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PROFESSOR: All right.
So where were we last time?
So we had derived the Fermion Lagrangian
which is at the top of the board,
and really the only difference between this
and what we talked about earlier is that we had to perform
the multipole expansion.
So we distinguish between collinear momenta that
are scaling like a collinear momentum and momentum that
are scaling like an ultrasoft, and only the collinear pieces
were showing up in here, and same with the collinear gauge
field.
It was only showing up in here, whereas both were showing up
in here.
There's only a single type of derivative
here, because ultrasoft and collinear
momenta are the same size in the end component.
But they're different in these other components,
so we had these labels in this label operator
that picked out the collinear momenta.
So we could define collinear covariate derivatives,
if you like, with this subscript n,
whereas this D is like a full derivative that involves
both types of gauge field.
And then at the end of lecture, it was kind of rushed,
but I was talking about the collinear gluon Lagrangian,
and I said there is a set of replacement rules
that we could make to effectively do the same thing
that we did here.
And now I've just written out for you
with that Lagrangian is with those replacement rules.
So curly D is basically just taking all the derivatives
and taking only the leading order pieces.
So we take the collinear pieces in this component
and this component.
We take the ultrasoft pieces in this component.
And then if you wrote the original Lagrangian
as a commutator of two covariant derivatives,
you'd just replace it by the leading order pieces,
and that's what the leading order action will be.
Then, we had to think about gauge fixing.
Since this is the collinear gluon Lagrangian,
we should think about collinear gauge fixing.
This here as a general covariant gauge
fixing with parameter tau, and then there's
a corresponding ghost term.
And in this Lagrangian, the usual way it would look,
this would be i partial, and then
I said that, because we don't want the collinear gauge fixing
term to break ultrasoft gauge invariance,
we're going to turn that i partial into
a covariant derivative under--
we're going to include this piece here
at the n dot a ultrasoft to make it ultrasoft gauge
invariant at lowest order.
OK?
So that's really the only--
other than just doing this most naive thing by just replacing
derivatives by covariant derivatives,
and you might think, well, I'll just keep
this as a partial derivative.
Since it's to doing gauge fixing,
I don't need to make it covariant,
but we do want to make it covariant under the ultrasofts,
and that's why I write this curly D ultrasoft.
OK.
So that gives you the leading order Lagrangian.
Once you put together this with what
you want for the ultrasoft, then you
would have the full leading order Lagrangian,
and the ultrasoft part is actually very simple.
So for the ultrasoft part, we just
take a full QCD action for the ultrasoft field.
So this is just q ultrasoft bar i
D slash ultrasoft q ultrasoft.
And likewise for the gluon piece, it's just QCD,
and we would do the gauge fixing without thinking
about any complications, like these ones,
and these are just involving ultrasoft fields, so just
QCD, just ultrasoft fields.
OK?
So the only place--
and you take everything here together--
the only place that the ultrasoft and collinear fields
talk to each other in the Lagrangian
is in this single component n dot partial, n dot a ultrasoft.
And that comes about basically because of the power counting,
that this is really the only place that these two things can
interact.
And we'll talk about the implications of that later on,
but that's the leading order Lagrangian.
So once you have this piece, and this piece is pretty straight
forward with this additional complication
of worrying about what gauge symmetry means, which
we'll talk more about later.
But we had to be careful not to break it,
when we introduced this term, because we wanted
to in some sense have a gauge fixing both for the collinear
gluon and a separate gauge fixing for the ultrasoft gluon,
and that's why we wanted to do this.
And then this piece here is simple.
It's just QCD, because it's, if you'd like,
it's the lowest energy mode, and it
doesn't know about any of the complications
that we had for the collinear modes.
OK?
So any questions about this so far?
OK.
So everything that we did in deriving these actions
is tree level.
All the steps that we did were tree level.
So you can ask, if I start to do loops,
will there be some Wilson coefficient
that shows up somewhere here?
Will I generate some new operators
that I don't see here?
Those are reasonable questions, and that's
what we're going to address next.
So to go further, we'll use symmetries,
and we're actually going to consider
three different symmetries, gauge symmetry, which I've
been promising you for a while, reparameterization invariance,
and spin symmetry.
Where I'll put a question mark by this last one,
because we need to answer the question whether there even
is a spin symmetry.
These two here, this number one and number two,
will turn out to be quite important.
Number three is not so important.
So reparameterization invariance here
will be like reparameterization invariance in HQET,
except now it's different.
We've introduced the parameters and then n bar,
and we'll have to see what kind of symmetries
we have with respect to that choice of basis factors
that we made.
But it otherwise will be analogous to our discussion
of HQET.
So let's actually first dispense with the one that in some sense
is the least important, this number three.
So first, let's revisit our spinners a little bit.
So if I put together the information
that we talked about when we derived
the equation at the top of the board there for the Lagrangian,
we worked out at a tree level we have this formula that
relates the fields.
So from that formula, if we just project onto the spinner
pieces, we can write down a formula
that relates the spinners.
So throwing away the gauge fields,
we have in momentum space the u of p, the spinner,
would be related to whatever spinner
we have for this cn field which I call Un by that formula.
And then if you take this formula,
you also see that if I hit it with a projector,
n slash n bar slash over 4, if I hit the u with a projector,
it's going to kill this piece.
Because the n slash n bar slash can be pushed through the p
perp slash, then n bar squared is 0,
so that kills that second piece.
So we also have this formula.
And then once you have this formula,
you have the formulas that we wanted for that.
OK?
