Several questions are about the limit $\hbar\rightarrow 0$, e.g.

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.

  • $\begingroup$ Consider to include what precise observable/statement is considered in the limit $\hbar\to 0$. The conclusion depends on it. $\endgroup$
    – Qmechanic
    Commented Jul 25, 2021 at 5:25
  • $\begingroup$ @Qmechanic I cannot answer that question because I do not know which observable is provable (see edit). I gave a few examples of observables, is it alrigt if i get an answer for any one of them? thanks. $\endgroup$
    – Erik Wang
    Commented Jul 25, 2021 at 5:50
  • 1
    $\begingroup$ Related meta discussion physics.meta.stackexchange.com/q/13713/2451. $\endgroup$
    – Qmechanic
    Commented Jul 26, 2021 at 4:45
  • $\begingroup$ I voted up your answer for your effort. $\endgroup$
    – Sebastiano
    Commented Jul 28, 2021 at 11:38
  • $\begingroup$ The observable statement: In the classical limit it is possible to measure the position without disturbing the momentum, and vice versa. That allows to find classical-like sharp values for the both position and momentum distributions (but the wave functions still remain complex valued). $\endgroup$
    – Hulkster
    Commented Jan 25, 2022 at 13:03

1 Answer 1


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.

  • $\begingroup$ Why is the phase of $\psi$ solving the HJ equation be meaningful? How do you relate the phase of $\psi$ with any classical observable? If I remember correctly the $S$ in the HJ equation is the classical action with a free endpoint $\endgroup$
    – FrodCube
    Commented Jul 26, 2021 at 15:18
  • $\begingroup$ The question - at least the part that I identified that I was answering - asks how to recognize the classical limit. This is it. The Schrodinger equation has a real form, as given, and under the conditions mentioned, the real form makes clear that the dynamics of the particle are given exactly by the classical equations. What more do you want? @FrodCube As noted in the answer, $S$ is Hamilton's principle function, which has known relations to a couple of variants of "action" within classical physics. $\endgroup$
    – Brick
    Commented Jul 26, 2021 at 15:27
  • $\begingroup$ If you drop the $Q$ term, then the right side of the HJ equation is the (classical) total energy, which, of course, is a classical observable. @FrodCube. Once you're classical, you can also compute momentum from that (since we also have $V$), and with initial conditions you can compute position as a function of time. That's all standard classical physics at that point, so I don't understand what you meant by relate "with any classical observable" in context of the comment. $\endgroup$
    – Brick
    Commented Jul 26, 2021 at 15:50
  • $\begingroup$ maybe my question is stupid. I know your derivation, I've seen it in several books. I just don't have an intuition (maybe I knew it at some point) for why the phase of a wavefunction is related to the action. $\endgroup$
    – FrodCube
    Commented Jul 26, 2021 at 17:32
  • $\begingroup$ @FrodCube I guess I prefer to think about it the other way around. The action is fundamental, and putting it in the exponential makes the operator definition $p \rightarrow \hat{p} = -i\hbar \partial_x$ work out. The complex version of the equation is more popular for both historical and practical reasons, but (as a matter of opinion) I think the real version is more closely tied to intuitive fundamentals. But anyway, we're probably wandering into a new question at this point.... $\endgroup$
    – Brick
    Commented Jul 26, 2021 at 17:39

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