Past exam of the advanced Quantum Condensed Matter Physics course of the University of Cambridge / 2013 / Paper 1 / 1 / Solution

The static Lindhard dielectric function describes the real, zero-frequency linear response of a uniform free electron gas. Here ${ϵ}_1=\mathrm{Re}ϵ$, $\mathbf{r}$ is the position operator, and $\mathbf{q}$ is the perturbation wavevector, $q=|\mathbf{q}|$, $V$ the normalization volume, $e>0$ the elementary charge, and ${ϵ}_0$ the vacuum permittivity. The normalized one-electron states are ${\psi }_{\mathbf{k}}=V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$ with $E_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$; $f(E)$ is their Fermi-Dirac distribution occupation. Both momentum sums exclude spin; the prefactor two counts the two equal spin populations. The density-operator matrix element is ${\delta }_{{\mathbf{k}}^{\prime },\mathbf{k}+\mathbf{q}}$. Coincident energy denominators are interpreted by the static limit or Cauchy principal value.

A weak total electric potential ${\varphi }_{\mathrm{tot}}$ perturbs an electron by $-e{\varphi }_{\mathrm{tot}}$. First-order state mixing produces $\delta n=e{\Pi }_0(q){\varphi }_{\mathrm{tot}}$, whereThe induced charge is $-e\delta n$. Self-consistent Poisson equation screening then gives ${\varphi }_{\mathrm{ext}}=ϵ{\varphi }_{\mathrm{tot}}$, with $ϵ=1+e^2{\Pi }_0/({ϵ}_0{q}^2)$. This is the random phase approximation: use the independent-electron response to the self-consistent total field.^[ref]

At zero temperature $f(E_{\mathbf{k}})=\Theta (k_F-k)$. Shifting the integration variable in the second occupation term givesThe two angular integrals are equal. Using the supplied integral, with $x=q/(2k_F)$, givesThe limits are $F(0)=1$, $F(1)=1/2$ and $F(x)\sim 1/(3x^2)$ for large $x$. The first recovers Thomas-Fermi screening; the logarithm produces the Kohn singularity of static Lindhard screening at $q=2k_F$.

Use the Fourier transform convention $\rho (\mathbf{q})=\int d^{3}r{e}^{-i\mathbf{q}\cdot \mathbf{r}}\rho (\mathbf{r})$. A point charge $Ze$ has total screened charge $Ze/ϵ(q)$ and induced charge $-Ze[1-ϵ(q{)}^{-1}]$. Angular inversion therefore gives the radial induced-charge screening kernelIn particular, ${\rho }_{\mathrm{ind}}(\mathbf{q}=0)=-Ze$: the total screening charge is negative although its local density can oscillate in sign.

Integrate the sine transform by parts twice; $g(0)=g(\infty )=0$ removes the boundary terms. For the Friedel density tail from a derivative singularity, write $Q=2k_F$ and $g^{\prime \prime }(q)\sim C/(q-Q)$. ThenIndeed, near $Q$ the principal value transform is ${\int }_{-\infty }^{\infty }\sin [(Q+s)r]ds/s=\pi \cos (Qr)$ for $r>0$. Smooth terms do not remove this leading oscillatory long-range contribution. Thus ${\rho }_{\mathrm{ind}}\propto \cos (2k_{F}r)/r^3$.

For a three-dimensional, single-valley gas with two spin states, the Friedel oscillation wavelength from carrier density isLow-temperature scanning tunneling spectroscopy near exposed dopants can map the oscillatory local density of states, with a Fourier feature near $2k_F$ at the Fermi energy. An energy-resolved tunnelling image measures local spectral weight, not directly the bulk energy-integrated charge: comparing occupied-bias maps and accounting for the surface geometry is necessary. In actual GaAs, the background relative permittivity and band effective mass change screening strengths; they do not change this wavelength at the stated density.