But this actually, this spinner here,
this Un is not exactly the same as the spinner
that we talked about earlier.
So let me come back.
Let me come to that in a minute.
So first thing you might consider
is whether, when I take this cn field,
and I take the Lagrangian up at the top of the board,
do I just get the collinear propagator that we
talked about?
And indeed, you do.
It's kind of obvious for the momentum-dependent parts,
and really you might only worry about the spin.
And so if you consider this formula,
and you consider the sum over spins of u u bar which
is what's going to appear in the numerator of the propagator
for the Fermion when you're driving the Fermion propagator,
you get a sum over the physical spins.
Then, from this formula, this is a projector on something
you know how to do the spin sum for which is the full theory
spinner.
And that spin sum is p slash, so this
is like p slash sandwiched between projectors.
And you could work out that that's exactly the numerator
that we had before after a little bit of Dirac algebra.
So that part works as expected.
If we take this Lagrangian, and we work out
what the propagator is, we get exactly the propagator
we got from expansion.
So quantizing lc0 gives us that propagator.
But the situation is not quite the same for the spinner.
And in some sense, this is not--
this point is not absolutely crucial,
but actually, there's a little simplification
I want to do in order to discuss the spin symmetry,
and in order to do that at this point,
it's important to understand this point.
So that I'm going through this in some sense,
because then it'll be very easy to discuss what spin symmetry
we have in the theory.
So this guy is actually not equal to our expanded spinner
which putting in some normalization I
could write like this.
So this is what we got by expanding,
something like this which is very simple where this u is
equal to 1, 0 or 0, 1.
in the Dirac representation.
But that's actually not what we get
if we just take the formula up here and use this.
So let's see what we do get.
So if you use the formula up there,
then you have the following.
So here's the full theory spinner
with a conventional normalization.
So this thing and this thing here is the projector.
OK?
So you could work out what that product is, and it turns out
that you can write it in the following way which
is very closely related but not precisely the same as what
we had before.
So you can write it in what looks like the same form
as we have over here, except this curly U, curly capital U,
is a more complicated object.
And it's just whatever I get by multiplying these two
things out which turns out to be something
I can write in that form.
So it's some two-component spinner,
but it's got momentum dependence, unlike our simple
1, 0 and 0,1.
But everything we said about Un really depended only
on the fact that it could be written in this form in terms
of some two-component spinner--
the fact that n slash on it was 0,
the fact that it had projection relation.
So these formulas here, if you have a formula like these,
these formulas here are true.
OK?
So actually, it would be true whatever spinner we have there.
So why should I want this U twiddle spinner
and rather than the U spinner?
That actually has to do with this reparameterization
invariant, so it'll become clear when we talk about that.
But these extra terms in U relative to those
for the other guy, the simple U, actually
ensure the proper structure under reparameterizations.
And basically, it'll become clear in a moment,
but basically, if we wanted to get this U spinner,
we should have a slightly different projector,
which I'll call Pn prime.
So we could have used a different projector which
is this one, and then we could have come up
with another projector which was the Pn bar projector which
would satisfy that the sum is 1, if we wanted.
And this projector here, when acting on the full theory
field, would have given something
that would have been proportional
to this combination over here.
OK?
So you just have to believe me.
I don't want to go through the algebra,
or you can check it yourself.
But this projector here is not invariant under that symmetry
of reparameterization variance.
When we talk about RPI, it'll be clear
why we want a projector which is this projector and not
the slightly different projector which
has this extra n slash over 2.
OK.
But nevertheless, the important point
that I wanted to emphasize is really
that we have this way of decomposing
the spinner, the true spinner, for our field cn
in terms of a two-component object, U twiddle.
OK?
So we are able to do that.
And if you want to talk about spin symmetry of the theory,
it's easier if you use a two-component notation.
So the reason that I wanted to go through this
is really to have on the board this equation,
which I can then use to motivate writing down
to two-component version of cn.
And once I have a two-component version of SCET,
then it's very easy to see what the spin symmetry is.
So let's write down a two-component version
of our collinear quark Lagrangian.
You can do the same thing, of course, in HQET right
down the two-component version rather than
a four-component version.
If you have a four-component version,
it has this projection relation.
If you have the two-component version,
the projection relation is built in.
The reason to consider the four-component version
is if you want a couple this object
to four-component fields, like the ultrasoft field,
then it's, of course, a nice thing
to have a four-component version,
but some things are easier in two components.
So take cn, and write it as follows, where
this phi n has two components.
And I've set things up, so the dimensions of cn
are equal to the dimensions of phi n.
OK?
So I can take that formula, plug it into our SCET Lagrangian,
and then I can, using the Dirac representation for the gamma
matrices, write out a Lagrangian for this phi n.
So that requires doing a bit of algebra,
which I will take you through.
And then we get an equivalent Lagrangian
but in terms of this field, and it looks as follows.
OK.
So it's almost independent of spin but not quite.
There's this sigma 3 sitting there.
If sigma 3 wasn't there, it'd be like HQET,
where you have an SU2 symmetry.
The fact that sigma 3 is there means
you don't have an SU2 symmetry.
And really, all you have is a U1 symmetry,
and that U1 really corresponds just to helicity.
So in four-component notation, that U1 would be the following.
It's projecting a spin operator onto perpendicular indices
anti-symmetric in both of them, and in the two components,
that just becomes a sigma 3.
So obviously, if we do an exponential rotation
with respect to sigma 3, sigma 3 commutes with sigma 3,
so does the identity.
And so we could have a rotation of this guy by that,
and that's the only spin symmetry
that you have is this helicity.
