Stanford University. OK. The question arose about the paradox of, if two particles, both fermions, make up a boson,

can we have a boson, two bosons, which are, let's say, in the same state, in the same state? Perhaps a more general question. The wave function of bosons, the wave function of bosons,

is supposed to be symmetric on the interchange of the two bosons. That means, for example, if you had two bosons, and the wave function describing them was a function of, let's say, x and y. This doesn't mean the two coordinates x and y in the x and y direction. X and y are simply now standing for particle number one and

particle number two. If they're bosons, then the rule is that that must equal psi of y and x. That's another way of saying you can't distinguish between the two particles. There's no difference between saying you have a two boson system

in the state psi of x and y, and saying you have two bosons in the state psi of y and x, it's interchange symmetric between them. And in that case, you can certainly have two particles in the same state. For example, a special case of this would be that psi is a product function.

Psi of x and y is a product of, I don't wanna use the label psi again, because I've used psi to label two particle wave functions. But let's invent another thing to describe one particle wave functions. Let's just call it phi. And psi of x and y could be phi of x times phi of y.

This would correspond to a boson with two bosons with wave function phi, with the same wave function, in other words, in the same quantum state. That's what this would mean, a two particle state. And each particle is in the state phi.

This, of course, is obviously symmetric. It is the same as phi of y and x, sorry, phi of y times phi of x. Wave functions are not operators, they commute. Wave functions are just ordinary numbers, they're not matrices. They're just functions of x, and

they satisfy all of the algebraic properties of perfectly ordinary functions. And so, what did I write? Yeah, I wrote the right thing. So yes, two bosons can be in the same state, and the wave function be symmetric. What about two fermions? The rule for two fermions, if we have two fermions,

is that the wave function has to change sign. Now, that's a very odd thing, but nevertheless, that is the mathematical rule, the abstract mathematical rule for fermions. When you interchange them, the sign of the wave function changes.

Now, okay, so that's the difference between them. Then, can you have two fermions in the same state? No, because phi of x and y is not equal to minus phi of y. And you see what I mean, phi of x times phi of y is not equal to phi of y and phi of x with a minus sign.

All right, now, supposing I have a wave function for two particles, can I make, and it's not symmetric, and it's not anti-symmetric. Can I make something out of it which is symmetric or

something which is anti-symmetric? So let me show you how you do that. Supposing I had some wave function which was neither symmetric nor anti-symmetric, let's just call it phi of x and y. When you interchange x and y, it neither changes sign nor

does it stay the same, so it's neither symmetric nor anti-symmetric. All right, if you want to make something symmetric, which is an appropriate wave function for two bosons, you just add psi of y and x. Now this is obviously symmetric.

If you interchange x and y, this one becomes this one, this one becomes this one, and nothing happens to that. Now this is a symmetric wave function. This is a possible wave function for bosons for any psi of x and y. And if we're talking about fermions, we can do the same thing,

except that we put a minus sign there. Now what happens if we interchange x and y? This one goes to this one, this one goes to this one, but the whole thing will change sign. The whole thing will change sign if we interchange x and y. Okay, so this is a wave function for fermions.

Symmetrized, this is called symmetrizing and anti-symmetrizing a wave function for two particles. Okay, now let's suppose we have four particles all together. Four fermions, okay? Four fermions, what are the rules for four fermions? And for simplicity, let's suppose there are two different kinds of fermions.

Let's say there is electrons and protons, and we're talking about hydrogen atoms. Now, the rule, okay, so let's see, what is the rule? We have a four body wave function. Let's now let electrons be called x and protons be called y.

So the wave function is a function of two x's, let's call them x1 and x2, y1 and y2, it's a function of four variables. What are the rules?

The rules are, first of all, that the wave function has to change sign if you interchange x1 and x2, that's the fact that the electrons are fermions. It must also change sign when you interchange y1 and y2.

Incidentally, let's see, let's do this right. It must also change sign when you interchange y1 and y2.

Yeah, that's the rule, those are the only rules. Sign change when you interchange x1 and x2, that's because electrons are fermions. And a sign change when you interchange y1 and y2 because protons are fermions, okay? So let's see what kind of wave functions we can build. Let's start with a hydrogen atom,

a hydrogen atom located at the origin, right? A hydrogen atom located at the origin has a wave function which is a function of its electron and its proton.

So one now picks out one of the two hydrogen atoms. One now stands for one of the two hydrogen atoms. And this might be some wave function governing one of the hydrogen atoms over here, but supposing I want to shift its location.

How would I shift its location? Well, it's easy to shift its location. You just shift the origin of both x and y. I guess I don't really have to do it. Okay, I don't need to shift its origin.

And then we have another pair, and the other pair is describing a hydrogen atom at some other location. All right, since it's a hydrogen atom at another location, it has a different wave function. The wave function may be the same except that it's been translated, but

it is different, phi of x2 and y2. Now in general, this is neither symmetric with respect to the x's or anti-symmetric with respect to the x's or for that matter with respect to the y's.

So it's not a good wave function for fermions. Supposing I want to make it anti-symmetric on the interchange of the electrons, that's very easy. We just subtract off here, psi of, I just interchange x1 and x2, leaving y1 and y2 alone.

All right, so let's do that first. Minus psi of x2, y1, phi of x1, y2.

It is now anti-symmetric with respect to the x's. If I interchange the two x's, the whole thing changes sign. But it's not anti-symmetric with respect to the y's, okay? How do I make it anti-symmetric with respect to the y's? The answer is take the whole thing that I have here,

interchange the two y's and put a minus sign in. All right, so we put in minus psi of x1 and y2, phi of x2, y2. That's exactly this, sorry, y1.

This and this are the same, except that y2 and y1 have been interchanged. And I put a minus sign in. And then I want to anti-symmetrize this one with respect to the exchange of the y's, so that becomes plus, so I have to change the sign. Psi of x2, y2, phi of x1, y1.

Okay, so this is now anti-symmetric with respect to both the x's and the y's. If you interchange in this expression x1 and x2, it'll go to this one and

change sign. If you interchange x1 and x2 here, this one, and so forth. The whole thing is fully anti-symmetric, yeah. My problem is if you take those two hydrogen atoms and put them closer and closer, there's nothing preventing you from doing that because there are bosons, right?

There may be forces between them. There may be, the Pauli principle does create a kind of force between the atoms. It will make it very, very, it'll make it essentially impossible to put two atoms right on top of each other, but you'll simply attribute that to a force.

But will you be able to put two atoms into the same momentum state? Momentum states are spread out all over the place. They are spread out in space. So when you put two atoms into the same momentum state, you're not ramming them in on top of each other, and it doesn't cost a lot of energy.

So yeah, the question of whether you can stick two things on top of each other is a different question than whether they're fermions and bosons. It can even happen that billiard balls can be bosons. You can't stick them on top of each other. They have hard cores. Nevertheless, they can be bosons.

You can't put them at the same point. It just costs too much energy, but you can put them into the same momentum. If your wave functions are eigenvalue, eigenfunctions of momentum, then couldn't you have the two hydrogen atoms in this same eigenstate of momentum?

