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 $m\frac{d}{dt}x(t)=p(t)$.

  • 1
    $\begingroup$ Use \langle,\rangle for braces: $\langle x,y\rangle$ $\endgroup$
    – FShrike
    Jan 23, 2022 at 14:50
  • 5
    $\begingroup$ This better fits in physics stack exchange. $\endgroup$
    – Peter
    Jan 23, 2022 at 14:51
  • 1
    $\begingroup$ And has been asked on that site: physics.stackexchange.com/questions/61569/… and physics.stackexchange.com/questions/102082/… $\endgroup$
    – Joe
    Jan 23, 2022 at 14:54
  • $\begingroup$ @Peter I think this question is about some suitable property of Fourier transform. $\endgroup$
    – Hulkster
    Jan 23, 2022 at 20:13
  • $\begingroup$ @Joe Thanks but those answers are not close enough. $\endgroup$
    – Hulkster
    Jan 23, 2022 at 20:14

1 Answer 1


The uncertainty principle is part and parcel of Fourier analysis and cannot be evaded. In fact, nondimensionalization of ℏ by absorbing it into the units of x and p makes the ℏ/2 on the right hand side of the inequality equal to 1! It cannot go away. (In signal processing this amounts to the Gabor limit).

The only way to ignore it is to make it less significant, in macroscopic systems with a huge characteristic action S. In such systems, the decohered wavefunction still has $$ \frac{\Delta x}{\sqrt S} \frac{\Delta p}{\sqrt S} \geq \frac{\hbar}{S} , $$ so the widths of the x and p spreads (the variances) can still not go to zero, but, on that scale (of S) they can be very-very small and let the distributions appear sharply localized, on the scale of S, a macroscopic lab action quantity, localized in units suitable for S. See this answer, or else.

That is to say, the distributions of x and p are not arbitrarily sharp on the scale of $\sqrt \hbar$, but they can be quite sharp on the scale of $\sqrt S$, and may be taken to be δ-functions in classical and classical statistical mechanics.

The deeper technical feature of this limit, whose subtleties are often swept under the rug, is that observable quantities of interest are analytic functions in $\hbar/S$, and so all orders $O(\hbar/S)$ may be dropped consistently in most expressions, as routine in deformation quantization, linked above.

  • $\begingroup$ Thanks for your efforts (+1). I'll try to digest it. $\endgroup$
    – Hulkster
    Jan 25, 2022 at 21:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .