Stanford University. First, we start with symmetries. And I'll remind you of a few things about symmetries, because supersymmetry, if nothing else, is a symmetry. Symmetries always relate things.

And what they do is they relate things which are related by being transformed in some way. A particular symmetry is left-right symmetry. My left hand and my right hand are symmetric.

And if the world and I and everything about me has left-right symmetry, then my right hand weighs exactly the same thing as my left hand, would be exactly as strong as my left hand, and would be in every respect identical to my left hand.

And in that sense, symmetries relate objects, objects which may be different but symmetric. In particular, those objects that are related by a symmetry generally have the same mass. So let's talk a little bit about the mathematics

of symmetries. We've already talked about some symmetries, in particular things like rotation symmetry in space. And now I'm thinking about continuous symmetries. Whether supersymmetry is a continuous symmetry or not is a question which has no answer.

It has no answer because it doesn't fall into the category really of the things where the question actually makes sense. But let's just review a little bit about symmetries. Symmetries are operations on a system which are described quantum mechanically by, first of all, operations,

but operations are described by operators. Operators on state vectors. We start with a state vector, whatever it happens to be, I don't care what kind of system it is, it's described by a state vector, and we apply some symmetry operation.

The most well-known symmetry operation that I can think of is rotation. Actually, I can think of an even simpler one, just translation of space. Then the operation would simply move an object from one place to another. But let's take the case of rotation as a more representative example for the moment.

What's the mathematical symbol to represent a transformation? A unitary operator. A unitary matrix, or a unitary operator. This unitary operator here could represent rotation in space.

You would have to indicate what the axis was, and you would have to indicate what the angle was. But once you did so, there would be a unitary operator. That unitary operator acts on any state to give a new one, which is a rotated version of the same thing. Yeah?

But it operates on states, not on space. It's not rotating states. It's the operation of rotating an object. It rotates, it acts on objects. That's right, it acts on states, but states mean collections of objects. And this is the operation which rotates everything.

It just rotates everything in space. So it takes whatever the state is. I'll give an example. The state might consist of the mathematical representation of a particle over here and a particle over here. So this is a state which represents two particles, one

over here and one over here. A rotation about an axis, a rotation, incidentally, always takes place about an axis. Let's say about this axis, it will simply give you a new state in which this particle has been rotated to here and this one to here, and so it's a new state with a particle over here and a particle over here.

But isn't the state vector an element of a Hilbert space? So that's an infinite dimensional vector. It is. And are we rotating the Hilbert space, or are we rotating three dimensional? Both. It's the rotation in Hilbert space that represents the rotation in space. It's like a next step up. Yeah, right.

That's right. Both. All right. But this is the rotation that we do on the Hilbert space. Now, it's not strictly a rotation. It's a unitary transformation, which is similar to a rotation. So it's both a transformation of the Hilbert space,

and it represents the act of rotation in space. All right, now, as I said, u represents a rotation about an axis, about a certain angle. There's particular transformations, which are especially interesting to us.

They're the infinite test. Oh, well, before that, let's talk about, very briefly, about the properties that a symmetry operation has to satisfy. Most important is that it has to be unitary. Unitary means that its Hermitian conjugate

is its own inverse. This is the unique property, which says that state vectors maintain their length. It's actually just probability, conservation of probability, that this vector should have the same length in the Hilbert

space, the same norm in the Hilbert space as the original. For every vector, that's the condition that you be unitary. That's why it's called unitary. Well, it's called unitary. All right, now, there's a special family of these, which are very small rotations, rotations by very small angles.

Now, every place I say rotation, you could substitute other interesting group operations. But let's stick for the moment with just ordinary rotations. The infinitely small rotations, infinitely small means by an infinitely small angle.

A rotation by no angle at all, in other words, no rotation, that's not represented by u equals 0. u equals 0 is not unitary. It's represented by u equals 1. So an infinitely small rotation is just u equals 1.

Well, supposing you want to rotate a little bit, not quite 1, then you have to add something. And if the angle is very small, let's say the angle is epsilon, epsilon is a small number, then you have to add epsilon times something. And in order to make it unitary,

you have to add i times a Hermitian operator. Call it L. L remembers the angle of rotation is epsilon, but L has to be indexed by a direction of space. So this could be rotation about the x-axis, for example.

It could be rotation about the y-axis, or it could be rotation about the z-axis. Or it could be rotation about some other axis. Or a sum, perhaps, x, y, and z. Well, if you sum them, you wind up getting a rotation about some axis, which, yeah, intermediate axis, that's right.

So there's three basic ones out of which you can build all the small rotations. An arbitrary axis is just a linear combination of rotation about x, y, and z. All right, that's the, and it's easy to check that if L is Hermitian, then U is unitary.

At least to the leading order in epsilon. We're gonna study things to orders, to lowest orders in epsilon. Now, one of the most important things about a group of symmetries is the commutator algebra. There's a mathematical concept, the commutator algebra of the generators.

These are called the generators. These are called the infinitesimal generators, the Ls. And they satisfy a commutator algebra. I'll tell you first what a commutator algebra is, and then I'll show you how it's connected

to these transformations. First of all, mathematically, what is a commutator algebra? It's basically, you take the independent generators, in this case, rotation about the x, y, and z axis, you call them Lx, Ly, and Lz, or better yet, just L sub i.

And a commutator algebra is just a closed set of relations that if you commute any one of the Ls with any other one of the Ls, you get back something proportional to an L again. There are some constants that appear here, all right?

So for example, let me write down exactly what you would have for Lx, Ly. You would have Lz here, but that's not quite right. What am I missing? I, I, and then cycle it through. Ly, Lz equals I, Lx, and so forth,

but you can summarize them all by writing i epsilon ijk Lk Li Lj. This just says epsilon ijk is just a symbol, which is zero if any of the two of them are the same,

i, j, and k, and it's either plus one or minus one, depending on whether this is an even or odd permutation of one, two, and three. All right, and all it says is Lx with Ly, commutator Lx with Ly is I, Lz, and so forth. Now what's the meaning, what's the meaning,

the geometric meaning of these commutator algebras? What are they saying, and why is it important that the generators close under commutation? The right word is they close under commutation, which means the commutator of any two of them is another one. This is called a commutator algebra.

