There is a well-known argument that if we write the wavefunction as $\psi = A \exp(iS/\hbar)$, where $A$ and $S$ are real, and substitute this into the Schrodinger equation and take the limit $h \to 0$, then we will see that $S$ satisfies the Hamilton-Jacobi equation (for example see http://physics.bu.edu/~rebbi/hamilton_jacobi.pdf).

I understand this, however I feel that I don't understand the claim that this shows that quantum mechanics reduces to classical mechanics in the $\hbar \to 0$ limit. I am confused because I would think that in order to show that QM reduces to CM we would need to show that as $\hbar \to 0$, $|\psi(x,t)|^2$ becomes very narrow and that its center moves in a classical trajectory, ie $|\psi(x,t)|^2=\delta(x-x_\text{classical}(t))$. And it seems that the above argument does not at all show this. In fact, my understanding is that all that matters for the physical measurement of position is $|\psi|^2$ (since this gives the probability distribution) and hence the phase factor $\exp(iS/h)$ seems to not matter at all.

Moreover, some books (see pg 282 of http://www.scribd.com/doc/37204824/Introduction-to-Quantum-Mechanics-Schrodinger-Equation-and-Path-Integral-Tqw-darksiderg#download or pgs 50-52 of Landau and Lifshitz) give a further argument to the one mentioned above. They further say that if $\psi = A \exp(iS/h)$, then $|\psi|^2 = A^2$ satisfies the classical continuity equation for a fluid with a corresponding velocity $dS/dt$, which in the $h \to 0$ limit is equal to the classical velocity.

This argument makes more sense to me. However, I still have some questions about this. (1) I know that there are stationary states whose modulus squared does not evolve in time, which seems to contradict this interpretation of a fluid flowing with velocity v. (2) The fluid interpretation seems to perhaps suggest to me that the wavefunction reduces in the classical limit more to a wave than to a particle. (3) This doesn't show that the wavefunction is narrow.


3 Answers 3


The subtlety is that an arbitrary wavefunction doesn't reduce to a point of the classical phase space in the limit $\hbar \to 0$ (thinking about phase space makes more sense since in the classical limit one should have definite coordinates and momenta).

So one could ask, which wavefunctions do. And the answer is that the classical limit is built on the so-called coherent states -- the states that minimize the uncertainty relation (though I don't know any mathematical theorem proving that it's always true in the general case, but in all known examples it is indeed so). States close to the coherent ones can be thought of as some "quantum fuzz", corresponding to the quasiclassical corrections of higher orders in $\hbar$.

Example of this for the harmonic oscillator can be found in Landau Lifshits.

Regarding the fluid argument. About your remark (1): the $|\psi|^2$ for the stationary state is indeed stationary, but it still satisfies the continuity equation since the current is zero for such states. Your remarks (2) and (3) are quite right because, as I already said, the classical limit can't be sensibly taken for arbitrary states, it is built from coherent states.

And also I must admit that the given fluid argument indeed doesn't provide any classical-limit manifestation. It's just an illustration that "everything behaves reasonably well" to convince readers that everything is OK and to presumably drive their attention away from the hard and subtle point -- it often happens in $\it{physics}$ books, probably unintentionally :). The problem of a nice classical limit description is actually an open one (though often underestimated), leading to rather deep questions, like the systematic way to obtain the symplectic geometry from the classical limit. In my opinion it is also connected to the problem of quantum reduction (known also as the "wave function collapse").


Good discussion. As per the comment of U. Klein above the $\hbar \rightarrow 0$ limit of quantum theory is not what people might think. It is natural, and intuitive, as explained above, to assume that the classical limit is a property of a certain class of states.

As it happens, that view is incorrect. You can actually obtain an exact recovery of Hamiltonian Classical Point Mechanics for any value of $\hbar$ using a different wave-equation: $$i\hbar\frac{d}{dt}|\psi\rangle = [H(\langle q\rangle,\langle p \rangle)\hat{1} + H_q(\langle q\rangle,\langle p \rangle)(\hat{q}-\langle q\rangle) + H_p(\langle q\rangle,\langle p \rangle)(\hat{p}-\langle p\rangle)]|\psi\rangle$$ This wave-equation propagates any state along classical trajectories and is unique (i.e. you can derive it and show it is unique). The nonlinearity is essential and comes from the presence of the expectation values in the parameters multiplying each of the three operator terms.

When one becomes familiar with this result then it becomes clear why there is so much confusion.

People thought that the classical limit should be contained within quantum theory.

The simple truth is that it is not. The limit does not exist within the theory.

It involves a very specific equation which is outside linear quantum mechanics. However, it is simple, since this equation produces the expected Ehrenfest relations: $$\frac{d \langle q \rangle}{dt} =+H_p(\langle q\rangle,\langle p \rangle)$$ and $$\frac{d \langle p \rangle}{dt} =-H_q(\langle q\rangle,\langle p \rangle)$$ which is the classical limit of wave-packets following classical paths.

The relevant equation was first derived in K.R.W. Jones (1991), "The Classical Schroedinger Equation" UM-P-91/45 (CSE) http://arxiv.org/abs/1212.6786. A simpler version was published in 1992: K.R.W. Jones (1992), "Classical mechanics as an example of generalized quantum mechanics" Phys. Rev. D45, R2590-R2594. http://link.aps.org/doi/10.1103/PhysRevD.45.R2590. The original pre-print containing the derivation and proof of uniqueness was posted on arXiv a few days ago (see link above).

The fact that the equation is nonlinear may explain why the literature missed it for so long. It is simple, but it is not trivial, nor is it obvious until you know it.

The derivation of the CSE involves group theory and an unusual nonlinear representation of the Heisenberg--Weyl group. The mathematical system of nonlinear quantum theory involved is that discovered first by Weinberg and (independently) by Jones.

A general prescription of how to take a classical limit using a dimensionless parameter $\lambda \rightarrow 0$ is given in the paper: K.R.W. Jones (1993), "A general method for deforming quantum dynamics into classical dynamics while keeping $\hbar$ fixed" Phys. Rev. A48, 822-825.

There you will find the general argument of how to do a classical limit consistently so that the phase space, trajectories and all other properties are recovered.

It is surprising that the physics community has not gotten to grips with this yet. However, the question is an excellent one and subtle indeed since the mathematics involved did not exist prior to 1989. The connection between this area and the Feynman path integral is particularly interesting.

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    $\begingroup$ Dear Kingsley Jones: For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links. $\endgroup$
    – Qmechanic
    Commented Jan 5, 2013 at 20:34
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    $\begingroup$ Sorry, added the actual equation and explained it more fully in self-contained fashion. $\endgroup$ Commented Jan 5, 2013 at 20:52

The main result of the paper "What is the limit ℏ→0 of quantum theory?" is that the classical limit of quantum theory is not classical mechanics but a classical statistical theory. My "technical paper" has been written with the idea in mind to contribute to the understanding (the interpretation) of quantum theory. The final conclusion of the paper presents - in my opinion- a strong argument in favor of the statistical interpretation of quantum theory. Motivation and conclusions are discussed in more detail in sections I and VIII of the paper. I will be happy to answer specific questions but please note that almost all answers I am able to give may be found on my website "http://statintquant.net" (I am just updating this, will be finished soon).

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    $\begingroup$ I've merged you unregistered accounts again. $\endgroup$ Commented Nov 13, 2012 at 13:34
  • $\begingroup$ -1 for not explaining anything. And why are you posting a new answer instead of editing your previous post? $\endgroup$ Commented Jul 5, 2013 at 10:56

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