Ideal Fermi Gas

The ideal Fermi gas is a system of noninteracting identical fermions. Its Hamiltonian is a sum of one-particle energies, yet at nonzero bulk density its ground state has kinetic energy and pressure because Pauli exclusion forces particles to fill distinct modes.

For a uniform gas in a cubic box of volume $V=L^3$,

$$H = \sum_{\mathbf k,a} \epsilon_{\mathbf k} c_{\mathbf k a}^\dagger c_{\mathbf k a}, \qquad \epsilon_{\mathbf k} = \frac{\hbar^2k^2}{2m}.$$

The index $a=1,\ldots,g$ labels independent spin or internal modes. The number operator is

$$N = \sum_{\mathbf k,a} c_{\mathbf k a}^\dagger c_{\mathbf k a}.$$

The model is exactly solvable because every occupation number is conserved and satisfies

$$n_{\mathbf k a} \in \{0,1\}.$$

The ideal gas is the reference point for electrons in simple metals, cold fermionic atoms, nuclear and astrophysical estimates, Fermi-liquid theory, and any calculation organized around a filled Fermi sea.

We consider three spatial dimensions, a periodic cubic box, temperature $T\geq0$, and a nonrelativistic quadratic dispersion. The $g$ internal components have the same mass and dispersion and share a chemical potential; their mean populations are equal. The thermodynamic limit keeps total density $n=N/V$ fixed. Here $N$ denotes a grand-canonical mean when particle exchange is allowed. The finite canonical example compares exact number constraints with a grand ensemble; the Fermi-scale reference compares dimensions and component conventions.

Helpful background. Fermi–Dirac statistics gives the mode probabilities; the thermodynamic limit explains the continuum state count.

Fermionic modes in a periodic box

Periodic boundary conditions give one-particle modes

$$\phi_{\mathbf k a}(\mathbf r) = \frac{1}{\sqrt V} e^{i\mathbf k\cdot\mathbf r} \chi_a,$$

with

$$\mathbf k = \frac{2\pi}{L} \mathbf n, \qquad \mathbf n \in \mathbb Z^3.$$

The many-body eigenstates are fermionic occupation-number states

$$\lvert\{n_{\mathbf k a}\}\rangle.$$

Their particle number and energy are

$$N = \sum_{\mathbf k,a} n_{\mathbf k a},$$

and

$$E = \sum_{\mathbf k,a} \epsilon_{\mathbf k} n_{\mathbf k a}.$$

There is no interparticle potential energy. For a finite box, up to $g$ particles can occupy the zero-momentum shell with zero kinetic energy in this convention; filling beyond it forces occupation of excited shells. The positive bulk ground-state energy follows at fixed nonzero density.

The internal label is part of a complete one-particle mode. Fermions may share a spatial orbital when their internal states differ; exclusion forbids sharing the same spatial-and-internal mode. Fermionic Fock space already incorporates antisymmetry, so introducing particle labels and antisymmetrizing the occupation configurations again would double-count states.

Energy-zero covariance

A constant one-particle energy shift $\Delta_0$ gives

$$\epsilon_{\mathbf k}'=\epsilon_{\mathbf k}+\Delta_0, \qquad H'=H+\Delta_0N.$$

The same grand-canonical state requires $\mu'=\mu+\Delta_0$, which leaves $H'-\mu'N=H-\mu N$ unchanged. At fixed $N$, energy and Helmholtz free energy increase by $N\Delta_0$, while occupations and entropy are unchanged. Pressure is unchanged if $\Delta_0$ is volume independent. Thus the pressure–energy relation below uses kinetic energy measured from $\epsilon_{\mathbf0}=0$, not an arbitrarily shifted total energy.

Grand Partition Function

Define

$$z \equiv e^{\beta\mu}, \qquad \beta = \frac{1}{k_{\mathrm B}T}.$$

One fermionic mode contributes

$$\xi_{\mathbf k a} = 1+ze^{-\beta\epsilon_{\mathbf k}}.$$

Therefore

$$\Xi = \prod_{\mathbf k,a} \left( 1+ze^{-\beta\epsilon_{\mathbf k}} \right),$$

and

$$\ln\Xi = \sum_{\mathbf k,a} \ln \left( 1+ze^{-\beta\epsilon_{\mathbf k}} \right).$$

Unlike the bosonic product, every finite fermionic mode factor is finite for every finite real $\mu$. Infinite-volume expressions still require a thermodynamic limit and a controlled density of states.

