= Solution
In the convention of <1/b/iii>, the invariant <decay amplitude> is $\mathcal M_{s_1s_2}=-g\bar u_\chi^{s_2}(p_2)u_\Psi^{s_1}(p_1)$. Including the common overall $i$ in the definition changes no squared <decay amplitude>. An unpolarized ensemble is an incoherent equal mixture of the two initial <spin> states; hence take a <spin average> over $s_1$ and a <spin sum> over the unobserved final $s_2$. Different final <spin> states are orthogonal outcomes, so their probabilities, rather than their <decay amplitudes>, add.
Using the <Dirac adjoint> and the <fermion spin sum> gives
$$
\begin{aligned}
\overline{|\mathcal M|^2}
&=\frac{g^2}{2}\sum_{s_1,s_2}
(\bar u_2u_1)(\bar u_1u_2)\\
&=\frac{g^2}{2}\operatorname{tr}[(\not p_2+m)(\not p_1+M)]
=2g^2(p_1\cdot p_2+mM).
\end{aligned}
$$
The <gamma-matrix trace> identities used here are $\operatorname{tr}\gamma^\mu=0$ and $\operatorname{tr}(\gamma^\mu\gamma^\nu)=4\eta^{\mu\nu}$. The <mass-shell conditions> $p_1^2=M^2$ and $p_2^2=p_3^2=m^2$, together with <four-momentum conservation>, imply
$$
m^2=(p_1-p_2)^2=M^2+m^2-2p_1\cdot p_2,
\qquad p_1\cdot p_2=\frac{M^2}{2}.
$$
Therefore
$$
\boxed{\overline{|\mathcal M|^2}=g^2M(M+2m).}
$$
In the parent's rest frame this also says $E_2=E_3=M/2$.
Back to article page
