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


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Aug 9, 2017 at 18:02
  • $\begingroup$ 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 $\exp\left(\frac{i}{2}(H^{-1})^{k\ell}[\phi_0]j_kj_{\ell} \right).$ $\endgroup$
    – Qmechanic
    Aug 9, 2017 at 18:11
  • 1
    $\begingroup$ 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). $\endgroup$
    – Qmechanic
    Nov 3, 2019 at 9:48

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


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