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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?

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1 Answer 1

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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.

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    $\begingroup$ It is possible to disregard diagrams with tadpoles (when a single gluon runs in a loop that begins and ends in the same interaction vertex), the justification for that is normal ordering. It is also possible to not avoid them (they will be effectively removed by renormalization). In either case you will get the same physics $\endgroup$ Feb 3, 2021 at 1:29
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Feb 3, 2021 at 4:30
  • $\begingroup$ 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 $S_k$? $\endgroup$
    – Sito
    Feb 3, 2021 at 9:36
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Feb 3, 2021 at 9:38

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