And so because of the coordinates we're using,
this corresponds to the spin along the direction of motion
which is the three direction, if you like,
which sometimes we denote by just saying
it's along the direction n which is then more independent of how
we pick our axes.
So this symmetry here is actually
related to what you would call the chirality in QCD.
So it's not really a new symmetry.
So this is just related to the chiral rotation.
So if you look at gamma 5 times c and gamma 5
in our representation would be 0, 1, 1 0.
And then if I write it out in terms
of this two-component thing, then that
is just giving me 1 over root 2, and it's swapping up and down.
And so that just means that phi n has gone to sigma 3 phi n.
So multiplying by gamma 5 is actually the same
as multiplying by sigma 3 in the two-component notation.
OK?
So this is not really new symmetry, and actually
all the usual things that you would
say about chiral rotations in QCD would apply here too.
So chiral rotations, of course, are not
exact chiral rotations are broken by masses.
Chiral rotations are broken by non-perturbative effects.
You can worry about anomalies.
They are useful in perturbation theory for quantifying
operators, and that remains true here,
but if the collinear fields were non-perturbative,
then you should worry about those other things as well.
OK.
So spin symmetry, there's not really anything new
there to talk about, but along the way,
we saw we could write SCET in two-component form
which is kind of nice.
So let's talk about something that's more important which
is gauge symmetry.
Is there any questions about the spin before we talk about
gauge symmetry?
AUDIENCE: So the extra term you got in determining the theta n
bar, that is why you had the minus n slash [INAUDIBLE]
and why you don't have the full SU2?
PROFESSOR: Yeah.
I looked at it once, but it's hard for me
to remember the answer.
I think not, but I think if you do the other version,
then you just get something more complicated here
but still breaks SU2.
Yeah.
AUDIENCE: Is it--
I'm just looking for [INAUDIBLE]..
PROFESSOR: So they're two disconnected fact.
The thing that I wanted to motivate
that I could use this formula, and I wanted
to be honest with you about where that came from,
and that's why I told you this other fact.
But I could have jumped right here
and said, remember, and glossed over it,
and that would have been fine for this discussion.
So they're in some sense two disconnected things, at least
in my mind.
AUDIENCE: Do you have some intuition
for what theta n bar is?
PROFESSOR: For theta n bar?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Oh.
Yeah.
It's really just saying--
my intuition for it is really just
that, when you're producing energetic particles,
it's what we did at the beginning.
When you're producing energetic particles,
these are the spin components that you have at lowest order,
and that leads to, if you like--
the fact that you have this form,
and you have this projection relation
is kind of non-trivial, even though you
don't have a spin symmetry.
There is something to it.
It's not like a SU2 symmetry of the spin,
but it does, for example, when you go and look
at form factors, it does lead to non-trivial relations
for those form factors.
So I don't call it a symmetry.
I don't think of it as a symmetry,
because it's not a formal group theory statement of Lagrangian,
but I think it's basically--
well, OK.
Let me not speculate any further.
It caused some confusion in the literature actually as well.
All right.
So what about gauge symmetry?
So we're doing a non-abelian gauge theory.
A gauge transformation is something like this.
And in order to just talk about gauge symmetry, what
we need to do is we need to talk about the gauge symmetries that
are related to the two type of gluons that we have.
We have a collinear gluon.
We have an ultrasoft gluon.
You know that the field acting like the connection
for that gauge symmetry.
So there should be, in some sense,
a gauge symmetry associated to the ultrasoft gluon
and a separate one associated with collinear gluon,
if the meaning of these things as gauge fields
is going to have any sense to it.
So rather than have just this set of gauge transformations
that we can make in QCD, we should
have some more complicated structure of gauge symmetry
in the effective theory, because we
have a more complicated structure of gauge fields.
So in order to talk about U's, we're
going to want to divide them into camps.
And the useful thing to think about
is that you want to have U's that, first of all,
leave you inside the effective theory.
And really what this means is that you
want talking about gauge symmetry
to commute with your discussion of talking
about power counting.
So by way of example of this, say
you had a gauge transformation that was large,
that made a formula like that one, that i partial on it
was scaling like q.
OK?
So this would be a very large momentum associated
to that spacetime dependence.
Then, if you talked about the gauge transformation which
would be U times c, even if this guy has collinear momentum
scaling, the new guy would not because of this fact.
OK?
So if we want to talk about gauge symmetries,
we want to talk about symmetries that take us to the fields
that we've already decided to have a certain power counting.
And we don't want the gauge symmetry to mess
that up, so we're going to demand that that's true.
So given that, we can think about dividing up the gauge
symmetry into pieces, into pieces
that have different scaling in terms
of how big the transformation are,
how the coordinates are behaving.
And it's convenient for this discussion
to divide them into three parts.
A global transformation, where this guy is not
spacetime dependent, a collinear transformation,
which I'll call Uc, where i partial on Uc
scales like a collinear momentum,
and then an ultrasoft U Us, where the i partial scales like
an ultrasoft momentum.
And basically, what we want to do
is we want to connect this gauge transformation to the gauge
field for the collinears and this transformation
to the gauge field for the softs.
OK?
So one complication that we have to deal with
is this momentum labels that we were talking about.
Because the way that we distinguish
between these types of momenta and these types of momenta
was dividing things into these labels and residuals.
Gauge transformations are something
that's very nice in position space but not so nice
in momentum space, and that's because gauge transformations
are local transformations.
Local means local in x not in P. So
what is simple multiplication in position space
becomes convolutions in momentum space.
So if you go over to this hybrid notation that we developed,
then you would have the following.
At this field, if I do a transformation,
it goes into something like that.
Where the way that you should think of what this sum over q
is in the following way.
But if you have a transformation in position space
that looks like that, then in a momentum space,
that simple product is convolution.
OK?
So this is one dimension, and you'd have that correspondence.
And so this complication that we have here just correspond
to the complication that you'd have for any gauge
symmetry that you want to write in momentum space.
But we can be slick about this and introduce a notation
and then dispense with it, because it's really
not a technical complication.
It's just a notational complication.
So let's do that in the following way.
Let's let Uc which is a matrix with two indices
be defined by this Uc of bl minus ql.
So I'll say define a matrix.
Then we can use a matrix notation for the convolutions,
where we sum over repeated indices,
and that takes care of this notational complication.
So what is meant by this is that the Pl, ql entry of this matrix
is a number, and that number is given by this thing
that we call Uc Pl minus ql.
OK.
So then the formula up here could just
be written in terms of that matrix
as a kind of matrix multiplication.
OK.
So what are the transformations once we adopt that?
Let me write them down and then talk about them.
So cn of x goes to Uc of x cn of x, where you should understand
this thing is a matrix that has two indices, like this,
also dependent on spacetime.
And I just dot over the repeated indices in this thing which
is like a vector.
So this is like a matrix, and this like a vector.
That makes it look like it's the usual notation which
is the point of introducing this notation.
So for the collinear gauge field,
it's going to be basically that it,
as we said over here, that it becomes the gauge
field of that transformation, and so that's
what this would be.
That would be the standard gauge transformation,
and then there's one little complication here, which is I
wrote curly D ultrasoft, the ultrasoft derivative,
the thing that has the n dot a ultrasoft field in it.
So the reason that I wanted to do that
is because I'm thinking here of the ultrasoft field.
When I'm thinking about collinear fields,
I'm thinking of the ultrasoft field as a background field.
And if I was to do a background field gauge transformation,
then I would have a covariant derivative
here under the background.
So this n dot a ultrasoft is like a background field
to the collinear gauge field.
So this, if you want to make an analogy of what this is like,
it's like the quantum gauge transformation of a field that
has a background.
If you look up how background field gauge
works and background field transformations,
you'd have the same thing.
You'd have something covariant under the background,
and that's what we're going to do here as well.
Yep?
AUDIENCE: Kind of a silly question.
So UC of x is a matrix.
So what exactly UC of p [INAUDIBLE]??
PROFESSOR: Well, no.
This is just a number.
So just think of this as like this is some vector,
and this is some other vector.
The difference of these are some label on this thing.
Right?
And for each value of this, I get a different number,
and then for each value it's based on,
I get a the different number, but this thing
is just a number.
And I'm saying think of it like a matrix in the sense of,
because we are thinking of these as discrete--
AUDIENCE: But isn't UC [INAUDIBLE]??
PROFESSOR: Oh, yeah.
Yeah.
That's another matrix.
It's an orthogonal matrix.
It's also a matrix in that space.
Yeah.
So here, I'm defining a matrix in a new space which
is this momentum space, and that's just really
my way of making transformations in momentum space look
like as simple as they do in position space
by just having a matrix notation for the convolutions.
AUDIENCE: So that number is essentially a matrix.
PROFESSOR: It's a matrix, yeah, in that space.
Good point.
What about the ultrasoft fields?
Well, ultrasoft fields shouldn't transform under collinear gauge
transformations because of exactly the same logic
that I gave you a minute ago about why I shouldn't
be able to inject a hard momentum like this one
into the effective theory.
If I want my ultrasoft fields to remain ultrasoft,
then I can't let them transform, because that would
spoil their power counting.
So q ultrasoft under this Uc transformation
just goes to q ultrasoft, and A ultrasoft
has to go to A ultrasoft.
So these things should not be touched by that transformation.
AUDIENCE: So why cannot make the ultrasoft and the collinear
talk to the--
PROFESSOR: Well, we don't want that.
You could think of trying to develop effective theory, where
they would mix.
Right?
But whenever you have a Lagrangian that
mixes two fields, the first thing you try to do
is to force them not to mix.
So it's better, say I set the theory up so that they mixed.
The first thing I would try to do to that Lagrangian
is make field redefinitions to go
to some form of the Lagrangian that wouldn't mix.
OK?
So I prefer, if you like, to develop the one that
doesn't mix right from the start,
and you could view it that way.
So you could always think of I'm going
to tell you like two sets of transformations.
You could always think of formulating linear combinations
of those transformations.
Say these are the fundamental ones, not the ones
you wrote down but these other ones.
Right?
And then by some manipulations, one
could go to the ones I'm talking about.
All right.
So now, the other set of transformations that we want
are these ultrasoft ones, and for that type
of transformation, we don't have the same type of problem
with the collinear.
We can't say the collinear fields don't transform,
because that type of transformation
doesn't spoil the collinear field being collinear.
We can always add ultrasoft momentum to it.
So these fields cn and An will transform
like background fields--
sorry, like quantum fields in a background.
So from the point of view of the cn fields
which are the shorter distance fields,
you can really think of like what
we're doing here is exactly like what
you would do if you had quantum and background gauge
transformations.
And the physics of thinking of the background as a longer
wavelength mode is exactly the right way
of thinking about how these fields which
are shorter wavelength think about the ultrasoft fields.
So what that means is that they transform
like matter fields in the appropriate representation.
OK?
