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For a two-dimensional energy band, a Fermi contour is the set of wavevectors at the Fermi energy. It is the two-dimensional counterpart of a Fermi surface. At a regular value it consists locally of curves; at a band extremum or a crossing it can acquire isolated points or intersecting branches. At zero temperature, each band's occupied region satisfies $E_n(\mathbf{k})<E_F$, so the contours must be labelled by band when determining occupations. The Fermi contours of an isotropic Rashba band illustrate why two contours need not belong to two different bands.

For the Rashba effect with $k_R>0$, put $D=\sqrt{k_R^2+2m{E}_F/{\hslash }^2}$. If $-E_R<E_F<0$, both Fermi contours belong to the lower branch and bound the singly occupied annulus $k_R-D<k<k_R+D$; the central disk and exterior are empty. At $E_F=0$, the lower occupied disk has outer radius $2k_R$, while the origin is a degenerate point at the Fermi energy. For $E_F>0$, the lower Fermi contour has radius $k_{-}=D+k_R$ and the upper Fermi contour radius $k_{+}=D-k_R$. The inner disk has two occupied spin-split eigenstates at each momentum, the intervening annulus has one, and the exterior is empty. Consequently $k_{-}-k_{+}=2k_R$, whereas below the crossing the two lower-branch radii add to $2k_R$. At zero temperature, changing occupation values only on these Fermi contours affects a null set and leaves the integrated carrier density unchanged.

The Fermi-Dirac distribution follows from Fermi-Dirac statistics and is $n(E)=[e^{\beta (E-\mu )}+1{]}^{-1}$.

The mean occupation of a fermionic one-particle state of energy $E$ isAt zero temperature this becomes the step function $1_{E<E_F}$.

Only single-particle states within order $k_{B}T$ of the Fermi energy can change occupation at low temperature. Their number is order $g(E_F)k_{B}T$, and their excitation energy is order $k_{B}T$, giving $U-U_0\propto T^2$ and electronic heat capacity proportional to $T$.

The Fermi energy is the highest occupied one-particle energy in the ground state of a noninteracting Fermi gas at zero temperature, measured here relative to the lowest one-particle energy. The Pauli exclusion principle allows only one fermion in each complete one-particle quantum state, including its spin, so adding more particles fills progressively higher energy eigenstates. A cold gas can therefore have substantial kinetic energy even when its temperature is zero. In the thermodynamic limit, with the same energy reference, $E_F=\mu (T=0)$ where $\mu$ is the chemical potential.^[ref]

For a uniform, three-dimensional gas of noninteracting electrons, with spin one-half, equal populations of the two spin states, no applied magnetic field, and nonrelativistic particles of mass $m=m_e$,Here $n$ is the total number density, $V$ the volume, $p_F$ the Fermi momentum, $\hslash =h/(2\pi )$ the reduced Planck constant, and $k_B$ the Boltzmann constant. The electron Fermi temperature $T_F$ expresses the same energy scale in kelvin; it is not the actual gas temperature. Strong degeneracy of a degenerate electron gas means $T\ll T_F$. The momentum formula counts states and survives changing the energy–momentum relation, whereas $E_F=p_F^2/(2m)$ requires $p_F\ll mc$.^[ref]

The reference matters: the displayed $E_F$ excludes rest energy and sets the zero at $p=0$. Including rest energy shifts both the one-particle energy and the zero-temperature chemical potential by $m{c}^2$, leaving occupations unchanged. More generally, adding a constant $C$ to every one-particle energy also adds $C$ to $\mu$; the Fermi-Dirac distribution depends only on their difference. In a finite quantum system the chemical potential at zero temperature can lie between the last occupied and first empty energy eigenstates; its equality with the filling edge is a continuum statement.

At positive temperature, $E_F$ remains the zero-temperature scale associated with the specified number density. The Fermi level $\mu (T)$ must instead satisfy the particle-number conditionwhere $g(E)$ is the extensive density of states, including both spin states. Smearing the Fermi-Dirac distribution while arbitrarily holding $\mu =E_F$ generally changes $N$. This distinction matters when interpreting electron heat capacity and any experiment that infers temperature from an occupation profile.^[ref]

Choose $n=10^{28}{\mathrm{m}}^{-3}$ for a uniform noninteracting electron model with two equally populated spin states. This is an original worked example, not a measured number density assigned to an actual material. Using $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$, $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesThe conversion uses $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$: an electronvolt is an energy unit, while a joule is its SI counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse length, $\hslash n^{1/3}$ has units of momentum, $p_F^2/m$ has units of energy, and $E_F/k_B$ has units of temperature. The ratio $p_F/(m_{e}c)\simeq 0.002574$ verifies the nonrelativistic limit for this model, where $c$ is the speed of light.

