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For an isotropic completely degenerate nonrelativistic Fermi gas in classical-trajectory collisionless kinetic theory, the parallel marginal distribution is $F(u)=3n(v_F^2-u^2)1_{|u|<v_F}/(4v_F^3)$. Integrating $F^{\prime }(u)/(u-\omega /k)$ yields the electron dielectric function in the title for fixed ions. Its real roots outside the occupied interval have no Landau damping. Quantum recoil, collisions and finite-temperature tails require an enlarged model.

For a homogeneous collisionless stellar model with its mean field subtracted, the response to an imposed Newtonian gravitational potential is ${\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{tot}}={\tilde{\Phi }}_{\mathbf{k}}^{\mathrm{ext}}/{ϵ}_{\mathbf{k}}$, whereThe phase-space distribution function $f_0$ is mass-normalized, and $\mathbf{k}\ne 0$. Causal analytic continuation uses the Landau contour wherever the equilibrium permits it. Zeros describe collective Landau modes. This response function is a gravitational analogue of the electrostatic dielectric function; its susceptibility sign reflects attraction.

Collisionless damping of a collective plasma perturbation by phase mixing and resonant exchange with particles near its phase velocity. In linear electrostatic response, the imaginary part of the dielectric function at real frequency is proportional to the derivative of the equilibrium phase-space distribution function at resonance. A velocity plateau removes this contribution on an interval; compact velocity support also permits undamped waves whose phase velocities lie outside that support.

For a fixed allowed band/subband pair, parabolic dispersions give $E_c(\mathbf{k})-E_v(\mathbf{k})=E_{j,n}^{\mathrm{edge}}+{\hslash }^2{k}_{\parallel }^2/(2{\mu }_j)$. If the photon energy exceeds the edge, the contour is a circle of radius $k_{\parallel }$ and the gradient magnitude is ${\hslash }^2{k}_{\parallel }/{\mu }_j$. ThusBelow threshold the contour is absent. With the conventional $d^{2}k/(2\pi {)}^2$ normalization, the two-dimensional joint density of states per area and independent channel isWith a locally energy-independent optical matrix element, each allowed pair therefore adds a constant step to the continuum dielectric function in the approximation of the supplied equation. Different hole families need not have equal step heights: reduced masses, spin multiplicities and Bloch-state matrix elements differ. Frequency-dependent overall optical prefactors, nonparabolicity, broadening and excitonic Coulomb enhancement can modify the literal step shape outside this approximation.

The absorptive dielectric function counts optically accessible electron-hole pairs. In the electric dipole approximation the perturbation is proportional to $\mathbf{a}\cdot \hat{\mathbf{p}}$, with $\mathbf{a}$ the light polarization and $\hat{\mathbf{p}}$ the electronic momentum operator. The squared matrix element between occupied valence and empty conduction states is the transition strength; it includes polarization and band-symmetry selection rules. The sums enumerate initial and final subbands. photon momentum is negligible on the crystal-momentum scale, so the transition is approximately vertical in the in-plane wavevector $\mathbf{k}$.

For each subband pair, $C$ is the constant-transition-energy contour in the two-dimensional Brillouin zone. Its line element is $dl$, and the reciprocal gradient weights the states satisfying energy conservation. Specifically, the level-set representation of the joint density of states isA small energy gradient means many nearly resonant states; a vanishing gradient can produce a Van Hove singularity or an edge. The displayed expression suppresses overall optical prefactors and assumes the momentum matrix element is approximately constant along each contour. More generally it stays inside that integral. Occupation factors $f_v-f_c$ are also implicit; at low temperature in an undoped well the relevant valence states are full and the conduction states empty. This is the independent-particle continuum contribution from Fermi's golden rule. Bound excitons add optical resonances below the corresponding free-pair thresholds.

The GaAs four-band model is rooted in bonding and spin-orbit coupling. Near the zone centre, the conduction edge is predominantly s-like and antibonding, while the upper valence manifold is predominantly p-like. Combining orbital angular momentum $l=1$ with spin $s=1/2$ splits the six p-derived states into a $j=3/2$ quartet and a lower $j=1/2$ doublet. The latter is the split-off valence band. Within the upper quartet, the $m_j=\pm 3/2$ and $m_j=\pm 1/2$ branches, defined relative to a propagation or quantization direction, have different curvatures and are called heavy holes and light holes. They coincide at the unstrained bulk zone centre but need not remain degenerate in a well. The conduction edge is a spin doublet. Thus “four bands” counts conduction, heavy-hole, light-hole and split-off branches, including their spin multiplicities; it does not count four individual states. Positive hole masses describe the downward valence curvature, whereas the conduction curvature is upward.

