# Perturbation theory in strong interaction regime

In summary: So in principle, for strong coupling systems perturbation theory does not give you the actual solution, but only an approximation.
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?

if we could build up all of them to infinite order, we would obtain the exact Green's function on the model.

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!##).

If we go to infinite order of perturbation, do we still have the exact Green's function of the problem?

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.

But I can't still see why going to infinite order is not giving the exact solution of the problem. I mean, if we are at zero temperature, you use the Gell-Mann Low theorem to perturb the system from the non-interacting ground state to the interacting one. This expansion is made by exact calculation of the S matrix, which is an exponential operator and you just express the exponential as a series, where the different contractions give you different diagrams. During all this treatment and derivation of the diagrammatic expansion for the Green function, its never said that V has to be necesarilly small compared to H0.

Further and for the sake of completeness, which are the available methods in many body theory to treat strong correlated systems apart from perturbation theory and RG? The RG methods also derive an approximation of the solution or they give in fact the actual solution?

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.

## What is perturbation theory in the strong interaction regime?

Perturbation theory in the strong interaction regime is a mathematical framework used to study the behavior of particles and their interactions in the presence of strong forces. It involves expanding the equations of motion in terms of a small parameter and solving them iteratively to obtain approximate solutions.

## Why is perturbation theory necessary in the strong interaction regime?

In the strong interaction regime, the interactions between particles are very strong and cannot be treated exactly using traditional methods. Perturbation theory allows us to make approximations and calculate the effects of small perturbations on the system.

## What are the limitations of perturbation theory in the strong interaction regime?

Perturbation theory is only valid when the perturbation is small compared to the overall system. In the strong interaction regime, where the interactions are strong, this may not be the case. In such situations, perturbation theory may not provide accurate results.

## How is perturbation theory used in the study of strong interactions?

Perturbation theory is used to calculate the scattering amplitudes and cross-sections of particles in the strong interaction regime. It is also used to study the decay rates of unstable particles and to understand the behavior of particles in high energy collisions.

## What are some applications of perturbation theory in the strong interaction regime?

Perturbation theory has many applications in nuclear and particle physics, such as in the study of quarks and gluons, the behavior of hadrons, and the properties of nuclear matter. It is also used in the development of theoretical models and to make predictions for experimental results.

• Atomic and Condensed Matter
Replies
9
Views
800
• High Energy, Nuclear, Particle Physics
Replies
8
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
9
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
1
Views
903
• Quantum Physics
Replies
14
Views
2K
• Quantum Physics
Replies
8
Views
981
• High Energy, Nuclear, Particle Physics
Replies
26
Views
4K
• High Energy, Nuclear, Particle Physics
Replies
5
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
1
Views
2K
• Quantum Physics
Replies
6
Views
745