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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?

Tobias Fünke
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eternalstudent
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    An easy but unsatisfactory answer is: Because it works. – Tobias Fünke Jul 05 '22 at 13:14
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    Does this answer your question? Name of concept: Replace classical variables by quantum operators – Roger V. Jul 05 '22 at 13:16
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    You can read about Heisenberg's reasoning for the development of matrix mechanics on wikipedia: https://en.wikipedia.org/wiki/Matrix_mechanics#Heisenberg's_reasoning. See also https://en.wikipedia.org/wiki/Heisenberg%27s_entryway_to_matrix_mechanics – Andrew Jul 05 '22 at 13:16
  • But rigorously speaking, shouldn't we prove that this the only possible way to make the expected position and momentum to behave like their classical counterparts? Otherwise I don't see why the results obtained as such would represent reality at quantum scales ? – eternalstudent Jul 05 '22 at 13:18
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    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. – Andrew Jul 05 '22 at 13:21
  • To add to @Andrew comment: Moreover, there is no a-priori reason why anything that you write down on a sheet of paper would correspond to reality. An instrumentalist or anti-realist would argue that the only physical thing are measurement results. Everything else is a way to make predictions, which can work or not, but does not tell you anything about the 'nature' of the world. – Tobias Fünke Jul 05 '22 at 13:23
  • thank you for your comments, what I mean is that for example if you take the Hamiltonian of the classical harmonic oscillator and replace x and p by the corresponding operators, there is no proof, in principle, that there is a system in the reality that obeys this new Hamiltonian. The conditions in Ehrenfest theorem are basically necessary (to match experience at large scales) but not sufficient. Sorry if this doesn't make sense – eternalstudent Jul 05 '22 at 13:31
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    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. – Cosmas Zachos Jul 05 '22 at 13:55

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