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 http://www.blau.itp.unibe.ch/lecturesPI.pdf 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 http://imotiro.org/repositorio/howto/artigoshistoricosordemcronologica/1948c%20-FEYNMAN%201948C%20Invention%20of%20the%20path%20integral%20formalism%20for%20quantum%20mechanics.pdf) 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 $\hbar \to0$ 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" ?

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/32110/2451 $\endgroup$ – Qmechanic Jul 17 '12 at 21:18
  • $\begingroup$ Feynman simply means that the coordinates $x_i$ that contribute most strongly to the integral will be the ones with highest probability should one inquire about the particle's position at some intermediate time. $\endgroup$ – Emilio Pisanty Aug 11 '12 at 23:29
  • $\begingroup$ Cross-posted from mathoverflow.net/q/102415 $\endgroup$ – Qmechanic Oct 26 '12 at 21:26

The semiclassical limit you're describing says that the amplitude for a particle to get from here to there in a set time is equal to the exponential of the classical action for the corresponding classical trajectory. In symbols this reads $$\langle x_b|U(T)|x_a\rangle=\int \mathcal{D}\varphi e^{iS[\phi]/\hbar} \approx e^{{i}S[\varphi_\textrm{cl}(x_a,x_b,T)]/\hbar}.$$ In a general quantum state, however, particles are not "here" and don't end up "there": they have an initial probability amplitude $\langle x|\psi(0)\rangle$ for being at each position $x$ at time $t=0$ and will have a final probability amplitude $\langle x|\psi(T)\rangle$ for being at position $x$ at time $T$. To apply the approximation, you pull out the propagator and insert a resolution of the identity: $$\langle x|\psi(T)\rangle=\int dy\langle x|U(T)|y\rangle\langle y|\psi(0)\rangle=\int dye^{{i}S[\varphi_\textrm{cl}(y,x,T)]/\hbar}\langle y|\psi(0)\rangle.$$

To get a full semiclassical limit, you also need a semiclassical initial state (since otherwise you've obviously got no hope!). You take, then, a state with (relatively) sharply defined position and momentum (of course, the state will occupy some finite region of phase space but you usually can assume, in these circumstances, that it is small enough), and this will make the amplitudes for points outside the classical trajectory interfere destructively and vanish.


So how does this happen? For one, $y$ must be close to the initial position, $y_0$ in order to contribute to the integral. For small displacements of the endpoints, then, the action along the classical trajectory varies as $$\delta S=p_{\varphi,x}\delta x-p_{\varphi,y}\delta y$$ (cf. Lanczos, The Variational Principles of Mechanics, 4th edition, Dover, eqs 53.3 and 68.1, or simply do the standard integration by parts and set $\int\delta L dx=0$ along the classical trajectory). The main contribution of the initial state to the phase is of the form $e^{ip_\textrm{cl}y}$, which means that the integral has more or less the form, up to a phase, $$\langle x|\psi(T)\rangle\approx \int_{y_0-\Delta x/2}^{y_0+\Delta x/2}e^{i(p_{\varphi,y}-p_\textrm{cl})y/\hbar}dy.$$

Here the momentum $p_{\varphi,y}$ is determined by $x$ and (to leading order) $y_0$, since there is a unique classical path that connects them. This momentum must match (to precision $\Delta p\approx \hbar/\Delta x$, which we assume negligible in this semiclassical limit) the classical momentum of the initial state, $p_\textrm{cl}$, and therefore only those $x$'s on the trajectory determined by the initial state will have nonzero amplitudes.

  • $\begingroup$ Thank you for your answer. I don't understand why "this will make the amplitudes for points outside the classical trajectory interfere destructively and vanish." Could you please explain this. Also, please see my new edit to the question. Thank you. $\endgroup$ – dab Aug 10 '12 at 23:18
  • $\begingroup$ Thank you very much for your reply. There are a few things which I could use some more details on to better understand your answer. Rather than getting into detailed questions, I was wondering if you know of a textbook or paper which goes through your argument. Thank you. $\endgroup$ – dab Aug 12 '12 at 22:59
  • 1
    $\begingroup$ I don't really know any references with this material but I would very much like to see one. I'm also rather grateful to you for forcing me to work this out clearly. $\endgroup$ – Emilio Pisanty Aug 14 '12 at 2:47

It's NOT true in general that the semiclassical limit is always dominated by stationary solutions of the original bare Lagrangian. Instead, the best quasiclassical limit looks more like the coarse grained consistent histories framework aka decoherent histories along the dynamically determined pointer basis.


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