Degenerate Fermi Gas

A Fermi gas is deeply degenerate when its temperature is much smaller than its Fermi temperature:

$$\theta \equiv \frac{T}{T_{\mathrm F}} \ll 1.$$

The dimensionless temperature $\theta$ is the central control parameter. It compares the thermal energy $k_{\mathrm B}T$ with the Fermi energy:

$$k_{\mathrm B}T_{\mathrm F} = \epsilon_{\mathrm F}.$$

Deep degeneracy does not mean that every particle is motionless. Even at $T=0$, the Pauli exclusion principle forces fermions to occupy a filled region of momentum space with a nonzero kinetic energy and pressure. Nor does degenerate mean that the many-body ground state must have several states of exactly equal energy. In this context, quantum degeneracy means that the phase-space occupation is too large for classical dilute statistics and Fermi–Dirac statistics controls the macroscopic state.

The organizing fact is simple:

At $T\ll T_{\mathrm F}$, almost all one-particle states far below the Fermi energy remain filled, almost all states far above it remain empty, and thermal changes in occupation concentrate in a thin energy shell of width $O(k_{\mathrm B}T)$; outside it they are exponentially small.

That thin shell explains why:

• the heat capacity and entropy are proportional to $T/T_{\mathrm F}$;
• scattering is restricted by occupied final states;
• density fluctuations are suppressed;
• the zero-temperature pressure remains large;
• low-energy response is concentrated near the Fermi surface.

The ideal-gas derivation gives the equation of state. Here we explain why the thin thermal shell controls fluctuations, collisions, and heat capacity, then apply that reasoning to atoms, electrons, and white dwarfs.

Helpful background. The ideal Fermi gas supplies the state count and equation of state; Fermi–Dirac statistics supplies the occupation law.

A filled sphere of quadratic modes

Unless stated otherwise, use:

• a uniform three-dimensional gas;
• noninteracting spinless or internally degenerate fermions;
• nonrelativistic dispersion;
• volume $V$ and density $n=N/V$;
• $g$ equally populated components with identical dispersions and a common chemical potential;
• the thermodynamic limit;
• fixed $N$ and $V$ when differentiating the low-temperature energy.

The one-particle dispersion is

$$\epsilon_k = \frac{\hbar^2k^2}{2m}.$$

At $T=0$, all modes inside the Fermi sphere are occupied:

$$f_{\mathrm F}(\epsilon) = \Theta(\epsilon_{\mathrm F}-\epsilon).$$

For the uniform three-dimensional model,

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

and

$$v_{\mathrm F} = \frac{\hbar k_{\mathrm F}}{m}.$$

In the grand-canonical formulas below, $N$ denotes the mean particle number. The kinetic band bottom is the energy zero. Fermi scales gives the formulas in other dimensions and for unequal component populations.

Temperature, phase-space density, and the Fermi edge

The thermal wavelength $\lambda_T=\sqrt{2\pi\hbar^2/(mk_{\mathrm B}T)}$ gives

$$\frac{n\lambda_T^3}{g}=\frac{4}{3\sqrt\pi} \left(\frac{T_{\mathrm F}}{T}\right)^{3/2}.$$

Thus $T\ll T_{\mathrm F}$ means a large phase-space density per component. The opposite limit is the Maxwell–Boltzmann regime. With the band bottom at zero, $\mu\simeq\epsilon_{\mathrm F}>0$ also implies $z=e^{\beta\mu}\gg1$; a small-fugacity expansion is inappropriate. A shift of the energy zero changes $z$, but not the physical degeneracy.

The equilibrium occupation

$$f(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/(k_{\mathrm B}T)}+1}$$

changes from nearly one to nearly zero over a width of order $k_{\mathrm B}T$. How small $T/T_{\mathrm F}$ must be depends on the accuracy wanted for a particular observable.

