I've been working on the one-loop corrections, and encountered the following diagrams:

enter image description here

[a] and [b] come from the Yukawa Lagrangian $\phi\bar\psi\psi$. We can assign the momentum $k$ to one of the internal propagators in the loop, so the other part has the momentum $p+k$.

I'm confused about how this is done for [c] and [d] for the $\phi^3$ theory. For [c], I think the process is the same as [a], but in a solution of my exercise, the integral was written as

$$ \int\frac{d^4k}{(2\pi)^4}\frac{i}{(k-p/2)^2-m^2+i\epsilon}\frac{i}{(k+p/2)^2-m^2+i\epsilon} $$

Therefore I wonder if we should split the momentum $p$ for the $\phi^3$ case, if so, why do we do that?

For graph [d], does my momenta assignment look correct? I think I'm contradicting myself because it seems like we also have $p-k$ on the external leg and $p+k$ in the loop. The solution didn't care about the external momentum $p$.


1 Answer 1

  1. Each vertex obeys momentum conservation. In particular, the external legs obey total momentum conservation.

    Therefore the external momentum $p=0$ is zero in OP's self-loop diagram [d].

  2. Concerning OP's first question: One is allowed to shift the loop-momentum/integration variable $k$ if the integral is convergent (e.g. via Wick rotation + dimensional regularization).


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