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I'm working with this interaction Hamiltonian density

$$ H_{int}(x) = ig\bar{\Psi}_{\nu_e}(x)\gamma^\rho P_L \Psi_e(x)V_\rho(x) + igV^\dagger_\rho(x)\bar{\Psi}_e(x)\gamma^\rho P_L \Psi_{\nu_e} $$ where $V_\rho(x)$, $\Psi_{\nu_e}$ ,$\Psi_{e}$ are the quantum fields describing respectively $W^\pm$ bosons, $\bar{\nu}_e,\nu_e$ and $e^\pm$ and $P_L = (1+\gamma_5)/2$.

I want to compute the S-matrix element up to second order in g for the process $e^- e^+ \rightarrow W^- W^+$ and the unpolarized $|M|^2$ assuming electron as massless particle. This is the feynman diagram

fey

where $\Delta_{\nu_e}$ is the propagator for the neutrino field.

Using Feynman rules and summing over polarizations I get $$ |M|^2 = \frac{g^4}{(2\pi)^6}\frac{Tr[p_2\!\!\!/\ \gamma^\nu(p_1\!\!\!/ - q_1\!\!\!/)\gamma^\mu p_1\!\!\!/\ P_R \gamma^\rho(p_1\!\!\!/ - q_1\!\!\!/)\gamma^\sigma]}{16 q_1^0 q_2^0 p_1^0 p_2^0}\Pi_{\nu\sigma}(q_2)\Pi_{\mu\rho}(q_1) $$ where $p\!\!\!/ = p^\mu\gamma_\mu$ and $\Pi_{\mu\nu}(p) = \eta_{\mu\nu} + (p_\mu p _\nu)/m_W^2$.

How can I simplify this product of the trace with two $\Pi$s? Which's the easiest way to compute it?

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