The Thermally Active Shell

Differentiate the occupation:

$$-\frac{\partial f}{\partial\epsilon} = \frac{1}{4k_{\mathrm B}T} \operatorname{sech}^2 \left[ \frac{\epsilon-\mu}{2k_{\mathrm B}T} \right].$$

This function is sharply localized near $\epsilon=\mu$ and has unit integral:

$$\int_{-\infty}^{\infty} d\epsilon\, \left( -\frac{\partial f}{\partial\epsilon} \right) = 1.$$

As $T\to0$,

$$-\frac{\partial f}{\partial\epsilon} \longrightarrow \delta(\epsilon-\epsilon_{\mathrm F})$$

in the distributional sense at fixed nonzero density. The displayed unit area is over the full mathematical energy line; on the physical band $\epsilon\ge0$ the area is $f(0)$, exponentially close to one when $\mu\gg k_{\mathrm B}T$. Low-temperature response integrals therefore sample the density of states and matrix elements at the Fermi energy.

Near $k_{\mathrm F}$,

$$\epsilon_k-\epsilon_{\mathrm F} \approx \hbar v_{\mathrm F} (k-k_{\mathrm F}).$$

An energy window

$$\delta\epsilon \sim k_{\mathrm B}T$$

corresponds to a momentum-shell thickness

$$\delta k \sim \frac{k_{\mathrm B}T}{\hbar v_{\mathrm F}}.$$

For a quadratic dispersion,

$$\frac{\delta k}{k_{\mathrm F}} \sim \frac{1}{2} \frac{T}{T_{\mathrm F}}.$$

At $T\ll T_{\mathrm F}$, holes just inside $k_{\mathrm F}$ and particles just outside it occupy a shell with $\delta k/k_{\mathrm F}\sim T/(2T_{\mathrm F})$. The drawn annulus is schematic, not a hard cutoff; the Fermi–Dirac tail extends beyond it. The dashed circle marks the zero-temperature surface.

The volume fraction of a thin spherical shell scales as

$$\frac{ 4\pi k_{\mathrm F}^2\delta k }{ (4\pi/3)k_{\mathrm F}^3 } \sim 3\frac{\delta k}{k_{\mathrm F}} \sim O\left( \frac{T}{T_{\mathrm F}} \right).$$

This estimates the fraction of modes with appreciable thermal rearrangement. The shell has no sharp outer edge; changing the chosen width changes its order-one numerical coefficient.

Particles and Holes

A fixed-$N$ thermal excitation removes a fermion from an occupied state below the Fermi surface and places it in an empty state above:

$$\text{filled state below} \longrightarrow \text{hole below} + \text{particle above}.$$

The hole is not a missing microscopic species. It is a useful description of the changed occupation relative to the filled reference sea. A particle and a hole near the Fermi surface can carry small excitation energy even though their individual momenta are of order $\hbar k_{\mathrm F}$.

Particle–Hole Excitations develops the operator state, signs of hole momentum and charge, continuum boundaries, and the way density and spin probes sum over these promotions.

This distinction is essential:

• the momentum of a thermally active fermion need not be small;
• the distance from the Fermi surface is small;
• low excitation energy means $|\epsilon-\mu|\lesssim k_{\mathrm B}T$, not $k\approx0$.

Pauli Blocking of Final States

Pauli exclusion constrains occupation numbers:

$$n_i \in \{0,1\}.$$

In a transition rate, this becomes a final-state availability factor. For distinct modes, an occupation-diagonal state with uncorrelated initial occupations gives a one-particle transition from $i$ to $f$ the statistical weight

$$f_i (1-f_f).$$

If the final mode is already occupied, $f_f=1$ and the process is blocked.

For two-body scattering,

$$1+2 \longrightarrow 3+4,$$

the corresponding factorized occupation weight for four distinct modes is

$$f_1f_2 (1-f_3) (1-f_4).$$

In a correlated state the exact statistical weight involves joint occupations such as $\langle n_1n_2(1-n_3)(1-n_4)\rangle$, rather than a product of means. Energy and momentum conservation must be imposed in addition. At low temperature, most candidate final states below the Fermi surface are occupied, so the available scattering phase space collapses toward the thin thermal shell.

What blocking does not mean

Pauli blocking is not a repulsive force. It does not add a new potential to the Hamiltonian. It is a restriction on antisymmetric many-body states and on the availability of final modes.

It also does not mean that all collisions vanish:

• excited particles can scatter into available modes;
• different internal components can interact in the $s$ wave;
• impurities need not share the same blocked sea;
• collective processes can redistribute several excitations;
• finite temperature creates holes below the Fermi surface.

For a regular three-dimensional Fermi liquid, quasiparticle collision rates near the surface typically scale as $T^2$. Dimension and interaction kinematics matter; two-dimensional rates can contain logarithms. Momentum-conserving collisions alone do not imply electrical resistance in a translationally invariant Galilean gas. The exactly ideal gas has no collisions with which to equilibrate.

Two distinct antisymmetry effects

Do not conflate:

1. partial-wave restriction: identical spin-polarized fermions have no $s$-wave collision channel at ultralow energy;
2. many-body final-state blocking: occupied modes suppress otherwise allowed transitions in a filled Fermi sea.

The first is a two-body exchange-symmetry statement. The second depends on the many-body occupation distribution.

Suppressed Occupation Fluctuations

For one fermionic mode in any state, the projector identity $n_i^2=n_i$ gives

$$\operatorname{Var}(n_i) = f_i(1-f_i).$$

Deep in the Fermi sea, $f_i\approx1$ and the variance is small. Far above it, $f_i\approx0$ and the variance is also small. Fluctuations are concentrated near $f_i=1/2$, inside the thermal shell.

Summing over modes gives

$$\operatorname{Var}(N) = \sum_i f_i(1-f_i)$$

in the grand-canonical ideal gas. Since

$$f(1-f) = k_{\mathrm B}T \left( -\frac{\partial f}{\partial\epsilon} \right),$$

the low-temperature scaling is

$$\operatorname{Var}(N) \sim k_{\mathrm B}T D(\epsilon_{\mathrm F}),$$

where $D(\epsilon_{\mathrm F})$ is the total density of states. Since $D(\epsilon_{\mathrm F})=3N/(2\epsilon_{\mathrm F})$ in this model,

$$\frac{\operatorname{Var}(N)}{N}\simeq\frac32\frac{T}{T_{\mathrm F}}, \qquad \frac{\sqrt{\operatorname{Var}(N)}}{N}\simeq \sqrt{\frac{3T}{2NT_{\mathrm F}}}.$$

The first ratio is the Fano factor; the second is the relative root-mean-square fluctuation. A fixed-number ensemble has zero global number variance. A spatial subregion can retain quantum number fluctuations even at zero temperature, because its number operator need not commute with the Hamiltonian.

The exact relation between number fluctuations and compressibility depends on the ensemble and volume under discussion. For the response relation see Fluctuations and Susceptibilities.

Degeneracy Pressure

At $T=0$, a nonrelativistic ideal Fermi gas has

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

and

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

This degeneracy pressure survives at zero temperature. It is not thermal pressure and not the result of pairwise repulsion.

Momentum-flux interpretation

For an isotropic gas with momentum $\mathbf p$, energy $\epsilon(p)$, and speed

$$v(p) = \frac{d\epsilon}{dp},$$

the pressure can be written

$$P = \frac{g}{3} \int \frac{d^3p}{(2\pi\hbar)^3} \, p\,v(p) f(p).$$

For

$$\epsilon(p) = \frac{p^2}{2m},$$

one has

$$p\,v(p) = \frac{p^2}{m} = 2\epsilon(p),$$

so

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

The filled sea carries momentum flux even at $T=0$.

Density-scaling interpretation

Because

$$k_{\mathrm F} \propto n^{1/3},$$

the Fermi energy scales as

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

Therefore

$$\frac{U_0}{V} \propto n^{5/3}$$

and

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

Compression raises the momentum required to fit every fermion into a distinct state.

Compressibility at Zero Temperature

Since

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

one finds

$$\frac{\partial P_0}{\partial n} = \frac{2}{3} \epsilon_{\mathrm F}.$$

The zero-temperature isothermal compressibility is

$$\kappa_T = \frac{1}{n} \left( \frac{\partial n}{\partial P} \right)_T = \frac{3}{2n\epsilon_{\mathrm F}}.$$

The gas is compressible, but not infinitely soft. Interactions and band structure alter this value in real systems.

