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This is in some sense a follow-up question to my previous question Why is it OK to keep the quadratic term in the small $\hbar$ approximation?. I understand how we can expand the action around a stationary point. This way we obtain a semiclassical expansion. I have read it in Coleman's Aspects of Symmetry that "If there are several stationary points, in general one has to sum over all of them".

I do not see why. I mean, we can always make the expansio around the stationary point of our choosing, isn't it? why do we have to sum all the expansions?

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Yossarian
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  • The path integral is taken over all paths. If you throw away the parts that are not stationary as not contributing, you are left with a sum over stationary paths. – By Symmetry Aug 08 '16 at 17:10

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  1. A path integral is in principle a weighted sum over all possible histories. Its precise mathematical definition is in general an open problem. A path integral is typically evaluated as an instanton sum $$Z~=~\sum_j Z_j ,\tag{1}$$ where each instanton $\phi_j$ has a fluctuation path integral $$Z_j~:=~\int\!{\cal D}\eta~ \exp\left\{\frac{i}{\hbar}S[\phi_j\!+\!\eta]\right\} \tag{2}$$ that is viewed as a perturbative expansion in $\hbar$.

  2. Now OP is essentially pondering if the instanton sum (1) amounts to overcounting the histories? The answer is No, because starting from one instanton sector, all other instantons are non-perturbative effects, which cannot be reproduced with the perturbative fluctuation path integral (2). See also this related Phys.SE post.

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Qmechanic
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