What does Ehrenfest's theorem actually mean?

Question

I am told that Ehrenfest's theorem, applied to a physical observable $\hat A$, is: $$\frac{d\langle\hat A\rangle}{dt}= \frac{i}{\bar h}\langle[\hat H,\hat A]\rangle$$ I don't understand how to use this equation or what it means intuitively.

Answer 1

The equations describes the time evolution of an expectation value of an operator (which is the expectation value of the value measured when doing many measurements on identically prepared systems). The expectation values are taken with respect to the state of the system at time $t$. This answer will show how to use the Ehrenfest theorem by applying it to an example. As for the intuition: Intuition is always a dangerous business with respect to quantum mechanics.

In some cases, you can use the Ehrenfest theorem to calculate the evolution of expectation values without solving the Schrödinger equation, if the commutator results in a sufficiently simple operator, or at least get results for the behaviour of sharp wave packets. Consider for example the Hamiltonian of a particle in a potential $V(x)$ $$H = \frac{p^2}{2m} + V(x).$$ Then you can compute $[H, x]$ and $[H, p]$: $$[H, x] = \frac{1}{2m} [p^2, x] = \frac{p}{2m} [p,x] + \frac{1}{2m} [p,x] p = -i\hbar \frac{p}{m},$$ $$[H, p] = [V(x), p] = [V(x), -i\hbar\nabla] = i\hbar \nabla V(x).$$ Plugging this in the Ehrenfest theorem gives the following equations of motion for the expectation values of $x$ and $p$: $$d_t \left< p \right> = - \left<\nabla V(x)\right> $$ $$d_t \left< x \right> = \frac{p}{m} $$ Those are almost the classical equations of motion for position and momentum. The only difference is that the "force" $-\nabla V(x)$ is averaged over the state and not taken at the central position, for a wave packet which is small compared to the variation scale of $V(x)$ this therefore leads to the classical equations of motion in good approximation. So we have used the Ehrenfest theorem to show how and under which circumstances Newtonian mechanics arise from quantum mechanics.

There is a special case where the resulting equations of motion can be solved without reference to the state, namely the harmonic oscillator $V(x) = \frac 1 2 k (x-x_0)^2$, here we get $$\left< - \nabla V(x)\right> = -k \left<x - x_0\right> = -k\big(\left<x\right> - x_0\big),$$ so the equations of motion for the averages are exactly the classical equations of motion for the harmonic oscillator.