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At $U=0$, use fermionic Matsubara frequencies ${\omega }_r=(2r+1)\pi /\beta$. The fermionic Gaussian integral has inverse propagator $-i{\omega }_r+{\xi }_{\mathbf{k}}$. With its thermal normalization, the determinant for one single-particle level is $1+e^{-\beta \xi }$, as also follows from its empty and occupied states. Two spin species therefore giveThe expression labeled Helmholtz free energy in the PDF is the grand potential, since the trace contains $H-\mu \hat{N}$. The thermodynamic Helmholtz free energy at a prescribed electron number $N_e$ is $F={\Omega }_0+\mu N_e$, with $\mu$ fixed by $N_e=2{\sum }_{\mathbf{k}}{n}_F({ϵ}_{\mathbf{k}})$. The Matsubara determinant requires a normalization prescription; a raw infinite sum of logarithms is not an ordinary convergent sum.

The $-\mu \psi$ term means that the generator is the grand potential, rather than just the physical energy. With boundary variations vanishing, the Hamiltonian functional isIts functional derivative isso the Gross–Pitaevskii equation is $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$. Treating $\psi$ and ${\psi }^{\ast }$ as independent variables in this variation gives the factor $g$, not $2g$.

For repulsive interactions $g>0$ and $\mu >0$, complete the square in the local energy:The gradient contribution is nonnegative, so the homogeneous ground state has arbitrary constant complex argument andThe infinite-volume constant may be subtracted to define a relative grand potential; it does not affect the functional derivative. At fixed particle number one instead minimizes the physical energy $H+\mu N$, with $\mu$ as its Lagrange multiplier. For $g>0$, $\mu \le 0$ the grand-potential minimum over $n\ge 0$ is the vacuum $n=0$. For $g<0$ the displayed unconstrained functional is unbounded below; the positive-density stable uniform state assumed in the following parts requires repulsion.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

Differentiating the grand potential givesA reversible adiabatic change in an isolated gas keeps $S$ and $N$ fixed. To infer constant $y$, one must check that this ratio is one-to-one, rather than merely assuming it.

Put $q=y/k$, $n(x)=1/(1+e^{x-q})$, $F_j={\int }_0^{\infty }{x}^{j}n(x)dx$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(x)(1-n(x))dx$ for $j=0,1,2$. Integration by parts gives $M_1=b{F}_a$ and $M_2=(b+1)F_{a+1}$, andThe strict inequality is the Cauchy-Schwarz inequality for the positive weight $x^{a}n(x)(1-n(x))$, since $x$ is not constant on its support. This proves strict monotonicity of Fermi gas entropy per particle and hence $\mu /T$ is constant. Then $N=kV{T}^b{f}^{\prime }(y)$ gives $V{T}^b$ constant, and $P=k{T}^{b+1}f(y)$ givesThis adiabatic scaling of a Fermi gas with power-law density of states concerns positive temperature; the zero-temperature limit is a separate limiting case.

For noninteracting fermions, each single-particle level has occupation $n_i\in \{0,1\}$ by the Pauli exclusion principle. A microstate has $E={\sum }_i{ϵ}_i{n}_i$, $N={\sum }_i{n}_i$, so its grand-canonical weight is the product of its independent level weights. The grand potential is $\Omega =-kT\log \mathcal{Z}$; the particle number, entropy and pressure follow from $N=-{\partial }_{\mu }\Omega$, $S=-{\partial }_T\Omega$ and $P=-{\partial }_V\Omega$. This occupation description is the basis for the product, continuum and adiabatic calculations.

Summing separately over the two allowed occupations of every level givesThusThe factorization uses the absence of interactions; each level, including any degeneracy labels, is counted separately.

The grand potential is

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.