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HONG LIU: OK, let us start.
So first, let me remind you what we did
at the end of the last lecture.
So we can see there's a large N matrix field theory.
And then we saw that when you write down your Feynman
diagrams to calculate things, then
there's a difference now between the order of contractions.
Which is now refracted whether you
have a Planar diagram or non-planar diagram, et cetera.
And that in turn, also affects your end counting.
So at the end, we discussed two observations.
One observation we said, is for the example to be considered is
the non-planar diagram, even though it cannot be drawn
on the plane without crossing lines,
can actually be drawn on the torus without crossing lines.
So the diagram can actually be straightened out on the torus.
And another observation, is that the power of N
is related to the number of faces you have in your diagram
after you have straightened it out.
So now, I'm going to generalize these two observations.
So first, I will tell you a fact.
Many of you may already know this.
So I'll first tell you a fact that any orientable two
dimensional surface is classified topologically
by integer, which I will call h.
So this h is called a genus.
So heuristically, this genus is equal to the number of holes
a surface has.
So for example, if you have a plane,
the plane topologically is actually
equivalent to a sphere, so when I
draw a sphere, that means a plane,
because if you identify the point
on the plane of the infinity, then it becomes a sphere.
So topologically, a plane is no different from a sphere.
So this genus 0.
So this is plane.
So h is equal to 0.
And then for torus, then there's one hole,
and this is just genus 1.
So this is torus.
You can also draw surfaces with as many holes as you want.
So this is surface with two holes.
A surface with two holes, and so this is genius 2.
So the remarkable thing is that this actually
classifies the topology of all two-dimensional surfaces.
And this so-called topological invariant,
there's a topological number a so-called Euler number,
which is defined to be Chi is equal to 2 minus 2h.
So if h labels the topology, and so the Chi
is related to h in this way.
So any two surfaces with the same topology
will have the same Chi, because if they have the same h,
they have the same Chi.
So Chi is what we call the topological invariants.
So this is just a mathematical fact.
So now I'm going to make two claims.
So these two claims are in some sense,
still evident after you have studied a little bit.
I'm not going to prove it here, will just make the claim,
and then I will leave the task to familiarize these two
claims to yourself.
But as I said, these two claims are actually self-evident,
if you just do a little bit of studying.
So claim one-- so this is regarding
the structural Feynman diagrams.
So remember, the example theory we conceded last time
was something like this, which phi is [INAUDIBLE] matrix.
So if you can look at the Feynman diagrams
of the theories, the claim is the following.
for any non-planar diagram, there
exists integer h so that the diagram
can be straightened out.
Straightened out just means non-crossing on a genus h
surface, but not on a surface with a smaller genus.
So actually, you can classify all the non-planar diagrams
by drawing it on a genus h surface
on two-dimensional surfaces with non-trivial topology.
And then you should be able to find the number h, which
is the lowest genus you need to make this diagram
to be non-crossing, instead of crossing.
Last time, we saw an example which you can stretch it out
on the torus, but it was a more complicated diagram
and torus would not be enough.
You would need, say, a genus 2 surface or higher genus
surface.
So you can easily convince yourself
that this is doable just by some practice.
Also, this is very reasonable.
Do you have any questions regarding this claim?
Good.
So this was claim one.
The claim two-- and this is a generalization
of the second observation.
It's a claim of the generalization
of the first observation of last time,
and claim two will be the generalization
of the second observation.
For any diagram, the power of n coming from contracting
propagators is given by the number of faces
on such a genus h surface.
As we explained last time, the number of faces
just means the number of these connected regions
in the diagram.
But if you are stretching it out,
then there's an ambiguous way you
can count the number of disconnected regions
In the diagram.
And that number is the power of n coming from the contracting
propagators.
So this claim is also self-evident
because each power of n comes from a single connected line.
And essentially, a single connected line
will be circles some face and so the number of independents
inside the face of course, are just the number of power n.
Any questions regarding the second claim?
Good.
So based on these two claims, now immediately write down
the independents.
So from this, we can find a vacuum diagram.
Again, we have been only talking about the vacuum diagram.
Vacuum diagram means the diagram has no external lags,
has the full independence.
So from here, we immediately conclude
that the vacuum diagram has the following, g squared
and dependence.
So g squared is just the carpeting constant.
So remember, each propagator is proportional to g squared
and that each vertex is one over g squared.
Just from there.
So let's just take an arbitrary vacuum diagram.