Why is this important? Well, what it really has to do with is the following operation. You take, you take a rotation represented by a u, then you take another rotation about some other axis by some other angle, let's start it out, u,

and then you multiply it by v. What is that? That corresponds to a rotation about an axis followed by another rotation about another axis. Then you undo the first one, so far what we've done is rotate, rotate, then rotate back along the original axis

by minus the angle that we originally rotated, and then undo v. Does that give back no rotation at all? Does it give back no rotation? No, the answer is no. If you rotate about an axis

and then rotate about another axis and then undo the first one and undo the second one, it is not the original configuration. It is not equal to the identity. It is itself a small, another rotation. Just, most of you have seen a demonstration of this sort of thing.

What happens if you rotate a thing about the x-axis, then rotate about the y, I can't do this, but you rotate about the x-axis, rotate about the y-axis, now I've forgotten which the x-axis is, and do it again and come back. You come back to some other configuration. Consider an airplane in roll, pitch, and yaw. What's that? Consider an airplane in roll, pitch, and yaw,

that makes it easy. I hope that was via an airplane where I can sense the commutator. All right, let's check what you get, in other words, what you get on the right-hand side is some other rotation.

I won't give it a name, it's another u, it's another u, but let's see what happens if we do it for small rotations. Let's check, in fact, exactly what happens for small rotations, and let's take the case of a rotation about x, a rotation about y, reverse the rotation about x, and then reverse the one about y,

but each one of them being small rotations, and let's see what we get. You know, I think that there's something that you know so well that you haven't mentioned, but when you did the u-dagger u equals i, that could also be written as u-dagger equals u-inverse.

It does. And so when you see something like you just wrote up there, you're doing the inverses of daggers. Good, so let me rewrite it then the way you would like. We started with u, then we did v, and then we undid u, I wrote u-dagger, you would like me to write u-inverse,

I know you do, so do I. That's the same operation, same operation, and incidentally, if we were just rotating about a given axis, let's say one axis, the x-axis, we're never gonna go from x to y, what would happen?

Rotation by x, rotation by another x, then undo the first one, then undo the second one, what do we come back to? No rotation at all, okay? So if we do everything in one plane, we don't get anything interesting. It's if we do it in two planes that we get something. The best thing you can do with that is you can put one of the v,

the second v on the other side, and u-inverse v-dagger, u equals v, the loss of x. Let's see what happens. We start with u, that's one plus i epsilon lx. That's rotation about an angle of epsilon. Then we're gonna rotate about the y-axis,

and I'm gonna use a different angle. I'm gonna call the angle delta now. One plus i delta ly. Then we're gonna undo the first one. That's one minus i epsilon lx. Incidentally, for small rotations,

the infinitesimal, the inverse operator is just one minus i epsilon instead of one plus. And then finally, we undo the last one. One minus i delta lx.

Delta ly. And let's see if we can figure out what we're going to get. There's a lot of terms when you multiply these all out. And some of them will get mixed up with others. Some of them will be order one.

In fact, there'll only be one which is order one. It's one times one times one times one. The next thing will be order delta. Let's see what's there to order delta. Just delta by itself. Well, there's gonna be i delta of y times the rest of the ones, right? So there'll be a minus i delta of y

times one times one times one. But then there'll be plus i delta y times one times one times one. So they'll cancel. Same thing of the i epsilon x. There'll be one times minus i epsilon x times one times one.

But then there'll be the opposite from this one over here. So the things to order delta and the things to order epsilon will cancel. Nothing left of them. There are things of order delta squared and things of order epsilon squared. I don't wanna keep track of them.

You can keep track of them another day. What I wanna keep track of is the things of order epsilon times delta. The things which are sensitive to the fact that I rotated about two different axes. One of them, the epsilon kept track of the rotation about the x-axis.

The delta kept track of the thing about the y-axis. So let's see what's there, what remains there to order delta times epsilon. Here we're going to, delta times epsilon will always involve L y times L x.

How many of them are there altogether? How many delta times epsilons are there? There's one over here, there's one over here, and what have I missed? I have missed this times this, right?

This one times this one. Third turn times the fourth turn. Third turn times the fourth turn. Third turn times the fourth turn, yeah. Third turn times the fourth turn.

First turn times the fourth turn, you got that? Yes, Dad. We have the first turn times the fourth turn? First turn times the fourth turn.

It's getting too complicated, I need to clean it up. Oh, there are two turns and two, that's the point. Yeah, right. Good, good, thank you. Here's a delta times an epsilon

with a sign which is L y times L x, right? Right, L y times L x times minus i times minus i I think is minus one, right? Okay, and then there is L y times L x over here

and that comes in with what sign? With the opposite sign, I don't like that. Okay, let's see what happens. So that cancels this? Does, same sign or opposite sign?

Help me, help me. Same sign. No, look at the i, i is a minus, right? Minus i, minus times minus is plus.

Wait, wait, sorry. Minus times minus is plus over here and then we also have L y times L x over here. The minus times the minus is a plus times the i and i. Right, so it's times two. So it's a minus. Yeah. Yeah, so it's minus two. Minus two, right?

Yeah. Yeah, minus two. And now we have, that's L y times x, now we have L x times L y. L x to the left, L y to the right. What do we have for that? I hope it's twice plus two.

That all right? Yeah. There are two terms with minus sign, two terms with plus sign. The terms with minus sign are L y times L x and the term with plus sign are L x times L y, I hope. So let me just check it. L y times L x is minus times minus

and L y and L y times L x is plus times plus. Those have the same sign, all right? Then there's L y, then there's L x times L y and that has the opposite sign. So there are two L y L x's and two L x L y's.

If L x and L y commuted, then this would cancel, okay? Then this would cancel. But the fact that these rotations don't cancel, the fact that a rotation, another rotation followed by back again, don't cancel are an indication

that this should not be zero. In other words, the commutator between L x and L y is the thing which keeps track of the fact that rotations in different orders don't cancel out. That's what the commutator really is. The commutator is the non-cancellation of operations

which, well, which don't cancel. That's what this is indicating here. So what does this tell you? This tells you the commutator of, this is the commutator, this is twice the commutator of L x with L y,

twice the commutator of L x with L y, that's what's left here, plus one. There's also the one. You know, I think that we're off by a factor of two because there are really three terms with L y, L x,

and one term with L x, L y. That's what I get. So what happened? Yeah, I think that's right. That's what, yeah, yeah, that's what I thought. So what is it, just one times?

Yeah, that seems right. There are three with one sign, no, three with one order and one with the other order. Remind me, just, just. Yeah, three with L y, L x. It's because all the L y's are on the left of all the L x's.

So if you go to the L y, you have two L x's to the right. OK? You go to the L, if you were looking for, but it isn't. I've done this at home a number of times just to make sure I was ready for you today.

Ah, L x, L y, right. So that's this one. And then there are three others. This L y has two L x's to the right. That L y only has one L x to the right. Do you get this?

OK, good enough, good enough. I'm blind by now, I can't do it. Anyway, this is what you get. You get something proportional to the commutator here. So in other words, the net answer, there was an epsilon times delta, epsilon times delta.

No two. L x times L y, or the commutator of L x times L y, this becomes one plus i epsilon delta L z. In other words, this combination of rotations about x, y, then back again along x,

then back again along y give you a small rotation by an angle epsilon times delta about the z-axis. This is what the commutators tell you. So whenever you have a group of transformations, in particular a continuous group of transformations,

the whole structure of the group is contained in these infinitesimal generators and especially in their commutator algebra. The commutator algebra is sort of the group multiplication table, but in the form of these infinitesimal generators.

All right, that's a... If you switch the u dagger and v dagger, then it will equal the identity, correct? Yeah, if you switch the middle. Yeah, if you switch the middle. So does this get more complicated if you have four instead of three,

you have x, y, and then there's z, which is the third. If you go to four, now it has to be... It does get more complicated, but it does get more complicated. But if you know the commutator algebra here, you can in principle figure out how to compound together many, many little transformations.

This was just one example. You can obviously build up any transformation out of lots of little ones. So if you know how the little ones combine, that's very much knowing the properties of the group, knowing the algebra and the whole structure of the group. All right, so that's what a commutator algebra is. It's a convenient form

for representing the whole structure of the group and how the operations intertwine among themselves. Now, next. Next statement, things that I'll remind you of.

Incidentally, in the case of rotations, L is of course just the angular momentum. I picked it just... There are other transformations of space besides rotations, incidentally. There's translations in space. Let me fill out the rest of the group

of transformations of space. These are the rotations, and they're rotations, for example, about the origin. The other things you can do with space is to translate. Translations are generated again by their unitary transformations, but what are the infinitesimal generators for translation, for spatial translation?

The momentum. That's what momentum is. A translation, let's say about the x-axis, where you push everything along the x-axis by an infinitesimal amount. That's again of the form one plus i times a parameter.

The parameter now would correspond to the distance that you push things, the distance that you translate. Again, let's call it epsilon. It's not an angle now. It's a distance times the x component of momentum. This would be a translation along the x-axis,

translation along the y-axis, one plus i epsilon p y, and so forth. So this would be the group of translations. This would be the group of rotations. What about commutators between rotations and translations?

Would you expect them to commute or not? Think so? How about commutation, first of all, among just translations? You translate something this way, then you translate it this way, then you translate it back this way, and then you translate it back this way. Translation along x, y, invert the x, and then back.

What do you get? Well, when you translate something or when you rotate something, you first have to apply a force to move it. Well, forget, we're not talking about, we're not talking about literally moving things. We're talking about just imaginary transformations.

No force here. All I'm saying is that you have to do it, then you have to undo what you did to make it do it. So you've got actually four actions involved. In other words, you have to start it to stop it, and then you have to start it to stop it again to undo it. These are not actual motions of a system.

They're just asking, what would the state of a world be like if you displaced everything? They're not saying, displace it. They're saying, what would the state be if you did displace it? A little bit of a difference. Mathematics, not physics, yeah.

Just a mathematical description of a displaced thing. If you displace along an axis, and then you displace along another axis, and then you displace back along the first axis, and then you displace back along the second axis, you come back to the same place. That's the property of Euclidean space.

So what does that tell you about commutators of momenta? They're zero. Commutators of momenta are zero. Now, let's try to see what would happen if we take a point,

first translate it. We translate by one unit in this direction. Now, rotate by an angle, let's say by an angle theta.

So we rotate from here up to, let's say, over here, by this angle theta, that's the angle theta, and then we translate back, translate along the same axis back, and then undo the rotation.

Where do we come to? Do we come back to the same point? No. We come back to about over here someplace. So it's clear that translations and rotations don't commute. Translations and rotations don't commute.

And if you work it out, it's not hard to work out. You do little translations, little infinitesimal translations, and you can work out what the commutation relations are between the rotation generators and the translation generators. I'll tell you what they are. They're just commutator of LX.

Well, let's just do it in two dimensions. Li with Pj is i epsilon ijk Pk. That's it, but this is not important.

This is not important. What's important is the idea that everything is kept track of by these commutation relations. All right, that's the first thing about groups. Groups are really commutator algebras in disguise. Or vice versa.

Next thing about groups in quantum mechanics, first of all, they're symmetries, but they tell you something about the energies of different configurations. Naively, it's clear. Naively, it's clear what they say. They say that if you perform one of these symmetry operations, and you really have a symmetry, it should not change the energy of a system.

If you have an object, whatever that object is, and it has a certain energy, and you just rotate it, it shouldn't change its energy, right? Or if you just translate it, it shouldn't change its energy. So if you have a legitimate symmetry, then the conclusion should be

that when you perform a symmetry operation, the energy of a state shouldn't change. Let's see what that says. See, mathematically, if we can see what that says, what the requirement is.

Okay, so let's suppose we have some state, any state, and it happens to have energy E. Let's say it has energy E. It's an eigenvector of the energy, has a definite energy, let's call it E. And now let's apply a small transformation to it.

Let's call it U. Let's take U. U is, it doesn't have to be small, just a transformation of the appropriate kind, translation, rotation, isospin transformation, anything that corresponds to a symmetry of the system.

And now ask what the energy of the resulting state is. We find the energy of the resulting state. Well, let's first of all say what this means. What this means, that it's a state of given energy, means that it's an eigenvector of the Hamiltonian with eigenvalue E, right? That's the definition in quantum mechanics

of saying a thing has a definite energy, that the Hamiltonian acts on it to just give the eigenvalue back. It's an eigenvector of energy E. Okay, now let's ask about this state. If the original state was an eigenvector of the energy, and we have a symmetry, then the resulting state

must also be an eigenvector of the energy, right? So that must say that the Hamiltonian on this state must be equal to the energy times the same state, U E.

But this is also equal to H U E. What did I just do? I just said that the energy times U times E

can be replaced by, put the Hamiltonian over here, why? Because H on E just gives E times E. All right, so now I just subtract. H U minus U H on E equals zero.

That's what the conclusion is if a symmetry operation doesn't change the energy, right? So what was the first relation that you had up there? H or U times? Okay, let's go through it again.

Let me go through it again. This is tricky. We start by saying we have an eigenvector of the energy. H on E equals E on E. This says that E, the eigenvector E,

is an eigenvector of the Hamiltonian with energy E. Okay, step number one, all right? Now, I assert that U times E has exactly the same energy. Why? Because U is a symmetry. It rotates or it does something

that's just really a coordinate. It translates, it rotates, it doesn't do anything important to the system. This must have the same energy, and so this must also be an eigenvector of the Hamiltonian. Let's write that. H U E must equal E U E.

Just let me draw. E is just a number, except if it's in a vector, then it's a vector. H times this vector here must equal E times the same vector.

Okay, that's the statement that U times E is also an eigenvector of the energy. Okay, now, E is just a number, as Michael said, so it can be brought inside here. Doesn't matter which side of U it's on. It's just a number.

U times E. Fair enough? I didn't do anything. I just interchanged the numerical number E with the operator U. But now, E times the vector E is H times the vector E,

so I can say that this is U H E equals this one over here. And now I come to the marvelous conclusion by transposing that H times U

minus U times H acting on E is equal to zero for an eigenvector of energy E.

But this has to be true not for one particular eigenvector of the energy, but for all eigenvectors of the energy. For any eigenvector of the energy, this must be true. If the energy eigenstates are a complete basis of states, then it follows mathematically

that H times U minus U times H must itself be zero. If something is true, if something gives zero on every vector or every eigenvector of a particular operator, then that thing must be zero. So, the conclusion is if you have a symmetry, it means that the symmetry operations

commute with the Hamiltonian. All right, so that's the signal, or that's the conclusion, not the conclusion, that's the criteria for whether a set of transformations like U are symmetries or not, they must commute with the Hamiltonian.

All right, so here is every symmetry, whatever it is, commutes with the Hamiltonian, and that's equivalent to saying that the symmetry operations do not change the energy of a state.

All right, so as an example, take a state of a spin pointing in a certain direction, rotate it, what happens to the energy of it? The answer is nothing. Why not?

Because rotation is a symmetry. Put the spin in a magnetic field. What happens if you rotate the spin in a magnetic field? It will change its energy, but the reason, of course, is because you're not really doing a symmetry operation, you're rotating one thing relative to a thing

which you're not rotating. What would happen if you did a true symmetry operation? That would mean rotate the spin and also rotate the magnetic field. That would not change the energy, okay? Same thing, for example, you'll have a particle, let's say an electron,

here's an electron in the field of a proton. What happens if you translate the electron? Does the energy of the electron stay the same? Of course not. The electron has to do work to move it against the proton so you say, oh, translation of the electron is not a symmetry. No, translation of an electron by itself is not a symmetry,

but translation of the whole thing is a symmetry. All right, so translation being a symmetry, rotation being a symmetry, says that those operations commute with the Hamiltonian. Now, if every U commutes with the Hamiltonian,

let's come back to the generators. That says that the generators commute with the Hamiltonian. If every operator of the form one plus i epsilon L commutes with the Hamiltonian,

one commutes with everything, the commutator of one with anything is zero, i and epsilon don't matter here, it says that all of the components of the generators of a symmetry commute with the Hamiltonian. What else can you say

that commuting with the Hamiltonian means? If you go back to your quantum mechanics, yeah. The time derivative of the thing which commutes with the Hamiltonian is zero. So, one says that symmetries imply conservation laws

or conservation laws imply symmetries, it's not important which. All right, so this is, I've simply repeated everything that we've talked about before about symmetries, but mostly to remind you of that. Momentum conservation,

angular momentum conservation, all associated with symmetries. And of course, there are plenty of other symmetries around. Now. So, it looks like the external field has to have that symmetry. Either the external field has the symmetry or you have to rotate the external field, right. Right, okay.

But important thing is that rotation does not change the energy. Or not just rotation, but symmetry operations don't change the energy. That's the reason that the different components, let's say we have electrons in orbit

around a, in a particular orbit around a nucleus. Okay, so the orbits are characterized by a total angular momentum and the z-component of angular momentum. The different z-components of angular momentum

can be rotated into each other. They get mixed up with each other under rotation. The total angular momentum, L, doesn't change under rotation. It's the total magnitude of the angular momentum. But the various z-components of the angular momentum do rotate into each other. And the symmetry under rotation tells you

that all the different states of an atom at a given L, but for different M. Everybody know what M is for an atom? M is the magnetic quantum number. It really corresponds to the z-component of angular momentum. The different components, z-components, have the same energy as a consequence of rotational symmetry.

All right, all of the symmetries that we've discussed up till now have one characteristic in common. When they act on bosons, they give back bosons. When they act on fermions, they give back fermions.

What happens to a boson? What happens to an electron if you rotate it? It stays an electron. It doesn't turn into a photon. Quarks have symmetries under the color transformation. Remember the color transformations?

Mixes up the three colors of quark. Takes a quark into a quark. It doesn't take a quark into a gluon. A gluon is a boson. A quark is a fermion. In fact, all of the symmetries we've talked about up till now don't change the spin of a particle. It would take scalar particles to scalar particles,

half-spin particles to half-spin particles. Generally speaking, the symmetries we've talked about don't do anything to the spin, don't do anything to the type of particle, don't do anything to the charge of a particle, keep the charge of a particle unchanged,

and especially don't do anything to the mass of a particle. In some way, the symmetries tell you that there are equalities between masses of different states. For example, they do tell you that the various z-components of angular momentum all have the same energy

because you can rotate one into the other, z-components of angular momentum in an atom, as long as the atom is not in an external magnetic field or electric field. All right, supersymmetry is a new kind of symmetry. It's a really crazy kind of symmetry, which, when it acts,

it takes a fermion into a boson or a boson into a fermion. Just like the rotations tell you that there have to be these multiplets with different z-components of angular momentum, that if there's a z-component of angular momentum which is plus a half, then rotation symmetry tells you

there has to be one of minus a half with the same energy. If there's a spin up with one unit, then it's enough to tell you there'll be a spin down with one unit, but if you look carefully at the mathematics, you'll also find out that there has to be one of zero in between, all of exactly the same energy. That's what a symmetry does for you.