And then you would have two fermions in the same state, wouldn't you? No, no, no, no, no. Just take any wave function like this. It's anti-symmetric with exchanges to x. It's anti-symmetric with respect to exchanges of y. What about if I interchange one and two?

One and two is interchanging the two atoms, right? What happens then? It's symmetric. It's symmetric. Yeah, so it's symmetric with respect to interchange of the two atoms. And why?

Because when you interchange the x's, you get a minus sign. When you interchange the y's, you get a minus sign. Minus times minus is plus. So let's see, yeah. If I interchange one and two, x1, y1 will become x2, y2 with the same sign.

X2, y2 will become x1, y1, again with the same sign. So this one will, these two will transform into each other by simultaneous changes of everything with a one or everything with a two. And that's exchanging the two atoms.

Likewise, let's see, we have psi of x2, y1, but here we have psi of x1, y2. Both of them have minus signs, so these will go into each other. So the whole thing is symmetric with respect to the interchange of atom one and atom two.

That doesn't mean that it doesn't, this whole story here is independent of the question of energies. It may be impossible for other reasons to put two things on top of each other in space, the reasons being that it may be very strong repulsive forces.

But a good example, yeah. We just wrote down an appropriate wave function for two atoms, for two bosonic atoms. No, these may or may not be in momentum space. They may be anything. They may be anything.

Well, I mean, what is, are we saying that if we have two bosonic atoms, that that's what their state will be? Yes. Okay. Yes. If we have a bosonic, if we have an atom with a wave function psi of x1 and y1, and another atom, if we didn't know anything about this atom over here,

we might write down a wave function for the electron and proton in the other atom. And now we want to write down a wave function which is appropriate for the two atom system, all right? The wave function for the two atom system has to be anti-symmetric when you interchange any x's, anti-symmetric when you change any y.

And so I just basically use this rule, except I use it twice. Once to exchange the x's, all right, gives me the top, and then take the whole thing and interchange the y's and change the sign. The result is, quite clearly, if you look at it carefully,

interchange symmetric under the exchange of the two atoms, one and two, atom one and atom two. So, that's, and, and that, at the end of the day, at the end of the day, that is the rule that a bosonic atom and another bosonic atom, when you interchange them,

the wave function has to have the same sign. Okay, now what kind of states can you put two, yeah, momentum states you can put two atoms into in the same wave function. That doesn't cost a great deal of energy because they're hardly ever

on top of each other. But it is true that if you try to stick two atoms on top of each other, it's very hard to do because it's very, very costly in energy. Let's see, what would happen, in fact, what would happen, okay,

let's see what would happen if we made both atoms with the same wave function. Not one atom with wave function phi and one atom with wave function psi. But two with the same, what happens then?

Let's see, okay, so let's, okay, so there are two atoms with the same wave function. We have to subtract, all right, so this is this.

Psi of x1, y2, psi, psi. Is that zero or not? I don't think so. It's not zero, is it?

Is it zero? Yeah, x1, y1, where else do you see? Yeah, yeah, yeah, no, no, no, it's not zero. No, this one and this one are the same, okay? So it's just twice this, two is irrelevant.

We don't care about a factor of two, and twice this. This is it, and this is not zero, okay? This is two atoms in the same state. It's not zero. Okay, some magic is going on here. You can't put two electrons in the same state.

Meaning to say the electron wave function has to be anti-symmetric. The proton wave function has to be anti-symmetric. Both are true here, but yet the two atoms are in the same state. So ponder that a little bit. Write down that wave function and look at it and of course,

the point is, among other things, that in the atom, the electron is not in a well-defined position. It has a probability of being many places in the atom. You put two atoms in the same place, it's quite true. The two electrons can't be in the same place, that's true. Two electrons cannot be in the same place.

Let's see, what would happen? Here's our wave function for the two atoms. Perfectly good, anti-symmetric and so forth. Now, supposing we tried to stick the two electrons into the same place. That would mean setting x1 equal to x2, then it would be zero.

All right, so although the two atoms are in the same state, it's still true that in that state, if you try to find the two electrons in the same place, let's just see, what would that be? The two electrons in the same place would be psi of x1, y1, psi of x1, y2.

I've just set the two x's equal to each other. Minus psi of x1, y1, psi of x1, y2. So I forced the two electrons to be in the same place, and guess what?

I think they cancel. Yes, they do, okay? So as I said, ponder a little bit, and you'll get familiar with it. And you'll see how it can possibly be the two fermions can be a boson.

Does that finally mean that whereas you can exchange two atoms, two hydrogen atoms in different states, you can't- So this, before I set x1 equal to x2 here, these two atoms were in the same state. The atoms were in the same state, but the electrons and

the protons were never in the same state. Okay, the atoms aren't in the protons, they're not in the protons. Whereas if they're both in the protons, you can't have a second case. When you did this operation of anti-symmetrizing, you basically projected out the pieces of the wave function,

where the electrons were at the same point, then where the protons were at the same point. So nothing was left of that piece of the wave function. It's still zero if you go back to feet. If you go back to what? To, I think, feet, and the atoms are in different states, but

you put the electrons in the same state. Yeah, yeah. Just any time you try to put electrons in the same place, or in the same momentum, it's going to be zero, right? Okay, let's come back to spin a little bit. Spin is more fun than we've dealt with so far.

The reason I'm spending time with spin is because the mathematics of spin is what you really have to know, to know such things as the mathematics of isospin, and the mathematics of color, and the mathematics of all of the symmetries of particle physics. The mathematics is the same. The physics, as was pointed out here, the analogies between

these different kinds of conserved quantities are formal mathematical analogies, but nevertheless, let's study the mathematics of spin a little more, a little bit more about the Dirac equation, just a bit, and then I want to

introduce the concept of isotopic spin, or isospin, which is a concept which I think dates back to the 30s. Okay, so let's start with half-spin.

The half-spin, the electron is a half-spin particle, so I'll call a particle an electron. I could call it a proton, I could call it a muon, any one of the half-spin particles, but let's just call it an electron. It has two possible states. If two possible values for the angular momentum along the z-axis.

Along the z-axis, we've called L sub z, we've called that M. Not quite, not quite, not quite, not quite. There's an h-bar in there, right?

But from now on, we'll set the h-bar equal to you know what. Okay, so for the electron, this is not the full angular momentum, incidentally, of the electron. In fact, I'm using a really weird notation. Usually L is used for orbital angular momentum, and S is used for spin.

So let's revert to that. S for spin, S for spin, but the mathematics of it is exactly the same as the mathematics of the Ls. In fact, we could write it down over here. Let's write it down over here. The mathematics of angular momentum, but

now I'm calling it S for spin instead of L. I don't know what L stands for. It stands for angular momentum. All right, so what was the commutation relations we wrote down last time, where Lx with Ly equals ih-bar Lz, and now it becomes Sx. Where Sy is equal to ih-bar, but I'll set h-bar equal to 1.

iSz, and cycle down, you know what to do next. SySz equals iSx and so forth, all right?

Cyclic permutations, all right? One other thing, we have S plus and minus, just to remind you. S plus and minus is just Sx plus minus iSy. These operators raise the value of the z component or

lower it depending on whether there's a plus sign or minus sign. And so they're raising and lowering operators for spin angular momentum. All I've done is change notation from L to S. Intrinsic spin. Intrinsic spin, it doesn't, it could be any spin.