Let me call this amplitude to be A. And A should
be proportional to g squared to the power
E. E should be the number of propagators,
because each propagator gives you a factor of g squared.
And also minus V. And V is the number
of vertices below to the number of vertices.
And each vertex gives you one over g squared,
so that should give you E minus V.
And then we just multiply N to the power of F,
and F is the number of faces.
So without doing any calculation,
so this has essentially characterized the N and the g
and the accompanying dependents of any diagrams.
Now, if you look at this expression,
you say we are doomed, because this kind of expression
does not have a sensible [INAUDIBLE]
So remember our goal, the original goal of the 't Hooft
is that you treat this N as a parameter
and then you want to take it N to be large
and then to it's ranking, one over N. Doing expansion in one
over N means you are expanding around the including infinity.
You want to expand around the including infinity.
But if you look at this expression
and then go to infinity, it's not a well-defined limit,
because I can draw a [INAUDIBLE] Feynman diagram, which
F can be as large as possible.
So you just have sufficiently many vertices,
and sufficient to be mainly propogators, F is unbounded.
So then, there's another well-defined angle
to infinity limits.
So this expression is not a sensible N of infinity limit.
If you don't have a sensible N infinity limit,
then you cannot talk about doing a longer expansion,
and you cannot even define the limit we just finished.
Yes?
AUDIENCE: I remember people do partial sum to-- can
they do the same thing here?
Put the large in the denominator pattern from series
of singular bubbles and--
HONG LIU: Yeah.
AUDIENCE: Like a random face approximation--
HONG LIU: Yes.
AUDIENCE: Somehow they can infinite a sum
over an infinite number of A, and put the margin
into like 1 minus N something for the denominator--
HONG LIU: Yeah.
So people certainly have been talking about expansion.
And certainly, if this just failed here,
we will not be talking about this.
I'm just setting up a target, which
I'm going to shoot it down.
Right.
Yeah, I would just say if you do the margin expansion,
this probably would be the place which would turn you back.
You say, ah, there's no margin limits,
and that's from another problem.
But hopefully, that's no ordinary person.
And so remember, there are several nice mathematical
tricks you have to go through.
First, you have to invent the stop line notation so that you
can count the end very easily.
Yeah, first you have to come up with this margin idea.
I think maybe not due to him, maybe other people
already considered similar things like right here.
But first you have to think about the double line notation
to make your computation easier, and then even after you reach
the staff and then you need to know this kind of topology,
et cetera.
After you reach this staff, still, there
is a roadblock here.
But 't Hooft found a very simple way to go around it.
Because when I say this limit it is not well-defined,
I made the assumption that when I go to infinity limit,
the g is kept fixed.
So that's the reason that this will blow up.
But then 't Hooft came up with a different image.
He said when you can see that N goes infinity,
but at the same time, you can see
that g squared goes to zero.
At the same time, you can see that g squared goes to zero.
Because the problem with this, is
that if you have a diagram with lots of faces
and if you multiply by a finite number,
if you take g equal to 0, then you have infinity
multiply something potentially goes to 0,
maybe you will get something finite.
One second.
So you said let's consider this limit,
so that the product of them is kept finite.
Let's consider this limit.
So in this limit, the end counting will be different.
So this A will be g squared N, E minus l, because we
want to keep this fixed.
So we put an N factor here, then we take the N factor out.
N plus F plus V minus E. And then
let me just write this slightly differently.
So this is Lambda.
If I use this notation, Lambda now is finite.
So this is E minus V. So let me just
write it as Lambda L minus 1 N to the power of Chi.
Now, let me explain [INAUDIBLE].
First, L is equal to E minus V plus 1,
is the number of loops in the diagram.
So do you guys remember this formula, why this is true?
The reason this is true is very easy to understand.
So if you look at the Feynman diagram, the number of loops
is exactly the same as the number
of undetermined momentum.
And each propagator will carry momentum, and at each vertex,
you have momentum conservation.
But then this overall momentum conservation,
which is guaranteed, so the number of independent momentum
is E minus V and then plus 1.
Yeah, it's just E minus V minus 1,
because there's only V minus 1 independent momentum
constraints.
So this is the number of independent momentums,
and so this is the number of loops.
And the Chi is just defined to be the number of faces
plus the number of vertices minus the number of edge.
Yes?
AUDIENCE: Is there any-- in general, is there
a relation between E, V, and F, like are they completely
independent to each other?
HONG LIU: E, V, and F?
AUDIENCE: Yeah.
