The role of estimates in field theories

Question

I have been taking a look at some papers in constructive quantum field theory and I got the impression that there is a systematic of estimating things like e.g the effective action or the free energy of a theory. I could cite a lot of examples of papers to justify this statement, but to make things more focused I am just giving one example which is Lohmann's "Some Improved Nonperturbative Bounds for Fermionic Expansions". There is no special reason why I chose this paper, however. I think it just captures the idea of the question, as many others would do.

I will focus on section 4.2 of the linked paper, where a single scale Fermi system is addressed. There, the author obtain bounds for the effective action: $$\Omega_{C}(W;\eta) = \log \int e^{W(\psi+\eta)}d\mu_{C}(\psi). \label{1}\tag{1}$$ This is the typical analysis I have been finding in the literature. However, I do not understand what is the idea behind these estimates. For a non-expert it is really difficult to understand what people really want with these bounds, because the basic problems to which these may apply are usually not stated. I was hoping you could help me with the following questions:

• What is the point of estimating $\Omega_{C}(W;\eta)$? What does one want with these estimates?
• Most of the time, $\Omega_{C}(W;\eta)$ is defined over a finite lattice or regularized theory. In this particular case, a Grassmann algebra with finitely many generators. Isn't it obvious that $\Omega_{C}(W;\eta)$ will converge, then?
• These estimates are done for a single scale. How can we pass to other scales? Is it by using the renormalization group? And if so, how one does that?

Thanks in advance!

Answer 1

Q1: What does one want with these estimates?

The overall motivation of studies such as this is to show that interactions between the fermions, if sufficiently weak, do not open a gap in the single-particle excitation spectrum. This means that if you prepare the system in the ground state (lowest energy state) and then search for the energy difference with the first excited state (= the excitation energy), then this energy vanishes in the limit that the system size $L$ goes to infinity (thermodynamic limit).

For non-interacting fermions the excitation energy in $d$ dimensions vanishes as $1/L^d$, so this is a gapless system in the thermodynamic limit. Interactions may change that, the superconducting instability is one effect that introduces a nonzero gap in the thermodynamic limit no matter how small the attractive interaction (Cooper instability).

The instability is signaled by a vanishingly small convergence radius of the expansion of the free energy in powers of the interaction strength. So if you can show that the radius of convergence is nonzero, you can be certain that the system remains gapless if the interaction is sufficiently weak.

Q2: Isn't it obvious that the perturbation series converges on a finite lattice?

What you want to show is that the perturbation series converges if you take the $L\rightarrow\infty$ thermodynamic limit at fixed nonzero interaction strength.

Q3: What is the role of the renormalisation group?

Momentum space is decomposed into a hierarchy of shells, corresponding to details on finer and finer length scales. The Green functions are computed for each shell and the effective interaction at each scale follows. This is explained in some detail on page 8 and following of Construction of a 2D Fermi liquid (2006).