I'm currently studying QFT and QCD for the first time and I have a question about the colour charge indices given below. I was asked the following question:

(a) Derive the Feynman rule for the 3-gluon vertex $V\left[A_{\mu_1}^{a_1}\left(p_1\right), A_{\mu_2}^{a_2}\left(p_2\right), A_{\mu_3}^{a_3}\left(p_3\right)\right]$ from $$ \mathcal{L}_{\mathrm{QCD}}=-\frac{1}{4} F_{\mu \nu}^a F^{a, \mu \nu}+\ldots, \quad F_{\mu \nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a-g_{\mathrm{s}} f^{a b c} A_\mu^b A_\nu^c. $$

I was given the correct working but I'm confused about the below section of it, specifically how the colour indices d and e seem to become b and c when collecting the terms. I'm sure it's probably straightforward and I'm missing something obvious but any help would be appreciated.

$$ \begin{aligned} & i L_{Q C D}=-\frac{i}{4} F_{\mu v}^a F^{a, \mu v}+\ldots \\ & =-\frac{i}{4}\left(\partial_\mu A_\nu^a-\partial_\nu A_\mu^a-g_s f^{a b c} A_\mu^b A_\nu^c\right)\left(\partial^\mu A^{a, \nu}-\partial^v A^{a, \mu}-g_s f^{a d e} A^{d, \mu} A^{e, v}\right) \\ & \end{aligned} $$

Collecting terms with AAA: $$ \begin{aligned} i L_{Q C D}= & -i\left(\partial_\mu A_\nu^a-\partial_\nu A_\mu^a\right)\left(-g_s f^{a d e} A^{d, \mu} A^{e, \nu}\right) \\ & -\frac{i}{4}\left(-g_s f^{a b c} A_\mu^b A_v^c\right)\left(\partial^\mu A^{a, v}-\partial^v A^{a, \mu}\right)+\ldots \\ = & \frac{i g_s}{2}\left(\partial_\mu A_v^a-\partial_v A_\mu^a\right) f^{a b c} A^{b, \mu} A^{c, v}+\ldots \end{aligned} $$

  • $\begingroup$ Since $X^{abcbc}=X^{dede}$, $X^{abcbc}+X^{adede}=2X^{abcbc}$. (I'm taking $X^{afghi}:=f^{afg}A^{h,\,\mu}A^{i,\,\nu}$, hiding Greek indices.) $\endgroup$
    – J.G.
    Dec 29, 2023 at 11:43

1 Answer 1


Since d and e are being summed over (and it’s not multiplying other terms with b and c indices which could cause semantic issues), it’s valid to rename them as b and c.

  • $\begingroup$ Ah it's a dummy index thing then. Is it convention in QFT that two identical indices imply summation even when they're both upper? I'm just used to upper/lower summation. $\endgroup$
    – Aidan
    Dec 29, 2023 at 13:45
  • $\begingroup$ @Aidan Yes, this implies summation. $\endgroup$
    – Ghoster
    Dec 29, 2023 at 18:37

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