 Solution

The initial quantum state is ${\psi }_1$, up to global phase, and ${\psi }_1=({\chi }_{+}+{\chi }_{-})/\sqrt{2}$. Applying unitary time evolution givesSince $J_3{\psi }_1=(\hslash /2){\psi }_1$ and $J_3{\psi }_2=-(\hslash /2){\psi }_2$, the Born rule givesThe other outcome has probability ${\sin }^2(\omega t)$, verifying that the two probabilities sum to one. This is a two-level quantum return probability.