I mean, it could be a spin which is due to the rotating basketball, but let's say intrinsic spin. Intrinsic spin, yeah. Okay, let's focus on half spin. Half spin is the case where Sz or

M can take on the value minus a half and plus a half, and that's all. That's a two state system. Incidentally, I'm also forgetting completely everything else about the electron. We're forgetting, for example, about where the electron is,

what its momentum is, and all that sort of stuff. Just forget it for the moment, and you're just concentrating on the spin. All right, so concentrating on the spin, the quantum mechanics is the quantum mechanics of a system with two orthogonal states, and only two orthogonal states.

It doesn't matter where it came from. In quantum mechanics, the space of states is just the space of states of a two component or a two state system. And you can always write it as a column vector, let's call it alpha and beta.

Where alpha is the amplitude that the spin along the z-axis is up, and beta is the amplitude that it's down. So you can always write the state of the system, the ket vector. Let's call it psi, which is alpha up plus beta down along the z-axis.

You can just symbolically represent it by constructing a little column vector and putting an alpha there and a beta there. Alpha star alpha is the probability that the electron is up. Beta star beta is the probability that it's down, and that's it.

In the same notation, operators become matrices. Okay, so operators become matrices. Let's see what we can learn about the matrices. These components of the spin have to be two by two matrices. Why two by two? Because the space of states is two dimensional.

So they have to be two by two matrices, and they have to satisfy this algebra. That's it. If you can find two by two matrices which satisfy this algebra. Oh, one other thing. We might want Sz to be diagonal.

No, no, no, not the dimensions of space. The, no, no, no, no. The dimensions of spin space. Spin space. Spin space, yeah. Not the dimensions of space. No, what I said to do is to forget about space. Forget for the moment the position of the electron. Forget its momentum. We're thinking now of an electron that somebody has nailed down to the wall so

that it can't move around, and all you can do is measure its spin. It has two distinct states if you measure the z component of its spin. That's it. Only two states, and so let's label them up and down.

Any quantum state of the spin is a linear superposition of them with complex coefficients, and the probabilities are, the probability for up is alpha star alpha, the probability for down is beta star beta. It's just a change of notation, a common change of notation to replace

this clumsy thing here by just writing a little column with an alpha and a beta, and remembering that the lower component stands for down, the upper component stands for up. In the same way, operators become two by two matrices, and

the two by two matrices just act on these two component vectors. Yes? No, I said I will choose SZ to be diagonal, SZ. It doesn't matter.

You just search around and see if you can find matrices, two by two matrices, which satisfy this rule. Yes, it's an operator. It's the angular momentum, it's an observable. It is an observable, it's an operator, and it becomes a matrix.

All operators become matrices that act on vectors to give other vectors. Same thing we've done over in the past. So let's see if we can find, well, of course we can, otherwise we wouldn't be doing this.

Three matrices which satisfy these commutation relations, X, Y, Z.

So two by two matrices, two by two matrices. All right, they are not unique. They are not unique, and the choice, the choice of them really corresponds to a choice of orientation of axes.

But there is a choice of them which is particularly useful for us. They're all equivalent incidentally, as I say. They're related just by rotation of the spatial axis, real space axes. And it is not important what choice you use. You get the same answers. But I will choose the following choice.

And you can check that these matrices satisfy Sx is equal to 0, 1, 1, 0. Sy is equal to minus i, i, 0, 0. Sz is equal to 1, 0, 0, minus 1.

You will never find three matrices which have these rule which, no, sorry. One half of this, one half of that, one half of that, okay? Check it. Just go through it.

I will not do it on the blackboard. It takes two minutes to do, well maybe three, three minutes to do. And you'll find that these matrices do satisfy these commutation relations here. They are the Pauli spin matrices. The Pauli matrices are these things without the one half in front of them, okay?

But the real angular momentum matrices which satisfy these rules here are half of them. Without the half, they're called Pauli matrices. With the half, they're just called spin matrices. I think they're properly called spin matrices with the half.

So this is a representation, I would say this is a representation of the angular momentum matrices. A particular two component representation of the angular momentum, of the half spin angular momentum matrices.

Okay, let's now see if we can find the states. That means the states means the column vectors which correspond to sigma z being up and sigma z being down, all right?

Or sz being up and sz being down. Those are the two possible orthogonal states, eigenstates of sz. All right, this is very easy. We just say, what does it mean to say a state whose sz is plus or minus?

It means an eigenvector of sz with eigenvalue plus one or minus one. We expect, incidentally, that the components of spin have value plus and minus a half, all right? Plus and minus a half, but that means that the s's should have value plus one and minus one.

Of course, that's true. These diagonal elements here are the eigenvalues of s. So let's just be very pedantic. Being very pedantic, we're looking for vectors, let's call them alpha beta, which have the property, minus one, that this is either equal to plus or minus, let's say plus first, alpha beta.

This would be an eigenvalue in which sigma z equals plus one. What about sigma z equals minus one? That would be one, zero, zero, minus one, alpha beta equals minus alpha beta.

That's the meaning of an eigenvalue and an eigenvector. If we find solutions of these, the solutions will correspond to the states in which the spin is up and the spin is down. Let's just check.

Let's check that out. Okay, so it's completely obvious. Well, let's just work it out. The top entry here by matrix multiplication is one times alpha plus zero times beta. So just multiplying out the matrices, this is going to be one times alpha plus

zero times beta in the top and zero times alpha and minus beta in the bottom. Well, obviously this is not equal to this unless what? Unless beta is zero. If beta is zero, then we've found a solution to the eigenvalue or the eigenvector problem, which corresponds to an eigenvalue plus one.

And surprise, surprise, it corresponds to a vector which only has an entry in the upper place. The upper place stood for spin up.

What shall I choose alpha to be? I can choose it to be one. Why one? Because I want the sums of the probabilities to be one. So alpha star alpha plus beta star beta should be one. And a convenient choice is just to choose alpha equal to one.

Now what if I, let's put back what I erased. What I erased is that this thing times alpha beta was alpha minus beta. Supposing I want to get a minus sign, then what do I do? Choose alpha equal to zero.

If alpha is equal to zero, this is true. And that corresponds to the vector zero one. Again, no big surprise. It corresponds to the state in which the electron is definitely down and has no probability of being up.

Okay, let's go a little bit farther. What about, we've done this many times in these series of classes, but just spend a minute or two doing it again. What about wave functions, state vectors, whatever we want to call them,

ket vectors corresponding to the possibility that the x component of spin. We're now going to do a different experiment. We're going to measure instead of the z component of the spin, we're going to measure the x component of the spin. Again, the two possibilities are plus a half and minus a half.

What are the eigenvectors which correspond to the two possible measurements for that case? All right, we're measuring sigma x and not the, or sx and not the, all right. In that case, we want to ask zero, one, one, zero,

alpha beta equals alpha beta. That's the eigenvector with eigenvalue plus one, and the eigenvector with eigenvalue minus one will have to satisfy one, one, if such things exist, alpha beta equal minus alpha beta.

