I understand that in the limit that $\hbar$ goes to zero, the Feynman path integral is dominated by the classical path, and then using the stationary phase approximation we can derive an approximation for the propagator which is a function of the classical trajectory (see this document, pg 46).

I am under the impression that this further implies that the particle follows the classical trajectory but I don't understand how the above mentioned fact implies this.

The propagator describes the time-evolution of the wavefunction, so I would think that this classical limit form of the propagator should give a time-evolution in which the wavefunction follows the classical trajectory, but I have not been able to find such work. Moreover, even this statement itself is problematic since the wavefunction describes a probability distribution and not a single trajectory.

$\textbf{New Edit:}$ In section 7 of Feynman's paper introducing the path integral (see here) he discusses the classical limit. It appears that the key to understanding why the fact that the classical path dominates the path integral further implies that the particle follows the classical trajectory may be found in Feynman's remark on pg 21: "Now we ask, as $\hbar → 0$ what values of the intermediate coordinates $x_i$ contribute most strongly to the integral? These will be the values most likely to be found by experiment and therefore will determine, in the limit, the classical path." However, I don't understand why "These will be the values most likely to be found by experiment" ?

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    $\begingroup$ This seems more off-topic for this forum and should be posted under physics-stackexchange. Anyway, the reasoning is precisely the Least Action Principle, and in fact, if you google this, you will find your answer!! $\endgroup$ Jul 17, 2012 at 2:21
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    $\begingroup$ I don't understand why it is off-topic, care to explain? $\endgroup$ Jul 17, 2012 at 7:56
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    $\begingroup$ It's a physics question, and is answered in the standard literature. $\endgroup$ Jul 18, 2012 at 0:06
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    $\begingroup$ The standard literature (i.e. see Feynman and Hibbs Section 2.3) explains how in the classical limit, the classical path dominates the path integral. But this seems to stop short of actually showing that the particle travels via the classical trajectory (right?). If there is any literature that you know of that explains this further question, could you please post a reference or link? Thank you very much. $\endgroup$
    – dab
    Jul 18, 2012 at 17:16
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    $\begingroup$ I agree with Gerig that this would be much better suited in a physics rather than math site, but disagree that it being answered in the standard literature is a point against it. Those answers are almost always hand-wavy, and in any case this is a question that plenty of grad students will still have after taking their grad courses. $\endgroup$ Aug 5, 2012 at 18:43

2 Answers 2


Things stay in this way. Consider the action of a given particle that appears in the path integral. We consider the simplest case $$ L=\frac{\dot x^2}{2}-V(x) $$ and so, a functional Taylor expansion around the extremum $x_c(t)$ will give $$ S[x(t)]=S[x_c(t)]+\int dt_1dt_2\frac{1}{2}\left.\frac{\delta^2 S}{\delta x(t_1)\delta x(t_2)}\right|_{x(t)=x_c(t)}(x(t_1)-x_c(t_1))(x(t_2)-x_c(t_2))+\ldots $$ and we have applied the fact that one has $\left.\frac{\delta S}{\delta x(t)}\right|_{x(t)=x_c(t)}=0$. So, considering that you are left with a Gaussian integral that can be computed, your are left with a leading order term given by $$ G(t_b-t_a,x_a,x_b)\approx N(t_a-t_b,x_a,x_b)e^{\frac{i}{\hbar}S[x_c]}. $$ Incidentally, this is exactly what gives Thomas-Fermi approximation through Weyl calculus at leading order (see my preceding answer and refs. therein). Now, if you look at the Schroedinger equation for this solution, you will notice that this is what one expects from it just solving Hamilton-Jacobi equation for the classical particle. This can be shown quite easily. Consider for the sake of simplicity the one-dimensional case $$ -\frac{\hbar^2}{2}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial\psi}{\partial t} $$ and write the solution exactly in the form given above. Substitute it into the equation and impose that Hamilton-Jacobi equation holds $$ \frac{1}{2}|\nabla S|^2+V(x)=\frac{\partial S}{\partial t}. $$ You can see that both solutions agree neglecting higher order derivatives and a possible Heaviside function. This means that, at this order, the description using waves or classical paths is perfectly identical. This situation is not different from the case of geometric optics and a full wave equation description. You can describe your light waves as rays exactly as in quantum mechanics your particles become classical ones and you can describe them with paths.

  • $\begingroup$ Thank you very much. A few follow-up questions please: 1) Is there any literature/links which give more details about your answer? 2) How do we get the given G(tb−ta,xa,xb)? (Do we bring e^iℏS[xc] out of the integral and then evaluate the remaining integral? How so?) 2) Having G(tb−ta,xa,xb), do we then compute $\psi(x,t)$ from it? What would $\psi(x,t)$ be? 3) Do we get to this argument: physics.bu.edu/~rebbi/hamilton_jacobi.pdf This link is still confusing since doesn't only $abs(\psi)$ matter for the particle's location? $\endgroup$
    – dab
    Jul 18, 2012 at 17:08
  • $\begingroup$ 1) These are standard arguments in physics literature. A very good book is books.google.it/…. 2) In order to evaluate $G(t_b-t_a,x_a,x_b)$ you have to be able to solve classical equations of motion. Then, you are able to get also the Gaussian integral in a closed form. 3) Having $G$ you can get $\psi$. Check the book I gave and refs therein. 4) In geometric optics you can describe light as ray or wave. Check averages of operators. $\endgroup$
    – Jon
    Jul 18, 2012 at 17:20
  • $\begingroup$ @dab: I recommend looking up Ehrenfest's theorem first, which is basically just this general result restricted to a point particle. It should help you understand the power and limitations of this reasoning about the quantum-classical transition without the additional mathematical overhead. $\endgroup$ Aug 4, 2012 at 21:05
  • $\begingroup$ @Jon Is there chance you could please see my related question physics.stackexchange.com/questions/33767/… Thank you $\endgroup$
    – dab
    Aug 9, 2012 at 2:11

Physics note: for any non-zero value of $\hbar$, and for any chaotic potential $V(x)$, a wavefunction which is localized around a classical configuration $x_a$ (which could be a point particle's position, an entire field configuration, etc.) will still evolve to a state which is not localized around any configuration $x_b$ within a few multiples of the Lyapunov time. (The number of multiplies depending only logarithmically on the smallness of $\hbar$.) Since most real-life potentials are chaotic, you will find that the wavefunction of isolated macroscopic objects evolves to highly non-classical states in a relatively short time.

My advisor wrote a colorful paper on this a while back, pointing out that even a massively macroscopic variable like the orientation of Saturn's moon Hyperion would find itself in a grossly non-classical state on human time-scales: Why We Don't Need Quantum Planetary Dynamics: Decoherence and the Correspondence Principle for Chaotic Systems.

Whether or not this is a problem is disputed. For physicists who think more needs to be said to explain classicality, the most popular answer involves decoherence.


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