quantum field theory - 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.

Answer

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.

Thus in general, the perturbation series does not give full perturbative information. Every summation procedure needs to make additional assumptions about the solution; it will resum the series correctly when these assumptions are satisfied but in general not otherwise.

In many toy instances one can prove that the assumptions of Watson's Borel summation theorem can be shown to hold; then Borel summation works. But it is known not to work in other cases, e.g., in the (frequent) presence of renormalons.

In 4D relativistic quantum field theory it is not known of any resummation method whether it will work. The most powerful resummation technique, based on resurgent transseries has the most promise.