Degenerate Fermi Gas at Zero Temperature

Cool an ideal gas of fermions toward absolute zero and something strange happens: it refuses to sit still. A classical gas would freeze into a single low-energy heap, but fermions cannot — the Pauli principle forbids two of them from sharing a single-particle state. So even at $T=0$ the particles are forced to stack into successively higher momentum states, filling every level from the bottom up to a sharp ceiling. That ceiling is the Fermi energy $\varepsilon_F$ , the filled region in momentum space is the Fermi sphere, and the whole construction is the ground state of the degenerate Fermi gas. This article establishes that ground state and its exact zero-temperature thermodynamics. We take the $T\to0$ limit of the mean occupation from aq-sm-fermi-dirac-distribution, whose Fermi function $\bar n=1/(e^{\beta(\varepsilon-\mu)}+1)$ collapses to a step $\Theta(\varepsilon_F-\varepsilon)$ : every state below $\varepsilon_F$ singly occupied, every state above empty. Then a single counting argument — how many single-particle states fit inside a sphere of radius $k_F$ in $k$ -space — fixes everything. The density of states $g(\varepsilon)\propto\sqrt\varepsilon$ that we plug into the moments is owned by aq-sm-density-of-states-free-particles, and the resulting energy scale $\varepsilon_F(n)$ is developed in aq-sm-fermi-energy-momentum-temperature; here we take both as given and read off the two owned results of this article: the filled-sphere count $N=g_sVk_F^3/6\pi^2$ and the exact ground-state energy $E_0=\tfrac35 N\varepsilon_F$ . The payoff is enormous. The Fermi temperature $T_F=\varepsilon_F/k_B$ of metallic electrons reaches $10^4$ – $10^5$ K, so room-temperature electrons are deeply degenerate — the ground-state picture, not thermal physics, governs them. The gas stores a colossal zero-point kinetic energy and pushes back with a degeneracy pressure that supports white dwarfs against gravity, even when completely cold.

Key takeaways

At $T=0$ the Fermi-Dirac occupation is a sharp step $\bar n(\varepsilon)=\Theta(\varepsilon_F-\varepsilon)$ : all states below $\varepsilon_F$ filled, all above empty — the Fermi sea.
In $k$ -space the filled states form the Fermi sphere of radius $k_F$ ; counting states inside gives $N=g_sVk_F^3/6\pi^2$ , hence $k_F=(3\pi^2 n)^{1/3}$ for spin- $\tfrac12$ fermions ( $g_s=2$ ).
The energy per particle is a pure number times the Fermi energy: $E_0/N=\tfrac35\varepsilon_F$ , an entirely quantum zero-point energy that survives at $T=0$ .
The cold gas exerts a finite degeneracy pressure $P_0=\tfrac25 n\varepsilon_F=\tfrac23 E_0/V$ — a consequence of exclusion, not of temperature.
Key idea: the $T=0$ ground state is the filled Fermi sphere; counting states inside it ( $N=g_sVk_F^3/6\pi^2$ ) and taking moments of the $\sqrt\varepsilon$ density of states gives $E_0=\tfrac35 N\varepsilon_F$ exactly — every ground-state property follows from this one geometric picture.

The core insight

At $T=0$ the Fermi function is a step $\Theta(\varepsilon_F-\varepsilon)$ , so the particles fill a sphere of radius $k_F$ in momentum space. Requiring the sphere to hold exactly $N$ particles fixes $k_F=(3\pi^2 n)^{1/3}$ (for $g_s=2$ ), and integrating the free-particle energy $\hbar^2k^2/2m$ over the filled sphere gives the exact ground-state energy $E_0=\tfrac35 N\varepsilon_F$ . Everything cold about a Fermi gas — its enormous mean energy, its degeneracy pressure, its stiffness — is the arithmetic of a filled ball.

Where this lives today

The degenerate Fermi gas is the working model for the electrons in every metal and the matter in every compact star. In a metal the conduction electrons form a Fermi sea with $T_F\sim10^4$ – $10^5$ K, so at ordinary temperatures $T/T_F\ll1$ and only the thin shell of width $\sim k_BT$ around the Fermi surface can be thermally excited — the reason the electronic heat capacity is a small linear-in- $T$ term rather than the classical $\tfrac32 Nk_B$ , computed in aq-sm-electronic-heat-capacity-metals via the aq-sm-sommerfeld-expansion built on this exact ground state. In astrophysics, electron degeneracy pressure supports white dwarfs and neutron degeneracy supports neutron stars; the softening of the equation of state from $n^{5/3}$ to the relativistic $n^{4/3}$ produces the Chandrasekhar mass $\approx1.4\,M_\odot$ . And in the laboratory, trapped $^{40}$ K and $^6$ Li atoms cooled below their Fermi temperature realize the textbook ground state directly: experiments image the sharp Fermi surface in time-of-flight, measure the $\tfrac35\varepsilon_F$ energy per particle, and observe Pauli blocking of collisions.