These are the eigenvalue equations for plus and minus. Here, I've put in sigma x instead of sigma z, all right, sigma x. Okay, so what is the top equation? Let's do the top one first.

What does the top equation say? Alpha equals beta, right. Let's be pedantic again. Zero times alpha, one times beta. So, working out the matrix product here or the matrix times vector, it gives us beta alpha.

This matrix just interchanges top and bottom. One times alpha, sorry, zero times alpha, one times beta. Same thing on the bottom row. And so, is there a solution of this equation? Well, yes, all it says is beta equals alpha.

In fact, the bottom equation says alpha equals beta, which is the same as beta equals alpha. So, we solve the equation by setting alpha equal to beta. A convenient choice, a convenient choice, they're all equivalent to each other, but a convenient choice is that we choose one over square root of two,

one over square root of two, and why one over square root of two? So again, so that the total probability adds up to one. Yes, yes, yes, yes, yes.

If we had included the one-half, we would be looking for eigenvalue one-half. So, to look for eigenvalues of Sx, we just look for twice the eigenvalue for sigma. Good point. Okay, all right, let's see what the other possibility is. The other possibility is 0110 alpha beta, which happens to be beta alpha.

That's by multiplication here. That should now equal minus alpha beta. That's the other eigenvalue. Take this away for a minute, the thing in the middle here. This just becomes the eigenvalue equation, or the eigenvector equation,

for an eigenvector with eigenvalue minus one, all right? And what does that require? That says beta equals minus alpha, and it also says alpha equals minus beta. If beta equals minus alpha, then alpha equals minus beta. And so all we have to do is put into the entries here.

Sorry, not this, but alpha beta here is equal to one over square root of two minus one over square root of two. All right, so what does this say? It says that if we take a linear superposition of states

where the electron is up along the z-axis and down along the z-axis, we superpose those states in the quantum linear superposition. We make an electron which is, let's use the right word, which is polarized along the positive x-axis.

If we make the same kind of linear combination with a minus sign, Then we get one which is polarized along the minus x-axis. All right, so we've now accounted for electrons which are up along the z-axis, down along the z-axis. Up along the x-axis, down along the x-axis.

And finally, there's the issue of the y, electrons polarized along the y-axis. I'm not going to work it out. It's just minus i, sorry, minus i, i alpha beta.

You have to solve this one, and this one equals minus. And the two solutions are very easy to do. The two solutions are alpha beta equals one i.

That's for the plus sign. I think that's for the plus sign. Let's see. I think that's for the plus sign, and the other one is one minus i, one minus i.

Yeah, sorry, excuse me, one over square root of two again. Notice that there's more information in these complex coefficients than just the fact that the probability of the z components of spin is a half a half. Both, this has probability a half a half.

So did the previous case, where the electron was along the x-axis. The fact that there's an i here is something new. It tells us that the electron is along the y-axis, all right, so this way. So now we've accounted for electrons which are polarized along all six directions.

What about a general direction? An electron whose spin is pointing along another axis. Well, for that, I'm not going to go through it now. You take linear combinations of these. We won't do it now. We'll do it another time. It's not something I intended to do this evening.

All right, so that's quantum mechanics of a- Are you guessing that the minus sign goes on the other eigenvalue? I may have a sign wrong. Do I? Well, I'm thinking that minus i over square root of two goes with the first one because of the minus i and the- You may be right.

I think you're probably right. So let's see, so this one should be minus then and this one should be plus. I think that may be right. Do I have it in my notes? Well, I don't have it in my notes. All right, that's spin a half. That's the whole theory of spin a half.

All wave functions are all state vectors corresponding to all possible directions of spin are linear combinations of the two states in which the spin is up along the z-axis or down along the z-axis. No, I take it back.

Do you take it back? Yeah, b equals ia. Okay, and it's plus, huh? Minus. I never remember. What about spin one? We've done this before. I mean, we've done this a lot of times before. And so I've made fast work of it. But now let's do spin one.

Spin one is also interesting. Again, there are many, many choices of the matrices. And they're all equivalent, physically equivalent, representing the angular momentum. But I'm gonna show you one. All right, spin one, the z component of angular momentum has three states.

Spin one is the integer angular momentum. It corresponds to bosons. And it's the case where the angular momentum can be zero. The z component can be zero, one, or minus one.

So there's three possibilities. Evidently, the space of states is three dimensional now. Not two dimensional, three dimensional. Again, we want to find spin matrices, which satisfy three by three matrices, which satisfy exactly the same algebra.

The algebra means the set of commutation relations. Sx with Sy equals Isz and so forth. Do there exist, I've left out the h-bar because I've set h-bar equal to one.

Do there exist three by three matrices which satisfy this? They're better because we sort of worked out the existence of a three state system by assuming these commutation relations. Yes, I will just show you a representation.

And you can work it out and see if you can prove the commutation relations. They're three by three matrices. Let's see, I think it's S, sorry, Sz, not z, but Sz, or S.

Spin along the z direction is equal to i times zero, one, zero, minus one, zero, zero, zero, zero, zero.

It sort of looks familiar, but not familiar. It's three by three, but it kind of looks like Sy over there, but it doesn't have the factor of two. It's different. S, let's see what's next. Sy is equal to i, zero,

zero, one, zero, zero, zero, minus one, zero, zero. Notice that this one has zeroes in the third row and the third column. This one has zeroes in the second row and the second column.

And Sx has zeroes in the first row and the first column. And it's zero, zero, zero, let's see, Sx. Zero, zero, one, zero, minus one, zero. There's a real symmetry between them.

Each one has a one and a minus one. They're all anti-symmetric. They're all anti-symmetric. Each one has a one and a minus one. And it has zeroes all along the row and diagonal that match the letter here. If z stands for three, then the third row and third column are zero.

If y stands for two, the second row and second column are zero and so forth. So they're easy to remember. The only trouble I ever have is remembering the signs. And again, you check those by checking the commutation relations. Okay, and they're not completely unique. But let's see if we can find the eigenstates of Sz.

This time Sz is not diagonal. I have not chosen Sz to be diagonal. But nevertheless, we can certainly work out what the eigenstates are.

They're not too hard to guess. So we want to solve Sz on, well, we want to take this matrix. And now multiply it not by alpha and beta, but by alpha, beta, and gamma.

How many, what should the eigenvalues be? One, zero, and minus one. Those are the three states of z component. Let's try for one first.

I hope I haven't written down. Yeah, good news, I do. No, yeah, yeah, it's just, incidentally, this corner of it over here is very similar to the Sy over here.

So it's gonna be similar. I think it's one minus i and zero.

So here's an eigenvector. I think this is a true relationship. You can check it later. Here is an eigenvector with eigenvalue plus one. It corresponds to the state where S sub z, where m is equal to plus one.

I didn't give it to you, Sz, in the simplest possible form. I gave it to you in a very symmetric form, where x and y and z look very similar to each other. But nevertheless, the right way to find the state which corresponds to m equals

plus one is just to find the eigenvector with eigenvalue plus one. What about the eigenvalue minus one? That turns out to be this one, minus.

