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Is is because of Ehrenfest theorem that implies that the Hamiltonian constructed as such by replacing will lead the expected position and momentum to behave like their classical counterparts? Or is there another reason?

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    $\begingroup$ An easy but unsatisfactory answer is: Because it works. $\endgroup$ Jul 5, 2022 at 13:14
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    $\begingroup$ Does this answer your question? Name of concept: Replace classical variables by quantum operators $\endgroup$
    – Roger V.
    Jul 5, 2022 at 13:16
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    $\begingroup$ You can read about Heisenberg's reasoning for the development of matrix mechanics on wikipedia: en.wikipedia.org/wiki/…. See also en.wikipedia.org/wiki/… $\endgroup$
    – Andrew
    Jul 5, 2022 at 13:16
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    $\begingroup$ It's not the only way to represent position and momentum quantum mechanically. In the path integral formalism, there are no operators. You have to distinguish the representation of physics from actual physics. There are always multiple mathematical representations of the same physics. $\endgroup$
    – Andrew
    Jul 5, 2022 at 13:21
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    $\begingroup$ It doesn't make sense indeed. In deformation quantization, you do not replace classical phase space variables with operators: you just tweak their composition laws to something different (*-multiplication) which is consistent and agrees with observed phenomena at small scales, in contrast to classical mechanics. $\endgroup$ Jul 5, 2022 at 13:55

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