= Solution
Square the operator solutions, retaining the symmetrized cross term rather than treating $x'(0)$ and $p'(0)$ as commuting variables. The initial symmetrized expectation vanishes, so
$$
\boxed{\begin{aligned}
\langle x'^2\rangle_t
&=\frac{\cos^2\omega t}{2\alpha}
+\frac{\hbar^2\alpha}{2m^2\omega^2}\sin^2\omega t,\\
\langle p'^2\rangle_t
&=\frac{\hbar^2\alpha}{2}\cos^2\omega t
+\frac{m^2\omega^2}{2\alpha}\sin^2\omega t.
\end{aligned}}
$$
The means remain zero. These are also the <variances>. Unless $\alpha=m\omega/\hbar$, the <Gaussian> state breathes: its <position> and <momentum> widths exchange periodically while the mean <energy> remains constant.
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