Subject: renormalizability
From: <Saved by Microsoft Internet Explorer 5>
Date: Wed, 14 Jan 2004 14:13:56 -0600

Renormalizability

John Baez

August 14, 1999
I assume you're reading this because you want to see
how to actually tell if a given quantum field theory is renormalizable. I'm
going to do an example where I consider a quantum field theory whose Lagrangian
looks like this:

psi* (D + m) psi + c psi* psi psi* psi

Here psi is a fermion field, D is the Dirac operator, and c is a number
called a coupling constant. I've left out a lot of indices to make things easy
to read. (Don't forget, I'm trying to make things seem simple; I'm not trying to
be really precise.) The first term is called the "free Lagrangian" and the part
that looks like psi* D psi is called the "kinetic part". The second term is an
"interaction Lagrangian" which corresponds, in the language of Feynman diagrams,
to a vertex like this:
\ /
\ /
\/
/\
/ \
/ \

Two fermions go in and bounce off each other; two fermions come out! It's
a 4-valent vertex because we've raised psi to the fourth power in the
interaction Lagrangian. By the way, I'm only assuming the fermions are massless
to reduce the clutter in my notation.

The Criterion

Okay, so how do we tell if an interaction Lagrangian like this gives us a
renormalizable theory? Well, it's actually really easy to tell if an interaction
is "superficially" renormalizable: it's just a matter of some dimensional
analysis. "Superficial" renormalizability can be spoiled by other, deeper
problems like "anomalies". On the other hand, theories that fail to be
superficially renormalizable can sometimes be saved by subtler considerations,
usually involving symmetry. Nonetheless checking superficial renormalizability
is a pretty way to start checking if your quantum field theory is really
renormalizable.

And here's how you do it. You just use the following criterion:

An interaction is superficially renormalizable if its coupling
constant has dimensions of length^d with d less than or equal to zero, but not
if d is greater than zero.
(In fact, if d is less than zero, the
interaction is superficially "superrenormalizable", which is even better than
renormalizable.)

Let's see how this goes in practice. Suppose we have a fermion field psi in 4
dimensions. Then psi must have dimensions of length^{-3/2}. To see this,
remember that the action is the integral of the Lagrangian, and it includes a
term like this coming from the kinetic part of the Lagrangian:

integral psi* D psi d^4 x

This must be have dimensions of length^0 since the action of a quantum field
theory must always be dimensionless if we set hbar and c equal to 1, as we
always do. The Dirac operator D has dimensions of length^{-1} since it involves
a first derivative, while d^4 x has dimensions of length^4. Thus the only way to
get the above integral to have dimensions of length^0 is to give psi dimensions
of length^{-3/2}. Okay?

This lets us see that a 4-fermion interaction in 4 dimensions can't be
superficially renormalizable. The action for this sort of interaction is just
the integral of the above "interaction Lagrangian":

c integral psi* psi psi* psi d^4 x

The integral has dimensions of length^{-2}, so to make it dimensionless the
coupling constant in front must have dimensions of length^2. 2 is greater than
zero, so...

BZZZZZT! YOU LOSE! THIS INTERACTION IS NOT RENORMALIZABLE!

Of course, we've been assuming that spacetime is 4-dimensional. In
general it gets easier to find renormalizable field theories in lower
dimensions. To see why, let's let's go down to 3-dimensional spacetime and
repeat the above analysis.

In 3-dimensional spacetime the action for the free massless fermion field
looks like this:

integral psi* D psi d^3 x

To make this dimensionless, we need to give the fermion field psi dimensions
of length^{-1}, so now the integral

integral psi* psi psi* psi d^3 x

has dimensions of length^{-1}, so the coupling constant must have dimensions
of length^1. 1 is greater than zero, so...

BZZZZZT! YOU LOSE AGAIN! THIS INTERACTION IS NOT RENORMALIZABLE!

But don't give up hope yet! In 4 dimensions the coupling constant
had dimensions of length^2, while in 3 it had dimensions of length, so we should
soldier on and try working in 2-dimensional spacetime!

So let's go down to 2-dimensional spacetime. To make

integral psi* D psi d^2 x

dimensionless, we need to give the fermion field dimensions of length^{-1/2},
so now the interaction

integral psi* psi psi* psi d^2 x

has dimensions of length^0, so its coupling constant must have dimensions of
length^0. 0 equals 0, so...

BINGO! YOU WIN! THIS INTERACTION IS RENORMALIZABLE! YOU CAN PICK
UP YOUR NOBEL PRIZE BY THE DOOR AS YOU LEAVE!

See, it's pretty easy to play this game. I urge you to write down some
Lagrangians and see which ones are superficially renormalizable. This will give
you a nice (though superficial) feel for why quantum field works the way it
does. In particular, pay attention to how it gets harder to find renormalizable
interactions in higher dimensions, and how it's almost impossible in dimensions
greater than 4.

Why the Criterion Works

Now if you have the slightest bit of curiosity whatsoever, you will not rest
satisfied knowing this criterion for superficial renormalizability. You will
want to know why it works!

As I explained elsewhere, the
basic reason it works is the following. Suppose a particular bare coupling
constant c in the Lagrangian has dimensions of length^d. And suppose we do our
calculations with a cutoff, so we ignore all effects at length scales below the
distance D. Then we expect that the physical coupling constant c', measured at
the length scale D', varies in the following way:

c'/c ~ (D'/D)^{-d}

If d is negative, this means the physical coupling constant gets small at
small length scales. This means perturbation theory works well, so our theory is
nice - it's superrenormalizable. If d is positive, the physical coupling
constant gets BIG at small length scales. This is nasty: perturbation theory
breaks down, and our theory is nonrenormalizable. If d = 0, our theory is said
to be renormalizable, but only more detailed study can tell if it's actually
nice or nasty.

But the real question is this: why does the physical coupling constant depend
on distance in the first place???

So let me explain this.

The lowbrow way to explain it is to talk about "screening": an electron is
surrounded by a cloud of virtual electron-positron pairs; this cloud gets
polarized and "screens" the electron charge just like any other polarizable
medium would, meaning that the charge you observe decreases as you go farther
from the electron. And this cloud also affects the observed mass of the
electron, and so on: all sorts of physical constants differ from bare ones in a
scale-dependent way, thanks to the effect of virtual particles.

But you've probably heard this before, so if you're still not happy, maybe
you need a slightly more mathematical way of seeing how this stuff works. As
usual, the more mathematical way bears no resemblance to the lowbrow handwaving
explanation... or at least, that's what it seems like at first! Eventually one
sees they are the same thing.

Okay, let's consider a particular quantum field theory. Since we've already
talked about it, lets use the theory with the 4-fermion interaction - the psi^4
theory, whose Lagrangian looks like this:

psi* (D + m) psi + c psi* psi psi* psi

Let's think about how the bare coupling constant c differs from the physical
coupling constant c' in this theory. I've already told you
what happens: since c has dimensions of length^2, we get a relationship
roughly like

c'/c ~ (D'/D)^{-2}

where D is the cutoff distance and D' is the length scale at which the
physical coupling constant is defined.

My job now is to explain WHY this happens.

First, let me rewrite the above formula in terms of momenta rather than
distances:

c'/c ~ (P'/P)^2

where P is the cutoff momentum and P' is the momentum at which we measure the
physical coupling constant. Remember, ignoring all particles with momentum
greater than P is the same as ignoring all effects happening at length scales
less than D = 1/P, and measuring the physical coupling constant at momentum P'
is like measuring it at length scale D' = 1/P'.

So how do we measure the physical coupling constant? Neglecting various
important subtleties, let me just say this: we do so by scattering 2 fermions
off each other. Here's one way they can bounce off each other:
\ /
\ /
\/
/\
/ \
/ \

This process happens with an amplitude proportional to the bare coupling
constant c, because this Feynman diagram corresponds to the term c psi* psi psi*
psi in the Lagrangian of our theory.

If this were the *only* way 2 fermions could bounce off each other, the
physical coupling constant would equal the bare coupling constant.

But it's not! There are lots of other ways that involve virtual particles! To
figure these out, we just draw all Feynman diagrams with 2 particles coming in,
2 going out, and a bunch of 4-valent vertices where interactions occur. All of
these contribute to the physical coupling constant.

For example, our theory gives us a process like this:
\ /
\ __ /
\/ \/
/\__/\
/ \
/ \

Scattering by means of 2 virtual fermions!

To evaluate the amplitude of this process we use the usual "Feynman rules".
You only need a vague grasp of what these are to follow what I'm going to say.
We treat the momenta of the particles coming in and going out as fixed, and we
integrate over the momenta of the virtual particles, subject to the constraint
that momentum is conserved at each vertex: the momentum coming in equals the
momentum going out. We get a factor of the bare coupling constant c for each
vertex and we get a factor of

1/(-ip^a gamma_a + m)

for each edge, where p is the 4-momentum of the particle corresponding to
that edge, and gamma_a are the gamma matrices. This funny expression is called
the "propagator" or "Green's function" for a fermion of mass m - you get it by
inverting the operator (D + m). Take my word for this, please! Actually, all we
need to know about the propagator is that it decays roughly like

1/|p|

when p is large. It's detailed form won't matter here.

Okay, now for the crucial point: when we do the integral to compute the
amplitude of this process, it diverges! To see this, note that we are really
integrating over one 4-momentum p: there are two virtual particles, but the
momentum of one determines the momentum of the other, since momentum is
conserved at each vertex. But our integral contains two factors that behave
roughly like 1/|p| when p is large, one for each virtual particle. So our
integral looks roughly like

integral |p|^{-2} d^4 p

This diverges when we integrate over arbitrarily large p.

Okay, now what? Well, to get a finite answer, we put in a momentum cutoff: we
only integrate over momenta p with |p| smaller than some number P. If we do
this, the integral above becomes finite. However, as we increase P, the integral
gets bigger. At what rate? Here's where the dimensional analysis comes in: we've
got this integral

integral |p|^{-2} d^4 p

with dimensions of momentum^2, so if we do the integral up to momenta with
|p| = P, we expect that the answer scales like P^2.

Thus the physical coupling constant, which involves the amplitudes of both
\ /
\ /
\/
/\
/ \
/ \

and
\ /
\ __ /
\/ \/
/\__/\
/ \
/ \

is different from the bare coupling constant. It depends on the momenta of
the fermions we're scattering. Moreover, it grows roughly like P^2 as we
increase the momentum cutoff P.

Dimensional analysis thus suggests that

c'/c ~ (P/P')^2

I haven't proved this to you. But hopefully it seems not wholly unreasonable.
And, of course, it's exactly what our criterion says should happen.

mailto:[email-masked]
© 1999 John Baez

home