= Solution
At leading order in <quantum electrodynamics>, the electron-positron pair annihilates into one virtual <photon>, which produces the muon-antimuon pair. This is an s-channel <Feynman diagram> with $q=p_-+p_+=k_-+k_+$ and $q^2=s$. No QED vertex changes an electron directly into a muon, so there is no corresponding single-photon t-channel flavor-changing diagram.
\Image[/past-exam-of-the-gauge-field-theory-course-of-the-university-of-cambridge/2015/paper-1-muon-pair-diagram.png]
{title=Leading-order electron-positron annihilation into a muon pair through a virtual photon}
{description=Original s-channel diagram. Arrows follow fermion-number flow; each labelled external four-momentum refers to the physical incoming or outgoing particle, and the virtual photon carries q from left to right.}
Each charged-fermion vertex contributes $-ie\gamma^\mu$, the internal photon contributes $-ig_{\mu\nu}/(q^2+i0)$ in Feynman gauge, and external legs supply incoming $u(p_-),\bar v(p_+)$ and outgoing $\bar u(k_-),v(k_+)$. Their ordered contraction gives the stated amplitude. The overall convention-dependent phase does not affect its squared modulus.
For unpolarized scattering, average over the two spin choices of each incoming particle and sum over outgoing spins. The identities $\sum u\bar u=\not{p}+m$ and $\sum v\bar v=\not{p}-m$, together with $\gamma^0(\gamma^\mu)^\dagger\gamma^0=\gamma^\mu$, convert the products into <fermion spin sums>:
$$
\boxed{\overline{|\mathcal M|^2}=\frac{e^4}{4s^2}
\operatorname{tr}\big[(\not{p}_+-m_e)\gamma_\mu(\not{p}_-+m_e)\gamma_\nu\big]
\operatorname{tr}\big[(\not{k}_-+m_\mu)\gamma^\mu(\not{k}_+-m_\mu)\gamma^\nu\big].}
$$
The factor $1/4$ averages the initial spins; there is no additional average over the final ones. In the massless limit the first trace is $4(p_{+\mu}p_{-\nu}+p_{+\nu}p_{-\mu}-g_{\mu\nu}p_+\cdot p_-)$, with the analogous expression for the outgoing current. Contracting gives
$$
\overline{|\mathcal M|^2}=\frac{8e^4}{s^2}
\left[(p_+\cdot k_-)(p_-\cdot k_+)+(p_+\cdot k_+)(p_-\cdot k_-)\right].
$$
In the <centre-of-mass frame>, every external energy is $E_{\rm cm}/2$. Choose $p_-$ along $+z$ and let $k_-$ make angle $\theta$ with it. The dot products are $p_-\cdot k_-=p_+\cdot k_+=s(1-\cos\theta)/4$ and $p_-\cdot k_+=p_+\cdot k_-=s(1+\cos\theta)/4$. Hence
$$
\boxed{\overline{|\mathcal M|^2}=e^4(1+\cos^2\theta).}
$$
The <electron-positron annihilation into a muon pair> cross-section is therefore
$$
\frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4s}(1+\cos^2\theta).
$$
Using $d\Omega=2\pi\,d(\cos\theta)$ gives
$$
\boxed{\sigma=\frac{4\pi\alpha^2}{3s}=\frac{4\pi\alpha^2}{3E_{\rm cm}^2}.}
$$
In ordinary units this becomes $4\pi\alpha^2(\hbar c)^2/(3E_{\rm cm}^2)$. It is the leading high-energy QED result: finite masses, radiative corrections and weak-interaction exchange modify the prediction outside that approximation.
Cross-section measurements can test spin assignments when the interaction and charges are specified. For example, keeping spin-half electrons but replacing the final muons by minimally coupled charged spin-zero particles replaces the fermion current by a momentum-difference current. In the massless limit, <scalar-pair production by fermion annihilation> gives $d\sigma/d\Omega=\alpha^2\sin^2\theta/(8s)$ and total $\pi\alpha^2/(3s)$. Its central angular peak differs from the fermionic $1+\cos^2\theta$ distribution.
Thus angular distributions, polarization information and suitably modeled rates distinguish spin hypotheses. A single total rate without assumptions about the couplings or the initial species does not independently establish both spins; changing interactions can mimic normalizations. The predicted fermion result is a testable consequence of the spin-half QED model, rather than a model-independent inference of spin from any one observed cross-section.
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