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In classical mechanics, usual observables of a system are its position $x(t)$ and its momentum $p(t)$, which are just symbols holding the unique position/momentum value at a given time.

I wonder why these classic observables become (self-adjoint) operators in quantum mechanics \begin{equation} \begin{split} x(t) \quad\longleftrightarrow\quad &\hat{x}\;,\quad \langle x|\hat{x}|\varphi\rangle = x\langle x|\varphi\rangle \\ p(t) \quad\longleftrightarrow\quad &\hat{p}\;,\quad \langle x|\hat{p}|\varphi\rangle = -i\hbar \frac{\partial \langle x|\varphi\rangle}{\partial x} \end{split} \end{equation} which operate on a wave function $\langle x|\varphi\rangle$.

I mean that I could not find any clear mathematical explanations of this after hours of googling and looking at the numerous other related (sometimes duplicate) forum's threads. This correspondence is usually given "as is" in most books and tutorials.

From this duplicate thread, my guess is that is has something to do with the underlying algebra. So to make a sound answer to that question would be first to understand the algebra beneath classical dynamics and how it is changed in quantum dynamics.

Unfortunately, I cannot post on this thread. Could someone help me to understand this change from classic to quantum?

Cosmas Zachos
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deb2014
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    Well, in general "why" questions are a bit "problematic" in physics. An answer to your query "I wonder why these classic observables become (self-adjoint) operators in quantum mechanics" can indeed be given as: We do so, because it works, i.e. because the predictions we phrase in the language of quantum mechanics agree with experiments. – Tobias Fünke Feb 16 '23 at 11:22
  • Yes, indeed, many physics laws derive from this (e.g. Born rule) : "it just works". Maybe I should have posted this question on the math forum. – deb2014 Feb 16 '23 at 12:16
  • My comment above is the answer to a question I've understood from your post (perhaps it is just me, tho). So if you actually want to ask something else, it might be worth to rephrase the question a bit... – Tobias Fünke Feb 16 '23 at 12:29
  • I think the question "why classic observables become operator in quantum mechanics ?" together with the first insights I gave are clear enough about the kind of answer I expect. Make something even clearer is beginning to answer the question, which I could do by mentioning (as far as I digged in the meanwhile) that classic observable are functions from the phase space to real space and that they form a Poisson Algebra ... where it becomes very mathematic, hence my suggestion to ask this question on the math forum. – deb2014 Feb 16 '23 at 12:48
  • You are discussing quantization maps, a subject well-represented on this site. Mathematicians will only nitpick and demand correct gambits. Your $\leftrightarrow$ gambit is flawed. Each operator has a phase space classical limit, but each classical phase-space function may well map to an infinity of different operators. ... – Cosmas Zachos Feb 16 '23 at 14:50
  • Linked. – Cosmas Zachos Feb 16 '23 at 14:57
  • Quantization is a mystery. – Cosmas Zachos Feb 16 '23 at 15:03
  • Thanks for your comments. Indeed, what am I looking for seems to be called "geometric quantization". I found here an overview with several references. But it involves quite a few mathematical objects I am not familiar with. As far as I understand, Geometric quantization comes after the Physics and fails to make a satisfying correspondence between classical and quantum observables in all cases. – deb2014 Feb 18 '23 at 22:07
  • @deb2014 you might be interested in this post and other links therein. – ZeroTheHero Mar 02 '23 at 01:10

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