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For a smooth single-spin density of states $g$ near a positive Fermi energy and fixed total particle number, the Sommerfeld expansion gives $\mu ={ϵ}_F-({\pi }^2/6)(g_F^{\prime }/g_F)(k_{B}T{)}^2+O(T^4)$. The spin contribution to the total-moment magnetic susceptibility, defined by differentiating with respect to $B$, is $2{\mu }_B^2\int g(E)[-n^{\prime }(E)]dE$. Expanding about $\mu$ and including the shift of chemical potential gives the displayed formula. The $g_F^{\prime \prime }$ term is thermal broadening; the $-g_F^{\prime 2}/g_F$ term enforces fixed particle number. A band edge or singular density of states within $k_{B}T$ of ${ϵ}_F$ requires a separate analysis. Orbital magnetic response is excluded.

Consider an ideal uniform graphene sheet connected to an electron reservoir, with an insulated gate. Let $c_g>0$ be geometric capacitance per unit area, $e>0$ the elementary charge, and $V_0$ the neutral-point offset. Neglect temperature, trapped electric charge and spatial inhomogeneity. A change in gate voltage pays for both the electrostatic potential drop, $en/c_g$, and the change in the sheet's Fermi energy, $E_F/e$. Here $E_F$ is measured relative to the band crossing; it is negative for excess electron holes.

Combine this relation with the carrier density in a graphene cone. Write $u=|E_F|/e$, measured in volts, $V=|V_g-V_0|$, and $\alpha =e^3/[\pi (\hslash v_F{)}^2{c}_g]$. ThenThe positive root is unique because $u+\alpha u^2$ is strictly increasing for $u\ge 0$. The often useful geometric approximation $n\simeq c_g(V_g-V_0)/e$ neglects the Fermi energy term. It overestimates $|n|$ in this model, especially near neutrality where relatively few states are available.

For a deliberately chosen model with $c_g=0.010F/{\mathrm{m}}^2$, $v_F=10^{6}m/s$, $V_0=0$ and $V_g=1\mathrm{V}$, the formulas give $E_F\simeq 0.252\mathrm{eV}$ and $n\simeq 4.67\times 10^{12}\mathrm{c}{\mathrm{m}}^{-2}$. The purely geometric value would be $6.24\times 10^{12}\mathrm{c}{\mathrm{m}}^{-2}$. These are chosen-model predictions, not measurements of the historical devices. Numerical conversion uses the exact Planck constant and elementary charge.^[ref]

Integrating the hydrostatic pressure support equation with the radiative-zero white-dwarf envelope law yields $T=4GM(1/r-1/R)/(17\mathcal{R})$. Matching $T=T_c$ at $r_b$ gives $s=R-r_b$ as displayed. A strongly degenerate bulk core has thermal energy small compared with its Fermi energy, making $q\ll 1$ and hence $s\ll R$. The inequality $T_c\gg T_{\mathrm{eff}}$ alone does not establish geometrical thinness.

For a two-dimensional Fermi gas of $N$ spin-one-half particles with Fermi energy $E_F$, the constant density of states makes the low-temperature internal energy correction and heat capacity at constant areaThus its low-temperature heat capacity obeys a power law with exponent one.

For a three-dimensional ideal gas with quadratic dispersion and fixed particle number, the total density of states at the Fermi energy is $g(E_F)=3N/(2E_F)$. The Sommerfeld coefficient then gives $C_V=({\pi }^2/2)N{k}_B(T/T_F)$ to leading order, where $T_F=E_F/k_B$. The fixed-number energy correction is $\Delta U=({\pi }^2/4)N(k_{B}T{)}^2/E_F$. Only states near the Fermi surface change occupation, explaining the small active fraction.

A quadratic-dispersion Fermi gas in two dimensions has constant density of states $g$. At zero temperature, $N=g{E}_F$ and $U=g{E}_F^2/2$. Hence the mean energy is half the Fermi energy, independent of a common spin-degeneracy factor.

A mobility edge separates localized from extended energy eigenstates on the energy axis. Three-dimensional disordered models can have both. In the infinite one-dimensional tight-binding model with independent continuous random site energies, all states are localized for any nonzero disorder, with probability one. This statement excludes specially correlated disorder.^[ref]

A mobility edge is not a band gap: states can exist on both sides of $E_c$, but differ in spatial extent. For electrons, the location of the Fermi energy relative to a mobility edge matters for low-temperature electrical conductivity. A three-dimensional sample need not have an extended interval at every disorder strength. In particular, the finite one-dimensional calculations below neither locate a three-dimensional mobility edge nor establish a nonzero universal critical disorder.

