 Solution

Let ${\theta }^{\star }$ be the angle from the incoming up quark, along the proton direction, to the outgoing charged antilepton in the W boson rest frame. The initial and final spin projections in ii are $-1$ and $+1$. The Wigner small-d matrix therefore givesAfter squaring and normalizing over $c=\cos {\theta }^{\star }$, the charged-lepton angular distribution isIts integral is one since ${\int }_{-1}^1(1-c{)}^{2}dc=8/3$. The charged antilepton is preferentially emitted toward the antiproton. Measuring the angle from the incoming down antiquark instead replaces $c$ by $-c$ and produces $(1+c{)}^2$; the axis convention is part of the answer. Per unit solid angle the normalized density is $3(1-\cos {\theta }^{\star }{)}^2/(16\pi )$.