Does the $\frac12mv^2$ law apply to quantum mechanics? Consider the classical Hamiltonian for a spring: 
\begin{equation}
H = \frac{1}{2}\frac{p^2}{m} + \frac{1}{2}kx^2
\end{equation} 
This is one of those simple cases where when you work out the math we find 
\begin{equation}
m\ddot{x} = -kx
\end{equation}
and it makes plain and obvious that the $\frac{1}{2}\frac{p^2}{m}$ term has now turned into $m\ddot{x}$, Newton's law. So it's clear here how the $ \frac{1}{2}\frac{p^2}{m}$ corresponds to the $m\ddot{x}$ term and the $-kx$ is consistent with Newtons law.
My Question:
Does this relationship between $ \frac{1}{2}\frac{p^2}{m}$ and $m\ddot{x}$ hold for the quantum mechanical operator? By the operator, I mean this Hamiltonian 
\begin{equation}
\hat{H} = \frac{1}{2}\frac{\hat{p}^2}{m} + \hat{V}
\end{equation} 
Its obviously the same format, but my suspicion is that it doesn't carry the same relationship to Newtons laws because of the fundamental differences between classical and quantum mechanics. If I am correct that the quantum mechanical operator is unrelated to Newton's law, can someone explain why given they are of the same format? 
 A: By Ehrenfest's theorem you have
$$i\hbar\frac{\text d}{\text dt}E_\omega[q] = E_\omega[[q,H]]$$
and
$$i\hbar\frac{\text d}{\text dt}E_\omega[p] = E_\omega[[p,H]]$$
where $E_\omega$ indicates the expectation value over the state $\omega$. Simple computations show that the first equation gives
$$i\hbar\frac{\text d}{\text dt}E_\omega[q] = \frac{i\hbar}m E_\omega[p]$$
while the second one gives
$$i\hbar\frac{\text d}{\text dt}E_\omega[p] = -i\hbar kE_\omega[q].$$
Setting $x:=E_\omega[q]$ and $\pi := E_\omega[p]$ you see that you get
$$\dot x = \frac\pi m\qquad\text{and}\qquad\dot \pi = -kx,$$
whence
$$m\ddot x = -kx.$$
More generally, given a generic potential $V$ which is differentiable, the above equations generalise to
$$m\ddot x = -U',$$
where $U':=\frac i\hbar E_\omega[[p,V(q)]]$, which coincides with $V'(x)$ only when $V'$ is linear in its argument.
A: Yes, it holds referring to the Heisenberg evolution of operators and in the specific case of the harmonic oscillator:
$$\hat{x}(t) := U(t)^\dagger \hat{x} U(t)$$
where $U(t) := e^{-it\hat{H}}$ (here $\hbar:=1$).
One has
$$\frac{d^2}{dt^2} \hat{x}(t) =\frac{d}{dt} \frac{d}{dt}  U(t)^\dagger \hat{x} U(t) = \frac{d}{dt}  U(t)^\dagger i [\hat{H}, \hat{x}] U(t) =
U(t)^\dagger i^2 [\hat{H},[\hat{H}, \hat{x}]] U(t)\:. $$
Using the explicit form of $\hat{H}$ of the harmonic oscillator and the canonical commutation relations, $[\hat{x}, \hat{p}]= iI$, you have $[\hat{H},[\hat{H}, \hat{x}]]= \frac{k}{m}\hat{x}$ so that,
$$U(t)^\dagger i^2 [\hat{H},[\hat{H}, \hat{x}]] U(t)= -\frac{k}{m} U(t)^\dagger \hat{x} U(t) = -\frac{k}{m} \hat{x}(t)$$
so that
$$m\frac{d^2}{dt^2} \hat{x}(t) = -k \hat{x}(t)\:.$$
This result does not hold for more complicated forms of $V$ as you can see by direct inspection.
