Why is it necessary in Fermi liquid theory to introduce quasiparticles? I understand the notion of system where someone can turn on the interactions slowly (i.e., adiabatically), but I do not understand why there is any difference between a Fermi liquid and a system with just "normal" electrons when the interaction is not turned on adiabatically.


When Lev Landau first derived the Fermi liquid, he defined it as a many-body system of interacting electrons whose interacting eigenstates can be adiabatically connected to the eigenstates of the Fermi gas. In other words, Landau defined the eigenstates of the Fermi liquid as having a one-to-one correspondence with the eigenstates of the non-interacting system. It is for this reason that we can write down the free energy as

$$ \delta F=\frac{1}{V}\sum_{k}(\epsilon-\mu)\delta n_k+\frac{1}{2V}\sum_k f_{kk'}\delta n_k\delta n_{k'} $$

where $\delta n_k=n_k-n_k^0$, where $n_k$ is the distribution function of the interacting system and $n_k^0$ is the distribution of the non-interacting system. If there are no quasiparticles, this entire description fails by definition. The advantage of this is that we can define a lot of the thermodynamics of the interacting system with nearly the same formalism as the non-interacting case, except with re-defined parameters. For example, in a Fermi liquid, the quasiparticles have a mass $m^*$ that can be related to the bare electron mass via

$$ \frac{m^*}{m}=1+\frac{F_1^s}{3} $$

where $F_1^s$ is the $\ell=1$ channel of the symmetric Landau parameter after I expand the free energy in terms of spherical harmonics.

Another way to see this is by looking at the quasiparticle weight, $Z$:

$$ Z=\left(1-\frac{\partial \Re \Sigma}{\partial \omega}\bigg|_{\omega=0}\right) $$ where $\Re \Sigma$ is the real component of the fermionic self energy. If $0<Z<1$, then I have a finite quasiparticle lifetime. If $Z\rightarrow 0$, then I have an interacting fermionic system without well-defined quasiparticles. One example of such a model is a strange metal, which is a metallic state which lacks a quasiparticle description.

| cite | improve this answer | |
  • $\begingroup$ So because the electrons don't t have the same eigenstates (when considering the interactoins) as they have in a Fermi gas, the (electric)quasiparticles where defined as particles with the same eigenstates as electrons in a Fermi gas (but with different mass and energy). Is that correct? $\endgroup$ – gamma Aug 19 '19 at 18:08
  • 1
    $\begingroup$ @gamma The point is that I turn on the interaction "infinitely slowly" in the Fermi liquid. Therefore, I can build a one-to-one correspondence between each quasiparticle and each bare particle. You could also say that the Fermi liquid retains the same number of degrees of freedom (DoF) as the free Fermi gas. If this one-to-one correspondence between quasiparticles and particles is lost, we have either gained or loss some DoF (like in a superconductor), and thus we can no longer treat the effective excitations in the liquid state as "renormalized" fundamental particles. $\endgroup$ – Joshuah Heath Aug 19 '19 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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