= Solution
In the <Schrödinger picture>, a time-independent <Hamiltonian (quantum mechanics)> gives $|\psi_S(t)\rangle=U(t)|\psi(0)\rangle$, where $U=e^{-iHt/\hbar}$. The <Heisenberg picture> transfers this evolution to each <observable>:
$$
A_H(t)=U^\dagger(t)A_S(t)U(t),\qquad |\psi_H\rangle=|\psi(0)\rangle.
$$
Both pictures give the same <probabilities> and <expectation values>. Differentiating yields the <Heisenberg equation of motion>
$$
\boxed{\frac{dA_H}{dt}=\frac{i}{\hbar}[H_H,A_H]+U^\dagger\frac{\partial A_S}{\partial t}U.}
$$
The last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the <observable> is conserved.
The corresponding classical equation is $dA/dt=\{A,H\}_{\mathrm{PB}}+\partial A/\partial t$. The <correspondence principle> relates the <Poisson bracket> to $[A,H]/(i\hbar)$, with suitable operator ordering and a <classical limit>. For $H=p^2/(2m)+V(x)$, the <canonical commutation relation> gives
$$
\dot x_H=\frac{p_H}{m},\qquad \dot p_H=-V'(x_H).
$$
Taking means gives <Ehrenfest theorem>. Its second equation contains $-\langle V'(x)\rangle$, which need not equal $-V'(\langle x\rangle)$. A narrow <wave packet> in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.
For a <quantum harmonic oscillator>, direct solution gives $x_H(t)=x_H(0)\cos\omega t+p_H(0)\sin\omega t/(m\omega)$ and $p_H(t)=p_H(0)\cos\omega t-m\omega x_H(0)\sin\omega t$. These have the classical form as operator identities. A <coherent state> supplies a localized center moving on the classical orbit, whereas an <energy eigenstate> has stationary density and $\langle x\rangle=0$, even at large excitation. Thus large quantum numbers alone do not create a localized classical trajectory. Classical behavior also concerns state preparation, relative fluctuations and the resolution of the measurements.
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