5

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
  • 220,844
Sergio Denti
  • 61

2 Answers2

5

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.

5

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
  • 6,362