# On shell and off shell simultaneously?

I am considering the following one loop virtual correction in the DIS process:

where I have a quark of momentum $p$ coming in, emitting a gluon before interacting with a photon of momentum $q$ to produce a fermionic propagator with momentum $p+q$. My question is, in the red box, I have an on shell initial or final state quark $p^2=0$ but in the green box I have an off shell fermionic quark propagator $p^2 \neq 0$.

So, in my equations, in particular upon evaluation of the loop integral $$\int d^D l \frac{\text{Tr}( \not p \gamma^{\nu} (\not p + \not q) \dots)}{p^2 (p+q)^2 (p-l)^2}$$ where the denominators are all off shell terms, in simplifying the numerator (the trace results in dot products of all the momenta scales in the problem) would I use $p^2=0$ or $p^2 \neq 0$?

• I'm curious where you got this diagram from? One usually considers connected amputated diagrams. Yours is not amputated. – Your Majesty Feb 28 '16 at 11:51
• I should have maybe drawn it with a cut through the propagator $p+q$. I want to compute the hadronic tensor $W^{\mu \nu}$ for this diagram which is the discontinuity of the forward scattering process I showed. Does that make more sense? – CAF Feb 28 '16 at 11:53
• The amputation should be wrt the self energy of the external leg. Read p19 in eg google.se/url?sa=t&source=web&rct=j&url=http://… – Your Majesty Feb 28 '16 at 15:48
• Ah I see, so p.19 had the same problem as I had - but they just said it gives infinity - so in my case do i get an infinity too and so I just don't bother continuing on with the calculation? – CAF Feb 28 '16 at 16:45
• What Faq is telling you is, basically, that the question is wrong. You shouldn't be calculating this diagram, because of the amputated legs prescription. What you can do is separately calculate the gluon's contribution to the quark propagator and then use that propagator in your tree level diagram. – Javier Feb 28 '16 at 19:27

Considering the free quark propagator in isolation and then looking at the $1$PI correction associated with the gluonic correction (the self energy) , we see that this vanishes in dimensional regularisation with the result that the diagram as given in the opening post does not contribute to the hadronic tensor.