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

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.

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.
• @korni1990 in momentum space the vertex with momenta $k_1, k_2, k_3$, which we will take to be outcoming, to each vertex one has to assign $-\lambda (k_2 \cdot k_3) + \text{permutations}$., where the conservation of momenta has to be imposed – spiridon_the_sun_rotator Aug 12 at 14:23