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A fully spin-polarized sea of composite fermions has a Fermi wavevector $\sqrt{4\pi n_s}$. Near half filling its semiclassical orbit radius depends on the residual effective field, not the full applied field. Matching an orbit diameter to a patterned length can produce transport resonances outside an incompressible fractional Hall state. Multiple pattern periods and random/unpatterned controls test that geometric interpretation. Density inhomogeneity and a soft potential can shift or broaden the response.^[ref]

For two isotropic electron-like bands with equal spin degeneracy, conserving total carrier density conserves the sum of their occupied momentum-space volumes. If both uncoupled radii equal $k_0$,Only when the upper pocket is empty does this imply a single Fermi wavevector $k_F=2^{1/3}{k}_0$; a nonzero mixing matrix element alone does not imply that emptying.

For a counterexample at arbitrarily weak coupling, take $E_a=A{k}^2$, $E_b=C+B{k}^2$, $A>B>0$, and $C=(A-B)k_0^2$. At the uncoupled chemical potential ${\mu }_0=A{k}_0^2$, both pockets have radius $k_0$. Let $\mu ={\mu }_0+\delta \mu$ and $y=k_F^2-k_0^2$. The characteristic equation at the Fermi level becomesFixed density cancels the first-order chemical-potential shift. Thus $\delta \mu =O({\alpha }^2)$ and $k_{F,\mp }=k_0\pm \alpha /(2k_0\sqrt{AB})+O({\alpha }^2)$: both pockets remain occupied at small coupling. This expansion assumes fixed $A,B>0$ before taking $\alpha \to 0$.

The wavelength associated with a specified Fermi wavevector. A local-density response normally requires variation on a scale large compared with $k_F^{-1}$; numerical factors such as $2\pi$ must be kept explicit when comparing a variation scale with the full wavelength.

The Fermi wavevector is the radius of the occupied sphere in wavevector coordinates for an isotropic Fermi gas. In the uniform two-component electron model, $k_F=(3{\pi }^{2}n{)}^{1/3}$ and $E_F={\hslash }^2{k}_F^2/(2m_e)$ for nonrelativistic particles. Its units are inverse length, whereas the Fermi momentum has units of momentum.

Mix a parabolic energy band ${ϵ}_k={\hslash }^2{k}^2/(2m)$ with a perfectly flat band at $E_d>0$ through momentum-independent $V$. At zero coupling suppose the two orbital sectors have equal populations, with the dispersive sector filled to $k_0=\sqrt{2m{E}_d}/\hslash$. At fixed total density and sufficiently low temperature, nonzero coupling fills only the monotonic lower hybridized branch, whose Fermi wavevector is $k_F=2^{1/3}{k}_0$, provided the available momentum-space states accommodate it. Its small slope at that radius gives the displayed ratio to $v_F^{(1)}=\hslash k_0/m$ for $|V|\ll E_d$. The result uses the singular perfectly flat limit and an empty upper band; it is not implied by Fermi-volume conservation in two hybridized electron bands for arbitrary pairs of dispersive bands. The zero-temperature limit must resolve the small hybridization energy scale.

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

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.

Assume equal excess electron and electron hole number densities $N_t$, with no additional doping imbalance. At zero temperature, both occupied carrier spheres have the same Fermi wavevector $k_F$ because their spin counting is the same:For $k<k_F$, the conduction band is occupied and the valence band has electron holes, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-optical gain interval is the supremumEquivalently $E_{\mathrm{Fc}}=E_g+{\hslash }^2{k}_F^2/(2m_c^{\ast })$ and $E_{\mathrm{Fv}}=-{\hslash }^2{k}_F^2/(2m_v^{\ast })$. With the supplied effective masses,Using the vacuum wavelength $\lambda =hc/(\hslash \omega )$ reverses the endpoints:These boundaries are idealized and open: the lower-energy edge has zero joint density of states, and the upper edge is the zero-temperature occupation discontinuity, not a finite-width gain peak. The calculation gives positive material optical gain, not the stronger requirement of exceeding the losses at laser threshold.

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.

For an isotropic free-electron gas at zero temperature, the Fermi wavevector is the radius of the occupied region in wavevector space. Its energy is $E_F={\hslash }^2{k}_F^2/(2m)$. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice; it contains one distinct crystal momentum per primitive cell and per band, with spin counted separately.

For a two-dimensional sample of area $A$, allowed wavevectors have density $A/(2\pi {)}^2$. Including two spin states, filling a circle gives $N=2[A/(2\pi {)}^2]\pi k_F^2$. Dividing by area therefore yieldsThe square lattice has one atom per area $a^2$, so for valence $z$ the free-electron density is $z/a^2$. Its Brillouin zone has $|k_x|,|k_y|\le \pi /a$. ThusThe monovalent Fermi circle lies inside the square. The divalent circle crosses the faces but misses the corners; its area equals the zone's area. These are the free-electron contours before distortion by the periodic potential.

At a Bragg plane, the periodic potential mixes degenerate plane waves differing by a reciprocal lattice vector. The two standing-wave combinations have different charge densities relative to the ions and split in energy, opening the gap. In a two-wave approximation, the boundary energies are $E_B\pm E_g/2$, with $E_g=2|V_G|$.

