Lie groups.

So mom tells us, look, commute Lx and Ly and see what you get. Let's commute Lx and Ly and see what we get. OK. Well, what do we need to do? We need to know the commutation relations between

x, y, and z on the one hand and px, py, and pz on the other hand. The rule, I'll write it over here, any x with y or z or anything like that is always equal to zero. x with itself is equal to zero, x with y, x with z is equal to zero.

They commute. Ps commute with each other. Px, py, and pz all commute with each other. Which things don't commute? The things that don't commute are x with px. And that's equal to i, strictly i times h bar.

Same with y, y with py is i and z with pz is i. And we'll write it down. How about x with py? That commutes.

Multiplication by x and differentiation with respect to y commute with each other. And you can check that. OK, so let's calculate the commutator of Lx with Ly. That's the commutator of y, p, z minus z, py, comma,

comma for commutator, z, p, x minus x, p, z. Now I'm not going to go through the rules for commutators, either you remember them or you don't. But it's by inspection. You can more or less see what to do.

Who in this side doesn't commute with who on this side? It's a commutator of this thing with this thing. What on this side doesn't commute with something on this side? Well, y commutes with z, y commutes with px, and so forth, but pz does not commute with z.

So there's two things which don't commute. And let's see over here, z doesn't commute with pz. Those are the only combinations. If we were to expand this out, the only combinations

which don't commute are pz, did I get it right, with z, and z, did I get that right? z, p, x minus x, no, no, what have I written here?

What am I doing? I'm commuting Lx, that looks right, y, p, z minus z, p, y, with Ly, which is z, p, x minus x, p, z. Yeah, that's right.

Yeah. OK? OK. All right, from this term over here, you get a commutator of pz with z. What's the commutator of pz with z? Minus i. Minus i. So there's a minus i, and that gets rid of pz and z,

and you're left over with y times, no, sorry, you're left with pz with pz, and that leaves over y and px. From this with this, you get y times px. Does it matter which order you write down y and px?

No. OK, so there's minus i, y times px, and then there's the other one, which is commutator of z with pz is again i, and now we get plus i, x, p, y, I believe.

Yeah, x times p, y. Where is it? x times p, y from here, which is our

friend i times Lz. So this is an example of commuting generators and discovering new generators. Not a big surprise, but nevertheless, we see it

before our eyes. And so now we can write down the general list of commutation relations. Notice that by commuting to angular momentum, we got a third one. There's only three. There's no way we're going to get more than three generators out of this sequence of commutations.

And so the Lie algebra is just going to be the Lie algebra of the three components of the angular momentum. Let's write it down. This one says that Lx with Ly equals Ilz, and then the

trick is always the same. Just cycle, LL equals IL, LL equals IL, brackets, brackets, comma, x, y, z, y, z, x, z, x, y.

Those are our basic commutation relations of angular momentum. If you continue commuting, you won't get anything new.

So we've discovered a genuine Lie algebra.

Oh, you want to prove that Lz is not a linear commutation. I suppose we do. But yeah, I think we can prove that. Let's see. What can we do? Let me not answer that right now.

It gets me off track. Let's come back to it. It should be very easy. I don't want to get my head confused. Of course, the z component of the angular momentum is

not likely to be a linear combination of x and y components. But yes, you're right. You're right. We do have the obligation of showing that Lz is not You know what? I think we can probably show that Lz, from these commutation relations, I believe we could show that Lz

does not commute with any linear combination of Lx and Ly. If Lz was a linear combination of Lx and Ly, it would commute with that linear combination. I think we can prove it.

But what does that mean? What does it mean? What does it mean that Lx is a? Yes, they are perpendicular components of some three-dimensional vector r cross p. But I don't want to use that.

I think the right thing to do is exactly, as this gentleman said, is to directly prove algebraically that Lz is not a linear combination of the others. I think we can prove that. Yeah? What's that?

A Lie algebra is an algebra where the rule for multiplication is commutation. OK, that's what it means. But don't worry about the name. The name is just a collection of objects whose commutation relations close.

