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
Best Answer
When considering real emission diagrams, one should always check the kinematics first. Your first process $g \rightarrow g+q$ is a perfect example (I use your link here to diagram C without the emitted photon). Gluons are massless, quarks are not and our theory (QCD) is Lorentz invariant. So we choose the rest frame of the initial quark and innediately see, that no gluon can be emitted without violating energy-momentum conservation.
A process like this never happens. That's why there is an additional photon emitted in the diagram linked above and thats why none of your other described processes work. Without further interaction, a gluon (or a photon in QED) cannot simply be emitted by a quark.
Consider QED, e.g. Cerenkov radiation. This phenomenon only seems to be just an electron emitting a photon. However, this can only happen in the vicinity of a nucleus (diagrammatically, this is implemented as an infinitely heavy "background", meaning that the mass $M$ of the nucleus is much larger than the electron mass: $M\gg m_e$.
(This is my old answer conserning loop corrections. I just leave it here, though it does not answer the asked questions.)
By sqaring the Feynman rule (as you put it) you draw a diagram, where you connect one gluon and one quark line. You thus have a loop diagram as a correction to the quark propagator. You can easily implement this, by multiplying with a delta function, that connects two indices, one of each $T$. E.G. $g_s^2T^a_{ij}T^a_{km}\delta_{jk} = 2 \pi \alpha_s(\delta_{jk}\delta_{im}-\frac{1}{N_c}\delta_{ij}\delta_{km})\delta_{jk} = 2 \pi \alpha_s (N_c \delta_{im}-\frac{1}{N_c}\delta_{im})=4\pi \alpha_s C_F\delta_{im}\ .$
The problem is similar here. You calculate a correction to the gluon propagator with a closed quark loop. Writing the terms in correct order, you see that, since it is a closed fermion line, the firs and last indices are the same, hence you calculate a trace.
Its the same thin again (here you calculate a gluon loop as a correction to the gluon propagator). You have to let go of the notion "squaring Feynman rules". Draw the diagram and add an index to each line. Apply the Feynman rules and you are done (modulo some $f$ acrobatics ;) ).