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How to "guess" Feynman rules from a Lagrangian?

+ 2 like - 0 dislike
1628 views

Hello everyone.

I'd like to know is there a way to just "guess" Feynman rule in QFT from a Lagrangian. I have a lot of simple effective Lagrangians and I have no intention to derive them everytime from Wick's theorem.

It seems to me that sometimes it's possible to not derive them but just guess. Could you recommend me such quasi-mnemonic technique?

Examples of Lagrangians:

\(L = \partial_{\mu} \phi \partial^{\mu} \phi + \lambda \phi F_{\mu\nu} F^{\mu\nu}\)

\(L = \partial_{\mu} \phi \partial^{\mu} \phi + \lambda \phi \epsilon^{\mu \nu \rho \sigma} F_{\mu\nu} F_{\rho\sigma}\),

where phi is a scalar in the first Lagrangian and a pseudo-scalar in the second one respectively. 

Thank you.

asked Oct 23, 2014 in Theoretical Physics by Kirill [ no revision ]
Most voted comments show all comments

Just think about expanding the exponential around the gaussian. Check out Srednicki's book for millions of examples.

Split the Lagrangian into a quadratic part and call the remaining terms interaction terms. Then you introduce vertices for each of the interaction terms with the propagators coming from the quadratic piece. As Ryan points out, you need to work one example to completion from one of the books. Then, the Feynman rules follow more or less by inspection.

Is $\lambda \phi F^2$ an interaction? Then where is a kinetic term for $F$ (something like $F^2$)? Or $F$ is an external field?

Yes, this seems weird to me too. I have not seen such a term before.

A nonquadratic term is necessarily an interaction.

Most recent comments show all comments

This theory was used by experimenters to reproduce simple properties of \(\pi^0 \rightarrow \gamma \gamma \) decay as an effective Lagrangian.

Thus, your Lagrangian misses the field kinetic term $\propto F^2$ as well as the pion mass term $m^2 \phi ^2$.

1 Answer

+ 4 like - 0 dislike

Hi. You cannot "guess" the Feynman rules just by looking the Lagrangian. You cannot guess the form of the propagators and correlation functions from the Lagrangian unless you use the machinery developed over the past 60 years either by taking the Lagrangian and using Schwinger-Dyson equations or by taking directly the path integral and using the standard rules you can find in any textbook (where, as mentioned by Ryan, you take the saddle point approximation for Z[J,λ]). What you can see straight from the Lagrangian is what interactions you will have in your theory. Thus, for your examples I would just take the interaction terms and make the appropriate variations to see what I get. By the way, just keep in mind that if you work in 4d your interaction terms are non-renormalizable while, I  think, your kinetic terms should include a covariant derivative \(D_i = \partial_i -iQA_i\) or something like that.

*Some lecturers seem to be guessing the rules since most examples done are pretty much standard and in some sense just by looking at the Lagrangian you can indeed write down the rules.

answered Oct 25, 2014 by conformal_gk (3,535 points) [ revision history ]
edited Oct 25, 2014 by conformal_gk

That' right. Without interaction, the diagrams are just free lines of all fields in any combination.

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