= Solution
The <Hamiltonian operator> $H=\omega J_3$ is already diagonal in the <angular momentum eigenstate> basis of part (b). Its three distinct <energy eigenvalues> and <stationary states> are
$$
\boxed{E_\mu=\hbar\omega\mu,\qquad
|\Psi_\mu(t)\rangle=e^{-i\omega\mu t}|1,\mu\rangle,\qquad \mu=1,0,-1.}
$$
Thus the three column <eigenvectors> are $(1,0,0)^{\mathsf T}$, $(0,1,0)^{\mathsf T}$ and $(0,0,1)^{\mathsf T}$, with <energies> $+\hbar\omega$, zero and $-\hbar\omega$, respectively. Each is nondegenerate since $\omega>0$. Any nonzero scalar multiple gives the same <stationary state> after normalization; a <quantum superposition> of different <energies> is generally not stationary.
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