# Avoiding unnecessary calculations when deriving Feynman rules

I'm trying to understand some principle behind deriving Feynman rules for QCD, i.e. avoid unnecessary calculations.

## Setup

Assuming we have a gloun field $$A_\mu^a$$, fermion fields $$\psi$$ and ghosts $$c^a$$, consider the action \begin{aligned} S=&\int d^4x\Big[-\frac{1}{4}\left(F_{\mu \nu}^{a}\right)^{2}-\frac{1}{2 \xi}\left(\partial_{\mu} A_{\mu}^{a}\right)^{2}+\left(\partial_{\mu} \bar{c}^{a}\right)\left(\delta^{a c} \partial_{\mu}+g f^{a b c} A_{\mu}^{b}\right) c^{c} \\ &+\bar{\psi}_{i}\left(\delta_{i j} i \partial_\mu\gamma^\mu+g A^{a} T_{i j}^{a}-m \delta_{i j}\right) \psi_{j}\Big], \end{aligned} where $$T^a$$ are elements of $$su(N)$$. This action can be split into a free and interaction part. The interaction part takes the form \begin{aligned} S_{\mathrm{int}}&=\int d^4x \Big[-g f^{a b c}\left(\partial_{\mu} A_{\nu}^{a}\right) A_{\mu}^{b} A_{\nu}^{c}-\frac{1}{4} g^{2}\left(f^{e a b} A_{\mu}^{a} A_{\nu}^{b}\right)\left(f^{e c d} A_{\mu}^{c} A_{\nu}^{d}\right)\\&+g f^{a b c}\left(\partial_{\mu} \bar{c}^{a}\right) A_{\mu}^{b} c^{c} +g A_{\mu}^{a} \bar{\psi}_{i} \gamma^{\mu} T_{i j}^{a} \psi_{j}\Big]\\ &=:S_{AAA} + S_{AAAA} +S_{Ac}+S_{A\bar\psi\psi}. \end{aligned} The generating functional is then defined as $$W[J_\mu^a, \eta_i,\bar\eta_i,\xi^a,\bar\xi^a]= \exp\left(iS_{\mathrm{int}}\left[\frac{\delta}{i\delta J_\mu^a}, -\frac{\delta}{i\delta \eta_i},\dots\right]\right)W_0[J_\mu^a, \eta_i,\bar\eta_i,\xi^a,\bar\xi^a],\tag{1}$$ where we have introduced source functions for the different fields ($$A^a_\mu \leftrightarrow J^a_\mu$$, $$\psi_i \leftrightarrow \bar\eta_i$$ and $$c^a \leftrightarrow \bar\xi^a$$).

The three-gluon vertex is then given by $$\left\langle 0\left|A_{\mu}^{a}\left(x_{1}\right) A_{\nu}^{b}\left(x_{2}\right) A_{\rho}^{c}\left(x_{3}\right)\right| 0\right\rangle=\left.\frac{\delta}{i \delta J_{\mu}^{a}\left(x_{1}\right)} \frac{\delta}{i \delta J_{\nu}^{b}\left(x_{2}\right)} \frac{\delta}{i \delta J_{\rho}^{c}\left(x_{3}\right)} W\left[J_{\mu}^{a},\dots\right]\right|_{J=0}.\tag{2}$$ To calculate $$(2)$$ I think one would expand the exponential in $$(1)$$ and then apply all the operators in $$S_{\mathrm{int}}$$ to $$W_0$$, i.e. we calculate explicitly $$iS_{\mathrm{int}}W_0$$. Then we apply the three functional derivatives of $$(2)$$ and we are done.

## Question

My question is about computing $$iS_{\mathrm{int}}W_0$$. Do we really need to compute the whole $$iS_{\mathrm{int}}W_0 = iS_{AAA}W_0 + iS_{AAAA}W_0 +iS_{Ac}W_0+iS_{A\bar\psi\psi}W_0$$ if I'm only interested in the three-gluon vertex? It seems that calculating $$iS_{AAA}W_0$$ should be enough, i.e. the other terms don't contribute anything. The thing is, I don't know if this is true, and if it is, what the general principle behind this is. If I'm interested in the four-gluon vertex, do I just need $$iS_{AAAA}W_0$$, etc?

What OP suggests is true if we are only calculating the connected three-gluon vertex function to first-order $${\cal O}(g)$$ in the coupling constant $$g$$. To higher order in the coupling constant $$g$$, the other interaction vertices (i.e. quartic, quark, ghost, counterterms) contribute as well. In perturbation theory, we should include all possible connected Feynman diagrams with 3 external gluon legs and all possible interactions.
• $\uparrow$ Right. Feb 3, 2021 at 4:30
• Thanks to both of you for the answers. Just to be sure, when you, @Qmechanic, say "To higher order in the coupling constant $g$", do you mean that we expand $\exp(iS_{\mathrm{int}})$ to higher orders, which then would lead to mixing of the different $S_k$?
• $\uparrow$ Yes. Feb 3, 2021 at 9:38