At finite volume and $T>0$, the quadratic spectrum also makes the full infinite-mode product converge: high-energy contributions are exponentially suppressed. A numerical mode cutoff controls approximation error; it is not an additional physical ultraviolet cutoff for this model.

The mode occupation is

$$\overline n_{\mathbf k a} = \frac{1}{ z^{-1}e^{\beta\epsilon_{\mathbf k}}+1 }.$$

For an exactly fixed particle number, extract the coefficient of $z^N$:

$$Z_N(\beta)=[z^N]\prod_\alpha \left(1+ze^{-\beta\epsilon_\alpha}\right),$$

where $\alpha$ labels complete modes. This constrains the joint occupations; it does not give independent canonical mode ensembles. The following finite spectrum makes the distinction visible before taking a continuum limit.

Finite canonical example

Retain two spatial orbitals with energies $0$ and $\Delta>0$, each with internal labels $\uparrow$ and $\downarrow$. This finite-spectrum illustration has four complete modes; it is not the full periodic-box spectrum. With exactly $N=2$, its $\binom42=6$ Slater determinants fall into three groups.

Occupation pattern	Energy	Degeneracy
both internal states in the lower orbital	$0$	$1$
one fermion in each spatial orbital	$\Delta$	$4$
both internal states in the upper orbital	$2\Delta$	$1$

Writing $x=e^{-\beta\Delta}$,

$$\begin{aligned} \Xi&=(1+z)^2(1+zx)^2,\\ Z_2&=1+4x+x^2,\\ U&=\Delta\frac{4x+2x^2}{1+4x+x^2}. \end{aligned}$$

The limits are $U\to0$ as $T\to0$ and $U\to\Delta$ as $T\to\infty$. Both particles occupy the lower spatial orbital in the ground state, but their complete modes differ by internal label. There is no thermodynamic Fermi surface or phase transition in this finite example.

At $\beta\Delta=\ln4$, the canonical result is $U/\Delta=6/11$. For the grand ensemble tuned to the same mean number $\langle N\rangle=2$, particle–hole symmetry gives $\mu=\Delta/2$. Each upper mode then has mean occupation $1/3$, giving $U/\Delta=2/3$. Equal mean number does not make the ensembles identical: the grand ensemble also includes the sectors $N=0,1,3,4$.

Momentum-Space State Counting

One momentum mode per internal state occupies volume

$$\left( \frac{2\pi}{L} \right)^3$$

in $\mathbf k$-space. Including all $g$ internal modes, the leading continuum count of one-particle states with magnitude below $k$ is

$$\mathcal N(k) = g \frac{V}{(2\pi)^3} \frac{4\pi k^3}{3} = \frac{gVk^3}{6\pi^2}.$$

At zero temperature and exactly fixed $N$, the lowest $N$ modes are occupied, with possible degeneracy in a partially filled shell. For the isotropic dispersion, they fill a sphere of radius $k_{\mathrm F}$ in momentum space.

At $T=0$, the occupied states fill a sphere; the upper panel is a planar cross-section. The lower panel plots $\epsilon_{\mathrm F}D(\epsilon)/N=(3/2)\sqrt{\epsilon/\epsilon_{\mathrm F}}$. Its shaded area from zero to one is exactly one, expressing the same total state count. Internal degeneracy is included in $N$ and $D$.

Fermi Momentum

Equating the state count to the particle number gives

$$N = \frac{gVk_{\mathrm F}^3}{6\pi^2}.$$

Thus

$$n = \frac{gk_{\mathrm F}^3}{6\pi^2},$$

and

$$k_{\mathrm F} = \left( \frac{6\pi^2n}{g} \right)^{1/3}.$$

For spin-$1/2$ fermions with two equally populated internal states, $g=2$:

$$k_{\mathrm F} = \left( 3\pi^2n \right)^{1/3}.$$

The density $n$ here is the total density summed over internal states. Using the density per spin component in the same formula without adjusting $g$ double-counts the degeneracy.

