Draw the lowest-order Feynman diagrams for the pair annihilation process
$$
\mathrm{e}^{+}\left(p_1, s_1\right)+\mathrm{e}^{-}\left(p_2, s_2\right) \rightarrow \gamma\left(k_1, \varepsilon_1\right)+\gamma\left(k_2, \varepsilon_2\right)
$$
Check your answer with Fig. 6.13. Use the Feynman rules to show that
$$
-i \mathscr{K}=(i e)^2 \bar{v}^{\left(s_1\right)}\left(p_1\right)\left(k_1^* \frac{i}{p_1-k_1-m} k_2^*+\xi_2^* \frac{i}{p_2-k_1-m} \xi_1^*\right) u^{\left(s_2\right)}\left(p_2\right) .
$$
Verify that, in the high-energy limit, the spin-averaged rate is given by
$$
\overline{|\mathscr{R}|^2}=2 e^4\left(\frac{u}{t}+\frac{t}{u}\right) \text {. }
$$
The $\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \gamma$ cross section has both forward and backward peaks, corresponding to the $t$ - and $u$-channel exchanged electrons being almost on mass shell. Result $(6.117)$ can also be obtained by crossing the amplitude for Compton scattering. To go from $\gamma \mathrm{e}^{-} \rightarrow \gamma \mathrm{e}^{-}$to $\mathrm{e}^{+} \mathrm{e}^{-} \rightarrow \gamma \gamma$, we simple "cross" the ingoing photon with the outgoing electron,
$$
\begin{aligned}
& k, \varepsilon \rightarrow-k_2, \varepsilon_2^* \\
& p^{\prime} \rightarrow-p_1 \quad \text { and } \quad \bar{u}^{\left(s^{\prime}\right)}\left(p^{\prime}\right) \rightarrow \bar{v}^{\left(s_1\right)}\left(p_1\right) .
\end{aligned}
$$
The initial electron and outgoing photon are unaltered:
$$
u^{(s)}(p) \equiv u^{\left(s_2\right)}\left(p_2\right), \quad k^{\prime}, \varepsilon^{\prime *} \rightarrow k_1, \varepsilon_1^* .
$$
Making these substitutions in (6.103) and (6.104) gives the pair annihilation amplitude (6.117).