Why the Heat Capacity Is Small

A classical monatomic gas has an order-$Nk_{\mathrm B}$ heat capacity because essentially every particle can explore thermally different energies. In a degenerate Fermi gas, particles deep in the sea cannot change state under a low-energy perturbation because nearby final modes are occupied.

The number of thermally active modes scales as

$$N_{\mathrm{active}} \sim D(\epsilon_{\mathrm F}) k_{\mathrm B}T.$$

Each active particle–hole excitation carries energy of order $k_{\mathrm B}T$. Therefore

$$U(T)-U(0) \sim D(\epsilon_{\mathrm F}) (k_{\mathrm B}T)^2,$$

and

$$C_V \sim D(\epsilon_{\mathrm F}) k_{\mathrm B}^2T.$$

The full Sommerfeld calculation fixes the coefficient. The underlying state-counting measure is developed on Density of States: First Encounter:

$$C_V = \frac{\pi^2}{3} D(\epsilon_{\mathrm F}) k_{\mathrm B}^2T + O(T^3).$$

For the uniform three-dimensional quadratic gas,

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

so

$$C_V = \frac{\pi^2}{2} Nk_{\mathrm B} \frac{T}{T_{\mathrm F}} + O\left[Nk_{\mathrm B}(T/T_{\mathrm F})^3\right].$$

The low-temperature entropy has the same leading coefficient:

$$S = \frac{\pi^2}{2} Nk_{\mathrm B} \frac{T}{T_{\mathrm F}} + O\left[Nk_{\mathrm B}(T/T_{\mathrm F})^3\right].$$

These bulk entropy and heat-capacity densities vanish as $T\to0$. A finite system with an exactly degenerate ground level can retain a residual entropy; that is a separate finite-size question.

Leading Low-Temperature Corrections

For the three-dimensional nonrelativistic ideal gas at fixed density, let

$$\theta = \frac{T}{T_{\mathrm F}}.$$

The leading results are

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

and

$$P(T) = P_0 \left[ 1 + \frac{5\pi^2}{12} \theta^2 + O(\theta^4) \right].$$

The energy and pressure corrections begin at order $\theta^2$, while $C_V$ and $S$ begin at order $\theta$. Differentiation lowers the power by one.

These coefficients are not universal for every fermionic system. They depend on the dispersion, density of states, dimension, fixed variables, and interaction corrections. Sommerfeld Expansion derives these coefficients and states the regularity assumptions.

Electrons in metals

The free-electron numerical example gives a Fermi temperature of tens of thousands of kelvin for a typical metallic density. Room temperature can therefore be small on the electronic scale. The thermal shell explains a small electronic heat capacity, even though the occupied electrons carry large kinetic energies.

For a real material, use its band dispersions and Fermi surface. Several bands, anisotropic velocities, interactions, and phonons can change the coefficients and relaxation mechanisms. A free-electron estimate is a useful benchmark, not a complete transport model.

Application: Ultracold Fermionic Atoms

Cold-atom experiments can tune particle number, internal composition, trap geometry, and interactions. The natural thermometer is often the reduced temperature

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

because an absolute temperature in nanokelvin says little without the density or trap scale.

For a homogeneous spin-polarized potassium-40 gas with illustrative density

$$n = 10^{18}\ \mathrm{m}^{-3},$$

the ideal-gas scales are approximately

$$k_{\mathrm F} \approx 3.9\times10^6\ \mathrm{m}^{-1},$$

and

$$T_{\mathrm F} \approx 92\ \mathrm{nK}.$$

A temperature of $20\ \mathrm{nK}$ would correspond to

$$\frac{T}{T_{\mathrm F}} \approx 0.22.$$

This is a degenerate gas, but not an arbitrarily accurate zero-temperature approximation: the leading energy correction is about $20\%$ at $T/T_{\mathrm F}=0.22$, whereas the chemical-potential correction is about $4\%$.