A collection of objects, when you keep commuting them, you don't get new ones. That's called a Lie algebra. And it's just a name. Lie, of course, was a mathematician.

I don't know, was algebra a mathematician? Well, it might have been, I don't know. Al-Jabir, yeah, I think he was. What's that? What's that?

I think it's an Arabic word, yeah. I think it was. I have a feeling his name was just Al-Jabir. But I'm not sure. I remember reading that. Or maybe I don't remember reading it. OK, so where are we? We have this nice collection of algebraic relations,

including three more. But I'll just write them as one more. The commutator of any component of L with a Hamiltonian is equal to 0, which means that the three components of angular momentum

are conserved. OK, what can we do with that? Well, a lot. This is a good time to remember what

we learned about the harmonic oscillator by having a similar kind of algebraic structure. The similar kind of algebraic structure that we had were the commutation relations of creation operators, annihilation operators, and the number operator.

We had closed commutation relations of a very similar kind, not exactly the same kind, but closed commutation relations. And they were quite powerful. We deduced the entire spectrum of the harmonic oscillator. We did it from constructing creation and annihilation

operators. It was much easier than solving any Schrodinger equation. We're going to do the same thing here. We're going to construct creation and annihilation operators. But the creation and annihilation operators are not going to change the energy. What they're going to change is the z component

of angular momentum. Now, why do I pick the z component of angular momentum? That's arbitrary. Remember, in quantum mechanics, if you want to actually work with things, you usually have to pick a basis. You usually have to pick a preferred set of operators

which you regard as defining a basis. Often it doesn't matter if there's the x basis or the x representation. Not a basis, a representation. In quantum mechanics of a particle, there's the position representation. There's the momentum representation.

In studying spins, we have the sigma z representation and the sigma x representation and so forth. It really doesn't matter for the purposes of the invariant physics which basis you use. But it's common to pick one and work with it and keep with it. We're going to do the same thing here.

We're going to pick out the z component of angular momentum to focus on. There's nothing special about it. We could have picked the x and we could have picked the y. But it's useful to pick the z component. No more useful than x and y. But let's do it. So we're going to work with the z component of angular

momentum and its eigenvectors. We're going to work with z component of angular momentum and its eigenvectors. And see what we can find out about the entire spectrum.

Lz has eigenvectors. It's a Hermitian operator. How many it has, I don't know. We'll find out as we go along. And let's call the eigenvalue of Lz little m.

It's the same eigenvalue or the same notation that I used in two dimensions. In two dimensions where there was only the z component of angular momentum, I call the eigenvalues m. And I'll continue to do that. Again, it's a historical notation.

It had to do with putting atoms in magnetic fields. You put an atom in a magnetic field and you study its spectroscopy. And arbitrarily, the z direction was chosen for the direction of the magnetic field. And the angular momentum about the z-axis was called the magnetic quantum number.

But for our purposes now, m is just an eigenvalue of L sub z. This is the equation that says that the eigenvector m is an eigenvector of L sub z with eigenvalue little m. Sort of redundant notation.

I've used the same notation for the eigenvector and the eigenvalue. And here's where mathematicians go crazy because they don't like you to use the same notation for two things, but we'll do it anyway. OK, so what we want to find out is what

are the possible values of m. The possible values of m will be exactly the same as the possible eigenvalues of L sub x and L sub y because there's nothing special about L sub z. But nevertheless, let's see what we can find out. Now in the case of the harmonic oscillator,

we figured stuff out by inventing creation and annihilation operators, or operators which raised and lowered the energy. We're now going to invent operators which raise and lower m. They raise and lower the magnetic quantum number. And how do you do it? You do it in a way very similar to the harmonic oscillator.

Yeah? When you refer to, we did the harmonic oscillator, you talked about the classical harmonic oscillator. We did. The quantum harmonic oscillator. OK, we never did that. We didn't? We didn't do that? Not in the last class, but it's in the book.

