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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 the free electron model, the Fermi energy is the highest occupied one-electron energy at zero temperature, measured relative to the bottom of the free-electron band. The occupied wavevectors fill a disk in two dimensions or a sphere in three dimensions. The radius is the Fermi wavevector $k_F$, withEach allowed wavevector has two spin states. At nonzero temperature, the Fermi-Dirac distribution rounds the occupation boundary; the zero-temperature Fermi energy remains a useful density-defined scale and should not be defined as the thermal energy $k_{\mathrm{B}}T$.

Substitution in part (a) givesandThe conversion uses $1\mathrm{eV}=1.60218\times 10^{-19}\mathrm{J}$. These are the predictions of the stipulated two-dimensional free electron model, with the vacuum electron mass and one carrier per inferred square cell.

With one valence electron per atom, the electron number density is $n=4/a^3$. For a spin-degenerate, three-dimensional free electron model at zero temperature,Substitution givesCopper has the larger Fermi wavevector and Fermi energy because its smaller cell gives a larger electron density. The energies are measured from the bottom of the free-electron dispersion, not from vacuum. Treating these real metals as monovalent free-electron systems is the model assumption, not a statement that their filled d bands are absent.

For area $A$, allowed wavevectors have density $A/(2\pi {)}^2$. Include two electron spin states and fill a disk of radius $k_F$ at zero temperature. ThenThe free electron model has parabolic dispersion $E={\hslash }^2{k}^2/(2m_e)$, with its minimum chosen as zero, soThese are the Fermi wavevector and Fermi energy for a spin-degenerate two-dimensional free electron gas. An imposed spin polarization or additional valley degeneracy would change the state count.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

In the free electron model, conduction electrons occupy independent plane-wave states with energy ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ in a uniform mean background. Crystal-periodic band structure and explicit electron–electron correlations are omitted. Electrons still obey the Pauli exclusion principle and Fermi–Dirac statistics; free does not mean distinguishable classical particles. A scattering time can be added phenomenologically to describe transport.

The Fermi energy is the highest occupied single-particle energy at zero temperature, measured here from the bottom of the parabolic band. The Fermi temperature is the corresponding scale $T_F=E_F/k_B$, not the actual thermal temperature of the sample. Metallic electron densities are large, so filling all distinct states up to the required momentum gives Fermi energies of several electronvolts, equivalent to tens of thousands of kelvin or more. At ordinary temperatures, only a narrow region around the Fermi surface has appreciably changing occupations.

A room-temperature metal and its lattice are nevertheless at room temperature, not at $T_F$. High occupied kinetic energies are present even in the zero-temperature ground state and cannot all be liberated as heat without violating the allowed occupancy structure. A metal below skin temperature often feels cold because it transfers heat rapidly out of the skin; its thermal conductivity and thermal effusivity matter, not the value of its Fermi temperature. At the same actual temperature, a poorly conducting material can therefore feel warmer while neither material is thermodynamically hotter.

For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times 10^8$ m${}^{-1}$. The free-electron model with $m=m_e$ givesThe given $10^3\Omega$ is a sheet resistivity, not a bulk resistivity in ohm-metres. The two-dimensional Drude sheet conductance is $n{e}^2\tau /m_e$, soThus $v_F=\hslash k_F/m_e=1.589\times 10^4$ m/s andThe sample temperature is not given. This is specifically the Fermi-surface mean free path, appropriate in the degenerate regime $T\ll 8.3$ K for the bare-mass model. At higher temperatures the characteristic thermal speed should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^{\ast }$, the stated results scale as $E_F,T_F\propto 1/m^{\ast }$, $\tau \propto m^{\ast }$ and $v_F\propto 1/m^{\ast }$; the inferred ${\ell }_F=\hslash k_F/(n{e}^2{\rho }_{\square })$ is mass independent under these same Drude assumptions. No semiconductor effective mass or valley degeneracy is silently introduced.

The density of states $g(E)$ counts one-electron states per energy interval, including both electron spin orientations. It may be a total or a per-volume quantity; $U$ must have the matching normalization. The Fermi-Dirac distributionis the occupation probability of such a state. In the fixed-volume free electron model, the integration variable $E$ and the density of states do not depend on $T$. The occupation depends explicitly on $T$ and implicitly through the chemical potential $\mu (T)$ needed to preserve electron number. Thermal expansion or a changing electronic band structure would require additional effects outside this model.

