At a temperature \(T \ll T_F\), the quasiparticles will be excited at an average energy of \(\epsilon = k_B T\) above the Fermi surface. Therefore, their average lifetime will be proportional to \(1 / T^2\), meaning that the conductivity of a Fermi Liquid scales as: \[\sigma_{\mathrm{FL}} \sim \frac{1}{T^2} \Leftrightarrow \rho \sim T^2\]

Before concluding that conductivity scales as \(1 / T^2\), there’s one element left to show. A non-infinite conductivity requires both a decay time and an energy loss mechanism. In the current form, the energy the quasiparticle loses goes back to the Fermi sea, effectively accelerating it. What leads to energy loss is that fraction of collisions that bring the crystal momentum of the quasiparticle outside the first Brillouin zone, also known as Umklapp scattering\cite{pal2012}. In essence, whenever the Fermi sea accelerates to a crystal momentum that’s outside the first Brillouin zone, this crystal momentum gets translated by a lattice translation vector back into the first Brillouin zone. The energy lost by the Fermi sea is then given to the crystal lattice.¹

Therefore, the \(1 / T^2\) scaling of conductivity comes from combining two processes. The first process is one that is biased towards keeping the Fermi sphere intact, resulting in a decay time of quasiparticles with an energy \(\epsilon\) above the Fermi energy proportional to \(1 / \epsilon^2\). The second process deals with the loss of energy of the Fermi sphere itself, which happens via Umklapp scattering.

Quasiparticle–impurity scattering

In a perfect crystal, the scattering between quasiparticles and individual crystal sites is already accounted for in the previous argument, with energy transfer happening via Umklapp scattering. However, crystal in nature are not perfect. The impurities, be it lattice defects or other elements act as individual scattering centres for the quasiparticles.\cite{nozieres1999}

Assuming that the matrix element of quasiparticle–impurity scattering is roughly constant, the scattering cross–section of this process is proportional to the density of defects \(n_\mathrm{d}\). Since the density of defects is weakly temperature dependent (most defects being topologically protected), this process gives a conductivity (and thus resistivity) which is constant in temperature.

Therefore, combining the two mechanisms outlined so far gives a Fermi Liquid resistivity of \[\rho = \rho_0 + A T^2 \mathrm{ ,}\] form which is valid at low temperatures, \(T \ll T_F\). At higher temperatures with a large population of phonons, quasiparticle–phonon scattering starts to become important as well, dominating over the electronic contribution.

Quasiparticle–phonon scattering

From equipartition, at high temperatures there’s an energy proportional to \(k T\) in each phonon mode. Assuming that the scattering cross section of each quasiparticle is proportional to the energy available in each phonon mode, we get a lifetime \(\tau_\mathrm{phonon}\) proportional to \(1 / T\). A more careful calculation\cite{Lawrence_Wilkins_1972} starting from the Boltzmann transport equation gives the same scaling.

Therefore, we can conclude the following: the resistivity, inversely proportional to the conductivity, has the following behaviour: \[\rho = \rho_\mathrm{impurity} + A_\textrm{qp-qp} T^2 \quad T \ll T_F\] \[\rho = \rho_\mathrm{impurity} + B_\textrm{qp-ph} T \quad T \gg T_F\]