Fermi Energy, Temperature, and Velocity

The Fermi energy is the one-particle energy at $k_{\mathrm F}$:

$$\epsilon_{\mathrm F} = \frac{\hbar^2k_{\mathrm F}^2}{2m}.$$

The Fermi temperature is

$$T_{\mathrm F} \equiv \frac{\epsilon_{\mathrm F}}{k_{\mathrm B}},$$

and the Fermi velocity is

$$v_{\mathrm F} = \frac{1}{\hbar} \left. \frac{d\epsilon_k}{dk} \right|_{k=k_{\mathrm F}} = \frac{\hbar k_{\mathrm F}}{m}.$$

These are density scales, not independent parameters. For the uniform ideal gas,

$$\epsilon_{\mathrm F} \propto n^{2/3}.$$

Three-Dimensional Density of States

Using

$$k = \frac{\sqrt{2m\epsilon}}{\hbar},$$

the integrated state count is

$$\mathcal N(\epsilon) = \frac{gV}{6\pi^2} \left( \frac{2m\epsilon}{\hbar^2} \right)^{3/2}.$$

The density of states is

$$D(\epsilon) = \frac{gV}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt\epsilon.$$

It includes the internal degeneracy $g$. If a quoted density of states is per spin or per component, $g$ must not be inserted again.

At the Fermi energy,

$$D(\epsilon_{\mathrm F}) = \frac{3N}{2\epsilon_{\mathrm F}}.$$

Per unit volume,

$$\frac{D(\epsilon_{\mathrm F})}{V} = \frac{3n}{2\epsilon_{\mathrm F}}.$$

The density of states near $\epsilon_{\mathrm F}$ controls low-temperature heat capacity, compressibility, and many weak-response coefficients.

Zero-Temperature Occupation

At fixed density, the chemical potential approaches the Fermi energy as $T\to0$:

$$\mu(0) = \epsilon_{\mathrm F}.$$

The occupation becomes

$$\overline n_{\mathbf k a} = \Theta \left( k_{\mathrm F}-k \right),$$

away from the boundary. Equivalently,

$$\overline n(\epsilon) = \Theta \left( \epsilon_{\mathrm F}-\epsilon \right).$$

The filled region is the Fermi sea. The sphere $k=k_{\mathrm F}$ is the Fermi surface of this isotropic continuum model.

The Fermi surface is a surface in momentum space, not a physical boundary in the box.

Ground-State Energy

At zero temperature,

$$U_0 = \int_0^{\epsilon_{\mathrm F}} d\epsilon\, D(\epsilon)\epsilon.$$

Since $D(\epsilon)\propto\sqrt\epsilon$,

$$U_0 = \frac{3}{5} N\epsilon_{\mathrm F}.$$

Thus the mean energy per particle is

$$\frac{U_0}{N} = \frac{3}{5} \epsilon_{\mathrm F}.$$

The ground-state energy is nonzero even without interactions. Pauli exclusion forces particles into modes with increasing kinetic energy.

Degeneracy Pressure

For a nonrelativistic quadratic dispersion in three dimensions,

$$P = \frac{2U}{3V}.$$

At zero temperature,

$$P_0 = \frac{2}{5} n\epsilon_{\mathrm F}.$$

Because $\epsilon_{\mathrm F}\propto n^{2/3}$,

$$P_0 \propto n^{5/3}.$$

It can also be derived directly from the ground-state energy: at fixed $N$ and $g$, $\epsilon_{\mathrm F}\propto V^{-2/3}$, so

$$P_0=-\left(\frac{\partial U_0}{\partial V}\right)_N =\frac{2U_0}{3V}.$$

The same pressure–energy relation follows at finite temperature because each periodic-box kinetic level scales as $V^{-2/3}$. At finite volume pressure is defined by this derivative; $P=-\Omega_{\mathrm G}/V$ is a bulk extensivity relation. This pressure persists at $T=0$. It is called degeneracy pressure and follows from the density dependence of the allowed fermionic kinetic states, not from thermal motion or a pairwise repulsive force.

Zero-Temperature Compressibility

Differentiating

$$P_0 = \frac{2}{5}n\epsilon_{\mathrm F}$$

at fixed particle species gives

$$\frac{dP_0}{dn} = \frac{2}{3} \epsilon_{\mathrm F}.$$

The zero-temperature compressibility is therefore

$$\kappa_0 = \frac{1}{n} \left( \frac{\partial n}{\partial P} \right)_{T=0} = \frac{3}{2n\epsilon_{\mathrm F}}.$$

Susceptibilities distinguishes this isothermal normalization from $\partial n/\partial\mu$ and from the retarded density response, including the required order of limits.