The physical picture

Pour fermions into a box one at a time. The first drops into the lowest single-particle state; the next cannot join it (beyond the $g_s$ -fold spin degeneracy), so it takes the next state up; and so on. Like water filling a basin, the particles settle into the lowest available levels, and when all $N$ are in, the "water line" is the Fermi energy $\varepsilon_F$ . Because a free particle's energy depends only on the magnitude of its wavevector, $\varepsilon=\hbar^2k^2/2m$ , the occupied states fill a ball in $k$ -space — the Fermi sphere — bounded by the Fermi surface at $|\mathbf k|=k_F$ . The step-function occupation is the smoking gun of degeneracy. At $T=0$ there is no thermal energy to promote a particle across the finite gap to an empty state above the surface, so the distribution is perfectly sharp. The first figure is the hero: the $T=0$ step, and beside it the filled Fermi sphere it describes.

Figure

The T=0 Fermi step and the filled Fermi sphere it describes

The punchline: a sharp step at $\varepsilon_F$ in energy is exactly a solidly filled ball of radius $k_F$ in momentum space. There is no gradient, no tail — the ground state is a filled sphere.

The ground-state energy is purely kinetic and purely quantum. Even with no interactions and no temperature, antisymmetry forces most particles into high-momentum states. Averaging the energy over the filled sphere gives each particle $\tfrac35\varepsilon_F$ on average — a huge number for a "cold" gas. The second figure shows why the average is $\tfrac35$ and not $\tfrac12$ : the states pile up as $\sqrt\varepsilon$ , so there are far more high-energy states than low-energy ones, dragging the mean upward.

Figure

Why the mean energy is three-fifths of the Fermi energy: the sqrt-epsilon density of states weights the top heavily

Because $g(\varepsilon)\propto\sqrt\varepsilon$ rises toward the surface, the mean occupied energy is pulled well above the midpoint — the exact moment ratio gives $\tfrac35$ .

Density controls everything. Since $\varepsilon_F\propto n^{2/3}$ , compressing the gas raises the Fermi energy — and hence the pressure — steeply. The cold Fermi gas resists compression far more stiffly than a classical gas at the same temperature, and that stiffness is what holds up a white dwarf.

The core equations

The starting point is the zero-temperature occupation. Taking $T\to0$ in the Fermi function (owned by aq-sm-fermi-dirac-distribution) sends the smooth shelf to a sharp step at the chemical potential, which at $T=0$ is the Fermi energy.

\bar n(\varepsilon)=\frac{1}{e^{(\varepsilon-\mu)/k_BT}+1}\ \xrightarrow{\ T\to0\ }\ \Theta(\varepsilon_F-\varepsilon),\qquad \varepsilon_F\equiv\mu(T{=}0) \qquad (1)

where: $\bar n(\varepsilon)$ — mean occupation of a state of energy $\varepsilon$ ; $\Theta$ — Heaviside step; $\varepsilon_F$ — Fermi energy, the highest occupied energy at $T=0$ . Physical reading: at $T=0$ occupation is all-or-nothing, so counting occupied states is a pure geometry problem — the volume of a sphere in $k$ -space. Counting the states inside the Fermi sphere fixes its radius. Each single-particle $k$ -state occupies a cell of volume $(2\pi/L)^3$ , so the number of states with $|\mathbf k|<k_F$ , times the spin degeneracy, must equal $N$ .

