# Perturbation theory in strong interaction regime

1. May 24, 2015

In QFT, we can expand the propagator and obtain the diagrammatic expansion to build up the Green's function. If we have a hamiltonian of the type $$H = H_{0}+V$$, where V is the perturbation, we can build up the Feynman diagrams,and if we could build up all of them to infinite order, we would obtain the exact Green's function on the model.

However, my question is related with the fact that, in all the formal development of the diagrammatic expansion, V is not assumed to be "small" compared with H0, I mean is a perturbation which in principle is always assumed to be small respect to H0. But what happens if V becomes larger and of the same magnitude as H0? If we go to infinite order of perturbation, do we still have the exact Green's function of the problem?

2. May 24, 2015

### Orodruin

Staff Emeritus
This is not true. The perturbation series always diverges at some level, since it is an asymptotic series. This is essentially due to the number of diagrams growing faster than the suppression from the powers of the coupling ($\sim \alpha^n$ vs $\sim n!$).

Depending on the coupling strength and (to some extent) on the available vertices, the asymptotic series is going to start diverging at different orders in perturbation theory. If the coupling constant is strong you will essentially never have a good approximation.

3. May 24, 2015

Some real physical effects are essentially non-perturbative and never show up at any order in perturbation theory. For example suppose you try to do a power series expansion of $\exp(-1/g^2)$ around $g = 0$. You'll find that you can't; this function has an essential singularity at $g = 0$. And some physical effects are proportional to $\exp(-1/g^2)$ where $g$ is a coupling. Such effects--for example instantons and tunneling--will never show up in the perturbative power series expansion around $g=0$. A classic example is a harmonic oscillator with a $\lambda x^4$ perturbation. For $\lambda < 0$ the system because unstable due to tunneling, but perturbation theory never detects this.