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Ideal Fermi Gas

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

For a uniform gas in a cubic box of volume V=L3V=L^3,

H=∑k,aϵkcka†cka,ϵk=ℏ2k22m.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,…,ga=1,\ldots,g labels independent spin or internal modes. The number operator is

N=∑k,acka†cka.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

nka∈{0,1}.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.

This page owns the full uniform-gas thermodynamic derivation. The Ideal Fermi Gas dossier owns the compact baseline record, quantity-specific exactness audit, finite-volume checks, model variants, and numerical benchmark handoff.

Unless stated otherwise, this page uses:

  • three spatial dimensions;
  • a uniform cubic box;
  • periodic boundary conditions;
  • nonrelativistic dispersion;
  • internal degeneracy gg;
  • fixed number density n=N/Vn=N/V;
  • the thermodynamic limit.

Periodic boundary conditions give one-particle modes

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

with

k=2πLn,n∈Z3.\mathbf k = \frac{2\pi}{L} \mathbf n, \qquad \mathbf n \in \mathbb Z^3.

The many-body eigenstates are fermionic occupation-number states

∣{nka}⟩.\lvert\{n_{\mathbf k a}\}\rangle.

Their particle number and energy are

N=∑k,anka,N = \sum_{\mathbf k,a} n_{\mathbf k a},

and

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

No potential energy appears because the particles do not interact. The nonzero ground-state energy is entirely kinetic and statistical in origin.

Define

z≡eβμ,β=1kBT.z \equiv e^{\beta\mu}, \qquad \beta = \frac{1}{k_{\mathrm B}T}.

One fermionic mode contributes

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

Therefore

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

and

ln⁡Ξ=∑k,aln⁡(1+ze−βϵk).\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 μ\mu. Infinite-volume expressions still require a thermodynamic limit and a controlled density of states.

The mode occupation is

n‾ka=1z−1eβϵk+1.\overline n_{\mathbf k a} = \frac{1}{ z^{-1}e^{\beta\epsilon_{\mathbf k}}+1 }.

One momentum mode per internal state occupies volume

(2πL)3\left( \frac{2\pi}{L} \right)^3

in k\mathbf k-space. Including all gg internal modes, the number of one-particle states with magnitude below kk is

N(k)=gV(2π)34πk33=gVk36π2.\mathcal N(k) = g \frac{V}{(2\pi)^3} \frac{4\pi k^3}{3} = \frac{gVk^3}{6\pi^2}.

At zero temperature, the lowest NN modes are occupied. For the isotropic dispersion, they fill a sphere of radius kFk_{\mathrm F} in momentum space.

Cross-section of the filled Fermi sea and the occupied part of the three-dimensional density of states up to the Fermi energy

At T=0T=0, every complete one-particle mode inside the Fermi sphere is occupied and every mode outside is empty. The same state count is the area under D(ϵ)D(\epsilon) up to ϵF\epsilon_{\mathrm F}; thermal excitations affect only a narrow shell near the boundary when T≪TFT\ll T_{\mathrm F}.

Equating the state count to the particle number gives

N=gVkF36π2.N = \frac{gVk_{\mathrm F}^3}{6\pi^2}.

Thus

n=gkF36π2,n = \frac{gk_{\mathrm F}^3}{6\pi^2},

and

kF=(6π2ng)1/3.k_{\mathrm F} = \left( \frac{6\pi^2n}{g} \right)^{1/3}.

For spin-1/21/2 fermions with two equally populated internal states, g=2g=2:

kF=(3π2n)1/3.k_{\mathrm F} = \left( 3\pi^2n \right)^{1/3}.

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

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

ϵF=ℏ2kF22m.\epsilon_{\mathrm F} = \frac{\hbar^2k_{\mathrm F}^2}{2m}.

