# Why does both the electron and the antineutrino have spin only $+1/2$ in the Wu experiment?

In ${\rm Co}\rightarrow {\rm Ni}^* + e^- + \bar{\nu}_e$, the Wu experiment, it is said that $J_{co}$ = 5, $J_{Ni}$ = 4 and hence J of "$e^- + \bar{\nu}_e$" system is 1. If orbital angular momentum is zero, its S=J=1. Because of this we say both electron and anti-neutrino has $S_z$=1/2. But I guess, if S=1, $S_z$ of this total system($e^- + \bar{\nu}_e$) can also be 1,0 or -1. This leads to $S_z$ of electron and neutrino to have values $S_z$=0,1/2 or -1/2. Then, why do we take only $S_z$ = +1/2 for both electron and anti-neutrino?

• What you say seems correct, can you maybe post or link the text where this is stated? The only caveat I can think of is that they might actually be talking about cases where also $J_{z \rm Co}$ and $J_{z \rm Ni}$ were measured... – Void Aug 27 '17 at 12:22
• master.particles.nl/LectureNotes/2011-CP.pdf is one of such pdf I just found. Actually I posted this question from a book which is not available online. So, in the pdf attached, on page number 5, where they explained Wu experiment, can I say that electron has $S_z$ = -1/2 and anti-neutrino has +1/2 and so the total $S_z$ of this system is 0? – kg__ Aug 27 '17 at 13:03

The point is that the $^{60}$Co nuclei beam in the Wu experiment was polarized by a strong magnetic field so that all the spin was in the direction of the magnetic field (the $z$ direction): $J_{z \rm Co} = J_{\rm Co} = 5$.
However, since $J_{z}$ is conserved in the decay, you also have to have $J_{ze^-} + J_{z\nu} + J_{z \rm Ni} = J_{z \rm Co} = 5$. There is only one possibility how to achieve this with the $J_{\rm Ni}=4,J_{\nu}=1/2,J_{e^-}=1/2$ particles, and that is $J_{z\rm Ni}=4,J_{z\nu}=1/2,J_{ze^-}=1/2$.