Equivalently,

$$\left( \frac{\partial n}{\partial\mu} \right)_{T=0} = \frac{D(\epsilon_{\mathrm F})}{V} = \frac{3n}{2\epsilon_{\mathrm F}}.$$

Finite-Temperature Number and Energy

At positive temperature,

$$N = \int_0^\infty d\epsilon\, D(\epsilon) \frac{1}{ e^{\beta(\epsilon-\mu)}+1 },$$

and

$$U = \int_0^\infty d\epsilon\, D(\epsilon) \frac{\epsilon}{ e^{\beta(\epsilon-\mu)}+1 }.$$

At fixed $n$, the number equation determines $\mu(T)$. Treating $\mu$ as fixed at $\epsilon_{\mathrm F}$ for all temperatures violates the fixed-density constraint.

The pressure remains related to energy by

$$P = \frac{2U}{3V}$$

for the uniform ideal gas with quadratic dispersion at every temperature.

Thermal Wavelength and Polylogarithm Form

Use the thermal wavelength

$$\lambda_T = \sqrt{ \frac{2\pi\hbar^2}{mk_{\mathrm B}T} }.$$

The continuum grand partition function is

$$\ln\Xi = - \frac{gV}{\lambda_T^3} \operatorname{Li}_{5/2}(-z).$$

The number equation becomes

$$n = - \frac{g}{\lambda_T^3} \operatorname{Li}_{3/2}(-z),$$

and the pressure is

$$P = - \frac{gk_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(-z).$$

For $s>0$ the real Fermi integral defines the relevant branch at every $z>0$:

$$-\operatorname{Li}_s(-z)=\frac1{\Gamma(s)}\int_0^\infty \frac{x^{s-1}\,dx}{z^{-1}e^x+1}>0.$$

The power series in $z$ used below does not converge for a deeply degenerate gas with $z>1$; use this integral or an appropriate asymptotic evaluation there.

Some references define normalized Fermi integrals with indices shifted by one. Stating the polylogarithm form prevents an otherwise common index-convention ambiguity.

With $\theta=T/T_{\mathrm F}$, the density and Fermi scales give

$$\frac{n\lambda_T^3}{g} =\frac{4}{3\sqrt\pi}\theta^{-3/2}.$$

Thus $\theta\ll1$ and $n\lambda_T^3/g\gg1$ describe the same degenerate regime for this balanced three-dimensional quadratic gas. The numerical relation changes with dimension or dispersion.

Dilute pressure correction

When the phase-space density per internal component is small, $n\lambda_T^3/g\ll1$, the equation of state becomes

$$\frac{P}{nk_{\mathrm B}T}=1+\frac{n\lambda_T^3}{2^{5/2}g} +O\!\left[\left(\frac{n\lambda_T^3}{g}\right)^2\right].$$

Pauli exclusion raises the pressure relative to the classical gas at the same density and temperature. The classical-limit derivation obtains this statistical correction by series inversion; it is not an interparticle repulsive potential.

Degenerate Regime

The gas is quantum degenerate when

$$T \ll T_{\mathrm F}.$$

Most modes far below $\epsilon_{\mathrm F}$ remain occupied, and most modes far above remain empty. Only states within an energy window of order $k_{\mathrm B}T$ around the Fermi energy change occupation appreciably.

The fraction of particles participating in thermal excitations is therefore of order

$$\frac{T}{T_{\mathrm F}},$$

not order one. This is why the low-temperature heat capacity is linear in $T$ rather than approaching the classical value $3Nk_{\mathrm B}/2$.

Sommerfeld Expansion Preview

For a smooth function $\varphi(\epsilon)$ and $k_{\mathrm B}T\ll\mu$,

$$\int_0^\infty d\epsilon\, \varphi(\epsilon) f_{\mathrm F}(\epsilon) ={} \int_0^\mu d\epsilon\, \varphi(\epsilon) + \frac{\pi^2}{6} (k_{\mathrm B}T)^2 \varphi'(\mu) + O(T^4).$$

This is the leading Sommerfeld expansion. That page derives the higher terms and specifies the regularity and endpoint conditions. Here it is used only to state the ideal-gas low-temperature results.

