# Does the $\frac12mv^2$ law apply to quantum mechanics?

Consider the classical Hamiltonian for a spring: $$H = \frac{1}{2}\frac{p^2}{m} + \frac{1}{2}kx^2$$ This is one of those simple cases where when you work out the math we find $$m\ddot{x} = -kx$$ 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

$$\hat{H} = \frac{1}{2}\frac{\hat{p}^2}{m} + \hat{V}$$

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?

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.
• 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? – Phoenix87 Mar 9 '15 at 12:46
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.