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$^1$ $$\begin{align}\frac{i}{\hbar}S[\phi]~=~&\frac{i}{\hbar}S[\phi_0] ~+~ \frac{i}{2}H_{k\ell}[\phi_0]~\eta^k\eta^{\ell} \cr &~+~ {\cal O}(\sqrt{\hbar}),\end{align}\tag{4} $$ 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.


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

  • 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 '19 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.