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.

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    $\begingroup$ In an actual diagram, the fields are not necessary on shell if they are internal lines. $\endgroup$ – Anonjohn Sep 10 at 16:47
  • $\begingroup$ So the fact that the fields are massless or not doesn't change the answer? $\endgroup$ – Slayer147 Sep 10 at 17:43
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    $\begingroup$ @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. $\endgroup$ – Dvij Mankad Sep 10 at 17:52
  • $\begingroup$ 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 $\endgroup$ – Slayer147 Sep 10 at 18:18

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