Let me explain in details. Let $\Psi=\Psi(x,t)$ be the wave function of a particle moving in a unidimensional space. Is there a way of writing $\Psi(x,t)$ so that $|\Psi(x,t)|^2$ represents the probability density of finding a particle in classical mechanics (using a Dirac delta function, perhaps)?

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    $\begingroup$ Is there anyone who can confirm if this is possible? Each answer says something different. $\endgroup$ Commented Jul 25, 2022 at 10:07
  • $\begingroup$ @ Balázs Börcsök Yes, by and large, something of the sort is possible, but in the language of phase-space QM. The Wigner function "morphs" into the Liouville density (for One particle!). The problem is one of language: if you use languages comparing apples with oranges, you are bound to get caught up in the miscommunication. If you compare both theories, either in phase space, or in Hilbert space, both available, the comparisons/bridges work out nicely... $\endgroup$ Commented Jul 2, 2023 at 22:10

7 Answers 7


Sure you can! This is actually a simple but very interesting result, and it is usually shown in quantum mechanics courses. It's called the Ehrenfest theorem, and I won't prove it here but I'll copy the result from Sakurai Modern Quantum Mechanics (1991). You can check the mathematical details there, or in many other books.

If you have a hamiltonian with the form $$H = \frac{p^2}{2\,m}+V(x)$$ you can prove that, in the Heisenberg picture, $$m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\nabla V(x) .$$ If you now take the expectation value of that equation (for certain state kets), you get $$m \frac{\mathrm{d}^2\langle x \rangle}{\mathrm{d}t^2} = \frac{\mathrm{d}\langle p \rangle}{\mathrm{d}t} = -\langle \nabla V(x) \rangle .$$ This result is valid in both Heisenberg and Schrödinger's picture.

If you want to recover the classical limit, you need to say that the area where the wavefunction is significantly nonzero is much smaller than the scale of variations of the potential. In that case, you can identify the center of the wavefunction with the position of the particle, and $\langle \nabla V(x) \rangle $ turns into $\nabla V(\langle x \rangle) $.

What this means, conceptually, is that the center of the wavefunction will move according to the classical laws if you can't "see" that your object/particle it's not a material point, and if your potential is also classical, in that it doesn't have variations that are comparable to the "size" of the wavefunction.


The short answer: No, does not exist any wavefunction in Hilbert space which reproduces classical mechanics.

The classical limit of quantum mechanics is studied with some deep in Ballentine textbook. For instance, section 14.1 is devoted to the Ehrenfest theorem and it is shown that the theorem is neither necessary nor sufficient to define the classical regime.

The paper What is the limit $\hbar \rightarrow 0$ of quantum theory? (Accepted for publication in the American Journal of Physics) shows that Schrödinger's equation for a single particle moving in an external potential does not lead to Newton's equation of motion for the particle in the general case. Page 9 of this more recent article precisely deals with the question of why no wavefunction in the Hilbert space can give a classical delta function probability.

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    $\begingroup$ Dear @juanrga, 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. Also it is frown upon to post nearly identical answers to similar posts. $\endgroup$
    – Qmechanic
    Commented Oct 29, 2012 at 21:40


Sorry, but I think there is an error in your answer:

The spatial extent of the particle wave-function, must be much smaller (and not longer) than the variation length-scale of the potential, to transform $\langle \nabla V(x)\rangle$ turns into $\nabla V\left(\langle x\rangle \right).$

Only in this case, it is possible to make a Taylor series of $V(X)),$ because $V(X)$ is slowly varying in the domain where the wave function is not null, and you can take the mean expectation :

$$\nabla{\mathrm i}\, V(X) = \nabla{\mathrm i}\, V(\langle X\rangle ) + (X_j - \langle X_j\rangle ) \nabla j \,\nabla{\mathrm i} \,V(\langle x\rangle )\, +\,\textrm{negligible higher order terms in}\,\, (X_j - \langle X_j\rangle)$$

So, $\langle \nabla{\mathrm i}\, V(X)\rangle = \nabla{\mathrm i} \,V(\langle X\rangle),$ because $\langle X_j - \langle X_j\rangle \,\rangle = 0\;.$


You can recover Schroedingers equation from the path integral formulation of Quantum mechanics by Feynman. In the path integral picture the classical trajectories are the stationary points of the integrand. So in the stationary phase approximation, they are the contribution of $0$-th order in $\hbar$. Of course that is not a direct relation between the Schroedinger equation and classical trajectories.


