# Renormalization Conditions

1. Aug 20, 2006

### fliptomato

Hi everyone! I have a few questions regarding renormalization in QFT.

1. In Peskin chapter 10, he renormalizes $$\phi^4$$ theory using the renormalization conditions in equation (10.19), which basically say that the propagator has a pole at $$p^2=m^2$$ and that the 4-point interaction is exact for $$s=4m^2$$. These are reasonable assumptions (I think). However, in equation (12.30) of chapter 12, he introduces a different set of renormalization conditions defined at a spacelike momentum. I.e. the propagator is defined at $$p^2=-M^2$$ and the four point function is defined at $$s=t=u=-M^2$$. These are unphysical values, why are these renormalization conditions valid (or reasonable)? Why not use $$+M^2$$ and physically accessible conditions?

2. In chapter 12 section 4 he describes the renormalization of local operators. Is it correct to define a local operator as one that is roughly of the form $$\phi(x)^n$$? In the diagrams for the Greens function with a local operator on page 431, the diagrams being summed have different numbers of legs! (Similar to page 601-603) I don't quite understand what's going on here and why these diagrams with different in/out states can be summed together.

3. In chapter 11, p. 355, why is it acceptable to use the "tadpole diagram = 0" renormalization condition in place of the usual one for the propagator? How is this equivalent to the propagator condition?

Thanks very much,
Flip

Last edited by a moderator: Aug 21, 2006
2. Aug 20, 2006

### fliptomato

It looks like LaTeX failed... I'll repost without LaTeX:

Hi everyone! I have a few questions regarding renormalization in QFT.

1. In Peskin chapter 10, he renormalizes phi^4 theory using the renormalization conditions in equation (10.19), which basically say that the propagator has a pole at p^2 = m^2 and that the 4-point interaction is exact for s=4m^2, t=u=0. These are reasonable assumptions (I think). However, in equation (12.30) of chapter 12, he introduces a different set of renormalization conditions defined at a spacelike momentum. I.e. the propagator is defined at p^2 = -M^2 nd the four point function is defined at s = t = u = -M^2. These are unphysical values, why are these renormalization conditions valid (or reasonable)? Why not use physically accessible conditions?

2. In chapter 12 section 4 he describes the renormalization of local operators. Is it correct to define a local operator as one that is composed of a product of fields at the same spacetie point? In the diagrams for the Greens function with a local operator on page 431, the diagrams being summed have different numbers of legs! (Similar to page 601-603) I don't quite understand what's going on here and why these diagrams with different in/out states can be summed together.

3. In chapter 11, p. 355, why is it acceptable to use the "tadpole diagram = 0" renormalization condition in place of the usual one for the propagator? How is this equivalent to the proagator condition?

Thanks very much,
Flip

3. Aug 21, 2006

### fliptomato

If it helps makes my questions clearer, I've posted them online with images and LaTeX:

"[URL [Broken]
http://fliptomato.wordpress.com/2006/08/21/help-renormalization-conditions/ [Broken]

Thanks!

Last edited by a moderator: May 2, 2017
4. Aug 21, 2006

### Physics Monkey

Hi flip,

I don't have my Peskin on hand, so the best I can do at the moment is some general remarks.

As you pointed out, P&S originally introduce renormalization with what could reasonably be called "physical" renormalization conditions. However, it's important to realize that these choices are basically arbitrary. The thing you have to always remember is that your original theory is defined by the couplings and masses that appear in the bare Lagrangian (along with a cutoff). In other words, you can think of your quantum field theory as being defined by some lattice formulation, say, of the path integral. If you can adjust the bare parameters in your Lagrangian so that the continuum limit gives finite answers for the Green's functions, then you can say the theory is solved and renormalizable. The important point is that your theory depends only on those bare parameters, so that any separation of the Lagrangian into "unperturbed" and counter term pieces must produce the same physics regardless of your choice of renormalization scheme.

Now, when first discussing renormalization and for certain elementary physical applications, it is often desirable to parameterize the theory in terms of physically measured variables. In this case the physical mass and physical coupling will appear in the "unperturbed" piece of the Lagrangian. The (usually infinite) counter terms are then adjusted so that the propagator always has a pole at the physical mass, etc. You know this story already.

However, this "old fashioned" view is clearly overly restrictive because as I have just explained, none of the parameters appearing in the "unperturbed" Lagrangian need have any direct physical significance. Again, this is because the division between "unperturbed" Lagrangian and counter terms is arbitrary; it is only the full Lagrangian which determines the physics. Personally, this point took me a while to really digest, mostly because renormalization seems to be taught in such a mysterious way.

There are many reasons, some physical and some technical, why it might be advantageous to use some other renormalization scheme. On a technical leverl, calculations may simply be easier with unphysical choices. For example, it is often convenient (it simplifies the algebra) to set all three kinematical variables (s,t,u) equal when defining your renormalization scheme. A very popular scheme called minimal subtraction (MS) which is used with dimensional regularization defines the counterterms to cancel just the poles at the physical dimension. A more physical reason for using MS, say, is that that the pole structure of your theory might not be what you expect. This might arise because of massless particles (infrared divergences) such as the photons of QED. As a final example, consider QCD. There are many more hadrons and mesons than parameters in your Lagrangian. Indeed, observable particles do not appear to have their own fields in the basic QCD Lagrangian, so in this case it is impossible to give the parameters in the Lagrangian direct physical meaning.

I hope this wasn't just a bunch of stuff you've heard already. I will also try to answer your other questions once I can actually dig up my P&S to have a look. Also, I very highly recommend the book "Renormalization" by John Collins for a super clear exposition of all these topics.