,\end{aligned}$$ one obtains the result $$(\bar{\chi}\gamma^\mu (1-\gamma_5) \psi)(\bar{\psi} \gamma_\mu (1-\gamma_5) \chi) = -(\bar{\chi}\gamma^\mu (1-\gamma_5) \chi)(\bar{\psi}\gamma_\mu (1-\gamma_5) \psi)$$ and, after reverting to the original spinors, $$(\bar{u}(\nu_b)\gamma^\mu (1-\gamma_5) u(\nu_a))(\bar{u}(\nu_a) \gamma_\mu (1-\gamma_5) u(\nu_b)) = -(\bar{u}(\nu_b)\gamma^\mu (1-\gamma_5) u(\nu_b))(\bar{u}(\nu_a)\gamma_\mu (1-\gamma_5) u(\nu_a)) \, \, \, .$$ Of course, the wavefunction of fermions is antisymmetric under particle exchange, hence $$-(\bar{u}(\nu_b)\gamma^\mu (1-\gamma_5) u(\nu_b))(\bar{u}(\nu_a)\gamma_\mu (1-\gamma_5) u(\nu_a)) = + (\bar{u}(\nu_a)\gamma^\mu (1-\gamma_5) u(\nu_a)) (\bar{u}(\nu_b)\gamma_\mu (1-\gamma_5) u(\nu_b)) \, \, \, .$$

Inserting this relation into Eq. (\[eq:EQ140\]), one obtains $$\begin{aligned}
\mathscr{T} = \frac{G_F}{2\sqrt{2}} &\big( \bar{u}(\nu_a) \gamma^\mu (1 - \gamma_5) u(\nu_a) \bar{u}(\nu_b) \gamma_\mu (1 - \gamma_5) u(\nu_b)\\
&+ \bar{u}(\nu_a)\gamma^\mu (1-\gamma_5) u(\nu_a) \bar{u}(\nu_b)\gamma_\mu (1-\gamma_5) u(\nu_b) \big)\\
= \frac{G_F}{\sqrt{2}} &\bar{u}(\nu_a) \gamma^\mu (1 - \gamma_5) u(\nu_a) \bar{u}(\nu_b) \gamma_\mu (1 - \gamma_5) u(\nu_b) \, \, \, .\end{aligned}$$ Therefore, the amplitude for the scattering of neutrini of the same flavour is twice the result of Eq. (\[eq:EQ100\]), i.e., twice the scattering amplitude for the scattering of neutrini of different flavour. Consequently, there is no need to repeat the calculation of Section \[sec:D1\] in order to obtain the result of this section; all one needs to do is multiply the DCS of Eq. (\[eq:EQ130\]) by $4$. Finally, $$\label{eq:EQ160}
\frac{d\sigma}{d\Omega} = \frac{4 G^2_F s}{8 \pi^2} = \frac{\sigma_0}{2 \pi} \, \, 