Why couldn't the decay $\pi^- \to e^- + \bar\nu_e$ occur if electrons were massless?

Question

If we assume that electrons (just like neutrinos) are massless, why can’t the decay $\pi^- \rightarrow e^- + \bar{\nu}_e$ occur under the weak interaction?

Answer 1

Since the spin of the charged $\pi$ is $0$, the spins of the daughter particles need to add up to $0$ as well, i.e., their spins need to be anti-parallel. That's nothing else than the conservation of angular momentum.

Assuming the anti-neutrino to be massless, it is always right-handed. Right-handed means that the momentum vector and the spin vector are parallel, while left-handed means that the momentum vector and the spin vector are anti-parallel. This is well defined for a massless particle, since it travels at the speed of light, and there is no inertial frame in which the momentum vector would switch direction.

Since the anti-neutrino is right-handed, the electron would need to be right-handed as well to conserve linear momentum and angular momentum (spin). But the decay of the $\pi$ happens via the weak interaction, i.e. via the $W$ boson. Since the $W$ boson is known to couple only to left-handed particles, the decay would be forbidden.

Since the electron is not massless, it has a small left-handed component. The decay is suppressed, but not forbidden. The heavier muon has a larger left-handed component, and its decay is less suppressed. Hence, pions usually decay into muons, although they have less phase space available.

Answer 2

pfnuesel's answer is absolutely correct and very satisfying to understand, and you should read it before this one.

There is, however, a route which would permit $\pi\to e+\nu$ even if the electron were massless. The conserved quantity which suppresses that decay is not spin, but total angular momentum. There exists an electron+neutrino wavefunction with spin $S=1$, orbital angular momentum $L=1$, and total angular momentum $\vec J = \vec L + \vec S$ of magnitude $J=0$; this wavefunction is available for pion decay and competes with the $S=0, L=0$ channel described by pfnuesel.

However, like all high-$L$ decays, this one is suppressed because the overlap of a wavefunction with nonzero $L$ and the parent particle with radius $R$ is proportional to $R^L$.