All right, let's talk about rotations of space in this language. Rotations of space are asymmetry and they can act on various things, but one of the things they can act on is

vectors. Those vectors might be a vector field or they might just be some vector in space, velocity vector, whatever. Rotations act on vectors and they rotate vectors from one vector to another.

Rotations can be combined. You can take an object, rotate it about an axis by a certain angle and then rotate it by another axis about an angle and the result is an overall rotation about some other axis by some other angle.

So in abstract sense, the rotations do form a group. Let's just think about it for a minute. Do the rotation operations have a group structure? Imagine taking a set of coordinates.

Incidentally, how do we parametrize rotations? Rotations can be parametrized as vectors. We won't worry about the fact that rotating for the moment by more than two pi brings you back to this. Supposing you don't worry about that for a moment. Rotations can be, first of all, they have an axis.

A rotation is about an axis. Let's take all rotations to always be right-handed, in other words, rotations right-handed. What happens if we have a rotation which is left-handed? We can either think of it as a rotation, a right-handed rotation by negative angle or we can think of it as a

right-handed rotation by the opposite axis. Doesn't matter. You get the same thing. All right. So rotation has an axis. Let's call that a unit vector N. And it has an angle that you rotate by, theta.

How many parameters does it take to describe a unit vector? It takes two parameters to describe a unit vector, longitude and latitude on the sphere. Unit vector is an arrow pointing to a unit sphere. So that's two angles and another angle for how much you rotate about.

That's three angles or three parameters to describe a rotation. It's a three-parameter family of objects. Now, when you rotate an object, if you rotate an object, a rigid object, imagine you have a rigid object and you rotate it about some axis and then you rotate it about some other axis, the result is always equivalent to a

rotation by yet some other axis and some other angle. So the product of two rotations, one done sequentially, one and then another, is always another rotation. So the answer is yes, there is a product structure to rotations, number one.

There is an identity operation for rotations, just no rotation at all. There is an inverse. Any rotation about any axis, the inverse of it is just a rotation by the opposite angle about the same axis or you

can, if you like, think of it the same rotation about the opposite axis. But let's not get into that. There's always, for any rotation about any axis, there's always exactly the opposite one which undoes it. So there's an inverse structure. Sorry, there's an inverse structure. What else do we need for a group?

And it also happens to be associative, yeah, it happens to be associative. That's a true fact, it's not quite as obvious. Okay, so rotations are a group. Have we failed to say anything about rotations? Oh, good.

It's non-abelian. This is the fact that a rotation about the X axis followed by a rotation about the Y axis is not the same as interchanging the two orders of rotation. Go home and do that if you haven't done it before. Rotate about the X axis, then the Y axis, and then do it in the opposite order, you won't get back to the same thing.

The smaller the rotations are, the closer it is to being a billion. Yes. Sequential rotations not simultaneously? No, no, sequential, one after another. Yeah, group action always corresponds to sequential, right. That is correct, the smaller the rotations, in a certain

sense, the closer it is to commuting, but the group is definitely a non-commutative or a non-abelian group. It's our first non-abelian group. Let's talk about its representations. Where we're going, incidentally, is, if we get to it tonight,

is the study of color. Color is a symmetry group and it's represented as a symmetry. The color of, I'm talking now not about the color of your shirt, I'm talking about the color of quarks, the red, green, and blue. It's not up and down, it's not a two state system, it's

interesting are the ones which mix up the different colors of quarks. So much of physics has that pattern of group symmetries

that we need to understand it before we go on. Okay, rotations. Rotations, how do we construct a representation, a matrix representation of rotations? And the easiest way to do it is to take the components of a vector. Let the basic objects be the components of an ordinary

vector. Call a vector V and it has components V, X, Y, and Z or VI, let's call them VI. I can equal 1, 2, and 3 or X, Y, and Z and so forth.

Now, when you rotate either, you can either think of rotating the axes or rotating physically the vector about some direction. Of course, the components VX and VY and VZ change and they change by being multiplied by a rotation matrix.

Alright, so for every rotation about any angle, about any axis, there is a rotation matrix. The rotation matrix depends on the angle of rotation and the axis and it is a matrix, IJ.

How do you find the new resulting set of components? You multiply the matrix by VJ and you get, let's call it V prime I. Again, summed over J. This is matrix times vector gives transformed vector.

Where these are rotation matrices. Now, do you know the properties of rotation matrices? I believe we've gone through them but I will go through them again.