Low-Temperature Chemical Potential

At fixed density in the three-dimensional ideal gas,

$$\mu(T) = \epsilon_{\mathrm F} \left[ 1 - \frac{\pi^2}{12} \left( \frac{T}{T_{\mathrm F}} \right)^2 + O \left( \frac{T^4}{T_{\mathrm F}^4} \right) \right].$$

The chemical potential decreases quadratically from $\epsilon_{\mathrm F}$ at low temperature.

This coefficient depends on the energy dependence of the three-dimensional free-particle density of states. It is not the same for every band structure or dimension.

Low-Temperature Energy and Heat Capacity

At fixed density,

$$\frac{U}{N} = \frac{3}{5} \epsilon_{\mathrm F} \left[ 1 + \frac{5\pi^2}{12} \left( \frac{T}{T_{\mathrm F}} \right)^2 + O \left( \frac{T^4}{T_{\mathrm F}^4} \right) \right].$$

Differentiating gives

$$\frac{C_V}{N} = \frac{\pi^2}{2} k_{\mathrm B} \frac{T}{T_{\mathrm F}} + O \left( k_{\mathrm B}\frac{T^3}{T_{\mathrm F}^3} \right).$$

The entropy has the same leading coefficient:

$$\frac{S}{N} = \frac{\pi^2}{2} k_{\mathrm B} \frac{T}{T_{\mathrm F}} + O \left( k_{\mathrm B}\frac{T^3}{T_{\mathrm F}^3} \right).$$

Only the thin thermally active shell near the Fermi surface contributes at low temperature.

Low-Temperature Pressure

Using $P=2U/(3V)$,

$$P ={} \frac{2}{5} n\epsilon_{\mathrm F} \Biggl[ 1 + \frac{5\pi^2}{12} \left( \frac{T}{T_{\mathrm F}} \right)^2 + O \left( \frac{T^4}{T_{\mathrm F}^4} \right) \Biggr].$$

Thermal pressure is a correction to the already nonzero degeneracy pressure.

Number Fluctuations

For independent grand-canonical modes,

$$\operatorname{Var}(N) = \sum_{\mathbf k,a} \overline n_{\mathbf k a} \left( 1- \overline n_{\mathbf k a} \right).$$

At zero temperature, modes away from the Fermi surface have occupations zero or one and therefore no mode-number variance. At low temperature, fluctuations are concentrated near $\epsilon_{\mathrm F}$.

The fluctuation–response relation is

$$\left( \frac{\partial\overline N}{\partial\mu} \right)_{T,V} = \beta \operatorname{Var}(N).$$

A fixed-$N$ canonical gas has no total-number fluctuation and has correlations among mode occupations. Ensemble equivalence for bulk thermodynamics does not make every global fluctuation identical.

Pauli Blocking

The ideal gas has no collisions by itself. When weak scattering is added and a kinetic factorization is justified, a final fermionic mode contributes an availability factor

$$1-f_{\mathrm F}(\epsilon_f).$$

At low temperature, modes deep inside the Fermi sea are already occupied and unavailable as final states. This suppresses scattering and relaxation channels whose final states would lie below the Fermi surface.

Pauli blocking follows from antisymmetry. It should not be described as an additional force between particles.

Equal-point exchange hole

For the normally ordered pair density, identical-component coincidence is forbidden: $g_{aa}^{(2)}(0)=0$. In the balanced grand-canonical Gaussian state, distinct components are independent. With $\hat n=\sum_a\hat n_a$ and equal mean component densities, the unresolved coincidence is

$$g^{(2)}(0) =\frac{\langle:\!\hat n(\mathbf r)^2\!:\rangle}{n^2} =1-\frac1g.$$

The same-component zero follows from antisymmetry alone. The total-density result also assumes independent balanced components; a general finite canonical thermal state is not Gaussian, and fixed-number or other component correlations can change it. The exchange hole is not evidence for a repulsive term in the Hamiltonian. Field-operator applications and structure factors develop the corresponding correlation observables.

Finite Size and Shell Structure

In a finite box, the allowed momenta are discrete. The ground state fills complete shells of equal $k^2$ when possible. If the last shell is partially filled, the ground state can be degenerate.

Consequences include:

• stepwise changes in addition energy;
• shell-dependent chemical potentials;
• finite-size oscillations in energy and response;
• sensitivity to boundary conditions;
• no perfectly sharp continuum Fermi surface.