At $T\ll T_F$, only states within order $k_{B}T$ of the Fermi energy change their occupation. There are order $g(E_F)k_{B}T$ such states, each carrying an excitation energy of order $k_{B}T$. Thus $U(T)-U(0)\propto g(E_F)(k_{B}T{)}^2$ and the electronic heat capacity is proportional to $g(E_F)k_B^{2}T$. Using the supplied coefficient,The shell argument fixes the scaling, not the numerical coefficient; the controlled Sommerfeld expansion fixes that coefficient at fixed electron number. Filled states far below $E_F$ cannot accept arbitrary thermal excitations because of the Pauli exclusion principle.

Let $g_v(E)$ denote the spin-summed density of states per volume. An electron moment with magnitude the Bohr magneton ${\mu }_B$ has Zeeman energies $\mp {\mu }_{B}B$. At low temperature, the two spin populations areThe common chemical potential changes only at order $B^2$, so it does not affect this linear imbalance. The magnetization is $M={\mu }_B(n_{+}-n_{-})={\mu }_B^2{g}_v(E_F)B$. Define SI volume magnetic susceptibility by $M={\chi }_{p}H$ and take the weak applied field as $B\simeq {\mu }_{0}H$. Then Pauli paramagnetism givesThe last equality uses the three-dimensional parabolic free electron model, $g_v(E_F)=3n/(2E_F)$. This susceptibility is nearly independent of temperature for $T\ll T_F$, unlike the Curie law for localized independent moments. The displayed result excludes orbital diamagnetism and interaction enhancements. If susceptibility instead denotes $\partial M/\partial B$, omit ${\mu }_0$; confusing that convention with $\partial M/\partial H$ changes the units.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

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.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

The free electron model replaces the detailed periodic potential by a constant background, treating conduction electrons as mobile particles with approximately parabolic dispersion $E={\hslash }^2{k}^2/(2m)$. It is useful to distinguish the classical Drude model from the quantum free electron gas. They share a simple transport picture, but differ fundamentally in occupation statistics.

With a momentum-relaxation time $\tau$, both lead in a simple isotropic one-band approximation toThis explains approximately Ohmic longitudinal conduction and a carrier-density scale from the Hall effect. The relaxation time is phenomenological: introducing it does not itself predict scattering from defects or phonons. A four-probe resistance measurement separates the sample's longitudinal response from current-contact voltage drops; field and current reversals isolate the transverse signal.

The classical model incorrectly assigns roughly $3N{k}_B/2$ of electronic heat capacity, far too large in a degenerate metal. Fermi-Dirac statistics fixes this: most occupied states cannot change occupation after a small thermal perturbation because nearby low-energy states are already filled. Only a fraction of order $k_{B}T/E_F$ near the Fermi energy participates. For $T\ll T_F$,The quantum model also gives a finite Fermi velocity even at zero temperature and Pauli paramagnetism. Under predominantly elastic scattering, its electronic heat and charge conductivities obey the Wiedemann-Franz law,This is a quantum success; classical equipartition does not give the correct small heat capacity and Lorenz number together.

Its limitations are equally instructive. Without the periodic potential it has no band gaps, so it cannot distinguish a filled-band insulator from a metal or explain semiconductor doping. The simplest single-electron Hall formula cannot explain positive Hall coefficients; band curvature and electron holes, or a multiband transport treatment, are needed. It neglects much Fermi-surface anisotropy and detailed optical interband absorption. Noninteracting electrons alone do not explain ferromagnetism or superconductivity, and lattice thermal behavior needs phonons. The nearly-free electron model, which allows reciprocal-lattice Fourier components of the potential to mix plane waves, provides a controlled first correction near Bragg planes.

Its main success is a simple quantitative baseline for degenerate conduction electrons; its main failure is treating the crystal and interactions as too featureless. A fit of conductivity alone is not a test of the whole model: compare heat capacity, Hall sign, optical response and temperature dependence using independently constrained parameters.

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 $gf$.