I understand that we can use the Heisenberg picture to show, for a Hamiltonian of the form
$$
\hat{H}=\frac{\hat{P}^{2}}{2m}+\hat{V}(\hat{X})
$$
the Ehrenfest theorem:
$$
m\partial_{t}\langle \hat{X}\rangle=\langle \hat{P}\rangle\ \text{ and } \partial_{t}\langle \hat{P}\rangle=-\langle \nabla\hat{V}(\hat{X})\rangle
$$
thus we return to the classical equations of motion if we let $\langle \hat{X}\rangle$ correspond to the classically measured position and $\langle \hat{P}\rangle$ correspond to the classically measured momentum. I don't understand why this means it is necessary for $\langle \hat{X}\rangle$ correspond to the classically measured position and $\langle \hat{P}\rangle$ correspond to the classically measured position. It seems like the expectation values could still obey this relation without corresponding to the classical values. Any idea?
Best Answer
In general, there is no such thing as a "classically measured position" for a generic quantum system/state. Some situations are simply not well-modeled by classical physics, and Ehrenfest's theorem itself is not about the classical limit of quantum physics. No one is saying that there is a general link between quantum expectation values and classical measurements.
What you're looking for is the correspondence principle: There is a certain class of quantum states (heuristically those with "large quantum numbers", in modern approaches technically often coherent states with high particle number) for which the uncertainties of the operators get small enough - compared to a relevant quantity such as the precision of the measurement apparatus - that the quantum nature of the states becomes invisible and their expectation value hence effectively the sole possible result of measurement. It is for these "corresponding states" that Ehrenfest's theorem implies that the classically measured values obey the same equation of motion as the quantum expectation values.