N=g_s\,\frac{V}{(2\pi)^3}\int_{|\mathbf k|<k_F}\!\mathrm d^3k=g_s\,\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k_F^3=\frac{g_sV\,k_F^3}{6\pi^2} \qquad (2)

where: $g_s=2s+1$ — spin degeneracy ( $=2$ for electrons); $V$ — volume; $k_F$ — Fermi wavevector. Physical reading: this is the owned counting result of the article — the filled sphere holds exactly $N$ particles, and inverting gives the Fermi wavevector purely from the density. Solving Eq. (2) for $k_F$ gives the Fermi wavevector as a function of density alone — no temperature appears.

k_F=\left(\frac{6\pi^2 n}{g_s}\right)^{1/3}\ \xrightarrow{\ g_s=2\ }\ (3\pi^2 n)^{1/3},\qquad n=\frac{N}{V} \qquad (3)

where: $n$ — number density. Physical reading: doubling the density raises $k_F$ by $2^{1/3}$ ; the Fermi energy $\varepsilon_F=\hbar^2k_F^2/2m$ (its density scaling developed in aq-sm-fermi-energy-momentum-temperature) then rises by $2^{2/3}$ . The ground-state energy is the sum of the kinetic energies of all filled states. Integrating $\hbar^2k^2/2m$ over the filled sphere — equivalently, taking the first moment of the $\sqrt\varepsilon$ density of states owned by aq-sm-density-of-states-free-particles — gives a strikingly clean result.

E_0=g_s\frac{V}{(2\pi)^3}\int_{|\mathbf k|<k_F}\!\frac{\hbar^2k^2}{2m}\,\mathrm d^3k=\int_0^{\varepsilon_F}\!\varepsilon\,g(\varepsilon)\,\mathrm d\varepsilon=\frac{3}{5}N\varepsilon_F \qquad (4)

where: $E_0$ — total (zero-point) kinetic energy of the filled Fermi sphere; $g(\varepsilon)=\tfrac{3N}{2}\sqrt\varepsilon/\varepsilon_F^{3/2}$ . Physical reading: this is the second owned result — the energy per particle is $\tfrac35\varepsilon_F$ , a pure number set by the geometry of a filled ball, independent of $g_s$ , $V$ , or the density of states prefactor. Because $\varepsilon_F$ carries all the volume dependence at fixed $N$ , the cold gas has a finite pressure. Any non-relativistic ideal gas obeys $PV=\tfrac23 E$ , so the ground-state energy immediately yields the degeneracy pressure.

P_0=\frac{2}{3}\frac{E_0}{V}=\frac{2}{5}n\,\varepsilon_F=\frac{(3\pi^2)^{2/3}}{5}\frac{\hbar^2}{m}\,n^{5/3} \qquad (5)

where: $P_0$ — zero-temperature pressure, finite despite $T=0$ . Physical reading: this "degeneracy pressure" is quantum, not thermal — its steep $n^{5/3}$ growth is the mechanical backbone of white-dwarf stability, developed fully in aq-sm-degeneracy-pressure-fermions.

Derivations

Derivation ACounting the Fermi sphere and the exact ground-state energy

The goal is to derive the two owned results of this article — the filled-sphere particle count $N=g_sVk_F^3/6\pi^2$ (hence $k_F=(3\pi^2 n)^{1/3}$ for $g_s=2$ ) and the exact ground-state energy $E_0=\tfrac35 N\varepsilon_F$ — from the single fact that at $T=0$ the occupation is a step. The strategy is pure counting: convert the $k$ -space sum to an integral using the uniform mode density $V/(2\pi)^3$ (owned by aq-sm-density-of-states-k-space), fill the sphere, and take the ratio of the energy moment to the number moment so that all prefactors cancel. Watch the one subtlety: the answer $\tfrac35$ must be independent of $g_s$ , $V$ , and $m$ — it is a property of the geometry, not the material.

Step 1. Convert the sum over occupied states to an integral. At $T=0$ every $k$ -state with $|\mathbf k|<k_F$ is filled once per spin, so $N=g_s\sum_{|\mathbf k|<k_F}1\to g_s\dfrac{V}{(2\pi)^3}\displaystyle\int_{|\mathbf k|<k_F}\mathrm d^3k$ . Why: each allowed $\mathbf k$ occupies a cell of volume $(2\pi/L)^3$ in a box of side $L$ , so the number of states in a $k$ -space region is its volume times $V/(2\pi)^3$ .

Step 2. Do the sphere integral. $\displaystyle\int_{|\mathbf k|<k_F}\mathrm d^3k=\tfrac43\pi k_F^3$ , so $N=g_s\dfrac{V}{(2\pi)^3}\tfrac43\pi k_F^3=\dfrac{g_sV k_F^3}{6\pi^2}$ . Inverting, $k_F=(6\pi^2 n/g_s)^{1/3}$ , which for $g_s=2$ is $(3\pi^2 n)^{1/3}$ . Why: the $\tfrac43\pi k_F^3$ is just the volume of the filled ball; the factor $(2\pi)^3=8\pi^3$ combines with $\tfrac43\pi$ to give the clean $6\pi^2$ .

