 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 partonic centre-of-momentum frame. Then the up quark moves along $+z$ with spin projection $-1/2$, while the down antiquark moves along $-z$ with helicity $+1/2$ and also has spin projection $-1/2$ along $+z$. ThusThe 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 states, but these do not participate in this massless 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 direction their combined angular-momentum projection is $+1$, because the oppositely moving neutrino has spin projection $+1/2$ on that axis. Finite masses, noncollinear production and sea contributions modify the ideal configuration.