New answers tagged non-perturbative
1) $\exp(-1/g)$ is not necessarily related to bound states. In the standard QM double well problem it is the splitting, not the binding energy, that is $O(\exp(-1/g))$. In conformal field theories instantons can give $\exp(-1/g)$ effects even though there are no bound states at all. 2) Instantons are one source of $\exp(-1/g)$ effects, but there are others. ...
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