graph of Fig. \[fig:NuNu\](c) is relevant. As the neutrini are solely left-handed, the averaging over the spin orientations in the initial state is not applicable (the average of one value is the value itself). The square of the scattering amplitude $\lvert \mathscr{T} \rvert^2$ is obtained after multiplying $\mathscr{T}$ with its complex conjugate. However, given that $\mathscr{T}$ is one (complex) element (i.e., not a matrix), this operation is equivalent to using the Hermitian of $\mathscr{T}$, namely $\mathscr{T}^\dagger$, in the product. $$\lvert \mathscr{T} \rvert^2 \coloneqq \mathscr{T} \mathscr{T}^\dagger = \frac{G_F^2}{8} \, {\rm Tr} [{\ensuremath{\mathrlap{\!\not{\phantom{k}}}k}}^\prime \gamma^\mu (1-\gamma_5) {\ensuremath{\mathrlap{\!\not{\phantom{k}}}k}} \gamma^\nu (1-\gamma_5)] \, {\rm Tr} [{\ensuremath{\mathrlap{\!\not{\phantom{p}}}p}}^\prime \gamma_\mu (1-\gamma_5) {\ensuremath{\mathrlap{\!\not{\phantom{p}}}p}} \gamma_\nu (1-\gamma_5)]$$ The application of the first trace operator yields the projectile-related tensor $$\label{eq:EQ110}
K^{\mu \nu} = 8 \left( k^{\prime \, \mu} k^\nu + k^{\prime \, \nu} k^\mu + \frac{q^2}{2} g^{\mu\nu} - i \epsilon^{\mu \nu a b} k_a k^\prime_b \right) \, \, \, .$$ The target-related tensor, resulting from the application of the second trace operator, is of similar structure $$P_{\mu \nu} = 8 \left( p^\prime_\mu p_\nu + p^\prime_\nu p_\mu + \frac{q^2}{2} g_{\mu\nu} - i \epsilon_{\mu \nu c d} p^c p^{\prime \, d} \right) \, \, \, .$$

It can be easily shown that $q_\mu K^{\mu \nu} 