HONG LIU: Yeah, that I'm going to explain.
Any other questions?
AUDIENCE: So the reason that you can take the F
to infinity limits without taking it
to 0, is that why we can't apply this to QCD to get a--
HONG LIU: No, because we apply this.
AUDIENCE: But if g is not small?
HONG LIU: It doesn't matter.
So this is an expansion scheme.
And you can apply it to QCD, then
you can ask whether this particular expansion scheme is
a good approximation or not.
That's a separate question.
And the focus is pretty good, it's not that bad.
Yeah, focus of each squaring is finite.
AUDIENCE: But it's [INAUDIBLE].
HONG LIU: No, here it does not tell you
if lambda has to be small.
Lambda can be very big here.
It just tells you the lambda in this limit has to be finite.
Lambda can be convening and it can be very big.
And actually, in the future, we will take lambda
to go to infinity.
So now, in order to have a well-defined limit
still depends on this chi.
So now, you ask why this works because we still have this chi,
but now again, we need another piece of mathematics.
First, when you draw, let me remind
you some diagram we had last time,
so this is the simplest diagram.
And there's a lot of non-planar diagrams
which you can draw on the torus, which
is the non-planar version of this, et cetera.
And you can also have more complicated diagrams.
So suppose you are on the torus, I
can consider more complicated diagrams, like that.
For example, such diagrams.
So if you think about such diagrams, then in a sense,
each Feynman diagram can be considered
as a partition of the surface.
So if you draw a diagram of the surface into polygons.
Yeah, it's very clear here.
I drew this now with wavy lines, just make them straight.
And this one topologically, I can just draw like this,
and then I have one, one, and the other part, similarly here.
So each Feynman diagram can be concealed
if you just partition whatever surface you draw the diagram.
Is this clear to you?
So this is a very important point.
Yes?
AUDIENCE: Can you repeat that?
HONG LIU: Yes.
Look at this diagram.
So this is a Feynman diagram drawn on the torus.
Does this look like a partition of torus into polygons?
Yes, so that's what I'm talking about.
And the statement that this would apply for any Feynman
diagrams.
Yes, so this is like a partition of a sphere.
You separate a sphere into three regions, one, two, three.
Any questions about this thing?
Yes, do you have a question?
AUDIENCE: Yeah, in that case the third one isn't abandoned,
is it?
The third--
HONG LIU: So this one I am going to show you,
you should view the plane as a sphere.
AUDIENCE: Oh, OK.
HONG LIU: Topologically, it's the same as a sphere.
So now, once you recognize this, now you
can use a famous theorem due to Euler.
Some of you could have learned it in junior in high school,
because the-- so this theorem, given a surface composed
of polygons-- so if you're not familiar with this,
you can go to Wikipedia-- with F faces, E edges, and V vertices.
Suppose you have a surface like this, and then
this particular combination chi, which
is defined to be F plus V minus E is precisely equal
to 2 minus h.
So this is the-- this is why this is called Euler's number.
And they only depend on the topology.
So this combination only depends on the topology of the surface
and nothing else.
So we can imagine 't Hooft a very good high school student.
So he already knew this.
So now, we can rewrite this thing
as A, reading script A lambda to the power L minus 1 N
to the power 2 minus 2h.
So remember h is greater than zero.
So now this expression has a well-defined limit.
So now this has a well-defined analogy and limit.
So to the leading order, and in particular,
this N dependence only depends on the topology of the diagram.
Only depend on the topology of the diagram.
For example, to leading order in large N,
then the leading order will be given by the planar diagrams.
Because that should be the diagrams with h
equal zero because h is not negative.
So the leading term is given by h zero.
So all of the planar diagrams in this limit,
in this so-called 't Hooft limit,
will have N dependence which is in N squared.
N squared.
So now if you go back to your loads,
above the fourth diagram we started last time,
above the two planar diagrams we started last time,
you can immediately tell that indeed it
is N squared in this limit.
Then you still haven't explained in terms of lambda.
So you can do Feynman diagrams, et cetera.
Say lambda just depends on the number of loops.
So if you have one loop, then start it
with a lambda equal to zero, start it with some constant.
If it's two loops, then lambda, three loops, lambda squared,
et cetera.
So the sum of four-- sum of all planar
diagrams, we have this structure.
So you can imagine you can sum all this,
if you're powerful enough.
And then you can write it as N squared, then
some function of lambda.
So the planar diagram would be just N squared time
some function of lambda.