And there is a video that covers it somewhere back in history. We never did the harmonic oscillator. OK, in that case, we'll come back to the harmonic oscillator. And instead of telling you how much this is like the harmonic oscillator, I will tell you how much the harmonic oscillator is like angular momentum.

Good. This is self-contained. It doesn't require us to know about it. I thought that was one of your main motivations for this second course, because you said, gee, we didn't get to cover the harmonic oscillator. I had forgotten.

Well, it's in our book. But I'll send the link out to the video. Good. The book will open. It's up. You'll find out. OK, do not worry about the harmonic oscillator now. It's much more complicated than this.

So we've picked out Lz as a little bit special, not physically, but mathematically for our purposes. And let's invent operators Lx plus i Ly, and also Lx minus

i Ly, and call them L plus and L minus. And what I want to figure out is the commutation relations of L plus and L minus with Lz.

For the moment, I don't really care too much about the commutator of L plus with L minus. I'll tell you now, it's Lz. But that's not so important. What I want is the commutation relations of L plus with L minus. So how do we get them? Well, we're going to get them from the last two of

these equations. Let's rewrite them. Let's see, how did I write them? Lx with Lz, that's this one over here.

I've changed the order of x and z. So that's minus i Ly. Do I have that right? Yeah. That's this one over here.

And since I changed the order of x and z, that changes the sign. Then there's the other one, commutator of Ly with Lz. And that is for i Ly with Lz. We have to do a little bit of work. Here's Ly with Lz.

Multiply it by i, and we get minus Lx. And now we're just going to add and subtract these two equations. Adding and subtracting them will give us the commutation relations between L plus and L minus on

the one hand and Lz. Minus Lx. OK, we add them and subtract them. If we add them, we get L plus commuted with Lz. And if we subtract them, we get L minus. We do the summation on this side also.

And here's what we get. I'll just write it out. We don't need to belabor it. The commutator of L plus with Lz is minus L plus.

And the commutator of L minus with Lz is L minus. Not minus L minus, but plus L minus. L minus. Those are the two basic commutation relations that follow from these.

Now I understand this is very abstract. How did anybody know to do this? Well, I'm not even sure who did it. It's probably Dirac. Dirac had a weird mind.

He liked to commute. I can't tell you exactly why he did this, but once you do it, you say, wow, OK, that's very clever.

So this is just another way of writing the Lie algebra. It's probably true that commutator of L plus with L minus is Lz. Or maybe twice Lz. You can work that out. But we won't use it. Now what I'm going to prove is the following very interesting little theorem.

Supposing we have an eigenvector of Lz. Supposing we have one. We found one somehow. Now I'm going to show you how to find another one with a different eigenvalue. We're going to do it by applying one of these

equations, which one are we going to do? L plus. Let's do L plus. We're going to take this equation for L plus and multiply it by the eigenvector m. By this little eigenvector m. So let's work it out. L plus L minus, sorry, L plus Lz minus Lz L plus.

That's the commutator. Acting on the vector m is equal to minus L plus acting

on the vector m. OK, so what do we know? Let's take this first term here on the left.

What does Lz do when it hits m? Multiplies by m. That's because m is an eigenvector of Lz. So the first term becomes m times L plus m. I'm going to take L plus and m, and I'm going to put

them in a red circle. Because I want you to think of it as a unit. Now let's add this over to the left side here.

We're going to take this term and move it to the right, and we're going to take this one and move it to the left. So let's take this one and move it over to the left. That will give us plus L plus m. That's also exactly the same thing that appears in the red circle.

So what do we get? We get m plus 1 times the object in the red circle. Let's just draw a red circle there. That's the left-hand side with this term

transposed to the left. Now let's transpose the other term to the right. That's equal to Lz times L plus m. But L plus m is again exactly the thing in the red circle.

