Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 40 / 2 / iii / Solution

For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.

The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with $E_q^2=|\mathbf{q}{|}^2+M^2$ and $E_l^2=|\mathbf{l}{|}^2+m^2$.

Equivalently, in the centre-of-momentum frame set $s=-(p+k{)}^2$. Integrating the energy delta function in the relativistic two-body phase space givesFor this elastic process $k_f=k_i$, so integrate $\overline{|T{|}^2}/(64{\pi }^{2}s)$ over the full solid angle. This supplies the requested prescription without evaluating the angular integral.