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IAIN STEWART: All right.
So let's get started.
So last time, we had talked about factorization
in the effective theory.
And there is one type of factorization,
which is this hard-collinear factorization, which
is a factorization between the low energy physics described
by the operator and the Wilson coefficients.
And we decided we wanted to think
about that in the following way, as having convolutions
between variables that appear in the Wilson coefficients
and variables that appear in the operator.
And the way that we can think about that
is just what the most general possible thing is that we can
write down for the Wilson coefficients and of course,
it can depend on large momenta.
And so large momenta here includes the label operator.
So large momenta always show up in Wilson coefficients
and in this case, that includes momenta
that are the large component, the order lambda 0
component of collinear fields.
So this is hard-collinear factorization
and it just comes out in a natural, standard way
in this effective theory because we've
set up the effective theory to have the right low energy
degrees of freedom.
So even though it's a little more complicated
than just a simple product, we see
what the variables are that can connect the two things just
by power counting, essentially.
The large momenta are the ones that are order lambda 0.
We can do this generically if we recall the definition
of our Wilson line.
We can see how this would carry over
to a generic case in the following way.
So that was one way of defining the Wilson line.
And as an operator, we had relations as well.
So we had this relation and in particular, we
could take this to any power and that just takes the p bar
to any power like that.
So you can really think of this as given
any function of this operator i m bar
dot d you can always write that as a Wilson line,
a function of p bar, and a Wilson line.
And what you want to do is you want to stick these Wilson
lines in the operator and you want
to put this function in the Wilson coefficient.
So you could think, lets me start with operator.
I can throw in i m bar dot d's because these guys are
order lambda 0, these are collinear derivatives.
I could just allow it to stick as many of those
as I like in my operator.
But given that I put in any function of them,
I could always do this and the right way
of thinking about it is that this function here
is determined by matching.
It's your Wilson coefficient.
And that's what we were effectively doing up here.
But in some sense of our general discussion
you can see that even if you had multiple places in the operator
where you could insert these derivatives, then
you would just do this.
And what would happen is at the end of the day,
you get a function of all the possible large momenta
that you can form from the operator.
So maybe one more line.
So think of it, you can always separate out.
Make a split like we did over there
with the integral over some variable
and then leave something that you can stick in your operator,
make it like this.
So this could go in the operator and then
that's the coefficient.
AUDIENCE: If I stick this in [INAUDIBLE]??
IAIN STEWART: Well, here I didn't write it.
So if I wanted to put in i m bar dot
d's, I'd stick them between.
Right?
If I wanted to use the formula.
If I wanted to use this kind of logic
I wouldn't have written this and I
would have said let me stick an arbitrary function of i m bar
dot d in here.
AUDIENCE: But Between before you even write--
IAIN STEWART: It has to be between them because
of the gauge invariance.
AUDIENCE: Right.
So before you write down the Wilson line.
So say you start and say, OK--
IAIN STEWART: Oh, yeah, so there's still--
right.
Right, right, right.
This Wilson line here came from this h.
So that's still going to be true.
And what I'm saying, if you think about this operator,
I could dress it up by putting any function of i
m bar dot d's right in here because that would still be
same order and gauge invariant.
AUDIENCE: OK, but this an alternative--
IAIN STEWART: But then if I use this formula,
it would sort of push this w.
That would cancel there and then this w comes back
and then you get the if.
OK.
So in general, the right way of thinking about this
is as follows.
We can encode this in some notation
by just setting up a convenient set of building blocks
which are gauge invariant objects
under the collinear gauge transformations.
And so we need to have a fermion field,
but we know that the fermion field,
generically we can make it gauge invariant by multiplying
by Wilson line like that.
And because of this, what we were just discussing here,
generically, we can be in our Wilson coefficient,
sensitive to the momentum of this object.
And so we can denote that by defining the following thing,
which is just a chi field that carries some momentum which
is the overall large momentum of this product of fields.
And sometimes this guy goes by the name of the quark jet field
because if you were to produce a quark in the hard scattering
process, the quark would be represented in your operator
by this w dagger c.
The Wilson coefficients would talk to that quark
through the large momentum and that quark,
in the low energy theory, would evolve into a jet.
So it goes by that name.
You could also think of it as a parton field
because as it turns out and as we'll talk about later on,
if you were to think about the parton distribution function
and what pulls a quark out of the proton,
it's exactly this operator.
And then that quark's momentum, what
momentum that quark carries is exactly picked out
by this delta function as well.
So these are the objects you'll usually
want to work in terms of.
We can do something similar for the gluon, which is a curly b
and I'm going to define it.
If we wanted to get a gluon, the natural thing
is to use a field strength because then you get a gluon
without any derivatives.
But I want my object to be dimension one,
so I'm going to define it in the following way.
So here's a field strength commutator
of covariant derivatives, these are collinear derivatives.
And I throw Wilson lines around it to make it gauge invariant,
and then to make it dimension one I throw in a 1 over p bar.
And if you start expanding this out,
then the first term will be the a n perp gluon
and so you have this b just has a n perp gluon in the same way
that the first kind of order term in this chi
is just to the fermion if I set the Wilson line to 1.
OK?
So it's starting out as a perpendicular gluon
but it's dressed up in a way that makes it gauge invariant
in dimension one.
And then just like we did here, we can put a subscript on it
with a omega to say that we fixed the large momentum of it
and sometimes there's a convention that that's done,
you have to decide if you want the Wilson line--
I mean, the delta function to be a w minus p bar or w
minus p bar dagger.
Anyway, it doesn't really matter that much because it's just
a sign of what you mean by a w.
If it's an outgoing gluon, this is a more convenient
convention.
OK, and so this delta function here and this p bar
here, when I put these square brackets, what
I mean by that is that they don't know how to act outside.
They just act on the operators inside.
So these are objects that exist by themselves
and they don't care about other things that
are multiplying them later.
So that's the gluon analog of the quark.
So it turns out that we can show the following.
That if you want to build operators
that are subleading order, a complete set
of things to do that is as follows.
It's this chi n, the bn perp, and then you
could also need p perps.
And then you could also have ultrasoft fields,
where it's really kind of similar to how
you're used to building operators in effective field
theories.
But if you just want to talk about the collinear sector,
the only things you need are this chi n this bn field,
and then p perps and no other things.
AUDIENCE: That's to go just one order?
IAIN STEWART: All orders.
AUDIENCE: So why aren't there any--
IAIN STEWART: I'll talk about it.
Yeah.
I'll show you.
AUDIENCE: OK.
IAIN STEWART: Yeah.
I mean, intuitively you would think
these are the physical degrees of freedom, right?
There's two of them so that's intuitive.
But I'll show you how you get rid of all the other ones.
It kind of matches up with what you want.
You could think that these are the physical gluons that you're
producing and your operators can be set up so that there's not
any spare use components.
OK.
So we'll get there.
So let me introduce a bit of notation.
I have to do.
So let's consider this thing.
A covariant derivative sandwiched
with Wilson lines on either side and this i dn
here is like p bar plus.
So it just involves the collinear gluon field.
So then if I have that operator and I just take n bar i dn,
then that's p bar.
if I take i dn perp mu, then you can
show that this guy is p perp u plus g bn perp mu.
So this one is just a relation we talked about before.
This one is not as obvious, but if you take this guy,
you can let this derivative p perp act on the Wilson line
or act through.
If it acts through, that's this term.
If it acts on, then I can manipulate the operator
so that it is exactly this form.
And part of that comes from the fact that--
I can explain it to you here.
I have it in my notes and you can look at it later,
but let me just explain it in words.
If you looked at the term in this commutator
that was other order, where the i m bar dot d was sitting here.
It would hit the Wilson line.
You'd get 0.
So that term was just put in to make
it look like a field strength.
OK?
So really you have i m bar dot d dn perp, without the comma,
without the brackets.
But then this combination here, you can use the identity
that we had with it and basically if you
push this guy through here, you're
canceling the p bar so it's just giving you the Wilson line.
So that's basically how you go from here to here.
OK?
So basically what this is saying is
I don't need to consider covariant derivatives because I
can instead consider p perps and b
perps and that's equally good.
OK.
Now if you do a similar type of thing in the n component,
then you can derive a similar type of object.
So this is the same object as we have
over here, but instead of having an i dn perp,
I'd have an n dot d.
So d dot n perp replaced n dot d.
And that looks like it could be something
that you'd build operators out of and and also, furthermore,
why not have operators that depend on this n dot partial?
OK?
So for p bar, we don't have to worry about that.
So OK, I said we'll include this, we'll include that.
I didn't say we have include this, this, and this.
So there's three things I have to argue away here.
The derivatives that are p bars, those just
go into the Wilson coefficient.
So if we had p bar chi n comma omega, for example,
then that's omega chi n comma omega
and this is put into the Wilson coefficient.
So that's why we don't have to worry about having p bars.
In some sense, we do have p bars.
They're all in the Wilson coefficients.
So it's really these other two that we
have to worry more about and those actually
can be simplified using the equation of motion.
So if you have i n dot p partial on chi n, then
equation of motion when you write it out
in terms of these objects, it has some form
and basically, you can just get rid of those terms.
So the equation of motion allows us to get rid of i n dot
partials that are on chi n's.
So that's why we don't have to worry about those.
And this is just like saying that in our leading order
action, there wasn't i n dot partial.
But that's just like saying there
was time derivatives in our leading order
action in some standard effective field theory.
But then in the higher dimension operators,
you can always use the equations of motion
to get rid of those time derivatives.
And here we're using the equation of motion
to get rid of i n dot partial.
So it appears in the leading order action,
but then we don't have to have it in any other subleading
operator and that's why it's not one of the ones that's
included in the list.
And if you had i n dot partial on curly b,
that's also part of the equations
of motion of the gluon field.
Now when you do the gluon equations of motion,
there's components because it's a vector
and one of the other components allows
you to get rid of the n dot b.
So there's another term that you can rearrange,
and I won't write it out because the equation is rather messy
but give you some idea.
There's another component that looks
like this where I can get rid of all the n dot curly b's
using the gluon equation.
So the gluon equations of motions
allows me to get rid of both of these things
and basically after I've done that, I just
have the objects that I've told you we can use.
So after using equations of motion,
we can get down to those objects.
And any other thing that you might dream up
can be reduced to these objects.
So I'm not saying that I went through a complete list here.
For example, what if you had a commutator of 2 d n perps?
Right?
You might say, oh, that's some new thing.
It's one of these.
You can also reduce that too.
So let me list that one just to give you
some idea that there's others you might think of dreaming up.
So you can reduce all of this to be that.
OK, so for the collinear sector this
is enough to build higher dimension operators, just
these three.
And then for the ultrasoft sector, so I guess this is two.
We do need ultrasoft derivatives and ultrasoft field strengths
and ultrasoft quarks.
So this part is really just similar to the story
that you'd have for a standard low energy effective field
theory.
Like integrating out a massive particle,
you can use the equation of motion.
One thing that is worth commenting about
is the connection between one and two.
So one is collinear and two is ultrasoft
and prior you might think, well, they're totally independent.
But we saw last time that reparamaterization invariance
connects them.
So if I've decided that this is the type of basis
I want to use for my operators, then
what is the reparameterization connection?
