Ehrenfest theorem and coherent states

[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

 

[blue_leaf77]

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.

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.

 

[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.

As early as 1926, at the very beginning of quantum mechanics, Schrödinger was interested in studying quantum states, which mimic their classical counterparts through the time evolution of the position operator$Q(t)=e^{{\small \frac i \hbar Ht}}Qe^{{\small-\frac i \hbar Ht}}$. In this relation, $H = \frac{P^2}{ 2m} + V (Q)$ is the quantum Hamiltonian of the system. Schrödinger understood classical behavior to mean that the average or expected value of the position operator,$q(t) = \langle \mbox{coherent state}|Q(t)|\mbox{coherent state} \rangle$ in the desired state, would obey the classical equation of motion $m \ddot{\bar{q}}(t) +\frac{\overline{\partial V}}{\partial q}=0.$

Thanks

 

[A. Neumaier]

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.

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.

 

[A. Neumaier]

For the hydrogen atom, the correspondence between classicality and coherent states is already not so simple.
Some random links of many possible:
http://arxiv.org/abs/quant-ph/0101092
http://www.optics.rochester.edu/users/stroud/publications/gaeta901.pdf
http://arxiv.org/abs/quant-ph/9708010