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Your claim "for the QHO you can derive that $\langle \hat{x}(t)\rangle =0$ for all $t$" applies only to expectation values of energy eigenstates $|n\rangle$, but not for the general case of an arbitrary state vector $$|\psi\rangle = \sum\limits_{n=0}^\infty c_n |n\rangle, \qquad \sum\limits_{n=0}^\infty |c_n|^2 =1.$$ A well known case, where the expectation value of the position operator $$ \hat{x}(t) = e^{i \hat{H} t/\hbar} \hat{x}(0) e^{-i \hat{H} t/\hbar}= \hat{x}(0) \cos \omega t +\frac{\hat{p}(0)}{m \omega} \sin \omega t $$ of the harmonic oscillator performs a "classical" oscillation is the coherent state $$ |z\rangle= e^{-|z|^2/2} e^{z a^\dagger} |0\rangle = \sum\limits_{n=0}^\infty\frac{e^{-|z|^2/2} z^n}{\sqrt{n!}} |n\rangle, \qquad z \in \mathbb{C}. $$