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In QFT is said that the renormalized Dyson series is only asymptotic. But to be able to say it is necessary to know to what function of $g$ (the coupling constant) the Dyson series is asymptotic.

For example, suppose that some transition amplitude $A(g)$ is given perturbatively by a series of powers of $g$. In order to prove that this series is asymptotic to $A(g)$ I need to know the value of $A(g)$ non-perturbatively, but this is not possible since the only way $A(g)$ is given is the Dyson series.

Qmechanic
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Sergio Denti
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2 Answers2

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Suppose that $A(g)$ is given by some perturbative expansion around $g=0$: $$A(g) = \sum_n c_n g^n.$$ Then the statement that this expansion is asymptotic means that the radius of convergence is zero: for fixed $g$, no matter how small, the limit $$\lim_{N \to \infty} \; \sum_{n < N} c_n g^n$$ diverges. Typically it's enough to know that large-$n$ behaviour of the coefficients $c_n$ to establish this. In QFT, sometimes you can estimate the magnitude of the $c_n$ (e.g. you count the number of Feynman diagrams at $n$-loop order multiplied by the typical contribution of a single diagram). For instance, if $$c_n \sim n!$$ you can prove that the series is asymptotic. You do this using the tools you learned in undergrad calculus (for example the ratio test).

Notice that this is a really fast growing series: if for instance $c_n \sim a^n$ for some constant $a$, then you have convergence inside a disk $|g| \leq 1/a$. So you need even faster than exponential growth of the $c_n$ to have an asymptotic series.

  • There’s also an interesting and unexpected fact: the same asymptotic expansion can approximate two different functions. In QFT, this corresponds to the possibility of having inequivalent nonperturbative theories agreeing with the same perturbation theory. – Prof. Legolasov Nov 13 '17 at 16:05
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By the Borel-Ritt Theorem, any (formal) power series $\sum_{n=0}^{\infty} c_n g^n$ is the asymptotic expansion at zero of some $C^{\infty}$ function $A(g)$. Of course the latter is very non-unique. You raised a good point about what is this $A(g)$ that the series should be asymptotic to? or since we know the existence of $A(g)$, the question should rather be what is the right $A(g)$? i.e. one needs some uniqueness, via imposing some extra conditions. It is known that for scalar fields in 2d and 3d ($\phi^4$ model) the perturbative series has zero radius of convergence but is also asymptotic to a non-perturbative definition of $A(g)$ obtained via the hard estimates of constructive quantum field theory. Moreover, the non-perturbative $A(g)$ is the Borel sum of the (moderately) renormalized perturbative series. This is one way of restoring uniqueness. The two references on this are

  1. Eckmann, Magnen, Sénéor, "Decay properties and Borel summability for the Schwinger functions in $P(\Phi)_2$ theories", CMP 1975.
  2. Magnen, Sénéor, "Phase space cell expansion and borel summability for the Euclidean $\varphi_{3}^{4}$ theory", CMP 1977.
Abdelmalek Abdesselam
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    @AccidentalFourierTransform: it's pretty clear from the context but I added the word "formal" anyway. – Abdelmalek Abdesselam Nov 13 '17 at 23:25
  • Sure, it's clear, but hey, it won't hurt to be precise (although admittedly slightly pedantic). Not everyone knows as much as you; a single word may help. – AccidentalFourierTransform Nov 14 '17 at 12:00
  • Tank you so much for the informations. Are there non-perturbative estimates even for more realistic QFT such as QED and QCD ? – Sergio Denti Nov 15 '17 at 13:49
  • I would Have another problem related to the previous one : I don't understand why going to high perturbative orders ( high pwers of g )it also goes towards great virtual energies ( Folland " QFT a tourist guide for mathematicians " pag. 253 : "...high order Feynman diagrams in QED describe electromagnetic processes involving large number of virtual particles and hence large virtual energies ... ". Even a feynman diagram with only one loop implies arbitrarily large energies with respect to the virtual particle. – Sergio Denti Nov 15 '17 at 14:01
  • @SergioDenti: "more realistic" implies that $\phi^4$ type scalar models are not realistic which is a rather misleading statement. Since 2012 we know there is particle in nature described by a scalar field. For QED in 3d see this article by Dimock and references therein https://arxiv.org/abs/1512.04373 for Yang-Mills in 4d with group SU(2) see this article https://projecteuclid.org/euclid.cmp/1104253284 and cited work of Balaban therein. Presumably the same methods should work for SU(3), i.e., QCD. – Abdelmalek Abdesselam Nov 15 '17 at 19:45