The Fermi temperature is

TF≡ϵFkB,T_{\mathrm F} \equiv \frac{\epsilon_{\mathrm F}}{k_{\mathrm B}},

and the Fermi velocity is

vF=1ℏdϵkdk∣k=kF=ℏkFm.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,

ϵF∝n2/3.\epsilon_{\mathrm F} \propto n^{2/3}.

Using

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

the integrated state count is

N(ϵ)=gV6π2(2mϵℏ2)3/2.\mathcal N(\epsilon) = \frac{gV}{6\pi^2} \left( \frac{2m\epsilon}{\hbar^2} \right)^{3/2}.

The density of states is

D(ϵ)=gV4π2(2mℏ2)3/2ϵ.D(\epsilon) = \frac{gV}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt\epsilon.

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

At the Fermi energy,

D(ϵF)=3N2ϵF.D(\epsilon_{\mathrm F}) = \frac{3N}{2\epsilon_{\mathrm F}}.

Per unit volume,

D(ϵF)V=3n2ϵF.\frac{D(\epsilon_{\mathrm F})}{V} = \frac{3n}{2\epsilon_{\mathrm F}}.

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

At fixed density, the chemical potential approaches the Fermi energy as T→0T\to0:

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

The occupation becomes

n‾ka=Θ(kF−k),\overline n_{\mathbf k a} = \Theta \left( k_{\mathrm F}-k \right),

away from the boundary. Equivalently,

n‾(ϵ)=Θ(ϵF−ϵ).\overline n(\epsilon) = \Theta \left( \epsilon_{\mathrm F}-\epsilon \right).

The filled region is the Fermi sea. The sphere k=kFk=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.

At zero temperature,

U0=∫0ϵFdϵ D(ϵ)ϵ.U_0 = \int_0^{\epsilon_{\mathrm F}} d\epsilon\, D(\epsilon)\epsilon.

Since D(ϵ)∝ϵD(\epsilon)\propto\sqrt\epsilon,

U0=35NϵF.U_0 = \frac{3}{5} N\epsilon_{\mathrm F}.

Thus the mean energy per particle is

U0N=35ϵF.\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.

For a nonrelativistic quadratic dispersion in three dimensions,

P=2U3V.P = \frac{2U}{3V}.

At zero temperature,

P0=25nϵF.P_0 = \frac{2}{5} n\epsilon_{\mathrm F}.

Because ϵF∝n2/3\epsilon_{\mathrm F}\propto n^{2/3},

P0∝n5/3.P_0 \propto n^{5/3}.

This pressure persists at T=0T=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.

Differentiating

P0=25nϵFP_0 = \frac{2}{5}n\epsilon_{\mathrm F}

at fixed particle species gives

dP0dn=23ϵF.\frac{dP_0}{dn} = \frac{2}{3} \epsilon_{\mathrm F}.

The zero-temperature compressibility is therefore

κ0=1n(∂n∂P)T=0=32nϵF.\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 ∂n/∂μ\partial n/\partial\mu and from the retarded density response, including the required order of limits.

Equivalently,

(∂n∂μ)T=0=D(ϵF)V=3n2ϵF.\left( \frac{\partial n}{\partial\mu} \right)_{T=0} = \frac{D(\epsilon_{\mathrm F})}{V} = \frac{3n}{2\epsilon_{\mathrm F}}.

At positive temperature,

N=∫0∞dϵ D(ϵ)1eβ(ϵ−μ)+1,N = \int_0^\infty d\epsilon\, D(\epsilon) \frac{1}{ e^{\beta(\epsilon-\mu)}+1 },

and

U=∫0∞dϵ D(ϵ)ϵeβ(ϵ−μ)+1.U = \int_0^\infty d\epsilon\, D(\epsilon) \frac{\epsilon}{ e^{\beta(\epsilon-\mu)}+1 }.

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

The pressure remains related to energy by

P=2U3VP = \frac{2U}{3V}

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

Use the thermal wavelength

λT=2πℏ2mkBT.\lambda_T = \sqrt{ \frac{2\pi\hbar^2}{mk_{\mathrm B}T} }.