So with that in hand, we can write down
how things transform under the ultrasofts.
A Fermion just transforms like that on the left,
and this, unlike our notation over there,
this is one number for all entries in the vector.
This guy doesn't have any indices.
OK?
So this is like transforming all the terms in the vector
by the same overall color matrix, overall number
in the momentum matrix base.
So this guy should transform like an adjoint,
and you can read the adjoint as a U on each side.
So there's two ways of writing the transformation here.
You could write it as like this.
If you wrote down an adjoint transformation,
then this guy would have two indices A and B,
and you'd have a transformation like that,
where this is in the adjoint.
But if you want to write it in terms of the U ultrasoft
that's e to the ita, which is in the fundamental,
then you write it like this, and those are equivalent.
So why are they equivalent?
So they're equivalent because--
well, OK.
Maybe I'll tell you later.
Yeah?
AUDIENCE: I'm always confused about why n dot is a background
field to A ultrasoft, because it actually
has the same size of momentum, so its wavelength
is the same size.
PROFESSOR: The wavelength is not the same size.
The component's power counting is the same size,
but the n dot An field carries momentum which have
P squared which is much larger.
AUDIENCE: Right.
P squared [INAUDIBLE] wavelength, right?
PROFESSOR: So I'm talking about the wavelength which
is related to that P squared.
So you think about the n dot An field carries a large P minus
momentum still.
Right?
Even though it has a small--
it carries in order a 1 P minus momentum.
So if it's confusing, think about just the P
minus momentum.
The n dot An field, right, has P minus, which is
the what are lambda to the 0.
So just let's think about wavelength with respect
to P minus.
AUDIENCE: That's definitely true, yeah.
PROFESSOR: Right.
Whereas the n dot A ultrasoft field carries P minus, which
is what are lambda squared, so just thinking about those two.
AUDIENCE: You were saying conflating [INAUDIBLE]
PROFESSOR: Yeah.
Yeah.
All right.
We have room here to squeeze the final transformation.
So the final transformations are just what you'd expect,
that the ultrasoft fields transform as regular gauge
fields.
So let me just squeeze that in.
All right.
So there's the regular transformations
that you'd have for a gauge field.
Yeah?
AUDIENCE: So I just want to clarify.
These two analogies for background field
transformations, they're not disconnected.
Right?
PROFESSOR: They're not disconnected
in the sense-- that's right.
Yeah.
So if you take the background field notation,
and you just really think about that,
then this is how you would transform the quantum
field into the background gauge transformation.
And this is how you would transform the quantum field
under the quantum gauge transformation.
Yeah.
That's right.
Yeah.
AUDIENCE: So even if you made these to separate the gauge
transformation for just [INAUDIBLE]
they still have the same quantum numbers?
PROFESSOR: Yeah.
That's right.
AUDIENCE: Are they going to meet at some level?
PROFESSOR: We're going to make sure they don't.
AUDIENCE: OK.
PROFESSOR: Yeah.
In some sense, we really want-- the picture is this one
that we had earlier, where we really
want the collinear fields to live up here,
so we had these hyperbolas.
Right?
We want this guy to live up here and this guy
to live down there in momentum space.
So this was P minus.
This is P plus.
So we want them to live in different places.
They have the same quantum numbers,
except we want them in momentum space
to live in different places and to describe
different degrees of freedom.
Right?
So everything that we did here is exactly related
to this picture, that we want to talk about this guy
as having larger momentum, talk about gauge transformations
that shuffle things around with components
that have large momentum versus small momentum.
All right.
So we're going to hypothesize that these gauge
transformations are the fundamental symmetry
of the effective theory.
And I tried to motivate them physically
by this analogy with background field gauge
and thinking about a longer wavelength mode with respect
to a shorter wavelength mode which is in this picture
exactly the fact that this hyperbola is
higher than this one.
Maybe the hyperbola should stay on the board.
OK?
So they multipole, where the multipole expansion
that we did between these fields and the statement
that this is a longer wavelength mode
is exactly the fact that there's a separation
between these guys.
So we'll take these to be the fundamental transformations
that do not get corrected by power corrections.
And I'm not claiming that that's necessarily
a unique way of getting to the effective theory.
You could think of there might be
an example in the literature of where people were thinking
of other gauge transformations.
These are the ones that people talk about now,
but in the early days, I think, there
is an example in the literature, where
people are thinking of other transformations,
and you could prove that they could
be connected to these ones.
But we'll just take the ones that
are the most useful from the start
and not worry too much about what we could have done.
All right.
So there's no mixing in the power counting,
and that's something that we like.
Set up that way.
If there was a mixing, if we connected things
that were different sizes in lambda,
then you could say, well, that the power counting,
that the gauge transformation induces power suppressed terms.
And you could end up with some situation
where you're trying to connect things at different orders,
and we don't have that here.
And that's because, if we want to think
about this thing as a symmetry, it should really
be a symmetry of the leading order action
that shouldn't require some other sub-leading terms
necessarily to compensate for it, at least
for a gauge symmetry like this.
We'd like to think that what the meaning of gauge symmetry is
and the redundancy encoded in gauge symmetry
is all something that you're talking
about at the order of the leading order action.
And you're not talking about redundancies
in fields in sub-leading order, et cetera.
OK.
So let's do an example, now that we
have this, of how it comes in.
And I want to come back to our example
of our heavy-to-light current and talk about gauge symmetry
there.
If you just took the tree level result
that we had in that case, then we
had something that just involved these two guys,
and this hv field was ultrasoft.