Observable signatures

• The momentum distribution approaches one particle per mode inside the Fermi surface for a single component.
• The cloud remains larger than a classical gas would at the same very low temperature because of Fermi pressure.
• Thermal mode fluctuations are suppressed by $f(1-f)$; measured spatial fluctuations also depend on the probe volume and resolution.
• Elastic scattering and evaporative cooling can slow as final states become blocked.
• Changing the number of spin components changes both collision channels and the density assigned to each Fermi sea.

In a harmonic trap, density and local Fermi energy vary with position. Quantum Gases in Traps derives the global trap Fermi scale and local-density profiles.

Interaction scale

Degeneracy and interaction strength are independent questions. For a short-range two-component gas, a common interaction parameter is

$$k_{\mathrm F}a_s.$$

One can have:

• a weakly interacting degenerate gas;
• a strongly interacting degenerate gas;
• a paired superfluid with a gap;
• a classical but interacting gas.

The condition $T/T_{\mathrm F}\ll1$ alone does not make the ideal-gas approximation accurate.

Application: White Dwarfs

A white dwarf contains ions carrying most of the mass and electrons supplying much of the pressure support. Charge neutrality gives approximately

$$n_e = \frac{\rho}{\mu_e m_u},$$

where $\rho$ is the mass density, $m_u$ is the atomic mass unit, and $\mu_e$ is the mean mass per electron in units of $m_u$.

The star can be hot in everyday units while still satisfying

$$k_{\mathrm B}T \ll \epsilon_{\mathrm F,e}.$$

Here $\epsilon_{\mathrm F,e}$ is kinetic energy above the rest-energy threshold. Its electrons are then deeply degenerate.

Nonrelativistic pressure

For spin-$1/2$ electrons,

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

In the nonrelativistic regime,

$$P_{\mathrm{NR}} = \frac{\hbar^2}{5m_e} (3\pi^2)^{2/3} n_e^{5/3}.$$

The exponent $5/3$ gives the familiar polytropic scaling. A rough hydrostatic estimate uses

$$P_{\mathrm{grav}} \sim \frac{GM^2}{R^4}$$

and

$$n_e \sim \frac{M}{ \mu_e m_u R^3 }.$$

Balancing $P_{\mathrm{NR}}$ against gravity gives

$$R \propto M^{-1/3}$$

at fixed composition, up to structure constants. More massive nonrelativistic white dwarfs are smaller.

Relativistic crossover

The nonrelativistic condition is

$$p_{\mathrm F} \ll m_ec.$$

As density grows, $p_{\mathrm F}$ becomes relativistic. For an ultrarelativistic ideal electron gas,

$$\epsilon(p) \approx pc,$$

and

$$P_{\mathrm{UR}} = \frac{\hbar c}{4} (3\pi^2)^{1/3} n_e^{4/3}.$$

The exponent changes from $5/3$ to $4/3$. Both ultrarelativistic degeneracy pressure and the characteristic gravitational pressure then scale as $R^{-4}$, so the radius drops out of the leading balance. The resulting mass scale is

$$M_{\mathrm{Ch}} \sim \frac{ (\hbar c/G)^{3/2} }{ (\mu_e m_u)^2 },$$

times a dimensionless structure coefficient.

This scaling argument explains why a limiting mass appears, but it is not a precision white-dwarf calculation. Composition, Coulomb corrections, inverse beta processes, finite temperature, rotation, magnetic fields, stellar structure, and general relativity refine the result. The zero-temperature, nonrotating ideal-electron model with Newtonian hydrostatic equilibrium gives approximately

$$M_{\mathrm{Ch}} \approx \frac{5.83}{\mu_e^2} M_\odot,$$

which is about $1.46M_\odot$ for $\mu_e=2$.

Degenerate Does Not Mean Ideal

Real degenerate fermions often interact. The ideal Fermi gas remains useful because it supplies $k_{\mathrm F}$, $\epsilon_{\mathrm F}$, and the phase-space geometry against which interactions are measured.

Fermi-liquid regime

In a conventional Fermi liquid:

• low-energy excitations are quasiparticles near a Fermi surface;
• the heat capacity remains linear in $T$;
• an effective mass replaces the bare mass in some coefficients;
• compressibility and spin response contain interaction parameters;
• quasiparticle lifetimes grow as the surface is approached.