As I said, supersymmetry is a very strange, new construction, which, when it acts, takes a fermion to a boson. The mathematics of it is very weird, very strange, very non-classical, extremely non-classical,

and absolutely impossible to visualize in any sense, though I will tell you a little bit about the mathematics of it. Why is it important, incidentally? It's important because if you have a theory that enjoys the symmetry, that has this supersymmetry,

then there will always be, for every fermion, a boson with exactly the same mass, and with every boson, a fermion of exactly the same mass. That's exactly the situation that we were looking for where bosons and fermions might be able to exactly cancel out infinities

or large, large renormalization effects that we discussed for a couple of times already. This is very, very hard to motivate. This is hard to motivate because it is so completely unintuitive, very unintuitive.

These are symmetries, as I said, which take fermions into bosons. The generators of them are called Q. Is it Q? There are several generators called Q. We'll just call them Q for the moment, and they take a boson into a fermion,

something totally new. These Qs, let's see, I'm trying to figure out how to say this right.

How can we make an operator which takes a fermion into a boson or a boson into a fermion? Let's think about that for a moment. Q is an operator which can take a fermion into a boson and a boson into a fermion. What kind of operator could do that? Well, it's very easy, actually.

It has to operate on the spin and the charge. Well, what does it have to do? If you have a boson, it has to remove the boson and put back a fermion, right? If there's a fermion there, it has to remove the fermion and put back a boson.

I'll construct for you a Q that does exactly that. We have creation and annihilation operators for fermions and bosons, right? All right, what do we want? Q, which takes a boson to a fermion, it has to annihilate a boson, so let's put an annihilation operator, let's see.

I always forget, is A a creation? A is a? A dagger is creation. A dagger is creation, right? So A is annihilation. This annihilates a boson. So just for the simplest thing, this annihilates a boson, any annihilation operator,

and then what do we want to follow it by? A creation operator for a fermion. What do we call creation operators for fermions? I can't remember. C, C dagger, right? C dagger. C dagger. Here's an operator which whenever it acts on a boson,

will always give a fermion, right? What happens if it acts on a fermion? It gives nothing, right? It just annihilates the fermion. So let's add something to it.

Let's add to it A dagger times C. This annihilates a fermion and creates a boson, right? So what does this object do to a boson? Well, this eats the boson and spits out a fermion,

so it creates a fermion. What about this piece of it? This piece of it annihilates a fermion, but there is no fermion, so this piece of it is dead. This relationship is correct. What happens if the same operator acts on a boson? A dagger, oh, this is a boson on a fermion,

plus C dagger A. What does that do on a fermion? Well, this one annihilates a boson, but there is no boson there, so nothing. This one annihilates the fermion and puts back a boson, so this one puts in a boson.

So yes, we can make operators which do have the property that they intertwine between fermions and bosons. Superpositions?

Superpositions. Does this imply you can get a superposition particle of a fermion and a boson? Well, if it acts on a superposition, it will produce a superposition, but if it acts on a boson, it makes a fermion. If it acts on a fermion, it makes a boson, and it won't make a superposition by itself. It's not a stateless operator.

I think what he was asking, thinking, was this is some kind of a superposition state, but really, it's just an operator. This is an operator. Now, this is an operator. If it acts on a general state, it will make a superposition, but if it acts on a boson, it makes a fermion.

If it acts on a fermion, it makes a boson. What happens if it acts on the vacuum? What does it do if it acts on a vacuum? It's a vacuum. It's a vacuum. It's a vacuum, doesn't it? Doesn't it?

No. It doesn't make the vacuum. It just makes zero. Why does it make zero? Let's try it on a vacuum.

A annihilates a boson, but there is no boson. C annihilates a fermion, but there is no fermion. You just get zero. Not the vacuum, but just zero, the zero vector. So the one exception to the rule, the rule being that when it acts on a fermion,

it creates a boson. When it acts on a boson, it creates a fermion. The one exception to this rule, I don't know if it's an exception to the rule, just a funny exception, is that when it acts on the vacuum, it just gives zero. So this is a class of operators which

has the properties of a supersymmetry generator. Supersymmetry generators are of this form. They involve fermions and bosons in a way that is mixed together like this. Now, the question is, here we can.

The question, though, about the dimension of that vector. The dimension of what vector? The state vector. For bosons, the boson state vector is usually dimensionality from a fermion state vector,

isn't it? OK, now what do you mean by dimensionality? Well, let's say a half-spin particle would have two dimensions. Two components, two components, yes, yes. So for example, yes. So for example, yeah, good, good, good, good.

You're now telling me a little more about this operator Q. So let's suppose it acts on a boson, and the boson is spin zero. It creates a fermion. But a fermion has a spin 1.5, and a spin 1.5 particle has two components. So does it create the spin up, or does it

create the spin down? Well, the answer must be, if we're going to make any sense out of this, that there can't be just a single Q, there has to be two of them. One which takes the boson into the up fermion, and one which takes the boson into the down fermion.

There has to be another index. Qs always have indices, and the index is always a half-spin index. It itself is a half-spin index because it has to be able to change the spin by half a unit. Now, what happens if a Q with a spin i acts on a fermion?

A fermion also has to have an index. It's either up or down. So we ought to put indices on these fermion vectors here. What happens if it acts on a fj?

Well, basically the answer is, if i is the same as j, it makes the boson. If i is not the same as j, it makes nothing. So whatever these operators are here, whatever these symmetry operators, the generators, these are the generators of supersymmetry.

The generators of supersymmetry, not only they're fermionic. They're fermionic in the sense that they have an odd number of fermion operators in them. They have an odd number of fermion operators. Every interesting quantity that we've

been experiencing up till now, things that we measure and so forth, always have even number of fermion operators in them. Supergenerators have odd number of fermion operators because they've got to change the fermion number. And they also have indices. The indices are spin indices. All right, really understanding supersymmetry and fermions

and fermionic symmetries. Why do I call them fermionic symmetries? Because the generators themselves have odd numbers of fermions in them. They change a fermion into a boson. Really understanding that the mathematics of it

really does entail a generalization of arithmetic, of the notion of numbers. Let's just remind ourselves of one thing about fermion fields and boson fields. Or just fermion operators, creation and annihilation

operators, good enough. Fields are made up out of creation and annihilation operators. Let's just remind ourselves what we know about creation and annihilation operators for bosons. Creation and annihilation operators for bosons have commutation relations. The commutation relations are commutator

of a with a. And this could be any creation operator, any bosonic creation operator. These are annihilation operators, aren't they? What's the commutation relation of an a with an a? Zero. In fact, they could be for different bosons, even so.