And what about the eigenvalue zero? Oh, incidentally, you should put a one over square root of two if you want them to add up to, right. Okay, what about the eigenvalue zero? Anybody see what the eigenvector with eigenvalue zero is? Zero, zero, one.

This one is obvious. Take the top row, multiply it this, you get zero because there's nothing to match this one. Likewise here, this is just plain zero, big fat zero.

Okay, what do these things mean? These mean that the three states, m equal to plus one, m equal to minus one, and m equal to zero, correspond to the vectors one, minus i, zero. One, i, zero, zero, zero, one.

These you might want to put a square root of two if you like to make the probability add up to one. All right, so again, we really do find, and you'll find there are no other eigenvectors,

no other eigenvalues, that's it. And these are the mathematical representations of the very abstract things we did last time. One of the marvelous things is that these square roots of two really do turn up in experimental predictions, and we're going to do a couple of them, but not with ordinary spin, but with isotopic spin.

I'm going to show you how those square roots of two turn up in various experimental quantities, but not now. Can you read the second column that those are both one over the square root of two? No, i over the square root of two. That's i over the square root of two, and the first one is one and i.

And minus i over the square root of two. One over the square root of two, minus i over the square root of two and zero. One over root two, i over root two and zero. This is, I think, m equals one, this is m equals minus one, and this is m equals zero.

Now, as I said, there are other representations. There are other triplets of matrices which satisfy the same algebra, but they are, in fact, physically completely equivalent. And I'm not going to go into the reasons for that now.

These are a good set of matrices to work with. They're fairly elegant, and we'll use them, we'll use them more than once. Okay, another question, yeah.

Supposing we now return to the real electron which can move around in space. It can move around in space, so it has a position, or a momentum, and a spin. Incidentally, spin and position are things which can simultaneously be measured.

Spin and momentum are things which can simultaneously be measured. What are the things which can't simultaneously be measured? Well, position and momentum, but also the different components of the spin. They don't commute with each other, but spin and momentum do commute with each other.

All right, how do we represent the wave function or the state vector of a electron given the fact that it also has a position? It has a position and a spin. And the answer is we turn these two component, they're called spinners.

They're called spinners. This is what a spinner is. The right-hand side, a two-component object like this, which just represents up-spin, down-spin, is called a spinner, all right? We just turn the spinner components into functions of position, all right? So let's imagine that now.

The electron is free to move around. The full state of the electron is described not just by a simple function of x, but by two functions of x. Let's call it psi up and psi down of x.

And if we like, we can just make a symbolic notation where we put them in a column. What's the meaning of this? The meaning of this is the following, that you don't just ask what's the probability that the electron is up or the electron is down. You ask a more refined question.

The more refined question is if the electron is found at position x, what is the probability that it's up? That may depend on x. So the probability that the electron is up might depend on where you look, all right?

Another way of saying it is classically, just corresponds to the classical statement, that the spin of the electron could vary from place to place. Quantum mechanically, what varies is the probability to find it up, the probability to find it down. And so at every point in space, each point in space,

you have a spinner, and the spinner tells you at that point in space, what's the probability the electron is up? Well, it's psi star psi up, psi up of x. That's the probability that if you look at point x and discover an electron there,

and then measure the spin, the z component of the spin, this is the probability it will be up. And psi star down of x, psi down of x is the probability that it's down. But at each point x, you have all of this apparatus here in exactly this form.

And that's the whole story of what an electron is. An electron is a particle described by a wave function which has two components. The two components being the amplitude that the electron is up or down.

What about a spin-one particle? Exactly the same. A spin-one, well, not exactly the same, almost exactly the same. But the spin-one particle, and I'm not going to call it psi, let's just give it another name, phi.

Phi is a more common name for bosons, and spin-one particles are always bosons. So, it becomes phi one of x, phi two of x, phi three of x, where at each point of space,

at each point of space, this phi one, phi two, and phi three replace alpha, beta, and gamma from before. Psi three of x, yeah, okay, you understand. So, it's just reproducing.

exactly the same structure over and over in space. And that's what spinners are. Yeah? I'm not sure. Is psi 1 the probability that it's spin 1? No, no, no. OK, so let's go back. Let's go back.

Good. That's a good question. Let's ask, what is the probability? Here's our wave function. What is the probability that the electron has spin plus along the z-axis and spin minus along the z-axis, and spin 0? OK, let's work it out. Here's what we have to do.

Remember those eigenvectors, the eigenvectors of spin? For the spin 1, not for the spin 0. Where are they?

OK, for m equals 1, that's spin up along the z-axis. The eigenvector was 1 over square root of 2, minus i over square root of 2, and 0. This is the eigenvector for m equals 1. Now, I'll remind you what the rule is

for calculating the probability for this condition given this wave function. The rule is take the inner product of this with a complex conjugate of this.

We take the inner product of this with the bra vector corresponding to this ket vector, which is its complex conjugate, and then square it, times its complex conjugate. So what we do, here's the rule, standard quantum mechanical rule, we take the inner product, we'll call

this m equals 1, we take that inner product, and then square it, or take its absolute value squared, it times its complex conjugate. How about this bra vector that I've called m equals 1?

Bra vectors are always complex conjugates of ket vectors. So that means the bra vector can be thought of as a row vector, that's just a convention, 1 over the square root of 2 plus i over the square root of 2 and 0. Plus i because it's the complex conjugate in the

second place here, i over the square root of 2. We take the inner product of that with phi 1, phi 2, phi 3, what is that?

i over the square root of 2, phi 1, no, 1 over the square root of 2, phi 1, right? Phi 1 over the square root of 2 plus i phi 2 over the square root of 2, and that's it.

1 over the square root of 2, phi 1, phi 2 over the square root of 2 times i. And then we multiply this by its complex conjugate. So what do we get? What's the complex conjugate, what does this times its complex conjugate give?

I think it's just phi 1 squared plus phi 2 squared over 2, right? This times its complex conjugate will give phi 1 squared plus phi 2 squared over 2. So the probability that the electron is up along the z

axis appears to be phi 1 squared plus phi 2 squared over 2. What about the probability that it's down along the z axis? I think it's the same. I don't think so. Is it?

No, I don't think so. It can't be negative. All right, I think it's the same. For this state, it's the same. What about the probability that the z component of spin is 0?

That's phi 3 squared. In that case, the m equals 0 state is 1, 1, 0. And the inner product is just phi 3. Sorry, 0, 0, 1 minus state, 0, 0, 1.

And the inner product is just phi 3. So the probability that the electron is in the 0 angular momentum state is phi 3 squared. OK? So here are the probabilities. Looks like it's phi 1 squared over 2 plus phi 1 squared

plus phi 2 squared over 2. Same thing, phi 1 squared plus phi 2 squared over 2 and phi 3 squared, comma, comma. Those are the three probabilities. They add up to 1, phi 1 squared plus phi 2 squared

plus phi 3 squared. Sorry, this should really be phi 1 squared means phi 1 times phi 1 star. No, yes, it does. It does.

It does. I think I'll leave this here for you to work out in detail. I hate doing things on the blackboard when I don't have them in my notes. I tend to make mistakes, and I probably have made a mistake. But you get the principle. The principle is take the inner product of these things with,

oh, I know what I did. Yeah, the reason I got these two to be equal is because I assumed that phi 1, phi 2 were real. They don't have to be real. If they're not real, you'll get different answers for these two. Check it out. Try it out and see what you get. OK. But the principle is always the same.