So if you are powerful enough to compute this f0 lambda exactly,
then you can say you have solved the planar.
You have solved the large N limit of this series.
Unfortunately for lambda being in gauge theory we cannot do
that.
We don't know how to compute this.
We can on only compute perturbatively which
actually does not work for QCD.
Yes?
AUDIENCE: But N going to be infinite.
It just seems like-- why does this overcome the same problem
that we had before?
HONG LIU: Because this is a specific power.
Here there's no specific power.
That F can be as large as you want.
AUDIENCE: But wasn't the other problem also
that N was going to infinity?
HONG LIU: No that's another problem.
Because here, there is a specification limit, when
you take N going to infinity.
In there, there's no specific limit.
There's no limit when you take N go to infinity.
No, the limit is N squared.
I can say-- this is your essentially,
your vacuum energy, right?
I can say E0, divided by N squared
has a well defined limit.
So the key is that this is a specific dependence on N.
But there it's unbounded.
AUDIENCE: I guess another question
is, we didn't assume that lambda is a small prohibiter.
HONG LIU: No we did not assume lambda.
AUDIENCE: So isn't this kind of perturbative analysis,
saying that that series converges to some function,
is that OK?
HONG LIU: It's a very good question, and it is OK.
But the reasoning is more complicated.
The reasoning is more complicated.
The keys are falling.
What I'm writing here you can understand
using two perspectives.
Just one is heuristic.
Just one say, suppose you can only do perturbation series,
and that would be the thing you compute.
And then say, if you are powerful enough
to solve the theory with perturbity,
then you're guaranteed just to find some general functions.
Yeah but whether this series actually
converges is an important mathematical question,
of course, one has to answer.
It turns out, actually this is convergent.
You can mathematically prove this is convergent,
for very simple reasons.
Yeah this is a side remark, but let me mention it
because this is a cool fact.
So if you do just lambda of g 5, 4 series,
just do the standard Feynman diagram
calculation, input the basic series,
the series is divergent.
The series is divergent, no matter how
g is small, just as the radius of convergence
is 0 for any non-zero value.
Yeah the radius of convergence is 0 in terms of g,
no matter how small g is.
The reason that it's divergent is
because the number of diagrams.
So when you go to Nth order, it increases exponentially in N.
So the coefficient of g-- say to the power
N-- the coefficient can become huge.
Because of the number of diagrams
to increase exponentially in N. So you think,
I have g to the power N times N factorial, something like that.
And that series is not convergent.
But the thing that is convergent here,
because can see the planar diagram, the number of planar
diagram is very, very small.
It only increases with N as a power of N,
rather than as a factorial of N. And so you
can make lambda small enough, then this can be convergent.
Yeah.
AUDIENCE: Didn't we already see at the end of the last lecture
that only planar diagrams keep the biggest contribution, so
why is that equal to--
HONG LIU: Sorry?
AUDIENCE: So at the end of the last lecture, we saw that--
HONG LIU: No not necessarily.
That is only for the two diagrams.
If you look at this thing, I can draw arbitrary complicated
non-planar diagrams, with a very large F. They only
depend on F. It does not depend on anything else,
if you do this.
OK so in general, then of course if you
include the higher non-planar diagram, et cetera--
so in general the vacuum energy, which you would normally
call log z.
OK log z is, essentially, the sum of all vacuum diagrams.
So this is the partition functions,
so it's a path integral.
So log z, then we'll have the expansion from h
equal to 0 to infinity, N two the power 2
minus 2h, F h lambda.
OK so at each genus level, you will
have some function of lambda.
Yeah the leading, order we just showed is f0.
And then if you add the Taurus now--
we'll add to the torus now our on non-planar diagram.
It's order one.
And then, for the two genus, it's
f over n squared, et cetera.
So let me just write down z explicitly.
Z is the partition function, i S phi.
So if you compute this, with the right boundary condition--
if you compute this path integral,
with the proper boundary conditions,
then that gives you the vacuum diagrams.
The log z is the sum of all connected vacuum diagrams.
I should say, the sum of all connected vacuum diagrams.
Any questions regarding this?
Good.
Now there is a heuristic way we can understand this expansion.
So it's actually a heuristic way to understand
this one-way expansion.
So let's just look at this path integral.
So let's look at the Lagrangian.
So the Lagrangian I wrote there.
In this 't Hooft limit, I can write it
as minus N divided by lambda.
So I want to write things in terms of lambda.
So I multiply the pre-factor N downstairs, and then upstairs.
