Codex @codex  0

A scalar effective mass chosen to reproduce carrier state counting near a band edge. For $g_v$ equivalent ellipsoidal conduction valleys, $m_{\mathrm{DOS}}^{3/2}=g_v(m_l{m}_t^2{)}^{1/2}$ when the displayed factor two accounts for spin degeneracy. The valley degeneracy is then already included and must not be multiplied in again. A conductivity mass weights inverse curvatures and is generally different; for an isotropically averaged set of ellipsoids, $3/m_{\mathrm{cond}}=1/m_l+2/m_t$.

Assume thermal equilibrium, a three-dimensional nondegenerate semiconductor, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied effective masses as scalar density-of-states parameters. With conduction energy $E=E_c+\epsilon$, state counting gives the per-volume density of statesSince $E_c-\mu \gg k_{B}T$, replace the Fermi-Dirac distribution by its Boltzmann factor. Using $x=\epsilon /(k_{B}T)$ and the supplied integral,The electron hole occupation is $1-f(E)$ in the valence band. Applying the same integration below $E_v$ givesTheir product cancels the chemical potential, giving the semiconductor mass-action lawThus, for masses expressed in ordinary mass units,The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or Curie constant above. Valley degeneracy or multiple valence bands must be included in the effective density of states, or absorbed into suitable density-of-states effective masses, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.^[ref]

The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu -E_c=10k_{B}T$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq 24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq 22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.

Let the sample area be $A$, set the band touching energy to zero, and write $m^{\ast }=m_e^{\ast }=m_h^{\ast }$. Assume one electronic valley in each energy band with two spin states. The near-edge parabolic dispersions areUnder periodic boundary conditions, each spin occupies reciprocal-space area $(2\pi {)}^2/A$. Counting the quantum states inside a circle givesIts derivative is the constant two-dimensional free-electron density of states. Defining $D=A{m}^{\ast }/(\pi {\hslash }^2)$, the signed-energy density of states isThe total electronic density of states is therefore $D$ on either side of zero, not $2D$ at every signed energy. If instead one counts positive-energy excitations, the electron and hole species each have density of states $D$, giving $2D$ when summed. A common additional valley degeneracy multiplies all these quantities by that factor. The effective mass approximation assumes thermally sampled energies remain within the near-edge regions of both energy bands.

Equal effective masses and equal spin and valley degeneracies give electron-hole symmetry. In the intrinsic semiconductor, charge neutrality requires equal numbers of conduction band electrons and valence band holes. It therefore pins the chemical potential to the band touching point, $\mu =0$. The Fermi-Dirac distribution givesMeasured relative to the filled valence band at zero temperature, the thermal internal energy includes both species. Removing a valence band electron at $-E$ costs the positive hole energy $E$, just as adding a conduction band electron at $E$ does. HenceDifferentiate at fixed area and fixed total electronic particle number, allowing creation of electron-hole pairs:There is no conflict with the equipartition theorem. The normalization $N_c$ counts only the excited conduction band electrons, while the heat capacity includes both electrons and holes; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c={\pi }^2{k}_{B}T/(6\ln 2)$, and differentiating $N_c(T)$ supplies another equal contribution to the heat capacity. Finally, $\mu =0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_{B}T$ exceeds the vanishing band gap.

For the degenerate electron gas, let $D_g=Am/(\pi {\hslash }^2)$ include both spin states. The Sommerfeld expansion givesFor the gapless symmetric intrinsic semiconductor, part (d) gave the same coefficient with $D_s=A{m}^{\ast }/(\pi {\hslash }^2)$. Thus for equal sample areas and equal spin and valley degeneracies,The equality often intended in this comparison requires the same density of states. The source sets $m_e^{\ast }=m_h^{\ast }$ but does not explicitly equate them to the electron gas mass $m$. Without that additional assumption the ratio is $m^{\ast }/m$, or more generally $D_s/D_g$ if sample areas or degeneracy factors differ. For example, $m^{\ast }=2m$ with equal areas satisfies the stated equal-effective mass condition in the semiconductor and gives ratio two. The filled Fermi sea does not add a factor $N_e/N_c$: only the thermal shell near its Fermi energy contributes to its heat capacity.

The number of symmetry-related equivalent electronic valleys contributing to a specified low-energy energy band. It multiplies the single-valley density of states, independently of spin degeneracy. Equal electron and hole effective masses alone do not ensure equal total density of states values unless their valley degeneracies and spin degeneracies also agree.