= Solution
The initial <quantum state> is a <coherent state>. Since $a|\alpha\rangle=\alpha|\alpha\rangle$ and the state is normalized, its mean <occupation number> is $N=\langle a^\dagger a\rangle=|\alpha|^2$. Each <quantum harmonic oscillator> <energy eigenstate> acquires the phase $e^{-i\omega(n+1/2)t}$, so
$$
|\alpha(t)\rangle=e^{-i\omega t/2}|\alpha e^{-i\omega t}\rangle,
\qquad \langle a\rangle_t=\alpha e^{-i\omega t}.
$$
The prefactor is a <global phase> and does not change an <expectation value>. Inverting the definitions of the <annihilation operator> and <creation operator> gives $x=\sqrt{\hbar/(2m\omega)}(a+a^\dagger)$. If $\alpha=\sqrt N e^{i\varphi}$, then
$$
\boxed{\langle x\rangle_t=\sqrt{\frac{2\hbar N}{m\omega}}\cos(\omega t-\varphi).}
$$
For a classical <harmonic oscillator> with amplitude $A$, the conserved <energy> is $E=m\omega^2A^2/2$. Setting $E=\hbar\omega N$ gives exactly the same amplitude, $A=\sqrt{2\hbar N/(m\omega)}$. Thus these <coherent-state oscillator expectation values> follow the classical orbit.
The full quantum mean <energy> is nevertheless $\langle H\rangle=\hbar\omega(N+1/2)$: the extra <zero-point energy> belongs to the nonzero position and momentum uncertainties required by the <Heisenberg uncertainty relation>, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A <coherent state> follows a classical mean trajectory while retaining finite position and momentum uncertainties.
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