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Symmetry of the excitation spectrum under exchange of conduction band electrons and valence band holes, after choosing a central zero of energy. In a two-band model with equal effective masses and equal degeneracy factors, charge neutrality in an intrinsic semiconductor fixes the chemical potential at that central energy. Indeed the Fermi-Dirac distribution obeys $1-f(-E;\mu =0)=f(E;\mu =0)$, so excitation populations are equal at every positive energy.

At fixed finite $L$, the Helmholtz free energy of a periodic Potts chain receives its first thermal correction from a pair of domain walls. The two-domain-wall degeneracy of a Potts ring divided by the ground state degeneracy factor supplies the coefficient. Expanding the exact canonical partition function proves the formula. The expansion is not uniform in $L$: taking the thermodynamic limit first gives $F/L=-J-{\beta }^{-1}\log [1+(Q-1)e^{-\beta J}]$, whose leading correction is of order $e^{-\beta J}$.

Assume $Q\ge 2$ and $L\ge 2$; the trivial exceptions are described below. In this periodic Potts chain, every satisfied bond contributes $-J$ to the energy, and no bond can contribute less. All $L$ bonds can be satisfied simultaneously by choosing the same state at every site. Conversely, satisfying every bond forces this uniform configuration. Therefore the ground state energy and degeneracy factor areThe $Q$ choices are distinct configurations on the labelled lattice: a global relabelling of the states is a symmetry, not an identification of microstates.

The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix byHere ${\delta }_{ab}$ is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition functionHence, writing $I$ for the identity matrix and $P$ for the all-ones matrix,Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. ThusMore explicitly, for $u=(1,\dots ,1{)}^T$, $Pu=Qu$. If ${\sum }_a{v}_a=0$, then $Pv=0$. These complementary eigenspaces have dimensions $1$ and $Q-1$. Equivalently, $P/Q$ is the orthogonal projector onto the linear span of $u$. The eigenvalues and their eigenvalue multiplicities are thereforeSince the two matrices share these eigenvectors, the exact finite-ring canonical partition function isBoth terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

Put $x=e^{-\beta J}$. Factoring out the ground state Boltzmann factor from the exact canonical partition function givesAt fixed finite $L$, the binomial theorem givesTaking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,The constant term is the ground state energy. The term $-{\beta }^{-1}\log Q$ is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:Each pair of domain walls costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of domain walls, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta )$.

This is an expansion for fixed $L$, not a uniform thermodynamic limit. In particular, the small correction requires $\left(\genfrac{}{}{0ex}{}{L}{2}\right)(Q-1)x^2\ll 1$, with higher terms also controlled as $x\to 0$ at fixed $L,Q$. Taking $L\to \infty$ first instead selects ${\lambda }_{+}$ and givesIts leading correction is of order $x$. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.

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 degeneracy factor $g_A$ counts internal states with the same energy, including the allowed spin states. The chemical potential ${\mu }_A$ measures the free-energy cost of changing the number of particles of species $A$; in this convention the rest energy remains in the exponential of the nonrelativistic Maxwell--Boltzmann number density.

The chemical potential is fixed by the densities of conserved charges and by chemical equilibrium between reactions. For a reaction ${\sum }_A{\nu }_{A}A=0$, chemical equilibrium requires ${\sum }_A{\nu }_A{\mu }_A=0$. Equivalently, when these are the only constraints, ${\mu }_A$ is a linear combination of the chemical potentials of the conserved charges carried by $A$. An unconstrained species has zero chemical potential. In particular, the photon chemical potential vanishes in thermal equilibrium when photon-number-changing reactions are effective.

Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

Let $N_u,N_l$ be the upper- and lower-level populations and $g_u,g_l$ their degeneracy factors. At thermal equilibrium, the Boltzmann distribution givesNo degeneracy factors are supplied, so the numerical answers use equal degeneracy factors. Stating the ratio as upper population divided by lower population removes an otherwise arbitrary inversion.

(a) At $T=500\mathrm{K}$,The two populations are nearly equal because the level separation is much smaller than the thermal energy $k_{B}T$.

(b) At $T=2\mathrm{mK}=0.002\mathrm{K}$,Almost every atom occupies the lower level. If the levels have unequal degeneracy factors, each boxed ratio must be multiplied by $g_u/g_l$; their individual values cannot be inferred from the paper.

For $Q,L\ge 2$, a periodic Potts chain has $Q$ uniform ground states with energy $-JL$. A single domain wall cannot close around the ring. Its first excited states have two domain walls and energy $-JL+2J$. Choose the wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways and assign distinct states to the two resulting arcs in $Q(Q-1)$ ways. This proves the displayed degeneracy factor without identifying globally relabelled configurations.