Invariant length, yeah, let's just see what that says. Let's work out the properties of rotation matrices. This is another group, of course, another group and another representation. This is a three-dimensional representation of the rotation group.

It is not the only representation of the rotation group. There's many of them but this is the three-dimensional three-by-three matrix representation of the rotation group and it's easy to think about because you can just visualize it in terms of the components of a three vector, of a spatial vector.

All right, if R rotates V and gives V prime and R is truly a rotation of the vector, then it doesn't change the length of the vector. That's the property of a rotation. It changes nobody's length. Incidentally, we take the components of the vector to be

real numbers for the moment, real numbers and the matrices R are also composed out of real elements. So what is the condition? The condition is, the condition for this to be a rotation is that for every possible V and V prime, V prime being

the result of rotation, that VI VI, this means the sum, V X squared plus V Y squared plus V Z squared, the length, the square of the length of V should be the same as the square of the length of V prime.

Let's work that out. Let's see what that says. We can work that out in detail. This is R I J, let's put V J here. Doesn't matter. We're summing. It doesn't matter what we call it, V J, same thing.

R I J V J is equal to V prime I, right? Oh, I don't know. Maybe I want that I here. It doesn't matter. This is V prime I, right? Let's write the same equation again.

R I K, V K equals V prime I. I've changed. You never, this dummy index, the summation index, I don't want to conflate with this one here. But this one is the same as this one.

Now, what I want to do is I want to form V prime I, V prime I. So I want to multiply this by this and sum over I. And what will I get? I will get V prime I, V prime I equals R I J, V J, R I K, V K with every repeated index summed over.

Or equivalently, just write it a little differently, R I J, R I K, V J, V K. This should be the same as what?

This should be the same as V I, V I. That's this over here is equal to this over here. What is the condition that this is equal to this for every possible V?

Let's write it in two steps. It is that R I J, R I K is equal to delta J K. In other words, if this were true, that R I J, R I K

summed over I was the Kronecker delta, then it would tell you to set J equal to K and this would just become V squared. So that's the condition for a rotation matrix. And it can be written in a different way. It can be written that the transpose R times transpose is

equal to one. This is the unit operator and this is the representation for R times its transpose. So a rotation matrix satisfies, let's write it over here, that R transpose times R is equal to one.

Does everybody remember what transpose is? Transpose is just interchanging rows and columns. R transpose R is equal to one. Now actually we immediately now know what the inverse of any rotation matrix is. It's just a transpose.

The rotation group has this property and that's basically its defining property. R transpose, that's the defining property of the matrices, of rotation matrices. If you construct for every rotation such a matrix and you multiply the matrices, you will find the matrix

representation of the multiplication table of rotations. I'm just doing this as an example. So there are three by three representations, three

dimensional representations of the rotation group. Can we think of the three dimensional representations of the rotation group as in any way being related to the quantum states of something? Well, we've studied spin.

Spin describes the angular momentum of a particle and so forth and you can rotate the spin of a particle. You can take a particle and you can rotate it. You can put it in a magnetic field, rotate the magnetic field and it will mix up the states. It will mix up the states.

We discovered that there were various angular momentum possibilities for the spin. There was spin 1.5, spin 1, spin 3.5, there was also spin 0, spin 0, spin 1.5, spin 1. How many states does a spin 1 particle have?

Three. What happens when you take the state of a spin, what does that mean? That means that a spin 1 particle, just the spin of it, nothing but the spin of it, can be described by a column vector with three entries representing the amplitude for any one of the three states of the spin.

Supposing you want to rotate that spin. Well, to rotate that spin you must multiply it by some kind of rotation, some sort of operator that rotates it. Rotate the spin state. It's got three states. The rotation matrices or the rotation operators which act

on the state vector to mix up the components must also be a three-dimensional representation of the rotation group. In fact, it's just that. So it's often one speaks about spin 1 particles being vector particles.

What that means is that the three states can be thought of as being in one-to-one correspondence with the three components of a vector and they mix up among themselves the same way as the components of the quantum states of a spin 1 particle. All right. But I'm interested now in a spin-a-half particle.

Well, I have a spin-a-half particle and I want to know how the state... Anybody want to take a rest? Let's take a little rest. Yeah. We've gone through a lot of material. I've tried to keep it as simple as I possibly can.

