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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.

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    $\begingroup$ 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. $\endgroup$ – Andrew Jun 30 '14 at 6:44
  • $\begingroup$ @Andrew thanks, I will look at those sections ASAP $\endgroup$ – gn0m0n Jun 30 '14 at 7:03
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    $\begingroup$ Some explanations are given here and Prof. Carl Bender's Mathematical Physics course, which exactly deals with asymptotic series, might be helpful. $\endgroup$ – Dilaton Jul 9 '14 at 7:49
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The ambiguity that people normally refer to is due to the lack of Borel summability of the perturbation series.

Consider a series of the form

$$A(g) = \sum_{n=1}^{\infty}{(-1)^n g^n (n!)} $$

If the coefficients $b_n$ are of order $1$, this series is obviously divergent. But we can compute its Borel sum. First compute the Borel transform:

$$\mathcal{B}A(t) = \sum_{n=1}^{\infty}{ (-1)^n t^n }$$

We may then compute the Borel sum:

$$ S(g) = \int_0^{\infty}{dt\, e^{-t} \mathcal{B}A(t g) }= \int_0^{\infty}{dt\, e^{-t} \sum_{n=1}^{\infty}{(-1)^n (tg)^n }}$$

We now pull the sum out of the integral, assuming the series can be integrated term by term. This, technically speaking, is a wrong thing to do since this series is not uniformly convergent -- in fact, it's not convergent at all -- but we'll do it anyway with the proviso that we'll simply define the answer to be what we get by doing this. I'm the one defining how to sum up this series, so I'm entitled.

$$S(g) = \int_0^{\infty}{dt\, \frac{e^{-t}}{1 + tg} } $$

And as long as $g > 0$, this expression is obviously finite. The procedure worked and we were able to find a nice well defined number to assign to this series.

You'll notice however that the fact that the original series was alternating was crucial for obtaining this result. Let's repeat it for a non alternating series, which is more representative of the kinds of series you might find when adding up Feynman diagrams in a field theory.

$$S(g) = \int_0^{\infty}{dt\, e^{-t} \sum_{n=1}^{\infty}{ (tg)^n }} = \int_0^{\infty}{dt\, \frac{e^{-t}}{1 - tg} } = \frac{1}{g} \int_0^{\infty}{dt\, \frac{e^{-\frac{t}{g}}}{1 - t} }$$

That is, for $g > 0$ the integrant has a pole on the positive real axis. You can still do the integral but you must choose which in which direction you must do your excursion into the complex plane. This is an ambiguity.

Notice that the residue at the $t = 1$ pole in the above is

$$Res(S,1) = -\frac{e^{-\frac{t}{g}}}{g} $$

Curiouser and curiouser! The imaginary part you find in the Borel sum is nowhere to be found in the original series -- since every term is real -- and, further, it is non perturbative in $g$! Hmm... that's suggestive.

It turns out that if you perform your calculations carefully enough and any analytic continuations are carried out consistently, then the imaginary contributions associated with instantons cancel out the imaginary contributions from the failure of the perturbation series to be Borel summable. The full expression for the energy eigenvalues therefore suffers from no ambiguities. This phenomenon where high orders in perturbation theory (about the trivial vacuum) somehow encode information about low orders around instanton solutions has been termed "resurgence". Key words in the literature also include "trans series", "Stokes wedges" and "Lefschetz thimbles".

Refs:

http://arxiv.org/abs/1210.2423

http://arxiv.org/abs/1210.3646

http://arxiv.org/abs/1411.3585

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Maybe this would be better as a comment, since it is not a full answer, but I don't have enough reputation for that. The most important ambiguity is that there is an infinite number of functions that have the same asymptotic expansion. As an example, if $f(g)$ has some asymptotic expansion in $g$ as $g \to 0$ than $f(g) + e^{-1/g^2}$ has exactly the same asymptotic expansion. First, notice that this looks like a toy version of an instanton contribution. Second, the fact that there are more than one functions with the same asymptotic exanpsions means that in order to actually "sum" the whole asymptotic series, one needs additional assumptions.

For example, if I know that the function that the asymptotic series approximates is a Steiljes function, than I know that if I construct Padé approximants from the Taylor like series, than I will be able to get an (exact) upper and lower limit on the function, from a divergent asymptotic series. What Stieljes functions and Padé approximants are is treated in the excellent lectures by Carl Bender, that was suggested by one of the comments. Additionally if I know that the coefficients in the expansion diverge slowly enough (I think $(2n)!$, but I am not sure I remember correctly) than I can not only give an upper and lower limit, but the Padé approximants actually converge to the function in question.

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