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Asymptoticity of Pertubative Expansion of QFT

+ 14 like - 0 dislike

It seems to be lore that the perturbative expansion of quantum field theories is generally asymptotic. I have seen two arguments.

i)There is the Dyson instability argument as in QED, that is showing the partition function is nonanalytic around the expansion point, by analyzing the ground state or instantons or somesuch. This is a wonderful argument but it requires some non-trivial knowledge about the behavior of your QFT which may not be available.

ii) There is some attempt at a generic argument which merely counts the number of Feynmann diagrams at each order, says this grows like $n!$ where is $n$ is the order of the expansion. and so our series looks like $\sum n!\lambda^n$, which is asymptotic. This is of course wholly unsatisfactory since it ignores interference among the terms (even granting the presumption that all the diagrams are of the same order, which feels right). It is true the series is still asymptotic if we take the diagrams to have random phase, but this ignores the possibility of a more sinister conspiracy among the diagrams. And we know that diagrams love to conspire against us.

So is there any more wholesome treatment of the properties of the perturbative expansion of QFT? I came to thinking about this while considering the properties of various $1/N$ expansions so anything known in particular about these would be nice.

This post has been migrated from (A51.SE)
asked Dec 12, 2011 in Theoretical Physics by BebopButUnsteady (325 points) [ no revision ]
It strongly depends on what you mean. For example, if your function is $(1+x)^{-2}$, its Taylor series has a finite convergence radius $x<1$, so no term-by-term summation works at large $x>>1$. But if you manage to guess this function right in the initial approximation (or sum up the Taylor terms into a finite formula $f=(1+x)^{-2}$), then the series asypmtoticity becomes irrelevant; nobody cares, the searched value can be calculated. Similarly for $exp(-C/g)$. If your initial approximation contains it right, there is no need to expand it in asymptotic series, and it may be still the same QFT.

This post has been migrated from (A51.SE)

1 Answer

+ 6 like - 0 dislike

You pretty much never expect a perturbation expansion of a generic theory to be convergent. There's a nice connection between the divergence of the perturbation expansion and nonperturbative effects (like instantons) leading to nonanalyticity at zero coupling (i.e., $e^{-C/g}$ effects). Mariño's notes here seem like a nice discussion with good references.

This post has been migrated from (A51.SE)
answered Dec 12, 2011 by Aaron (420 points) [ no revision ]
Thank you, that is an excellent set of notes.

This post has been migrated from (A51.SE)

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