## Renormalizability

#### November 14, 2006

I assume you've looked at my webpage on renormalization, and 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:

ψ* (∂ + m)ψ + c ψ* ψ ψ* ψ

Here ψ is a fermion field. ∂ is my feeble attempt to draw the Dirac operator, which should really have a slash through it. ψ* is my feeble attempt to draw a ψ with a bar over it. m is a number called the mass and c is a number called a coupling constant. I've left out a lot of indices to make things easy to read. After all, I'm trying to make things seem simple; I'm not trying to be really precise. The first term is called the "free Lagrangian". The part that looks like ψ* ∂ ψ is called the "kinetic term" and part that looks like ψ* mψ is called the "mass term" since m is the mass of our fermion. The second term is the "interaction Lagrangian". In the language of Feynman diagrams it corresponds to a interaction like this:

```     \    /
\  /
\/
/\
/  \
/    \
```
Two fermions go in and bounce off each other; two fermions come out! A total of four edges meet at the vertex in the picture, because we've raised ψ to the fourth power in the interaction Lagrangian.

### 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 lengthd 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 ψ in 4 dimensions. Then ψ 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:

∫ ψ* ∂ ψ d4x

This integral must have dimensions of length0 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 ∂ has dimensions of length-1 since it involves a first derivative, while d4x has dimensions of length4. Thus the only way to get the above integral to have dimensions of length0 is to give ψ 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 ∫ ψ* ψ ψ* ψ d4x

The integral has dimensions of length-2 so to make it dimensionless the coupling constant in front must have dimensions of length2. 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:

∫ ψ* ∂ ψ d3x

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

∫ ψ* ψ ψ* ψ d3x

has dimensions of length-1, so the coupling constant must have dimensions of length1. 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 length2, 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

∫ ψ* ∂ ψ d2x

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

∫ ψ* ψ ψ* ψ d2x

has dimensions of length0, so its coupling constant must have dimensions of length0. 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 lengthd. 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 ψ4 theory, whose Lagrangian looks like this:

ψ* (∂ + m)ψ + c ψ* ψ ψ* ψ

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 length2, 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 ψ* ψ ψ* ψ 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/(-ipa γa + m)

for each edge, where p is the 4-momentum of the particle corresponding to that edge, and γa 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. Its 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

∫ |p|-2 d4p

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

∫ |p|-2 d4p

with dimensions of momentum2, so if we do the integral up to momenta with |p| = P, we expect that the answer scales like P2.

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 P2 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.