Incidentally, we've encountered two different kinds of groups. Discrete groups and continuous groups. The group composed of 1 and p has only two elements. Only two elements to the group and it's sort of

discrete. You can either flip or not flip. That's all there is. The rotation group or the multiplication by e to the i phi is continuous. Phi can be a continuous variable. So there's a continuous infinity of elements and of course in the rotation group there are also a continuous infinity

of elements. So the kind of transformations and the kind of symmetries that we can think about in physics break up into two kinds. The discrete symmetries and the continuous symmetries. Rotation being a continuous symmetry, reflection being a

discrete symmetry. The groups that are associated with continuous symmetries are usually called Lie groups, L-R-E. They were first studied in mathematics by a mathematician named Lie, L-R-E, a German mathematician I guess in the

19th century. Our interest will mainly be in continuous groups. Well, now and then we might think of a discrete group.

Okay, where were we? Yeah, I said that for a spin one particle which has three states, the mathematics of rotating around those states is just the mathematics of the three dimensional representation of the rotation group. But then you can ask what about spin-a-half particles? Well, first of all, what about spin zero particles?

What happens to a spin zero when you rotate it? Nothing. So the symmetry group as applied to a spin zero particle, the rotation group, really only has one element and it's

the unit element and nothing else and its representation is just completely represented by nothing more than the unit operator. It's a completely trivial system, the spin zero particle. The spin-a-half particle also already has two states. Now, given a state of the spin-a-half particle, what

does it mean to be given a state? It means one zero, that's to be identified with up and

zero-one is to be identified with down and a general quantum state has a complex number here, let's call it alpha-one and alpha-two. Two complex numbers, what do we specify about them? We specify nothing more than that the sums of the

squared magnitude of them, alpha-one star alpha-one plus alpha-two star alpha-two should be equal to one. That's all. Other than that, it's free to choose alpha-one and alpha-two. If we choose alpha-one equal to one and alpha-two equal to zero, then we are talking about a spin pointing in

the up direction. Now, up of course is relative to some definition of an axis, but let's pick our axis vertical, Z axis, then this state represents an up spin, this state represents a down spin, and in general this combination here

represents a spin state in some direction, spin oriented in some direction which is determined by alpha-one and alpha-two. Now, we're not going to do the algebra figuring all that out, but we can't ask what if we take a given spin

state, alpha-one and alpha-two, and we rotate it. We just rotate our coordinates, or we physically rotate all the spin states. We physically rotate the spin to another axis or just whatever the configuration was, we rotate it.

What do we get? Well, we have to operate on this to get something new. Alpha-one prime, alpha-two prime. What do we operate on? States, to get new states, we operate with operators, all

with matrices again, matrices in the matrix representation. So, this is going to be some element, let's call it U, U11, U12, U21, U22.

Every rotation can be represented as a two-by-two matrix acting on spinners, on half-spin states. Let's see what the properties of such a matrix.

Now, this is just another representation of the rotation group, and it's a two-by-two representation. There's a subtlety about this, which I'm not going to get into, but some of you may know it, that the two-valuedness of these representations is not important. This is going to represent a rotation of the components of a

spinner, of a half-spin particle, when you rotate the directions of space. All right, so what's the first thing that's important? Well, first of all, when you act on a state, you should get another state. If a state is characterized by the sums of the squares of

the elements being equal to one, what does that mean? That means alpha one star, alpha one plus alpha two star, alpha two equals one, then the same should be true after rotation. So this should equal, well, first of all, it should be

equal one, but it should also equal alpha one prime star alpha one prime plus dot, dot, dot, plus the same thing for two. Well, it's very much similar in pattern to the fact that the sums of the squares of the components of a vector

should be conserved under a rotation. The only differences here with dealing with complex numbers here. The half-spin system is described in terms of complex components here. Incidentally, let's count components.

How many components does it take to pick a direction? Two, right? How many independent components are there here? Four.

They're complex numbers and they're each a pair of them. But we always know we don't care about the phase of these things. The phase of them doesn't matter. So dividing through by the phase and forgetting by the overall phase, that cuts it down to three. There are only really three independent numbers describing

such a spinner. But what else do we require? Alpha one star, alpha one plus, and that's one equation. So really there are only two independent things here. Two independent things, even though there are four complex numbers. One of the parameters doesn't matter.

The overall phase, it never affects anything. And the other is the length of the vector, the sums of the squares of them are always set equal to one. So given any pair of components or given any spinner like this, there's always some direction that it's polarized in, that it's along, that its axis is along.

