= Solution
The <Hamiltonian operator> is
$$
\boxed{H=\frac{p_x^2}{2m}+V(x)=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x).}
$$
Since multiplication by $x$ commutes with multiplication by $V(x)$, part (b) gives $[x,H]=i\hbar p_x/m$. Similarly $[p_x,H]=-i\hbar V'(x)$. Substituting into part (c) gives <Ehrenfest theorem>:
$$
\boxed{\frac{d\langle x\rangle}{dt}=\frac{\langle p_x\rangle}{m},\qquad \frac{d\langle p_x\rangle}{dt}=-\langle V'(x)\rangle.}
$$
Thus the mean <momentum> changes according to the mean force, and the <derivative> of mean <position> is mean <velocity>. These exact quantum equations resemble Hamilton's classical equations, but $\langle V'(x)\rangle$ generally differs from $V'(\langle x\rangle)$. The classical point-particle equation for the center is exact for potentials at most quadratic and is an approximation for a localized packet when the force varies little across it. A spatially broad packet in a nonlinear potential need not follow a classical trajectory. Even where its center follows classical motion, its width and measurement uncertainties remain quantum properties.
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