So g squared N give me lambda.
And then I have a trace, et cetera.
So now it's easy to see that you're leading order,
this things should give you order N squared.
Because there is already a factor of N here, and the trace
is the sum of N things.
So generically, this should be of the order N squared.
So now we-- a little bit leap of faith--
say supposing the large N limit, since the leading order
is order N squared, you can argue that actually all the N
squared is the expansion parameter,
if you do scatter point approximation.
And then naturally, you will see the power will
be given by 1 over N squared.
any questions regarding this?
Yes?
AUDIENCE: What does N over lambda factor to?
HONG LIU: It's--
AUDIENCE: OK.
HONG LIU: So clearly this discussion
actually does not-- so when I say clearly here,
it requires a little bit of practice.
But clearly, our discussion only depend
on the matrix nature of the Lagrangian and the fields.
So I'm going to make a claim.
It says for any Lagrangian, of matrix valued fields
of the form L, which is N divided by some coupling
constant, times a trace of something.
Doesn't matter what you put inside the trace
here, as far as you have a single trace.
Such a series will always have the expansion like this.
It will always have an expansion like this.
So let me just summarize.
In the 't Hooft limit-- by 't Hooft limit,
I always mean this form-- the coupling constant is defined
such that you have some coupling concept here,
and then you have over N factor.
And then of course, you can also have some coupling constant
inside here.
It doesn't matter.
As far as those coupling constants are independent of N.
And as far as things inside the trace are independent of N.
So for such a kind of series, the 1 over N expansion
is equal to the topological expansion.
It's the expansion in terms of topology of Feynman diagrams.
So this is a very, very beautiful--
and as I said, it's [INAUDIBLE], because in principle, it
puts a very simple structure into something that's,
in principle, very complicated.
Yeah, any questions regarding this?
Yes?
AUDIENCE: Maybe it's not can I just understand it
in such a way that he kind of treats the Feynman
diagram as a triangulation of different spaces.
HONG LIU: Yeah for example, you can think of from that point.
Yeah, so we use that to derive, to use this formula.
Yeah and that picture will be very useful in a little bit
from now on.
Keep that picture in mind.
The Feynman diagram is like the partition of some surfaces.
And that will be very useful later.
AUDIENCE: Does there ever arise a situation
in which you care not about two surfaces,
but Feynman diagrams on three surfaces or something
like that?
Because this asks the question, you
don't necessarily have to consider
the topology of two surfaces.
Are there any situations in which it's more complicated?
HONG LIU: Yeah but we always draw Feynman diagrams
on the paper, which is two dimensional.
Yeah it's enough.
Two dimensions is enough.
You don't need to go to three dimensions.
Yeah and this structure only comes
when you go to two dimensions, because if you
go to three dimension, of course, they don't cross.
In three dimension, you can no longer
distinguish planar or non-planar diagram.
AUDIENCE: Well if you did-- OK.
HONG LIU: Good any other questions?
AUDIENCE: Why is it always orientable surfaces?
HONG LIU: That's a good question.
It's because those lines are orientable because when
we draw the double roation, so you
have this two, double rotation.
So essentially those lines are orientable.
They have a direction.
And essentially, this is, of course, one and two.
Yeah so those Feynman diagrams actually have a direction,
have a sense of orientation.
So I'm going to mention, by passing in nature,
but let me just mention also now.
So if it's not [INAUDIBLE] matrix,
say if it's a real symmetric matrix, then the then
there's no difference between two index.
And then there's no orientation.
And that would be related to m orientable surfaces.
And then you need to slightly generalize this.
Any other questions?
AUDIENCE: Now you mentioned that it
has something to do with string theory,
but does that it has anything to do with scatter particles--
HONG LIU: Yeah, so we're going to talk about that.
No, we're going to talk about that.
Good?
So now let me talk about general observables.
I think we're a little bit short on time,
if we want to reach the punchline today.
So right now we only have looked at the vacuum diagrams.
So now let's look at the general observables.
Before talking about general observables,
let me just again make a side remark,
Which is the gauge versus global symmetries.
So in the example we talk about here--
let me call this a, equation a.
Then let me write down another equation
b, which is a Yang-Mills theory.
So the difference between a and b-- so a is this guy.
So the difference between a and b,
is that a, as we discussed last time,
is invariant on the global symmetry,
is a U(N) global symmetry.
It's that phi is invariant under the acting of a unitary matrix.
But this U must be constant.
