There is a wellknown principle that one can recover classical mechanics from quantum mechanics in the limit as $\hbar$ goes to zero. I am looking for the strongest statement one can make concerning this principle. (Ideally, I would love to see something like: in the limit as $\hbar$ goes to zero, the position wavefunction reduces to a delta function and Schrödinger's equation / Feynman's path integral reduces to the Newtonian/Lagrangian/Hamiltonian equations of motion).

1$\begingroup$ I don't think this question is meant for this site. Anyway, taking $\hbar\to 0$ (or really $S/\hbar\to\infty$ for classical action $S$) does give you all that, stemming from $[x,p]=\pm i\hbar$ (I always forget the sign). Ehrenfest's theorem comes into play here. In particular, for the path integral, this webpage should help: scholarpedia.org/article/Path_integral (the leading contributions to the path integral come from approximately extremum paths of the classical equations of motion). $\endgroup$– Chris GerigJul 15, 2012 at 23:41

2$\begingroup$ In the limit, the commutator bracket reduces to the Poisson bracket, and Schrodinger's equation reduces to the Hamiltonian formulation of classical mechanics. This is more or less worked out in this blog post: qchu.wordpress.com/2011/08/14/… . Is this the kind of thing you mean? (The path integral reduces to the Lagrangian formulation of classical mechanics, but I don't know enough about either to say much more about this.) $\endgroup$– Qiaochu YuanJul 16, 2012 at 0:09

6$\begingroup$ Note also that in the semiclassical limit $\hbar \to 0$, most wavefunctions $\psi\rangle$ (i.e. quantum pure states) do not converge to delta functions $\delta_{(q,p)}$ (classical pure states), but instead to superpositions of delta functions $\rho(q,p)$ (classical mixed states), with $\rho$ being the limit of the Wigner transform of $\psi\rangle$. (Density matrices (quantum mixed states) will also converge to classical mixed states, of course.) $\endgroup$– Terry TaoJul 16, 2012 at 1:55

1$\begingroup$ In classical limit of the wave functions corresponds to Lagrangian submanifolds. See some comments at mathoverflow.net/questions/87723/… $\endgroup$– Alexander ChervovJul 16, 2012 at 10:39

1$\begingroup$ If one wants to explain, why we don't see superpositions of macroscopic objects, one needs decoherence theory. See physics.stackexchange.com/questions/32112/… . $\endgroup$– jjcaleJul 16, 2012 at 21:29
3 Answers
There are two different views about the semiclassical limit in quantum mechanics, the first is based on a somewhat shaky ground due to the fact that the existence of the Feynman integral is not proved yet. On the other side, Wiener integral, its imaginary time counterpart does exist and one could pretend to work things out from this and then move to the Feynman integral. The other approach relies on substantial mathematical theorems due to Elliott Lieb and Barry Simon in the '70 and is essentially valid for manybody physics. These latter results make the limit $\hbar\rightarrow 0$ and $N\rightarrow\infty$ equivalent while the former is not really a physical limit due to the fact that Planck constant is never zero.
Starting from Feynman path integral, the standard formulation applies to a mechanial problem described from a Lagrangian $L$, normally $L=\frac{\dot x^2}{2}V(x)$ but one can extend this to more general cases, and then the postulate is that, given a path $x(t)$, this must contribute to the full quantum mechanical amplitude of a particle going from the point $x_a$ to $x_b$ with a term $e^{\frac{i}{\hbar}S}$ being $S=\int_{t_a}^{t_b}dtL(\dot x,x,t)$ the action. All the possible paths contribute and so, the full amplitude will be given by the formal writing $$ A(x_a,x_b)\sim\int[dx(t)]e^{\frac{i}{\hbar}\int_{t_a}^{t_b}dtL(\dot x,x,t)}. $$ Be warned that this integral is not proved to exist yet, but the Wiener counterpart, that can be obtained changing $t\rightarrow it$, exists and describes Brownian motion. Now, if you take the formal limit $\hbar\rightarrow 0$ to this integral you will immediately recognize the conditions to apply the stationary phase method to it. This implies that the functional must have an extremum and this can be obtained by pretending that $$ \delta S=\delta \int_{t_a}^{t_b}dtL(\dot x,x,t)=0 $$ that is, the paths that give the greatest contribution are the classical ones and one recover the classical limit as a variational principle as learned from standard textbooks.
While this is a quite common approach, to extend what really happens to a macroscopic system that we can see to respect all the laws of classical mechanics, we have to turn our attention to the limit of a large number of particles $N\rightarrow\infty$. In this case one has more rigorous results. These are due to Lieb and Simon as already said above. They published two papers about
Lieb E. H. and Simon B. 1973 Phys. Rev. Lett. 31, 681.
Lieb E. H. and Simon B. 1977 Adv. in Math. 23, 22.
In the first paper, their theorem 4 states
Theorem: For $\lambda < Z$, let $E_N^0$ and $\rho_N^0(x)$ denote the groundstate energy and oneelectron distribution function for N spin$\frac{1}{2}$ electrons obeying the Pauli principle and interacting with $k$ nuclei as described above. Then (a) $N^{\frac{7}{3}}E_N^0\rightarrow E_1$, as $N\rightarrow\infty$; (b) $N^{2}\rho_N^0(N^{\frac{1}{3}}x)\rightarrow\rho_1(x)$ as $N\rightarrow\infty$, where convergence in (b) means that for any domain $D\subset R^3$, the expected fraction of electrons in $N^{\frac{1}{3}}D$ approaches $\int_D\rho_1 (x)d^3x$.
Where $\rho_1(x)$ and $E_1$ refer to the ThomasFermi distribution and the corresponding energy. This theorem states that the limit $N\rightarrow\infty$ for a quantum system, under some mild conditions, is the ThomasFermi distribution. A system with this distribution is a classical system. The fact that a system with a ThomasFermi distribution is a classical one can be seen through the following two references:
W. Thirring(Ed.), The Stability of Matter: From Atoms to Stars  Selecta of E. Lieb, SpringerVerlag (1997).
L. Hörmander, Comm. Pure. Appl. Math. 32, 359 (1979).
The second paper just gives the mathematical support to derive ThomasFermi approximation as the leading order of a classical expansion for $\hbar\rightarrow 0$ that I will not present here.

