I have been reading Srednicki from the beginning and doing all the exercises, and I hit a big roadblock at Q10.4, as I can't seem to figure out what Srednicki is doing in his solution. Luckily, I found an additional resource (http://hep.ucsb.edu/people/cag/qft/QFT-10.pdf), which details a formula for finding vertex factors for generic interaction terms in scalar field theory. What I am wondering is how I would go about deriving this formula? It's on page 18 of the source I linked, along with a brief (but not sufficient, at least for me,) explanation of why it is true. I'll repeat it here:

$$ \text{Vertex Factor} = i \prod_{i} \frac{\delta}{\delta\widetilde{\phi}(k_{i})} \mathcal{L}_{int} $$

Where the $\frac{\delta}{\delta\tilde{\phi}}$ is a functional derivative, and $\mathcal{L}_{int}$ is the interacting part of the Lagrangian, for example a term like $\frac{1}{3!}g\phi^{3}$. The vertex factor is the factor applied to a Feynman diagram in momentum space at a given vertex (ie. $iZ_{g}g$ for $\phi^{3}$ theory or $iZ_{\lambda}\lambda$ for $\phi^{4}$ theory).

In exercise 10.4 of Srednicki, the interaction term that was giving me trouble was $\frac{1}{2}g\phi\partial^{\mu}\phi\partial_{\mu}\phi$; but sure enough the above formula spat out the correct vertex factor after some algebra. My question then is the following:

  1. How would I derive this formula?
  2. Does it have a more general form for multiple scalar fields/gauge fields/fermion fields? For example, the vertex factor for the theory for $\mathcal{L_{int}} = g\chi\phi^{\dagger}\phi$ given by $i(\frac{\delta}{\delta\tilde{\chi}(k_{3})}\frac{\delta}{\delta\tilde{\phi}^{\dagger}(k_{1})}\frac{\delta}{\delta\tilde{\phi}(k_{2})})\mathcal{L_{int}} = ig$, which is actually the correct vertex factor. So it seems quite general for scalar fields, but I don't know if it applies to fermion/gauge fields.
  • $\begingroup$ Go through the derivation of the Feynman rules and watch where the coupling constant comes in. $\endgroup$ – Javier Oct 2 '17 at 21:23
  • $\begingroup$ @Javier I did this for my example of $\mathcal{L_{int}} = g\chi\phi^{\dagger}\phi$, which is how I arrived at my answer originally. I didn't notice a particular pattern that would have led me to this formula when I did it then (I did not even use a Fourier transform, which is required in the formula). In particular, in my problem Lagrangian (the one with derivative interactions), deriving the Feynman rules was much more difficult for me. I do however find it very easy to just write down the vertex factor for simple interaction terms by doing this. $\endgroup$ – Jay Oct 2 '17 at 21:31

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