You can rewrite what we said last time in terms
of these curly d's because it basically just means
moving the Wilson lines that we had in this formula last time.
They were around the ultrasoft operator last time
and if I just move them over to the collinear operator,
then I get this curly d.
Right?
So the RPI connection is connecting
the curly d collinear to the d ultrasoft
and then you can write this guy as the p perp plus the b perp.
And likewise if you do the same for the n bar sector
and you move the Wilson lines from this term to that term,
then it looks like these two combinations.
So that's just rewriting what we had before but now in terms
of this type of notation.
OK?
So that's enough to build operators at higher order
and then we write down Wilson coefficients
for those operators that are functions of a large momentum
and then we start doing physics with them.
So any questions about that?
AUDIENCE: For the equations of motion,
you always just use the leading order equation of motion--
IAIN STEWART: Yeah.
AUDIENCE: --into a higher order term?
IAIN STEWART: Yeah, these are the L0 equations of the motion.
That's what I'm doing here or writing here.
Yeah, there's one more actually.
There's three but it's only needed at some very high order.
OK.
So the next thing I want to talk about
before we start doing explicit examples
and going through processes is how loops
work in this effective theory.
And we're going to have to come back and talk about our grid
that we have for the split up of momenta and how
should we actually think about it in practice
and then we'll also deal with how matching and running work.
So I'm going to do this in the context of an example
and again, I'm just going to pick the simplest example that
has only one jet just to make our lives a little bit simpler.
So we'll consider our heavy, light current
and I think actually once you see
how it works in this example, you'll
understand what all the general features are of doing loops
in SCET.
So we had operators that we constructed,
lowest order operator.
And let me write it in this way which
was prior to making our field redefinition.
I could just write it this way if I want.
And gamma for beta s gamma is a tensor operator.
And there's a photon field, which is a field strength,
f mu nu.
So let's just think of that as all part of gamma.
OK, and that's also appearing here
and I could use the spin structure properties
of these things to reduce the sigma and mu nu
but that's not really going to be part of our story
so let's not bother with that.
So what I do?
I just compute the QCD loops and the SCET loops
and I compare them.
And if SCET is the right effective theory
for this limit where I have an energetic photon
and am back to back with an energetic strange quark,
then I should match all the infrared divergences
in that QCD computation, I should
be able to extract a matching coefficient,
I should be able to determine what
the C is from that calculation, and I should
be able to run the operator.
I should be able to make this into m s bar,
do some renormalization of this operator,
and then do some renormalization group evolution of that Wilson
coefficient.
Well, let's just think about computing the graphs.
So we have to decide, when we compute the graphs,
how to regulate them?
Right?
And we need to use the same infrared regulator
in the full theory and the effective theory.
So here's how I'm going to regulate them.
We could do this different ways and it's
useful to understand that various answers that we get
are independent of the regulator.
So I'm going to take p squared not equal to 0
for the strange quark.
So there's infrared divergences associated
with the strange quirk and I'm going
to take p squared not equal to 0 for them.
I'm going to use dim reg for the heavy quark.
So I could take the b quark to also be offshell.
That would make the formulas even more complicated.
So instead of doing that, I'm just
going to allow epsilon to regulate the b quark.
These guys are pretty easy to track
so that won't be a problem.
And I'm going to use Feynman gauge.
I don't have to do that, but that's a nice gauge
for doing calculations.
So what are the diagrams where I have my photon
and I have a vertex diagram like this
and then I have wave function renormalization?
And so let me, in the usual kind of cavalier way,
denote the wave function renormalization, which
is just the multiplication by the appropriate z factors.
My diagram's like that.
And this is a standard QCD calculation
and we can carry it out.
And what does the answer look like?
So this diagram here has double logs
and single logs that involve p squared, which
is our IR regulator, and then it has some terms that are finite,
which I'll just note by--
so p here is the momentum of our strange quark going out
and pb is the momentum of our b quark coming in.
And we're not going to talk much about these finite terms
but what I mean by finite here is
terms with no IR divergences.
The IR divergences are these single logs and double logs
and then there's some remainder that I can write that way.
I'm expanding it for small p squared.
So p squared is not equal to 0 but I
take the limit p squared goes to 0
and then that limit, these are the IR singularities.
And then there's some function of b dot pb, which
is just an order one thing.
And then b squared, of course, is pb squared.
So that's sort of the remaining kinematic variables
this could possibly depend on.
AUDIENCE: [INAUDIBLE]?
IAIN STEWART: No.
AUDIENCE: Is it finite?
IAIN STEWART: It's really finite.
So yeah.
Yeah, so let's see.
Yeah, I guess I carried out the so there
is a z ten for the tensor and when I write this--
yeah, let's see.
So, OK.
Let me do this.
I think it's better.
So I added up all these graphs and I want to add one more.
So there's a counterterm for the tensor field.
So let me change what I was going
to say and do it this way.
So the sum of these four graphs is this.
OK?
So there's no UV divergences.
And I'll just tell you what the z factors for these three are,
then you could figure out what this graph is
by subtracting the two.
So the tensor current in QCD has a z factor, looks like that.
There is a z for the heavy quark.
And if I include the finite residue as well as
the divergent pieces, this looks like this.
I think I'm going to have to make one
adjustment to my formula here.
Let me just fix something here.
Yeah.
So if I want it to be the way I say, then what do I have to do?
So this guy should be three halves
and there will be one more divergence.
I think that's right.
So this 2 over epsilon ir here is
that 1 over epsilon ir and the UV renormalization
is taken care of once I have the z tensor.
There's divergences in this diagram in these ones.
but there's one left over, and that's
taken care of by z tensor.
So there's no 1 over epsilon UVs after I take care of this guy.
So that's just by the definition of what z tensor should be
and then everything else is as I wrote.