$\begingroup$ Please see my followup question: mathoverflow.net/questions/102415/… Thank you. $\endgroup$– dabJul 17, 2012 at 15:32
The question can be seen from two points: rather elementary one and hot research topic. Let me comment on both.
The elementary ones:
1) As it was mentioned by Jon and Chris by the stationary phase approximation you can immediately see that Feynman path integral for h>0 corresponds to extrema of the Lagrangian  which are precisely the classical equations of motion in Lagrange's description. (In particular case L=kineticEnery  PotentialEnergy you will get Newton's equation  this is subject of classical mechanics textbooks).
2) In the Heisenberg picture of QM one considers the equations of motion in the form
$d/dt \hat O = [\hat H, \hat O] $
The commutator of operators corresponds to Poisson brackets in the classical mechanics. So classical limit of this equation is:
$d/dt O = ${ $ H, O $ }
Which are Hamilton equations of motion in classical mechanics. So we see that quantum motion > classical motion for h>0.
PS It might be worth to remind here the connection between Heisenberg picture and Schrodinger's: $d/dt \Psi = H\Psi$. This is purely linear algebra: if you consider the evolution on vector space V given by this equation then operators (i.e. V\otimes V^*) will evolve according to d/dt O = [H, O].
3) Now about what is the classical limit of wave function. The main surprise that it does not correspond to delta functions as naively expected. The first approximation is that classical limit corresponds to Lagrangian submanifolds.
Let me first give example: consider the wave function which is $\Psi(x)= \delta(xx_0)$ (it is nonnormalized, but still) naively it corresponds to particle which have coordinate equal to $x_0$. The classical limit of this would be a line on the phase space (p,q) which is $q=x_0$.
Let me repeat from Coherent states vs quantization of Lagrangian submanifold
Consider sumanifold defined by the equations $H_i=0$. Consider "corresponding" quantum hamiltonians $\hat H_i $, consider vector $\psi$ in the Hilbert space such that $\hat H_i \psi = 0$. This $\psi$ we are talking about. Why it is important "Lagrangian" ? It is easy. If $A \psi =0$ and $B\psi = 0$ then it is true for commutator $[A,B]\psi = 0$. In classical limit commutator correspond to Poisson bracket so we see that even if we start from $H_i$ which is not close with respect to Poisson bracket we must close it  so we get coisotropic submanifold. Lagrangian  just restiction on the dimension  that it should be of minimal possible dimension  so after quantization we may expect finite dimensional subspace (in the best case 1dimensional).
Exercise: derive from this general prescription an example above.
Current research
It is fruitful line of research to think about the correspondence between quantum and classical realms. Not just the limit, but sometimes one may hope to go in opposite direction and to completely describe quantum objects in terms of classical one. Sometimes it is subject of deepest and fascinating conjectures:
http://arxiv.org/abs/math/0512169
Automorphisms of the Weyl algebra
Alexei BelovKanel, Maxim Kontsevich
Abstract:
We discuss a conjecture which says that the automorphism group of the Weyl algebra in characteristic zero is canonically isomorphic to the automorphism group of the corresponding Poisson algebra of classical polynomial symbols. ...
So the conjecture says that classical object  automorphisms of Poisson algebra $\{ p_i, q_k \} =\delta_{ik}$ is exactly the same as quantum object  automorphism of Weyl algebra ($[p_i, q_k]=\delta_{ik}$)
It is related to famous Jacobian conjecture, see http://arxiv.org/abs/math/0512171 by the same authors.

$\begingroup$ some problem with typing Poisson brackets { } it seems \{ does not work in math mode ? $\endgroup$ Jul 16, 2012 at 16:27

$\begingroup$ Please see my followup question: mathoverflow.net/questions/102415/… Thank you. $\endgroup$– dabJul 17, 2012 at 15:32
Even though the $\hbar \rightarrow 0$ is commonly referred as the classical limit of quantum mechanics (QM), the truth is that QM doesn't have such a welldefined classical limit. Moreover, any argument on $\hbar \rightarrow 0$ is not valid as $\hbar$ is not a dimensionless parameter $^{*}$. For example, in special relativity $v/c \rightarrow 0$ represents the classical limit of the theory, and so does $r/r_c \rightarrow 0$ in General relativity ($r_c$ is the Schwarzschild radius), by which you can restore Newton's law of gravity. But the story is totally different for QM as there is no such a counterpart by which you can restore the classical limit.
($^{*}$To be clear, the fact that $\hbar$ isn't dimensionless means that depending on the units you choose you can make it arbitrarily small/large, thus rendering it invalid as a classical limit.)

5$\begingroup$ When you write $v/c\rightarrow 0$ as a legitimate expression of the nonrelativistic limit, you are comparing a typical speed $v$ of your system with $c$. You likewise, of course, have to compare a typical action $S$ of your system with $\hbar $ to assess how close you are to the classical limit $\hbar /S\rightarrow 0$. This is understood when people write $\hbar \rightarrow 0$. $\endgroup$ Jan 3, 2021 at 5:27