Step 3. Write the energy as the same integral weighted by $\varepsilon=\hbar^2k^2/2m$ . $E_0=g_s\dfrac{V}{(2\pi)^3}\displaystyle\int_0^{k_F}\dfrac{\hbar^2k^2}{2m}\,4\pi k^2\,\mathrm dk=g_s\dfrac{V}{(2\pi)^3}\dfrac{4\pi\hbar^2}{2m}\dfrac{k_F^5}{5}=\dfrac{g_sV\hbar^2 k_F^5}{20\pi^2 m}$ . Why: the shell at radius $k$ has volume $4\pi k^2\,\mathrm dk$ and each state there carries energy $\hbar^2k^2/2m$ ; the integral $\int_0^{k_F}k^4\,\mathrm dk=k_F^5/5$ supplies the $\tfrac15$ .

Step 4. Take the ratio to eliminate every prefactor. $\dfrac{E_0}{N}=\dfrac{g_sV\hbar^2 k_F^5/20\pi^2 m}{g_sV k_F^3/6\pi^2}=\dfrac{6}{20}\dfrac{\hbar^2 k_F^2}{m}=\dfrac{3}{10}\dfrac{\hbar^2k_F^2}{m}=\dfrac{3}{5}\dfrac{\hbar^2k_F^2}{2m}=\dfrac35\varepsilon_F$ . Why: $g_s$ , $V$ , $m$ , and the numerical prefactors all cancel in the ratio, leaving the pure geometric number $\tfrac35$ times $\varepsilon_F=\hbar^2k_F^2/2m$ . Equivalently, the same $\tfrac35$ is the moment ratio $\int_0^{\varepsilon_F}\varepsilon^{3/2}\mathrm d\varepsilon\,/\,(\varepsilon_F\int_0^{\varepsilon_F}\varepsilon^{1/2}\mathrm d\varepsilon)=\tfrac{2/5}{2/3}=\tfrac35$ of the $\sqrt\varepsilon$ density of states. Verified: the sphere count $N=g_sVk_F^3/6\pi^2$ , the inversion $k_F=(3\pi^2 n)^{1/3}$ at $g_s=2$ , the energy integral $E_0=g_sV\hbar^2k_F^5/20\pi^2 m$ , and both the $k$ -space and DOS moment ratios giving $E_0/N=\tfrac35\varepsilon_F$ are checked symbolically in verify_degenerate-fermi-gas-zero-temperature.py.

Go deeper · Derivation B · 1 minDegeneracy pressure at absolute zero

Derivation BDegeneracy pressure at absolute zero

The goal is to show that the cold gas exerts a finite pressure $P_0=\tfrac25 n\varepsilon_F=\tfrac23 E_0/V$ , a purely quantum "degeneracy pressure" with no thermal component. The route is mechanical: since $S=0$ at $T=0$ , the pressure is the adiabatic derivative $P_0=-(\partial E_0/\partial V)_N$ , and the only volume dependence sits in $\varepsilon_F\propto V^{-2/3}$ . Watch the interpretation: this pressure does not vanish as $T\to0$ — it is the mechanical face of the Pauli principle, and its steep density scaling is what makes it astrophysically decisive. The full stellar treatment is forward-teased to aq-sm-degeneracy-pressure-fermions; here we establish only the equation of state.

Step 1. Identify the pressure with the adiabatic energy derivative. At $T=0$ the entropy is fixed at $S=0$ , so $P_0=-\left(\dfrac{\partial E_0}{\partial V}\right)_{N,S}$ . Why: the first law $\mathrm dE=T\,\mathrm dS-P\,\mathrm dV$ with $\mathrm dS=0$ gives $P=-(\partial E/\partial V)_S$ directly.

Step 2. Extract the volume dependence. With $n=N/V$ , $\varepsilon_F=\dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}\propto n^{2/3}\propto V^{-2/3}$ , so $E_0=\tfrac35 N\varepsilon_F=C\,V^{-2/3}$ at fixed $N$ , where $C$ collects the $V$ -independent constants. Why: only $\varepsilon_F$ depends on $V$ at fixed $N$ ; the number of particles and the $\tfrac35$ do not.

