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

Question

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?

Answer 1

QCD in addition to the quark-gluon vertex, which has a direct analogue in QED - electron-photon vertex, has additional vertices, which appear in the case of non-abelian gauge group:

• 3-gluon vertex
• 4-gluon vertex
• ghost vertex (if one doesn't work in axial or any gauge, where the ghosts decouple)

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.