# D'Alembert operator interaction term in QFT Lagrangian

I'm trying to understand how to find the Feynman rules (and use them to calculate loop diagrams) for this Lagrangian (found on the Saclay lectures): $$\mathcal{L}=\mathcal{L}_\text{kin}-\frac{\tilde{C_4}}{4!}\phi^4-\frac{\tilde{C_6}}{4!M^2}\phi^3\square\phi.\tag{2.6}$$ Now, the only thing I know about derivative interactions is that, when writing a field in momentum space, one has a $$e^{ipx}$$ factor, and so a term like $$\partial^\mu\phi$$ becomes $$(ip^\mu)$$ in the Feynman rule. Since here there is a $$\partial^\mu\partial_\mu\phi=\square\phi$$ term, I expect the Feynman diagram to be modified by $$(-p^2)$$. This should be everything, because the numerical factor should still be $$4!$$ (I have 4 fields that can all be the "derivate one", and once that's chose, I have a $$3!$$ factor for the other 3 fields).
So my rules would be: $$\phi^4\longrightarrow-i\tilde{C_4}$$ $$\phi^3\square\phi\longrightarrow i\tilde{C_6}\frac{p^2}{M^2}$$

Something tells me that I'm wrong, because the 1-loop correction to the propagator is written as $$-\tilde{C_4}\text{(integral of propagator)}+\frac{\tilde{C_6}}{4M^2}2(k^2+m^2)\text{(integral of propagator)}\tag{2.21}$$ Why are there the "2" and "4" factors? But most importantly, how can the mass $$m^2$$ be there?

I think he has a typo in $$(2.21)$$. In the first line the $$m^2$$ at the numerator of the second integral should actually be a $$p^2$$. You can see that in the second line a $$p^2$$ appears out of nowhere.

Anyway, the derivative is only acting on one field, so the $$p^2$$ in the Feynman rule is the momentum of one of the legs. When you take the momentum of the two legs that enter in the loop, you will get the $$k^2$$ terms, while when you take the momentum of the external legs you will get the $$p^2$$. The factor $$2$$ comes from the fact that you have $$4$$ legs, two internal and two external.

EDIT: for clarification, the full Feynman Rule is

$$\phi^3\square\phi\to -3!(p_1^2+p_2^2+p_3^2+p_4^2)$$

where $$p_i$$ is the momentum of the $$i$$ leg.

• And what about the factor $4$ in the denominator? Is it another symmetry factor? Jun 11, 2022 at 14:38
• @MauroGiliberti When you compute the Feynman rule of that operator you get a $3!$ from the three non-differentiated fields, so you are left with a $1/4$ from the $1/4!$ of the definition of the operator that is not cancelled. Jun 11, 2022 at 14:59
• But shouldn't one also consider that the differentiated field can be chosen as one of 4 fields, so a factor of $1/4$? Or is the differentiated field "different" from the others in that when I compute the Feynman rule, I have to treat it like another, separate, field? Jun 11, 2022 at 15:12
• @MauroGiliberti The fact that you can choose the differentiated field in 4 ways is what afterwards gives you the $2(k^2+p^2)$. I have added the full Feynman rule to the answer, maybe that's more clear Jun 11, 2022 at 15:20