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The three-dimensional static screening function is finite at $q=2k_F$, but its derivative diverges logarithmically. Expanding the explicit static Lindhard dielectric function gives the title's nonanalytic term. The singularity comes from particle–hole excitations connecting opposite parts of the Fermi surface. It produces a Friedel density tail from a derivative singularity; analogous electronic response can cause a Kohn anomaly in a coupled lattice vibration.

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.