= Solution
A spin-zero <pion> has no preferred emission direction. Choose the <antineutrino> <momentum> along $+z$. A massless <antineutrino> has positive <helicity>, so its <spin> points along $+z$. The <muon> moves along $-z$, and total $J_z=0$ requires its <spin> to point along $-z$:
$$
\begin{array}{c|cc}
&\mu^-&\bar\nu_\mu\\\hline
\text{Momentum}&-p\widehat{\mathbf z}&+p\widehat{\mathbf z}\\
\text{Spin}&-\tfrac12\widehat{\mathbf z}&+\tfrac12\widehat{\mathbf z}\\
\text{Helicity}&+\tfrac12&+\tfrac12
\end{array}
$$
\Image[/past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021-pion-spin-configurations.png]
{title=Allowed spin and momentum orientations in negative-pion decay}
The reversed configuration is obtained by rotating the entire event and is equally allowed. The two spins are opposite, while each is aligned with its own particle's <momentum>; \b[both outgoing helicities are positive]. A negative-helicity <muon> with this fixed antineutrino direction would give a nonzero total spin projection and is forbidden for this two-body decay of a spin-zero parent. The required positive-helicity charged lepton is the suppressed component of the <weak charged current>.
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