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Take the quantum electrodynamics interaction density to be ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. The momentum-space Feynman-gauge photon propagator, with Lorentz indices $\mu ,\nu$ and momentum $k$, isThe electron-photon vertex from the QED Feynman rules iswith its spinor indices connecting the incident and outgoing fermion lines. The muon-photon vertex has the same coupling. The Feynman i-epsilon prescription specifies the pole boundary condition; using a different overall charge-sign convention changes both vertices consistently and leaves the squared amplitude unchanged.

For the electron-positron annihilation into a muon pair, the exchanged photon has $k=p+q$ and $k^2=s$. Up to the conventional overall phase, the scattering amplitude is

Substitute the preceding spin average into the supplied differential scattering cross-section. Since $e^4=16{\pi }^2{\alpha }^2$,The angular integral is $2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair isThe outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form givesThe integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.