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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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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.

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upvoted, but there's some ugly typos in the math that you should correct... – wsc Nov 30 '11 at 5:44
@wsc Thanks for the heads-up! – Arnoques Nov 30 '11 at 10:30
Sakurai treatment is far from rigorous or complete. As shown in Ballentine textbook (see my answer) there exist systems that satisfy Ehrenfest theorem exactly but are not classical and there exist classical systems that do not satisfy the theorem. – juanrga Oct 29 '12 at 20:37

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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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. – Qmechanic Oct 29 '12 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 $\langle X_j - \langle X_j\rangle \,\rangle = 0\;.$

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Good catch! Thanks for the correction. That's what I meant, but I wrote the wrong word. The next paragraph was correct, luckily. It's great that you expanded my explanation with the maths for it. Upvoted! – Arnoques Dec 6 '11 at 21:27

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.

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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.

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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… . – Heidar Nov 30 '11 at 7:39
@Heidar Thank you for pointing out my mistake. I found the textbook, refreshed my memory and changed my answer accordingly. – Misha Nov 30 '11 at 9:14
@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 $\Psi(x,t)$? – Physicist Student Dec 1 '11 at 2:18
You could better ask this at . 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. – Misha Dec 1 '11 at 5:18

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