The thing in the red circle is a certain ket vector. So what does this say? This says that there's another eigenvector of Lz. This equation is an eigenvector equation for Lz. Lz acting on the red eigenvector gives us m plus 1

times the red eigenvector. What does this mean? It means we've discovered another eigenvector of Lz with eigenvalue m plus 1. We can call the thing in the red bracket here, we can just call it the eigenvector m plus 1.

We found a trick for taking any eigenvector of Lz and promoting it, lifting it up one unit, and increasing the eigenvalue by one unit. What about if we did the same thing with L minus?

If we did the same thing with L minus, we would again discover another eigenvector with one lower unit. So we've used a simple algebra of commutators to generate a spectrum, a spectrum of values of Lz.

And what's more, they're separated by an integer. They're separated by integers. So we've discovered the following fact, that if we plot on the vertical axis here, the possible values, the values of m, or Lz, the possible values of Lz, if we

find one at some value of m, first of all, let's lay some integers down. 0, 1, 2, 3, 0, minus 1, minus 2, minus 3.

Let's suppose we found an eigenvector, an eigenvalue of m. Nothing yet has told us it must be an integer. So it could be anywheres, it could be over here. It could be at 3 quarters, or it could be at 7 eighths,

or it could be pi quarters, or it could be a half. It could be anywheres. Not over here. Let's put it over here. Then? Just one question. I thought the way it was defined, it had to be an integer, going all the way back. In fact, we're going to find it has to be, yeah.

You're right. You're right. Except, not quite. What I mean is, I thought you already said that. Yeah. We could go back to what I said earlier and use that. I don't want to. I don't want to, because I want to show you another

argument for it. Now, the other argument will not prove it's an integer, it'll prove something a little bit looser. And the looser fact is in fact the correct fact. But we'll see. So the first thing is, we know that there's got to be another one above it. One unit above it.

And another one one unit below it. Now, there's one exception to this statement that when we apply L plus, we will necessarily get a new eigenvector with one unit up. What's the other alternative possibility? That you get zero. That you get just plain zero.

No vector at all. So this could terminate. Conceivably, it could terminate. I wish I had a green pen. I don't. It could terminate somewheres. Let's say it terminates here. And by the same logic, it could terminate from below if

L minus killed the vector. So we will generate a family which may or may not also terminate somewheres. Now, the first thing which is true is if it terminates,

if it terminates, then there's a relationship about where it terminates from above and below. And the reason is very simple. M is the eigenvalue of Lz. Remember, we have rotation invariance. Supposing I use my rotation invariance to completely

rotate by 180 degrees. If I rotate by 180 degrees, then Lz becomes minus Lz. If there's really rotational symmetry, then it must mean the spectrum, the possible eigenvalues of Lz must be the

same as the eigenvalues of minus Lz. It's just looking at the same system except standing on your head. And if there's rotation invariance, you should get exactly the same spectrum. So from that, you can conclude that this red dot and this red dot have to be symmetrically placed.

There's only two possibilities. If the red dots are symmetrically placed, and these are separated by integers, then there's only two possibilities. The first possibility is that the red dot is at an integer, let's call this, this is the integer m, and

0, 1, 2, 3, 1, 2, 3, down here. Symmetrically placed about the origin and separated by integers, that would be one possible spectrum.

1, 2, 3, 0, 1, 2, 3, if it terminates. There's another possibility, and the other possibility that still is consistent with the symmetric spectrum

and separation by integers, and that's the half integer option, is 0 again. 1, 2, whatever, whatever, start here.

Shift by an integer, shift by an integer, shift by an integer, let's see, here's 0, 1, 2, 3, 1, 2, 3, and it's also symmetrically placed about the origin. In this spectrum over here, if it's physically reasonable,

the values of the angular momentum, the z component of the angular momentum are half integers. Half integer, three halves, five halves, minus a half, minus three halves, minus five halves, ending somewheres. Or the other possibility is integer spectrum.

Now, you mentioned before that I've already sort of proved that the spectrum is integer, and I did. Somebody mentioned, and that's correct. That is correct for orbital angular momentum. Orbital angular momentum means r cross p, it means wave functions of particles that depend on position.