For a three-dimensional ideal degenerate electron gas, let $r=\hslash {\omega }_{\mathrm{pe}}/E_F$. If one heuristically defines a Debye length by replacing $k_{\mathrm{B}}T$ with the Fermi energy, then $k_F{\lambda }_D=\sqrt{2}/r$ and the associated electron count is the displayed expression. A count below one signals failure of the classical many-particle screening interpretation. It does not prohibit plasmons. The zero-temperature compressibility instead gives the Thomas-Fermi screening length squared $2{ϵ}_0{E}_F/(3n_e{e}^2)$.

In 2001, Andrew G. Truscott, Randall G. Hulet, and collaborators observed degeneracy pressure in trapped lithium. A Zeeman slower fed a dual magneto-optical trap; optical pumping prepared atoms for a cloverleaf apparatus for magnetic trapping of neutral atoms. Microwave evaporation removed energetic lithium-7 bosons; elastic collisions cooled lithium-6 fermions by sympathetic cooling.^[ref]

Sequential absorption imaging recorded both trapped clouds on a camera. Fits to axial profiles used the boson cloud as a thermometer. At common temperature, the nearly equal magnetic trapping potentials provided a comparison: the clouds had similar sizes near $T/T_F=1$, but the fermion cloud remained broader as cooling reached $T/T_F=0.25$. Figure 3 compared its radius with ideal Fermi gas and classical predictions.^[ref]

Varying the wait before imaging tested thermal equilibrium. Finite imaging resolution restricted the analysis to the axial direction; fits neglected interaction-induced density shifts. Cooling eventually suffered from the dwindling boson heat capacity. The result supports quantum occupation effects, without directly resolving individual occupied states or reaching zero temperature.^[ref]

These were neutral alkali atoms in a spatially varying trap. The Fermi energy of a model electron gas instead uses uniform electrons, two equally populated spin states, and the electron mass. The measured trapped-atom radii are not measurements of that electron model's $p_F$ or $E_F$; comparing their numerical Fermi energies requires the appropriate particle mass, internal-state populations, and confining potential.

The static Lindhard dielectric function describes the real, zero-frequency linear response of a uniform free electron gas. Here ${ϵ}_1=\mathrm{Re}ϵ$, $\mathbf{r}$ is the position operator, and $\mathbf{q}$ is the perturbation wavevector, $q=|\mathbf{q}|$, $V$ the normalization volume, $e>0$ the elementary charge, and ${ϵ}_0$ the vacuum permittivity. The normalized one-electron states are ${\psi }_{\mathbf{k}}=V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$ with $E_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$; $f(E)$ is their Fermi-Dirac distribution occupation. Both momentum sums exclude spin; the prefactor two counts the two equal spin populations. The density-operator matrix element is ${\delta }_{{\mathbf{k}}^{\prime },\mathbf{k}+\mathbf{q}}$. Coincident energy denominators are interpreted by the static limit or Cauchy principal value.

A weak total electric potential ${\varphi }_{\mathrm{tot}}$ perturbs an electron by $-e{\varphi }_{\mathrm{tot}}$. First-order state mixing produces $\delta n=e{\Pi }_0(q){\varphi }_{\mathrm{tot}}$, whereThe induced charge is $-e\delta n$. Self-consistent Poisson equation screening then gives ${\varphi }_{\mathrm{ext}}=ϵ{\varphi }_{\mathrm{tot}}$, with $ϵ=1+e^2{\Pi }_0/({ϵ}_0{q}^2)$. This is the random phase approximation: use the independent-electron response to the self-consistent total field.^[ref]

At zero temperature $f(E_{\mathbf{k}})=\Theta (k_F-k)$. Shifting the integration variable in the second occupation term givesThe two angular integrals are equal. Using the supplied integral, with $x=q/(2k_F)$, givesThe limits are $F(0)=1$, $F(1)=1/2$ and $F(x)\sim 1/(3x^2)$ for large $x$. The first recovers Thomas-Fermi screening; the logarithm produces the Kohn singularity of static Lindhard screening at $q=2k_F$.

