quantum mechanics - Ehrenfest Theorem and boundary Conditions

In what cases does Ehrenfests Theorem hold? If I look at the wavefunction of electrons in a squared box of length $L$ (with periodic boundary-conditions, $\Psi(0) = \Psi(L)$), then the solution to Schrödingers Equation are plain waves: $$ \Psi(x,t) = \frac{e^{i k x - \omega_k t}}{\sqrt{L}} $$

If I then compute the mean values for position, it is:

$$ \langle \hat{x} \rangle (t) = \frac{L}{2} \\ \frac{d}{dt} \langle \hat{x} \rangle (t) = 0 $$

But for momentum, I compute:

$$ \langle \hat{p} \rangle (t) = \hbar k \neq 0 $$

In this example Ehrenfests Theorem doesn't hold: $$ \frac{d}{dt} \langle \hat{x} \rangle (t) \neq \frac{1}{m}\langle \hat{p} \rangle (t)$$

Obviously because of the choice of my boundary-conditions. What did I do wrong? What other restrictions do I have to impose to make the Theorem applicable? It works very well If I assume the particle to be in a infinitely high potential box.

Answer

The reason for your conflicting results has to do with the subtleties of hermiticity on finite intervals.

Look carefully at the formal steps in the derivation of the Ehrenfest theorem: $$ \frac{d}{dt} \langle \psi(t) | x |\psi(t)\rangle = \langle \frac{d\psi}{dt} | x |\psi\rangle + \langle \psi | x |\frac{d\psi}{dt}\rangle = \frac{i}{\hbar} \left[ \langle H\psi | x|\psi\rangle - \langle \psi | x H |\psi\rangle\right] $$ Usually at this point one makes use of the hermiticity of $H$ and proceeds to rearrange the first term so as to obtain $$ \frac{i}{\hbar} \left[ \langle H\psi | x|\psi\rangle - \langle \psi | x H |\psi\rangle\right] = \frac{i}{\hbar} \langle \psi | [H, x] |\psi\rangle = …= \frac{1}{m} \langle \psi | p |\psi\rangle $$ And it is no problem to check that $H$ is indeed hermitic.

Here's the catch however: in this particular case it is hermitic on the space of functions periodic on $[0,L]$. But let us look closer at $$ \langle H\psi \;|\; x\;|\;\psi\rangle \equiv \langle H\psi \;|\; x\psi\rangle = \langle x\psi \;| \;H\psi\rangle^* $$ It is the matrix element of $H$ between one periodic function, $\psi$, and one that is no longer periodic, $x\psi$, which falls outside its proper domain. For this combo $H$ is no longer hermitic, as we can easily check by direct calculation: $$ \langle H\psi \;|\; x\psi\rangle \sim - \int_0^L{dx\; \frac{d^2\psi^*}{dx^2} x \psi} = - \int_0^L{dx\; \frac{d}{dx}\left(\frac{d\psi^*}{dx} x \psi\right)} + \int_0^L{dx\; \frac{d\psi^*}{dx} \frac{d}{dx}\left(x\psi\right)} = \\ = - \frac{d\psi^*}{dx} x \psi \big|_0^L + \int_0^L{dx\; \frac{d}{dx} \left[\psi^* \frac{d}{dx}(x\psi)\right]} - \int_0^L{dx\; \psi^*\frac{d^2}{dx^2}\left( x \psi\right) } = $$ $$ = \left[ \psi^* x \frac{d\psi}{dx} - \frac{d\psi^*}{dx} x \psi \right] \big|_0^L - \int_0^L{dx\; \psi^*\frac{d^2}{dx^2}\left( x \psi\right) } $$ where the first term on the last line was simplified based on the periodicity of $\psi$. But in what is left of it, the presence of $x$ spoils the periodicity and it no longer vanishes as one would naively hope.

Compare to the case when $\psi$ vanishes on the boundary, as for a particle in an infinite box: the boundary term disappears, or in other words, $x\psi$ is still in the domain of $H$ and the theorem is fine.