Feynman rules for interactions with derivatives: How exactly do the momentum factors appear?

Question

I know how to treat Feynman interactions without derivatives by Wick contraction. But now, take for example $$mathcal{L}_{int}=lambda phi (partial_{mu}phi)(partial^{mu}phi).$$
Now many books write that in momentum space the derivatives turn into momenta. While I can imagine this happening, I don't really know how to write this down explicitly. At what point do I consider the Fourier transform of the field? Am I still using Wick contractions, but now with the field depending on the momenta? I have not found a source doing this explicitly.

spiridon_the_sun_rotator · Accepted Answer

For your case, starting with this interaction term, let us substitute the expansion of $phi$ in Fourier modes:
$$
phi = sum_k phi_k e^{i kx}
$$
The action of derivative produces a factor of $ik$. Then, in the action you sum(integrate) over all $x$ :
$$
sum_x sum_{k_1, k_2, k_3} (ik_{2 mu}) (ik^{3 mu}) lambda phi_{k_1} phi_{k_2} phi_{k_3} e^{i (k_1 + k_2 + k_3) x} = 
 sum_{k_1, k_2, k_3} (ik_{2 mu}) (ik^{3 mu}) lambda phi_{k_1} phi_{k_2} phi_{k_3} delta (k_1 + k_2 + k_3) 
$$
Where in the last expression we have employed the well-known integral for exponent. The change from derivatives to momenta is simply results of change from positional basis, to momentum basis and has nothing to do with Wick theorem.

Feynman rules for interactions with derivatives: How exactly do the momentum factors appear?

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