The process

The last process of Part 2, and it is one you have already computed.

An electron meets a positron and they annihilate into two photons: $\mathrm{e}^+\mathrm{e}^- \to \gamma\gamma$. Nothing is left of the original particles — no muon pair, no scattered electron, just light.

Take lesson 2.5's Compton diagram, $\mathrm{e}^-\gamma \to \mathrm{e}^-\gamma$, and rotate it. Move the incoming photon into the final state and the outgoing electron into the initial one, and the picture becomes this one. Lesson 2.2 established what that costs: a substitution, not a calculation.

But the far end of the diagram is new. The two outgoing photons are identical particles — and photons are bosons.

Lesson 2.4 met identical particles for the first time, with two electrons in the final state, and found that Fermi statistics demanded an antisymmetric amplitude and delivered a relative minus sign. Bosons demand the opposite. The amplitude must be symmetric under exchanging the two photons, which means the same two diagrams now enter with a relative plus.

Same machinery, opposite sign. That single flip is what this lesson is about — and it is the last piece Part 2 needs, because with it every ingredient the part introduced has been used at least twice.