 Solution

Diagonalizing the $J_1$ matrix gives the normalized eigenvectorswith eigenvalues $+\hslash ,0,-\hslash$, respectively. The first projective measurement prepares $v_{+}$ up to global phase, regardless of the earlier state, conditional on obtaining $+\hslash$.

The unitary time evolution under $H=\omega J_3$ gives, with $\theta =\omega t$,The probability amplitudes for the second projective measurement areThe Born rule therefore gives the general spin-one transverse measurement after axial precession:At the specified times $t_k=\pi k/\omega$, $\theta =\pi k$, soFor even $k$, the result is certainly $+\hslash$; for odd $k$, it is certainly $-\hslash$. Although zero belongs to the spectrum of the observable, it never occurs at these times. Indeed, for even $k$ the evolved vector is $v_{+}$, while for odd $k$ it is $-v_{-}$, where the minus sign is a global phase.