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" ?
Best Answer
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.