But what are the rotation operators U? Well, first of all, the most important thing and the only important thing really is that they have to preserve the length of any given vector. Not the length in three-dimensional space, but the length

meaning the total probability here. That gives rise to a very, very similar condition to this over here. It gives rise to a condition which is very similar to R transpose times R equals one. I'm not going to work it out now. You can work it out yourselves.

It's the condition that U transpose complex conjugate times U is equal to one. It's just an extra little complex conjugate because when

we sum things, we sum them multiplying by their complex conjugate. This combination here, the transpose complex conjugated is called the Hermitian conjugate and it's represented by U dagger. So we've seen that before. U dagger is equal to the transpose complex conjugated.

Transpose means you flip things about the main diagonal. Complex conjugate, you just complex conjugate every element. U dagger is equal to U transpose.

I wrote it here. I don't need to write it again. So U dagger times U is equal to one. And in fact, that's all we're really going to need. There is a two by two representation of rotations in space.

They're complex matrices and they're unitary. This is the condition for a unitary matrix. How many components does a unitary matrix have? Can you figure that out? Well, it's a complex matrix.

Eight real components, right? Eight real components. Four complex components, but no, not quite. I mean, now there's a bunch of relationships here. Now, how many relationships is this? I heard the right answer. Four. Why is it four? Because this is a matrix equation and there's a separate

equation for each component. There are four components. This is four real equations here, four real equations. And so there are four real equations among eight unknowns.

There are four independent parameters of a unitary two by two matrix. Now, that's one too many because we've already found that rotations, how many parameters characterize the rotation? Three.

The unit vector, which is two, and the angle was three. So there's one too many components here, one too many degrees of freedom to actually describe the rotation group. But there was an obvious candidate for a, for one more,

well, I don't know how obvious it is, but there was a candidate for one more restriction that we can put on these unitary matrices, which will cut the number of parameters by one more. And that is to set the determinant of them equal to one.

Why the determinant? Well, first of all, you can see that the determinant of the product of two matrices is just the product of the determinants. Determinant of A times determinant of B is the

determinant of A times B. So if we were to set the determinant of every U equal to one, then at least that would be consistent because when we multiply two Us together to get another U, we would still have the determinant equal to one. So it is actually a theorem. This is very easy to prove.

You can prove this is a one-liner, basically, that if you were to take only the matrices, the unitary matrices, which have determinant one, that they form a closed group. Setting the determinant equal to one does not destroy the group property simply because, first of all, the unit

operator has determinant one. And since the product of determinants is the determinant of the product, there still is a product structure to the group. So one additional equation that one can impose without destroying the group property, let's just write it as

determinant of U, determinant I'll indicate by a pair of brackets like that, is equal to one. It's also true that the determinant of U dagger is equal to one. They're complex conjugates of each other. So the group of unitary matrices, two-by-two unitary

matrices, with the property that every determinant is equal to one is, first of all, a three-parameter group. It has the same number of parameters as the rotation group. In fact, it is the rotation group. There's a subtlety there, which I won't get into. It is the rotation group. It has the same multiplication table as the rotation group.

It is the rotation group. But it doesn't act on vectors. It acts on the states of a spin-a-half system. So we've discovered something. We've discovered that there are two-by-two representations of the rotation group.

Yeah. The determinant of the U dagger is the same as the... No, it's a complex conjugate of it if it's not one. If U is not one, it's a complex conjugate. But if you have the equation U dagger U equals I, then

ignoring the determinant condition... It's the determinant of U times the determinant of U dagger is one. That's just the determinant of U dagger. And it has to be one or minus one? No. It has to be, it can be E to the I theta and E to the minus I theta. Okay. But now we have fixed it.

We have fixed it so that the determinant is equal to one. All right. The group... All right. Definition. The group of N-by-N unitary matrices is called U-N. A special case is one-by-one unitary matrices.

That's U-1. That's that thing up there. All right. The group of unitary matrices with determinant one is called the special unitary group. S-U-N. S-U-N.

Special unitary group. It plays an enormous role in physics. S-U-N. Special unitary N-by-N matrices. Okay. So, this group here which acts on the two components of a spinner is S-U-2.

That's S-U-2. Let's go one step further before we turn to quarks. One step further. Is everybody happy with this? I was just wondering where the spin matrices that you discussed last semester would come in.

The spin matrices. The sigmas. You're talking about the sigmas. S-X commuted with S-Y equals I-S. Yeah. Yeah. Okay. So, yeah. That is a good question. I will tell you. I'll tell you right now then. Okay. Good. Let's do that right now.