Addition and removal thresholds

Order the complete one-particle energies, including internal multiplicity, as $\epsilon_{(1)}\leq\epsilon_{(2)}\leq\cdots$. Then

$$\begin{aligned} E_0(N)&=\sum_{j=1}^N\epsilon_{(j)},\\ \mu_-(N)&=E_0(N)-E_0(N-1)=\epsilon_{(N)},\\ \mu_+(N)&=E_0(N+1)-E_0(N)=\epsilon_{(N+1)}. \end{aligned}$$

At strict zero temperature, $\mu_-(N)<\mu<\mu_+(N)$ selects the $N$-particle sector in the grand ensemble. A closed shell has a gap and a nonempty interval. At an open shell the thresholds coincide, the canonical ground state can be degenerate, and a single grand-canonical step does not specify arbitrary partial filling. A limiting prescription, or an explicit fixed-$N$ state or mixture, is then needed.

When the continuum smooths the shells

Many occupied modes require $k_{\mathrm F}L\gg1$. The smooth density of states estimates a mean level spacing

$$\delta_{\mathrm F}\simeq\frac{1}{D(\epsilon_{\mathrm F})} =\frac{2\epsilon_{\mathrm F}}{3N}.$$

A continuum low-temperature calculation needs the window $\delta_{\mathrm F}\ll k_{\mathrm B}T\ll\epsilon_{\mathrm F}$. The first inequality is only a mean-spacing condition: cubic-box levels occur in degenerate arithmetic shells, so temperature must also smooth the actual shell gaps and oscillations to the required accuracy. Taking $T\to0$ first at finite $N$ resolves those shells instead of producing a smooth surface. Finite-Size Effects explains why a small box may have no broad smoothing window.

Numerical controls

The ideal Fermi gas benchmark provides reference values and low-temperature coefficients. Solve the number equation for $\mu(T)$ at each fixed density; its positive derivative, proportional to $\sum_\alpha f_\alpha(1-f_\alpha)$, makes the positive-temperature root unique. A finite retained mode set requires a target mean number strictly between zero and its capacity.

Use a branch-stable logistic occupation to avoid exponential overflow and a bracketed number solve. Report the number residual, momentum or energy quadrature cutoff, and asymptotic truncation separately. Test a decreasing sequence of $T/T_{\mathrm F}$ rather than treating a low-order Sommerfeld formula as exact, and obtain heat capacity from analytic derivatives or a convergence-tested fit instead of a noisy two-point difference. For a small system with exact $N$, use coefficient extraction or constrained enumeration rather than interpreting grand-canonical number fluctuations as solver error.

Example: Electrons in a Simple Metal

For free electrons, $g=2$ and

$$k_{\mathrm F} = (3\pi^2n)^{1/3}.$$

The ideal gas explains the existence of a Fermi energy, degeneracy pressure, and a heat capacity much smaller than the classical prediction at ordinary temperatures when $T\ll T_{\mathrm F}$.

Real electrons move in a crystal potential and interact. Band dispersions, effective masses, multiple Fermi-surface sheets, electron–electron interactions, and phonons modify the free-gas picture. For material-specific spectra and transport, see Fermi surfaces and the linked band-structure discussion.

Example: Ultracold Fermionic Atoms

For a dilute gas of neutral fermionic atoms, the internal-state degeneracy and population balance are experimentally controlled. In the weakly interacting limit, the ideal gas supplies $k_{\mathrm F}$, $\epsilon_{\mathrm F}$, $T_{\mathrm F}$, and the baseline density profile.

Interaction strength is often compared with the Fermi scale. Near a Feshbach resonance, pairing and strong correlations make the ideal model insufficient, but its Fermi units remain useful reference scales.

Example: Nuclear and Astrophysical Matter

Neutrons, protons, and electrons are fermions, so ideal-gas degeneracy pressure provides a first estimate for dense matter.

The nonrelativistic formula

$$P_0 = \frac{2}{5}n\epsilon_{\mathrm F}$$

fails when momenta become relativistic or when strong interactions dominate. White-dwarf electrons can require a relativistic Fermi gas, while neutron-star matter requires nuclear interactions, relativity, composition constraints, and gravity.

The ideal nonrelativistic gas is an organizing baseline, not a realistic equation of state for compact stars.

