• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,726 comments
1,470 users with positive rep
818 active unimported users
More ...

  How to "guess" Feynman rules from a Lagrangian?

+ 2 like - 0 dislike

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?

A nonquadratic term is necessarily an interaction.

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

Most recent comments show all comments

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

Would you be able to provide an example of a theory which has such interaction terms? Is there a well-known example or just toys?

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 (3,625 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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights