# Feynman rules for scalar field with second order derivatives in the interaction term

Given the interaction term with $$N$$ scalars $$\phi_i$$, each massless, what would be the Feynman rules for an interaction term in the action as

$$\int d^dx (\partial^2 \phi^i)\phi_i(\partial_\mu \phi^j)(\partial^\mu \phi_j).$$

I tried to expand each $$\phi(x)$$ as $$\phi (x) = \int \dfrac{d^dp}{(2 \pi)^d}e^{ipx}\phi(p)$$, but then I noted that the second derivative will give a $$p^2$$ term, which gives zero for the vertex because the fields are massless. I think there is something wrong with my reasoning, would be glad with someone could clarify.

• In an actual diagram, the fields are not necessary on shell if they are internal lines. – Anonjohn Sep 10 at 16:47
• So the fact that the fields are massless or not doesn't change the answer? – Slayer147 Sep 10 at 17:43
• @Slayer147 There would be IR divergences because of the masslessness but it would be the same story as in the case of QED I suppose, i.e., you regularize and then sum over all the soft photon modes before setting the regulator to zero. – Dvij Mankad Sep 10 at 17:52
• I got $(2\pi)^d \delta(p_1 + p_2 + p_3 + p_4) p_1^2 p_2 p_4$, should I symmetrize the answer? I think that would be the correct thing to do – Slayer147 Sep 10 at 18:18