The color factors in QCD tell us the relative strength of the coupling of a quark emitting a gluon, a gluon emitting a quark-antiquark pair or a gluon emitting two gluons. To calculate let them we need to apply the QCD Feynman rules to the simple one-vertex Diagrams mentioned above.
Note: the Feynman rule for a vertex involving two quarks and a gluon is: $-i g_s \gamma^{\mu}_{ij} T^{a}_{mk}$
- $q \rightarrow q + g$: Squaring the Feynman rule above gives a Matrix element proportional to $g_s^2 T^a_{ij} T^a_{km} = 4 \pi a_s \frac{1}{2} (\delta_{jk}\delta_{im} - \frac{1}{N_c}\delta_{ij}\delta_{km})$ where I have used the Fierz Identity. In the end I should get something like $4\pi a_s C_F \delta_{ij}$ where $C_F = \frac{N_C^2 - 1}{2N_C} = 4/3$, but I can't seem to manipulate the Delta functions in a way that would give me this result.
- $g \rightarrow q\bar{q}$: I'm generally confused how to calculate this vertex. The only Feynman rule I have is the on written down above. Exchanging the 4-momenta of the incoming quark in the above vertex with the gluon gives me sort of the process I want, but the quark should also become an antiquark. In the end I should get a contribtuin proportional to $g_s^2 T^a_{ij}T^b_{ij}$ which is a trace, not a matrix as above. How does the summation change from one diagram to the other?
- $g \rightarrow gg$: Using the 3-gluon vertex that is proportional to (when squared) $g_s^2 f^{abc} f^{dbc}$ where f is the structure factor of SU(3) one gets $4 \pi a_s C_A \delta^{ad}$ in the end, which is just doing some acrobatics with the structure function. However I'm also curious why squaring the structure function changes one of the indices ($a \rightarrow d$) in the right factor? Why is not $f^{abc}f^{abc}$? I seem to have some problem with the index notation here.
Cheers