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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?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/32112/2451 , physics.stackexchange.com/q/56151/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 24, 2022 at 13:42
  • $\begingroup$ Is there any point to the theorem that I'm missing then? or is the whole point that it looks similar to classical physics? $\endgroup$ Sep 25, 2022 at 17:44
  • $\begingroup$ Thought experiment: how does one measure something classically, given that we do not live in a universe that operates on classical mechanics? $\endgroup$ Sep 26, 2022 at 15:20
  • $\begingroup$ What does it mean "classically measured"? I would just say: expectations values evolve with the classical equations of motion derived from the same Hamiltonian (caveat: not exactly 100% correct, see reed.edu/physics/faculty/wheeler/documents/Quantum%20Mechanics/…). The point (apart from the caveat) is that there is nothing more than that, Ehrenfest says nothing about "measurement". $\endgroup$
    – Quillo
    Sep 28, 2022 at 12:09
  • $\begingroup$ Related and useful (Ehrenfest theorem) physics.stackexchange.com/a/800488/226902 $\endgroup$
    – Quillo
    Feb 7 at 2:54

3 Answers 3

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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.

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Expectation values of quantum mechanics correspond to classically measured values by construction. Indeed, if the classical theory were not a limiting case of the quantum one, this latter would be terribly wrong (it would contradict to wealth of experimental evidence from our everyday life.)

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  • $\begingroup$ How do you deal with the fact that expectation values for many systems don't match any eigenvalue. Eigenvalues are the energies that are supposed to be measured, not expectation values.? $\endgroup$ Sep 26, 2022 at 9:27
  • $\begingroup$ @untreated_paramediensis_karnik How do you deal with the fact that expectation values for many systems don't match any eigenvalue. I am not dealing with it (that is I am not expressing any opinions/original ideas) - quantum mechanics does, more precisely in this case the Ehrenfest theorem - what we measure classically is an average, not an eigenvalue. What might mislead you is that in QM one talks a lot about quantum measurement (wave function collapse, etc.) But what the Ehrenfest theorem deals with is really a more mundane problem - the limited precision. $\endgroup$
    – Roger V.
    Sep 26, 2022 at 9:38
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    $\begingroup$ My point is that when you perform a stern gerlach experiment, you measure either spin down or up, not the expectation value, which could be zero spin, I guess. I don't understand what is wrong with this line of thought. $\endgroup$ Sep 26, 2022 at 12:40
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    $\begingroup$ @untreated_paramediensis_karnik strictly speaking, Ehrefest theorem is for position and momentum, but your example is interesting. Can we call Stern-Gerlach a classical measurement? Is there spin in classical limit? I think no... but I am curious to hear your line on this. $\endgroup$
    – Roger V.
    Sep 26, 2022 at 12:47
  • $\begingroup$ Classical physics is not a limiting case of quantum physics, at least not to our current knowledge. It should be, but we can't actually prove it with our theories. This inability is called "the measurement problem". $\endgroup$
    – Juan Perez
    Jan 8, 2023 at 19:53
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Surely the simplest explanation is that quantum mechanics is based on probability theory, so that its results will peak at the classical value.

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