They commute with each other. What about a dagger with a dagger? Same. What about a with a dagger? Or is it a dagger with a? I think it's that way. What does that give? One.

That's the algebraic properties of bosonic upwards. In particular, this particular relation over here, let's take this one over here. This could actually stand for two different.

different creation operators, one let's say for a particle of one momentum, one for a particle of another momentum or for a particle at one place and a particle in another place. Let's take this relationship here and create a two particle state, two bosons, one with

label I and one with label J. As I said, label I and label J could represent momentum, they could represent position, they can represent anything you like. How are we going to create two of them? We're going to create by just applying the

creation operators. This is the state of two bosons, one of type I and one of type J. How does this differ from a state with the same exact particles but

in which they've been introduced in the opposite order? It doesn't differ, it's exactly the same thing. It's exactly the same thing and mathematically the commutation relations here tell you that a dagger i, a dagger j is the same as a dagger j ai, they're equal to each other. This is the symmetry of wave

functions under interchange of bosons. When you interchange the arguments, the arguments of the wave function or the labels, labeling the states here, these are related to each other just by interchanging i and j, it does nothing. It multiplies the state by the number one. That's the character of

bosons and the character bosons is indicated by commutation relations between fields or between things that make up the fields. What about C's? Okay, what about the commutator of two C's? Let's go after this one here first.

C, C, C dagger. Is that one? No, that's not the way we do things with

fermions. With fermions, it's anti-commutator. What does anti- commutator mean? It means C, C dagger plus C dagger C. C, C dagger plus C

dagger C, not minus. Let's see what that means. In fact, every place where you saw commutator, replace it by anti-commutator. What does anti-

commutator mean? Anti-commutator of two things, A with B, means AB plus BA. Okay? AB plus BA. Okay, first of all, let's see what this says.

What does this say about two particle states? First of all, suppose there's only one kind of fermion, just a C with no label. Anti-commutator of C dagger with C dagger just means C dagger C dagger plus C dagger C dagger. It just means

twice C dagger C dagger. That's equal to zero. So it just says when you multiply C dagger by itself, you get zero. That doesn't mean that C dagger is itself zero. C dagger creates a fermion. What does it mean to say that when you

square it, you get zero? It simply means you can't put two fermions into the same state. You simply can't put two fermions into the same state. That's the Pauli exclusion principle. So the Pauli exclusion principle is simply obtained

from the anti-commutation relations. Now, supposing you happen to have two different kinds of fermions, they might be the same kind of fermion, an electron, but they might be electron at one place, or an electron another place, or electron of one momentum, an electron or another momentum, or they might just be different spin states of the electron.

Let's suppose again there were two labels. Now let's let's make them fermions now. C dagger I, C dagger J. How does that relate to the state C dagger J, C dagger I, where I've interchanged the I and the J? Well, the

anti-commutation relations say that C dagger I, C dagger J, plus C dagger J, C dagger I equals zero. Another way of saying it is that C dagger I, C dagger J

is minus C dagger J, C dagger I. This is the anti-symmetry of the wave function of, or the anti-symmetry of, fermion states with respect to interchange of the labels of the fermions. So the character of fermions and the character

of bosons is encoded in these commutation or anti-commutation relations. Whatever fermions are, or whatever the mathematics of fermions are, it's related to the mathematics of bosons by replacing commutator by

anti-commutator, but particularly the property of the square of a fermion operator being zero is very, very different than the square of a boson operator. Now boson fields, boson fields, when you build up a large number of

bosons, behave very classically. When we have, you know, a large number of photons, we think of the fields as the electric and magnetic fields, and we think of them as ordinary numbers. We think of them as ordinary numbers, and

ordinary numbers commute. Ordinary numbers commute. What about fermions? Well, for fermions, you never build up large numbers of fermions in the same state. So field operators for fermions never get big. They never get big because their square is zero, so how can they ever get big? There's a whole

generalization of arithmetic, which I'll tell you a little bit about, just a little bit, in which the numbers of arithmetic are replaced by numbers which anti-commute instead of commute, and these numbers are extremely useful.

They're really just bookkeeping devices. They're bookkeeping devices the same way that fermion fields are bookkeeping devices. They keep track of various relationships. These are not real numbers. These are not numbers that you can measure. They're not numbers that the experiments can be answers, that

are called Grassman numbers, a whole theory of new kind of numbers. As I said, it's a generalization of arithmetic, and I'm just going to tell you a little bit about it, because fermion fields are really Grassman numbers. They're

really Grassman numbers, and the generators of these fermionic symmetries, these things which take fermions to bosons, are also Grassman numbers. Grassman numbers have the property that they anti-commute, so let me just tell

you a little bit about them. For our purposes now, they're really just a curiosity, but they really are at the heart of what supersymmetry is, so we have a collection of ordinary numbers to begin with. Let's call them ordinary numbers. Let's call them alpha sub i. i just labels which number

we're talking about, okay, and I'm just labeling them alpha sub i. Here's something which is true of any pair of ordinary numbers. Alpha sub i, alpha sub j is equal to alpha sub j alpha sub i. It's three times five is five times three, okay. Now we're going to invent these new numbers called

Grassman numbers. We're going to call them theta instead of alpha. Theta i, I don't know how many there are. There might be only one of them. There might be two of them. There might be three of them, however many there are. They have the properties that theta i anti-commutes with theta j. In other

words, theta i theta j is equal to minus theta j theta i. When you interchange, when you interchange them, they change sign. That's the character,

that's one of the characters of Grassman numbers. Now don't ask me what the value of theta i is. It's not seven, it's not six, it's just theta i. It doesn't have any ordinary numerical value. Another thing you can say about them is their squares are zero. Theta squared, which is equal to, let's

say, theta one squared. Theta one squared is just theta one theta one, but theta one theta one is just one half of theta one theta one plus theta one theta one. Theta one theta one plus theta one plus theta one, that's just

the anti-commutator of theta one with theta one. So theta squared are equal to zero. Now let's suppose for simplicity now, let's suppose there's only one

Grassman number. Let's just call it theta, one Grassman number. It's like just having one number in arithmetic, the number seven. Okay, oh we'll also add an ordinary number. We can have both ordinary numbers and Grassman numbers in the same arithmetic. I should tell you about, oh well I should, okay, so we should

add. If we take an ordinary number, a commuting number, let's call it an alpha, the rule is ordinary numbers commute with thetas. They don't anti-commute with thetas, they commute. Alright, they commute with themselves. The only thing

that anti-commutes is a Grassman number with itself. Grassman, no Grassman,

no Grass, like the kind you smoke. Grass, Grass, Grass, Grass, Grass, Grassman.

