Clarification of Ehrenfest theorem

Question

The page on Ehrenfest theorem in Wikipedia(https://en.wikipedia.org/wiki/Ehrenfest_theorem) says-

"Although, at first glance, it might appear that the Ehrenfest theorem is saying that the quantum mechanical expectation values obey Newton’s classical equations of motion, this is not actually the case. If the pair ${\displaystyle (\langle x\rangle ,\langle p\rangle )}{\displaystyle (\langle x\rangle ,\langle p\rangle )}$ were to satisfy Newton's second law, the right-hand side of the second equation would have to be

${\displaystyle -V'\left(\left\langle x\right\rangle \right),}$ which is typically not the same as ${\displaystyle -\left\langle V'(x)\right\rangle}$"

I consulted the paper mentioned in the reference,but that explains more of when these two are equal,or cases where equivalence of Ehrenfest theorem and Newton's law fails,and other more sophisticated things.

I understand that those two are not the same,but I think ${\displaystyle -\left\langle V'(x)\right\rangle}$" gives the value of force and not the other one${\displaystyle -V'\left(\left\langle x\right\rangle \right),}$because, we must take the derivative of potential energy and then take the expectation value.Why should taking the expectation of position and then calculating the derivative of potential at that "average" position should give us the value of force but not the other expression(${\displaystyle -\left\langle V'(x)\right\rangle}$)?

Answer 1

Your reference is not saying that either one or the other is "correct" as a definition of force. What they are saying is that, as a matter of mathematical equivalence, one expression will lead to mean values of x and p that behave in the same way as classical x and p (because they obey the same equations), whereas the other expression does not.

Ehrenfest's theorem says what it says irrespective of whether the terminology of "force" is brought in. What that theorem tells us is that mean values of x and p behave in ways that are reminiscent of, but not exactly the same as, classical x and p.

Answer 2

I don't think it particularly important that the classical potential is not exactly equal to the expectation of the quantum potential in all cases. Classical potential is defined in the first instance by integrating force. To derive the force from the potential is circular. Really we have two slightly different potentials, because the classical notion of a force does not apply well in the quantum domain.

A more general and correct derivation of Newton’s second law is possible by first proving conservation of momentum from particle interactions, since N2 & N3 are together equivalent to conservation of momentum together with a definition of classical force, bearing in mind that classical forces don't sit well in quantum theory and are replaced with particle interactions (although the word force is still used). Conservation of momentum appears from the derivation of Feynman rules from the interaction density in qed, and field theories generally.