= Massless quark Compton scattering
{title2=$\overline{|\mathcal M|^2}=-2e^4Q_q^4(s/u+u/s)$}
The two electromagnetic <tree-level Feynman diagrams> for $q\gamma\to q\gamma$ exchange a <quark> in the $s$ and $u$ channels. Their sum obeys both photon <Ward identities>. Averaging over the initial <quark> <spin>, <color charge> and two physical incoming <photon polarization vectors> gives $-2e^4Q_q^4(s/u+u/s)$ after final-state sums. The <color average> is one because the interaction preserves the single open <quark> colour index. The interference trace is proportional to $(p_1+p_2)\cdot(p_1-p_4)=(s+t+u)/2=0$ for massless external states.
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