Moller scattering
$e^-e^- \to e^-e^-$
Elastic scattering of two electrons. The simplest process involving identical fermions in the initial and final state, requiring antisymmetry under exchange of the two final-state electrons.
The t-channel and u-channel diagrams interfere with a relative minus sign from Fermi statistics. This antisymmetry is physically observable in the angular distribution.
Order $\alpha$ diagrams
2 diagrams.
Order $\alpha^{2}$ diagrams
3 diagrams.
From diagrams to cross section
How the individual diagram contributions combine into the physical result.
Fermi antisymmetry
$$i\mathcal{M} = i\mathcal{M}_t - i\mathcal{M}_u$$
The total amplitude must be antisymmetric under exchange of the two identical final-state electrons $p_1' \leftrightarrow p_2'$. The u-channel enters with a relative minus sign.
Square and average
$$\overline{|\mathcal{M}|^2} = \overline{|\mathcal{M}_t|^2} + \overline{|\mathcal{M}_u|^2} - 2\mathrm{Re}\,\overline{\mathcal{M}_t\mathcal{M}_u^*}$$
The minus sign from Fermi statistics makes the interference term appear with a minus in $|\mathcal{M}_t - \mathcal{M}_u|^2$, but when evaluated the $T_{tu}$ trace is itself negative, so the net contribution to the cross section is positive.
Full Moller amplitude (massless limit)
$$\overline{|\mathcal{M}|^2} = 2e^4\left[\frac{s^2+u^2}{t^2} + \frac{s^2+t^2}{u^2} + \frac{2s^2}{tu}\right]$$
Compare to Bhabha: the interference term has opposite sign (positive here). The cross section diverges at both $\theta \to 0$ and $\theta \to \pi$ (forward and backward peaks). At $\theta = 90°$ the interference term produces a minimum — the signature of Fermi antisymmetry.
Differential cross section
$$\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{2s}\left[\frac{s^2+u^2}{t^2} + \frac{s^2+t^2}{u^2} + \frac{2s^2}{tu}\right]$$
In the CM frame: $t = -(s/2)(1-\cos\theta)$, $u = -(s/2)(1+\cos\theta)$.