Step 3. Differentiate. $P_0=-\dfrac{\partial}{\partial V}\!\left(CV^{-2/3}\right)=\tfrac23\,CV^{-5/3}=\dfrac23\dfrac{E_0}{V}$ . Substituting $E_0/V=\tfrac35 n\varepsilon_F$ gives $P_0=\tfrac25 n\varepsilon_F$ , and with $\varepsilon_F\propto n^{2/3}$ this is $P_0=\dfrac{(3\pi^2)^{2/3}}{5}\dfrac{\hbar^2}{m}n^{5/3}$ . Why: the exponent $-\tfrac23$ produces the universal $PV=\tfrac23 E$ of a non-relativistic gas; the steep $n^{5/3}$ growth is what resists gravitational compression. Verified: the mechanical relation $P=-(\partial E/\partial V)_N=\tfrac23 E/V$ for an $\varepsilon\propto k^2$ gas, the identity $P_0=\tfrac25 n\varepsilon_F$ , and the $n^{5/3}$ form (with the ultrarelativistic $\tfrac34\varepsilon_F$ , $n^{4/3}$ cross-check) are verified in verify_degenerate-fermi-gas-zero-temperature.py.

Limiting cases

The $T=0$ ground state is itself a limit, and it is the reference point for several regimes.

Case	Result	Meaning
Strong degeneracy $T\ll T_F$	$\bar n\to\Theta(\varepsilon_F-\varepsilon)$	Occupation is a sharp step; finite- $T$ corrections are analytic in $(T/T_F)^2$ via the Sommerfeld expansion. Metals at 300 K live here.
Classical $T\gg T_F$	$\bar n\to e^{(\mu-\varepsilon)/k_BT}\ll1$	Occupations are tiny; Fermi-Dirac reduces to Maxwell-Boltzmann and the sphere picture is irrelevant ( $n\lambda_T^3\ll1$ ).
Ultra-relativistic $p_Fc\gg mc^2$	$E_0=\tfrac34 N\varepsilon_F,\ P_0=\tfrac14 n\varepsilon_F$	Linear dispersion $\varepsilon=\hbar kc$ softens the equation of state to $P\propto n^{4/3}$ — the origin of the Chandrasekhar mass.
Full spin polarization $g_s=1$	$k_F=(6\pi^2 n)^{1/3}$	One spin species holds all $N$ ; $k_F$ and $\varepsilon_F$ rise by $2^{1/3}$ and $2^{2/3}\approx1.59$ — polarizing costs kinetic energy.

Common trapDegeneracy pressure is not thermal pressure, and degenerate does not mean level-degenerate

Three traps live in this one word. First, the degeneracy pressure $P_0=\tfrac25 n\varepsilon_F$ does not vanish as $T\to0$ ; it is zero-point pressure from the Pauli principle, entirely non-thermal. A classical gas has $P=nk_BT\to0$ at $T=0$ ; the Fermi gas does not. Second, "degenerate" here means quantum-statistically degenerate ( $T\ll T_F$ , $n\lambda_T^3\gg1$ ) — it has nothing to do with energy levels sharing an eigenvalue. Third, $\mu(T{=}0)=\varepsilon_F$ but $\mu$ decreases with temperature, $\mu(T)\approx\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ ; do not use them interchangeably for $T>0$ .

Worked examples

Example 1 — Fill four fermions into oscillator-like levels and read off the ground state.

Step 1. Setup. Take single-particle levels with energies $\varepsilon_k=k$ (in units of some $\varepsilon_0$ ), each spin-degenerate ( $g_s=2$ ), and place $N=4$ spin- $\tfrac12$ fermions. Fill from the bottom, at most $2$ per level.

Step 2. Fill. Level $\varepsilon=0$ takes $2$ (spin up, spin down); level $\varepsilon=1$ takes the remaining $2$ . Levels $\varepsilon\ge2$ are empty. The highest occupied energy is $\varepsilon_F=\mathbf{1}$ (in units of $\varepsilon_0$ ).

Step 3. Ground-state energy by direct sum. $E_0=2(0)+2(1)=\mathbf{2\,\varepsilon_0}$ , so $E_0/N=2/4=\mathbf{0.5\,\varepsilon_0}$ . This discrete toy has $E_0/N=\tfrac12\varepsilon_F$ , not $\tfrac35$ — the $\tfrac35$ is the continuum result for a $\sqrt\varepsilon$ density of states, which this tiny system does not yet approximate.

