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QED Feynman Diagram: e+e- to mu+mu-, |M|^2 and the Cross Section

A tree-level QED (quantum electrodynamics) playground for $e^+e^- \to \mu^+\mu^-$. Panel A draws the $s$-channel Feynman diagram and reads off the Feynman-rule bookkeeping: each vertex carries a factor $e$, so a $V$-vertex amplitude scales as $\alpha^{V/2}$ and $|M|^2$ as $\alpha^V$; a one-loop variant inserts a vacuum-polarisation bubble (four vertices, $\alpha^4$, suppressed by $\alpha^2$ relative to the tree). Panel B plots the total cross section $\sigma(\sqrt{s})$ in nanobarns on log-log axes: it is exactly zero below $\sqrt{s} = 2 m_\mu$, rises through the muon-pair threshold to a peak, then falls as a straight power-law line ($\sigma \sim 1/s$). Panel C shows the Mandelstam invariants with the exact identity $s + t + u = 2 m_e^2 + 2 m_\mu^2$ and the differential cross section $d\sigma/d\Omega$, which is even in $\cos\theta$ (forward-backward symmetric) and tends to $1 + \cos^2\theta$ ultrarelativistically, becoming isotropic near threshold. The Mandelstam identity $s + t + u = 2 m_e^2 + 2 m_\mu^2$ holds, the cross section turns on sharply at the muon-pair threshold and falls as $1/s$, and the angular distribution is forward-backward symmetric, tending to $1 + \cos^2\theta$ ultrarelativistically.

Figure 1. The s-channel QED process e+e- to mu+mu-: two vertices give an amplitude proportional to alpha and a cross section proportional to alpha^2; it vanishes at the muon-pair threshold, peaks just above it and falls as 1/s, with a forward-backward-symmetric 1 + cos^2 angular distribution and the exact identity s + t + u = sum m_i^2. Method: closed-form tree-level QED matrix element and phase space; Canvas2D, deterministic.
sqrt(s) (GeV)1.00
scattering angle (deg)60
diagram order
show channelss

WHAT TO TRY

  • Slide sqrt(s) up the resonance: the s-channel cross section for e+e- to mu+mu- peaks and the angular distribution sharpens. Center-of-mass energy is the single knob that sets the rate.
  • Switch the diagram order from tree (two vertices, alpha^2) to one loop (four vertices, alpha^4): the vacuum-polarization bubble is suppressed by alpha^2, which is why tree level is such a good first approximation.
  • Toggle the channels (s, t, both) and sweep the scattering angle: the s and t amplitudes interfere, and the Mandelstam check s + t + u = sum of masses-squared is the built-in kinematic test.