Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 1 / b

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal{M}$ as the product of the stated Feynman rules.

Suppressing external colour indices, the quark graphs areThe massless Dirac equations give $\bar{u}/p_1=0$ and $/p_{2}v=0$. ConsequentlyThe last equality uses $\bar{u}/qv=0$. The quark-exchange sum is therefore $-g^2\bar{u}/k_1[T^a,T^b]v$.

For the gluon graph, use all incoming momenta $(k_1,k_2,-q)$ in the three-gluon tensor:Its double contraction is $k_{1\mu }{k}_{2\nu }{V}^{\mu \nu \rho }=s(k_2-k_1{)}^{\rho }/2$. The vertex and gluon propagator giveCombining the graphs yieldsThus the requested spinor proportionality factor is $-g^2\bar{u}(p_1)/k_{1}v(p_2)$ when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal{M}$, rather than $i\mathcal{M}$, the entire result gains one factor of $i$.

The double-longitudinal Ward identity for gluon fusion uses the defining Lie algebra identity $[T^a,T^b]=i{f}^{abc}{T}^c$ makes the boxed amplitude vanish:In QED, both external gauge bosons are photons, the charge matrices commute, and there is no three-photon vertex. The two fermion-exchange graphs then cancel without a third graph. In QCD, omitting the three-gluon interaction would leave a generally nonzero colour commutator: gauge-boson self-interaction is essential to the cancellation.

This is a tree-level manifestation of a Ward identity and, more generally in a gauge-fixed non-Abelian theory, the Slavnov-Taylor identities. They connect the matter representation, the structure constants and the gauge interactions, ensuring that auxiliary longitudinal polarizations do not enter physical probabilities. With all other external legs physical and on shell, replacing one external massless gauge-boson polarization by its momentum leaves zero. The double-contraction result proved here is one consequence, not by itself a proof that every open-index amplitude tensor is transverse off shell. At loop order the corresponding non-Abelian identities also involve Faddeev-Popov ghosts and a consistent symmetry-preserving renormalization.