# Can we get complete non-perturbative information of the interacting system by computing perturbation to all orders?

As we know, the perturbative expansion of interacting QFT or QM is not a convergent series but an asymptotic series that is generally 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 tricks like Borel summation, Padé approximation and etc. to sum a divergent series to restore original non-perturbative information. This trick is widely used in computing the critical exponent of $$\phi^4$$ etc.

My questions:

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

2. Is it true that, in principle, non-perturbative information like instanton and vortex can also be achieved by the 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 the 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 analytically, $$Z(g)= \frac{e^{\frac{1}{8g}}K_{1/4}\left(\frac{1}{8g}\right)}{2\sqrt{\pi g}} \tag{3}$$ 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.

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

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

Then compute the Borel Sum

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

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

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

• 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.
– user178876
Aug 15, 2018 at 23:49
• @marmot It's not a Taylor expansion but a Asymptotic expansion. You can get this result by resummation Aug 15, 2018 at 23:53
• @marmot You can see my updated version's example. Aug 16, 2018 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). Aug 16, 2018 at 2:16
• @maplemaple Please stop making trivial edits to bump the question into the front page. Thank you. Aug 17, 2018 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.