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Is it non-trivial that the coupling constant $g$ in gluon self-interaction terms is the same as the coupling constant $g$ in gluon-fermion interaction term in Yang-Mills theory?

Pure Yang-Mills theory is well- defined itself. But it seems to me that it contains more information due to the same $g$ in the minimal coupling between the Yang-Mills field and fermions.

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    $\begingroup$ Posit different couplings, and observe gauge invariance failure. $\endgroup$ Commented Aug 17, 2023 at 13:42

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In annihilation of fermion-antifermion into gluons, there are three diagrams, the usual two that appear in the analogous QED process (the $t$ and $u$ channels).

t and u channel diagrams

However, because of the existence of the three-gluon coupling, there is also a new $s$-channel diagram.

s channel

[Diagrams from Peskin and Schroeder, Introduction to Quantum Field Theory, chapter 16]

If we call the coupling between the fermions and the gauge bosons $g_{f}$ and the pure gauge coupling $g_{b}$, the first two diagrams are $\mathcal{O}(g_{f}^{2})$, while the third is $\mathcal{O}(g_{f}g_{b})$. To have a well-defined theory, these three diagrams have to add up to a gauge-invariant, unitary (meaning with no production of unphysical timelike gauge field modes with negative norm) matrix element $\mathcal{M}$. In order for this to happen, there have to be cancelations between the $\mathcal{O}(g_{f}^{2})$ and $\mathcal{O}(g_{f}g_{b})$ parts of $\mathcal{M}$, which is not generally going to be possible if $g_{f}$ and $g_{b}$ are independent parameters. This requires that the coupling to the fermions be through a representation of the gauge group, with the same coupling $g$ as in the pure gauge sector.

Incidentally, it is also possible to use a similar argument to show that the three- and four-gluon vertices also have to have the same couplings $g$ and $g^{2}$. In this case, there needs to be cancelations among these four diagrams in order to ensure unitarity.

gluon-gluon scattering

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