Increasing this chosen number density eightfold doubles the Fermi momentum and quadruples the Fermi energy, giving approximately $6.7701\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a momentum radius, but kinetic energy of nonrelativistic particles grows as its square. This makes Fermi energy a useful bridge between a count of particles and a mechanical energy scale.

At fixed sheet density $n_s$, each resolved Landau level gains degeneracy $eB/h$ per unit area as the magnetic field grows. A partly filled ideal level has chemical potential at its energy; when it empties into lower levels, the chemical potential drops. This produces an inverse-field-periodic sawtooth in the clean zero-temperature model. At fixed field and increasing density the corresponding chemical potential is an upward staircase with density increment $eB/h$. Fixed reservoir chemical potential is a different boundary condition, in which the density responds instead. spin counting, disorder and temperature qualify these ideal pictures.

The Fermi level is the chemical potential controlling electron occupations. For an ideal Fermi gas at zero temperature it equals the Fermi energy; at nonzero temperature the chemical potential need not equal that zero-temperature energy.

The motion of particles occupying the momentum sea of a Fermi gas, especially nucleons in an atomic nucleus. It persists at zero temperature and broadens scattering kinematics relative to a stationary-nucleon approximation.

The 2007 paper Comparison of the Hanbury Brown-Twiss effect for bosons and fermions by T. Jeltes and others compares polarized helium-3 fermions with helium-4 bosons. The chemical atoms occupy a metastable atomic level. Switching off magnetic trapping of neutral atoms releases them onto a microchannel plate detector with a delay-line anode, 63 cm below. Position and arrival-time readouts give three-dimensional pair separations by time-of-flight atom detection. About 1000 clouds per isotope supply histograms; normalization removes the broad cloud shape.

The pair detection correlation dips below one for helium-3 and rises above one for helium-4. The same apparatus, matched temperatures near $0.5\mu \mathrm{K}$, nearly identical trapping potentials, negligible interactions, and magnetic field gradient checks of spin polarization support an exchange-statistics interpretation. Finite detector resolution suppresses contrast; trap switch-off and resolution uncertainties limit quantitative agreement. A diverging laser lens deepens the dip, but its aberrations preclude a precise lens-model test. This detects exchange correlations, rather than measuring occupation of one selected spin orbital.^[ref]

The Fermi sea is the zero-temperature many-fermion quantum state with every one-particle state below the Fermi energy occupied and every state above it empty. For a uniform isotropic Fermi gas, the occupied region in momentum space is a solid sphere; its boundary is the Fermi surface. These are momenta, not positions: filling the sphere does not put the particles in a ball in ordinary space. Opposite momentum contributions cancel, so the gas at rest has zero total momentum even though its total kinetic energy is positive.

The filled states are forced upward in energy by the Pauli exclusion principle. This is different from the zero-point energy of one particle confined by a boundary. At fixed number density, the Fermi energy remains positive as the box grows to the thermodynamic limit, while the lowest box confinement energy tends to zero. Low-energy excitations move particles from occupied states near the Fermi surface into nearby empty states; deep states have no nearby available destination. This explains why only a small fraction of electrons contribute to the low-temperature electron heat capacity.^[ref]

For an isotropic three-dimensional free electron gas at zero temperature, occupied wavevectors fill a sphere of radius $k_F$. Including two spin states, $n=k_F^3/(3{\pi }^2)$. The volume in reciprocal space counts occupied states; folding an extended-zone sphere into a first Brillouin zone does not remove states outside that first zone.

The Fermi temperature expresses a Fermi energy as a temperature scale, $T_F=E_F/k_B$, where $k_B$ is the Boltzmann constant. In the convention $E_F>0$ measured from the relevant band minimum, $T\ll T_F$ is the regime of a strongly degenerate Fermi gas. For an interacting Fermi liquid, an effective Fermi scale can be introduced through its quasiparticle dispersion; its convention should be specified.

The Fermi window measures sensitivity of occupation to energy near the chemical potential. It is positive, even about $\mu$, has width of order $k_{B}T$, and integrates to one over the full energy line. Its centred variance is ${\pi }^2(k_{B}T{)}^2/3$. A band restricted to nonnegative energy contains weight $n_F(0,\mu )$ instead of exactly one; extending to the full line is exponentially accurate when $\mu \gg k_{B}T$. At zero temperature the window tends to a Dirac delta function at the chemical potential.

A ferromagnetic spin chain favors aligned neighbouring spins. Its conventional long-wavelength magnons have quadratic dispersion.

For a dispersion relation $E(k)=E_0+t\cos (ka)$, $t<0$, occupied states at zero temperature form the interval $|k|<k_F$ in the first Brillouin zone. Its occupied fraction is $f=2k_F/(2\pi /a)=k_{F}a/\pi$. A spin-degenerate band therefore holds $2f$ electrons per site. If the band is independent of the other two wavevector coordinates, its Fermi surface consists of two planes. For $t>0$ the occupied region is instead near the zone edges, bounded by $|k|=\pi (1-f)/a$.

A battery's charge reservoir means a finite amount of electric charge can pass through the external electrical circuit before its usable chemical state is depleted. It does not mean a large net electric charge is sitting inside the casing. For one net cell chemical reaction transferring $n$ electrons per reaction event, an ideal reaction inventory obeyswhere $F$ is the Faraday constant and $\xi$ the extent of reaction in moles, assuming all the electric current serves that reaction. At fixed temperature and pressure, its reversible electromotive force obeys ${\Delta }_{r}G=-nF\mathcal{E}$: a decrease in Gibbs free energy provides the maximum reversible electrical work per mole. The electromotive force also depends on composition; the Daniell cell illustrates this with thermodynamic activity.^[ref]

As an original finite-reservoir exercise, stipulate usable $Q_{\max }=360\mathrm{C}$ and hold the preceding $3\mathrm{V}$, $1\Omega$, $4\Omega$ parameters constant until an abrupt cutoff. Then $t=Q_{\max }/I=600\mathrm{s}$, available electrical work is $\mathcal{E}{Q}_{\max }=1080\mathrm{J}$, load work is $V_L{Q}_{\max }=864\mathrm{J}$, and internal dissipation is $216\mathrm{J}$. This explicitly finite idealization predicts ten minutes, not an experimentally measured runtime. Real accessible charge depends on discharge conditions and the chosen terminal-voltage cutoff. Charge alone is also not an energy: converting it to delivered work requires the terminal electric potential difference throughout discharge.

An internal harmonic vibration spectrum with lowest positive frequency ${\omega }_{\min }$ has the displayed low-temperature thermal phonon population, where $g_{\min }$ is its degeneracy. The continuum bulk law fails when thermally occupied wavelengths reach the sample size. Mechanical boundary conditions determine ${\omega }_{\min }$; rigid translation of the entire solid is not an internal phonon mode.

A finite rod generally contains both exponential branches. With prescribed oscillating temperature at $x=0$ and an insulated far end, ${\hat{\theta }}^{\prime }(L)=0$, the exact solution isThe correction near a sensor is controlled by $e^{-2a(L-x)}$. The important separation is between the farther sensor and the end, not merely between the heater and the end. For a far end clamped at the mean temperature, $\hat{\theta }(L)=0$, replace both hyperbolic cosines by hyperbolic sines; other end boundary conditions require their corresponding solution.

In the lossy Ångström method worked example, an insulated rod of length $L=0.080\mathrm{m}$ gives $D=0.354866$ and $\Delta \varphi =0.615651\mathrm{rad}$. Incorrectly applying the semi-infinite formula returns $1.29418\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $29.4\%$ high. At $L=0.300\mathrm{m}$, it returns $9.99877\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $0.0123\%$ low. Side loss remains real in both cases; the error arises because the measured spatial response no longer has one exponential complex wavenumber.

Choose the period and sensor spacing so that the lag and attenuation are resolvable while the downstream signal remains larger than sensor noise. Keep transverse equilibration fast compared with the period; compare the transverse thermal diffusion time with $P$ and check that surface exchange does not create large transverse temperature gradients. A short period reduces far-end influence but can expose sensor response times; a long period increases penetration and finite-end bias. Calibrate both sensors together, exchange their positions to check channel bias, compare successive cycles, and repeat at several spacings and periods. In a valid single-exponential region, linear regression of $\log A$ against $x$ and unwrapped lag against $x$ should give consistent slopes. These are proposed diagnostic controls. If they fail, fit the finite-rod or coupled-sensor model rather than interpreting every delay as intrinsic thermal diffusivity.