|
ResistivityVsTemperature< Superconductivity.ChargeTransferInsulators | Index | Superconductivity.DopedCuprate > The big surprise is that normal state resistivity at optimal doping is linear in all cuprates. Why is a T-linear behavior so surprising? First if all impurity scattering would be independent of temperature (Matthiesen rule), hence it must be negligible in the relevant range. But the standard metal behavior is quadratic, {$$ \begin{equation}\rho=\frac {m^*}{ne^2\tau} = AT^2\end{equation}$$} not linear, as predicted by simple Fermi liquid theory. This amounts to saying that the Drude (Markovian) scattering probability (the frequency {$\tau^{-1}$}) for a metal is proportional to {$T^2$} at very low temperature. This implies that electron-phonon scattering, scaling as {$T^5$}, cannot be the dominant cause of low temperature resistivity. Indeed acoustic phonons population gives {$1/\tau_{\mathrm{ph}}\propto T^3$}, and furthermore, the tiny deviation imparted by each scattered phonon, of wavevector {$q=k_{{\mathrm B}}T/c$}, on the electron wavevector {$k_{\mathrm{F}}$} reduces the Drude probability by a factor {$1/\tau = 1/\tau_{\mathrm{ph}}(1-\cos\theta)$}, proportional to the square of the deviation angle, which is {$\theta\approx q/k_{\mathrm{F}} \approx k_{{\mathrm B}}T/ck_{\mathrm{F}}$}. Electron-electron scattering thus becomes dominant at sufficiently low temperature, and the probability is proportional to the fraction of empty final states times the fraction of filled initial states available for elastic scattering, i.e. on the crust of thickness {$k_{\mathrm{B}}T$} at the Fermi energy. Both fractions are {$k_{\mathrm{B}}T/E_{\mathrm{F}}$}, yielding {$\rho\propto T^2$}. The explanation is still not clear, although, apparently, all cuprates at optimal doping reach the so-called Planck regime, where the scattering frequency value obeys {$1/\tau(T) = k_{{\mathrm B}} T/h$}. < Superconductivity.ChargeTransferInsulators | Index | Superconductivity.DopedCuprate > |