I've seen that for the QCD process $gg \rightarrow u\bar{u}$, where $g$ is a gluon and $u, \bar{u}$ are the up quark and the corresponding antiquark, there is s, t and u channels.

I perfectly understand the existence of s and t channels but I don't get how is u channel possible. For $e^+e^- \rightarrow e^+e^-$ there isn't u channel due to positron and electron are not identical, so why is this channel in the QCD process if quark and antiquark are not identical? Is not the same scheme?


Is not the same.

For e+e- -> e+e- the t-channel is a photon exchange, that takes into account electric charge conservation. This last constraint rules out the u-channel.

For gg-> quark anti-quark, you have a t-channel exchanging a quark. And you have a u-channel also by exchanging the initial legs. Note that these are different amplitudes. On the t-channel you can have a vertex involving (g(p1_in),u(p1_out),intermediate_quark) while on the u-channel you would have this vertex with (g(p2_in),u(p1_out),intermediate_quark). Here I have denoted the momentum as p_i.

  • $\begingroup$ So the key is that I can exchange a quark or antiquark in t-channel and this difference respect to my QED example (where photon and anti-photon are the same particle) implies the existence of u-channel? So, u-channel = t-channel with the anti-mediator? $\endgroup$ – Vicky Feb 16 at 0:44
  • $\begingroup$ The best way to approach this is to write all the diagrams that are consistent (preserve charge conservation for example) with the feynman rules of the theory to the order you are working, doing this notice that you can exchange the initial legs (gluons) while on the QED case you cannot. When building the feynman diagrams, remember the usual convention of drawing arrows on the fermion lines which track the conservation of charge. $\endgroup$ – Alejandro Celis Feb 16 at 10:51

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