Ambiguity in Asymptotic Perturbative Series and Instantons

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I know there are a number of questions about the asymptoticity of perturbative series and about instantons on StackExchange (e.g. Instantons and Non Perturbative Amplitudes in Gravity from user566, Instantons and Borel Resummation asked by felix, and How can an asymptotic expansion give an extremely accurate predication, as in QED? asked by yonni). Reading them was helpful but left me with two short questions:

1) What is meant by "ambiguity" in this context? Several posters use the term in alluding to the problems in asymptotic series. Does it have a technical meaning here?

2) How can we see that the instantons would "correct" the series in the full theory?

Perhaps the only thing to do is to read these notes ("Instantons and large N" by Marino) (which I plan on doing) but I was wondering if someone could give a quick answer for #1 and perhaps a clever or intuitive way of making #2 plausible.

This post imported from StackExchange Physics at 2014-07-01 10:32 (UCT), posted by SE-user gn0m0n
Great question!! The quick and dirty way to see what is going on is to consider what in your notes is called the 'toy integral', so I would read sections 2.3 and 4.2 first. Note that eqn 2.32 is an instanton solution in the toy model. The ambiguity is the branch cut talked about in 2.3. You can also see what is going on in eqn 2.38: the perturbation series doesn't converge because of factorial growth of the size of the terms in the series. Also note that in QFT instantons are not the only nonperturbative features that perturbation theory misses, there are also renormalons.

This post imported from StackExchange Physics at 2014-07-01 10:32 (UCT), posted by SE-user Andrew
@Andrew thanks, I will look at those sections ASAP

This post imported from StackExchange Physics at 2014-07-01 10:32 (UCT), posted by SE-user gn0m0n

Maybe Carl Bender's Mathematical Physics course is helpful, it exactly deals with the topic of asymptotic series.

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Sometimes an explicit example works better for me than anything else. So here goes.

Consider the sum $S=\sum_{m=0}^\infty m! x^m$. Typically one gets such a series in some perturbative expansion by expanding the integrand in some integral in some small parameter $x$. Otherwise, one carries out Borel resummation to obtain an alternate representation for S. This is a classic asymptotic series. Suppose we are interested in obtaining the sum when $x=x_0$. Consider the partial sums $S_N=\sum_{m=0}^N m! x_0^m$. The error in the truncation in an asymptotic series is given by the next term in the sequence, $(N+1)! \ x_0^{N+1}$ in this case -- we can try to minimise the error by choosing $N$ suitably. Using Stirling's formula for the factorial, and then minimising w.r.t. $N$, one sees that the error is minimum when $N=N^*\sim 1/x_0$ and the error from this $\sim e^{-1/x_0}$. Truncating the asymptotic series at $N^*$  is called the superasymptotic truncation by Berry. This simple example teaches us a couple of things: (i) The truncation point gets larger as $x_0$ gets smaller and (ii) The truncation error is non-analytic in $x_0$.

In physical applications, $x_0$, is typically some coupling constant. The truncation point (or the point where perturbation theory in $x_0$ breaks down) could appear at some high order. The non-analytic behaviour in some cases (see comment by Andrew above) are typical of instanton contributions.

answered Jul 9, 2014 by (1,545 points)

Thank you!

I don't see an option to choose your answer... maybe there isn't one on overflow??

@gn0m0n Welcome to PhysicsOverflow :-). We have disabled the feature of accepting an answer, because the majority of our members thinks it is not really needed. To upvote, you need 50 rep, but if you have SE posts you would like to have here, you can request them to be imported by following the link "Import SE questions" in them menu bar below the PhysicsOverflow logo. This would give you some deserved rep too ...

a very belated 'thanks, Dilaton' :)

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Ambiguity refers to the fact that a Taylor series with zero radius of convergence does not determine the function. For example, the functions $g(x)=ce^{-1/x^2}$ (augmented by $g(0)=0$) are infinitely often differentiable and have the same asymptotic expansion at $x=0$ as the zero function. Thus one can add to any interpretation of the asymptotic series an arbitrary function of the above form (and in fact many other similar functions) without changing the asymptotic series.

Most of these modifications do not make sense physically, but those corresponding to instantons do.

answered Jul 12, 2014 by (15,488 points)

Not sure if I didn't see this earlier or what, but thanks for your answer.

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