= Solution
{c}
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.
\Image[/past-exam-of-the-advanced-quantum-field-theory-course-of-the-university-of-oxford/2022-annihilation-exchange-channels.png]
{title=Annihilation and exchange channels with labelled external momenta}
{description=Original <Feynman diagrams>, with arrows denoting <fermion flow>. The left panel with $V=g$ and $f'=d$ is the distinct-<quark flavor> process. Both panels with $V=\gamma$ describe <Bhabha scattering>; both with $V=g$ and $f'=f=u$ describe identical-<quark flavor> scattering. The wavy internal line stands for the indicated vector mediator.}
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> is
$$
\mathcal M_s=\frac{g^2}{s}(T^a)_{ji}(T^a)_{kl}
[\bar v(p_2)\gamma^\mu u(p_1)][\bar u(p_3)\gamma_\mu v(p_4)].
$$
The <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 $\sum u\bar u=\sum v\bar v=\not p$, and the initial <spin average> contributes $1/4$. The resulting two <gamma matrix> <traces> contract to
$$
\begin{aligned}
\frac14\operatorname{tr}(\not p_2\gamma^\mu\not p_1\gamma^\nu)
\operatorname{tr}(\not p_3\gamma_\mu\not p_4\gamma_\nu)
&=8[(p_2\cdot p_3)(p_1\cdot p_4)+(p_2\cdot p_4)(p_1\cdot p_3)]\\
&=2(t^2+u^2).
\end{aligned}
$$
The initial <color average> contributes $1/N_c^2$. Summing final <color charges> gives the <distinct-flavor quark scattering color factor>
$$
K_{ss}=\frac1{N_c^2}\sum_{a,b}\operatorname{tr}(T^aT^b)\operatorname{tr}(T^aT^b)
=\frac{N_c^2-1}{4N_c^2}.
$$
Therefore
$$
\boxed{\overline{|\mathcal M_s|^2}=g^4\frac{N_c^2-1}{2N_c^2}\frac{t^2+u^2}{s^2}
=\frac{4g^4}{9}\frac{t^2+u^2}{s^2},\qquad C=\frac{4g^4}{9}.}
$$
Equivalently $C=64\pi^2\alpha_s^2/9$ for the <QCD coupling> $\alpha_s=g^2/(4\pi)$. There is no initial <gluon> <spin average>: the <gluon> is internal.
Back to article page
