I am a super beginner when it comes to Solid State Physics and when wanting to learn more on the subject, I end up reading on Landau's Fermi liquid theory that supposedly justifies the quasi-free electron model I was taught at the undergraduate level.

I have spanned now few .pdf lectures that I could find online and find them very confusing as I don't even understand really the meaning of the first sentence which more or less always starts by:

in a Fermi gas, the excitations satisfy blablaba...

Basically, the problematic is presented as being that of interactions between electrons (or phonons) in a metal and to account for them in the statistical thermodynamic treatment of charge carriers in metals to rationalize, among other things, the effective mass one needs to consider when looking for quantitative agreement between the quasi-free electron model and experiments.

My questions are super simple here:

  • What is an excitation here? I do know what is an excitation if I have a ground state: it is an excited state full stop. But the terminology used in Fermi liquid theory suggests excitations of the equilibrium state...what is that?

  • Why do we even look at these excitation things? I come from a soft matter background and if you have to deal with interactions, you do plenty of things like cluster expansions, path integral formulations followed by Saddle point or perturbative expansions, Gibbs-Bogoliubov etc... So why is it apparently more clever to look at these objects?

  • From what I understand, the logic seems to be inspired by particle physics or rather by what I learnt in basic QFT where a vacuum is the ground state of the field. Then having "particles" means that you have excitations of this ground state. The ingenious part of Landau's theory seems here to look at the statistics of excitations of the Fermi sea (instead of the statistics of the particles in the sea) and assume that they will still follow a Fermi-Dirac distribution at equilibrium. The reason why it may be nice, is because interactions will only change the ground state (energy, pressure etc...) but not necessarily the way excitations equilibrate; which will always be Fermi-Dirac like...is that it?

I understand that there are multiple questions here but it is to be more specific about my primary question which is on the "logical reasoning behind the Fermi liquid theory".


1 Answer 1


You're correct, in an undergrad statistical mechanics course we are taught that we can study metals as a free electron gas, and indeed it gives you (roughly) the correct physical quantities, like specific heat and compressibility. There are two problems with this pictures. First, the electrons are contained in a lattice. Bloch's theorem tells us that free electrons in a lattice behave as if they we're in homogeneous space, except for some amplitudes, so we're fine. Therefore let's ignore the lattice (and the phonons too) for simplicity, the so called jellium model. We now have to explain the second problem, namely that the electron-electron coulomb interaction will not change radically the free electron conclusions. This is what Fermi's liquid theory is for, as you've noticed.

Now, all equilibrium thermodynamical properties are dictated by the ground state (or vacuum in QFT language) in zero temperature, or by the ground state plus excited states with statistical weight in non-zero temperature. Let me focus first on the $T=0$ regime and then we can discuss the more general case. In the free electron model the ground state is just the Fermi surface, and the excited states are electron-hole pairs created in the surface. What about the interacting case? The key here is Gell-Mann-Low Theorem, which says (roughly) that if the interaction is weak we can connect the ground state of the non-interacting theory adiabatically with the interacting one. See, in the interacting case the coulomb repulsion will force some electrons out of the Fermi surface in a very complicated way, making the new ground state difficult to analyse with usual perturbation theory. The excitations are indeed excited states full stop, is just that without the Fermi surface as reference we don't really know who they are. The theorem allows us to calculate everything in terms of the non-interacting system using the perturbative calculations for the Green's functions. Why care about excitations at all? Well, according to linear response theory, quantities such as magnetic susceptibility and conductivity can be framed as a scattering process between excited states. Using Green's functions we can calculate everything based on the non-interacting system which is great. So it's important to look at this thing in order to have measurable physical quantities.