There were the matrices S-1, 2, and 3. And they were actually just half the Pauli matrices.

I'll assume you remember what the Pauli matrices are. Okay. Think about group elements which are very close to the identity. That means rotations by very small angles. Rotations by very, very small angles.

The operators or the matrices should be very close to the unit operator. So, we should be able to write those, the special case of small rotations, we should be able to write them as having unitary matrices which are close to one. One plus something small.

And then to indicate small, let's put an epsilon there. Epsilon times some matrix. Epsilon times some matrix. Let's first ask what is the condition that, yeah, now we

want to know what's the condition that U is unitary? All right. So, let's find out. Let's multiply U times U dagger. U dagger. This is not V. This is U dagger. Let's multiply it by U dagger. One plus epsilon M dagger.

The Hermitian conjugate of one is just one. I'm taking epsilon, well, let's see. We'll take epsilon to be real for the moment. Let it be. It's just a small number. It's just a small number to indicate smallness. Anything complex can go into these M's here.

And this is going to be one plus M dagger. And this should be equal to one. All right. When we multiply this out, I'll only keep things to linear order in epsilon. We're going to drop things to quadratic order in epsilon. And that says that epsilon M plus M dagger is equal to zero.

We have one times one over here, which cancels this one. Quadratic things we don't care about. We have epsilon M plus M dagger is equal to zero, since the one's canceled. In other words, what it says is that whatever M is, it should be minus its Hermitian conjugate.

Now, this is called anti-Hermitian. The thing which has this property is called anti-Hermitian. Can we make something Hermitian out of something that's anti-Hermitian?

Just multiply it by I. Let's think about putting an I in here. Let's go from the beginning and put an I in here. What do we do over here? Minus I. Hermitian conjugation always involves complex conjugation, among other things. If we did that, then we would get M minus M dagger equals zero, because of this relative minus sign.

And then we would find that M equals M dagger. In other words, if we incorporated an I into the definition here, then M would be Hermitian. M is a Hermitian two-by-two matrix, right? It's a Hermitian two-by-two matrix.

How many Hermitian two-by-two matrices are there? How do you make a Hermitian two-by-two matrix? Oh, something else. What about the determinant? What's the condition that the determinant is equal to one? This is a little trickier.

Yeah, what is the condition? Oh, oh, oh, I know. Let's just write this in the following way. It's one plus I epsilon M minus M dagger, and then order epsilon squared.

So let's forget order epsilon squared. This is the determinant of one plus a small matrix. Do you know what the determinant of one plus a small matrix is? Anybody know?

It is one plus a small number. The question is, what is a small number? E to something. Could you approximate that? Well, you could. You could do that. It's easier than that, though. Yeah, let's do it. One, one, and then a small matrix, plus M1, plus M12, M21, and one plus M22.

I haven't put the epsilon in. Okay, but we'll just treat M as a small thing.

Let M be a small thing. Alright, what's the determinant? The determinant is this times this minus this times this. This times this is quadratic in small numbers. It doesn't matter. M11 times M22 is quadratic in small numbers, but there is a piece here

which is one times M22 plus M11 times one. What is that? That's M11 plus M22. Trace. The trace of M. So, the determinant of one plus a small matrix is basically just a trace of the small matrix.

So, yeah, the bottom line is the M's have to be traceless. Yeah, the bottom line is the M's have to be traceless. That we can prove. So, we have to construct a Hermitian small matrix which is traceless.

That's exactly what these sigmas are. They are the traceless two-by-two Hermitian matrices. Any traceless two-by-two Hermitian matrix can be written as a sum of sigma matrices.

So, these M's here are simply linear combinations of the sigmas. The particular linear combination that you choose will depend on the axis of rotation. If you want to make a small rotation about the X axis, you'll put sigma X here.

If you want to make a small rotation about the Y axis, you put sigma Y there and so forth. If you want to make a small rotation about the linear combination of the X and Y axes, in other words, the 45 degrees, you'll put the linear combination. That's the connection between the sigmas and the rotation matrices. They're the representations of the small rotations.

Okay, the last thing is how to combine. You know what? I think I'm going to wait until next week to talk about combining representations and just jump to quarks. Let's get off the mathematics for a moment or two.

We only have a moment or two. So, let's talk about quarks. Now, we're not interested in the spin of quarks. Well, we are interested in the spin of quarks. It's been a half, but that's not what we're talking about now.

