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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?

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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.

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  • $\begingroup$ Putting aside domain issues and regularity of the potential, it seems to me that the Hamilton equation $\dot p = -\nabla V$ holds in general in this approach, as $[\hat p,V(\hat q)] = -i\hbar \nabla V(\hat q)$, where $\nabla V(\hat q)$ is the "multiplication by $\nabla V$" operator. Am I overlooking something? $\endgroup$ – Phoenix87 Mar 9 '15 at 12:46
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    $\begingroup$ Just domain issues, which however could be very subtle. In general, the commutation relations of self-adjoint operators like the one you consider descend from the commutation relations of associated unitary groups only in a weak sense (as far as I know there is no mathematically satisfactory proof of Eherenfest's theorem though it seems physically obvious)... unless the generators belong to a common Lie algebra and a suitable domain (Garding / Nelson) is used. $\endgroup$ – Valter Moretti Mar 9 '15 at 13:40
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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.

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