Solution (source code)

= Solution

For <Electron>-<neutrino> scattering the <weak charged current> adds an <Electron> left-current. The supplied <Fierz rearrangement> has a minus sign for matrix components; the extra minus from ordering the external fermions cancels it. Accordingly the charged contribution is constructive in the <neutrino>-current/<Electron>-current ordering used in part (b):
$$
\boxed{\mathcal M_e=-\frac{G_F}{\sqrt2}
[\bar u_\nu(k')\Gamma_L^\alpha u_\nu(k)]
[\bar u_e(p')\gamma_\alpha((c_V+1)-(c_A+1)\gamma^5)u_e(p)].}
$$
Thus $\mathcal M_e$ is obtained from the physically consistent $\mathcal M_\mu$ by $c_V\mapsto c_V+1$, $c_A\mapsto c_A+1$. The right-<Electron> coupling $c_V-c_A$ is unchanged, while the left-<Electron> coupling $c_V+c_A$ increases by two. This is the <charged-current shift in neutrino-electron scattering>.

For the antineutrino, choose the external phase convention in which
$$
\boxed{\overline{\mathcal M}_e=+\frac{G_F}{\sqrt2}
[\bar v_\nu(k)\Gamma_L^\alpha v_\nu(k')]
[\bar u_e(p')\gamma_\alpha((c_V+1)-(c_A+1)\gamma^5)u_e(p)].}
$$
The reversed ordering of the antineutrino spinors implements <crossing symmetry>; the displayed overall sign can be changed by an external-state phase. Its <neutrino> trace has the opposite antisymmetric part, so the $s^2$ and $u^2$ weights exchange:
$$
\overline{|\overline{\mathcal M}_e|^2}
=4G_F^2\left[(c_V-c_A)^2s^2+(c_V+c_A+2)^2u^2\right].
$$
The phase-space integration already performed in part (b) immediately gives
$$
\boxed{\sigma_e=\frac{G_F^2s}{4\pi}
\left[(c_V+c_A+2)^2+\frac13(c_V-c_A)^2\right],}
$$
$$
\boxed{\overline\sigma_e=\frac{G_F^2s}{4\pi}
\left[(c_V-c_A)^2+\frac13(c_V+c_A+2)^2\right].}
$$
Equivalently, with $a=c_V+1,b=c_A+1$, these are $G_F^2s(a^2+ab+b^2)/(3\pi)$ and $G_F^2s(a^2-ab+b^2)/(3\pi)$. This is the <neutrino-antineutrino interchange of chiral scattering weights>. It uses the contact limit, not the full resonant $s$-channel propagator.

If the printed flavour-mismatched interaction is retained literally, its $\nu_\mu e$ amplitude already equals the displayed $\nu_e e$ amplitude. There is then no further flavour shift between those two processes, although antineutrino crossing still exchanges the weights. The distinction identified in part (b) is therefore necessary for the Standard Model interpretation of all three processes.