Step 4. Interpret. The exclusion principle alone forces $E_0>0$ : four distinguishable particles could all sit at $\varepsilon=0$ for $E_0=0$ , but fermions cannot. This is the microcosm of the Fermi sea — filling forced upward by antisymmetry, giving a finite zero-point energy even with no interactions and no temperature.

Example 2 — Fermi parameters of copper's conduction electrons.

Step 1. Setup. Copper contributes one $4s$ electron per atom, giving $n=8.47\times10^{28}\,\mathrm{m^{-3}}$ ; with $g_s=2$ , use $k_F=(3\pi^2 n)^{1/3}$ .

Step 2. Fermi wavevector. $3\pi^2 n=3\pi^2(8.47\times10^{28})=2.508\times10^{30}\,\mathrm{m^{-3}}$ , so $k_F=(2.508\times10^{30})^{1/3}=\mathbf{1.36\times10^{10}\,m^{-1}}$ .

Step 3. Fermi energy and temperature. $\varepsilon_F=\dfrac{\hbar^2k_F^2}{2m_e}=\dfrac{(1.0546\times10^{-34})^2(1.36\times10^{10})^2}{2(9.109\times10^{-31})}=1.127\times10^{-18}\,\mathrm J=\mathbf{7.03\ eV}$ , and $T_F=\varepsilon_F/k_B=\mathbf{8.16\times10^{4}\,K}$ .

Step 4. The degeneracy verdict. Room temperature is $300\,\mathrm K$ , so $T/T_F=300/(8.16\times10^4)=\mathbf{3.7\times10^{-3}}$ — the electron gas is deeply degenerate, sitting essentially in its ground state. Each electron carries $E_0/N=\tfrac35\varepsilon_F=\mathbf{4.2\ eV}$ of zero-point kinetic energy, and the Fermi velocity is $v_F=\hbar k_F/m_e=\mathbf{1.57\times10^{6}\,m/s}$ — about $0.5\%$ of the speed of light, all of it quantum zero-point motion, none of it thermal.

Origins · 1 minsubrahmanyan chandrasekhar and how this was discovered

subrahmanyan chandrasekhar

1910–1995 · The Indian-American astrophysicist who, on the voyage to England in 1930, applied the cold degenerate Fermi gas to white dwarfs. Combining the relativistic equation of state P₀∝n^{4/3} with hydrostatic equilibrium, he found a maximum mass — the Chandrasekhar limit ≈1.4 solar masses — above which electron degeneracy pressure cannot halt collapse. Ridiculed by Eddington at the time, the result won him the 1983 Nobel Prize and underlies the theory of neutron stars and Type Ia supernovae.

History

1926

— Fermi and, independently, Dirac derive the statistics of exclusion-obeying particles

1927

— Sommerfeld applies Fermi-Dirac statistics to metallic electrons, explaining the small electronic heat capacity

1928

— Bloch extends the free-electron gas to periodic lattices, founding band theory

1930

— Chandrasekhar computes the white-dwarf mass limit on his voyage to Cambridge

1932

— Landau independently proposes a maximum mass for degenerate stars

1967

— Pulsars are discovered, confirming neutron stars supported by neutron degeneracy pressure

1999

— DeMarco and Jin produce the first degenerate atomic Fermi gas of potassium-40

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §8.1; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.16–9.17; Landau & Lifshitz, Statistical Physics Part 1, §57; Ashcroft & Mermin, Solid State Physics, Ch. 2.
Builds on: the $T=0$ step from Fermi-Dirac Distribution; the master sum $N=\int g(\varepsilon)\bar n\,\mathrm d\varepsilon$ from Quantum Statistics of Ideal Gases: Occupation Numbers; the $\sqrt\varepsilon$ density of states from Density of States for Free Particles; the energy scale from Fermi Energy, Momentum and Fermi Temperature.
Leads to: the Sommerfeld Expansion (finite- $T$ corrections about this step) · Electronic Heat Capacity of Metals · Degeneracy Pressure of Fermions (the stellar equation of state) · Chemical Potential of a Fermi Gas vs Temperature.
Provenance: AllQFT `sm` export (Pathria & Beale, §8.1), aq-sm-degenerate-fermi-gas-zero-temperature, ref STAT.IQGAS.DFG0.1218. All counting, energy-moment, pressure, and Fermi-parameter claims python-verified (22 checks, all PASS). No base error found — the migrated formulae were algebra- and sign-consistent; the copper numbers $\varepsilon_F=7.03$ eV, $T_F=8.16\times10^4$ K, $v_F=1.57\times10^6$ m/s reproduce the Ashcroft-Mermin values.

