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The long-wavelength root of the degenerate-plasma dielectric response. Expanding its logarithm at large $s=\omega /(k{v}_F)$ gives $ϵ=1-{\omega }_p^2/{\omega }^2-3{\omega }_p^2{k}^2{v}_F^2/(5{\omega }^4)+\cdots$. The occupied Fermi sphere has $⟨v_{\parallel }^2⟩=v_F^2/5$, accounting for the dispersive coefficient. The static electron response instead defines the Thomas-Fermi screening length $v_F/(\sqrt{3}{\omega }_p)$.

Let $\kappa =k{V}_e/{\omega }_{\mathrm{p}e}$ and $s=\omega /(k{V}_e)$. The long-wavelength electron branch has $|s|\gg 1$. The convergent power seriesturns the degenerate-plasma dielectric response intoSubstitute ${\omega }^2={\omega }_{\mathrm{p}e}^2+a{k}^2{V}_e^2+O(k^4{V}_e^4/{\omega }_{\mathrm{p}e}^2)$. The coefficient of $k^2$ is $a-3/5$, so the degenerate-plasma Langmuir dispersion isThe Langmuir wave therefore has a finite frequency at zero wavenumber, with the first dispersive correction set by the Fermi velocity. Indeed the occupied Fermi sphere has $⟨v_{\parallel }^2⟩=V_e^2/5$, so the coefficient $3⟨v_{\parallel }^2⟩=3V_e^2/5$ agrees with the high-phase velocity velocity moment expansion.

Screening-length convention. The source defines ${\lambda }_{\mathrm{D}e}^{\mathrm{exam}}=\sqrt{3}{V}_e/{\omega }_{\mathrm{p}e}$, so its long-wavelength condition is equivalent to $\kappa \ll 1$. We retain that definition when expressing answers in its symbols. The actual electron Thomas-Fermi screening length obtained from the static response is insteadFor a concrete comparison, at $k={\omega }_{\mathrm{p}e}/V_e$, the static dielectric function is $4$; using the printed length as the physical Thomas-Fermi screening length would give $1+1/3=4/3$. Calling the printed length the Debye length thus differs from the physical degenerate-electron screening convention by a factor of three. It does not affect the boxed dimensional dispersion relation.

For the positive-frequency branch write $s=1+\delta$, $\delta >0$, and $Q=k^2{V}_e^2/(3{\omega }_{\mathrm{p}e}^2)={\kappa }^2/3$. The degenerate-plasma dielectric response is equivalent toAt large $Q$, a logarithmic divergence is needed to balance the right-hand side; hence $\delta \ll 1$. Keeping the constant as well as the large term givesMore precisely the relative correction is $O((1+Q)e^{-2(1+Q)})$. The additive constant $-2$ in the exponent fixes the leading prefactor $2e^{-2}$ and should not be discarded.

The negative-frequency branch follows from the even real dielectric function outside the occupied interval. ThusUsing the physical Thomas-Fermi screening length instead gives $\delta \sim 2e^{-2-2(k{\lambda }_{\mathrm{TF}}{)}^2}$. This is degenerate-plasma zero-point sound: the phase velocity lies exponentially close to, but strictly beyond, the edge of the Fermi sphere for every finite $k$ in this approximation. The limit is a property of the supplied classical-trajectory kinetic theory; applying it at microscopic wavelengths would require checking quantum recoil, which that equation omits.

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.

In a conductor, mobile electrons redistribute in response to an electric potential, producing an induced charge density that opposes its source. Finite electronic compressibility means a finite density change requires a change of local kinetic Fermi energy. Screening is therefore not perfect at every microscopic distance. On scales comparable to atomic structure or the Fermi wavelength, a continuum local-density model itself fails.

Thomas-Fermi screening treats electrons as a locally uniform degenerate free electron gas, replaces the ionic background by a smooth neutralizing charge, and neglects nonlocal response, exchange and correlation corrections. The electrochemical chemical potential is uniform in equilibrium. Because an electron has charge $-e$, its electrostatic energy is $-eV$, givingAt zero potential $\mu =E_{F0}$. Linearizing at fixed $\mu$, with $|eV|\ll E_{F0}$, givesInsert this in Poisson's equation to obtain $({\nabla }^2-{\xi }^{-2})V=-{\rho }_{\mathrm{free}}/{ϵ}_0$, where the Thomas-Fermi screening length isAway from a point source, direct radial differentiation shows ${\nabla }^2(e^{-r/\xi }/r)={\xi }^{-2}{e}^{-r/\xi }/r$. The decaying solution is therefore $V=C{e}^{-r/\xi }/r$. Integrating Poisson's equation over a small sphere containing charge $Q$ fixes $C=Q/(4\pi {ϵ}_0)$. For a sufficiently large source, the linearized formula is only a far-field approximation, not a justified description of its singular core.

For monovalent sodium, the electron number density is its atomic number density,This gives $E_{F0}=3.15\mathrm{eV}$ andRoom temperature is far below the corresponding Fermi temperature, so degeneracy is well justified, and sodium's simple conduction band makes a free-electron estimate useful. However $k_F\xi =0.615$ and the mean atomic spacing is $n_0^{-1/3}=0.340\mathrm{nm}$. The potential varies on a scale shorter than $k_F^{-1}$ and the atomic spacing, so the slowly varying local-density and continuum assumptions are not quantitatively controlled at the screening length itself. A nonlocal treatment includes Friedel oscillations, absent from the pure exponential.

Nearly unchanged interlayer force constants in two-layer and twenty-layer films are consistent with very local screening and bonding forces: reducing the distant bulk need not strongly change the force between adjacent layers. They also indicate that metallic electronic response can survive in the thin film. The observation does not by itself validate the numerical Thomas-Fermi approximation, establish an identical full phonon spectrum, or rule out surface-induced electronic changes; it directly constrains the measured local force constant.

Use Thomas-Fermi screening, treating a homogeneous three-dimensional conductor in static linear response. For a positive test charge the screened electric potential and bare Coulomb potential areHere ${\epsilon }_b$ is the background permittivity before the mobile carriers screen; it is ${ϵ}_0$ in the simplest free-electron model. Both curves have the same leading $1/r$ divergence near the origin, while the screened curve lies below the bare one and decays much faster.

A positive potential attracts electrons. If $g(E_F)$ is their total spin-summed density of states per volume and energy, $\delta n=eg(E_F)V$ and ${\rho }_{\mathrm{ind}}=-e^{2}g(E_F)V$. Inserting this charge into Poisson equation gives the Thomas-Fermi screening lengthA larger density of states lets the conductor rearrange more charge for the same potential change and shortens the screening length. For a parabolic free electron model, $g(E_F)=3n/(2E_F)$, hence ${\lambda }_{\mathrm{TF}}^2=2{\epsilon }_b{E}_F/(3n{e}^2)$.

The exponential form is the intended local screening model. A microscopic zero-temperature metal also has Friedel oscillations, so the exponential alone is not an exact description at every length scale.

The Thomas-Fermi screening length is the exponential decay scale of a statically screened potential in the local degenerate-electron approximation. For a three-dimensional parabolic free electron model, $g(E_F)=3n/(2E_F)$, so ${\lambda }_{\mathrm{TF}}^2=2{\epsilon }_b{E}_F/(3n{e}^2)$. At fixed mass and background permittivity it scales as $n^{-1/6}$, rather than the classical Debye density power at fixed temperature.