While I am trying to compute the amplitude for the following Feynman diagram


I realized that the external leg contraction of $g$ should carry group generator index $A$ or $B$, is that right? If so, what would their polarization sum be

$$ \sum _{\mathrm{polarization}} \epsilon _{\mu}^A \left( k_1 \right)^* \epsilon _{\nu}^B \left( k_2 \right) $$

Thanks for any hint!

  • $\begingroup$ You need also to take into account the colors of the quarks. And the polarization sum can be replaced by $-\eta_{\mu\nu}\delta^{AB}$. $\endgroup$ – user178876 Dec 20 '18 at 17:33
  • $\begingroup$ @marmot Could you please give me a hint on doing the calculation? Thanks! $\endgroup$ – zyy Dec 20 '18 at 18:08
  • $\begingroup$ Look up the Feynman rules for QCD, read off the matrix element and square it. Make sure that you take into account the colors of the quarks. Do not use explicit structure constants because when you square the diagram you will be able to use the group theoretical relation between the sum over the structure constants, and also that $c_1=\ell=$Dynkin index for the adjoint. $\endgroup$ – user178876 Dec 20 '18 at 18:14

No, there can only exist colorless final states, and so cannot carry a color index. The external leg factors only care about the spin of the particle. Namely, spin-0 particles get unity, spin-1/2 get the eigenspinors $u(p), \bar{u}(p)$ (or $v$), and spin-1 particles the polarization vectors.

The color index will only occur in the vertex factor, though when we draw the feynman diagram for a color triplet we usually write $i, j$ (fundamental color indices) on the "edge" of the propagator, though the indices only appear in the vertex factors.

  • $\begingroup$ Thanks for your answer! Then my further question is, who is going to balance out the indices in the structure constant $f^{A B C}$ incurred in the three gluon vertex? $\endgroup$ – zyy Dec 20 '18 at 21:55
  • $\begingroup$ When you square and average over color indices, you get things like $Tr(T^a T^b) \propto \delta_{ab}$. If you want a more in depth answer you can ask another question and post the link here. $\endgroup$ – InertialObserver Dec 20 '18 at 22:13
  • $\begingroup$ Also, if you believed I answered your question don't forget to accept it as the answer. $\endgroup$ – InertialObserver Dec 20 '18 at 22:14
  • $\begingroup$ I am sorry, I should have made the question clearer, in the equation, the superscripts for $\epsilon$ are group generator indices, meaning they correspond to different group generators. $\endgroup$ – zyy Dec 21 '18 at 16:15

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