# Why do Fermi liquids have $T^2$ resistivity?

I have often read that metals that are Fermi liquids should have a resistivity that varies with temperature like $\rho(T) = \rho(0) + a T^2$.

I guess the $T^2$ part is the resistance due to electron-electron interactions and the constant term is due to impurity scattering.

Is there a simple argument to show this? Or maybe you could point me to a nice reference?

Also, it seems that for electron-electron interactions to introduce a finite resistivity, some umklapp scattering is necessary (to break Galilean and translational invariance). Is this correct? Which of these symmetries (Galilean or translational) has to be broken?

• I am looking for a better answer, but my simple understanding is as follows: $\rho\sim\Im\Sigma\sim\omega^2\sim T^2$. And $\Im\Sigma\sim\omega^2$ is what that defines the Fermi liquid behavior. – Everett You Apr 3 '13 at 12:03
• The $T^2$ scaling needs both Umklapp and electron-electron scattering. Effectively, a $O(kT)$ vicinity of the Fermi surface for quasiparticles participates in the interactions which implies the scaling, arxiv.org/abs/1204.3591 . – Luboš Motl Apr 3 '13 at 12:04
• @EverettYou: That's what I was thinking too, but where does the umklapp come in? – jjj Apr 3 '13 at 12:45
• Has someone some good references about the computation of umklapp effect in the Fermi liquid theory? – Andy Bale Jan 31 '15 at 11:17
• There are some simple "phase-space" arguments to motivate the $T^2$ dependence; have you come across them, @jjj? – AlQuemist Aug 31 '17 at 8:43

How electron-electron interaction leads to a $T^{2}$ dependence can be explained by understanding the constraints placed on electron-electron scattering by conservation of momentum and the Exclusion Principle.

Consider the fermi surface of an electron gas in 3D. The Fermi surface is a sphere of radius $k_{f}$. At finite temperatures, electrons occupy states outside the Fermi surface governed by the Fermi Dirac equation, charactarized by a shell outside the Fermi sphere with a radius that is proportional to the temperature. There are, therefore, empty states within the Fermi sphere within a shell of the same radius.

If we turn on electron-electron interactions, at small interaction strengths, we can consider it as scattering of electrons between these states in the above non-interacting picture. Electrons, being Fermions, can only occupy states that are already not occupied, along with satisfying conservation of momentum. Thus, we have to pick two electrons, both of which are on the shells of radius proportional to T, on either side of the surface of radius $k_{f}$, so that one can scatter into an empty state outside the $k_{f}$ surface and the other into an empty state in the shell inside the $k_{f}$ surface. Thus, the probability of picking two such electrons is proportional to $T^2$.

Since the contribution to resistivity is proportional to the probability of these scattering events, these interactions lead to a $T^2$ dependence in resistivity.

There are more rigorous arguments but I think this gives an intuitive picture, valid in the context of weak interactions and low temperature.

Or maybe you could point me to a nice reference?

The details behind the following answer can be found in the following arXiv paper (and references therein) arXiv:1109.3050v1.

Is there a simple argument to show this?

It appears not but I can say the following. The conductivity due to electron-electron collisions is generally given by: $$\sigma = \frac{ n \ e^{2} \ \tau_{coll} }{ m } \tag{0}$$ where $\sigma$ is the electrical conductivity, $n$ is the electron number density, $e$ is the fundamental charge, $m$ is the electron mass, and $\tau_{coll}$ is the average collision time scale (or relaxation rate). Note that the resistivity, $\eta$, is just the inverse of the conductivity in the scalar approximation.

For a Landau-Fermi liquid, the average relaxation rate for electrons on a Fermi surface can be shown to be: $$\tau_{coll}^{-1} = \frac{ \alpha \ \left( m* \right)^{3} \ \left( k_{B} \ T \right)^{2} }{ 12 \ \pi \ \hbar^{6} } \ \langle \frac{ W\left( \theta, \phi \right) }{ \cos{\left( \theta/2 \right)} } \rangle \tag{1}$$ where $\alpha$ is the efficiency of momentum transfer to the ionic lattice as a dimensionless quantity satisfying $\alpha$ < 1, $k_{B}$ is the Boltzmann constant, $\hbar$ is the Planck constant, $W\left( \theta, \phi \right)$ is the transition probability for inelastic scattering.

Quoting from the referenced arXiv paper above:

However, the fact that a solid does not possess full translation symmetry has important consequences. Already in 1937 Baber demonstrated a mechanism for finite resistivity in a two-band model in which $s$ electrons are scattered from heavier $d$ holes by a screened Coulomb interaction... single band Umklapp processes allow momentum transfer to the crystal coordinate system...

where Umklapp processes refer to electron-phonon and/or phonon-phonon scattering in a lattice. The authors also show that the term in the angle brackets can be integrated to the following: $$\langle \frac{ W\left( \theta, \phi \right) }{ \cos{\left( \theta/2 \right)} } \rangle = 12 \lambda_{\tau}^{2} \frac{ \left( \pi \ \hbar \right)^{5} }{ \left( m* \right)^{3} \ \epsilon_{F}* } \tag{2}$$ where $\lambda_{\tau}$ is a dimensionless parameter describing the interaction effective in polaron-polaron scattering and $\epsilon_{F}*$ is the the Fermi energy of the polarons. After a little algebra, we can show that: $$\frac{ \hbar }{ \tau_{coll} } = \alpha \ \lambda_{\tau}^{2} \frac{ \pi }{ \epsilon_{F}* } \left( \pi \ k_{B} \ T \right)^{2} \tag{3}$$

Thus, the resistivity is proportional to $\eta \propto T^{2}$.