# Ehrenfest theorem and coherent states

1. Dec 31, 2015

### ShayanJ

From the Ehrenfest theorem, we know that the equation below is correct for any state $\psi$.
$m\frac{d^2}{dt^2}\langle x \rangle_{\psi} =-\langle \frac{\partial V(x)}{\partial x} \rangle_{\psi}$

But then one of the definitions of coherent states is states for which the expected value of position operator in that state, satisfies equation above. But by the Ehrenfest theorem, there should be no state that doesn't satisfy the equation above. So it seems I'm confused about one of the things above but I can't find which one it is! I'll appreciate any hint.
Thanks

2. Dec 31, 2015

### blue_leaf77

I don't think you can use that as one of the defining properties for a coherent state for it's indeed generally applicable for any state. This state being the eigenstate of a lowering operator is a sufficient definition. In fact you are not the first one to observe this agreement with the Ehrenfest theorem, see the sentence just before the last paragraph in the last page in http://www.fysik.su.se/~hansson/KFT2/extra notes/cstates copy.pdf.

3. Dec 31, 2015

### ShayanJ

The following quote is from the book "Coherent States in Quantum Physics" by Jean-Pierre Gazeau. This doesn't make sense because all states do obey that equation so that can't be an indication of classicality. I was partly confused by this.
Thanks

4. Dec 31, 2015

### A. Neumaier

With $V(\overline q)$ in place of $\overline V$, which is probably what was intended (and is different from the Ehrenfest theorem), it happens to be true for the harmonic oscillator, which was the only system for which Schroedinger considered coherent states.

5. Jan 1, 2016