Why is the WKB tunneling amplitude a non-perturbative result?

Question

The tunneling amplitude obtained from WKB aprroximation is given by $$|T(E)|=\exp\Big\{-\frac{1}{\hbar}\int\limits_{x_1}^{x_2}dx[2(V(x)-E)]^{1/2}\Big\}[1+O(\hbar)]$$ where $x_1$ and $x_2$ are the classical turning points at energy $E$.

Why is this referred to as a non-perturbative result? Why can't this phenomenon reveal itself, as it is often said, at any order in perturbation theory?

Addendum: Given a non-solvable potential, I can solve for the approximate energy eigenfunctions, and compute the tunneling amplitude as one does for a step potential. Will it not reveal the tunneling phenomenon?

Answer 1

Consider the function $\exp\Big(-\frac{1}{\hbar}f(x)\Big)$. Every derivative of it can be written in the form, \begin{equation} \frac{d^n}{d\hbar^n}\exp\Big(-\frac{1}{\hbar}f(x)\Big)=\frac{p_{2n}\Big[\hbar,f(x)\Big]}{\hbar^{2n}}\exp\Big(-\frac{1}{\hbar}f(x)\Big) \end{equation} where $p_{2n}$ is some polynomial in $\hbar$. In the limit $\hbar\rightarrow 0$ the exponent goes to zero faster than any inverse power increases. Therefore for any $n$, \begin{equation} \frac{d^n}{d\hbar^n}\exp\Big(-\frac{1}{\hbar}f(x)\Big)\Bigg|_{\hbar=0}=0 \end{equation} Therefore Taylor series in $\hbar$ is zero at every order and such contribution won't be seen in any order of perturbation theory in $\hbar$ i.e. is non-perturbative for such expansion.

Now we considered the expansion in $\hbar$. However note that $\hbar$ is either some dimensional quantity or set to $1$. What you actually have to consider is some dimensionless combination of the model parameters that usually involves what we call a coupling constant.

Answer 2

I agree with the answer OON. Of course for a potential barrier, we can make perturbative calculation of wave functions in and out, then take the particle current ratio we get tunneling amplitude to the perturbative order. But this result is obviously different from the one by WKB semiclassical tunneling one. As said by OON, the WKB one contains contributions from which don't show up in perturbative one. Actually semiclassical solution satisfies the equation of motion no matter how strong the coupling is. This is the foundation for instanton solutions.