 #1
CAF123
Gold Member
 2,914
 88
Main Question or Discussion Point
At next to leading order, a photon may couple to gluon via means of a quark loop insertion. Consider the process $$ \gamma(k_1) + g(p_1) \rightarrow \gamma(k_2) + g(p_2)$$ Apparantly there are eight diagrams at one loop contributing to this process and two are zero on grounds of colour constraints. The six non zero diagrams are shown in the image (the pencilled in momenta is due to me, time flows from left to right so I think we always have a ##k_1## and a ##p_1## on the left hand side and ##k_2## and ##p_2## on the right)
I understand that diagrams 3), 4), 5) and 6) are different but I don't see how diagrams 1) and 2) make sense. I redrew diagrams 5) and 6) to have the standard box structure of 3) and 4) (in 5) and 6) the two gluons are on vertices diagonally across from each other when drawn in this way) and with a different direction of route momentum this would ultimately give different integral representations for these two processes.
When I redraw 1) and 2) in a similar manner, I get ##p_2## as an initial state particle (i.e appearing on the left of the diagram) and ##p_1## as a final state one (i.e appearing on the right) so I don't see how this diagram contributes.
Thanks!
I understand that diagrams 3), 4), 5) and 6) are different but I don't see how diagrams 1) and 2) make sense. I redrew diagrams 5) and 6) to have the standard box structure of 3) and 4) (in 5) and 6) the two gluons are on vertices diagonally across from each other when drawn in this way) and with a different direction of route momentum this would ultimately give different integral representations for these two processes.
When I redraw 1) and 2) in a similar manner, I get ##p_2## as an initial state particle (i.e appearing on the left of the diagram) and ##p_1## as a final state one (i.e appearing on the right) so I don't see how this diagram contributes.
Thanks!
Attachments

24.8 KB Views: 535