# Feynman rules for space-dependent coupling

Let's say I have an effective action which looks like (I got this action from large $$N$$ method for $$\varphi^4$$ theory): $$\int \frac{d^4x}{2g}\phi^2(x)+\int d^4x \ \log(-\nabla^2+\mu^2+i\phi(x)).$$

I want to know what are the Feynman rules for this effective theory. For that I thought to expand the log term as: $$\ \log(-\nabla^2+\mu^2+i\phi(x))= \log(1+i(\nabla^2+\mu^2)^{-1}\phi(x))+\log(\nabla^2+\mu^2)=\sum_n \frac{(i(\nabla^2+\mu^2)^{-1}\phi(x))^n(-1)^{n+1}}{n}$$

I dropped the $$\log(\nabla^2+\mu^2)$$ since it's just an infinite constant. This sum contains all powers of interaction but now the vertex factor in not a constant. I tried to fourier this term to see what is the action in this space but I got an awful thing which I'm not sure of... how can I know what are the Feynman rules here?

• Looks like the effective action you wrote down just involves introducing Hubbard-Stratanovich auxillart field $\phi$ to decouple $\phi_{}^{4}$ term in $\phi_{}^{4}$ theory and doing gaussian for $\phi$ field path integral. Where is the information of large $N$ used? Further can Eq. (A3) of Qmechanic's answer be useful to develop perturbation theory? – Sunyam Jul 9 at 18:40
• True. I omitted the factor of N because I didn't see how this is relevant to the questions.... Qmechanic's answer is helpful to get to this effective action, but how does it relevant to feynman rules? – Noam Chai Jul 9 at 18:52
• Can't expanding $\log$ like you id did and treating everything apart from quadratic part using trick described in Qmechanic's answer useful? – Sunyam Jul 11 at 15:43