Only for a constant U is this a symmetry.
But if you have studied a little bit of gauge theory,
or if you have not studied gauge theory,
just take my word for it, the b is
invariant under a local symmetry, a local U(N)
symmetry.
He said A miu-- so A mu is what make up
the F-- U x A mu x, U dagger x minus i partial mu U x.
It doesn't matter.
The only thing I want to say is that this U x is arbitrary.
It can have arbitrary space time dependence.
Just like of the generalization of the QED,
it's the gauge symmetry.
So the key difference between the two.
the key difference between this local and the global symmetry,
are manifested in what kind of observables we can see there.
For example, for a--
AUDIENCE: I think that should be U dagger, partial mu and then
U dagger.
HONG LIU: HONG LIU: So this difference,
you can say what's the big deal?
in one case, this is constant, and this
is dependent on space time.
So the key difference between them
is that in the case for a, operator like this phi squared,
phi is a matrix.
So phi squared is a matrix.
Phi squared x is an allowed operator.
So this operator is not invariant under
this global U(N) symmetry.
But it doesn't matter because this is a global symmetry.
So this is an allowed operator.
But if you have gauge symmetry, all operators
must be gauge invariant.
That means that all operator must
be invariant under this kind of transformations.
So the analog of this is not allows operators.
So observables in the gauge theory are much more limited.
So we will be interested gauge theories.
We will be interested in gauge theories.
So that means we are always interested in observables,
which are invariant under the symmetry.
So we're interested in gauge theories.
So that means we're always interested in gauge invariant
operators.
So the kind of Lagrangian does not matter.
So you can have the gauge fields.
You can also have some other field,
say some matrix phi, et cetera.
As phi is the Lagrangian of this form, it's OK.
We always only consider the Lagrangian of that form.
But it doesn't have an arbitrary number of fields,
and with arbitrary kind of interactions.
So you start your theory.
So let's for simplicity, this can see the local operators.
In this kind of theory, then allowed, say local operators,
must always have some kind of trace in it.
Say you must have some form trace, F mu U, F mu U,
a trace phi squared, et cetera.
A trace phi to some power F N, phi to some power k, et cetera.
You can also have operators with more than one
trace, say trace phi squared, trace F mu U, F mu U.
So we are going to make a distinction
because the operator with only a single trace,
we call them single trace operator.
And the operator with more than one trace,
we call them multiple trace operators.
OK so the reason for this distinction will be clear soon.
So multiple trace operators-- it's self-evident again--
that multiple trace operators can always
be written as products of single trace operators.
AUDIENCE: I have a question.
Is it a possible case to have a local gauge invariant operator,
the F mu U times F mu U?
HONG LIU: Yeah I always can see the local gauge invariant
operators.
We can see the gauge theories.
AUDIENCE: So this combination, F mu U times F mu U, it's only--
HONG LIU: No this is gauge invariant.
AUDIENCE: Is that the only gauge invariant component?
HONG LIU: Sorry?
AUDIENCE: Is this is the only gauge invariant
component that we can use to construct the gauge
invariant local operators?
HONG LIU: Sorry, I don't quite understand what you mean.
No this is just one specific operator.
No, you can take F to the power N, an arbitrary number of-- as
far as they're inside the trace, it's always gauge invariant.
AUDIENCE: I see.
HONG LIU: I'm just writing down a particular example.
So just for notational simplicity,
I will just write-- so from now on I will just
denote the single trace operators collectively just
as O with some script n, which denotes
the different operators.
so n denotes different operators.
And then for the multiple trace operator, then you have O1, O2.
That would be a double trace operator.
And O1, O2, O3, say times O3 would
be a triple trace operator.
n just labels different operators.
I'm just using abstract notation.
And the reason for this-- a distinction will be clear soon.
Then for such gauge theories, the general observables,
in the quantum field theory is just correlation functions
of gauge invariant operators.
So by gauge invariant operators, you
can have local operators, non-local operators, et cetera.
So for simplicity, I will restrict my discussion only
on the local operators.
But local operators means that the fields
evaluated at a single point.
So the typical correlation functions,
then the typical observables will have this form,
will be just a product of some correlation functions,
a product of some operators, and I say their correlation
functions.
By c I mean the connected correlation functions.
So you can see immediately, these multi
trace operators are just the product
of a single trace operator.
And the correlation function of a multiple trace operator
can be obtained from those of a single trace one.
You just identify some of the acts.
Then that will be enough.
So we only need to talk about the correlation function
of single trace operators.
So now the question-- let me call this equation one.
So now the question follows what we discussed before,
is how do we decide the N dependence of the guy?
Previously we determined the N dependence
of this vacuum diagrams of this petition function.
But now want should be the N dependence
of general correlation functions?
One way to do it, you just start immediately calculating.
And then you can find some root, et cetera.
But actually there's a very simple trick.
There's a very simple trick to determine
the N dependence of this.
So now I will explain.
Now I will tell you.
So I will not give you any examples
because this trick is so nice, and it just works very easy.
And so the question, what is N dependence of one?
So this is the question we want to address.
so here is a very simple, beautiful, trick.
So let's consider the following generating functional.
So in quantum field theory, when we
talk about correlation functions,
it's always convenient to talk about the generating
functional.
So whatever is your field, you do the path integral
of all fields.
And then you look at the action.
So you have your regional action, which I call S0.
And then that's add those operators.
Ji x, O i x.
Yes so this is a standard story.
When you take the derivative over Ji,
then you will bring down a factor of O i.
Then that essentially give you the correlation function.
You have, for example, a correlation function,
the connected correlation function.
O n, the connected correlation function
would be just you take the derivative of log z.
And then delta J1 x delta Jn xn.
And then you set J equal to zero.
You set all the J equal to zero.
And that gives you the end point function.
I should write i here, so i to the power n.
So now here is the beautiful trick.
You can determine this in a single shot, the N
dependence of this guy.
And this simple trick is just to add N here.
You add N here.
In order to get the correlation function,
we need to divide it by N to the power N. Now you
wanted to get O i, you need to divide-- take the root of N
times Ji, so you also need to divide it by 1 over N.
But why does this help?
Why does this help?
It helps for the following reason.
Let me call this whole thing iS effective.
iS effective, so the key thing is
that this O i, single trace operators,
then this S effective then has the form
N times the trace something.
Because you already know S0, which is our starting point,
has this form.
And now the term you added in, precisely, also has this form,
is the N factor times something single trace.
So that means the whole thing still has this form.
Then now we can immediately conclude
this log z J1, Jn, must have this expansion.
h from 0 to infinity, and to the 2 minus 2h, f
h lambda J1, et cetera.
So adding that N is a powerful, powerful trick.
So now you can immediately, just from here,
we can immediately find out that for endpoint function,
connected endpoint function, the leading order is 2 minus n,
because the leading order is n squared.
Yeah it's 2 minus n.
And then suppressed by 1 over N squared, et cetera.
Good?
So for example, if you look at a 0 point function, which
is essentially the partition function, so this is order
N squared.
So this is what we found before, to leading order.
And if you look at the one point function,
some operator, then, will be order
N. If you look at the two point function, connected two point
function of some operator, so it would be order one.
And the three point function of some operator
will be order 1 over N, as the leading order.
And then all higher order just down by 1 over N
squared, compared to the leading order.
And again, the leading order is given by the planar diagrams.
Because of the leading order contribution to here,
in terms of this S effective is planar diagram.
And then they must be, under those things,
just obtained by through [INAUDIBLE], so they
must be planar diagrams.
so again this comes from planar diagrams.
Good?
So now let's talk about the physical implications of this.
So what does this mean?
So we have found out, this is our N dependence for our gauge
invariant operators.
So what does this mean?
Now let me talk about physical implications.
AUDIENCE: What did one of those [INAUDIBLE] have with--
HONG LIU: Yeah, I defined them without chi.
AUDIENCE: Then something like--
HONG LIU: You define something.
So when you write down your theory,
you define this 't Hooft limit.
Then everything is already in terms of lambda
or some other order one number.
So that can depend on those numbers in an arbitrary way.
It doesn't matter can depend on coupling constants
in an arbitrary, but it cannot have N dependence defined
inside the operator.
Once we introduce this 't Hooft limit,
then the operator can depend on the coupling in the 't Hooft
limit in an arbitrary way.
Because they're all just all the one constant.
Let's talk about physical implications of this.
It turns out, these simple and [INAUDIBLE] behavior
actually has a very simple physical picture behind it.
So first he said, in the large N limit-- so let's just
look at leading order behavior.
So in a large N limit, if we consider this state of O
i x acting on the vacuum.
So some single trace operator acting on the vacuum.
So i, again, just labels different state,
different operators.
These can be interpreted as creating a "single particle"
state.
I'm first describing the conclusion.
Then I will explain why.
So you can see that the state obtained
by adding a single trace operator on the vacuum,
then this can be interpreted as a single particle state.
So if you add on the double trace operator on the vacuum,
then this can be interpreted as two particle state.
Similarly, say O n acting on the vacuum
would be N particle states.
So why is this so?
Why this is so.
So I will support this statement using three arguments.
First remember O i O j, the connecting Green function
of any two operators of order one.
So we can actually just diagonalize them.
If you can just diagonalize them, so that O i O j
are proportional to delta i j.
So in some sense, a two point function,
those operators can be considered as independent.
And now the second statement.
So if you want to call this single particle,
this two particle, multi particle, then they
should be that they don't overlap.
Because a single particle cannot overlap with two particle,
et cetera.
And then to see the overlap between a single particle
with a multi particle, you look at these correlation functions.
So if you look at the overlap of a single particle state,
with a two particle state, say some double trace operator.
Let me just, to avoid confusion, let
me just use the inside-- use this notation
to see this as a single operator.
So you can see that the overlap we
saw of the single trace operator with this double trace
operator.
And they start off-- this whole thing
is like a three point function.
Just put these two over the same point.
So from our discussion here, is how you connect the Green
functions of order 1 over N.
So this goes to 0, compared to this overlap with itself.
So that means in the N goes to infinity limit,
there is no mixing between what we called single particle
state, single trace, and the multiple trace states.
So the third thing is that now let's
look at the two point function of two double trace operator.
So O1 O2 x, say O1 O2 and y.
OK double trace operators.
So let's look at the two point function
with double trace operators.
So there's an even contribution to this.
So let's include all diagrams.
Also these can all be connected So leading order
is a disconnected diagram, which is essentially
O1 x O1 y and O2 x and O2 y.
Because I have a diagonal next to them.
have And then plus the connected Green
functions, which is order 1 over N squared.
So if you see this leading order contribution,
it's just like two independent particle propagating.
Just like the product of two independent propagators.
So it's sensible to interpret this
as just the propagating of two particles.
So it's sensible to interpret this two particle
state, this double trace operator
as just creating some two particle state.
Yeah, again this goes to zero, in the large N limit.
So I should emphasize when we call this a single particle
state, it's not necessary they really
exist on a shell particle corresponding to this state.
We're just saying that the behavior of these states
can be interpreted as some kind of single particles.
The behavior you can just interpret as single particles.
It's not really necessary they exist
as stable on shell particles.
In certain cases, there might be.
There might be actual particles, actual stable,
on shell particles associated with these kind of states.
But for this interpretation to be true,
it does not have to be.
So in QCD actually sometimes-- so I just say they exist.
So we call them "glueball" state.
For example, in QCD, the analog of this kind of operator
can create some state, which they are short-lived.
They are not long-lived.
They quickly decay.
And so they're typically called the glueball state.
So from now on we just call them glueballs.
Call them glueballs.
So this is one of the first implication--
the first indication is that they're just
a single particles.
A single trace operator can be interpreted
as creating single particle states,
and the multiple trace operator can
be interpreted as creating multi particle states.
Another second, he said if this glueball operators, so
fluctuations of glueballs suppressed.
So let me explain what this means.
So let me suppose some single trace operator, O,
which has a non-zero expectation value,
suppose some state has a non-zero expectation value.
And then let's look at the variance of this operator,
the variance of the expectation value, which
is given by O squared, minus O squared for the fluctuation.
And this is, by definition, just a collected Green function of O
squared.
The is the full O squared.
This is the disconnected one.
So this is just a collected part of the O squared.
And this one, N dependence, we know
this is order one, order N to the power 0.
So that means that these to below this order
N. So the variance of this, compared to the expectation
value of this operator itself is surprising, the large N limit.
So essentially, in the large N limit,
so that means the variance provided by the operator
itself is 1 over N because of the 0 in the large N limit.
So assuming that if you have a two point function-- so
suppose each two point function, so each O2
have a non-zero expectation value,
then you can factorize this O1 O2, plus O1 O2.
So this is disconnected part.
And again this will be of order one.
But this part is of order N squared.
So essentially the disconnected the part is always factorized.
So the disconnected part is always factorized.
Yeah this connected part is always small,
compared to the disconnected part.
So this is like a classical theory.
AUDIENCE: We have just five minutes.
We can proceed before it's 5:00
HONG LIU: Yeah but it may go to 10 minutes.
It's just hard to say.
Yeah, let me do it next time.