So there's either divergences that
are associated to this strange quirk offshellness, that's
these two terms.
There's an IR divergence associated, a heavy quark going
onshell, that's that term.
And that's the sum of diagrams.
OK, so what about SCET?
So there's going to be, in SCET, collinear diagrams
and ultrasoft diagrams.
So I'm going to use Feynman gauge for everything again.
This is not something I have to do,
but this is what I'm going to do.
So let's start with the ultrasoft loops.
So there's a vertex graph.
So using the notation that we've adopted where the collinear
quarks are dashed and the heavy quarks are double lines,
we have a diagram that looks like that.
There's some free factor.
Let's focus on what the loop looks like.
So this loop here, the k that's going through this loop
is just a residual k.
There's no label k for this loop because it's
an ultrasoft gluon.
So when we write down all the terms in this loop,
it's just standard field theory.
There's nothing special about it.
Since I'm taking the strange quirk offshell,
the propagator that I get is this shifted iconal propagator
where basically the fact that it's offshell
it gives me this extra term there.
That's regulating some minor divergences.
And we want that because we want to regulate the IR divergences
in the same way in the full theory of the effective theory.
This is proportional to that and if I put in all the factors.
So it does have double logs of the p squared,
doesn't have single logs of the p squared,
and actually if we look at the double logs,
the coefficients also don't match
with what we had over there.
And so there's going to be some other diagrams that
are going to involve double logs of p squared as well.
One thing that we can note here is
that if you think about the scales in the problem,
remember that our loop integral was totally
homogeneous in the power counting,
the lambdas were totally homogeneous.
And if you think about p squared scaling
like lambda squared, which is natural size
for an external collinear momentum, all right.
If p squared scales like lambda squared then
so does p squared over m bar dot p.
And this is a dimension one thing
and this is the ultrasoft scale, right?
For some ultra soft momentum and so
if you want to look at these logarithms
and you ask, what scale is this effective field theory
diagram sensitive to?
It's sensitive to the ultrasoft scale.
So the logs are not large logs, they're
order 1 logs as long as mu squared
mu is of order lambda squared, which
is the scale for the ultrasoft momentum.
And that's what we expect from this ultrasoft diagram,
that it's telling us about physics at the ultra soft scale
and that's what we see setting things
up, doing the calculation.
So you could also think about a wave function diagram
with an ultra soft gluon, but this guy 0's
out since in Feynman gauge you just get n mu n mu, which is 0.
So there's no so z for the collinear quark
field from an ultrasoft loop is 0.
That wouldn't necessarily be true in some other gauge,
but in some other gauge this diagram would also change.
All the diagrams would change.
And so if there was a non-zero contribution in this diagram,
it would just be taking care of gauge invariance.
And then finally, there's ultrasoft loop
on the heavy quark.
And that's an HQET diagram, has nothing
to do with the collinear quark.
So from that diagram we would just
get the z factor, which is the appropriate z factor in HQET
with r regulator.
If I specify ultraviolet and infrared divergences,
that's this, and this infrared divergence here
is actually exactly the same as the one that we had over here.
OK, so the ultrasoft sector, there's
nothing really tricky about it.
It's just write down the diagrams, do the loops.
AUDIENCE: So for the [INAUDIBLE],,
isn't that IR divergences of those diagrams
cancel with every automation [INAUDIBLE]??
IAIN STEWART: So what are we looking at, right?
So it depends on what we're looking at here.
And so, yes, in general that would be true, right?
If you were calculating some cross-section, which
was IR finite cross-section and that,
of course, would depend on defining what you
mean by measuring this quark.
Right?
So the IR divergence would become some physical scale
like the mass of a jet.
The IR divergences would turn into something physical
if you put this into a physical cross-section.
And that's exactly basically what would happen.
These p squareds would become the mx squareds
that we talked about when we talked about beta s gamma.
But here what we're interested in doing
is a matching calculation.
So we fix the external state, still
have a particular number of partons,
and we want to compare the full theory calculation
with the effective theory.
The effective theory should reproduce
the infrared divergences.
So really all we care about is not
that this is infrared finite, but rather
that the effective theory has the same infrared divergences.
And we'll see that in the end when
you can think about is rather than canceling in
for divergences as you're thinking in the full theory,
we've matched the full theory onto the effective theory
and that matching gives us the Wilson coefficient.
And then we take effective theory and all the cancelations
that you're thinking about between real and virtual graphs
will occur in the effective theory too.
So you can just think about the effective theory
virtual-real graphs, then the cancelation will take place
there.
But then you're thinking about that cancelation
later, at a lower scale, which is what you actually
want to do.
Because what I just was telling you about IR
divergences becoming different things in the final state
becomes very trivial once you're in the effective theory
and we'll see sort of exactly how that works later on.
But first we have to talk about linear graphs.
So in the collinear graphs we had
graphs like this one where we can take a gluon out
of the vertex here.
That corresponds to taking it out of the Wilson line.
So let's label the graph in the following way.
This would be k plus p this will be p This will be k.
And if we follow our rules for what this is,
we would write a sum over labels and then an integral
over residuals.
And let me put the residual integral
in dimensional organization.
Let me try to squeeze in everything,
which won't be possible.
So I'm going to write out the components
to make clear whose residual on whose label.
So in the denominator there's one more term.
So the plus guys are always residual,
the perp and the minuses are always labeled.
The short way of saying what I'm trying to squeeze in here.
So the denominator has these three terms, this guy here,
this guy here, and this guy here.
And when I label the diagram like this,
you should think that each of these
has a label and residual part.
So k you can think of as a pair, k label, k residual for now.
And remember the importance of doing this.
The importance of doing this was related
to always being able to identify what the lowest order
term was on the right here.
OK?
And that actually becomes more important
once you start to think about diagrams where
you would add like an extra ultrasoft gluon somewhere
in this picture.
Then, of
Course when that ultrasoft gluon feeds its way
through this loop, you gotta make
sure that it's only the lowest order piece that's showing up.
In this case here, we just have a collinear loop
and we don't have any sort of real ultrasoft
momenta from ultrasoft particles besides the heavy quark, which
is just an external particle to the loop.
So what we want to do with this is we
want to turn it back into an integral.
Just let me say it this way.
And if it wasn't for these restrictions here,
then that's very easy, actually.
and then I'll talk about why it's true.
I claim that if we ignored the restrictions
and those restrictions are ensuring that we don't double
count between our collinear and our ultrasoft degrees
of freedom.
So they are important, but let's ignore them for a minute.
If we do ignore them, we would just get the following.
Where I just basically stop thinking about residuals
and labels, write everything as a full momentum,
and write down exactly the same thing I just wrote.
So this is a full k squared and this
is a full k plus p squared.
OK?
So what I'm going to do is I'm going to first ignore
these restrictions and then I'm going
to tell you how it would work, how this actually
does turn into this.
What are the rules for doing that.
And then I'll come back and I'll tell you
what these extra conditions do.
So really this looks like maybe it's trivial,
but we should think about it.
And it's almost trivial but not quite.
So really what we're doing here is
we're combining back together labels and residual momentum,
right?
And the place that we have to worry about that
is in the perp in the minus space.
And recall we had the grid and the grid
is sort of our way of guiding, our guidance
to see how to put things back together.
So we had vectors that lived in this space
and this is the label and then there's
the residual, if we want to point to some place
in that space.
This picture was like a Wilsonian effective field
theory because the picture makes you think of sharp edges.
But the real effective theory that we're doing
is a continuum one and so you have
to expand your brain a little bit
and think that each of the boxes in this picture
is actually an infinite space as well,
because the residual space doesn't have restrictions
like that that would spoil Lorentz symmetry.
So each grid point really is specifying an infinite space
of residual momenta.
And it's R4 or Minkowski space, so the momenta components
are real numbers and there's some rules.
So what are the rules?
So I'll tell you what the rules are without restrictions
for now and then we'll come back and I'll
tell you what the rules are with restrictions.
So rule number one is the simplest,
and it just says that say I had the following,
I'll use a one dimensional notation.
Say I had a sum over kl's and an integral over all kr's.
Well that's just the same as not having split it up
and doing an integral over everything.
Because we split this thing into boxes and if we really
integrate over all boxes and sum over all labels,
then we should just get back the full integration
over all the momenta.
So that's true for each, for minus and perp momenta.
But really what we have to do is we have to do this
with some integrand, right?
So the type of integrand that we have is following.
If you think about the components
that we're talking about here, which are the minus and perp
ones, then in our integrand up there,
there's no minus or perp residuals.
So it's just a function of the labels.
Right?
And so what that says effectively
is that this is a constant function in this box,
in each of the boxes.
And effectively, the way that you use this formula is you
do the following.
You say, well, if it's a constant function in the box,
I could evaluate it at a different point in the box
and I'd still get the same value.
So in particular, I could evaluate
it k label plus k residual and then I
can use that formula there to say
that this is just an integral with a continuous k
of the function evaluated at that continuous k.
So that's the way that number one works.
We are integrating functions that are constant in the boxes
and so then it's kind of trivial,
how to put the boxes back together.
Yeah?
AUDIENCE: So this d kl division doesn't have kl plus--?
IAIN STEWART: Yeah, so what I mean by it is--
yeah.
So I'm using here for each minus or perp momenta,
one dimensional notation.
Yeah, and so I have three of these.
I have to do this.
And for each one of them it's true that what I'm saying.
But of course, if I tried to write that on the board
then it would confuse the point, I think,
which is, in some sense, a simple point.
OK?
So this is, in some sense, saying
that this whole split up that we were doing
was not really needed.
We could've just written a continuous integral and that's
kind of what this is saying, right?
We could have just written a continuous interval
and not worried so much about all the split
up that we were doing.
The place that we have to be careful about
is these restrictions, and we'll come back to that.
And the other place that we have to be careful about
is when you have the multiple expansion.
But as long as you take care of that, then basically this
is always happening.
So the reason why this works is the following.
For every label loop momentum that there is in any diagram,
there's always going to be some corresponding residual that's
not specified by delta functions in terms of external momenta.
And effectively, therefore that we can absorb in order
to do what we just said.
So we can always go back from this discrete type notation
back to a continuous notation.
The discrete notation was just helping
us to set up the expansion and be careful about it,
but we can always go back to the continuous one
because there's always a kr that has this property.
OK.
So now in general though, you might have
some more complicated thing.
And if I'm going to give you some rules
I should give you a complete set.
So we have to append our list of rules by the following one.
So if I thought about doing what I just said over here
but I went to some higher order, then what could happen?
Well then these kr minus and kr our perps could show up.
They'd never show up in the denominator of our propagators
if they were just collinear propagators,
but they could show up at some point in the numerator.
And we need actually a rule like this one, which would be clear
if we were regulating d kr in dimensional regularization,
that the power of divergences are getting set to 0.
And that is basically to maintain the run symmetry
in the residual space we need a rule like this for j
greater than 0.
So that doesn't come into our calculation,
but included for completeness.
So when would these integral kr actually do
something non-trivial?
It would do something non-trivial
if we had an ultrasoft loops and collinear
loops at the same time.
So in the case where we just have collinear loops,
it's basically up to this issue about the restrictions
that we'll talk about.
It's basically that we just could
have done everything as continuous
and ignored this split.
But if we have both ultrasoft particles that
are participating through the loops and/or then in general
these will give non-trivial loop momenta
in the residual momenta.
And hence there will be some that we can't just
absorb in the fashion that we just said.
So there will be, in this situation, residual momenta.
Some residual momenta will be absorbed in the same way
to turn the integrals into continuous ones,
but other ones won't be absorbed.
And that's because the ultrasoft propagators rate
would involve the lr plus, the lr minus, and the lr perp
in the denominator, so we don't have this rule to apply.
We can't do what we said here with the constant boxes
because now the functions are depending on that variable.
So we just have to do the integral.
So you could have something that looks like just
in a schematic formula.
Let's have there be 2 kr and an lr.
We're in kind of an obvious notation.
I'm saying that the function could depend on lr residual,
it doesn't depend on kr residual.
So we absorb kr back into the sum to make it continuous
and then this integral we just have to do.
OK, but we're still, in the end, just doing an integral.
So this guy come from ultrasoft propagators, for example.
OK?
So those are the different cases that you can get.
But nevertheless, even if you have ultrasoft particles
and propagators floating around, there
will always be a residual momentum associated to
of what we were doing over here that you
can absorb in the same way I stated,
so let's proceed along these lines and see where it takes us
and then we'll come back and put in
this additional restrictions.
AUDIENCE: Do we ever have to do a discrete sum?
Like is there a sum over ll?
IAIN STEWART: Yeah, so you never have to do a discrete sum.
AUDIENCE: That's good.
IAIN STEWART: Yep.
The discrete sum is really just a way of thinking
and I'll show you in a minute that you even
can avoid thinking about it once you know what to do.
So it's guiding you towards the right answer
but it's not really something that you
have to think about this grid picture.
It's really just if you get confused,
you can always think about it but if you're not confused,
you don't have to think about it.
OK, so this guy is what we said, proportional to what we said.
It's just in this case, the reason why it was so simple
is because k residual is one component of this guy,
and then there's these other three, right?
And the other three, we just absorb
the sum and the integral back together
and there's nothing further to talk about.
So that's why this is, in some sense, very simple.
So we just do that, and what do we get?
Get some 1 over epsilons, we get some logs of p squared,
and I'm putting in even the constant term
since it's pretty simple.
4 minus pi squared over 6.
So if you think about where this came from,
there was a place that this Wilson line
came from was attaching this guy over here
and then integrating out that offshell propagator.
So this is a vertex diagram.
We had a vertex diagram, which is an ultrasoft gluon,
and now we've added an identical type up topology.
Except we drew it different because this line
was offshell so we drew it as a Wilson line
and that's the right way of thinking about it.
But if you think about where it came from,
it was the same topology and we've
added another vertex diagram.
So the effective theory has two type
of vertex diagrams, this collinear
one and the ultrasoft one and we're
going to add them together.
And in this one, you see the logs
are minimized at a different scale, mu squared
of order p squared, which is the right scale for a collinear
loop.
Because remember, the collinear particles
lived at a larger p squared than the ultrasoft ones.
So there's a larger scale that would minimize these loops,
it's a collinear scale.
So the effective theory is capturing the physics
at that scale.
Squeeze in another diagram here.
If we do the collinear wave function renormalization,
then this is non-zero.
This was 0 for the ultrasoft gluon
and but for the collinear gluon, it's non-zero
and it's exactly, actually, the same as the full theory.
It's the same as massless QCD.
So you can do the diagram using the effective theory
Feynman rules, then that's what you find.
But you could also understand why that's true.
And reason why it's true is nothing in this diagram really
specifies a frame.
We called all the particles collinear,
but it's not attached to anything.
So we could just take the whole diagram
and boost back to the frame where everything's kind of soft
and then we would usually be thinking about it in terms
of a full theory field.
So that's effectively why this diagram
like this that doesn't have any reference, unlike this one
which has a reference because it's
attached to the heavy quark.
That's why it's the same as QCD.
Yeah?
AUDIENCE: How do you know what the epsilon position
for the Wilson line is set?
IAIN STEWART: Oh, for this one?
AUDIENCE: m bar dot k.
IAIN STEWART: Yeah, for this one.
Yeah, for this guy it actually doesn't depend on the i epsilon
prescription up to a--
yeah, if I remember correctly.
Yeah.
That's true, I believe, when I'm being
a little bit cavalier with it but once I put it the zero bin
restrictions, I'm not sure if that's true anymore.
AUDIENCE: OK?
IAIN STEWART: Yeah.
I mean, really what solves that is that in a minute
I'm going to be talking about the fact
that there is a subtraction term here
and once you put the subtraction term in,
that i0 is not relevant.
But I think even if you do this diagram with arbitrary i0--
because what's happening is m bar dot k going to 0
is related to some of these 1 over epsilons
and we'll talk about that more in a minute.
Yeah, I'm not 100% sure.
It could be that there's some sign that might flip.
I'm not 100% sure.
AUDIENCE: Even if they get the wrong side,
the difference would be subtraction.
IAIN STEWART: I guess I know that--
yeah.
What I know is that--
yeah, let me answer your question later.
It'll be easier because I'm trying
to say a bunch of things that depend on something else
that I haven't explained yet.
So what are the other possible topologies we could write down?
So we could write down this one, where
we take two attachments in the Wilson line
and just loop them back up but that's
proportional m bar squared and so that's
0 in our Feynman gauge.
And likewise, there's a looping back up in the vertex
in the wave function renormalization,
but this guys scale is power law divergent and so we can just
set it to 0 and dim reg.
You don't have to worry about it.
OK, so that's all the diagrams.
Let's think about doing matching,
i.e let's think about comparing QCD and SCET by adding up
the diagrams.
In QCD we carried out the randomization,
we added the z for the tensor current.
Let me just write again the answer looked like.
In SCET we didn't carry out renormalization yet,
so let me call this the bare SCET result for now.
Once we add the ultrasoft and collinear diagrams together,
the logs of p squared match up exactly with the full theory.
So this is the first sign really that it makes sense
to be thinking about adding these loops.
Even though they were the same topology,
we are correctly reproducing those logs of p
squared in the full theory.
And then there's some other pieces.
I'll write out all the other pieces
so you see what they look like.
Well, maybe I won't I won't write the constant.
So there's all the effective field theory terms.
So these terms here we can match up with these terms here.
So that's good.
These terms here, which remember in the full theory were finite,
and these terms here, which in the effective theory,
are finite, the difference of those
is going to give the Wilson coefficient.
Now we said that the Wilson coefficient
could be a function of p bar, so what's going on with that?
Well, if you look at momentum conservation
in this process of beta s gamma--
so I probably should've said this earlier.
So when you look at beta s gamma,
if you look at momentum conservation then
the p minus of the strange quark has
to be equal to the p minus of the b quark,
but that's just mb.
OK, so actually p minus is equal to mb by kinematics.
So mb's in this result here you shouldn't think
of as p minuses, and that's this p bar that was in our Wilson
coefficient is just getting set to mb
because of some delta functions that are specifying kinematics.
So that leaves the 1 over epsilon terms.
And so what we'd like is that those terms are
associated to renormalization.
Of the effective theory, right?
I wrote that the effective theory was bare.
But if I want to do that, then I have
to ensure that all these epsilons that
are appearing here are really ultraviolet divergences.
If they're infrared divergences, then doing that
doesn't make sense.
And that's actually the remaining issue
that we have to deal with.
I just wrote epsilon, that means I'm ignorant to what they are.
And if I knew this one was epsilon
IR because it came from the wave function
renormalization of the heavy quark
and that was the same on both sides.
It was the same diagram, it was the wave function
renormalization diagram in the full and the effective theory.
So I could match up that one.
These ones just came out, but it turns out
that so far with what we've done, some of these epsilon
here are IR.
And so the IR divergences aren't matching up
and the reason is because we didn't
put in those restrictions on our sum over labels.
OK.
So we have these restrictions, k label not equal to zero
and k label not equal to minus pl.
Those are the restrictions that I'm talking about.
The place that those restrictions came from
was k was the momentum of the gluon.
kl not equal to 0 is saying that this is the restriction
that the gluon is collinear.
Because kl equals 0 is the ultrasoft gluon
and this is the restriction that the fermion is
collinear in the loop and that's why there was two of them.
So these are called zero bins.
Zero because it's where the ultrasoft momentum lives
and from the point of view of collinear, that's zero.
Imposing these restrictions is removing the zero
bin, if you like and what these restrictions do
is they avoid double counting and the way I've
said it that's, I think, clear.
So far in our calculation we haven't avoided double counting
and that's the problem.
OK.
So we have to modify our rule or we
extend our rule to include the case where
we have these restrictions.
In an extended version of rule two that house restrictions.
So really what we want to do is that
and we want to think about that as an integral.
So here's how we can manipulate these to think
about it as an integral.
That sum over all kl's, but then we'll
subtract the limit of this f where we take the f
and we let the kl go to the place.
So we integrate over everywhere, including the place
we don't want to go, and then we subtract it back.
So what is this fl of k?
This f of kl goes to 0.
it's defined by taking the scaling limit of the collinear
momenta towards the ultrasoft.
So you take your collinear momenta,
which are the minus in the perp here,
and you scale them towards an ultrasoft momentum in whatever
components, i.e you start counting the kn's as order
lambda squared and you keep the leading order piece
or you keep the piece that's the same size as this term.
And then that defines what this f is.
Once you take that limit and you expand,
then that's what the f is.
So what you're doing here by doing this procedure
is you're basically setting things up
so that this guy is integrated over in a way
that we can combine back into an integral.
But then we have to subtract an overlap
of when that integral would go into the ultrasoft region.
But the overlap we're subtracting
is also an integral, so we have a difference of integrals.
And you should think of the second integral
as integrating over the square where the zero bin was.
So if you think about our picture
where the collinears were up here, ultrasofts are down here
and you thought about there being some box,
you're taking this scaling limit when this guy goes down
into that box.
You add up all the boxes, that's this,
and then you subtract out that box again.
And that avoids having a double counting in that region.
So then this guy here, you can do the same kind of trick
as before.
So continuing with the equation.
kl is going to 0, well, we can think of it just as a function
of kr's.
And so effectively what we get in the end is
an integral over all k of f of k,
the full f, evaluated with a continuous momentum
minus some f that's expanded and avoids there
being overlap in this box.
So rather than having these discrete sum
with the restriction, we have a difference of integrals
and this subtraction term avoids the overlap in that region.
So all the discrete sums are good for is
a means of figuring out what limits you need to take
to generate these subtractions.
Once you've done that, everything's
a continuous integral.
And this is called the zero bin subtraction.
OK, so if you like, one way of phrasing what's going on
is that the collinear propagators are really
distributions.
They're distributions that know that they should
have a subtraction in order to not overlap the other region
where we have another degree of freedom.
So you can think about it that way
and having this sum it's just a way of encoding that.
But in the end, it looks like some kind of plus function
where you have a subtraction.
AUDIENCE: So it seems like in the second term
you would have to relearn the expression because initially
in the [INAUDIBLE] k you ignore kr minus and kr perps.
IAIN STEWART: I'll show you how it works in an example
up there.
So let's go to our example.
Yeah, so what you just said is not quite the way
you should think about it.