This guy was collinear.
So if you did a Uc transformation,
then under our rules, what you'd find is that the cn transforms,
and this guy doesn't, the h doesn't.
So that would not be gauge invariant.
But actually, we saw that this also wasn't the lowest order
current.
There was this additional thing, the Wilson line showed up,
and that Wilson line is going to fix this issue.
It's going to make it gauge invariant.
So we have to ask, how do Wilson lines transform
under gauge symmetries?
And that Wilson line in position space,
I called it something like this.
I can't remember if I used a twiddle or a bar.
We'll use a bar here for the position space Wilson line.
So in general, in just a regular gauge theory,
the Wilson line between two points
transforms at the endpoints.
So you get U of x.
Then, you get a U dagger of y.
So if you like, what we would say
for this Wilson line is that we have a U of x on one side
and a U dagger of minus infinity on the other side.
Minus infinity is really long wavelength physics,
and so we have to worry a little bit
about the overlap between our different types of gauge
transformations.
I divided them into three camps-- collinear, ultrasoft,
and the global, so let me do the following.
To avoid some double counting with what I called U global,
let's just simply take, since U global acts the same
everywhere, let's just simply take Uc dagger at minus
infinity to be 1.
So this is at some particular spacetime point.
We take it to be 1, and this U global
will transform at that point.
We then know there's no overlap between Uc and U global.
OK?
Then, that's enough to ensure that.
So with that, our guy under a collinear gauge transformation
just transforms on one side.
If you like, what's happening is that the coordinate here
corresponds to this very large momentum,
and we're stretching ourselves out to long distance
physics which is corresponding to the smaller momentum.
So when I think about transformations
that should be collinear, I want them to be associated to the x.
And I want to associate other types of transformations
to what's going on at infinity which is really
a spacetime of its own, which is this ultrasoft spacetime that's
sitting at the collinear infinity.
So that's why I want, actually, it to look like this.
AUDIENCE: [INAUDIBLE] just with the collinear fields?
PROFESSOR: Yeah.
AUDIENCE: OK.
PROFESSOR: Yeah.
So this guy here is a path-ordered exponential,
and it just involved the n bar guy.
Yeah.
So that's what it was in position space,
and in momentum space, they called it W,
and it was a sum over a bunch of things.
I'm not going to--
we had this formula.
Let me not write it all out again.
As a short form for that formula,
we can write the following formula
which is given our notation that we've
developed a useful way of thinking about it.
So if we formally expand out this exponential,
and we just keep the order of that expansion, where the P
bar acts on all the fields to the right, that will correctly
reproduce this denominator, the sum over permutations
is something that we still have to do.
It was 1 over n factorial, k factorial in here.
M factorial with this notation.
OK?
But this is like a little shorthand
for that messy formula.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah?
AUDIENCE: I think it's related to the gauge transformation
equation before.
PROFESSOR: Yeah.
AUDIENCE: How do you get the i epsilon prescription?
[INAUDIBLE]
PROFESSOR: So that actually is related to the fact that,
when you look at the--
for this Wilson line with the i epsilon,
it's not related to the gauge transformation at infinity,
but it's related to the fact that the path goes that way.
So that when you do the Fourier integral,
the fact that it goes from minus infinity to x will give the i0.
OK?
But for these Wilson lines here, if you
think about what's happening at the place
where the i0 would actually matter,
that's exactly the 0 bin.
So it's going to be a little more--
Yeah.
So you don't really care.
AUDIENCE: OK.
PROFESSOR: Yeah.
I can only say that--
we'll explain what I just said to Ilia more
when we come to it later on.
OK.
So this field here has--
I said, it's a momentum space, but then I wrote x.
So really what I mean by momentum space
is momentum is in the label momentum space.
There's some guy, and the labels are these labels on the fields
here.
Right?
So there's still an x, and that x corresponds
to the residual coordinate that we were using
to do the multipole expansion.
So in some sense, this isn't simply the Fourier transform
of that, because of that.
It's still knows about the separation
between the large momentum and the soft momentum
from that multipole expansion.
So this dependence on x allows us
to think about the multipole expansion on these Wilson
lines.
It encodes all the residual momenta,
and the fields in the Wilson line carry a residual momenta.
So we're just putting our notation to use here.
And if you really strictly want to connect to that formula
up there, then you would take the residual momenta to be 0.
And then the Fourier transform with respect
to just the large momentum Pl minus
gives a line which is just depending
on that one single coordinate.
So if we just had a coordinate which
is conjugate to the large momenta, which
is the thing that's showing up here
are the large momenta, then that would be the analog of what
we were talking about up here.
Where in general, this thing also
has a dependence on the residual x, which
we'll make use of in a minute.
OK.
So let's think about, so I told you
how this Wilson line transforms under Uc see.
How does it transform under U ultrasoft?
So under Uc, again using this matrix-type notation,
it's just Uc.
So this is a matrix multiplication
in momentum space.
What about under an ultrasoft transformation?
Well, under an ultrasoft transformation,
this should be a derived--
in some sense, under a collinear transformation,
it was a derived quantity, because we knew
of the gauge field transform.
The thing that gave us the transformation of the Wilson
line is just the transformation of the gauge field.
OK?
So that's what led to, actually, this form, once we
put in this fact.
So under the ultrasoft transformation,
we just take the transformation that we have for the collinear
gauge field, and then we derive how
the Wilson line should transform under the ultrasoft gauge
symmetry.
OK?
So how does this guy transform?
This guy transforms by putting U's on the left
and the right that are at x.
