Pair annihilation
$e^+e^- \to \gamma\gamma$
Electron-positron annihilation into two photons. Related to Compton scattering by crossing symmetry. The two tree diagrams correspond to the two time-orderings of photon emission.
Order $\alpha$ diagrams
2 diagrams.
From diagrams to cross section
How the individual diagram contributions combine into the physical result.
Sum both channels
$$i\mathcal{M} = i\mathcal{M}_t + i\mathcal{M}_u$$
Both diagrams must be summed. The identical outgoing photons (bosons) give a positive relative sign, unlike Moller where identical fermions give a minus.
Crossing relation to Compton
$$\overline{|\mathcal{M}(e^+e^- \to \gamma\gamma)|^2} = \overline{|\mathcal{M}(e^-\gamma \to e^-\gamma)|^2}\bigg|_{s\leftrightarrow t}$$
The squared amplitude is related to Compton scattering by crossing: exchange $s \leftrightarrow t$ in the Mandelstam variables. The mass terms cannot be dropped.
Total cross section
$$\sigma = \frac{\pi\alpha^2}{2m^2}\frac{1-\beta^2}{\beta}\left[(3-\beta^4)\ln\frac{1+\beta}{1-\beta}-2\beta(2-\beta^2)\right]$$
In the CM frame with $\beta$ the electron velocity. Near threshold ($\beta \to 0$): $\sigma \to \pi\alpha^2/m^2$. At high energy ($\beta \to 1$): $\sigma \sim (\pi\alpha^2/s)\ln(s/m^2)$.