= Solution
{c}
The initial <quantum state> is $\psi_1$, up to <global phase>, and $\psi_1=(\chi_++\chi_-)/\sqrt2$. Applying <unitary time evolution> gives
$$
\begin{aligned}
\Psi(t)&=\frac1{\sqrt2}\left(e^{-i\omega t}\chi_++e^{i\omega t}\chi_-\right)\\
&=\cos(\omega t)\psi_1-i\sin(\omega t)\psi_2.
\end{aligned}
$$
Since $J_3\psi_1=(\hbar/2)\psi_1$ and $J_3\psi_2=-(\hbar/2)\psi_2$, the <Born rule> gives
$$
\boxed{\Pr\left(J_3(t)=+\frac\hbar2\right)=|\langle\psi_1,\Psi(t)\rangle|^2=\cos^2(\omega t).}
$$
The other outcome has <probability> $\sin^2(\omega t)$, verifying that the two <probabilities> sum to one. This is a <two-level quantum return probability>.
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