Use the Fourier transform convention $\rho (\mathbf{q})=\int d^{3}r{e}^{-i\mathbf{q}\cdot \mathbf{r}}\rho (\mathbf{r})$. A point charge $Ze$ has total screened charge $Ze/ϵ(q)$ and induced charge $-Ze[1-ϵ(q{)}^{-1}]$. Angular inversion therefore gives the radial induced-charge screening kernelIn particular, ${\rho }_{\mathrm{ind}}(\mathbf{q}=0)=-Ze$: the total screening charge is negative although its local density can oscillate in sign.

Integrate the sine transform by parts twice; $g(0)=g(\infty )=0$ removes the boundary terms. For the Friedel density tail from a derivative singularity, write $Q=2k_F$ and $g^{\prime \prime }(q)\sim C/(q-Q)$. ThenIndeed, near $Q$ the principal value transform is ${\int }_{-\infty }^{\infty }\sin [(Q+s)r]ds/s=\pi \cos (Qr)$ for $r>0$. Smooth terms do not remove this leading oscillatory long-range contribution. Thus ${\rho }_{\mathrm{ind}}\propto \cos (2k_{F}r)/r^3$.

For a three-dimensional, single-valley gas with two spin states, the Friedel oscillation wavelength from carrier density isLow-temperature scanning tunneling spectroscopy near exposed dopants can map the oscillatory local density of states, with a Fourier feature near $2k_F$ at the Fermi energy. An energy-resolved tunnelling image measures local spectral weight, not directly the bulk energy-integrated charge: comparing occupied-bias maps and accounting for the surface geometry is necessary. In actual GaAs, the background relative permittivity and band effective mass change screening strengths; they do not change this wavelength at the stated density.

A uniform magnetic field quantizes the transverse cyclotron motion of an electron. With electron charge $-e$ and the Landau gauge for a uniform magnetic field $\mathbf{A}=(0,Bx,0)$, minimal coupling givesFor an eigenstate of $p_y$ with eigenvalue $\hslash k_y$, this is a quantum harmonic oscillator centered at $x_c=-\hslash k_y/(eB)$, with cyclotron frequency ${\omega }_c=eB/m$. Its Landau levels areThe oscillator width is set by the magnetic length ${\ell }_B=\sqrt{\hslash /(eB)}$. In three dimensions, the free motion parallel to the magnetic field adds ${\hslash }^2{k}_z^2/(2m)$. The electron spin adds a Zeeman splitting $\pm g_s{\mu }_{B}B/2$. In a crystal the appropriate band effective mass and effective spin factor replace the free-electron values.

Changing $k_y$ moves the orbit center without changing the energy. In a rectangle $L_x{L}_y$, adjacent allowed centers differ by $2\pi {\ell }_B^2/L_y$. Requiring the center to lie in the sample gives the degeneracy of a Landau level, per spin,This large state degeneracy, together with the increasing level spacing, rearranges the occupied states as the magnetic field changes. To resolve the Landau levels, thermal broadening $k_{B}T$ and disorder broadening should both be small compared with $\hslash {\omega }_c$.

The de Haas-van Alphen effect is an oscillation of magnetization as successive Landau levels pass the Fermi energy. The Shubnikov-de Haas effect is the corresponding oscillation in longitudinal magnetoresistance. For a simple orbit, successive crossings occur periodically in $1/B$; the general relation to the extremal orbit area is Onsager quantization. These measurements reveal the geometry and effective masses of a Fermi surface.

In a two-dimensional electron system, filling an integer number $\nu$ of resolved levels leads to the integer quantum Hall effect, withand very small longitudinal resistivity on a plateau. Disorder localization helps produce finite plateau widths. Cyclotron resonance provides another direct signature: radiation is absorbed at the adjacent-level spacing $\hslash {\omega }_c$.

Metallic atomic densities place very many electrons in a small volume. The Pauli exclusion principle prevents all these fermions occupying the lowest spatial state, so filling the available states produces a Fermi energy of order electronvolts. At room temperature, $k_{\mathrm{B}}T\simeq 0.026\mathrm{eV}$, much smaller than that scale. Equivalently, ordinary metals are at $T\ll T_F=E_F/k_{\mathrm{B}}$, where $T_F$ is the Fermi temperature. Their electrons are quantum degenerate even at ordinary laboratory temperatures.

