= Solution
Time dependence in <quantum mechanics> describes how a prepared <quantum state> changes and thereby changes the <probabilities> of later measurements. For an isolated system, the <Schrödinger picture> assigns this evolution to a state vector $|\psi(t)\rangle$, governed by the <Time-dependent Schrödinger equation>
$$
i\hbar\frac{d}{dt}|\psi(t)\rangle=\hat H(t)|\psi(t)\rangle.
$$
The <Hamiltonian> is a <self-adjoint operator>; it has units of <energy>, while $\hbar$ has units of <action>, so $\hat H/\hbar$ has units of inverse time. For a particle with <potential energy> $V(\mathbf r,t)$, this becomes
$$
i\hbar\partial_t\psi(\mathbf r,t)
=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r,t)\right]\psi(\mathbf r,t).
$$
Given a suitable initial <wavefunction> and the physical <boundary conditions>, this first-order equation determines its subsequent propagation. Its <linearity> means that the evolution of a <quantum superposition> is the corresponding <quantum superposition> of evolved states.
For a real <potential energy>, this equation and its <complex conjugate> imply the local <continuity equation>
$$
\partial_t\rho+\nabla\cdot\mathbf j=0,
\qquad \rho=|\psi|^2,
\qquad \mathbf j=\frac{\hbar}{m}\operatorname{Im}(\psi^*\nabla\psi).
$$
Here $\rho$ is the <probability density> and $\mathbf j$ the <probability current>. Vanishing net boundary flux preserves <wavefunction normalization>, as required by the <Born rule>.
The <time-evolution operator> packages the solution for every initial state:
$$
|\psi(t)\rangle=U(t,t_0)|\psi(t_0)\rangle,
\qquad i\hbar\partial_tU(t,t_0)=\hat H(t)U(t,t_0),
\qquad U(t_0,t_0)=I.
$$
Because $\hat H^\dagger=\hat H$, differentiation gives
$$
\frac d{dt}(U^\dagger U)
=\frac i\hbar U^\dagger\hat H U-\frac i\hbar U^\dagger\hat H U=0.
$$
Therefore $U^\dagger U=I$: this is <unitary time evolution>, preserving <inner products> and all state <norms>. It is reversible, with $U(t,t_0)^{-1}=U(t_0,t)=U(t,t_0)^\dagger$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.
For a time-independent <Hamiltonian>, the solution is the <operator exponential>
$$
\boxed{U(t,t_0)=\exp[-i\hat H(t-t_0)/\hbar].}
$$
If the <Hamiltonian> varies with time, the analogous simple exponential of its integral is valid when $[\hat H(t),\hat H(t')]=0$ for all relevant times. Otherwise <time ordering> is essential:
$$
U(t,t_0)=\mathcal T\exp\left[-\frac i\hbar\int_{t_0}^t\hat H(t')\,dt'\right].
$$
The meaning is explicit in the first terms of the ordered series,
$$
U=I-\frac i\hbar\int_{t_0}^t\hat H(t_1)\,dt_1
-\frac1{\hbar^2}\int_{t_0}^t dt_1\int_{t_0}^{t_1}dt_2\,\hat H(t_1)\hat H(t_2)+\cdots:
$$
the later-time operator stands on the left. In the position representation, the same propagation is expressed by the <quantum-mechanical propagator>
$$
K(\mathbf r,t;\mathbf r',t_0)=\langle\mathbf r|U(t,t_0)|\mathbf r'\rangle,
\qquad
\psi(\mathbf r,t)=\int K(\mathbf r,t;\mathbf r',t_0)\psi(\mathbf r',t_0)\,d^3r'.
$$
Its initial value is the <Dirac delta function> $\delta^3(\mathbf r-\mathbf r')$.
A time-independent <Hamiltonian> has <energy eigenstates> satisfying $\hat H|n\rangle=E_n|n\rangle$. A single such initial state evolves as
$$
|\psi_n(t)\rangle=e^{-iE_n(t-t_0)/\hbar}|n\rangle.
$$
It is a <stationary state>: its <global phase> changes, but its <probability density> and the distributions of time-independent <observables> do not. Its <energy> is sharp. Any <quantum superposition> entirely within one degenerate <energy eigenspace> also acquires one common <global phase> and is stationary. Stationarity does not require zero <probability current>; for instance, $u_n(\phi)=e^{in\phi}/\sqrt{2\pi}$ on the ring has a constant density and persistent circulation when $n\ne0$.
