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For an on-shell massless quark with incoming momentum $k$ and outgoing momentum $k+q$, average the initial spin and sum the final spin. The gamma matrix trace identities give $S^{\mu \nu }=\frac{1}{2}\mathrm{Tr}[/k{\gamma }^{\mu }(/k+/q){\gamma }^{\nu }]=2[k^{\mu }(k+q{)}^{\nu }+k^{\nu }(k+q{)}^{\mu }-g^{\mu \nu }k\cdot q]$. On shell $2k\cdot q=Q^2=-q^2$, and defining $T^{\mu \nu }=-g^{\mu \nu }+q^{\mu }{q}^{\nu }/q^2$ and $\tilde{k}=k-(k\cdot q/q^2)q$ yields the displayed transverse decomposition. With the hadronic-tensor prefactor $1/(4\pi )$, one-particle Lorentz-invariant phase space gives $w_f^{\mu \nu }=(Q_f^2/2){\delta }_{+}((k+q{)}^2)S^{\mu \nu }$. The antiquark has the same unpolarized tensor and squared electric charge.

For the massless contact interaction $-(G_F/\sqrt{2})[\bar{d}{\gamma }^{\alpha }(P+Q{\gamma }^5)u][\bar{c}{\gamma }_{\alpha }(P+Q{\gamma }^5)s]$ with real $P,Q$, spin and colour averaging givesThe gamma matrix trace identities give the averaged amplitude $4G_F^2[(C^2+4P^2{Q}^2)(p_2\cdot k_1)(p_1\cdot k_2)+(C^2-4P^2{Q}^2)(p_2\cdot k_2)(p_1\cdot k_1)]$, where $C=P^2+Q^2$. Substitute the massless Mandelstam variables and divide by $64{\pi }^{2}s$. The two singlet colour contractions give factor one after final summation and initial averaging. This is a low-energy Fermi interaction result, not its high-energy extrapolation.