Take the eigenvector, take its inner product with the actual wave function, and take its square times its complex conjugate. That's a probability. And notice, most important in this case is that it's a function of position. These phi's are functions of position, can vary from place to place.

So that's the basic idea of spin. It's an exercise to work it out for spin two, or for spin three halves. And it's significantly more complicated, but not in principle. Not in principle. And it's not hard to find the matrices, but a little harder.

All right, I very, very quickly want to come back to the Dirac equation. Let's see, how we, yeah, three halves should be four. Yeah, let's see, three halves, three halves is the case.

Plus a half, minus a half, three halves, minus three halves. So there are four states, spin two is five states. You have a question? Yeah. None of this discussion seems to be symmetrical with respect to time.

It's x, y, and z, but no t. Oh, that's true, yeah. So is all this non-relativistic? Spin is something like mass, best defined in the rest frame. Most easily defined in the rest frame.

So in that sense, what you do is you study it in the rest frame. And then you find the right way to Lorentz transform it to some other frame. But we won't do that now. It really doesn't change very much when you go to relativity.

And that's because the definition of spin is pretty much what you would get if you did all this in the rest frame of the particle. But yeah, we can simplify the story. So far, this is pretty non-relativistic. It won't change much in relativity, with one exception, which we won't do now.

Okay, let me come back to the Dirac equation and see what the Dirac equation has to do with spin. Again, spin is best defined in the rest frame.

But let's, okay, here's the Dirac equation. It had matrices, four by four matrices, and it looked like this.

I d psi by dt. Now, the psi's were four component objects. Psi one, psi two, psi three, psi four. Where did that come from? All right, I'll tell you in a moment where it came from, but let's write the equation. The equation is that this is equal to minus i alpha sub m,

m runs from one to three, the three directions of space, x, y, and z, times the derivative of psi with respect to xm. Sum over m, so this is alpha one, d psi by d x, plus alpha two,

d psi by d y, plus alpha three, d psi by d z, plus beta times the mass of the particle times psi. That was the Dirac equation that we wrote down a couple of lectures ago. And let me remind you what the conditions on the matrices alpha and

beta were in order, what was the condition? The condition was that the frequency, let me write this in a different way. This is frequency is equal to momentum alpha dot k, we can write it this way, or just alpha m k m.

This is the same as alpha one k one, or alpha x k x, plus alpha two k two, plus alpha three k three, which is the same as alpha x k x, alpha y k y, alpha z k z, plus beta m.

This is shorthand. This is shorthand. Do I have a, am I missing an i? Are those m's different? Yeah, this m's are an index, this is the pair next to the particle. Okay, let's not use m here, i, i, good, thank you.

Yeah, sorry, and this means sum over i, usual summation convention. Okay, when an index is repeated twice, it always means you sum over it.

And we required that omega squared is equal to k squared plus m squared. Remember what we did. We took omega squared by squaring this out, and required that we got k squared plus m squared.

What were the conditions for that? The conditions were conditions on the matrices alpha and beta. I'll just remind you what they were. They required every alpha squared to be two, I believe. Is it two?

I think it's two. No, no, one, one. Every alpha squared i for each i, alpha x squared equals alpha y squared equals alpha z squared.

No, no, beta is one matrix. Oh, beta is a matrix. Beta is a matrix. It requires beta squared to be equal to one. These things are just there so that you get the k squared plus m squared.

But then, in order that you don't get cross terms, in order that you don't get things like k times m when you square it out, or k1 times k2 and so forth, we require things like alpha x, alpha y, plus alpha y, alpha x, equals zero.

That would be, for example, the cross term between kx and ky. All right, so same thing for alpha x, alpha z. We can write this in a neato notation. Anti-commutator of alpha x, alpha y equals zero.

Anti-commutator of alpha x, alpha z equals zero. And anti-commutator of alpha z with alpha y equals zero. What about anti-commutator of alpha x with alpha x? Oh, sorry, and one more. Any alpha with beta should be also equal to zero.

What's the anti-commutator of beta with itself? Two, because it's beta times beta plus beta times beta. Okay, right. So we can summarize these commutation relations just by saying

that everybody's anti-commutator with himself is two. Everybody's anti-commutator with somebody else is zero. Simple anti-commutator. All right, it's a theorem. It's not hard to prove, but I'm not going to prove it. That the smallest dimension, the smallest number of components,

the smallest matrices which can solve this are four by four. If you didn't have the betas here, then you can find two by two matrices, which satisfy it. In fact, the sigma matrices, the Pauli matrices, satisfy it.

The Pauli matrices anti-commute with each other. That's something you can check. And their square is equal to one. So if you didn't have the extra beta matrix here, the Pauli matrices would solve these equations. But in two by two matrices, there are only three matrices

which anti-commute among themselves that way. And if you look for a fourth one, there just won't be one. You have to go to larger dimensional matrices. You try three by three matrices, there aren't. You try four by four matrices, and you find, yes, that there are

representations of this anti-commutation algebra here. I think I wrote down a representation. I'm going to write down another one tonight, which is a little different, but it's basically, again, it's equivalent.

So I'll give you an example of four matrices which satisfy exactly this algebra. Now, four by four matrices, these are four by four matrices,

but you can write them in block form. Writing them in block form means that you divide this up into two by two blocks. You have two by two blocks, each one of which is a two by two matrix.

That's just a simple way to write them down. And then for the alphas, for each alpha, alpha one, alpha two, alpha three, put in this two by two box over here the corresponding Pauli matrix, sigma i.

It's a two by two matrix, so it fills up four entries here. Down here, put the same Pauli matrix except for the minus sign, and zeros here. That's pretty easy.

I know, I told you, I used a different representation last time. They are equivalent. I think I put last time, I put the sigmas off diagonal, did I? Yeah, they are equivalent.

They're mathematically equivalent and there are similarity relations, similarity transforms which relate them, but I like this one better. I decided I like this one better. And what about beta? Beta is just in the same notation, same two by two block notation.

It's just one, now what's one? One is the two by two matrix, one, zero, zero, one. It really is one, the unit matrix and the unit matrix down here.

And that's it, those matrices have the desired property. Incidentally, when you multiply four by four matrices in this form, you can do it by tricks where really you multiply in the blocks two by two matrices. So it's an easy way to do this and it's no harder than working with two by two matrices,

but I'll leave that to your devices to check. These are the Dirac matrices, the alphas and the betas, or one particular realization of the alphas and the betas. I think last time you had beta, the bottom would be minus i.

Minus i here? No, that was last time. What's that? This is different, this is a different representation. The last time I had these off diagonal and I don't remember what I had for this one. I can't remember. I thought I had beta diagonal last time.

I thought I had beta diagonal. I can't remember. Yeah, yeah, yeah. Okay, this is a little bit different. And you get your choice. The physics of it will be exactly the same. They just correspond to choosing linear combinations of the four entries in the,

and there's no real difference. It's as physically interesting as rotating coordinates. What you're doing is rotating coordinates in this four dimensional space, which is not four dimensional space time. These four entries here do not have to do with the four. They're not direct, the same as the four dimensions of space time.

