= Solution
There is one <tree-level Feynman diagram>: the distinguishable <electron> and <muon> exchange a <photon> in the $t$ channel. There is no identical-particle exchange diagram.
\Image[/past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022-lepton-scattering.png]
{title=Electron-muon scattering by one-photon exchange}
Multiplying the external <Dirac spinors>, the two <Feynman vertices> and the <photon propagator> gives
$$
i\mathcal M=[\bar u_3(-ie\gamma^\mu)u_1]\frac{-ig_{\mu\nu}}{q^2+i\epsilon}[\bar u_4(-ie\gamma^\nu)u_2],\qquad
\boxed{\mathcal M=\frac{e^2}{q^2}[\bar u_3\gamma^\mu u_1][\bar u_4\gamma_\mu u_2].}
$$
The overall phase of this <scattering amplitude> is convention dependent. Four external <spin> labels with two choices each give $2^4=16$ possible configurations before taking a limit. For $E\gg m,M$, the vector <QED> interaction preserves <fermion chirality>, which becomes <helicity> for each external particle. Thus the surviving configurations, written as $(h_e,h_\mu)_{\rm in}\to(h_e,h_\mu)_{\rm out}$, are
$$
\boxed{LL\to LL,\quad LR\to LR,\quad RL\to RL,\quad RR\to RR.}
$$
Here $L,R$ specify <helicity> along each particle's own <momentum>. The other configurations require at least one <mass>-suppressed flip and vanish in the strictly massless limit.
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