Other Dimensions and Dispersions

The numerical coefficients above are specific to

$$d = 3, \qquad \epsilon_k = \frac{\hbar^2k^2}{2m}.$$

In another dimension, the $k$-space volume and density of states change. On a lattice, the Fermi sea follows a band dispersion and need not be spherical. For anisotropic effective masses, constant-energy surfaces are ellipsoids rather than spheres.

The robust procedure is:

1. Specify the dispersion and degeneracies.
2. Count states below the chemical potential.
3. Integrate the Fermi–Dirac occupations.

What the Ideal Model Omits

The ideal Fermi gas omits:

• interparticle interactions;
• self-energy shifts and finite lifetimes;
• pairing and superconductivity;
• screening and collective modes;
• crystal bands and lattice geometry;
• disorder;
• relativistic dispersion;
• collision rates and equilibration mechanisms.

Every free-mode occupation is conserved. Assigning an equilibrium ensemble therefore does not show that the isolated Hamiltonian thermalizes an arbitrary initial distribution.

An interacting Fermi liquid may retain a Fermi surface and quasiparticles, but its effective mass, compressibility, heat capacity, and response contain interaction corrections. Random Phase Approximation uses this ideal gas as the reference polarization and resums induced density feedback to describe screening and collective charge modes. A strongly correlated system may not admit a simple quasiparticle description at all.

Exercises

Derive the ground-state energy and pressure

Show that

$$U_0 = \frac{3}{5}N\epsilon_{\mathrm F}$$

and

$$P_0 = \frac{2}{5}n\epsilon_{\mathrm F}.$$

Solution

Write the density of states as

$$D(\epsilon) = A\epsilon^{1/2}.$$

The number is

$$N = A \int_0^{\epsilon_{\mathrm F}} d\epsilon\, \epsilon^{1/2} = \frac{2A}{3} \epsilon_{\mathrm F}^{3/2}.$$

The energy is

$$\begin{aligned} U_0 &= A \int_0^{\epsilon_{\mathrm F}} d\epsilon\, \epsilon^{3/2} \\ &= \frac{2A}{5} \epsilon_{\mathrm F}^{5/2} \\ &= \frac{3}{5} N\epsilon_{\mathrm F}. \end{aligned}$$

For a quadratic dispersion in three dimensions,

$$P_0 = \frac{2U_0}{3V} = \frac{2}{5} n\epsilon_{\mathrm F}.$$

Recover the linear heat capacity

Use

$$\frac{U}{N} = \frac{3}{5} \epsilon_{\mathrm F} \left[ 1+ \frac{5\pi^2}{12} \left( \frac{T}{T_{\mathrm F}} \right)^2 \right]$$

to derive the leading low-temperature heat capacity at fixed $N$ and $V$.

Solution

The temperature-dependent energy correction is

$$\begin{aligned} \Delta U &= N \frac{3}{5} \epsilon_{\mathrm F} \frac{5\pi^2}{12} \frac{T^2}{T_{\mathrm F}^2} \\ &= N \frac{\pi^2}{4} \frac{k_{\mathrm B}^2T^2}{ \epsilon_{\mathrm F} }. \end{aligned}$$

Differentiating gives

$$\begin{aligned} C_V &= \left( \frac{\partial U}{\partial T} \right)_{N,V} \\ &= N \frac{\pi^2}{2} \frac{k_{\mathrm B}^2T}{ \epsilon_{\mathrm F} } \\ &= N \frac{\pi^2}{2} k_{\mathrm B} \frac{T}{T_{\mathrm F}}. \end{aligned}$$

Polarization at fixed total density

Two independently conserved spin components have densities $n_\uparrow=n(1+p)/2$ and $n_\downarrow=n(1-p)/2$. At zero temperature, compare their total pressure with the unpolarized pressure $P_0(n,p=0)$. Why cannot one simply keep $g=2$ with a common Fermi radius when $p\neq0$?

Solution

Each component has $g=1$ and contributes a pressure proportional to $n_\sigma^{5/3}$. Therefore

$$\frac{P_0(n,p)}{P_0(n,0)} =\frac{(1+p)^{5/3}+(1-p)^{5/3}}2\geq1.$$

The inequality follows from convexity. Complete polarization gives a factor $2^{2/3}$: fewer internal modes are available at the same total density, so occupied momenta must extend farther. Unequal component populations require separate Fermi radii and, in equilibrium with separately conserved populations, separate chemical potentials.

