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?


1 Answer 1


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.

  • $\begingroup$ And what about the factor $4$ in the denominator? Is it another symmetry factor? $\endgroup$ Jun 11, 2022 at 14:38
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    $\begingroup$ @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. $\endgroup$
    – FrodCube
    Jun 11, 2022 at 14:59
  • $\begingroup$ 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? $\endgroup$ Jun 11, 2022 at 15:12
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    $\begingroup$ @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 $\endgroup$
    – FrodCube
    Jun 11, 2022 at 15:20

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