0
$\begingroup$

Given a Lagrangian density containing an interaction with derivates, it's easy how to guess the Feynman rules for vertexes. However i was wondering about propagators: in S-matrix expansion it's possible to have a contraction of a field with it's derivative, and in this case propagator would be different from the free-field one. I read in the answers of post here that propagators depend only on the free Lagrangian part, meaning, no new terms. However in interactions with no derivates there are not contractions between fields and fields derivative, so this would lead to new terms. This is were i am confused. (Note: i didn't made yet the path integral formulation of QFT).

Example The scalar QED interaction is like $$(\phi \partial _\mu \phi ^* - \partial _\mu \phi \phi ^*)A^\mu $$ so in T-product Wick 's theorems predicts contraction $contr [\phi (x) \partial _\mu \phi ^*(y)]$. This contraction is new respect to QED one, but vanishes as $\mu = i$ (i have an integral antysimmetric in $k_i$), however the $\mu = 0$ term is non-vanishing and gives rise to $2i \delta^3 (\vec x-\vec y)\cos(\omega (x_0-y_0))$ if integrated in $d^3 k$. I don't know if i am wrong with calcs or for every derivative-interaction i have also new contractions (and so feynman rules) to insert; in this case it would be $$contr [\phi (x) \partial _\mu \phi ^*(y)] \rightarrow i \frac{k^0}{k^2-m^2+i\epsilon}$$

$\endgroup$
  • 1
    $\begingroup$ I read this a couple times and I'm still not sure what your question is. What's wrong with contracting with a derivative? It looks like you know how it looks in momentum space. $\endgroup$ – octonion Jan 2 at 16:52
  • $\begingroup$ Hi, the problem is i have contraction with derivatives so new propagators, however in literature i never found a modification in propagators due to derivative interaction (as an example sQED, whose propagators are the usual ones and i read nowhere about a propagator due to contraction phi dphi) $\endgroup$ – Lenz Jan 3 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.