Can we get full non-perturbative information of interacting system by computing perturbation to all order?

As we know perturbative expansion of interacting QFT or QM is not a convergent series but an asymptotic series which generally is divergent. So we can't get arbitrary precision of an interacting theory by computing higher enough order and adding them directly.

However we also know that we can use some resummation trick like Borel summation, Padé approximation and so on to sum a divergent series to restore original non-perturbative information. This trick is widely used in computing critical exponent of $\phi^4$ etc.

My questions:

1. Although it's almost impossible to compute perturbation to all order, is it true that we can get arbitrary precision of interacting system (like QCD) by computing higher enough order and using resummation trick like Borel summation?

2. Is it true that in principle non-perturbative information like instanton, vortex can also be achieved by above methods?

There is a solid example: $0$-dim $\phi^4$ theory,

$$Z(g)\equiv\int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2 -gx^4/4}$$ From the definition of $Z(g)$ above, $Z(g)$ must be a finite number for $g>0$.

As usual we can compute this perturbatively,

$$Z(g)= \int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2}\sum_{n=0}^{\infty}\frac{1}{n!}(-gx^4/4)^n \sim \sum_{n=0}^{\infty} \int_{-\infty}^{\infty}\frac{dx}{\sqrt{2\pi}}e^{-x^2/2} \frac{1}{n!}(-gx^4/4)^n \tag{1}$$

Note: In principle we can't exchange integral and infinite summation. It's why asymptotic series is divergent.

$$Z(g)\sim \sum_{n=0}^{\infty} \frac{(-g)^n (4n)!}{n!16^n (2n)!} \tag{2}$$ It's a divergent asymptotic series.

In another way, $Z(g)$ can be directly solved, $$Z(g)= \frac{e^{\frac{1}{8g}}K_{1/4}(\frac{1}{8g})}{2\sqrt{\pi g}} \tag{3}$$ where $K_n(x)$ is the modified Bessel function of the second kind. We see obviously that $Z(g)$ is finite for $g>0$ and $g=0$ is an essential singularity.

But we can restore the exact solution $(3)$ by Borel resummation of divergent series $(2)$

First compute the Borel transform $$B(g)=\sum_{n=0}^{\infty} \frac{(-g)^n (4n)!}{(n!)^216^n (2n)!} = \frac{2K(\frac{-1+\sqrt{1+4g}}{2\sqrt{1+4g}})}{\pi (1+4g)^{1/4}}$$ where $K(x)$ is the complete elliptic integral of the first kind.

Then compute the Borel Sum

$$Z_B(g)=\int_0^{\infty}e^{-t}B(gt)dt=\frac{e^{\frac{1}{8g}}K_{1/4}(\frac{1}{8g})}{2\sqrt{\pi g}} \tag{4}$$

$$Z_B(g) = Z(g)$$

We see concretely by using the trick of Borel resummation, we can restore the exact solution from divergent asymptotic series.

• The answer is no. At least if you mean standard perturbation theory. Just do a Taylor expansion of $\exp(-1/g^2)$ around $g=0$ to see why. – marmot Aug 15 '18 at 23:49
• @marmot It's not a Taylor expansion but a Asymptotic expansion. You can get this result by resummation – maplemaple Aug 15 '18 at 23:53
• @marmot You can see my updated version's example. – maplemaple Aug 16 '18 at 0:29
• Note also that Borel resummation contains ambiguities whenever the Borel transform contains singularities along the positive real axis (which can be argued to be usually the case). – AccidentalFourierTransform Aug 16 '18 at 2:16
• @maplemaple Please stop making trivial edits to bump the question into the front page. Thank you. – AccidentalFourierTransform Aug 17 '18 at 0:52

Perturbation theory gives for the solution an asymptotic series in the coupling constant $g$. There are infinitely many functions having the same asymptotic series, since for example adding a function of $e^{-c/g^2}$ vanishing at zero will not change the asymptotic series.
2. When we perform Borel resummation, there is a phenomenon called "resurgence" where these exponential terms come in series which look just like the "vacuum" perturbation series but with a prefactor like $e^{-S_0/g^2}$ where $S_0$ is interpreted as the instanton action. The list of examples of this is growing every day. See this paper for instance (and those referencing it): https://arxiv.org/abs/1210.2423 . Presumably after you resum the series it converges to an exact answer, at least in cases with resurgence. I don't know any theorems though.