And how do we prove that? Remember how we proved it? Oh, we proved it by saying the wave function has to be single valued when you go all the ways around. So, on the face of it, it would seem that this is the only physical possibility.

In fact, this possibility here is also real, and it corresponds to spin angular momentum. But for the time being, let's just get rid of it. We'll come back to it. All right, there's a maximum.

Given a state like this, we can generate a few more. It may or may not terminate, but let's assume it terminates. We found a multiplet of states, a multiplet of states which are in some way related to each other, and

which have to have the same energy. Why do they have to have the same energy? Let's prove they have the same energy. And then we'll quit. That'll be enough for tonight. Let's prove one more thing. They have to have the same energy.

Let's suppose that we found the state m, and it has an energy equal to e times m. It's an eigenvector of the energy.

It has L sub z equal to m, and it has energy e. Now, let's go to m plus 1. m plus 1 is L plus times m. That's m plus 1.

And let's ask what its energy is. So what do we do? We apply the Hamiltonian to it and see what we get. But the Hamiltonian commutes. It's assumed that the Hamiltonian commutes with all the components of the angular momentum.

So that means we can write that this is the same as L plus h times m. But h times m is just e times m. What does this equation say?

It says h on m plus 1, that's L plus on m, is equal to e times m plus 1. We can do the same thing with m minus 1. That tells us that if we find an eigenvector of the z

component of angular momentum, which happens also to be an eigenvector of energy, then by raising and lowering, until we run out of possibilities, until we hit the end points, we create states of exactly the same energy.

This is the idea of degeneracy following from symmetry when symmetries don't commute with each other. This is what we talked about in the beginning. When symmetries don't commute with each other, the L's in this case, it tells us that there are degeneracies.

And the degeneracies are states of equal energy. And in this case, the states of equal energy are just found by going up and down this ladder in this way. Are these just states going to the different axes? Is the axis equivalent to the y axis? Yeah.

They're very closely related to the idea of taking a state and rotating it. For example, you could ask the question, what are the eigenvectors of L sub x? The eigenvectors of L sub x are not individual ones of these, but linear combinations of them.

Linear combinations of these are eigenvectors of Lx and Ly. And so in that sense, what we're finding here is essentially the eigenvectors rotated, the rotated versions of the L.

And that's kind of obvious, because when you apply an L, you're generating a rotation. So this is the quantum analog of saying that if you take an orbit and rotate it about an axis, you'll find a new orbit with a different angular momentum, but with the

same energy, different z component of the angular momentum. It's exactly what's going on here. Rotating the configuration is creating states of different z component of the angular momentum, but with

exactly the same energy. That's what's happening. Yeah. By applying the other rotation operators. Now, it's convenient to use these particular combinations, L plus and L minus, just because Dirac figured out how to do it this way.

So this is the trick of using raising and lowering operators. And this occurs all the time in quantum mechanics. Whenever you find an algebra like this, which closes on the commutation, that's a signal that you should be looking

for creation or raising and lowering operators and using the algebra to sort out the spectrum of the system, the spectrum of eigenvalues. That'll give zero.

By definition, that's the definition of the top level. There may not be a top. You may just go up and up and up and up and up. How do you define you reach the top? What happens if you know you reach the top? L plus gives zero.

Now, this can all be made concrete by specific wave functions so that this happens. I didn't want to write down complicated angular wave functions. One possibility is you never run out of anything. That would correspond to basically infinite angular momentum states, states of infinite angular, infinite

capacity, but typically real angular momentum states of atoms and so forth in some words. We'll work out more about angular momentum, a little more, and then we'll apply it to atoms and understand a little bit about atomic spectra, a little bit about the spectrum

of the hydrogen atom. But this is where the degeneracies or some of the degeneracies, well, it's where all of the exact degeneracies of the hydrogen atom come from, all of the equality of and these states are the orbital, are the

various orbital quantum numbers of the hydrogen atom. There's more, there's more, but OK.

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