Classical limit in quantum mechanics proof Several questions are about the limit $\hbar\rightarrow 0$, e.g.

*

*When does $\hbar \rightarrow 0$ provide a valid transition from quantum to classcial mechanics? When and why does it fail?


*Classical limit of quantum mechanics


*Classical Limit of Schrodinger Equation


*Semiclassical limit of Quantum Mechanics


*Why does the classical path give the dominant contribution in the path integral?


*How do you solve classical mechanics problems with quantum mechanics?
I read $\hbar \rightarrow 0$ in quantum mechanics. High upvoted answers say both that this limit is an acceptable way to recover Newton's laws of motion from Schroedinger equation (SE) (https://physics.stackexchange.com/a/108226/307786), and that it isn't (https://physics.stackexchange.com/a/42007/307786). Can someone provide a proof instead of examples?
I do not know exactly which observable statement I am considering in the limit. Something easy, hopefully. The first link I have states that in the limit, the energy spectrum of quantum harmonic oscillator becomes continuous. I will accept a proof that

*

*all bound states for any $V$ become continuous under the limit.


*Or a proof that the wave function takes a different classical meaning in the limit.


*Or that SE becomes an Euler-Lagrange (EL) equation or Hamilton equation or Newtonian equation ($F=ma$-like) with no complex numbers.


*Or that the solutions to SE look like delta functions in position space and momentum space simultaneously (because in center of mass (CM) there is no position or momentum uncertainty).
I got these ideas from What makes a theory "Quantum"?. I believe that a proof of one of these statements will imply most of the others. I cannot definitely say which one I want, because I do not know which ones are correct and provable. But I will accept a proof for any such argument.
 A: I'll respond to this part of the question:

Or that SE becomes an EL equation or Hamilton equation or Newtonian equation (F=ma-like) with no complex numbers.

Let $\psi$ be a solution to the Schrodinger equation
$$ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + V\psi$$
You can always write the complex-valued function in polar form $\psi = R e^{ iS/\hbar}$ for real functions $R$ and $S$.  If you substitute that into the Schrodinger equation and separate the real and imaginary parts, you get two coupled, real-valued PDEs
$$ \frac{\partial R}{\partial t} = -\frac{1}{2m} \left[ R \nabla^2S + 2 \nabla R \cdot \nabla S \right]$$
and
$$ \frac{\partial S}{\partial t} = - \left[ \frac{| \nabla S |^2}{2m} + V + Q \right] $$
where
$$ Q = -\frac{\hbar^2}{2m} \frac {\nabla^2 R}{R} $$
is the "quantum potential".
Note that the only term that has a factor of $\hbar$ in it is $Q$.  The first of the two equations, if you define $\rho = R^2 = \| \psi \|^2$, will give you the continuity equation from quantum mechanics.  The second is like the Hamilton-Jacobi equation for a classical particle with Hamilton's principle function $S$ except for the additional "quantum" term $Q$.
If you formally take $\hbar \rightarrow 0$, you recover the HJ equation for the classical particle exactly, which, I think, answers the part of your question up to the known fact that you can go back and forth between the EL and HJ equations once you're entirely within classical mechanics.
(Since $\hbar$ is a constant, rather than taking the limit that it goes to 0, I prefer to think of the limit that $|Q| \ll \left| |\nabla S|^2/(2m) + V \right|$, but the conclusion is the same.)
There is still the continuity equation, which I think becomes part of your remaining confusion.  The continuity equation is still valid.  In the classical context, it might more often be called a special case of the Fokker-Planck equation, in this case with 0 diffusion.  If you have some uncertainty (in the classical I-don't-know-everything-precisely sense rather than the quantum I-cannot-know-everything-precisely sense), then this equation tells you how to propagate that uncertainty forward.  What's critically different now is that the underlying dynamics don't depend on that uncertainty once the $Q$ term is ignored.  If you have no uncertainty, which is allowed in classical theory, this equation still works out in some generalize distribution sense  when the distribution goes to the limit of a delta function $\rho \sim \delta$ .
Some of the questions, answers, and papers that you're referencing are making that point explicit and separately from the "$\hbar \rightarrow 0$" limit.  That might be, at least in part, a difference in terminology. I'm willing to call the Fokker-Planck equation "classical" where others want to give it a distinguishing label like "stochastic classical" or "probabilistic classical".  Those in the latter camp (correctly) then point out that you also need to take the limit $\rho \rightarrow \delta$ to get to the "deterministic" classical case.
