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For density of states $AV{ϵ}^a$, $a>-1$, the grand canonical ensemble has $\log \mathcal{Z}=V{T}^{a+1}f(\mu /T)$. Strict monotonicity of Fermi gas entropy per particle makes $\mu /T$ constant at fixed entropy and particle number. Therefore $V{T}^{a+1}$ and $P{V}^{(a+2)/(a+1)}$ are constant along a reversible adiabatic process.

For a two-dimensional Fermi gas of $N$ spin-one-half particles with Fermi energy $E_F$, the constant density of states makes the low-temperature internal energy correction and heat capacity at constant areaThus its low-temperature heat capacity obeys a power law with exponent one.

The spin-independent term cancels in the signed sum. The linearized stationary equation becomesThe even powers vanish because the equation is odd under $M\mapsto -M$. Equivalently, the mean-field action has curvatureThus the unpolarized stationary point becomes unstable whenAt low temperature, the derivative of the Fermi distribution samples a narrow window around $\mu$, so ${\chi }_0\simeq \nu (\mu )$ for a smooth density of states. Measuring energy from the Fermi level gives $U_c\nu (0)\simeq 1$, the Stoner criterion in the spin-summed convention fixed above. This identifies local instability of the paramagnetic state; the detailed transition order also depends on higher terms in the action.

In the isotropic ordered phase of itinerant ferromagnetism, continuous spin-rotation symmetry leaves a degenerate direction of magnetization. Its transverse fluctuations produce one gapless, quadratically dispersing mean-field itinerant ferromagnetic Goldstone mode, $E(\mathbf{q})\propto q^2$. The two broken spin generators form a canonical pair, as for the type-B Goldstone boson of the localized ferromagnet. The system also remains metallic, with exchange-split Fermi surfaces and low-energy spin-conserving particle-hole excitations. Longitudinal-amplitude fluctuations and the spin-flip continuum are distinct from the long-wavelength magnon; their energy and damping scales depend on the band structure. Spin anisotropy, if added, could gap the transverse mode, but it is absent from the stated Hamiltonian.

Replace the sum by the integral against the density of states and set $x=ϵ/(kT)$, $y=\mu /T$:ThereforeFor the given $a>0$, the integral converges at zero and infinity for every finite $y$.

For an ultrarelativistic particle, $\epsilon =pc$. Including the two spin states of an electron, counting wavevectors in a sphere gives the density of statesThe grand canonical partition function of the ideal Fermi gas is thereforeSince the grand potential satisfies $pV=k_{B}T\log \mathcal{Z}$, integration by parts givesThusThis is the equation of state of an ideal ultrarelativistic gas.

Let $g(E)$ denote the total density of states, including both spin directions. The field lowers one spin energy by ${\mu }_{B}B$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is thereforeMultiplying by the Bohr magneton carried by each excess aligned electron givesThus, with this total-density convention,The zero-temperature magnetic susceptibility isThis weak positive response is Pauli paramagnetism. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.

For an ideal gas of identical bosons, the Bose-Einstein distribution gives the mean occupation of a discrete one-particle state $r$ aswhere $\mu$ is the chemical potential. If the spectrum is sufficiently dense to use a density of states $g(E)$, the discrete sum becomes

For $\alpha =1/2$, the preceding density of states hasWhen $T\gg T_D$, the Bose-Einstein distribution has the classical approximationEach of the $2n$ modes therefore contributes $k_{B}T$ to the energy, so the Dulong-Petit law gives

When $T\ll T_D$, the upper integration limit tends to infinity. The Bose integralthen yieldsHence the heat capacity at constant volume isso $d=4$. The high-temperature constant is the classical equipartition theorem result, while the low-temperature power law makes $C_V\to 0$ as required by the Third law of thermodynamics.

The single-spin density of states counts momentum states for one spin species. For degenerate up and down species, the spin-summed density of states is twice this value. Specify separately whether the density is per unit volume or extensive. In the BCS gap equation, the momentum sum labels opposite-spin pairs once and therefore uses the single-spin density per volume when the interaction has normalization $V/\mathcal{V}$.

For a spin-degenerate band, the total density of states includes both spin species and is twice the single-spin density of states. Per lattice site, $\nu (ϵ)=2N^{-1}{\sum }_k\delta (ϵ-{ϵ}_k)$. Using a separate explicit spin sum then requires a compensating factor one half when converting momentum sums. The normalization matters in the Stoner criterion.

In the homogeneous spin-vector mean-field scheme with field cost $M^2/(2U)$ and spin-summed density of states per site, the paramagnetic curvature is proportional to $U^{-1}-\int \nu (ϵ)[-n_F^{\prime }(ϵ)]dϵ$. It changes sign at $U{\chi }_0=1$. For a smooth low-temperature density of states, ${\chi }_0\simeq \nu (\mu )$, giving the familiar Fermi-level criterion. Interaction and spin-counting conventions must be specified before comparing coefficients.

For density of states proportional to ${ϵ}^a$, $a>-1$, and positive temperature, set $q=\mu /(kT)$, $n(x)=(1+e^{x-q}{)}^{-1}$, $b=a+1$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(1-n)dx$. Differentiation and integration by parts giveThe inequality is strict Cauchy-Schwarz inequality, since $x$ is not constant under this positive weight. Thus fixed entropy per particle forces fixed $\mu /T$. This is the injectivity step behind the adiabatic scaling of an ideal Fermi gas.

For spin-one-half particles in three dimensions with dispersion $\epsilon =pc$, the density of states isAt zero temperature,