The scaling survives, but coefficients are renormalized.

Pairing

An attractive interaction can produce Cooper pairing and a superfluid or superconducting gap. A fully gapped quasiparticle spectrum gives an exponentially suppressed contribution; nodes can instead produce power laws. Neither generally reproduces the ideal normal-gas result. Degeneracy is a prerequisite for much pairing physics, not a guarantee that the normal ideal state remains stable.

Strong correlations

Near unitarity, in narrow bands, or near correlation-driven transitions, the ideal quasiparticle picture can fail quantitatively or qualitatively. A small $T/T_{\mathrm F}$ does not by itself control the interaction expansion.

Equilibrium and Timescales

The Fermi–Dirac distribution describes equilibrium. An exactly noninteracting gas cannot redistribute occupations and establish thermal equilibrium from an arbitrary initial state. In practice, weak interactions, collisions with another component, external noise, or preparation dynamics provide equilibration.

Deep degeneracy creates a tension:

• interactions are needed for thermalization;
• Pauli blocking suppresses the available final states;
• equilibration can therefore become slow as $T/T_{\mathrm F}$ decreases.

The equilibrium distribution and the relaxation time are different questions.

Exercises

Relate phase-space density to reduced temperature

For a uniform three-dimensional ideal gas, derive

$$\frac{n\lambda_T^3}{g} = \frac{4}{3\sqrt\pi} \left( \frac{T_{\mathrm F}}{T} \right)^{3/2}.$$

Solution

Use

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

and

$$\lambda_T^3 = \left( \frac{2\pi\hbar^2}{mk_{\mathrm B}T} \right)^{3/2}.$$

Since

$$\epsilon_{\mathrm F} = \frac{\hbar^2k_{\mathrm F}^2}{2m} = k_{\mathrm B}T_{\mathrm F},$$

one has

$$\frac{\hbar^2k_{\mathrm F}^2}{mk_{\mathrm B}T} = 2\frac{T_{\mathrm F}}{T}.$$

Therefore

$$\begin{aligned} \frac{n\lambda_T^3}{g} &= \frac{1}{6\pi^2} \left( \frac{2\pi\hbar^2k_{\mathrm F}^2}{ mk_{\mathrm B}T } \right)^{3/2} \\ &= \frac{1}{6\pi^2} \left( 4\pi\frac{T_{\mathrm F}}{T} \right)^{3/2} \\ &= \frac{4}{3\sqrt\pi} \left( \frac{T_{\mathrm F}}{T} \right)^{3/2}. \end{aligned}$$

Analyze a blocked transition

In an ideal grand-canonical ensemble with independent occupations, a matrix element connects a pair in modes $1,2$ to modes $3,4$, with all four modes distinct. Write the statistical occupation factor and evaluate it at $T=0$ when the initial modes are occupied, $\epsilon_3<\epsilon_{\mathrm F}$, and $\epsilon_4>\epsilon_{\mathrm F}$.

Now consider instead the prepared, occupation-diagonal state

$$\rho=\frac12|1100\rangle\langle1100| +\frac12|0011\rangle\langle0011|,$$

where the entries give occupations of modes $1,2,3,4$ in that order. Compare the exact weight $\langle n_1n_2(1-n_3)(1-n_4)\rangle$ with the product of the four corresponding mean factors. Does diagonality suffice for factorization?

Solution

The statistical factor is

$$f_1f_2 (1-f_3) (1-f_4).$$

At $T=0$, an initial occupied pair has $f_1=f_2=1$. Since $\epsilon_3<\epsilon_{\mathrm F}$,

$$f_3 = 1,$$

so

$$1-f_3 = 0.$$

Even though mode $4$ is empty, the product vanishes:

$$f_1f_2(1-f_3)(1-f_4) = 0.$$

One occupied final mode is enough to block the process. Energy and momentum conservation could forbid it independently.

