1
$\begingroup$

The proof of the statement of Ehrenfest theorem in the Schrodinger picture does not depend on the state vector. However, Wikipedia claims that:

for states that are highly localized in space, the expected position and momentum will approximately follow classical trajectories

Here is my question: Is it not the case that by the proof of the Ehrenfest theorem, $ \bf for \ all \ \text{states}$ and potentials we have the first ordinal moment (which is the expectation values) of $ \nabla V$ and $ \hat x$ operators being related by $$ m \left(\frac{\mathrm{d}^2}{{\mathrm{d}t}^2} \left< \hat x \right> \right) ~=~ -\nabla V \,? $$

On the other hand it is possible to prepare systems in states that are initially localized, but get more and more de-localized with time which for the position operator could be described in terms of the second ordinal moments(the variances ) of the position operator as $$ \left<{\left(\Delta x\right)}^2\right>_t \left<{\left(\Delta x\right)}^2\right>_0 ~\le~ \frac{\hbar^2t^2}{4m^2} $$ Here, the subscript of the variances refer to the time variable.

So we can say that there are states in which we can prepare our systems to be in, whose probability distributions in the measured values of $ \hat x$ at time $t=0$ and at time $t>t_0$ can have the respective variance satisfying the uncertainty relation specified above. On the other hand for this system and for system in any state, the expectation values of $\hat x$ at any given time satisfies $$ m \left(\frac{\mathrm{d}^2}{{\mathrm{d}t}^2} \left< \hat x \right> \right) ~=~ -\nabla V \,. $$ Is there any error in what I have said so far?

So, is the popular account including the Wikipedia one conflating variance with expectations of the measured probability distributions of the appropriate observables?

$\endgroup$
  • $\begingroup$ Your question is twisting together 4 distinct concepts and misreads Wikipedia. Can you clarify your question by highlighting it? For a harmonic potential and a coherent state the space variance remains fixed. The inequality you wrote for the free wave packet is off. Its connection to the Ehrenfest acceleration is logic-challenged. What do semiclassical trajectories have to do with anything? $\endgroup$ – Cosmas Zachos Feb 11 at 23:50
  • $\begingroup$ The question seeks to clarify about the generality of the ehrenfest theorem. The deduction employed to derive it doesn't seem to rely on any particular class of states. Does this suggest that the relation between the expectation values of position and potential dictated by the ehrenfest theorem applicable for any possible state in which we might prepare our system and for any possible potential? I interpreted(English is not my first language) the wikipedia article to mean that it is strictly applicable to only certain class of states for which the semi-classical approximation is reliable. $\endgroup$ – Varun Feb 12 at 0:05
  • $\begingroup$ The Ehrenfest theorem applies to all states and potentials, basically: it is a theorem. All semiclassical considerations in the article are essentially footnotes and could be omitted without harm, if you are not interested in semiclassical systems. $\endgroup$ – Cosmas Zachos Feb 12 at 0:07
  • $\begingroup$ Thanks, now I see what the article meant. $\endgroup$ – Varun Feb 12 at 0:10

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.