Grassman must have been a mathematician, a German mathematician from the 19th century. Now he was

really smoking something when he invented this. A Grassman number times an ordinary number is a Grassman number. Right, right, this is, that's right, we

can easily prove that, that Grassman numbers times ordinary numbers are ordinary numbers. What about a Grassman number times a Grassman number? A Grassman number times a Grassman number is ordinary. What about an ordinary times a Grassman? It's a Grassman, right. So, ordinary times ordinary is ordinary, ordinary times

Grassman is Grassman, and Grassman times Grassman is ordinary. Okay, so it's like odd and even numbers. You know, odd times odd is even, even times even is even, odd times odd, odd times even is odd. Okay, so that's the property of Grassman numbers. Okay, what's that? Okay, so let's think, a

real, well yes, a real times a real is a real, a real times imaginary is imaginary, and imaginary times imaginary is real.

So yes, it's also true of imaginary and real numbers. But it's particularly true of these, now, what is unusual is that the square of any Grassman number is zero, because it anti-commutes with itself. What's that? Okay, yeah, I mean the cross product of A

times B is minus the cross product of B times A, that's true, that's true. But, yeah, I don't think the analogy goes too far, but yes, that's right. Okay, can't you think of this as a generalization of like imaginary numbers? It's just a generalization of numbers.

It's a generalization of numbers where you replace commuting, the, you know, one of the postulates of arithmetic is that multiplication commutes. Here, the postulate of arithmetic is that multiplication anti-commutes. Okay, let's take the

case of only one Grassman number, okay, then the number, and think about functions, let's think about functions now.

Of Grassman numbers, we'll have Grassman numbers and ordinary numbers. Ordinary numbers are ordinary, but there's only one Grassman number, okay, only one Grassman number. The functions of Grassman numbers are very limited, let's see what kind of functions we can build of Grassman numbers.

Well, we can start trying to build polynomials. The polynomial might be an ordinary number, any ordinary number, let's call it alpha, plus another, you can multiply Grassman numbers by ordinary numbers. So, the next thing you can build is beta times a Grassman number, there's

only one Grassman number now. Now, let's try to build a quadratic polynomial, let's call it plus alpha beta gamma times theta squared. Well, theta squared is zero, so we've run out. There are linear polynomials and that's it, those are the most general functions of a

single Grassman number. The space of functions is not very rich, okay, it consists of basically two functions, a constant, and a linear function of the Grassman number. No more, there are no more functions than that, so it's a pretty damn simple arithmetic, or a pretty damn, yeah.

Theta cubed is theta times theta squared, and if theta squared is zero, theta cubed is zero. So, everything beyond theta, theta is zero, that's it. There's not much of a space of functions, okay, let's

suppose that there are two Grassman variables, theta one and theta two, then what kind of functions are there? Well, again, there's alpha plus beta theta one, beta one theta one, let's call it. We can then add

plus beta two theta two. Beta one and beta two are just ordinary numbers. Ordinary numbers could be complex numbers, it's not important at the moment whether they're complex numbers or real numbers, whatever they happen to be. And then, we can have plus gamma theta one theta two. Theta one times theta two is not zero. Theta

one squared is zero, because it anti-commutes with itself, it's a commuting number, but it's surely not an ordinary number.

Why isn't it an ordinary number? Because it is not zero, but its square is zero. Theta one theta two, theta one theta two, that's zero. So, it's surely not an ordinary number, okay? It's not zero, but its square is zero.

But it does have the property that it commutes with this, alright? So, that's it. That's the last, that's the biggest function, that's the most complicated function you can make of two thetas. Incidentally, this is not equal to alpha plus beta one theta one plus beta two theta two plus gamma theta two theta

one. Why not? Because they anti-commute, and that means to make them equal, you would have to put a minus sign here. Question, you said you multiply an ordinary number times the theta, you get a theta. So, it seems like beta times theta would give you a third.

Beta, beta, you multiply beta times theta one, that seems like that would give you a different theta. But you're saying there's only two thetas in that equation. No, no, sorry. You can multiply a theta by a real number. Yeah, you can multiply it by a real number.

So, what should we say? We should say that thetas, there's only two of them up to, they really form a two-dimensional vector space. Right. So, apart from multiplication by an ordinary number, there are really only two of them.

Okay, so, the space of functions is very limited. There's not much structure to it. If there are three thetas, well, it gets a little bit longer. You can go up to beta three theta three, then you can have theta one theta two, theta one theta three, theta two theta three, and then finally theta one theta two theta three, and that's it.

Okay. So, polynomials in this theory are, polynomial functions of theta, polynomial functions of theta are very limited. And there are no non-polynomial functions. Okay. What about exponentials of theta? That's an interesting question.

Let's see what exponentials do. Let's see if e to the a, let's say there's only one theta. I think we can do this with just one theta. e to the a theta times e to the b theta. Oh, okay, let's do this first. e to the a theta times e to the b theta.

Is that equal to e to the a plus b, a to the a theta, you know, do things add in the exponent? Alright, so let's see. e to the a theta is one plus a theta plus a squared

theta squared over two factorial plus a cubed theta cubed over three factorial, but we're dead after theta. So e to the a theta is just one plus a theta. e to the b theta is one plus b theta. The product of these is one plus a theta plus b theta.

Plus ab theta squared, but theta squared is zero, so this is one plus a plus b theta, and that is just e to the a plus b theta.

So, exponentials exist, or the algebra of exponentials is exactly the same as the algebra of exponentials of ordinary things, but exponentials are no more complicated than this.

I'm curious, let's see if it works with two thetas. Two different thetas, let's try it. Okay, so we have e to the theta one, e to the theta two.

Okay, let's see what we get. We get one plus theta one times one plus theta two, which is equal to one plus theta one plus theta two plus theta one theta two.

I think that's right, isn't it? And that's it, that's as far as we go. Okay, let's see if that's equal to e, I hope it is, e to the theta one plus theta two. This should be equal to one plus theta one plus theta two plus theta one plus theta two squared over two factorial, right?

The problem with higher is we're not going to get anything. The reason we get something from here is because we have theta one times theta two. Theta one squared is zero, theta two squared is zero, but we do have twice theta one theta two over two factorial, which is exactly what appears here.