In the prepared mixture, the $|1100\rangle$ configuration permits this occupation change and the $|0011\rangle$ configuration does not. Thus

$$\langle n_1n_2(1-n_3)(1-n_4)\rangle=\frac12.$$

Every individual mean is $f_i=1/2$, so the factorized expression would instead give

$$f_1f_2(1-f_3)(1-f_4)=\frac1{16}.$$

Both configurations contain exactly two particles. Their classical occupation correlations survive even though $\rho$ has no off-diagonal entries. The product formula requires independence or a justified kinetic approximation; diagonality alone is insufficient. This comparison tests statistical availability, not whether a specified Hamiltonian allows an energy-conserving collision.

Why an ultrarelativistic mass scale appears

Repeat the white-dwarf scaling argument using $P_{\mathrm{UR}}\propto n_e^{4/3}$. Explain why the radius cancels.

Solution

The ultrarelativistic degeneracy pressure scales as

$$P_{\mathrm{UR}} \propto \left( \frac{M}{R^3} \right)^{4/3} = \frac{M^{4/3}}{R^4}.$$

The gravitational pressure scales as

$$P_{\mathrm{grav}} \propto \frac{GM^2}{R^4}.$$

Both contain $R^{-4}$. Their balance therefore constrains $M$ rather than determining $R$:

$$M^{4/3} \sim G M^2$$

after restoring the quantum, relativistic, and composition constants. Dimensional restoration gives

$$M \sim \frac{ (\hbar c/G)^{3/2} }{ (\mu_e m_u)^2 }.$$

This is the origin of the Chandrasekhar mass scale. The precise coefficient is a stellar-structure result, not fixed by scaling alone.

Compare two fluctuation measures

A grand-canonical ideal gas has $N=10^4$ and $T/T_{\mathrm F}=0.02$. Estimate its Fano factor and relative root-mean-square number fluctuation. What changes if the whole gas has fixed particle number?

Solution

The Fano factor is about $0.03$, whereas the relative root-mean-square fluctuation is $\sqrt{0.03/10^4}\simeq0.00173$, or $0.173\%$. Both global fluctuations vanish in a fixed-number ensemble. This does not set the particle-number fluctuations of a smaller spatial region to zero.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston (1976) — degenerate electrons, Sommerfeld theory, bands, and transport.
• S. Chandrasekhar, “On Stars, Their Evolution and Their Stability”, Nobel Lecture (1983) — historical and physical perspective on stellar stability and white dwarfs.
• S. Chandrasekhar, “The Highly Collapsed Configurations of a Stellar Mass”, Monthly Notices of the Royal Astronomical Society 91, 456–466 (1931) — relativistic degeneracy and compact-star structure.
• B. DeMarco and D. S. Jin, “Onset of Fermi Degeneracy in a Trapped Atomic Gas”, Science 285, 1703–1706 (1999) — early realization and thermometry of a trapped degenerate atomic Fermi gas.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003 reprint) — Fermi-gas response, particle–hole excitations, and interacting extensions.
• S. Giorgini, L. P. Pitaevskii, and S. Stringari, “Theory of Ultracold Atomic Fermi Gases”, Reviews of Modern Physics 80, 1215–1274 (2008) — authoritative review of trapped and interacting cold Fermi gases.
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980) — degenerate gases, thermodynamics, and quasiparticle reasoning.
• B. Mukherjee, Z. Yan, P. B. Patel, Z. Hadzibabic, T. Yefsah, J. Struck, and M. W. Zwierlein, “Homogeneous Atomic Fermi Gases”, Physical Review Letters 118, 123401 (2017) — observation of a saturated momentum distribution in a uniform gas.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 4th ed., Academic Press (2021) — ideal quantum gases and low-temperature expansions.
• C. Sanner, E. J. Su, A. Keshet, R. Gommers, Y. Shin, W. Huang, and W. Ketterle, “Suppression of Density Fluctuations in a Quantum Degenerate Fermi Gas”, Physical Review Letters 105, 040402 (2010) — Pauli suppression of atom-number fluctuations.
• A. G. Truscott, K. E. Strecker, W. I. McAlexander, G. B. Partridge, and R. G. Hulet, “Observation of Fermi Pressure in a Gas of Trapped Atoms”, Science 291, 2570–2572 (2001) — direct cold-atom manifestation of Fermi pressure.