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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.

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.

Ordinary doped semiconductor devices need a regime where their donor impurities or acceptor impurities supply mobile charge carriers while intrinsic thermal excitation remains sufficiently small.

At low temperature, donor impurities or acceptor impurities may no longer be ionized. Dopant freeze-out reduces the carrier density, raises electrical resistance and changes the intended carrier injection and p-n junction response. At high temperature the intrinsic carrier concentration grows aswhen band effective masses and band gap are treated as constant. Once it becomes comparable with the net doping density, it overwhelms the intended electron-hole imbalance; reverse leakage and dissipation increase and useful p/n control is lost.

The usual operating window lies between freeze-out and intrinsic conduction, with mobility and thermal management placing further practical limits. This is a statement about conventional doped-device operation, not a fundamental prohibition on engineered cryogenic or wide-bandgap devices.

In the cold solid, phonon scattering makes $\tau \propto T^{-1}$ and therefore conductivity falls as temperature rises, at roughly fixed carrier density. After melting and ionization, the character of scattering changes; a hot Coulomb plasma eventually follows an increasing $T^{3/2}$ branch. A minimum in the crossover region is natural, but a weakly coupled plasma formula is not quantitatively justified throughout the warm dense regime.

By National Institute of Standards and Technology.

Official text: www.nist.gov/pml/nanoscale-device-characterization-division/popular-links/hall-effect/resistivity-and-hall

Institutional measurement procedures covering contact geometry, reversal of electric current and magnetic field, and extraction of sheet carrier density and Hall mobility. Bulk density requires the conducting thickness.

Rex Tayloe's germanium laboratory treatment uses the Hall effect together with electrical conductivity to investigate electric charge carriers. A transverse voltage gives a sign-sensitive Hall response, while a longitudinal measurement supplies the conductivity. In a one-carrier model, combining the two separates carrier density from mobility: $R_H=1/(nq)$ and $\mu =|R_H|\sigma$. The interpretation requires a carrier model; simultaneous electron and hole contributions cannot generally be reduced to one unrestricted density estimate.^[ref]

At sufficiently low temperature, thermal ionization of shallow dopants becomes inefficient, reducing the mobile carrier density. An acceptor-doped sample may then have many neutral acceptors and few mobile electron holes. Extracting an activation energy requires a model of dopant ionization and compensation; the mobile density need not have the same simple activation exponent in every compensation regime.

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.

In a local London equation description, the measured London penetration depth satisfies ${\lambda }_L^2=m/({\mu }_0{n}_s{e}^2)$ when $n_s$ counts superconducting electrons and $m$ is their effective mass. Thus screening measures $n_s/m$. If instead one counts Cooper pairs, use pair mass $2m$ and pair electric charge $-2e$, giving pair density $n_s/2$ for the same screening length. Identifying this superconducting carrier density with the total conduction-electron density requires the additional assumption that the relevant carriers all participate in screening at the measurement temperature.

Xiaoge Liu is the first-listed author of the 2014 study of band nonparabolicity and the optical properties of indium tin oxide. The Stanford laboratory's alumni record also identifies Liu as a former doctoral student.^[ref][ref]

The physical issue can be expressed through the Drude model relationwhere $n$ is carrier density and $m^{\ast }$ the relevant effective mass. If $m^{\ast }$ were fixed, doubling $n$ would double ${\omega }_p^2$. A density-dependent mass changes that prediction. Combining electrical and optical measurements can therefore test a band model rather than treating every shift in plasma frequency as a direct measure of carrier density alone.