Why isn't the ground-state energy just zero at T=0?

For distinguishable classical particles it would be — all $N$ could pile into the lowest state. But fermions cannot: the Pauli principle allows at most $g_s$ per single-particle state, so filling $N$ of them forces particles into successively higher momentum states. The result is a huge zero-point kinetic energy $E_0=\tfrac35 N\varepsilon_F$ that exists with no interactions and no temperature. It is a direct, quantitative consequence of antisymmetry — exclusion converted into energy.

Is the degeneracy pressure a kind of thermal pressure?

No — it is the opposite. Thermal pressure $P=nk_BT$ vanishes at $T=0$ ; the degeneracy pressure $P_0=\tfrac25 n\varepsilon_F$ is maximal there and is entirely quantum. It comes from the volume dependence of the zero-point energy: squeezing the box raises $\varepsilon_F\propto V^{-2/3}$ , so the filled sphere pushes back. This is why a white dwarf, though "cold" in the thermal sense, resists gravity: its support is exclusion, not heat.

How can occupation be a sharp step when real metals are at 300 K?

Because $300\,\mathrm K$ is tiny compared to the Fermi temperature $T_F\sim8\times10^4\,\mathrm K$ . The step softens only over an energy width $\sim k_BT$ , which for copper is $\sim0.5\%$ of $\varepsilon_F$ . To the eye the occupation is still a step; only a thin shell near the surface is "active." That is exactly why the ground-state treatment works so well for metals and why finite- $T$ effects are captured by a systematic expansion in $(T/T_F)^2$ .

Where does the three-fifths come from, intuitively?

The density of states rises as $\sqrt\varepsilon$ , so there are far more states near the top of the sea than near the bottom. The mean occupied energy is therefore pulled well above the midpoint $\tfrac12\varepsilon_F$ . Precisely, $\langle\varepsilon\rangle=\int_0^{\varepsilon_F}\varepsilon\cdot\varepsilon^{1/2}\,\mathrm d\varepsilon/\int_0^{\varepsilon_F}\varepsilon^{1/2}\,\mathrm d\varepsilon=\tfrac{\varepsilon_F^{5/2}/(5/2)}{\varepsilon_F^{3/2}/(3/2)}\cdot\varepsilon_F^{-1}\cdot\varepsilon_F=\tfrac35\varepsilon_F$ . The crowded top does the work.

What changes when the fermions become relativistic?

In dense stellar matter the Fermi momentum can be large enough that $p_Fc\gg mc^2$ , and the dispersion switches from $\varepsilon=\hbar^2k^2/2m$ to $\varepsilon=\hbar kc$ . The energy moment then gives $E_0=\tfrac34 N\varepsilon_F$ and the pressure $P_0=\tfrac14 n\varepsilon_F\propto n^{4/3}$ — a softer equation of state. That softening is decisive: a softer $P(n)$ cannot always beat gravity, which is why there is a maximum mass, the Chandrasekhar limit $\approx1.4\,M_\odot$ .

Does this ground state assume the fermions don't interact?

Yes — it is the ideal-gas ground state, with independent modes and a filled Fermi sphere. Remarkably, Landau's Fermi-liquid theory shows the picture survives interactions: the low-energy excitations of a real metal are long-lived quasiparticles that still fill a Fermi sea up to a (renormalized) Fermi surface, which is why the free-electron model predicts the right $\varepsilon_F$ , heat capacity, and Fermi velocity for copper. Strong correlations (Mott insulators, heavy fermions) can break this, but the ideal ground state remains the reference.

Why is the Fermi velocity so large for a cold gas?

Because $v_F=\hbar k_F/m$ and $k_F$ is fixed by density, not temperature. For copper $v_F\approx1.57\times10^6\,\mathrm{m/s}$ — about $0.5\%$ of $c$ — even at $T=0$ . This is entirely zero-point motion forced by exclusion: the electrons at the Fermi surface are moving fast not because they are hot but because antisymmetry has pushed them to high momentum. It sets the scale for electrical conduction, since only these fast surface electrons carry current.