You should think about it that you've
given the effective theory f and now
I'm saying that that f, as given, with its momentum as
given still has an overlap with the ultrasoft region
that I want to subtract.
So I'm taking a limit of that f, I'm
not add anything back to it.
So this guy, integral dk.
So this is what we wrote before.
And now if I take the ultrasoft limit of it, of the k,
then I would write this.
OK?
So when I take the scaling limit,
if you think about k squared, k plus,
k minus, minus k perp squared, when I take the scaling
limit of all momentum being ultrasoft,
the components of the k are still homogeneous.
So there's nothing there to expand.
Some expansion happened in this k plus p term.
One of our n bar dot k changes its power counting,
but it's still n bar dot k.
There's nothing to expand there.
So this is what the subtraction looks
like and then the numerator, the n bar dot
k can be dropped relative to the n bar dot
p, which is large and external.
OK, so this is taking the ultrasoft limit of this.
AUDIENCE: Do you have to subtract off
the the ultrasoft limit of the quark as well?
IAIN STEWART: Yeah, so in general
I would have to subtract off the ultrasoft limit of the quark as
well.
And when I do that, what I find is
a term that's power suppressed and so I drop it.
But in general, I would have to do that as well.
That's right.
And so it looks like an ultrasoft diagram
except it's got this n bar dot k inside of the v dot k
that we had in the ultrasoft diagram.
And if you do a power counting with the loop momentum
scaling as ultrasoft, then you have this piece is of the order
the same size as this piece and that's
why you keep only that term.
AUDIENCE: Do you have to worry about a higher power?
IAIN STEWART: No.
So the prescription we have is that we drop the higher powers
AUDIENCE: But if I were to do say--
IAIN STEWART: Oh, if you did the higher power--
AUDIENCE: --lower power higher power zero bin?
IAIN STEWART: No.
So if we did the higher power, then this
would start out at higher power.
And then when we took the limit of it,
it would end up just starting at that power or higher.
AUDIENCE: Right, but there was a--
IAIN STEWART: Yeah, you don't have to.
No.
AUDIENCE: No?
IAIN STEWART: No.
Yeah.
AUDIENCE: Is there a reason?
IAIN STEWART: Yeah, so really what
you care about subtracting here are the log divergences
and that's what this minimal subtraction is doing.
By keeping the piece that's scaling the same way,
you're removing the log divergent pieces.
And it's the log divergent pieces which
are giving one of our epsilons.
The pieces that you would get from the higher expansion,
they would all be kind of like power law divergent terms
from the point of view of the power counting
and we just don't have to worry about those.
And another way of saying it is, it's not
that I'm removing absolutely this whole integrand
in that region, right?
There could still be a constant, for example,
that comes from that region.
But if there's a constant that comes from that region,
I don't care.
What I care about removing is any spare use IR singularities.
And for those I can make a minimal subtraction,
which is just the first term.
All right, so I want to finish this discussion.
So if we do this, we get an answer,
which I will try to write on the board for you.
So now I'm going to distinguish all the epsilons
and then we'll see what this subtraction does.
So if I was careful and I distinguished all the epsilons
in our original calculation, it'd actually look like this.
And then the subtraction piece gives an extra contribution
and it's actually scaleless in the n bar dot k here.
So there's a scaleless loop in this guy.
So it actually vanishes if the epsilon IR and the epsilon UV
are said to be equal.
But what it does is it converts the epsilon IRs that
are in the first expression into epsilon UVs, which
is what we want.
So once you add up these two things,
the epsilon IRs are canceling and the epsilon IRs
that were coming in the original formula, those
were coming about because of this bad behavior
as n bar dot k goes into the limit of n bar dot
k going small.
You can think about that roughly as where the ultrasoft is.
This is subtracting off that behavior and the remainder
then is coming from only having divergences for a n bar
dot k goes to infinity, which is a proper collinear ultraviolet
divergence not from n bar dot k going to 0.
So once you put the two together,
the epsilon IRs cancel and then we get exactly,
actually, the same expression we had before but where
all those 1 over epsilons are 1 over epsilon UVs.
AUDIENCE: So the epsilon has come from second term?
IAIN STEWART: From both terms.
So they both have epsilon IRs but they cancel between them.
AUDIENCE: OK.
IAIN STEWART: Yeah.
And the remainder is just epsilon UVs, so
all the epsilons that I wrote my earlier formula would
be now epsilon UVs once I take into account the subtraction.
So I could have just ignored the subtraction and that's
often what people do.
If they know that the zero bins are giving a scaleless
integral, they say, well, let's ignore the subtraction,
we'll just say that all the epsilons are UV
and the zero bin makes them UV.
But if we really want to look and see
that things are working properly,
we should take the subtraction and calculate it and make sure
that that's true.
But we could have just taken the answer
that I wrote down earlier and said those epsilons are
ultraviolet and let's throw them in a counterterm
and calculate an anomalous dimension.
So we'll proceed that way next time,
but we now know that actually they
are ultraviolet divergences.
So next time we'll take the ultraviolet divergences
and we'll define from them a counterterm
And we'll see how we get an anomalous dimension
and what kind of logs we sum by using that anomalous dimension.
So the zero bin that's scaleless in this particular example
is not always scaleless.
So sometimes it could give a nonce.
Depends on the problem you're dealing with.
So sometimes you can set things up so that it's scaleless
and then you just basically can ignore it.
But that's not always true, so you
do have to think about whether it's really going to be
true for what you're doing.
If it is true, then you can effectively
ignore it because it's sort of just making
the physics come out right, making sure there's no overlap.
But if your regulators set up so that it's scaleless,
you can just get around it.
But in general, that might not be true.
If you had more scales in the problem,
if you're doing some calculation that
had some jets of finite size then that
won't be true typically.