The <spin sum> becomes a <gamma-matrix trace>:
$$
\sum_{s,s'}|\lambda\bar u^s(p)v^{s'}(q)|^2
=\lambda^2\operatorname{tr}[(\not p+m)(\not q-m)]
=4\lambda^2(p\cdot q-m^2).
$$
Since $(p+q)^2=M^2$, $p\cdot q=(M^2-2m^2)/2$. Therefore the normalization requested in the paper gives
$$
\mathcal P=\frac14\sum_{s,s'}|\mathcal A_{s,s'}|^2
=\frac{\lambda^2}{2}(M^2-4m^2).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 / 3 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-301/3/a/iv/solution