A <Fierz rearrangement> puts the charged-current operator in the same current ordering as the neutral-current operator. Accounting for the interchange of fermionic fields, the combined amplitude is
$$
\mathcal M=\frac{G_F}{\sqrt2}
[\bar u(k')\gamma^\alpha(1-\gamma^5)u(k)]
[\bar u(p')\gamma_\alpha(C_V-C_A\gamma^5)u(p)],
\qquad C_V=c_V+1,\quad C_A=c_A+1.
$$
Sum over final spins and average over the initial electron spin. The <fermion spin sum> and <gamma-matrix trace> identities give
$$
\overline{|\mathcal M|^2}
=4G_F^2\left[(C_V+C_A)^2s^2+(C_V-C_A)^2u^2\right].
$$
For massless two-body scattering in the <centre-of-momentum frame>, $d\sigma/dt=\overline{|\mathcal M|^2}/(16\pi s^2)$ and $u=-s-t$. Integrating $-s\leq t\leq0$ therefore yields
$$
\begin{aligned}
\sigma
&=\frac{G_F^2s}{4\pi}\left[(c_V+c_A+2)^2+\frac13(c_V-c_A)^2\right]\\
&=\boxed{\frac{G_F^2s}{3\pi}\left(c_V^2+c_A^2+c_Vc_A+3c_V+3c_A+3\right)}.
\end{aligned}
$$
Consequently
$$
\boxed{H(s)=\frac{s}{3\pi},\qquad B=1,\quad C=1,\quad D=3,\quad E=3,\quad F=3.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-305/3/c/solution