I'm interested in the color of quarks. And the color of quarks is described by a three-state system, red, green, or blue. So, if I had a single quark, that quark could be in one of three states. Just as a spin could be in one of two states,

the quark color can either be red, green, or blue. And so, we could describe the state of a quark by three entries. The three entries will correspond to red, green, and blue. But, you know, an alpha one, an alpha two, an alpha three. And the squares of alphas will just be the probability

that it's a red quark, a green quark, or a blue quark. And what are the possible symmetries? Oh, and another thing we can say. Let me just go one step further and say we could also think of the three fields.

The field operator for a red quark, the field operator for a green quark, and the field operator for a blue quark, and assemble them also in a vector like this. What kind of symmetries can we imagine among the three quarks?

We can imagine, first of all, certain discrete symmetries. We can imagine the discrete symmetries which interchange them, interchange red and green, or interchange green and blue, or so forth. But we can imagine a larger family that includes that. Incidentally, these rotation matrices here happen to include the matrix one-one.

This matrix here is a special unitary matrix. So, among these rotation matrices is a matrix which interchanges up and down. This means that the interchange group is a subgroup

of the full rotation group. Happens to be a rotation group. I think we have to put an I in there some ways, don't we? Yeah, we have to put an I into a... Yeah.

Yeah, yeah. Um, that's right. Yeah, to give a determinant one. That's right. So, the permutation of two things is a special case of rotations. It's not...

Alright. Is there anything more complicated that we could do? Yeah, there is. The more complicated thing we can imagine is unitary... No, it's not obvious that there is such a symmetry in nature. Well, I will tell you there is such a symmetry in nature. And in fact, it's the whole point of having color here

is that it adds another symmetry to physics, important symmetry, that symmetry will play a very important role in the properties of quantum chromodynamics, but the symmetry is the multiplication of the three components, red, green and blue,

by special unitary three by three matrices. Alright, same mathematics has appeared here, except the mathematics involves three by three matrices. I won't fill in all the components, but, you know, it's U11, dot, dot, dot, dot, dot, dot, dot, dot.

Three by three matrices. Special unitary matrices, determinant one again. The special, that's called a special, that's called SU3. This is the group SU3.

Now, I'll tell you right now from the start, it is believed, in fact, it's mathematically absolutely required that physics is invariant under this SU3 operation. In other words, it's a kind of mixing up of the quarks, which, of the colors of the quarks, which is analogous to the mixing up of the

components of up and down, which is also analogous to the mixing up of the directions, the three directions of space. But this is not ordinary rotations. This is special unitary matrices. How many special unitary, let's count the number of special unitary matrices. Let's count the number of parameters. How many parameters to describe a special unitary

three by three matrix? Okay. Right. 18 minus nine minus one. Okay, why 18? You've got nine elements, nine complex elements, that's 18. Okay. You have the equations U dagger U

equals one. There are nine such equations, one for each entry in the matrix here. So that's 18 minus nine, and then we set the determinant equal to one. 18 minus nine minus one is eight independent, eight, it's an eight parameter group. Just like the rotation group is a three parameter group.

So there's an eight parameter family of rotations. I bet you're wondering what that eight has to do with the fact that there are eight gluons, aren't you? Yeah, it's the same fact. Yeah, it's the same fact. But we will come to it. Why aren't there nine gluons?

Because we threw away the determinant. That doesn't make any sense at all. But it's true anyway. Yep, it is true. Yeah, so special unitary three by three matrices are a particular representation of SU3,

but SU3 is the symmetry group, the basic symmetry group of quantum chromodynamics. And just like quantum electrodynamics was invariant under this operation here, and that told us that only in the Lagrangian only things with the same number

of psi's and psi daggers can enter, the idea that the Lagrangian is invariant under SU3 also has profound consequences for the structure, for the symmetries, for the conservation laws. And so the rule is every possible term in the Lagrangian of quantum chromodynamics

of quark physics should be invariant under this group here. We'll come back and we'll start to talk about how you construct quantum chromodynamics a little bit. I think we're probably finished for tonight. I'm certainly finished. I had intended to go a little bit further tonight

and tell you a little more, but I think that was quite a lot. And I don't know how many of you followed, how many of you didn't follow, but I hope you at least get the flavor, if not the color of what the... Okay. We'll go a little bit further with this, but then we'll get back

to some things that can be understood without too much mathematics. For more, please visit us at stanford.edu.