# Why does the classical path of a particle give the dominant contribution in the path integral?

Why is it that the classical path of a particle gives the dominant contribution in the quantum mechanical path integral? How do we understand this?

In the classical limit $$\hbar\to 0$$, this is just the WKB/stationary phase approximation.
1. Heuristically, near a stationary field configuration $$\phi_0$$ with $$\left. \frac{\delta S[\phi]}{\delta\phi}\right|_{\phi_0}~=~0\tag{1}$$ in field configuration space, the action $$S[\phi]~=~S[\phi_0]+{\cal O}\left((\phi-\phi_0)^2\right)\tag{2}$$ varies slowly, so the phase factors $$\exp\left(\frac{i}{\hbar}S[\phi]\right)$$ from neighboring field configurations sum up, and give a contribution; while away from a stationary field configuration $$\phi_0$$, the action varies rapidly, and the phases of neighboring field configurations are uncorrelated and cancel in average.
2. Perturbatively, near each stationary field configuration $$\phi_0$$, let us parametrize the field $$\phi^k~=~\phi^k_0+\sqrt{\hbar}\eta^k\tag{3}$$ in terms of a quantum fluctuation field $$\eta^k$$. Then the argument of the exponential reads $$\frac{i}{\hbar}S[\phi]~=~\frac{i}{\hbar}S[\phi_0] ~+~ \frac{i}{2}H_{k\ell}[\phi_0]~\eta^k\eta^{\ell} ~+~ {\cal O}(\sqrt{\hbar}), \qquad H_{k\ell}[\phi]~:=~ \frac{\delta^2 S[\phi]}{\delta\phi^k\delta\phi^{\ell}},\tag{4}$$ and the path/functional integral \begin{align}Z&~~~=~\int\!{\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left(\frac{i}{\hbar}S[\phi]\right) \cr &\stackrel{(3)+(4)}{=}~\sum_{\phi_0}\int\!{\cal D}\eta~\exp\left(\frac{i}{\hbar}S[\phi_0]+\frac{i}{2}H_{k\ell}[\phi_0]~\eta^k\eta^{\ell} + {\cal O}(\sqrt{\hbar})\right)\cr &\stackrel{\text{WKB}}{\sim}~\sum_{\phi_0}{\rm Det}\left(\frac{1}{i} H_{k\ell}[\phi_0]\right)^{-1/2}~\exp\left(\frac{i}{\hbar}S[\phi_0]\right) \quad\text{for}\quad\hbar~\to~0\end{align}\tag{5} becomes formally a sum over instantons $$\phi_0$$, i.e. classical field configurations.