# Is there any proof that any result from perturbation theory is necessary an asymptotic series?

I know that almost all the series coming from perturbation theory are divergent, such as those from eigenvalue problems or the S-matrix in quantum field theory. The lore is that the series are asymptotic series. However, how to prove that the series are asymptotic to the original solutions? One could probably use some resummation technique to construct a function that has the desired asymptotic series (by Borel-Ritt lemma). However, under what conditions is the function constructed by a resummation method, such as Borel summation, the same as the original solution? Thanks!

• Hard to tell, because in general there is no "original solution" -- there is no satisfactory/rigorous non-perturbative definition of a QFT. One assumes that the perturbative series is asymptotic to some non-perturbative object, but there is no real reason to assume such an object actually exists. One hopes there is, but nothing more. – AccidentalFourierTransform Aug 13 '18 at 1:27
• How about in ordinary undergraduate quantum mechanics? Is there any proof or counterexample in this settings? – Herman Chu Aug 13 '18 at 14:46

The argument I am aware of is quite heuristic but basically works as follows. At order $n$ in the perturbative expansion, there are roughly $n!$ number of Feynman graphs. Assuming they all contribute roughly equally to the amplitude, the coefficient of the $O(\lambda^n)$ term is $O(n!)$. This means that the perturbative sum has the form $$A = \sum_{n=0}^\infty \lambda^n n! c_n , \qquad c_n=O(1).$$ Such a sum is generally asymptotic.
Of course this is a very heuristic argument and can easily be circumvented by setting up special (though not rare) circumstances. For instance, in supersymmetric theories, there are still $n!$ factorial graphs at each order but fermionic diagrams often cancel bosonic ones so that $c_n$ is no longer $O(1)$ and the series can be summed.
• Asymptotic series often result from changing the orders of sums and integrals, eg the path integral and a sum for expansion of $\exp(iS)$. Can we say anything concrete about this? – innisfree Aug 13 '18 at 5:49
• Is it possible to show that $A(\lambda)$ satisfies the definition of asymptotic series? That means $A(\lambda)-\sum_{n=0}^{n=N-1}\lambda^n n! c_n=O(\lambda^N)$ for all $N$ when $\lambda$. In other words, the asymptotic series is converging for any fixed $N$ in the limit $\lambda \rightarrow 0$. – Herman Chu Aug 15 '18 at 2:02