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For a nonrelativistic zero-temperature free electron model with fixed particle number, electron degeneracy pressure obeys $P\propto (N/V{)}^{5/3}$. Consequently its bulk modulus is $K=-V(\partial P/\partial V{)}_N=5P/3=2n{E}_F/3$. At temperatures small compared with the electron Fermi temperature, the leading isothermal compressibility is $1/K$. This electronic contribution alone omits ionic binding and interaction corrections in a real metal.

The free electron model approximates conduction Electrons by noninteracting particles with the displayed parabolic dispersion relation, filling one-particle states according to Fermi-Dirac distribution. A uniform positive background supplies charge neutrality. Periodic boundary conditions give one spatial state per $k$-space volume $(2\pi {)}^d/V_d$ in $d$ dimensions; the physical spin degeneracy is two. The model captures basic state counting and degeneracy but omits the periodic lattice potential, interaction corrections and detailed electronic band structure.

An atom is monovalent in a specified bonding or conduction model when its relevant valence is one. In the simple metallic free electron model, one monovalent atom per primitive unit cell supplies one conduction Electron per cell. Counting the two spin states then fills half of a single energy band at zero temperature.

For fixed Electron number $N$, the specified electron degeneracy pressure scales as $P\propto V^{-5/3}$. Differentiating at fixed low temperature gives the degeneracy-pressure bulk modulus,A monovalent sample has one conduction Electron per atom. Its carrier density isusing the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$. Substitution with reduced Planck constant $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$ givesEquivalently, $P=2n{E}_F/5$ and $K_T=2n{E}_F/3$ for this free electron model. This is the assumed electronic contribution, not a calculation of the complete equation of state of real potassium: ionic cohesion and interaction corrections have been excluded as instructed. The supplied pressure is the zero-temperature form, appropriate to leading order for $T\ll T_F$.

Use the quantum free electron model: conduction Electrons move as independent particles in a three-dimensional box of volume $V$, with a uniform positive background replacing the ionic lattice potential. Neglect the detailed periodic potential and explicit electron-electron interactions; the interior one-electron dispersion relation is $ϵ={\hslash }^2{k}^2/(2m_e)$. The boundary confines the particles, and its detailed shape does not affect the bulk state count in a sufficiently large sample.

The Electrons are spin-$1/2$ fermions: each spatial quantum state accommodates two opposite spin states, with occupations given by the Fermi-Dirac distribution and the Pauli exclusion principle. A fixed number $N$ of conduction Electrons and thermal equilibrium are assumed. Collisions can be represented by a phenomenological momentum relaxation time for transport, but are unnecessary for the equilibrium state counting here. These assumptions distinguish the quantum model from a classical Maxwell-Boltzmann Electron gas.

The occupation probability of an individual one-electron state is the Fermi-Dirac distribution,At $T=0$, it is a step: one below $E_F$ and zero above it. Heating rounds the step over an energy range of order a few $k_{B}T$, creating empty states below and occupied states above the chemical potential. At every positive temperature, $f(\mu ,T)=1/2$.

For fixed $N,V$ in the three-dimensional free electron model, the chemical potential is determined by $N=\int g(ϵ)f(ϵ,T)dϵ$; it is not exactly constant. In the degenerate regime $T\ll T_F$, the Sommerfeld expansion givesIt is therefore reasonable in a simple low-temperature sketch to set $\mu \simeq E_F$, provided this approximation is stated. The figure includes the leading small shift. At much higher temperature the fixed-$E_F$ approximation fails and the distribution approaches the classical dilute limit. The plotted occupation is $f$, rather than the energy distribution $gf$.

The quantum free electron model treats conduction Electrons as independent particles moving in an approximately constant potential, with parabolic dispersion relation $E={\hslash }^2{k}^2/(2m)$. Fermi-Dirac statistics successfully explains important features such as a Fermi energy and a small electronic heat capacity, but a constant potential gives no band gaps and cannot by itself explain why different band fillings produce metals, semiconductors and band insulators. It also misses lattice-dependent effective masses, detailed directional responses and the topology of real Fermi surfaces. Scattering and finite electrical resistivity require additional physics.

