Past exam of the quantum condensed matter course of the University of Cambridge / 2008 / Section B / B2 / b / Solution

Thomas-Fermi screening treats the carrier density locally as that of a uniform electron gas whose chemical potential is shifted by the slowly varying electrostatic potential. Linearizing gives $\delta n=(\partial n/\partial \mu )e\phi$ for electrons. The resulting screened Poisson equation has inverse screening lengthFor a degenerate gas, $\partial n/\partial \mu =g_{\mathrm{vol}}(E_F)$ including both spins, giving Thomas-Fermi screening length $k_s^{-1}$. A nondegenerate semiconductor instead has $\partial n/\partial \mu =n/(k_{B}T)$, giving the Debye length $\sqrt{\epsilon k_{B}T/(n{e}^2)}$. Electrons and holes contribute additively to the charge susceptibility when both are present; the lattice permittivity replaces the vacuum value.

The approximation requires a potential varying slowly compared with microscopic electronic length scales and a small perturbation for the linearized result. It captures smooth electrostatic screening, but not the full nonlocal response or Friedel oscillations. Low carrier density or carrier freeze-out can make a semiconductor's screening much weaker than a metal's.