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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

The one-loop graph is a fermion line that emits and reabsorbs one internal photon:

The QED Feynman rules assign $(-ie{\gamma }^{\mu })(-ie{\gamma }^{\nu })$ to the two vertices, the Feynman-gauge photon propagator contracts $\mu$ and $\nu$ and contributes ${\delta }_{\mu \nu }/(p-k{)}^2$, and the internal Dirac propagator is $(-i/k+m)/(k^2+m^2)$. Hence