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Let $n=N/\mathcal{V}$ and $c_e=C_e/\mathcal{V}$ be the carrier density and electronic heat capacity per unit volume. In isotropic kinetic theory, the angular average $⟨v_x^2⟩=v_F^2/3$ gives the electronic thermal conductivitywhere ${\ell }_Q=v_F{\tau }_Q$ is the mean free path associated with the heat-current relaxation time. With $v_F^2=2E_F/m^{\ast }$,The Drude model gives the electrical conductivity $\sigma =n{e}^2{\tau }_{\sigma }/m^{\ast }$, where ${\tau }_{\sigma }$ is the momentum relaxation time. ThereforeWhen the same approximately elastic collisions relax heat and charge currents, ${\tau }_Q={\tau }_{\sigma }$. Using $A={\pi }^2/2$ gives the Wiedemann-Franz law$L_0$ is the Lorenz number. Degenerate electrons move at the Fermi velocity, not the classical thermal velocity; using classical equipartition would give the wrong coefficient.

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.

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.