For the final exciton comparison, define the reduced mass ${\mu }_j$ and screened exciton Rydberg $R_j^{\ast }$ byThe resulting $({\mu }_j/m_e,R_j^{\ast })$ are approximately $(0.05948,4.937\mathrm{meV})$ for heavy holes, $(0.03646,3.027\mathrm{meV})$ for light holes and $(0.04631,3.844\mathrm{meV})$ for split-off holes. In the ideal two-dimensional exciton model, the first two optically active internal states lie $4R_j^{\ast }$ and $4R_j^{\ast }/9$ below their continuum edge. Their separation is $32R_j^{\ast }/9$, independent of a common threshold offset. The corresponding three-dimensional separation would be only $3R_j^{\ast }/4$. The listed peak spacings specified for this theoretical spectrum giveHere each pair comprises the stronger lower-energy peak and its weaker nearby companion, in the order associated with the continuum edges calculated in part (c). The approximately 18-meV and 11-meV pair splittings support the two-dimensional series and are incompatible with the three-dimensional series at the supplied masses and permittivity. Equivalently, for a two-dimensional pair with energies $E_1,E_2$, the inferred edge is $E_2+(E_2-E_1)/8$; the observed pair splittings predict ground-state bindings around 20 and 12 meV, close to $4R_j^{\ast }$.

Absolute agreement should not be overstated. For example, the stated square-well parameters give the first heavy-hole edge $1.56390$ eV and the first light-hole edge $1.59161$ eV. Their ideal two-dimensional peak predictions are approximately $(1.54415,1.56170)$ eV and $(1.57950,1.59027)$ eV, respectively. The listed peaks are lower by about 1–2 meV across the families; rounded parameters and this simple effective-mass model do not reproduce every absolute energy exactly. No gap, width or mass has been silently adjusted to fit them. The peak-spacing comparison remains useful because it avoids a common edge offset. The split-off transitions predicted by the four-band data are discussed separately in part (c), rather than assigned fictitious extra peaks.

At 300 K, $k_{B}T\simeq 25.9$ meV, whereas the ideal ground-state binding energies are only about $19.7$, $12.1$ and $15.4$ meV; excited states are still more weakly bound. None of these series is expected to remain a clearly resolved, robust bound-exciton peak at room temperature in this simple estimate. Thermal ionization and phonon broadening work against their visibility. This is a scale argument, not a theorem that the exciton population or all Coulomb-enhanced absorption becomes exactly zero: actual observability also requires temperature-dependent gaps, linewidths, screening and carrier density, which are not supplied.

At zero magnetic field the two circular conductivities coincide, so an isotropic material has no magnetically induced circular birefringence. In the collisionless Drude model range the dielectric function is $ϵ={ϵ}_{\infty }-{\omega }_p^2/{\omega }^2$. A high carrier density gives a large plasma frequency, often placing the screened plasma edge ${\omega }_p/\sqrt{{ϵ}_{\infty }}$ in the infrared or visible range.

Below that edge the real dielectric function is negative and a transverse wave is evanescent in the bulk. The material is strongly reflecting, with penetration through a skin depth and losses set by finite $1/\tau$. Above the plasma edge propagation becomes possible, subject to absorption and interband transitions. Carrier density alone does not specify a complete optical spectrum: the statements assume the probe frequency relative to the plasma scale and the validity of the single-band Drude model.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

There is no Landau damping of this branch within the completely degenerate collisionless model. Its phase velocity obeys $\omega /k>V_e>V_i$, while each parallel marginal distribution vanishes outside $|u|\le V_{\alpha }$. There are no particles at the resonant velocity, and $F_{\alpha }^{\prime }(\omega /k)=0$. The Landau contour consequently contributes no imaginary part to the dielectric function; a real root is allowed with $\boxed{\gamma =0}$.

If a small nonzero temperature supplied a tail beyond the Fermi surface, the conclusion would change: the resonant derivative would generally be nonzero. For the convention $p=\gamma -i\omega$ and a simple weakly damped root, expanding the continued dielectric function would give $\gamma =-{ϵ}_I(\omega ,k)/{\partial }_{\omega }{ϵ}_R(\omega ,k)$. This conditional formula is not needed to manufacture damping for the exact cutoff. Neither collisional attenuation nor nonlinear evolution is included in the given dispersion relation.

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.

Use the electron electric charge $-e$, so $E=-{\partial }_x\phi$ gives acceleration $+(e/m_e){\partial }_x\phi$. The Vlasov equation, its average, and its linearized fluctuating part areThe quasilinear approximation retains the averaged quadratic term but drops products of fluctuations from the fluctuating equation. It freezes the slowly evolving $f_0$ during the fast response calculation. A real electrostatic potential has ${\phi }_{-k}={\phi }_k^{\ast }$, and the sum includes both signs of every nonzero wavenumber. Spatial averaging selects the opposite Fourier series harmonics; statistical homogeneity supplies the corresponding diagonal noise covariance.

Let $\hat{g}(p)={\int }_0^{\infty }{e}^{-pt}g(t)dt$. Zero initial $\delta f_k$ givesInsert this into the electrostatic self-consistency relation. Using ${\omega }_{pe}^2=4\pi e^2{n}_e/m_e$ and $p+ikv=ik(v-ip/k)$ givesThis derives the stated dielectric function with the Landau contour. A noise-free growing collective response is a zero of $ϵ$.

The Laplace transform of the white noise covariance is particularly simple. Writing $q=p^{\prime \ast }$,For clarity about contour signs, introduce the inverse-transform measuresThe first Bromwich inversion formula contour runs upward on $\mathrm{Re}p=\sigma$; its conjugate contour runs downward on $\mathrm{Re}q=\sigma$. Both measures therefore have the same positive orientation in their real frequency parameters. Combining the averaged equation with ${\widehat{\delta f}}_k$ yieldsAt a zero of $f_0^{\prime }$, the last expression means its continuous coefficient obtained before cancelling $f_0^{\prime }$. Substitution of the two inverse Laplace transforms and the transformed covariance givesThis is the paper's expression, with the orientation of $d{p}^{\prime \ast }$ made explicit. Reality follows by pairing $k$ and $-k$. The double integral retains switch-on transients; replacing it by a time-independent diffusion coefficient requires the additional long-time limit in b.

A linearly stable $f_0$ has no zeros of the causal dielectric function in $\mathrm{Re}p>0$. Hence the inverse Laplace transform contours can approach $\sigma =0^{+}$, retaining the causal boundary value at $p=-i\omega +0^{+}$. To obtain a stationary forced response, further assume that the excited collective perturbations actually damp. This is the decay condition in the question's long-time instruction.

A transparent evaluation of the double integral uses the stationary power spectral density. The white noise covariance $2A_k\delta (t-t^{\prime })$ has temporal spectrum $2A_k$. Linear filtering by $1/ϵ$ therefore givesThe causal solution for $\delta f_k$ in a gives, after the initial transients have decayed,The real part can be taken because the full $k$ sum pairs conjugates. Using the distribution theory identity $\mathrm{Re}{\int }_0^{\infty }{e}^{ia\tau }d\tau =\pi \delta (a)$, with a positive vanishing convergence factor, samples the temporal spectrum at $\omega =kv$:The factor $1/2$ from the one-sided time integral cancels the $2$ in the noise covariance. The result is nonnegative, as a stationary velocity-space diffusion coefficient should be.

Qualification of stability. Absence of growing modes alone does not guarantee this stationary limit. For example, the electron waterbag distribution $f_0(v)=n_e{1}_{|v|<V}/(2V)$ hasas undamped collective roots. White noise drives their field variance to grow instead of settling. Thus the formula requires decay of excited modes, not merely absence of exponentially growing ones. For weak Landau damping, the waiting time can be much longer than ${\omega }_{pe}^{-1}$, and must still be short compared with evolution of $f_0$ for the frozen-background quasilinear approximation to apply.

Set $\kappa =|k|$ and first work with a representative positive $k=\kappa$. For $\kappa {\lambda }_{\mathrm{De}}\ll 1$, the thermal core is strongly screened. However, the causal dielectric function has weakly damped Langmuir wave roots ${\omega }_s+i{\gamma }_s$, with $s=\pm 1$, ${\omega }_s\simeq s{\omega }_{pe}$ and ${\gamma }_s<0$. Their phase velocities $v_s={\omega }_s/\kappa$ are in the tail. Both roots must be retained for each spatial Fourier series harmonic.

Write $ϵ(-i\omega +0^{+},\kappa )={ϵ}_R(\omega )+i{ϵ}_I(\omega )$. The supplied approximation givesExpanding $ϵ({\omega }_s+i{\gamma }_s)=0$ to first order gives the Landau damping rateIn particular, stable damping means ${\omega }_s{f}_0^{\prime }(v_s)<0$. Near either root,The narrow Lorentzian resonance approximation then gives the leading resonant contributionThis is a formula under the integral over velocity; it is not a pointwise approximation away from the resonances. Insert it into the heating formula in c. For one spatial harmonic, the contribution from one root isbecauseThus the tiny tail slope cancels the tiny Landau damping rate: weak absorption is compensated by a large noise-driven wave intensity. The two temporal roots supply two equal halves of the answer. Negative $k$ gives the same result by the conjugate response of a real electric field, without altering the original convention that the spatial sum includes both signs. ThereforeAs in c, equality means the leading result in the stated asymptotic limit. This calculation assumes nonzero damping, observation times $t\gg {\max }_s|{\gamma }_s{|}^{-1}$, and forcing weak enough that $f_0$ changes negligibly during that relaxation. Taking zero damping before the long-time limit invalidates the stationary calculation.

In the periodic cube, $H_0=|\mathbf{v}{|}^2/2=(2\pi /L{)}^2|\mathbf{J}{|}^2/2$, so $\mathbf{\Omega }=(2\pi /L)\mathbf{v}$. The supplied action-angle variables are canonical: $d^3\theta d^{3}J=d^{3}x{d}^{3}v$. For $\mathbf{k}=2\pi \mathbf{n}/L$,Consequently c becomesFor each nonzero wavenumber, the Poisson equation for Newtonian gravity givesSubstitute the preceding response and add the externally imposed Newtonian gravitational potential:ThereforeThis gravitational dielectric function has the opposite susceptibility sign to the electron dielectric function, reflecting attractive rather than repulsive collective interactions. The homogeneous mean field is explicitly subtracted in this fictitious periodic model: the Poisson equation for Newtonian gravity acts on density fluctuations, and the $\mathbf{k}=0$ mode is omitted. Otherwise a nonzero uniform positive mass density would be incompatible with a periodic potential. This is the convention often called the Jeans swindle.

For analytic continuation, resolve velocity parallel to $\mathbf{k}$: let $u=\hat{\mathbf{k}}\cdot \mathbf{v}$ and $F(u)=\int f_0(u,{\mathbf{v}}_{\perp })d^2{v}_{\perp }$. ThenStart at $\mathrm{Im}\omega >0$ and deform the Landau contour continuously below the moving pole. At real $\omega$, the integral isFor $\mathrm{Im}\omega <0$, this causal continuation equals the undeformed real-axis integral plus $2\pi i{F}^{\prime }(\omega /k)$ wherever $F^{\prime }$ has the necessary analytic extension. It must not be replaced by the undeformed integral alone.

Zeros of ${ϵ}_{\mathbf{k}}$ are self-consistent collective Landau modes: with no external forcing, a nonzero Newtonian gravitational potential is possible at such a root. For time dependence $e^{-i\omega t}$, upper-half-plane roots grow, lower-half-plane roots describe Landau damping, and real roots are neutral. Damped Landau modes are poles of the continued response; they need not be regular discrete eigenfunctions of the original collisionless operator.

Regularity qualification. Continuation throughout the complex plane presupposes suitable analytic $F$. Arbitrary smooth $F$ need not allow it. For example, $F(u)=C\exp [-1/(1-u^2)]$ for $|u|<1$ and zero elsewhere is smooth but not analytic at $u=\pm 1$; its Cauchy integral cannot be holomorphically continued across those endpoints. The Landau contour gives the standard causal continuation on the domains where the equilibrium permits it, with possible singularities retained.

