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There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

Write $C_{ikjl}={\sum }_a{T}_{ik}^a{T}_{jl}^a$. Hermiticity converts its complex-conjugate factors into $T_{ki}^b{T}_{lj}^b$. Summing all external colors therefore givessince SU(3) has eight generators. This is a sum, before averaging the nine equally populated incoming color states.

For the spinor current, use $\bar{u}=u^{†}{\gamma }^0$, $({\gamma }^0{)}^2=1$ and the stated gamma-matrix adjoint identity:The spin sum completeness relation then closes each product of currents into a trace. Averaging two spins for each incoming quark, in addition to nine incoming colors, givesThe factor is $1/18$, rather than $2/9$: the latter omits the initial spin average.

To derive the four-gamma matrix trace, first anticommutation and cyclicity give $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$, with $\mathrm{Tr}1=4$. If $T=\mathrm{Tr}({\gamma }^{\lambda }{\gamma }^{\mu }{\gamma }^{\sigma }{\gamma }^{\nu })$, move ${\gamma }^{\lambda }$ successively through the next three matrices. The last term is $-T$ by cyclicity, soSubstitution yieldsConsequently the first momentum trace is $4(p_3^{\mu }{p}_1^{\nu }+p_3^{\nu }{p}_1^{\mu }-g^{\mu \nu }{p}_3\cdot p_1)$. Contracting both traces givesHere $s=(p_1+p_2{)}^2$, $u=(p_1-p_4{)}^2$ and all external masses vanish. Thus the distinct-flavor quark scattering color factor givesWe used $t=-s(1-\cos {\theta }^{\ast })/2$ and $u=-s(1+\cos {\theta }^{\ast })/2$. The forward divergence reflects massless $t$-channel exchange in this parton-level tree approximation; it is not an unrestricted prediction for an experimentally resolved jet cross section.

Finally, completeness of the SU(3) generators gives the fundamental-generator Fierz identityFor each distinct pair $i\ne j$, the allowed final colors are either $k=i,l=j$, with coefficient $-1/6$, or $k=j,l=i$, with coefficient $1/2$. If $i=j$, only $k=l=i$ is nonzero, with coefficient $1/3$. These are 15 ordered configurations: six unchanged-color cases, six exchanged-color cases and three all-equal cases. Their squared coefficients sum to $6/36+6/4+3/9=2$, independently checking the first result. Flavor remains tied to each external line; exchanging colors does not exchange the up and down quark species.