But every field in this exponential
has the same, U U dagger, multiply it,
think about expanding that out.
All the U U daggers that are next to each other cancel.
You just get an overall U U dagger that comes right
outside the exponential.
So this guy is just transforming with U ultrasoft of x.
Wilson line back again, U ultrasoft dagger of x,
and that's because this Wilson line with our notation
is really a local thing with respect to the coordinate x.
All the fields sit at x.
OK?
So it just transforms like that, because that's
how the gauge field inside it transforms.
So in some sense, both of these are
derived from the transformation of An,
but it's easier to think about how a Wilson line transforms
and then think about how it should transform
under collinear than just go through the An route
and impose this condition.
OK.
So let's come back now to our full current which was this,
that had the Wilson line in it.
And now if you ask how it transforms under Uc,
you see what's going to happen.
You get Uc dagger, Uc, and the ultrasoft guy
doesn't transform, and then these cancel.
So it's invariant under the transformations
that have momenta of order of the ultrasoft.
It'll have a momentum of what are the linear scale.
There's two objects in this thing that have momenta
that can be of that size.
The gluons, they're in W in this collinear quark field.
Both of those things transform, but that transformation
cancels between the two things.
AUDIENCE: [INAUDIBLE]
PROFESSOR: Whoops, that's important.
Thank you.
And then under an ultrasoft transformation,
we can also transform this guy, and then everybody's
transforming.
Gamma, of course, has not got color indices,
and again, it's gauge invariant.
OK?
Missed a gamma.
Gamma's not so important.
All right?
So this current that we derived earlier
is actually invariant under two different types of symmetries
which are really momentum space splitting of a standard gauge
symmetry into pieces where we have a large momenta and pieces
where you have these larger momenta, collinear
momenta and pieces where we have a smaller momenta,
this ultrasoft momenta.
But both of these types of momenta
are momenta that we're allowing inside the effective theory.
Yeah?
AUDIENCE: Just trying to remember
that the ultrasoft scale was the same for the [INAUDIBLE]..
PROFESSOR: Yeah.
The momentum of the ultrasoft--
the ultrasoft momentum of the collinear
quark, you pulled out mv, and then
the remaining part was ultrasoft in the discussion
for this particular thing.
All right.
So it's the Wilson line, the moral of this
is that what we were doing before, where we're integrating
out those collinear gluons, is actually
connected to gauge symmetry.
Because without that Wilson line,
we wouldn't have a result that's invariant under the type
of gauge symmetry we've been talking about.
So this Wilson line carries these n collinear gluons,
and those gluons would be the gluons that in the full theory
would give you a gauge symmetric result. They would combine.
They would be the attachments to the heavy quark that
combined with attachments to the light quark
to give you a result that's invariant,
if you looked at word identities, for example.
But in the effective theory, these are just
put into a Wilson line, because that's the leading order
way that they can show up, but still we
have a gauge invariant answer.
So the Wilson line is really needed for the gauge symmetry.
And finally, the ultrasofts, as can be clear from the notation,
the ultrasofts can just as well include
the global transformation, if you like.
So we don't really need three.
We could put the global back with the ultrasofts.
Then, we just have two.
All right.
So that's our symmetry.
That's an example of how it works.
Now, let's talk about covariant derivatives.
So gauge symmetry ties together regular derivatives
with the gauge field, and here, what it ties together
at the following things.
So because of the treatment of the A ultrasoft
field as a background, it actually ties together
both of these things.
And then it does what you expect in a collinear sector.
It ties together the derivative with the gauge field.
And if you're acting on ultrasoft fields,
then there's an ultrasoft covariant derivative
which is just like that.
So now, you can ask the question if I just
had this gauge symmetry, which is going
to force me to use covariant derivatives-- which I was
already thinking ahead to when I wrote down
the Lagrangian earlier.
Right?
We already wrote it in terms of these kind of objects,
so it seems like we're good to go.
So you can ask the question if I just
have power counting, the cn that we
decided we're going to use, and gauge symmetry, is that enough?
Am I done?
Do I have everything I need for the leading order Lagrangian?
So power counting told us that we had this type of derivative,
and that we could have two of these perp derivatives.
OK?
So those are the two type of terms we were getting.
There's no other order lambda squared field structures that
would have the right dimensions and that could do the job.
So these are really the only two that we have to think about.
So you could try to think about maybe
I can have something with some additional n bars floating
around, but really, it boils down--
since you need to get a lambda squared,
you need either two perp derivatives.
You need two of them, because you still have a rotation
symmetry around the direction of motion,
or you could have an n dot D.
But there is one operator that is not rolled out,
by gauge symmetry alone, and that is this operator.
When we wrote the operator, we had D perp slash D perp slash.
But nothing stops me from doing something similar in some way
by just taking the D perps and contracting them, like that.
That's a different operator than the D perp slash D perp
operator that we got before.
Oops, only one of those.
OK?
So this operator is a different operator.
In a priori, you could say, well,
do loop connections generate this operator which
is different than the D perp slash D perp slash operator.
The answer is no, actually, but so far what we've done,
it's not enough to see that.
So we need actually this other symmetry that I alluded to,
the reparameterization invariance,
and that will actually rule out this guy.
So what's RPI?
Well, when we formulated this theory,
we needed a direction for the collinear line
or for the collinear particles.
And then we needed an auxiliary vector n bar,
and we formulated the whole set up
in terms of this n and n bar.
But having vectors like that that we write down,
that breaks Lorentz invariance in the same way that
specifying v mu in HQET breaks Lorentz invariance.