The continuum grand partition function is

ln⁡Ξ=−gVλT3Li⁡5/2(−z).\ln\Xi = - \frac{gV}{\lambda_T^3} \operatorname{Li}_{5/2}(-z).

The number equation becomes

n=−gλT3Li⁡3/2(−z),n = - \frac{g}{\lambda_T^3} \operatorname{Li}_{3/2}(-z),

and the pressure is

P=−gkBTλT3Li⁡5/2(−z).P = - \frac{gk_{\mathrm B}T}{\lambda_T^3} \operatorname{Li}_{5/2}(-z).

The polylogarithms are negative for z>0z>0 in these combinations, so nn and PP are positive.

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

For z≪1z\ll1,

−Li⁡s(−z)=z−z22s+O(z3).-\operatorname{Li}_s(-z) = z - \frac{z^2}{2^s} + O(z^3).

Define the phase-space density per internal state,

x≡nλT3g.x \equiv \frac{n\lambda_T^3}{g}.

Then

x=z−z223/2+O(z3),x = z - \frac{z^2}{2^{3/2}} + O(z^3),

while

PλT3gkBT=z−z225/2+O(z3).\frac{P\lambda_T^3}{ gk_{\mathrm B}T } = z - \frac{z^2}{2^{5/2}} + O(z^3).

Eliminating zz gives

PnkBT=1+nλT325/2g+O[(nλT3g)2].\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].

The positive correction reflects Pauli exclusion. It is a statistical contribution, not a physical repulsive potential.

The gas is quantum degenerate when

T≪TF.T \ll T_{\mathrm F}.

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

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

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

not order one. This is why the low-temperature heat capacity is linear in TT rather than approaching the classical value 3NkB/23Nk_{\mathrm B}/2.

For a smooth function φ(ϵ)\varphi(\epsilon) and kBT≪μk_{\mathrm B}T\ll\mu,

∫0∞dϵ φ(ϵ)fF(ϵ)=∫0μdϵ φ(ϵ)+π26(kBT)2φ′(μ)+O(T4).\begin{aligned} \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). \end{aligned}

This is the leading Sommerfeld expansion. Its canonical page owns the derivation, higher terms, regularity conditions, and endpoint cautions. Here it is used only to state the ideal-gas low-temperature results.

At fixed density in the three-dimensional ideal gas,

μ(T)=ϵF[1−π212(TTF)2+O(T4TF4)].\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 ϵF\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.

At fixed density,

UN=35ϵF[1+5π212(TTF)2+O(T4TF4)].\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

CVN=π22kBTTF+O(T3TF3).\frac{C_V}{N} = \frac{\pi^2}{2} k_{\mathrm B} \frac{T}{T_{\mathrm F}} + O \left( \frac{T^3}{T_{\mathrm F}^3} \right).

The entropy has the same leading coefficient:

SN=π22kBTTF+O(T3TF3).\frac{S}{N} = \frac{\pi^2}{2} k_{\mathrm B} \frac{T}{T_{\mathrm F}} + O \left( \frac{T^3}{T_{\mathrm F}^3} \right).

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

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

P=25nϵF[1+5π212(TTF)2+O(T4TF4)].\begin{aligned} P ={}& \frac{2}{5} n\epsilon_{\mathrm F} \Biggl[ 1 + \frac{5\pi^2}{12} \left( \frac{T}{T_{\mathrm F}} \right)^2 \\ &\qquad + O \left( \frac{T^4}{T_{\mathrm F}^4} \right) \Biggr]. \end{aligned}

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

For independent grand-canonical modes,

Var⁡(N)=∑k,an‾ka(1−n‾ka).\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 ϵF\epsilon_{\mathrm F}.

The fluctuation–response relation is

(∂N‾∂μ)T,V=βVar⁡(N).\left( \frac{\partial\overline N}{\partial\mu} \right)_{T,V} = \beta \operatorname{Var}(N).