The boundary wavevector magnitude is $\pi /a$ along $[10]$ and $\sqrt{2}\pi /a$ along $[11]$. The free-electron boundary energies are thereforeWithin the stated direction-independent gap picture, the upper band's lowest face-edge energy is approximately $E_X+2\mathrm{eV}=6.18\mathrm{eV}$, while the lower band's corner-edge energy is approximately $E_M-2\mathrm{eV}=6.36\mathrm{eV}$. There is consequently a small overlap, about $0.18\mathrm{eV}$, rather than a global insulating gap. The divalent Fermi level lies in that overlap, producing occupied upper-band states and unfilled lower-band states.

Thus the intended isotropic-gap approximation predicts a metal, with small electron and hole pockets. Positive direct gaps can coexist with band overlap. A fully specified square-lattice potential would require simultaneous treatment of the corner degeneracies; a numerical boundary splitting alone does not uniquely determine an arbitrary two-dimensional band structure. The conclusion here uses the given simplified gap picture, rather than electron count alone.

For one conduction electron per atom, let $n_a$ be the atom density. State counting in a Bloch band gives the Fermi wavevector $k_F=(3{\pi }^2{n}_a{)}^{1/3}$. The Debye model preserves the three vibrational modes per atom by a spherical Debye wavevector cutoff $q_D=(6{\pi }^2{n}_a{)}^{1/3}$, soBoth wavevectors are of order the inverse atomic spacing. The numerical factor uses the spherical Debye approximation; a real zone boundary and the largest-frequency mode depend on direction and branch. Because $q_D<2k_F$, short-wavelength acoustic phonons can scatter Fermi surface electrons through substantial angles. For nearly elastic scattering within a spherical Fermi surface,Near room temperature, when a substantial fraction of the phonon spectrum is occupied, electron-phonon scattering is often the main source of electrical resistance. For $T\gg {\Theta }_D$, the phonon population and scattering rate are approximately proportional to $T$, giving $\rho \propto T$, $\sigma \propto T^{-1}$ and roughly constant $K_e$ when the Wiedemann-Franz law holds.

For small-angle electron-phonon transport at low temperature, thermally active acoustic modes instead haveThey predominantly scatter forward. The factor $1-\cos \theta$ makes such collisions inefficient at relaxing an electrical current, while a phonon energy of order $k_{B}T$ can still substantially change an electron's thermal excitation. Heat-current relaxation therefore need not share the charge-current relaxation time. In the clean three-dimensional acoustic-phonon model this gives ${\tau }_Q^{-1}\propto T^3$ but ${\tau }_{\sigma }^{-1}\propto T^5$, so the phonon-limited electrical resistivity scales as $T^5$ and $K_e\propto T^{-2}$. These laws describe the regime before residual elastic scattering takes over; at sufficiently low temperature impurities give nearly constant residual resistivity and $K_e\propto T$, restoring the usual Lorenz number.^[ref]

Phonons also transport heat themselves, with $K_{\mathrm{ph}}\simeq c_{\mathrm{ph}}{c}_s{\ell }_{\mathrm{ph}}/3$. At low temperature their population decreases and their phonon mean free path can become boundary- or defect-limited. At higher temperature Umklapp scattering can relax the phonon heat current. Such lattice conduction is an additional contribution, not the quantity constrained by the electronic Wiedemann-Franz law.

Let the common uncoupled Fermi wavevector be $k_0$. Equal carrier densities imply equal Fermi surface volumes, and hence equal radii for the spherical spin-degenerate electron pockets. The conserved total density isIf hybridisation leaves only one occupied electron-like band, state counting givesThis is the familiar single-large-Fermi-surface answer. It requires the other hybridised band to be empty.

For finite heavy-band mass, that conclusion does not follow from an arbitrary nonzero coupling. At sufficiently small $\alpha$, the upper band still has an occupied pocket, so there are two Fermi momenta, not one. The exact fixed-density statement is Fermi-volume conservation in two hybridized electron bands:For an explicit weak-coupling check, write $E_p=A{k}^2$, $E_h=C+B{k}^2$, with $0<B\ll A$, and enforce the uncoupled equality at $k_0$: $C=(A-B)k_0^2$, ${\mu }_0=A{k}_0^2$. Put $\mu ={\mu }_0+\delta \mu$ and $y=k_F^2-k_0^2$. The Fermi-surface equation isAt fixed density $\delta \mu =O({\alpha }^2)$, givingFor arbitrary coupling, put $q_{\pm }=k_{F,\mp }^2$. The roots of the same characteristic equation, before discarding unoccupied pockets, areChoose the chemical potential so that $[\max (q_{+},0){]}^{3/2}+[\max (q_{-},0){]}^{3/2}=2k_0^3$. This specifies the Fermi momentum of each pocket for any supplied curvatures and coupling.

The transition to one occupied pocket occurs at $\mu =E_{+}(0)$. Set $q_{\ast }=2^{2/3}{k}_0^2$; at the transition the two squared radii are $q_{\ast }$ and zero. Their sum and product giveFor $0<\alpha <{\alpha }_c$ there are two pockets; for $\alpha \ge {\alpha }_c$ the only nonzero radius is $2^{1/3}{k}_0$. The outer radius grows continuously and the inner one shrinks. The nearly flat band limit can make depopulation occur at a small coupling, but the mass ratio or a single-surface assumption is needed to turn the question into a unique finite-coupling numerical change.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.