= Solution
Neglect the participating <quark> and <lepton> <masses>, so that the <weak charged current> selects a left-<helicity> $u$ and a right-<helicity> $\bar d$. Choose $+z$ along the <proton> beam in the <parton model>[partonic] <centre-of-momentum frame>. Then the <up quark> moves along $+z$ with <spin angular momentum>[spin projection] $-1/2$, while the <down antiquark> moves along $-z$ with <helicity> $+1/2$ and also has <spin angular momentum>[spin projection] $-1/2$ along $+z$. Thus
$$
\boxed{J_z(W^+)=-1.}
$$
The collinear two-body initial state has no orbital projection about its collision axis. In this approximation the produced <W boson> therefore has the indicated transverse <polarization>; its longitudinal spin state is absent. The unpolarized <hadron> beams contain other constituent <spin angular momentum>[spin states], but these do not participate in this <massless particle>[massless] <weak charged current>[charged-current] subprocess.
In the final state the <Positron> has <helicity> $+1/2$ and the <electron neutrino> has <helicity> $-1/2$. Along the outgoing <charged lepton>[charged-lepton] direction their combined angular-momentum projection is $+1$, because the oppositely moving <neutrino> has <spin angular momentum>[spin projection] $+1/2$ on that axis. Finite <masses>, noncollinear production and sea contributions modify the ideal configuration.
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