A fixed-NN 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.

Scattering into a final fermionic mode carries an availability factor

1−fF(ϵf).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.

In a finite box, the allowed momenta are discrete. The ground state fills complete shells of equal k2k^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.

The continuum formulas require many occupied modes and observables coarse enough not to resolve individual level spacings. Finite-Size Effects gives the general shell, parity, boundary, and resolution audit; this page retains the Fermi-gas thermodynamics.

For free electrons, g=2g=2 and

kF=(3π2n)1/3.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≪TFT\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. Material-specific bands and transport belong to Quantum Matter.

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 kFk_{\mathrm F}, ϵF\epsilon_{\mathrm F}, TFT_{\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.

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

The nonrelativistic formula

P0=25nϵFP_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.

The numerical coefficients above are specific to

d=3,ϵk=ℏ2k22m.d = 3, \qquad \epsilon_k = \frac{\hbar^2k^2}{2m}.

In another dimension, the kk-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.

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.

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.

This page owns:

  • the uniform three-dimensional nonrelativistic ideal Fermi gas;
  • momentum-space state counting with internal degeneracy gg;
  • kFk_{\mathrm F}, ϵF\epsilon_{\mathrm F}, TFT_{\mathrm F}, and vFv_{\mathrm F} in this model;
  • the density of states and filled Fermi sea;
  • zero-temperature energy, pressure, and compressibility;
  • finite-temperature number, pressure, and energy integrals;
  • leading low-temperature thermodynamics;
  • finite-size, dimensional, relativistic, and interaction limitations.

Other pages own:

  • the one-mode Fermi–Dirac distribution and thermal window: Fermi–Dirac Statistics;
  • the T≫TFT\gg T_{\mathrm F} Maxwell–Boltzmann criterion and leading exchange correction: Classical Limit of Quantum Statistics;
  • a focused treatment of T≪TFT\ll T_{\mathrm F} and physical applications: Degenerate Fermi Gas;
  • dimension-by-dimension reference formulas: Fermi Momentum and Fermi Energy;
  • generic Fermi-surface geometry and low-energy kinematics: Fermi Surface;
  • the general low-temperature asymptotic method: Sommerfeld Expansion;
  • Fermi surfaces in bands and materials: Quantum Matter;
  • interacting quasiparticles and Fermi-liquid theory: Quasiparticles and Collective Modes;
  • direct screening and collective density response around the ideal reference: Random Phase Approximation;
  • relativistic degenerate matter: Relativistic QM and QFT.org.
  • Forgetting the internal degeneracy factor gg in the state count.
  • Counting gg twice when the density of states already includes it.
  • Using total density in a per-component formula, or vice versa.
  • Treating the Fermi surface as a surface in real space.
  • Calling degeneracy pressure a thermal pressure or a new repulsive force.
  • Setting the zero-temperature energy to zero because the particles do not interact.
  • Assuming μ=ϵF\mu=\epsilon_{\mathrm F} at every temperature.
  • Applying the three-dimensional free-particle density of states to a lattice band or trap.
  • Using Maxwell–Boltzmann statistics when T≪TFT\ll T_{\mathrm F}.
  • Assigning the classical heat capacity 3NkB/23Nk_{\mathrm B}/2 to a degenerate Fermi gas.
  • Applying Sommerfeld coefficients without checking the density of states and fixed variables.
  • Using the nonrelativistic pressure for ultrarelativistic fermions.
  • Treating an interacting electron, neutron, or cold-atom system as ideal without a controlled approximation.

Derive

kF=(6π2ng)1/3k_{\mathrm F} = \left( \frac{6\pi^2n}{g} \right)^{1/3}

for a three-dimensional periodic box.

Solution

Each momentum state occupies volume (2π/L)3(2\pi/L)^3 in k\mathbf k-space. Including gg internal modes, the number of states inside a sphere of radius kFk_{\mathrm F} is

N=gV(2π)34πkF33=gVkF36π2.\begin{aligned} N &= g \frac{V}{(2\pi)^3} \frac{4\pi k_{\mathrm F}^3}{3} \\ &= \frac{gVk_{\mathrm F}^3}{6\pi^2}. \end{aligned}

Dividing by VV gives

n=gkF36π2.n = \frac{gk_{\mathrm F}^3}{6\pi^2}.

Solving for kFk_{\mathrm F} gives the stated expression.

Derive the ground-state energy and pressure

Section titled “Derive the ground-state energy and pressure”

Show that

U0=35NϵFU_0 = \frac{3}{5}N\epsilon_{\mathrm F}

and

P0=25nϵF.P_0 = \frac{2}{5}n\epsilon_{\mathrm F}.
Solution

Write the density of states as

D(ϵ)=Aϵ1/2.D(\epsilon) = A\epsilon^{1/2}.

The number is

N=A∫0ϵFdϵ ϵ1/2=2A3ϵF3/2.N = A \int_0^{\epsilon_{\mathrm F}} d\epsilon\, \epsilon^{1/2} = \frac{2A}{3} \epsilon_{\mathrm F}^{3/2}.

The energy is

U0=A∫0ϵFdϵ ϵ3/2=2A5ϵF5/2=35NϵF.\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,

P0=2U03V=25nϵF.P_0 = \frac{2U_0}{3V} = \frac{2}{5} n\epsilon_{\mathrm F}.

Relate the density of states to particle number

Section titled “Relate the density of states to particle number”

Show that

D(ϵF)=3N2ϵF.D(\epsilon_{\mathrm F}) = \frac{3N}{2\epsilon_{\mathrm F}}.
Solution

Since

D(ϵ)=Aϵ1/2,D(\epsilon) = A\epsilon^{1/2},

the zero-temperature particle number is

N=∫0ϵFdϵ D(ϵ)=2A3ϵF3/2.\begin{aligned} N &= \int_0^{\epsilon_{\mathrm F}} d\epsilon\, D(\epsilon) \\ &= \frac{2A}{3} \epsilon_{\mathrm F}^{3/2}. \end{aligned}

But

D(ϵF)=AϵF1/2.D(\epsilon_{\mathrm F}) = A\epsilon_{\mathrm F}^{1/2}.

Eliminating AA gives

D(ϵF)=3N2ϵF.D(\epsilon_{\mathrm F}) = \frac{3N}{2\epsilon_{\mathrm F}}.

Use

UN=35ϵF[1+5π212(TTF)2]\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 NN and VV.

Solution

The temperature-dependent energy correction is

ΔU=N35ϵF5π212T2TF2=Nπ24kB2T2ϵF.\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

CV=(∂U∂T)N,V=Nπ22kB2TϵF=Nπ22kBTTF.\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}

Show that the dilute ideal Fermi gas obeys

PnkBT=1+nλT325/2g+O[(nλT3g)2].\begin{aligned} \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]. \end{aligned}
Solution

Let

x=nλT3g.x = \frac{n\lambda_T^3}{g}.

The number expansion is

x=z−z223/2+O(z3).x = z- \frac{z^2}{2^{3/2}} + O(z^3).

Inverting gives

z=x+x223/2+O(x3).z = x+ \frac{x^2}{2^{3/2}} + O(x^3).

The pressure expansion is

PλT3gkBT=z−z225/2+O(z3).\frac{P\lambda_T^3}{ gk_{\mathrm B}T } = z- \frac{z^2}{2^{5/2}} + O(z^3).

Substitution yields

PλT3gkBT=x+x225/2+O(x3).\frac{P\lambda_T^3}{ gk_{\mathrm B}T } = x+ \frac{x^2}{2^{5/2}} + O(x^3).

Dividing by xx gives the stated positive statistical correction.

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