Shift the energy zero

Let every one-particle energy increase by the same constant $\Delta_0$. Find the transformations of the canonical partition function, Helmholtz free energy, chemical potential, and grand partition function.

Solution

Every fixed-$N$ many-body energy gains $N\Delta_0$, so

$$Z_N'=e^{-\beta N\Delta_0}Z_N, \qquad F_N'=F_N+N\Delta_0.$$

Set $\mu'=\mu+\Delta_0$. Then $\epsilon_\alpha'-\mu'=\epsilon_\alpha-\mu$, and

$$\Xi'(T,V,\mu')=\Xi(T,V,\mu).$$

Occupations and entropy are unchanged. A volume-independent $\Delta_0$ also leaves pressure unchanged, while internal energy increases by $N\Delta_0$ in the canonical ensemble, or by $\langle N\rangle\Delta_0$ in the grand ensemble with the shifted chemical potential.

Number fluctuations in the active shell

Derive the grand-canonical fluctuation–response relation and estimate the leading low-temperature number variance of the three-dimensional gas. Take the continuum limit before the low-temperature limit, and evaluate the result at a chemical potential adjusted to the prescribed mean density.

Solution

Independence of the grand-canonical modes and differentiation of their occupations give

$$\operatorname{Var}(N)=\sum_\alpha f_\alpha(1-f_\alpha), \qquad \frac{\partial f_\alpha}{\partial\mu} =\beta f_\alpha(1-f_\alpha).$$

Hence, at fixed $T$ and $V$,

$$\operatorname{Var}(N)=k_{\mathrm B}T \left(\frac{\partial\langle N\rangle}{\partial\mu}\right)_{T,V}.$$

For $\theta=T/T_{\mathrm F}\ll1$, the leading derivative is the total density of states at the Fermi energy:

$$\left(\frac{\partial\langle N\rangle}{\partial\mu}\right)_{T,V} =\frac{3\langle N\rangle}{2\epsilon_{\mathrm F}} \left[1+O(\theta^2)\right].$$

Therefore

$$\operatorname{Var}(N)=\frac32\langle N\rangle\theta +O(\langle N\rangle\theta^3).$$

Only the thermal shell contributes. The relative root-mean-square fluctuation is asymptotically $\sqrt{3\theta/(2\langle N\rangle)}$, vanishing in the thermodynamic limit at fixed small $\theta$. This is a grand-ensemble statement; an exactly fixed-$N$ gas has zero total-number variance.

Diagnose an open shell

Use the ordered finite-volume spectrum to determine when a strict zero-temperature chemical-potential interval selects exactly $N$ particles. What changes if $\epsilon_{(N)}=\epsilon_{(N+1)}$?

Solution

The sector’s grand ground energy is $\Phi_N=E_0(N)-\mu N$. Comparing it with its neighbors gives

$$\begin{aligned} \Phi_N<\Phi_{N-1}&\iff\mu>\epsilon_{(N)},\\ \Phi_N<\Phi_{N+1}&\iff\mu<\epsilon_{(N+1)}. \end{aligned}$$

The ordered one-particle levels make $E_0(N)$ convex, so these inequalities also exclude every more distant number sector. A nonempty interval exists precisely when $\epsilon_{(N)}<\epsilon_{(N+1)}$. If the energies coincide, several particle numbers or configurations meet at the same grand energy. The canonical problem remains well defined, but a grand-canonical zero-temperature limit needs a prescription for partial filling.

Derive the leading exchange correction

Let $y=n\lambda_T^3/g\ll1$. Use the small-fugacity expansion to recover the first correction to the classical pressure and interpret its sign.

Solution

The number and pressure series are

$$\begin{aligned} y&=z-\frac{z^2}{2^{3/2}}+O(z^3),\\ \frac{P\lambda_T^3}{gk_{\mathrm B}T} &=z-\frac{z^2}{2^{5/2}}+O(z^3). \end{aligned}$$

Series inversion gives $z=y+y^2/2^{3/2}+O(y^3)$. Substitution yields

$$\frac{P\lambda_T^3}{gk_{\mathrm B}T} =y+\frac{y^2}{2^{5/2}}+O(y^3), \qquad \frac{P}{nk_{\mathrm B}T} =1+\frac{y}{2^{5/2}}+O(y^2).$$

The positive sign reflects the exclusion restriction on occupation. This statistical pressure correction does not imply a repulsive pair potential.

References

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• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987).
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