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  Deriving Feynman rules from Renormalized Lagrangian

+ 7 like - 0 dislike

In the context of Renormalized Pertubation Theory Peskin Schröder says:
The Lagrangian $$ \mathcal{L}=\frac{1}{2} (\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}\phi_r^4 + \frac{1}{2} \delta_Z(\partial_\mu\phi_r)^2 -\frac{1}{2}\delta_m^2\phi_r^2-\frac{\delta_\lambda}{4!}\phi_r^4 $$ gives the following set of Feynman rules:
------------>------------ = $\frac{i}{p^2-m^2+i\epsilon}$
------------X------------ = $i(p^2\delta_Z-\delta_m)$
and the two 4-vertices.

The question is: Why look the Feynman rules for the first and the fourth term of the Lagrangian look so different? I believe the answer is connected to the fact that one has to bring the kinetic term of the Lagrangian to its canonical form $\frac{1}{2} (\partial_\mu\phi_r)^2$ and has to interpret everything else as (possibly momentum dependent) vertices. How does this look in formulae?

This post imported from StackExchange Physics at 2014-06-06 20:05 (UCT), posted by SE-user quan
asked Jun 4, 2014 in Theoretical Physics by quan (35 points) [ no revision ]
retagged Jun 6, 2014

1 Answer

+ 3 like - 0 dislike

What is suggested in the question is essentially correct. When one defines some perturbative theory, one has to define which part of the Lagrangian is considered as a perturbation and which part is not. In the present case, the perturbation part is made of the terms 3,4,5,6. The Feynman rules for the "free propagators" always appear as inverse of the corresponding terms in the Lagrangian whereas the Feynman rules for the "interactions" always appear as the corresponding terms in the Lagrangian. The easiest way I find to remember this is the path integral derivation of the Feynman rules. Here is a finite dimensional analogue:  to compute perturbatively the integral $\int_{\mathbb{R}}dx e^{-\frac{1}{2} ax^2 + V(x)}$ where $V(x)$ is the perturbation simply means to expand $e^{V(x)}$ and to integrate each term of the expansion against the gaussian $e^{-\frac{1}{2}ax^2}$. It is obvious that the terms in $V(x)$ will appear to some positive power whereas $a$ will appear with some negative power because

$\int_{\mathbb{R}} dx e^{-\frac{1}{2}a x^2}x^{2n} = constant \frac{1}{a^n}$. 

answered Jun 7, 2014 by 40227 (5,140 points) [ revision history ]
Most voted comments show all comments

In QED, the renormalized Lagrangian is gauge invariant and so compatible with the "gauge principle".

The Lagrangian of QED in dimension 4 is the most general Lagrangian compatible with gauge invariance and renormalizability so it follows form the "gauge principle" and from the renormalizability. If one forgets the renormalizability condition, one can obviously write higher order gauge invariant terms in the Lagrangian.

@VladimirKalitvianski The issue is that you are writing off-topic comments. What could be your intention, if not to promote yourself? I am not deleting (or voting to delete) your comments yet, let's see what the rest have to say regarding the appropriateness of your comment in this case.

Hi @VladimirKalitvianski ,

I think it would really be worthwhile for you to finally go beyond the point of view many physicists held in the 1930/40s and familiarize yourself with the new RG and EFT point of view introduced by Wilson in the early 1970s. Lumo nicely wrote about this and you have even read  it (I assume) as you took part in the comment discussion


Before you do not understand this current point of view, which also gives a new view on the counterterm business and explains why it has to work when done right, you and the other people here are talking past each other at best ...

There is also nothing wrong with QED, which is just another effective theory with a specific domain of validity.

@dimension10: You are right, my comments are off topic. I will try to ask my questions as questions, OK?

Most recent comments show all comments

@40227: Thanks for your patience. That's right, the renormalizability is an essential part to "guess" the interaction Lagrangian. It means, we, doing by analogy with working cases with $\mathcal{L}_{int}=j_{ext}A$ for field or $\mathcal{L}_{int}=j\cdot A_{ext}$ for a charge motion and advancing $\mathcal{L}_{int}=jA$ in case of both $j$ and $A$ unknown, are making an error. We awkwardly repair this error  with counter-terms, which do not follow from the "gauge principle". I want to say that the gauge invariance has to be respected, of course, but it alone does not fix the interaction Lagrangian unambiguously. My idea is to formulate QED in such a way that no renormalization is necessary. You can consider this as fulfilling the subtractions exactly in the total Lagrangian, as in this toy model, so one starts from a physical Lagrangian with only physical coefficients and one obtains convergent perturbative series from the very beginning.

Seeing so many downvotes, I guess somebody does not like my idea of reformulation. Somebody thinks that QED is already OK and cannot be formulated differently. The latter is not proven so do not vote down. Otherwise prove the uniqueness of the current QED formulation, please.

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