Classical limit in quantum mechanics proof

Question

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

• When does $\hbar \rightarrow 0$ provide a valid transition from quantum to classcial mechanics? When and why does it fail?

• Classical limit of quantum mechanics

• Classical Limit of Schrodinger Equation

• Semiclassical limit of Quantum Mechanics

• Why does the classical path give the dominant contribution in the path integral?

• How do you solve classical mechanics problems with quantum mechanics?

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.

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.

Answer 2

In addition to Brick's answer, I think (looking at the comments) it will be worthwhile to motivate the appearance of the phase of the wave and why it is precisely this quantity (or a scalar multiple thereof) that we would expect to satisfy the Hamilton-Jacobi equation (HJE) in the classical limit.

It is helpful to compare the situation to the theory of electromagnetic waves. Here, light is fundamentally an electromagnetic wave and light propagation is a priori described within the theory of wave propagation. However, in certain regimes (the 'geometric optics limit') it is possible to describe the propagation using the notion of light rays that have particle-like trajectories. These rays correspond to the movement of individual 'points' on the wavefront of the wave, moving at all times perpendicular to the wavefront. Roughly speaking, such an approximation is valid whenever interference effects can be neglected.

The latter is of course precisely what we aim to do in the classical limit of QM. The wavefunction represents a wave, and in the classical limit we would like to identify the motion of an individual point on the wavefront with the motion of a classical particle, neglecting interference effectes. The classical limit of QM in this sense can thus be thought of as a kind of geometric optics limit. In fact, it was through reverse engineering this analogy that Schrödinger came up with his equation.

So the question is: given a wavefunction $\psi(\vec x,t)$, how do we obtain the motion of an individual point on its wavefront, i.e. the trajectory of a 'ray' of the wave? In analogy with the standard geometric optics limit, we assume the wavefunction to be of the general form

$$\psi(\vec x,t) = A(\vec x,t)e^{i B(\vec x,t)},$$

where $A(\vec x,t), B(\vec x,t)$ are both real and $A(\vec x,t)$ is slowly varying so that the latter may be treated effectively as a constant. This ansatz essentially says that we have a wavepacket centered around a single frequency. Since $A\approx $ constant, the wavefronts (surfaces of constant amplitude) are given by $B(\vec x,t) = $ constant. This is where the phase appears.

Now, in the HJ formulation of classical mechanics it is a well-known fact that the trajectories of a classical particle with Hamiltonian $H$ can be thought of as the rays (as in the EM analogy) of some corresponding time-evolving surface or 'wavefront', namely the surface $S(\vec x,t)=$ constant, where the function $S$ satisfies the HJE with the corresponding Hamiltonian. This suggests a correspondence $B\leftrightarrow S$, up to a possible rescaling (note that any scalar multiple of $B$ has the same level sets $B=$ constant).

Indeed, all of this means that in order to show that the Schrödinger equation with Hamiltonian $H$ reduces in the classical limit to Hamilton's equations with Hamiltonian $H$ (for the trajectory of the 'ray', interpretted as particle), it suffices to show that in the limit, $B$ or rather some scalar multiple of $B$ satisfies the HJE with the same Hamiltonian.

And that is what is nicely done in Brick's answer: it turns out that $S=\hbar B$ does the trick, justifying the conventional notation $\psi \sim e^{i S/\hbar}$.