They're just the smallest representation of this algebraic structure. Okay, that's what the Dirac equation is. Here it is, Dirac equation.

It can be summarized in a form like this where this really means just this. And here are the matrices. What I want to do, the only thing I want to point out about it tonight is that there are two kinds of two by two-ness to this.

One is these blocks and the other is within each block. This by, what shall I call it, this doubling of the two by two-ness corresponds to a doubling of the two by two-ness in physics. One of the two by two-nesses is positive and negative energy.

The other is spin. So there are particles of positive energy with all possible spins. There are positive particles with negative energy with all possible spins. All right, we can see this and that's why there are four by four matrices if you like.

Roughly speaking, there's an entry for positive energy with positive spin, positive energy with negative spin, negative energy with positive spin, negative energy with negative spin. Those are the four possibilities. That's why when I say positive spin and negative spin, I mean spin along the z-axis,

half spin along the z-axis. That's why there are four components to the Dirac equation. Positive and negative energy exactly as in the one-dimensional example that we described. And spin, up spin and down spin if you like. That's why there are four entries in the Dirac spinner, in the Dirac equation.

And perhaps the easiest way to see what's going on is to look at a Dirac particle with zero momentum. Just to take a particle with zero momentum, that means k is equal to zero.

In that case, we can just throw away the alphas altogether and we get a very simple equation. When the momentum is zero, that means that psi is constant in space is what it means. It has no derivatives with respect to space, k is absolutely zero.

In that case, the whole Dirac equation just becomes i d psi by dt

is equal to i m beta d psi. That's it. That's the Dirac equation for a particle at rest. Particle with momentum equals zero means a particle at rest.

If it's at rest, the wave function has no space dependence. It's just completely constant with respect to space. And this is the whole thing. And it tells you what the possible frequencies are. We can write it a different way. We can write it maybe.

Yeah. No i here. Right. No i there. Let's see what it says. So here's the way we'll think about it.

We'll take these psi. There are four entries. One, two, three, four. And we'll divide it, the same kind of division, into blocks. And each half of it, the upper half and the lower half, I'll put a thing with two entries.

We'll call this psi plus and psi minus. Does plus and minus mean positive and negative? No, they don't. They don't mean anything right now. Just plus and minus. Nothing special. But each psi plus, what is psi plus? This is equal. Psi plus is psi upside down.

And the bottom one here, I think, is something else. Two others. This corresponds to an electron. Okay. We're just writing the column vector in a notation,

which is similar to the way I divided up matrices. Divided up matrices by just drawing some red lines and doing two by two matrices in here. We'll divide up the spinners, the column vectors, by just dividing them in half.

And in each place, put a two-component object. So psi plus and psi minus are themselves two-component objects. Two two-component objects make a four-component object. It's just a pretty way to organize the Dirac indices,

the Dirac structures here. Okay. Then we can write this equation very nicely. This equation is pretty simple. What does it say? Beta is the matrix zero, one, one, zero.

What does this matrix do when it hits a two-component object? Psi plus, psi minus. What it does is it swaps the upper and lower one. When this matrix hits a vector like this, it just interchanges psi plus and psi minus.

We can also write this as a two-component object, psi plus, really a four-component object, but two two-component objects. And now we can see what the equation says. What the equation says is that i d by dt of psi plus

equals m psi minus and i d by dt of psi minus equals m psi plus.

That's the whole upshot of the Dirac equation for particles at rest. It's even simpler than that. You can, the next step,

anybody got a good idea what to do with this to simplify it? That's one thing you could do. Add them and subtract them. Add them and subtract them.

What happens if you add them? i d by dt of psi plus plus psi minus equals m psi plus plus psi minus.

That's adding them. And subtracting them, i d by dt of psi plus minus psi minus equals minus m psi plus minus psi minus.

The upshot here is that this really breaks up into uncoupled equations for two things. One thing is psi plus plus psi minus does not get mixed up with psi plus minus psi minus. And psi plus minus psi minus doesn't get mixed up with psi plus plus psi minus.

They satisfy equations which are completely independent of each other. How about the frequencies of this one? What can you say about the frequency of this one? i d by dt is just omega, right?

If the wave function has a definite frequency, then i d by dt just gives you its frequency. So this object here has a frequency. This one has a frequency omega which is just equal to m. This one has a frequency omega equal to minus m because of the minus sign here.

What does it say about the energies of these two solutions? One of them is a positive energy solution and one of them is a negative energy solution. So just exactly the same as the one-dimensional example we gave,

there are positive energy electrons and negative energy electrons. I don't expect you to remember all of this, and to be able to deal with all of this from one sitting with it. It's worth going back now, finding a little book on the Dirac equation and going through it. But basically, the two-by-two blockiness here has to do with positive and negative energy.

Psi plus plus sign minus are positive energy things. Psi plus minus sign minus are negative energy things. But that still leaves the two-by-two-ness of the individual entries here.

Each one of these is a two-by-two entry. So let's forget psi plus and psi minus. We've taken care of that by adding them and subtracting them. But now let's look at the entry of just one of them. It also is a two-by-two thing.

What is that two-by-two thing? It's the spin. It's the spin which is described by these Pauli matrices here. If we completely forgot about the two-by-two-ness associated with the upper and lower pieces here and just worried about the two-by-two-ness within here,

then that's the same as the two-by-two-ness within the two-by-two matrices here. That's the spin. So the Dirac equation predicts that positive energy electrons come in two different kinds.

The two different kinds have to do with the upper and lower components or the up and down components within a psi plus here. And that's the spin. It also predicts that there's a positive and negative energy-ness to it. That comes from here. That has to do with the existence of positrons.

Holes in the negative energy sea are positrons. That's what the four-by-four is about. You can either think about it as positive and negative energy spin up and spin down, or you can think about it as electrons and positrons, each of which are half-spin particles.

So, yeah. Can you portray the opposite way around where the big ones are spinning? Yep, yep. You're asking whether you can... Yeah, absolutely, you can. Yep, you can put the positive and negative energy-ness into the smaller matrices

and the spin into the bigger matrices. Yes, you can. Either way, they're kind of equivalent to each other, and they just correspond to reshuffling of the indices. Could you do that one more time on the positron explanation?

Yeah. You can see from here that there are both positive frequency and negative frequency solutions. The things which have positive frequency are the sums of these two, and the things which have negative frequency are the difference between the two. Negative frequency means negative energy. Positive frequency means positive energy.

So, if there are positive energy solutions, that means positive energy electrons. If there are negative energy solutions, that means negative energy electrons. What do you do with a negative energy electron? You fill it up. You fill up the Dirac sea with all states of negative energy.

You always lower the energy by putting in a particle of negative energy. If you want to lower the energy down to the absolute bottom, you fill every negative energy state. Among others, you fill the negative energy states which have no momentum. I just worked it out for zero momentum. But you fill all the negative energy states, and that's the vacuum.

You take a negative energy electron and stick it into a positive energy state. You've made an electron and a hole, or an electron and a positron. So, in exactly the same way as for the one-dimensional example,

the positive and negative energy electrons become electrons and positrons. The other doubling becomes the spin. So, they're half-spin particles which come in particle-antiparticle pairs, and that was the prediction of the Dirac equation. You know, one of these marvelous examples of just fiddling with abstract mathematics

diddle, diddle, diddle, diddle, and boy, and then out comes this very, very dramatic, incredible prediction, positive and negative energy particles with spin. And remember that spin had only been discovered as in a sort of empirical fact

from that chart up there a couple of years before. I mean, just a few years before, Dirac did this, very shortly, short time before. And so, yeah, it was a very dramatic event. Are you saying that negative energy electrons are a positron? No, I'm saying the absence of a negative energy electron is a positron.

We went through this before. You fill up all the negative energy particles and you call that the vacuum. You can't put more than one negative energy particle in a state. So, the vacuum takes all the negative energy, E less than zero, E greater than zero,

and fills them up. You put an electron into every one of those states. That has the lowest energy because all the negative energies are filled. That's the vacuum. That's empty space. Now you remove an electron from here and put it over here,

not necessarily at the same place, up here. What is that? Well, now there's a positive energy electron and the absence of a negative energy electron. The absence of a negative energy electron is a positive energy positron.

Okay? It's exactly the same as in the one-dimensional example that we did, where the absence of a negative energy electron, a hole in the Dirac sea, becomes a particle of opposite charge and, yeah.

Well, why don't the higher energy negative electrons above that fall down? Into here. Into there. They won't fall down to here. Right, but why don't they fall and fill that little gap? They do. That's called positron-electron annihilation and it creates photons. Well, the first thing it has to do is conserve energy. Now, how much energy does an electron and a positron have?

Both the positron and the electron have positive energy. Why does the positron have positive energy? Because it's the absence of a negative energy. Okay? It's the absence of a negative energy, so it has positive energy. All it is is a positive charge and a negative charge. When the positive charge and the negative charge come together,

you can either describe it as the annihilation of the positive charge with a negative charge, but energy has to be conserved, so photons have to go off. The other way to think about it is as an atom. You know, an excited electron dropping down to a lower state.

What happens then? Photons go off. Yeah? Just a question on the mathematics. So here you've very clearly shown how in a zero-momentum state with this particular representation of beta, you get positive and negative concentrations of the psi. To get the spin representations in a zero-momentum state,

would you take the different alpha, beta? I have a hard time seeing how you get that cleanly, because there's three alpha matrices. There's three alpha matrices. There's more. Oh, and those three alpha matrices become the three spin matrices.

But in a zero-momentum state, you don't see them. So if you look at spin in a zero-momentum state, how would the mathematics show that simply? Well, the zero-momentum positive energy state, let's say. Zero-energy positive, that would correspond to a spinner of the form,

let's call it psi plus, we're the same psi plus here. We want to set psi minus equal to zero, because we don't want any negative energy part of the solution. So that means psi plus psi plus. All right, well, alpha, for example, alpha is not quite, alpha is not really.

Are you asking me how you make the the spin operators from the Dirac operators? No, not really, I'm just asking how if there's a way, simply as you've shown the negative energy state, energy state, you show how you discover spin from this equation. Well, you've discovered that the positive energy particles come in a pair,

one which you can call up and one which you can call down. Now you want to see that that pair really has angular momentum. Yeah, that really has the same. Now you can see that these Pauli matrices do come into it.

But to see that it really corresponds to angular momentum, you have two ways of going about it. One is through the connection between angular momentum and rotation symmetry. And the other is through the fact that angular momentum is conserved, okay? So what you would want to show, let's use the conservation definition.

What you would want to show is that in any kind of collision, where an electron comes in and scatters off something else, that if you assign the electron an angular momentum

whose description is in terms of this two-foldness within here, that that angular momentum together with all the other angular momentum and the problem would be conserved. All right, this is a little too much for us to do tonight anymore. What we can do more easily is show the connection with rotation symmetry. But I don't want to do this, but I think we've all reached a certain limit by now.

So there happens to be another electron around. It can annihilate with the positron even if it doesn't have the same energy. Yes, yes. And then you just get a different number of photons. Or photons coming out with, I'll tell you what happens.

Okay, supposing you have two electrons with the same energy. Two electrons coming in in the center of mass frame, let's say. Well, I'm assuming that you had the particle pair and then another half electron happens to be around. Oh, it doesn't matter which electrons annihilate. I mean, the electrons don't remember which one.

Yeah, but when you kick this electron out of here, you can kick it up to any energy. It doesn't necessarily have the same energy as the positron. Yeah, when you hit that electron, you don't necessarily kick it up to the same energy. You just hit it. You hit it and it goes somewhere to some other energy.

And then there's no memory of which, let's suppose there was another electron around. The axi is filled and now a positron is formed and another electron.

Then this electron can fall into the hole or this electron can fall into the hole. In either case, you get photons out. I'm still unclear. So if you have, let's say that the positron was the eighth level or whatever, negative eight, right? Why would like negative seven and negative six drop into that?

And why would it have to be an electron? Oh, no, that can happen. This is a perfectly good thing to happen. But photons, all right, so let's see. We start, let's see what that says. We start with negative seven over here, a hole.

That means a positron, one positron, one E plus, a positive charged electron. Notation for a positron is E plus. Notation for an electron is E minus. All right, so we have one E plus with an energy equal to,

let's say, minus, with an energy equal to seven units, right? We took the electron with minus seven units and then removed it, which means we have plus seven units of positron. And the electron went someplace.

Now, let's not even worry about where the electron went. The electron, this became an electron up here someplace. Now you ask, why can't one of these fall down to here, okay? What would that mean? You would start with a positron and you would end with a positron, but you would end with a positron of a different energy.

This hole was a positron. Now an electron from here comes and falls into this hole. There's a negative energy electron, falls into this hole. That's what you're asking, okay? So you started with a positron, here it was, or the hole, and you ended with a positron.

It's just a positron of a different energy. So a positron has changed its energy. When a positron changes its energy, it can only do so by emitting photons, right? How do you make an electron change its energy? By emitting photons, slamming it into something and having it emit.

So a positron from here, under the right conditions, sorry, an electron from here can fall down to here. That's just another way of saying the positron scattered from one state to another and radiated some photons.

Okay, so wouldn't they just always, because there's always higher energy electrons, wouldn't they all fall down and the only thing you'd be left with is a positron? The reason it doesn't happen, so you're saying why is the positron stable, why doesn't that, no, the answer is because you have to conserve both energy and momentum.

And the conservation of energy and momentum doesn't allow it to happen. Okay, so we can work that out, but again, all right, so why don't you ask me that at the beginning of next week, and I'll show you why the, look, you could ask, you could have asked exactly the same thing about the original electron. Why can't an electron always lower its energy and radiate photons?

Why can't an energy, an electron just drop down in energy until it's the lowest energy electron? And the answer is because it can't simultaneously conserve energy and momentum and emit photons. Just, it just doesn't happen.

So I will get in touch with you and give you the exact schedule. We will definitely meet 10 times this quarter. Next quarter, I have less obligations, so I won't be disappearing on you next quarter.

Thank you for your patience. For more, please visit us at stanford.edu.