Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with spin degeneracy two. Their effective densities of states areFor the semiconductor mass-action law, integrating the square-root density of states against the Boltzmann tail gives $n=N_c{e}^{(\mu -E_c)/(k_{B}T)}$ and $p=N_v{e}^{(E_v-\mu )/(k_{B}T)}$. Multiplication cancels the chemical potential, givingIf there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi {\hslash }^2{)}^3$. No numerical mass or valley multiplicity has been invented.

When $\mu =E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive temperature. Define the Normalized complete Fermi-Dirac integralFor the carrier densities with chemical potential pinned to a band edge, with the two specified equal masses, put $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Direct Fermi-Dirac integration then givesThus $n\propto T^{3/2}$ exactly within the parabolic model. At low temperature, $p\simeq N(T)e^{-E_g/(k_{B}T)}$ is thermally activated; at $k_{B}T\gg E_g$, $p/n\to 1$ from below. Both vanish at zero temperature and increase at fixed chemical potential. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.

For the Pauli reduction of the pinned-edge mass-action product, the exact product relative to the classical law isIt approaches $f_0$ at low temperature and $f_0^2\simeq 0.58545$ in the formal high-temperature limit. The low-temperature activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-temperature limit here refers to the given unbounded parabolic bands, not a claim that real semiconductor band structures remain unchanged at arbitrarily high temperature.^[ref]

Finally, for a fixed positive $\mu$, the zero-temperature Fermi sphere has radius $k_{\mu }=\sqrt{2m\mu }/\hslash$, soThe low-temperature expansion at fixed $\mu$ gives the fixed-chemical-potential Sommerfeld density increase:The monotonic density at fixed chemical potential in a square-root density of states initially rises quadratically from $n(0)$. For $k_{B}T\gg \mu$, its leading behaviour crosses to $f_{0}N(T)\propto T^{3/2}$. The fixed chemical potential is what allows the particle number to change.

For a spin-degenerate band, each unit cell contributes two available states. An even electron count per cell can fill an integer number of bands; a band insulator additionally needs a global gap between the highest occupied and lowest unoccupied bands. Even filling alone is not sufficient when bands overlap.

For a free two-dimensional gas with spin degeneracy two,The reciprocal vectors of the periodic potential are $Q\hat{\mathbf{x}}$ and $Q\hat{\mathbf{y}}$. Its first Brillouin zone is the square $-Q/2\le k_x,k_y\le Q/2$, of side $Q$ and area $Q^2$; the real-space cell has side $a=2\pi /Q$.

At zero potential, the extended-zone Fermi surface is the circle of radius $k_F$. Since $\sqrt{2}<\alpha <2$, it crosses every square face but misses the corners. For Fermi-surface reconstruction in a weak square potential, weak positive potential opens gaps at the Bragg crossings: the second band has electron pockets at the two inequivalent edge centres $X,Y$, while unfilled regions of the first band form a hole pocket at the zone corner $M$. Opposite edges are identified: four edge half-pockets are two electron pockets, and four corner pieces are one hole pocket.

For the circular-cap carrier count in a square Brillouin zone, put $R=k_F$ and $s=Q/(2R)=\alpha /2$. Each circular cap outside a square face has areaNo two such caps overlap in the specified interval. The total second-band electron area is $A_e=4C$. The first-band hole area is the square area minus the occupied central-circle area, $A_h=Q^2-\pi R^2+4C$. Including spin,Thus at $\alpha =\sqrt{3}$,The carrier densities are the electron- and hole-pocket counts, not the total electron density. The carrier compensation and even band filling in a square lattice follows because their difference obeys $n_e-n_h=n-Q^2/(2{\pi }^2)$, so compensation occurs at $\alpha =\sqrt{\pi }$. These area formulas are the $V_0\to 0^{+}$ limit, before finite-gap reconstruction changes the pocket shapes.^[ref]

The filling per real-space cell is $\nu =n{a}^2=2\pi /{\alpha }^2$. In the given interval, the only even integer filling is $\nu =2$, again requiring $\alpha =\sqrt{\pi }$. At any other fixed filling a full gap cannot yield an ordinary spin-degenerate band insulator in this unchanged periodicity. In particular, the $\alpha =\sqrt{3}$ system retains excess electron carriers even after its holes disappear.

A rough gap threshold comes from comparing different symmetry directions, rather than only a direct zone-face gap. Define $E_Q={\hslash }^2{Q}^2/(2m)$. The zone-face two-plane-wave gap for a cosine potential follows because each cosine couples plane waves by $-V_0/2$, giving the one-dimensional zone-edge levels $E_Q/4\mp V_0/2$. The four-plane-wave splitting at a square-zone corner is the sum of two such pairs. The potential is separable in $x,y$, so to first orderThe indirect gap of a separable square cosine potential is consequentlyAt the compensated filling, this is $V_{0,\min }\sim (\pi /6)E_F^{(0)}$, with $E_F^{(0)}={\hslash }^2{k}_F^2/(2m)$. An insulator needs both the even filling and a positive indirect gap. At the touching threshold the gap is zero; a positive direct gap at every Bragg crossing is not by itself sufficient. The estimate neglects higher-plane-wave corrections and is not an exact critical amplitude.

Let $\alpha ={\hslash }^2/(2m^{\ast })$, so the initial conduction-electron kinetic energy is $E_1=\alpha k_1^2$. In the normal momentum-conserving process with the final hole fixed at zero wavevector, the two final conduction electrons obey ${\mathbf{k}}_1={\mathbf{k}}_2+{\mathbf{k}}_3$. Their total kinetic energy satisfiesEnergy conservation for impact ionization is $E_1=E_g+\alpha (k_2^2+k_3^2)$. Consequently $E_1\ge E_g+E_1/2$, givingThe bound is attained when ${\mathbf{k}}_2={\mathbf{k}}_3={\mathbf{k}}_1/2$; opposite spins permit the two electrons to share that wavevector. Fixing the hole momentum is essential: a mobile final hole gives a different threshold. This is the threshold for impact ionization with a hole at zero momentum.

The two hot electrons subsequently redistribute energy by electron-electron collisions and transfer it to the lattice by electron-phonon scattering. Optical-phonon emission can provide large discrete losses while the electrons are hot; acoustic phonons dominate later small energy exchanges. Electron-electron collisions alone conserve the electron system's total energy and do not cool it into the lattice. The electron-hole pair eventually recombines radiatively or through nonradiative channels, such as defect-assisted recombination or Auger recombination. Recombination is distinct from the preceding intraband cooling.