Now, in the free case the Green's function (in momentum-frequency space) is


where $\epsilon_k$ is the energy of the state with momentum $k$, and I'm ignoring the chemical potential for simplicity. The spectral function $A(\omega,k)=-\pi^{-1}\mathcal{Im}\{G\}$ will be just a Dirac delta, which indicates a particle behavior, that is energy with a definite momentum. Now, the interacting Green's function will be

$G(\omega,k)=\frac{Z_k}{\omega-E_k+i\gamma_k}$+ some background noise

Where $E_k$ is the energy evaluated in perturbation theory. The background noise will appear as a non-zero signal in every momentum, reflecting the lots of different electron-hole pairs, but the first part will give a spectral function which looks like a gaussian package centered at $E_k$ with a width $\gamma_k$. This looks almost like the particle distribution, hence quasi-particle, except that the width $\gamma_k$ implies that this quasi-particles eventually decay, by interaction, into the background noise. But if the width is very small then the decay times can be very long, and almost everything will seem like the free electron gas. Note that $E_k=v_F(k-k_F)$ near the Fermi surface, but $\gamma_k\propto (k-k_F)^2$, where $k_F$ is the Fermi momentum, and $v_F$ the Fermi velocity, so that for low excitations it is true that the width vanishes faster than the energy, therefore justifying the free electron gas. The weight $Z_k<1$ will tell you how much of the system behaves in this way.

It should be intuitive that this quasi-particles are still fermionic, but in any case in a finite temperature formalism, such as Matsubara's, we can strictly prove that they obey Fermi-Dirac statistics, and finite temperature proceeds smoothly from there.

So the ingenious part is that, if the perturbation is small, one can use the whole machinery from QFT to evaluate everything in terms of the non-interacting case, and the computations show that in the general case you get Green's functions that almost look like free particle ones, except for some renormalized energies and a decay probability.

Just for completeness, note that for high momentum excitations the width grows, so somewhere all those approximations break. Also, it could be the case that the coulomb repulsion is not a small perturbation, invalidating Gell-Mann-Low Theorem. In the strong interacting regime some different thing might appear, such as a Mott insulator.

  • $\begingroup$ Wow thanks, that's an awesome answer!! I will try to process a bit and accept your answer in few days if I haven't got any other reponse. $\endgroup$
    – gatsu
    Commented May 21, 2014 at 22:05
  • $\begingroup$ @gatsu, great to help, but accept the answer only if you think it resolves the question. If there is something that I did not cover there's no harm in inviting other people to have a go. Either way, I used mainly Fetter & Walecka "Quantum Theory of many-Particles Systems" as a basis, but if you're familiar with path integral I guess there's a lot more mileage in Altland & Simons "Condensed Matter Field Theory" $\endgroup$ Commented May 21, 2014 at 22:40
  • $\begingroup$ your answer is amply sufficient and definitely answers my questions and goes way beyond. The fact that I can't grasp everything of what you say is just symptomatic of my own ignorance on the subject than anything else...I have ordered the first book you talked about and it looks awesome. I therefore accept your answer. $\endgroup$
    – gatsu
    Commented May 22, 2014 at 18:10
  • $\begingroup$ Nice answer! Thanks. Could you please make a bit more clarification on your interacting Green's function? I suppose it ought to be of the form $\mathscr{G}(\vec{p},\mathrm{i}\omega_n)=\frac{1}{\mathrm{i}\omega_n-\xi_\vec{p}-\Sigma(\vec{p},\mathrm{i}\omega_n)}=\frac{1}{\mathrm{i}\omega_n-\Xi_\vec{p}-\rm{i}\cdot\rm{Im}\Sigma(\vec{p},\mathrm{i}\omega_n)}$ in Matsubara formalism, for instance. So where does $Z_\vec{k}$ and background noise come from? I haven't seen such "noise" in e.g. Mahan's book. Is it in the Fermi liquid theory? $\endgroup$
    – xiaohuamao
    Commented Oct 4, 2014 at 12:57
  • 1
    $\begingroup$ @huotuichang, sorry for the inaccuracy, you expression is indeed exact at all orders, I was trying to say that beyond first-order is customary to expand $\Sigma(p,i\omega_n)$ in order to separate the contribution to the poles and the ones that are absorbed in the renormalization of couplings, so it is just a different way to write things in order to clarify the physics. As you say it is just a variant that expresses the usual case for Fermi Liquids, although not necessary for all models. You are correct in the interpretation. $\endgroup$ Commented Oct 13, 2014 at 21:19

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