# Classical mechanics from Quantum mechanics

I'm looking at a way to prove that one recovers, under ad hoc assumptions, classical mechanics from quantum theory. Usually, we can find in textbooks that the propagator

$K(x,x_0;t)=\langle x|e^{-i \hat H t/\hbar}|x_0\rangle$

is given in the classical limit by using the path integral formulation

$K(x,x_0;t)\propto e^{i S[q_c(t)]}$

where $S[q]$ is the classical action and $q_c(t)$ is the classical path from $x_0$ to $x$ in a time t (which minimizes $S$).

But this is not very satisfying, as $K$ is a probability amplitude, which is really not what we would like in the classical limit. The least we would like to obtain is that the probability to be in $x$ at time $t$ is given by

$P_t(x)=\delta(x-q_c(t))$ $\,\,\,\,\,\,\,\,$ (1)

or something equivalent. My naive approach to recover this result would be the following. The propagator is not enough, as the "classicalness" of the motion should be given by the initial state. It seems natural to choose the initial wave-function to be a wave-packet peaked around the initial position (at x=0 from now on) with spread $\Delta x$ and with momentum $p$, for example

$\psi(x)=\dfrac{e^{-\frac{x^2}{2\Delta x^2}-i p x/\hbar}}{\pi^{1/4}\sqrt{\Delta x}}$.

We would say that the dynamics will be classical at least if $\Delta p\propto \hbar/\Delta x \ll p$. There is maybe other constraints, for instance the the dynamics is classical only at times $t$ long enough, but this is still not clear to me.

Finally, the classical probability is given by (or at least proportional to)

$P_t(x)=|\int d x_0 K(x,x_0;t) \psi(x_0)|^2\propto |\int d x_0 e^{i S[q_c(t)]} \psi(x_0)|^2$. $\,\,\,\,\,\,\,\,$ (2)

My question is : is there a way to show/prove that under these assumptions, we can get (1) from (2) for any hamiltonian ? I have looked at the simplest case of a free particle, and it seems to work (I still have some issues to get the final result, but my feeling is that it works). If it helps, I could post the calculation later. But a general proof would be great.

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One thing common in all proofs, is that one needs to look at the expectation values and needs to set the limit of h to 0. Check out section 31 'The motion of wave packets' in Dirac's Principles of Quantum Mechanics for the most general proof. While he doesn't use path integrals, his equations can easily be re-written in terms of path integrals. –  dj_mummy Oct 15 '13 at 9:31
You might be interested to read this motls.blogspot.com/2011/11/… Where Lubos shows how from an ensemble of photons one gets the classical electromagnetic wave . –  anna v Oct 15 '13 at 10:35
@Qmechanic : I had read these posts, but I don't think they answer exactly my question, which can be reformulated in a more general way (without reference to path integrals) : Is it possible to show that $|\psi(x,t)|$ gives a, under some particular initial conditions, a probability of the form of equation (1). But maybe I can use these posts to answer my question. –  Adam Oct 15 '13 at 20:16
@dj_mummy: Thanks for the reference, it is useful, even though it does not directly address my concerns (I don't want to work with the expectation values, but with the wave function and Born's rule). –  Adam Oct 15 '13 at 20:17

## 1 Answer

The question of how quantum mechanics reduces to classical mechanics in the limit $\hbar\to 0$ has been asked several times before, see. e.g. this, this and this Phys.SE posts, and links therein.

I) Let $\Delta t:=t_f-t_i$ and $\Delta x:=x_f-x_i$. The key fact is now that the Feynman propagator/kernel/amplitude $K(x_f,t_f;x_i,t_i)$ localizes to a delta function

$$\tag{A} K(x_f,t_f;x_i,t_i) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^{+}.$$

The delta distribution (A) has support at $$\tag{B} x_f~=~x_i.$$

Heuristically, eq. (A) follows because for sufficiently short times $|\Delta t| \ll \tau$, where $\tau$ is some characteristic time scale, the potential $V$ has (so to speak) no time to interact, and the kinetic term blows up, so that the Gaussian formula for a free particle can be used.

II) If there is no potential $V$ in the model, i.e. the model is a free particle, it is enough to assume $\hbar|\Delta t| \ll I$, to derive the limit (A), where $I$ is some characteristic scale (with dimension equal to a moment of inertia), so that in this case one may consider the limit $\hbar\to 0$ without going to short times.

III) More generally one may apply the WKB semiclassical methods. In the case of a harmonic oscillator, the amplitude becomes

$$\tag{C} K(x_f,t_f;x_i,t_i) ~\longrightarrow~\delta\left(\sqrt{(x_f^2+x_i^2) c-2x_fx_i }\right)~=~\delta\left(\sqrt{\left(x_f - \frac{1+s}{c}x_i\right)\left(x_f - \frac{1-s}{c}x_i\right)c }\right) \quad \text{for} \quad \hbar \to 0^{+},$$

where $c:=\cos \omega \Delta t$ and $s:=\sin \omega \Delta t$. The delta distribution (C) has support at $$\tag{D} x_f=\frac{1 \pm s}{c}x_i.$$

IV) The condition (B) for the free particle and (D) for the harmonic oscillator only make proper classical sense in the short time limit $|\Delta t| \ll \tau$. For large times $|\Delta t|$, the classical limit (B) and (D) are a remnant of quantum averaging procedure over many histories, which are not quite classical in the limit $\hbar\to 0$. A heuristic interpretation is possibly sort of a classical mixed state.

V) The probability

$$P(x_f,t_f;x_i,t_i)~=~|K(x_f,t_f;x_i,t_i)|^2$$

in the limit (A) of a square of the Dirac delta distribution is mathematically ill-defined, cf. e.g. this and this Phys.SE posts. Concerning normalization of the Feynman propagator and interpretation as probability, see also this Phys.SE post.

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Thanks for you answer, but it does not really address my question. I know how to recover the initial wave-function as $\Delta t\to 0$. I want to show that for any $\Delta t$, the probability of being at $x=x_f$ is given by equation (1) of the my question if the initial state is "classical enough". –  Adam Oct 15 '13 at 15:05
I updated the answer. –  Qmechanic Oct 15 '13 at 23:26