e⁺e⁻ → μ⁺μ⁻ Annihilation

4.1 The Process and Its Feynman Diagram

We compute the simplest QED scattering process: an electron and positron annihilate into a muon-antimuon pair. At tree level, there is exactly one Feynman diagram — the s-channel, where the $e^+e^-$ pair annihilates into a virtual photon which then creates the $\mu^+\mu^-$ pair.

We label the momenta as: $e^-(p_1) + e^+(p_2) \to \mu^-(p_3) + \mu^+(p_4)$. Since $m_\mu \gg m_e$, we work in the limit $m_e = 0$ but keep $m_\mu$ general initially, then take the high-energy limit $s \gg m_\mu^2$ at the end.

4.2 Writing the Amplitude from Feynman Rules

Reading the diagram from right to left, the QED Feynman rules give us:

Electron vertex: The incoming electron ($u(p_1)$) and positron ($\bar{v}(p_2)$) couple to the photon via $-ie\gamma^\mu$. The electron current is:

Photon propagator: The virtual photon with momentum $q = p_1 + p_2$ contributes (in Feynman gauge):

Muon vertex: The outgoing muon ($\bar{u}(p_3)$) and antimuon ($v(p_4)$) give:

Assembling the amplitude: Contracting everything together:

or equivalently:

4.3 Squaring the Amplitude and Spin Sums

For unpolarized beams and unobserved final-state spins, we compute the spin-averaged squared amplitude:

The factor 1/4 comes from averaging over the two spin states of each initial particle (electron and positron). Writing out $|\mathcal{M}|^2 = \mathcal{M}\mathcal{M}^*$:

Using the identity $[\bar{\psi}_1 \Gamma \psi_2]^* = \bar{\psi}_2 \gamma^0 \Gamma^\dagger \gamma^0 \psi_1 = \bar{\psi}_2 \bar{\Gamma} \psi_1$ where$\bar{\Gamma} = \gamma^0 \Gamma^\dagger \gamma^0$ (and $\overline{\gamma^\nu} = \gamma^\nu$), the spin sums factorize into two independent traces using the completeness relations:

The electron-side spin sum becomes a trace (setting $m_e = 0$):

The muon-side spin sum (keeping $m_\mu$):

The full spin-summed result is:

4.4 Evaluating the Traces

Electron trace: Using $\text{Tr}[\gamma^\alpha\gamma^\beta\gamma^\mu\gamma^\nu] = 4(g^{\alpha\beta}g^{\mu\nu} - g^{\alpha\mu}g^{\beta\nu} + g^{\alpha\nu}g^{\beta\mu})$:

Muon trace: Expanding $(\not{p}_3 + m_\mu)\gamma^\mu(\not{p}_4 - m_\mu)\gamma^\nu$, the terms with odd numbers of gamma matrices vanish in the trace. The $m_\mu^2$ term gives $-m_\mu^2 \text{Tr}[\gamma^\mu\gamma^\nu] = -4m_\mu^2 g^{\mu\nu}$. Thus:

Contracting the tensors: We compute $L_{e\,\mu\nu} L_\mu^{\mu\nu}$ by contracting term by term. Using$g_{\mu\nu}g^{\mu\nu} = 4$ in 4 dimensions:

After careful bookkeeping, the contraction simplifies to:

4.5 Result in Mandelstam Variables

The Mandelstam variables for this process are:

with the constraint $s + t + u = 2m_\mu^2$ (setting $m_e = 0$). The dot products become:

By momentum conservation, $p_2 \cdot p_3 = p_1 \cdot p_4$ and $p_2 \cdot p_4 = p_1 \cdot p_3$ (in the massless electron limit). Substituting into our trace result:

In the high-energy limit $s \gg m_\mu^2$, we set $m_\mu = 0$ everywhere and use $s + t + u = 0$:

4.6 Differential Cross Section in the CM Frame

In the center-of-mass frame with all particles massless, $|\vec{p}_f| = |\vec{p}_i|$ and the master formula gives:

We express $t$ and $u$ in terms of the scattering angle $\theta$. With $E = \sqrt{s}/2$ for each particle:

Therefore:

Putting it all together and writing $e^2 = 4\pi\alpha$:

This is one of the most celebrated results in QED. The $(1 + \cos^2\theta)$ angular dependence is a direct signature of spin-1/2 particles coupling to a spin-1 mediator.

4.7 Total Cross Section

Integrating over the full solid angle $d\Omega = \sin\theta\, d\theta\, d\phi$:

The $\phi$ integral gives $2\pi$. For the $\theta$ integral, let $x = \cos\theta$:

Therefore: