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In a uniform three-dimensional Hartree-Fock electron gas with repulsive contact coupling $\gamma$, a spin-independent neutralizing background can cancel the Hartree energy. The remaining same-spin exchange term is negative and quadratic in each spin density. Integrating the filled single-spin Fermi sphere gives $A=3{\hslash }^2(6{\pi }^2{)}^{2/3}/(10m)$. At fixed total density the exchange term favors spin polarization, whereas the kinetic term opposes it.

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 $v_F/(\sqrt{3}{\omega }_p)$.

Use normalized plane-wave spin orbitals $V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}{\chi }_{\sigma }$ and fill a separate Fermi sphere for each spin species. Each spin state counts once, soIts kinetic energy per volume isThe spin-independent contact interaction has uncancelled direct energy $E_H/V=\gamma n^2/2$, with $n=n_{\uparrow }+n_{\downarrow }$. By the imposed background assumption this direct contribution cancels. For the Fock term the contact delta function sets the two spatial positions equal; its spin factors then select equal spins. The local one-body density matrix for each spin has diagonal value $n_{\sigma }$, givingCombining these results gives the contact-interaction Hartree-Fock spin energyHere $\gamma$ has units of energy times volume. The result is the model with the stated background subtraction; an isolated contact-interacting gas without that subtraction retains its direct contribution.

At fixed total density, the unpolarized and fully polarized candidate energies arePolarization increases the kinetic energy because one spin species must occupy a larger Fermi sphere, but gains negative exchange energy. The fully polarized candidate wins this endpoint comparison ifUsing the defined $r_s$ givesThus the specified two-state comparison indeed predicts a polarized state below the unpolarized state's energy for $r_s>1.175$.

The endpoint and continuous polarization thresholds of a contact electron gas are different. To check the actual onset within the same uniform Hartree-Fock model, allow $P=(n_{\uparrow }-n_{\downarrow })/n$ to vary continuously. In units $2A(n/2{)}^{5/3}$,The unpolarized state loses local stability already at $r_s=10/9\simeq 1.111$. Full polarization becomes the minimizing endpoint only when $f^{\prime }(1)\le 0$, or $r_s\ge (5/6)2^{2/3}\simeq 1.323$. Between those values the optimum is partly polarized. Consequently $1.175$ is the requested endpoint crossing, not the exact onset of spontaneous polarization after all spin populations are allowed.

The Fermi sphere volume in reciprocal space isThe primitive real-space cell has volume $a^3/2$, so the first Brillouin zone has volumeEach band accommodates two electrons per primitive cell when spin is counted. Six electrons therefore require three such zone-volumes of occupied reciprocal-space states in the free-electron counting picture. The extended-zone sphere is not confined to the first zone; folding it produces multiple bands. It does not imply that the actual tungsten Fermi surface is spherical.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

In the one-band electron Hall effect model, $R_H=-1/(ne)$, with $e>0$. The measured Hall coefficient therefore givesA monovalent body-centered cubic lattice has two conduction electrons per conventional unit cell. Thus $n{a}^3=2$, andCounting two spin states within the Fermi sphere gives $k_F=(3{\pi }^{2}n{)}^{1/3}=9.0418\times 10^9{\mathrm{m}}^{-1}$. Using the specified effective mass in the dispersion,There is one mole of conduction electrons per mole of sodium. Its molar density of states at $E_F$ is $3N_A/(2E_F)$, givingUsing the volume density of states in c givesThe effective mass changes the dispersion and density of states; the electron Bohr magneton used for the spin coupling retains the bare electron mass. Replacing it by $e\hslash /(2m^{\ast })$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.

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.

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.

The two real branches start at $\omega =\pm {\omega }_{\mathrm{p}e}$, have zero initial slope, and approach the lines $\omega =\pm k{V}_e$ from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient $3V_e^2/5$ in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has $d{P}_e/d{n}_e=m_e{V}_e^2/3$, suggesting, at long wavelengths,Both species are degenerate here, so $V_i=(m_e/m_i)V_e$. The hierarchy $V_i\ll c_s\ll V_e$ explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at $u=c_s$, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order $\sqrt{m_e/m_i}$, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to $c_s^2$. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

Write the lattice reciprocal vector as $G=2\pi /a$ in one dimension. The nearly-free electron model starts from ${\epsilon }_k={\hslash }^2{k}^2/(2m_e)$ and couples plane waves separated by G. Near a zone boundary, the two-wave Hamiltonian isAt $k=G/2$ the free waves are degenerate and the periodic potential opens a gap $2|V_G|$. A reduced-zone drawing folds k modulo G into $[-G/2,G/2]$, displaying multiple bands at each reduced wavevector. The original plot shows free folded parabolas, avoided crossings and an allowed vertical optical transition. ^[ref]

Including two spin states, filling a free Fermi sphere gives $n=2(4\pi k_F^3/3)/(2\pi {)}^3=k_F^3/(3{\pi }^2)$. A monovalent BCC cell contributes two electrons, soThe sphere therefore lies inside every first-zone face, since the supplied distance is the nearest boundary.

For the shortest BCC reciprocal vector, $G=2\sqrt{2}\pi /a$. In the zero-potential folded-band limit, the energy of a vertical transition from k along G to the neighbouring free branch is ${\hslash }^2(G^2-2Gk)/(2m_e)$. It decreases over the occupied interval and is smallest at $k=k_F$. Thus the monovalent BCC direct optical threshold isThe final state is empty because $G-k_F>k_F$. Finite weak $V_G$ changes the energies and makes optical matrix elements nonzero; an exactly free electron gas by itself has no interband absorption. The expression is a leading weak-potential threshold, not a claim that every peak equals the onset.

Using the supplied lattice constants and bare-electron free bands gives $E_F=(3.24,2.12,1.85)\mathrm{eV}$ and thresholds $(2.06,1.35,1.18)\mathrm{eV}$ for Na,K,Rb. The source plot's broad higher-frequency features shift toward lower energy in the same order as a increases, qualitatively consistent with the predicted $a^{-2}$ scale. Their peak shapes, heights and exact positions need matrix elements, band corrections and scattering; low-frequency intraband conductivity is a separate contribution. No measured trace is reproduced.

Count two electron spin states in each reciprocal-space cell of volume $(2\pi {)}^3/V$. Filling a Fermi sphere at zero temperature givesThe free electron model then givesThe energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.