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This is a somehow follow-up question from this one about a derivative interaction: I'll use the Lagrangian and couplings of that question, but any theory with two quartic interactions (or interactions with the same number of fields) would suffice.

The Lagrangian I'm working with is this one: $$\mathcal{L}=\mathcal{L}_\text{kin}-\frac{\tilde{C_4}}{4!}\phi^4-\frac{\tilde{C_6}}{4!M^2}\phi^3\square\phi$$ and, learning from the accepted answer to the linked question, the Feynman Rules are $$\phi^4\longrightarrow-i\tilde{C_4}$$ $$\phi^3\square\phi\longrightarrow i\frac{\tilde{C_6}}{4M^2}\sum_{i=1}^4 p_i^2. $$ Now, can I simplify my calculations by assuming that there's only a single 4-vertex in my theory, and its Feynman Rule is $$-i\tilde{C_4}+i\frac{\tilde{C_6}}{4M^2}\sum_{i=1}^4 p_i^2.$$ My doubt comes from the fact that, when I have multiple vertices in a diagram (3-3 scattering at tree-level, or 2-2 scattering at 1-loop-level), I should consider all possible combinations of the two vertices, and I'm not sure if summing the two in a single one would mess with this.

Edit: after some checks with FeynCalc, it seems that I can indeed, at least with this particular example. Is this always the case?

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Feynman rules are just an organizational tool, and yes you are free to collect "similar looking vertices" into a single vertex if you like. The potential issue/reason that you wouldn't want to (in this case), is that the two operators you have written down seem to be coming from different orders in your power counting - i.e. from an Effective Field Theory perspective if $M$ is some large scale of physics you have integrated out, then usually you will be working to some fixed order in $M$ (i.e., leading order would be ignoring all $\frac{1}{M}$ terms). By collecting vertices of different power counting into a single vertex, you are messing up your power counting, which is usually not advisable.

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  • $\begingroup$ This Lagrangian is precisely an effective one, and I want to discard all terms smaller than $M^{-2}$. Can't I work with all the couplings and then ignore those with a power of $\tilde{C6}$ higher than 1? $\endgroup$ Jun 13, 2022 at 7:24
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    $\begingroup$ Yes - it's just that since you'll essentially always be computing things to some fixed order in $M$, it's nicer to keep the vertices separate since they contribute at different orders $\endgroup$ Jun 14, 2022 at 2:08

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