Finite quasiparticle lifetimes in Fermi Liquid Theory

Question

I am trying to clarify a conceptual issue about phenomenological Fermi liquid theory. My confusion can be explained using the following two sentences from Dupuis's many body theory notes, but the same sentiment is present in many other sources as well. The sentences are:

According to the adiabatic continuity assumption, as the interaction is slowly turned on we generate an (excited) eigenstate of the interacting system. However, because of the interactions the state under study is damped and acquires a finite lifetime.

There seems to be two competing concepts here, both of which seem central to FLT. On one hand,

1. Quasiparticles correspond to excited energy eigenstates of the interacting Fermi liquid. These eigenstates can be obtained by starting with a corresponding excited state of the free Fermi gas and adiabatically switching on the interactions. Landau's theory postulates adiabatic continuity, so that the interacting eigenstates stand in 1-1 correspondence with the free eigenstates and can therefore be labeled by the same quantum numbers.

On the other hand,

2. Whereas free particles do not interact, and an excited state of the free theory will persist indefinitely, quasiparticles of the interacting theory do interact with each other. While a quasiparticle will then in general decay, its lifetime will diverge as it approaches the Fermi surface.

These two notions of quasiparticles seem contradictory to me. If the quasiparticles are eigenstates of the interacting theory, then they should not decay. Conversely, if the quasiparticles do interact and decay, then how are they related to the free particle excitations, and how should I understand that the quasiparticles carry the same quantum numbers as the free particles?

EDIT: After talking to a friend, I think the answer lies in the fact that the adiabatic theorem does not hold for an eigenstate without a gap. If the system were gapped, and none of the energies crossed as the interaction were turned on, then eigenstates would necessarily evolve into eigenstates. But since the Fermi system is gapless, there's no reason that the adiabatically evolved eigenstates remain eigenstates. But it would be nice to have confirmation from someone more knowledgeable, and it's strange that this point is not discussed in any of the sources I've checked.

EDIT 2: Apologies for the multiple edits. After doing some research, I think that my previous edit was incorrect. As far as I can tell, the Gell-Mann and Low theorem guarantees that an infinitely slow adiabatic turning on of the interaction evolves eigenstates of the free theory into eigenstates of the interacting theory. The application to FLT seems immediate to me: if we start with an excited free particle state, and turn on the interactions infinitely slowly, we expect the state we obtain to be an eigenstate. But clearly this cannot be what we are actually doing in FLT, since the quasiparticles are not eigenstates. So how should I make sense of this?

Answer 1

I suggest thinking about Fermi liquid theory as two successive levels of approximation (which are unfortunately not clearly delineated in the sentences you quote):

1. In the limit of vanishing energy above the Fermi level, the eigenstates of the interacting theory have the same quantum numbers as free electrons, and the two are smoothly connected as the interactions are turned on.

2. For a small but nonvanishing energy above the Fermi level, this is no longer true. However, the first-order correction is to consider these adiabatically connected states as 'almost-eigenstates' which acquire a finite lifetime but are still nearly conserved (more precisely, they are conserved for long times compared to the energy of the excitation). In a spectral function picture, which is widely used and useful in this context, the quasiparticles are not Dirac delta function poles, as eigenstates with infinite lifetimes would be, but are broadened to sharp peaks with a finite width.

Also, I should point out that the idea of the adiabatic turn-on of the interactions is that it is slow enough to smoothly connect particles to quasiparticles, but not so slow that the quasiparticles can decay. So, it is not infinitely slow, and the Gell-Mann and Low theorem doesn't apply. You should remember that this is an argument, which is empirically successful in many systems, but it is definitely not a theorem and Fermi liquid theory does break down in various cases.

This answer does not come from a single source that I can point to. However, my recommended general exposition of the theory, if you can get it, is Pines and Nozieres' The theory of quantum liquids (Vol. 1), which I find is still unbeatable despite being nearly six decades old.

Edit: paraphrasing slightly, a question in the comments is: why work with quasiparticles, which are approximate eigenstates, instead of the exact excited state eigenstates? If we postulate a Fermi liquid-like theory in which the excitations are instead exact eigenstates, which might seem like the natural thing to do, what are we missing?

Well, the first thing to notice here is that this is equivalent to taking the approximation that the quasiparticle lifetimes are infinite. In other words, this corresponds to taking only approximation 1 above, and not going on to approximation 2, which will result in a restricted version of Fermi liquid theory that is only valid for energies very close to the Fermi level. Of course, for some purposes this might be enough. However, for experimental probes such as conductivity the finite quasiparticle lifetime plays an essential role, so extending the theory to account for these makes it much more powerful.

One could imagine that historically the theory was developed in this way- first considering the limit of quasiparticles that are completely stable, then generalizing to unstable but long-lived- but I have no idea whether this is actually how it happened.

Answer 2

Here's a heuristic argument from me.

I believe the problem here is that we are not allowed to assume the excited states are connected adiabatically because excited states are very degenerate in general (one can imagine in the non-interacting case at least, many particle-hole excitations and particle excitations having the same energy). However, the ground state is gapped (before taking the thermodynamic limit), so we are allowed assume the interacting ground state is adiabatically connected.

Still heuristically, one can argue using Fermi's golden rule that the higher excited states have more phase space to decay (there are more particle hole states allowed by momentum conservation).

The momentum being good quantum number is not very hard to imagine since total momentum is conserved classically in a system where pairwise Coulomb repulsion is introduced. More formally, since $\frac{d\langle P\rangle}{dt}$~$[P,H(\lambda)]$ by Ehrenfest theorem and that P is a constant of motion, $[P,H(\lambda)]=0$. Note here the quasiparticle states are not generally eigenstates of $\frac{a_k^\dagger a_k}{2m}$ for a specific $k$, but total momentum $P=\sum_k\frac{a_k^\dagger a_k}{2m}$(?), where $a_k$ is the non-interacting creation operator.

Finally, I suspect we do not use the exact interacting eigenstates because we can't really say anything about them (not adiabatically connected).

Ref: https://www.phys.lsu.edu/~jarrell/COURSES/SOLID_STATE/Chap6/chap6.pdf (this note by Landau(?) illustrates the argument with Fermi's golden rule.)