Paul Dirac states in his book the principles of quantum mechanics that “...classical mechanics may be regarded as the limiting case of quantum mechanics when h tends to zero.”
Fine, but how to actually show that is the case? The following discussion is my own understanding of this problem and how to state it in mathematical terms.
We know that in quantum mechanics both the position and momentum become probability distributions, $|\psi(x,t)|^2$ and $|\phi(p,t)|^2$, that are connected by the (suitably scaled) Fourier transformation $\psi(x) = (\mathcal{F}\phi)(p)$, where $p=\hbar x$. Also, the Heisenberg's uncertainty relation illustrates the mutual limit for those distributions, i.e. $\Delta(\psi)\Delta(\phi) \geq \hbar/2$. But I haven't found an explanation that shows why those distributions are allowed to become classical-like delta-distributions as $\hbar \rightarrow 0$. Maybe there is some obvious scaling argument why making the $\hbar$ smaller allows to squeeze both distributions in the Heisenberg's uncertainty relation?
More precisely: How to show (if possible?) that if $\hbar$ tends to zero, then the following functions $\psi, \phi \in L^2(\mathbb R, \mathbb C)$ can both be sharply localized? $$\psi(x,t)=\frac{1}{\sqrt{2\pi \hbar}}\phi(p,t)e^{ip \cdot x/\hbar}dp \, \,\, \,\, \,\, \,\, \,\text{and}\, \,\, \,\, \,\, \,\, \,\phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}}\psi(x,t)e^{-ip \cdot x/\hbar}dx.$$
Of course, the transition $\hbar \rightarrow 0$ only allows to pass the uncertainty principle. After that we are free to prepare a quantum state of a mass particle with approximately delta-distributions for both the position and momentum. Then the averaged equation of motion, i.e. the Ehrenfest theorem $m\frac{d}{dt}\langle x\rangle=\langle p\rangle$, where $\langle x\rangle = \int x|\psi(x,t)|^2 dx$ and $\langle p\rangle = \int p|\phi(x,t)|^2 dp$, is approximately the desired, point-like classical equation of motion $m\frac{d}{dt}x(t)=p(t)$.
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for braces: $\langle x,y\rangle$ $\endgroup$