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 $k_{\mathrm{B}}T$.

At fixed $N$, the Fermi energy scales as $E_{\mathrm{F}}\propto (N/V{)}^{2/3}$, so $\partial E_{\mathrm{F}}/\partial V=-(2/3)E_{\mathrm{F}}/V$. From $U=3N{E}_{\mathrm{F}}/5$,This is the nonrelativistic electron degeneracy pressure. Because $n{E}_{\mathrm{F}}\propto V^{-5/3}$,The units are energy per volume, hence pressure. Converting the Fermi energies to joules gives the ideal Fermi gas bulk modulusThese are free-electron model values. The complete mechanical bulk modulus of a real metal also involves the ionic lattice, electronic bands and interactions; the model's positive electron pressure does not by itself explain the equilibrium crystal volume.

For $T\ll T_F$, only electrons within an energy window of order $kT$ around the Fermi energy can change occupation appreciably, because of the Pauli exclusion principle. The active fraction is of order $kT/E_F=T/T_F$. Giving each of these electrons energy of order $kT$ estimates an excitation energy $U-U_0\sim N_{e}k{T}^2/T_F$, and thereforePer mole of conduction electrons, $N_{e}k=R$, which is the quoted $2RT/T_F$. Per mole of metal atoms with $z$ model conduction electrons each, the conduction-electron count in the molar electronic heat capacity instead gives $2zRT/T_F$. The numerical factor is only approximate: the Sommerfeld expansion gives $C_V=({\pi }^2/2)N_{e}kT/T_F$ for a three-dimensional free-electron gas.

Formal ionic counting gives La(III), N(−III), and therefore Re(VI), with a $5d^1$ configuration. An isolated localized $d^1$ center would carry a saturation magnetic moment of order $1{\mu }_{\mathrm{B}}$, much larger than the value quoted. It is not a $d^0$ ion made diamagnetic simply by an octahedral ligand field.

The spatially extended Re $5d$ orbitals overlap strongly with N $2p$ orbitals. Re–N $\sigma$ interactions raise the $e_g$-derived antibonding states, while $\pi$ interactions mix the $t_{2g}$ states with the nitrogen network. Repetition through corner-sharing octahedra turns these combinations into broad energy bands. The nominal $d$ electron occupies a delocalized, partially filled band rather than an isolated magnetic orbital.

Only electrons close to the Fermi energy change their occupations appreciably in an applied magnetic field. Consequently a Pauli paramagnetism has a small, nonzero field-induced magnetization:when $g$ is the density of states including both spin directions, expressed per formula unit and per unit energy. This band model explains a very small moment without requiring the absence of electrons in the Re-derived band. The reported material has metallic transport and a Pauli-like response.^[ref]

The description “saturated” in the question should not be interpreted as complete alignment of one localized spin per Re: a nearly linear metallic response over a laboratory field range need not be a true fully spin-polarized saturation moment.

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.

For a one-dimensional free-electron gas of length $L$, allowed wave vectors have spacing $2\pi /L$. Including two spin states and occupied wave vectors from $-k_F$ to $+k_F$,The ground-state kinetic energy is found by integrating rather than assigning every electron the Fermi energy:Charge neutrality gives $N_e=ZN$ because each positive ion has charge $+Ze$. Thus $K_e=ZN{E}_F/3$. Since the ion separation is $2R$, $L\simeq 2RN$ and $n_e=Z/(2R)$. These are one-dimensional number densities and state counts; using a three-dimensional $k_F\propto n^{1/3}$ relation here would change the model. The results concern the ground-state Fermi-gas contribution, neglecting electron interactions beyond the supplied electrostatic energy.

Using $c=2700\mathrm{m}{\mathrm{s}}^{-1}$ gives the electron-acoustic recoil lengthThe temperature ratio therefore fixesIt follows that $n=4/a^3=9.48\times 10^{28}{\mathrm{m}}^{-3}$, $k_F=1.4105\times 10^{10}{\mathrm{m}}^{-1}$ and $k_D=1.7771\times 10^{10}{\mathrm{m}}^{-1}$. Substitution givesThe resulting Fermi energy is about $7.58\mathrm{eV}$ and the temperature ratio is indeed $240$. These values follow from the supplied effective sound speed and free-electron/Debye approximations; they are not independent precision measurements of copper's parameters.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.