There is a way in which the Schrödinger equation on the limit $\hbar \to 0$ reduces to the Hamilton-Jacobi equation. In that limit, it turns out that it is not so much the wave function itself that satisfies the Hamilton-Jacobi equation but its phase. If a collision state of the form is considered:

$$\psi(\boldsymbol{x},t) \approx e^{iS(\boldsymbol{x},t)/\hbar}$$

Where $S(\boldsymbol{x},t)$ is the action of the particle, substituting this equation into the Schrödinger equation:

$$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\left( \frac{\partial^2 \psi}{\partial x^2}+\dots+\frac{\partial^2 \psi}{\partial z^2}\right) + V(\boldsymbol{x})\psi$$

results in the following equation for the action:

$$-\frac{\partial S}{\partial t} e^{iS/\hbar} = \frac{\hbar^2}{2m}\left[ \left( \frac{\partial S}{\partial x}\right)^2 +\dots + \left( \frac{\partial S}{\partial z}\right)^2\right]e^{iS/\hbar} + V(\boldsymbol{x})e^{iS/\hbar} - \frac{i\hbar}{2m}(\Delta S) e^{iS/\hbar}$$

By canceling the exponential factor and taking the real part of this equation, we have:

$$\frac{\partial S}{\partial t} + \frac{\hbar^2}{2m}\left[ \left( \frac{\partial S}{\partial x}\right)^2 +\dots + \left( \frac{\partial S}{\partial z}\right)^2\right] + V(\boldsymbol{x}) = 0$$

Which is the classical Hamilton-Jacobi equation for a particle in a potential. The imaginary part null, in the limit $\hbar \to 0$:

$$\frac{i\hbar}{2m} \Delta S = 0$$

If we start from the totally general expression:

$$\psi(\boldsymbol{x},t) = A(\boldsymbol{x},t)e^{iS(\boldsymbol{x},t)/\hbar}$$

in the limit $\hbar \to 0$, the additional terms cancel out and we simply arrive also at the Hamilton-Jacobi equation for the complex phase of the wave function.


More intuitive picture is in Arnoques answer, alternative and a bit more formal approach is to note that all QM equations of motion have their classical mechanics equivalent if you formulate them using commutators and then replace commutator with Poisson bracket ($\partial A/\partial t = [H,A]$ $\Rightarrow$ $ \partial a/\partial t = \{ H,a \}_{q,p} $, if you "hide" Planck constant). The commutator itself is of course zero in classical case, when operators reduce to numbers. Accordingly, all general system properties easily map from QM to CM.

And it may be shown (too much to copy, sorry) that a formal limit $\hbar\to0$ leads to exact equivalence between commutator and Poisson bracket.

Concerning the wavefunction, classical motion is definite. Instead of probability you have definite correspondence between $t$ and $x$. Indeed, you may formulate it saying that classical $|\Psi(x,t)|^2=\delta(x-x_c(t))$ where $x_c(t)$ is classical tragectory. To write $\Psi$ itself, you have to treat $\hbar\to0$ accurately to avoid divergent integrals. Normally, there is no reason to do this. And technically, there is no guarantee that a limit $\hbar\to0$ of some particular quantum state is a "normal" solution of classical problem.

  • $\begingroup$ Maybe I misunderstand what you mean, but the Poisson bracket between position and (canonical) momentum is not zero but actually $\{q_i,p_j\} = \delta_{ij}$. And I don't think that replacing operator commutators with Poisson brackets is the same as taking the physical classical limit, but rather an ad hoc procedure exploiting some formal structure. Luboš Motl had a nice exposition recently motls.blogspot.com/2011/11/… . $\endgroup$
    – Heidar
    Commented Nov 30, 2011 at 7:39
  • $\begingroup$ @Heidar Thank you for pointing out my mistake. I found the textbook, refreshed my memory and changed my answer accordingly. $\endgroup$
    – Misha
    Commented Nov 30, 2011 at 9:14
  • $\begingroup$ @Misha The delta function as the squared absolute value of the wavefunction is where I was trying to get at. With that formulation, what is the wavefunction $\Psi(x,t)$? $\endgroup$ Commented Dec 1, 2011 at 2:18
  • $\begingroup$ You could better ask this at math.stackexchange.com . There is no regular way. That thing is not a function nor even generalized function. When you find yourself in a trouble with delta function I recommend to go back to its most naive definition as a limit of funcional sequence and try to perform operation you want on that sequence. If the limit of the result is not self-consistant, this operation is wrong or too complicated. $\endgroup$
    – Misha
    Commented Dec 1, 2011 at 5:18

This is exactly what Feynman's path-integral does, many-body quantum effects reveals classical limit at high temperature! Instead of solving the Schrödinger equation, it solves Newton's equation by splitting the atom into beads and perturbing the system effectively to find out the thermal equilibrium state of the system. The method showed the isomorphism of quantum theory and classical statistical mechanics, which leads to an interesting point: the imaginary time is isomorphic to the inverse of temperature (see Wick-rotation). You may want to read Feynman's thesis. BTW, I personally started the topic with reading this paper: Feynman's derivation of the Schrödinger equation


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