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.

  • $\begingroup$ Do you have in mind particle in a real box with walls being a region of high potential, or is the box just a region of space delimited from the rest by imaginary walls in thought? Is the "periodic boundary condition" motivated physically (particle bound to a ring) or just mathematically (let us suppose $\psi(0)=\psi(L)$ and find the implications)? $\endgroup$ – Ján Lalinský Jul 14 '16 at 9:15
  • $\begingroup$ It is motivated mathematically (Solid state physicists do this all the time to calculate electron movements in crystals) $\endgroup$ – Quantumwhisp Jul 14 '16 at 18:50

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.

| cite | improve this answer | |
  • $\begingroup$ So you are saying, that the property of an Operator to be hermitic depends on the vectors that he is acting uppon? That seems strange to me now. $\endgroup$ – Quantumwhisp Jul 16 '16 at 8:11
  • $\begingroup$ Yes, strange but true. Another simple but famous example: the momentum operator for a particle in an infinitely deep box, see for instance reed.edu/physics/faculty/wheeler/documents/Quantum%20Mechanics/… $\endgroup$ – udrv Jul 17 '16 at 3:02

For a more or less clear answer to all this, that actually goes further (turns out that the extra non-Hermitian terms have their own life - follow well-defined patterns) see https://arxiv.org/abs/1605.06534

Konstantinou, et al. “Emergent Non-Hermitian Contributions to the Ehrenfest and Hellmann-Feynman Theorems.” [1402.1128] Long Short-Term Memory Based Recurrent Neural Network Architectures for Large Vocabulary Speech Recognition, 10 July 2016, arxiv.org/abs/1605.06534.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.