2
$\begingroup$

In modern terms, Landau's theory of Fermi liquids is understood as the fixed point of a Wilsonian RG as one scales towards the Fermi surface.

Shankar and others use the RG interpretation to explain why Landau stopped at pairwise quasi-particle interactions in his phenomenological theory (i.e., why he didn't also include an quasi-particle vertex with 6 legs, 8 legs, etc.) This is because such interactions are irrelevant operators as one flows towards the Fermi surface, while the 4-point q.p. vertex is only marginally-irrelevant (provided there's no BCS instability).

I have 2 questions:

1) If I start with the path integral for a normal fermi system and do the Wilsonian integration of modes all the way down to the fermi surface (pretend I can do this exactly), will the action at the end of the flow literally just consist of the dressed quasi-particle energies, i.e., the term quadratic in the electron fields, plus a quartic term whose coupling corresponds to the Landau f-function (i.e., the quasiparticle interaction)? Or would it also contain higher terms containing 6,8,10, etc. fermion fields with non-vanishing couplings? If it's the latter, how do I square this with the claim that the RG treatment "explains" why Landau stopped at pairwise q.p. interactions?

2) Imagine now I have a family of unitarily-equivalent starting actions. By construction, they all describe the same physics, but they might correspond to widely separated initial points in the coupling space.
If I now do the RG flow for this family of starting actions, will the trajectories start to converge as more and more modes are integrated? My naive guess is yes, since the Landau parameters can be put in 1-to-1 correspondence with observable quantities (e.g., specific heat, etc.).

$\endgroup$

1 Answer 1

1
$\begingroup$

First a clarification: the forward scattering vertex is exactly marginal, not marginally-irrelevant. The latter would imply that these vertices renormalize to 0, and there would not be any Landau parameters/functions to talk about. So the Fermi liquid fixed point is a deceptively simple example of an interacting fixed point.

I respond in the same order of the questions:

1) You have already answered the question...

This is because such interactions are irrelevant operators as one flows towards the Fermi surface

If one integrates out modes, all symmetry allowed vertices are generated. The idea of an infra-red stable fixed point does not imply an absence of irrelevant operators. The irrelevant operators are still present (albeit suppressed by inverse powers of the ultra-violet cutoff), but they cannot affect the scaling behavior of correlation functions. In fact marginal operators cannot affect scaling either (up to logarithms). So Landau's intuition was correct insofar as the identify of the most important terms of the hamiltonian that determines the low energy behavior of Fermi liquids.

2) During the course of RG flow all `irrelevant' information is lost. In this sense trajectories start to converge. For a Fermi surface problem this still implies infinitely many trajectories that don't converge, but flow in parallel. These trajectories are parameterized by the Landau functions. Notice that the Landau functions may only affect the pre-factors but not the scaling dimensions of observables.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.