# Is Compton Scattering the "Abelian limit" of $qg \rightarrow qg$?

I have calculated the average over initial and sum over final states of the squared amplitude for both Compton scattering $$e^-\gamma \rightarrow e^-\gamma$$ (QED) and quark-gluon scattering $$qg \rightarrow qg$$ (QCD).

Both these quantities are in agreement with the literature.

My understanding is that QED is the Abelian limit of QCD. Hence, I expect that to recover the QED analogue of a QCD scattering amplitude, I should be able to send the number of colours $$N_C \rightarrow 1$$.

However, I do not get that my $$qg \rightarrow qg$$ scattering amplitude reduces to $$e^-\gamma \rightarrow e^-\gamma$$ in the $$N_C \rightarrow 1$$ limit.

Is my understanding incorrect, or is something else going on here that I may naively be overlooking?

However, note that the $$3$$-gluon vertex and $$4$$-gluon vertex are proportional to the antisymmetric structure constant $$f^{abc}$$, which is absent in the abelian case. So these vertices vanish readily.
As for the quark-gluon vertex, it differs from the QED vertex $$i e \gamma^{\mu}$$ by the generator of the gauge group $$G$$ $$i g \gamma^{\mu} T^{a}$$. In the diagrams you will have a summation over the colors, the gluon propagator is proportional to $$\delta^{ab}$$, hence, the result will differ only by factors of: $$\sum_a \text{Tr} (T^{a} T^{a}) = C(G)$$ Where $$C$$ is a quadratic Casimir. For $$SU(N)$$ gauge group $$C(N) = C(adj) = N$$, so the case of QCD with $$N=1$$ will be in agreement with QED.