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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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