But now what happens when you go to theta one plus theta two cubed? That's going to have things like theta one cubed, that's zero. It's going to have things like theta two cubed, that's zero. It's going to have theta one squared times theta two, that's zero because theta one squared is zero.

In other words, there's always going to be some repetition of one of the thetas. And a repetition means that it's equal to zero, so this thing is not there. And yes, it does reproduce this. Interesting, just an interesting fact. Exponential functions exist, but they're extremely simple.

Doesn't the squared term there give you theta one theta two plus theta two theta one, which is zero? No, it doesn't, it doesn't, it doesn't. Let's see why that, that's a good point, why doesn't it?

Theta one plus theta two theta one, you know? You may be right, that looks right. Theta one, hmm. Theta one squared plus theta two squared plus theta one theta two plus theta two plus theta one.

Yeah, no, that's worrisome. Does theta one plus theta two give you another theta type of thing? Is the sum of two Grassman numbers another Grassman number? Yeah, sum of two Grassman numbers is another Grassman number.

So the question, yeah, yeah, yeah, yeah. Yeah, yeah, looks like the exponentials don't always make sense. I didn't know that. It looks like there is a failure of multiplication of, am I missing something?

I don't know, looks like it doesn't work. I actually didn't know that. I'm sure that's the first time I ever tried that. What's that?

Yeah, that third term is zero. No, but this is not zero. Theta one theta two is not zero. Is it possible that if you go to cubed, you might get that term? What's that? If you go to the cubic thing, you might get that term?

If you go to the cubic thing, I don't think so. Theta one, no, I don't think so. I don't think so. No, I think it fails. I think it really does fail. I'll check it. Just a dumb question. You have one plus theta one.

What does that mean? Because you have an ordinary number. Is that an ordinary number? No, one plus theta one is not an ordinary number. Yeah, you're allowed to mix them. You're allowed to add them. You're allowed to add them. You're allowed to add them, but it looks like you can't exponentiate them.

I'm confused by it. I thought you could, but... A little bit like quotations about different things or something. They are what they are. I don't think they're like anything else. I don't think like they're like anything else. They're a bookkeeping device, which is useful, but they're not like anything else.

They're Grassman numbers. Grassman, functions of Grassman numbers can be differentiated and integrated, which is a useful bookkeeping device. When I say they can be integrated, I mean you can define the integrals of a Grassman number

and the derivatives so that it satisfies certain properties. If you have the derivation, the derivative of a function of a Grassman number is straightforward. The derivative with respect to theta of one or any alpha is equal to zero.

The derivative of theta with respect to theta is one.

And so, for example, if we have just one theta in a problem, we have functions a plus b theta. The derivative of that with respect to theta is very straightforward. It's just beta. That's it. Supposing we have more than one theta.

This is easy. I mean, this is just ordinary derivative. And this is definition. This is definition. The derivative with respect to theta one of a function alpha... Well, there are two thetas now, just as an example.

Two thetas, alpha plus beta theta one plus beta two theta two plus finally gamma theta one theta two.

All right, you go through it just as if it was an ordinary function. Derivative with respect to theta one, you find something from here, just beta one. Nothing from here, nothing from here, and from here you have gamma theta two.

This is exactly what you would expect. The only surprise in the definition is that if I were to differentiate with respect to theta two. Nothing from the alpha, nothing from here, theta two, but what about here?

You might expect theta one, right? Okay, the rule is that the derivative operation is also like a Grassman number, and when you pass it through another Grassman number, it changes sign. The rule for Grassman numbers is when you pass them through each other, they change sign.

The derivative of a Grassman number is counted as a Grassman number with respect to sign. So that means when you pass it through theta one, you change sign, and this will give you minus gamma theta one.

So you treat derivatives the same way you treat the Grassman numbers themselves. A derivative with respect to a theta is also a Grassman kind of variable, and when it passes through any other Grassman variable, it changes sign. That's the only peculiar thing about differentiation with respect to Grassman numbers.

Incidentally, these rules ensure some of the rules of ordinary differentiation. For example, they ensure the rule that the derivative of a product is the first term times the derivative of the second one,

plus the second one times the derivative of the first one, the usual rule of calculus for products. If you didn't change the sign here, you'd get into trouble with that. So the calculus of Grassman variables is again simple.

You don't do anything unusual except to remember that whenever a derivative passes through a Grassman variable, the thing changes sign. We have to do integration of Grassman variables, which is too late for now.

We will be able to write a table, since there are so few Grassman functions, the table of Grassman integrals will be about that long. So we'll also define Grassman integrals. Why I'm doing this is just to show you that there does exist a sort of mathematical algebraic framework

in which supersymmetry and which fermion operators fit in to a new kind of arithmetic. It is essential to understand all of this, to understand supersymmetry at its deepest level. It looks like the product of any two Grassman numbers is zero.

No, theta one times theta two is not zero. Add theta one and theta two and square them, and if that's zero, you'll see the product is zero. No, it's not true that the square of...

Square that, and we know that's zero. You square anything at zero, right? Yes it is. And therefore the product of any two Grassman numbers is zero. No, no, no.

This, no, you see this is equal to theta one squared plus theta two squared plus theta one theta two plus theta two theta one. Right. Not quite true. Right. This plus this...

Well, yeah, the result is zero, but it doesn't prove that... Theta one theta two is equal to zero, it just proves that theta one theta two plus theta two theta one is equal to zero. Yeah, that's already a rule. So theta one times theta two is equal to minus theta two theta one, but theta one times theta two is not equal to zero.

Good. But I'm still puzzled about this exponential. I had always thought that I never used it for anything, and I don't know what it's good for, but I was hoping that exponentials existed. Any time you have the sum of the two things, powers of the sum of the two things, once you get to the square, it's all zero.

Yeah, but if you multiply e to the theta one times e to the theta two, it does look like you get left with a... It doesn't look like you have a source of theta one theta two. You do have a theta one theta two. I'm saying that this expansion of the exponential gives you square terms and higher, but square terms are automatically zero.

Yeah, so that means this is true, but that's not the same as this. So it's simply not the case that this is equal to this. Simply not the case.

But aren't you assuming that the expansion is valid? Aren't you assuming that the exponential expansion is valid? That may not be the case. Oh, yeah, yeah, yeah. That was the definition of the exponential. Yeah, I think there may be some other... There may be some other definition. I have to think about it. There may be another definition.

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