The electron and positron annihilate into a virtual photon, which then produces a new electron-positron pair. The virtual photon carries momentum $k = p + q$ with $k^2 = s$.
Homi Bhabha computed this process in 1936.
The computational chain
From diagram to number—every step of the amplitude calculation.
The diagram
An electron and positron annihilate at vertex 1 into a virtual photon with momentum $k = p + q$, which produces a new pair at vertex 2.
Feynman rules expression
$$i\mathcal{M}_s = \left[\bar{v}^r(q)(-ie\gamma^\mu)u^s(p)\right] \frac{-ig_{\mu\nu}}{(p+q)^2+i\epsilon} \left[\bar{u}^{s'}(p')(-ie\gamma^\nu)v^{r'}(q')\right]$$
Read off the Feynman rules: a spinor sandwich at each vertex (row vector $\times$ gamma matrix $\times$ column vector), connected by the photon propagator which contracts the Lorentz indices.
Simplify prefactors
$$i\mathcal{M}_s = \frac{ie^2}{s}\left[\bar{v}(q)\gamma^\mu u(p)\right]\left[\bar{u}(p')\gamma_\mu v(q')\right]$$
Pull out the coupling constants and contract the metric with the second gamma matrix. Here $s = (p+q)^2$ is the Mandelstam variable.
Spin sum and traces
$$\overline{|\mathcal{M}_s|^2} = \frac{e^4}{4s^2}\,\mathrm{Tr}\left[(\not{q}-m)\gamma^\mu(\not{p}+m)\gamma^\nu\right]\,\mathrm{Tr}\left[(\not{p}'+m)\gamma_\mu(\not{q}'-m)\gamma_\nu\right]$$
Square the amplitude, average over initial spins (factor 1/4), sum over final spins. The completeness relations $\sum_s u\bar{u} = \not{p}+m$ and $\sum_r v\bar{v} = \not{q}-m$ convert spinor products into traces of gamma matrices.
Evaluate traces
$$T_1^{\mu\nu} = 4\left(q^\mu p^\nu + q^\nu p^\mu - g^{\mu\nu}(q\cdot p + m^2)\right)$$
Apply the trace theorems: $\mathrm{Tr}[\gamma^\alpha\gamma^\beta] = 4g^{\alpha\beta}$, $\mathrm{Tr}[\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\delta] = 4(g^{\alpha\beta}g^{\gamma\delta} - g^{\alpha\gamma}g^{\beta\delta} + g^{\alpha\delta}g^{\beta\gamma})$, and traces of odd numbers of gamma matrices vanish.
Contract and express in Mandelstam variables
$$\overline{|\mathcal{M}_s|^2} = \frac{2e^4(t^2 + u^2)}{s^2} \quad\text{(massless limit)}$$
Contract the two trace tensors, substitute the Mandelstam variables $s = 2p{\cdot}q$, $t = -2p{\cdot}p'$, $u = -2p{\cdot}q'$ in the massless limit $m \to 0$.
Physical contribution
Dominant at high energies near threshold; gives the annihilation peak in e+e- colliders.