The nearly-free electron model introduces a weak periodic lattice potential. Bloch's theorem replaces individual free plane waves by Bloch states, and Fourier components of the potential mix plane waves whose wavevectors differ by a reciprocal lattice vector. Near degeneracy at a Brillouin zone boundary, this coupling opens a band gap; for a two-wave crossing its size is $2|V_G|$. Away from the crossing, energies remain close to free-particle values. Band curvature produces band effective masses, and band filling determines whether the Fermi energy lies inside an allowed band or a gap.

The improvement is the inclusion of lattice periodicity, rather than a switch from classical to quantum statistics: the quantum free electron model already uses Fermi-Dirac statistics. The strong delta barriers in the later numerical example are treated exactly by the Kronig-Penney model, rather than by assuming that their strength is perturbatively small.

In a metal, a partially filled energy band or overlapping energy bands supplies Electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile Electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the Electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One monovalent atom per unit cell suppliesElectrons. At zero temperature, the Pauli exclusion principle fills the lowest energy states, with two spin states per spatial wavevector. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space density of states gives, in one dimension,In two dimensions,In three dimensions,Each Fermi energy follows from the free electron model dispersion relation $E_F={\hslash }^2{k}_F^2/(2m)$. The Fermi wavevectors have units ${\mathrm{m}}^{-1}$ and the Fermi energies have units joules for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in electronvolts cannot be uniquely calculated. The spin factor of two is necessary here for actual monovalent crystals; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.

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.

Work at zero temperature and fixed total carrier density. Write $n_{\uparrow }=n/2+\delta$ and $n_{\downarrow }=n/2-\delta$. Let $g_s(E_F)$ be the single-spin density of states per unit volume in the unpolarised state. Since $d{\mathcal{E}}_s/d{n}_s=E_{Fs}$ and $d{E}_{Fs}/d{n}_s=1/g_s$, the linear kinetic-energy shifts cancel and the quadratic cost isThe repulsive interaction instead decreases under spin polarization:Hence the Stoner criterion for an opposite-spin contact repulsion isThe two forms differ only in whether the stated density of states counts one spin or both. For the three-dimensional free electron model, $g_{\mathrm{tot}}(E_F)=3n/(2E_F)$, soEquivalently, for fractional polarisation $\zeta =(n_{\uparrow }-n_{\downarrow })/n$, the kinetic-energy density is $\frac{3}{10}n{E}_F[(1+\zeta {)}^{5/3}+(1-\zeta {)}^{5/3}]$; its quadratic coefficient is $n{E}_F/3$, while the interaction contributes $-U{n}^2/4$. With number densities, $U$ has units of energy times volume and the displayed interaction is an energy density; multiplying by the sample volume does not change the threshold.

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two Electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

For an interaction-energy density $U{n}_{\uparrow }{n}_{\downarrow }$, let $n_{\uparrow }=n/2+\delta$, $n_{\downarrow }=n/2-\delta$ at fixed total carrier density. If $g_s(E_F)$ is the unpolarized single-spin density of states per unit volume, the kinetic-energy cost of the two occupation shifts is ${\delta }^2/g_s(E_F)$, whereas the interaction changes by $-U{\delta }^2$. ThereforeFor the three-dimensional free electron model, $g_{\mathrm{tot}}(E_F)=3n/(2E_F)$ and $U_c=4E_F/(3n)$. Both the interaction normalization and the spin counting in the density of states must be specified when quoting a Stoner criterion.

For a three-dimensional free electron model, the spin-summed density of states at the Fermi energy is $g(E_F)=3N/(2E_F)$. A shell estimate treats $N_{\mathrm{act}}\sim g(E_F)k_{B}T$ electrons as thermally active and assigns each an excitation energy of order $k_{B}T$. For the illustrative choice $3k_{B}T/2$, $\Delta U\sim (9/4)N{k}_B^2{T}^2/E_F$ and differentiation gives $C_V\sim (9/2)N{k}_{B}T/T_F$. The numerical coefficient is heuristic: the controlled Sommerfeld expansion instead givesBoth estimates express Pauli exclusion principle suppression of the classical heat capacity when $T\ll T_F$. A shell argument establishes scaling rather than a universal prefactor.

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.