An arbitrary normalized state can instead contain different <energy eigenvalues>:
$$
|\psi(t)\rangle=\sum_n c_ne^{-iE_n(t-t_0)/\hbar}|n\rangle,
\qquad \sum_n|c_n|^2=1.
$$
A <continuous spectrum> requires corresponding integrals. For the discrete case, an <energy measurement> gives $E_n$ with fixed <probability> $|c_n|^2$ (summed over degeneracies), but relative <quantum phases> change with time. Consequently
$$
|\psi(\mathbf r,t)|^2=\sum_{m,n}c_m^*c_n\phi_m^*(\mathbf r)\phi_n(\mathbf r)
e^{i(E_m-E_n)(t-t_0)/\hbar}.
$$
The <interference> terms can oscillate at <angular frequencies> $(E_m-E_n)/\hbar$. For two populated <energy levels>, this supplies a characteristic period $2\pi\hbar/|E_m-E_n|$ when the <interference> term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.
For a normalized state and an <observable> $\hat A(t)$, its <expectation value> is
$$
\langle A\rangle_t=\langle\psi(t)|\hat A(t)|\psi(t)\rangle.
$$
This is the statistical <mean> over repeated preparations and measurements. In an <eigenbasis> of <energy eigenstates> for a time-independent <Hamiltonian> and a time-independent $\hat A$,
$$
\langle A\rangle_t=\sum_{m,n}c_m^*c_n A_{mn}e^{i(E_m-E_n)(t-t_0)/\hbar}.
$$
Only <matrix elements> connecting different <energies> can generate these oscillations. If $[\hat A,\hat H]=0$, its <expectation value> is constant for every initial state, provided it has no explicit time dependence.
Differentiating the <expectation value> and substituting the <Schrödinger equation> for both the bra and ket proves the general <Ehrenfest theorem>:
$$
\begin{aligned}
\frac d{dt}\langle A\rangle
&=\langle\dot\psi|\hat A|\psi\rangle+
\left\langle\frac{\partial\hat A}{\partial t}\right\rangle+
\langle\psi|\hat A|\dot\psi\rangle\\
&=\frac i\hbar\langle\hat H\hat A-\hat A\hat H\rangle+
\left\langle\frac{\partial\hat A}{\partial t}\right\rangle.
\end{aligned}
$$
Equivalently,
$$
\boxed{\frac d{dt}\langle A\rangle
=\frac1{i\hbar}\langle[\hat A,\hat H]\rangle+
\left\langle\frac{\partial\hat A}{\partial t}\right\rangle.}
$$
These manipulations require the state to lie in the relevant <operator domains>. Taking $\hat A=\hat H$ gives $d\langle H\rangle/dt=\langle\partial_t\hat H\rangle$, so <energy> is conserved under a time-independent <Hamiltonian>.
For $\hat H=\hat p^2/(2m)+V(\hat x)$, the <canonical commutation relation> gives
$$
[\hat x,\hat H]=\frac{i\hbar}{m}\hat p,
\qquad [\hat p,\hat H]=-i\hbar V'(\hat x).
$$
Hence the particularly useful position-and-momentum form of <Ehrenfest theorem> is
$$
\boxed{\frac d{dt}\langle x\rangle=\frac{\langle p\rangle}{m},
\qquad \frac d{dt}\langle p\rangle=-\langle V'(x)\rangle.}
$$
It resembles <Newton's second law>, but the mean <force> generally differs from the <force> at the mean position. If a narrow <wave packet> samples a region where the <force> varies little, $\langle V'(x)\rangle\simeq V'(\langle x\rangle)$ gives approximate classical motion. For a quadratic <potential energy> this replacement is exact. In particular the <quantum harmonic oscillator> obeys
$$
\frac{d^2}{dt^2}\langle x\rangle=-\omega^2\langle x\rangle,
\qquad
\langle x\rangle_t=\langle x\rangle_{t_0}\cos\omega\tau+
\frac{\langle p\rangle_{t_0}}{m\omega}\sin\omega\tau,
\qquad \tau=t-t_0.
$$
The centroid follows the classical <harmonic oscillator>, although this does not specify the full evolving <wavefunction> or its width.
The <Heisenberg picture> expresses the same physics by assigning evolution to <observables> instead:
$$
\hat A_H(t)=U^\dagger(t,t_0)\hat A_S(t)U(t,t_0),
\qquad |\psi_H\rangle=|\psi(t_0)\rangle.
$$
The <expectation value> is unchanged, and differentiation yields $d\hat A_H/dt=(i/\hbar)[\hat H_H,\hat A_H]+U^\dagger(\partial_t\hat A_S)U$. Thus state propagation, evolving <expectation values>, and the classical correspondence in <Ehrenfest theorem> are mutually consistent descriptions of the same <unitary time evolution>.
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