Let $A_0=f_0(0)$, $\delta =v_2-v_1$, $V=v_0+\delta$ and $K=k^2{\lambda }_{\mathrm{De}}^2$. Use the Debye length convention of the paper, ${\lambda }_{\mathrm{De}}=v_0/{\omega }_{\mathrm{pe}}$. On the candidate intervals in a, the Landau contour contributes no imaginary part. The piecewise constant slopes giveSince $n_0=v_0{A}_0$, setting the dielectric function to zero yieldsThe absolute values express the real response on the allowed intervals. Solving this real equation inside a sloping portion of $f_0$ would not satisfy the imaginary part of the dielectric function.

An exact algebraic form will also make the number of undamped branches clear. DefineOn each allowed interval, the quantity inside the logarithm equals $u^2(u-v_2)/[(u+v_0)(u-v_1)(u-V)]$. Exponentiating and usinggives the cubic equationFor every $k>0$, the logarithm runs from $0$ to $+\infty$ on $(-\infty ,-v_0)$, from $+\infty$ to $-\infty$ on $(v_1,v_2)$, and from $+\infty$ to $0$ on $(V,\infty )$. The intermediate value theorem supplies a root in each interval, and the cubic equation permits at most three. Thus there is exactly one undamped branch in each interval.

For the two outer branches, $|u|\gg v_0$ at small $k$. An integration by parts and a large-$u$ expansion giveThere is no boundary contribution because the phase-space distribution function vanishes at both endpoints. Direct area integration givesThus the exact small-$k$ cutoff, retaining the paper's definition $n_0=v_0{A}_0$, is $\omega \to \pm {\omega }_{\mathrm{pe}}\sqrt{n_{\ast }/n_0}$. To leading order in the narrow-velocity plateau approximation this isThese are Langmuir waves. The difference between $n_{\ast }$ and the reference $n_0$ is only first order in $\delta /v_0$, so it does not alter the requested leading-order result.

For the finite-phase velocity branch put $u=v_1+\delta a$, with $0<a<1$. To leading order in $\delta /v_0$ and $\delta /v_1$, the real dispersion relation becomesAt long wavelength set $K=0$ to leading order. Then $a=v_1^2/v_0^2$, andFor comparison, the exact $k\to 0$ phase velocity from the cubic equation in b is $B/A$; expanding it gives precisely $v_1+\delta v_1^2/v_0^2+O({\delta }^2/v_0)$ at fixed $v_1/v_0$.

A single Maxwell-Boltzmann velocity distribution has a nonzero slope at every finite nonzero $u$ and has no velocity plateau. The would-be electron acoustic wave is therefore subject to strong Landau damping, so there is no corresponding completely undamped acoustic branch in a single Maxwellian electron population with immobile ions. At its only stationary point $u=0$, a Maxwellian has $\int f_0^{\prime }(v)/vdv<0$, making the static dielectric function positive rather than zero. This statement does not exclude acoustic modes of suitably distinct electron populations or an ion-supported mode when ion dynamics is retained.

A weak-fluctuation closure retaining the averaged quadratic fluctuation flux while linearizing the fluctuating kinetic equation about a slowly evolving mean phase-space distribution function. One solves the fast linear response with the mean frozen, then inserts its covariance into the slow averaged equation. A short-memory stationary limit can yield quasilinear diffusion, but finite-time responses retain memory. Self-consistent collective fields enter through the dielectric function in homogeneous electrostatic problems.

The static independent-electron density response of a uniform three-dimensional, two-spin Fermi gas, combined with self-consistent Poisson equation screening, gives the displayed dielectric function. For bare electron mass and vacuum background, ${\kappa }^2=m_e{e}^2{k}_F/({ϵ}_0{\pi }^2{\hslash }^2)$ andHere $F(0)=1$, $F(1)=1/2$ by continuous extension, and $F(x)\sim 1/(3x^2)$ at large $x$. A host dielectric background and band effective mass require replacing the vacuum parameters appropriately. This nonlocal response contains Friedel oscillations, which its long-wavelength Thomas-Fermi screening limit misses.

An interval of particle velocities on which an equilibrium phase-space distribution function is constant. Its zero derivative removes the resonant imaginary part of the linear dielectric function for waves with phase velocity inside the interval. This permits a completely undamped electron acoustic wave in suitable single-population, nonthermal electron distributions. Zero slope alone is necessary but does not guarantee that the real dispersion relation has a root.