# How do I assign momenta for internal loops of a Feynman diagram?

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

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).