jordi said:
Can you please elaborate?
Basically when you do perturbation theory you compute an expansion of the Correlation functions:
$$\mathcal{W}\left(x_1,...,x_n,\lambda\right) \rightarrow \sum_{n}\lambda^{n}\mathcal{G}_{n}\left(x_1,...,x_n\right)$$
It turns out that the ##\mathcal{G}_{n}\left(x_1,...,x_n\right)## functions can be computed from Fock space. So when you are doing perturbation theory you can ignore Haag's theorem.
The problem is that the series doesn't sum back up to the original ##\mathcal{W}\left(x_1,...,x_n\right)## and begins to diverge around ##n \approx \frac{1}{\lambda}##.
There are methods of getting around this such as Borel summation that try to modify the series. A simple case is that you alter the ##\lambda^{n}## to be ##\frac{\lambda^{n}}{n!}##. Let's make this replacement:
$$\mathcal{B(W)}\left(x_1,...,x_n,t\right) = \sum_{n}\frac{t^{n}}{n!}\mathcal{G}_{n}\left(x_1,...,x_n\right)$$
Note I've replaced ##\lambda## with ##t## to avoid confusion later.
Then you can recover the original function with the integral:
$$\mathcal{W}\left(x_1,...,x_n\right) = \int_{0}^{\infty}{e^{-\frac{t}{\lambda}}\mathcal{B(W)}\left(x_1,...,x_n,t\right)dt}$$
This works great in lower dimensions, but unfortunately in 4D in for example Yang-Mills theory the integral has poles for certain values of ##t##.
These poles prevent you from performing the integral. Of course one can shift the contour around the poles (typically there are infinitely many) however this introduces an ambiguity in the evaluation. This ambiguity is roughly ##\mathcal{O}\left(e^{\frac{1}{\lambda}}\right)##, so clearly a nonperturbative effect.
Some poles are instantons and others are renormalons.
Renormalons originate in chains of bubble graphs in perturbation theory. When you renormalize the theory such diagrams have their divergences canceled by counterterms, but the counterterms also contain finite parts that do not cancel infinities in graphs. These finite counterterm pieces end up contributing a ##n!## term to the perturbative series. It's this rapid growth term that causes the renormalon poles.
So note that the unrenormalized Feynman series lacks such renormalon poles. However due to being unrenormalized they diverge as you remove the cutoff and thus have no continuum limit (obviously since this is why you bother renormalizing at all).