So if you say that m mu nu is the set of Lorentz
transformations, the usual six generators that
are anti-symmetric, then the ones that are broken
are these ones.
And there's five of them, and the one that's not broken
is this one which corresponds to rotations about the three,
axis with the axis specified by n.
OK?
So those rotations would act in the components of these guys
that are 0, if you like.
So there's no issue there, and these guys
are the guys that are connected to the pieces that
were non-0 in general.
So there's going to be a larger, because there's
five things here and because there's two vectors,
it's going to be a larger set of reparameterization symmetries
than in HQET.
First, talk about just the n and n bar themselves.
There's three types of reparameterization invariance
that would leave n squared equals 0
and bar squared equals 0 and n dot n bar equals 2.
And those were the formulas that we were using over and over
again really.
Everything else was just convention.
OK?
So what are those three types?
Well, we could take n, and we could
change it by some amount in the perp direction
and leave n bar unchanged.
And since n bar dot something perp is 0,
that would satisfy this, or we could do the opposite.
That would be type two.
So don't change n, and change n bar.
And so this is two transformations
there's two things specified by this perp guy and two here.
And then there's one more which is called three,
and this one let me write it like this.
It's a simultaneous transformation
of these two guys, where I just do a multiplicative factor.
Well, a multiplicative factor is not
going to change the fact that's the square of something 0.
The place where the normalization comes in
is this n dot n bar, and if I just rescale them
both by an opposite amount, then that remains satisfied as well.
So I formulated this in terms of an infinite--
I'm going to think about this is an infinitesimal transformation
for delta perp, infinitesimal for epsilon perp.
I could also expand this exponential
and think of it as infinitesimal,
but the finite one's easier.
So let's just think of the finite transformation
in the last one.
Now, this is an effective theory.
So whenever we think about transformations,
we should think about power counting.
We just spent a lot of time thinking
about that for gauge symmetry, and we should think
about power counting here too.
Did you have a question?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah.
So what I was just about to talk about, how big things can be.
So the power counting that's the right power counting
for these guys, so these things are infinitesimal or finite,
but they also have some counting in a different space which
is the power counting space.
And the right power counting for these guys is as follows.
So this means I can make arbitrarily
large transformations in a power counting sense of type
two and type three.
There's no constraint.
That's what this means, so that maybe that's easier to swallow.
This one, there is a constraint, and the reason
that you can think of there being a constraint,
I can give you a simple example, and then you'll
see how you would derive the other things as well.
So think of n dot P. If you transform
n dot P, that becomes n dot--
think of this is n mu, P mu.
That becomes n mu, P mu plus delta perp dot P perp.
So this is order lambda squared, so this better
be order lambda squared.
But this momentum here is order lambda,
so you need to have a power counting for the delta perp
that makes it of order lambda.
In order to keep the transform guy the same size as the guy
you started with, because I don't
want to consider transformations that would take me away
from my power counting, that would violate it.
So imposing that this thing is of order lambda squared
says that this is true.
If you go through the same logic for the other guys,
the n bar component was order lambda 0,
so you can make the epsilon perp of order lambda 0.
It doesn't cost you anything, because there's no constraint.
You're not making something that would mess up
the power counting.
All right.
So what does this correspond to physically?
So type three is actually pretty simple.
Type three is like a boost along-- if you
think about our back-to-back vectors,
type three is like the analog of a boost,
but you're transforming it in the passive sense
of transforming the axis.
And what the outcome is is very simple.
It just implies that any operator that has an n
must have a corresponding n bar sitting next
to it, effectively.
So I'll give you some examples.
So really, if you have--
let's just say it this way-- if you have an n bar mu
in the numerator, then you either have a corresponding n
in the numerator, or--
well, you have an n bar in the denominator.
Since we could have n bars in the denominator and that
was showing up in some places.
Those are the two possible ways of compensating
for the transformation.
So it's just like a simple counting that you have.
So if you look at our lc 0, we had various terms.
One of them was n bar slash 1 over i n bar over D.
So here, we have an n bar in both the numerator
and the denominator, and that compensates
for the transformation.
And then in another term, we had an n bar slash n dot D,
and then that again is invariant in this type of transformation.
What about type one and two?
So type one is the following.
So think about these guys as in this kind of language
they would be back to back.
We're trying to describe the physics
in this cone for the collinear particles,
and n was the vector that was pointing inside that cone.
What this is saying, type one, is that I can rotate that guy,
and as long as I'm not rotating it by too much--
i.e.
I'm rotating it by a small amount
of order lambda, which you can think
of as staying inside the cone--
then I can describe everything equally well
by some other vector that lives in that cone,
and I can decompose the momenta and the modes
in terms of that vector, and it will work equally well.
OK.
So that's what type one is physically doing,
and so you're not making a large transformation here,
because you want to still be inside the cone.
You still want something that's pointing in the collinear
direction.
Type two is related to the fact that this n bar
vector was just an auxiliary vector that we
used to decompose things.
We didn't really care about it.
It didn't have a strong physical motivation.
It was just needed because we're using light cone coordinates.
So I can make a very large transformation of that guy.
So here's a large transformation,
and I can use some other guy.
And we gave you an example earlier, which was 3,
2, 2, 1 as a possible value for the n bar,
and that's something that you could think
of getting to by a finite type two reparameterization
transformation.
OK?
So that's the picture for what these transformations are,
and we will finish up talking about them next time
and will show that actually this additional term that we could
write down in the Lagrangian is actually
ruled out reparameterization symmetries.