Past exam of the particle and nuclear physics course of the University of Cambridge / 2022 / Section B / 6 / a / Solution

The charged pion has spin zero. In its rest frame, the outgoing muon or Positron and the neutrino travel in opposite directions. With a nearly massless left-helicity neutrino, total angular momentum zero requires the positive charged antilepton to have negative helicity. The weak charged current instead creates a massless positive antilepton with positive helicity. A nonzero charged-lepton mass supplies the required opposite-helicity component.

Thus the decay amplitude is proportional to $m_{\ell }$: this is helicity suppression. Explicitly, contracting the pion's axial-current matrix element, proportional to its four-momentum $p_{\pi }^{\mu }$, with the leptonic current and applying the Dirac equation leaves a factor $m_{\ell }$. The leptonic pseudoscalar decay width is, in the $f_{\pi }$ convention,The phase space advantage of the lighter Positron cannot offset its much stronger helicity suppression. The common factors cancel in the pion electron-to-muon decay ratio:This tree-level estimate explains why the muonic branch dominates by roughly four orders of magnitude. The paper's table gives an electronic branching fraction $1.230\times 10^{-4}$; electromagnetic corrections change the simple ratio by a few percent without changing the mechanism.