This is a simple question arisen with the evaluation of the gluon propagator in the Landau gauge for $SU(N)$ Yang-Mills theory. I have to evaluate the integral $$ \int d^4xe^{ipx}\langle T^cA^c_\mu(x),T^dA^d_\nu(0)\rangle, $$ being $T_c$ the generators of the group, that we know has the form $$ G\left(\eta_{\mu\nu}-\frac{p_\mu p_\nu}{p^2}\right)\Delta(x,0) $$ being $G$ the contribution due to the $SU(N)$ group in the fundamental representation. One assumes that the propagator $\Delta(x,0)$ is given (maybe). Now, my guess for $G$ is $$ G=\frac{N^2-1}{2N}. $$ Is this correct?
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$\begingroup$ You should be able to evaluate the propagator in perturbation theory. The tree level contribution should be trivial to evaluate. You can use it to derive $G$, provided this factor does not depend on the coupling constant (does it?). $\endgroup$– AccidentalFourierTransformCommented Jan 3, 2018 at 11:56
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$\begingroup$ I thought this was textbook matter. I agree with you that the propagator is known through perturbation theory. This is tree level and I guess no contribution from coupling is seen at this order. $\endgroup$– JonCommented Jan 3, 2018 at 11:59
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$\begingroup$ G should come from T, so why you mention fundamental reps? $\endgroup$– Rho PhiCommented Jan 3, 2018 at 13:13
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It kind of depends on what you mean by $T^c$. Let these matrices live in a representation $R$ of the gauge algebra.
Due to $\mathrm{SU}(N)$ invariance, we have $$ \langle A^a A^b\rangle\propto \delta^{ab} $$ and therefore the colour structure $G$ is given by $$ G=T^aT^a\equiv C(R)\times \text{identity matrix} $$ as per Casimir.
If $R$ is the fundamental representation, we have $C(F)=\frac{N^2-1}{2N}$. If $R$ is the adjoint representation, $C(A)=N$.
answered Jan 3, 2018 at 12:40
AccidentalFourierTransformAccidentalFourierTransform
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$\begingroup$ So, are you confirming that my conclusion is right? $\endgroup$– JonCommented Jan 3, 2018 at 14:02
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$\begingroup$ @Jon Yes. But only if $T^a$ live in the fundamental representation (as you say in the OP). In other representations, $G$ changes (it is representation-dependent). $\endgroup$ Commented Jan 3, 2018 at 14:20