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

Why is it that the classical path 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$$^1$$ \begin{align}\frac{i}{\hbar}S[\phi]~=~&\frac{i}{\hbar}S[\phi_0] ~+~ \frac{i}{2}H_{k\ell}[\phi_0]~\eta^k\eta^{\ell} \cr &~+~ {\cal O}(\sqrt{\hbar}),\end{align}\tag{4} where $$H_{k\ell}[\phi]~:=~ \frac{\delta^2 S[\phi]}{\delta\phi^k\delta\phi^{\ell}}\tag{5}$$ is the Hessian. 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~\cr &\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)\cr &\quad\text{for}\quad\hbar~\to~0\end{align}\tag{6} becomes formally a sum over instantons $$\phi_0$$, i.e. classical field configurations.

3. For a simple introduction into this topic with many pictures and almost no formulas, see e.g. this blog post by The Physics Mill.

--

$$^1$$Here we are using DeWitt condensed notation.

• Notes for later: Action: $S[\phi] +J_k\phi^k$ $=\frac{\hbar}{2}\eta^kH_{k\ell}[\phi_0]\eta^{\ell} +S_{\neq 2}[\phi_0,\sqrt{\hbar}\eta]+J_k\phi^k$ $=\frac{\hbar}{2}\eta^kH_{k\ell}[\phi_0]\eta^{\ell} +S_{\neq 2}[\phi_0, \frac{\hbar}{i}\frac{\delta}{\delta J_k}] +J_k\phi^k,$ where EOM $\left.\frac{\delta S[\phi]}{\delta \phi^k}\right|_{\phi=\phi_0}=0$ and $\phi_0$ are defined WITHOUT sources. Hm. Not a stationary point for $J\neq 0$, so WKB does not apply. NB: It is a bit delicate what $J$-dependence should be differentiated. Aug 9, 2017 at 18:02
• Action with only fluctuation source: $\frac{i}{\hbar}S[\phi] +j_k\eta^k$ $=\frac{i}{\hbar}S[\phi_0+\sqrt{\hbar}\eta] +j_k\eta^k$ $=\frac{i}{\hbar}S[\phi_0] +\frac{i}{2}H_{k\ell}[\phi_0]\eta^k\eta^{\ell} +\frac{i}{\hbar}S_{\rm int}[\phi_0,\sqrt{\hbar}\eta] +j_k\eta^k$ $=\frac{i}{\hbar}S[\phi_0] +\frac{i}{2}H_{k\ell}[\phi_0]\eta^k\eta^{\ell} +\frac{i}{\hbar}S_{\rm int}[\phi_0,\sqrt{\hbar}\frac{\delta}{\delta j}] +j_k\eta^k.$ Source term is suppressed with $\hbar$ so it doesn't change the stationary point $\phi_0$. Propagator $\exp\left(\frac{i}{2}(H^{-1})^{k\ell}[\phi_0]j_kj_{\ell} \right).$ Aug 9, 2017 at 18:11
• It is possible to calculate quantum corrections perturbatively in $\hbar$. For a single variable $\eta$ in 0D, one can use the formula $\int_{\mathbb{R}} \!d\eta ~\eta^n e^{-\frac{a}{2}\eta^2}~=~\left(\frac{2}{a}\right)^{\frac{n+1}{2}}\Gamma(\frac{n+1}{2})~=~(n-1)!!\sqrt{\frac{2\pi}{a^{n+1}}}$ if $n$ even (and 0 if $n$ odd). Nov 3, 2019 at 9:48

The contribution of paths deviating from the classical path are suppressed by interference.