• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

143 submissions , 120 unreviewed
3,899 questions , 1,377 unanswered
4,837 answers , 20,500 comments
1,470 users with positive rep
495 active unimported users
More ...

  Why is the Borel summation relevant for asymptotic series of physical observables?

+ 6 like - 0 dislike

A number of questions on this topic already exist. The most detailed one is probably this (http://www.physicsoverflow.org//26751/physics-of-asymptotic-series-and-resummation?show=26751#q26751). However, I still find some things confusing.

Assume that we have a (Gevrey-1) divergent series $\sum_n a_n g^n$ intended to represent a physical quantity. Naturally, we think that the full exact answer must be some function $f(g)$ with this series as an asymptotic expansion $f(g)\sim \sum_na_ng^n$. However, such a function is not unique. For instance $\tilde{f}(g)=f(g)+e^{-1/g}$ has the same asymptotic expansion at $g=0$.

Now, the Borel summation is a method to obtain a particular example of such function $\mathcal{B}f(g)\sim \sum_na_ng^n$. There are cases when the Borel resummation is ambiguous (due to singularities of the Borel transform on the real axis). But let us assume that in the case of interest there is no such problems and the Borel summation gives unique non-ambiguous result $\mathcal{B}f(g)$. In all the physical literature that I've been through it was implicitly assumed that when we are in such in a situation, i.e. when the Borel summation exists and is unique, it is the exact quantity that we are looking for. 

My question is: why is it so? Why are we supposed to prefer $\mathcal{B}f(g)$ over, say, $\mathcal{B}f(g)+e^{-1/g}$ when computing 'physical quantities'?

Maybe it is worth having some particular example. Consider integral

$$I(g)=\int_{0}^{\infty}dx\, e^{-x-gx^2}$$

which has the following asymptotic expansion

$$I(g)\sim \sum_{n=0}^\infty g^n \int_0^\infty dx\, e^{-x}\frac{(-1)^nx^{2n}}{n!}=\sum_{n=0}^\infty g^n \frac{(-1)^n(2n)!}{n!}$$

The series is sign-alternating and hence the Borel transform has no singularities on $\mathbb{R}_+$. In this case, the Borel sum $\mathcal{B}I(g)$ is indeed equal to the original integral. Written explicitly, this identity is

$$\int_0^\infty dx\, e^{-x} \sum_{n=0}^\infty g^n \frac{(-1)^nx^{2n}}{n!}=\int_0^\infty dx\, e^{-x} \sum_{n=0}^\infty g^n \frac{(-1)^nx^{n}(2n)!}{(n!)^2}$$

Looking at this particular example I have no temptation to add the non-perturbative term $e^{-1/g}$ as it clearly does not belong here. However, my impression is that this situation is general, relevant for the most 'physical quantities'. The only general argument that I can think of is that any 'physical quantity' can be represented as a some (path) integral for which our example is a toy model. Then the reasoning could be somehow generalized I suppose. However this argumentation seems too vague for me.

asked Oct 20, 2015 in Theoretical Physics by Weather Report (210 points) [ revision history ]

2 Answers

+ 4 like - 0 dislike

Borel summation is appropriate always when one can prove the analyticity assumption that the Borel integrals satisfy for complex $g$ (Watson's Theorem).

This is the case for asymptotic expansions of many integrals, and also in many applications in quantum mechanics and quantum field theory. (For 2D QFT, see, e.g., the second edition of the QFT book by Glimm and Jaffe.)

In cases where one cannot prove anything one may simply try Borel summation and hope for the best.

But Borel summation is not always appropriate; for example, instantons produce nonperturbative $e^{-O(1/g)}$ effects that do not show up at any order of perturbation theory. In 4D quantum field theory, there may be obstructions to Borel summability, generally discussed under the name renormalons. (Note: Suslov's papers referenced in the Wikipedia article linked to here are not uncontroversial.) Renormalons are also discussed in Vol. 2 of Weinberg's QFT book.

answered Oct 20, 2015 by Arnold Neumaier (12,335 points) [ revision history ]
edited Oct 20, 2015 by Arnold Neumaier

Thank you, that clarifies a lot! But I am a little confused about your comment on instantons and renormalons. Don't they reveal themselves as singularities of the Borel transform? I assumed it is always the case. So in this sense, the Borel summation (appropriately extended to account for ambiguities) is applicable to most physical situations?

@WeatherReport: I don't understand the limitations of the Borel transform well enough to fully answer your comment.

The point is that in order to be sure that some version of the Borel transform is applicable one must always analyse the source of the asymptotic series, and use the information gained in this way to choose the correct summation formula.

As regards QFT, I believe that Borel singularities correspond to renormalons, which in turn are related to Landau poles, while instanton contributions don't give singularities but exponential terms that must be considered separately. But better check this with the literature rather than trust my memory!

Personally, I have always understood it as a method of regularization similar to the zeta-function regularization of $\sum n$. When you can prove Watson's criteria, you can often sum directly or verify by other methods. In practice, I have seen it understood more as a "renormalization" procedure with some possible obscure consequences such as the ones mentioned by Arnold.

+ 0 like - 0 dislike

Borel summation is not relevant for asymptotic series of "physical observables", generally speaking. Your referencing to the path integral is more than vague and is not universal. As Arnold said, there may be the terms for which the Borel summation is useless. The simplest case is not QFT, but QM. An atom in an external electric field is unstable with respect to ionization and the lifetime is just this nasty term with ${\rm{e}}^{-{O(1/g)}}$ (or "tunnelling exponent"). Thus, perturbative series for the atomic energies in powers of a weak external field miss something and one should be more careful in order to deal with "better observables" (or "better mathematical expressions") for their expansions to make sense.

answered Feb 19 by Vladimir Kalitvianski (12 points) [ revision history ]

These exponential terms do not always (''generally speaking'') arise, but they arise in all cases where tunneling happens. In quantum field theory this is discussed in terms of instantons.

Arnold, I agree with you. However, tunneling terms in QFT are known since long ago and well before the instantons, if one remembers the Klein paradox, pair creation with a constant electric field (Schwinger's effect), etc.

Instantons is simply the modern term for tunnelling in QFT.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights