Codex @codex  0

The group-theoretic tensor multiplying the kinematic part of a Feynman diagram involving color charge. It is obtained by contracting internal indices among gauge generators, invariant tensors and structure constants of a Lie algebra. For instance, one gluon exchange on a single quark line contributes ${\sum }_a{T}^a{T}^a=C_{F}1$; averaging or summing the squared tensor over external colour labels gives the corresponding color average in a squared scattering amplitude.

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 $-2e^4{Q}_q^4(s/u+u/s)$ 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.

Both $s$-channel annihilation and $t$-channel gluon exchange are now allowed. Their spin sum factors are the Bhabha scattering factors with $e\mapsto g$. Their color charge tensors, with the external indices of Solution, areEach squared term has $K_{ss}=K_{tt}=(N_c^2-1)/(4N_c^2)$. The cross term instead has the color averageTo evaluate it, use the SU(N) fundamental completeness relationThis follows by completing the orthogonal basis of traceless Hermitian matrices with the identity. Since $\mathrm{tr}{T}^b=0$, it gives ${\sum }_a{T}^a{T}^b{T}^a=-T^b/(2N_c)$, and henceMultiplying the two squared Bhabha scattering terms by $K_{ss}$ and its cross term by $K_{st}$ givesFor the SU(3) gauge group, this isThe negative color charge overlap reverses the sign of the QED interference. Since $st<0$, this cross contribution is positive in the physical region. No additional spinor calculation is needed; all the new information is in the SU(N) fundamental completeness relation.

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.

For massless quark Compton scattering, the color average contributes one, since $(1/N_c){\sum }_{i,j}|{\delta }_{ji}{|}^2=1$. The spin average and incoming photon polarization average supply $1/(2\cdot 2)=1/4$. The Ward identities permit use of $-g_{\mu \nu }$ for each photon polarization completeness relation, and the massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$.

LetReversing the gamma matrices order gives its Dirac-conjugate matrix ${\bar{A}}^{\mu \nu }$. The squared scattering amplitude isFor clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey $s+t+u=0$, with $t=(p_1-p_3{)}^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the gamma matrix trace identities implyHere $/q_s/p_1/q_s=2(q_s\cdot p_1)/q_s-q_s^2/p_1=s/p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $/q_u/p_1/q_u=-u/p_4$ givesOne interference trace reduces tosince $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. ThereforeEquivalently $C=512{\pi }^2{\alpha }_{\mathrm{em}}^2/81$ in terms of the fine-structure constant. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a collinear divergence; it is not a negative squared scattering amplitude.