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In 2001, Andrew G. Truscott, Randall G. Hulet, and collaborators observed degeneracy pressure in trapped lithium. A Zeeman slower fed a dual magneto-optical trap; optical pumping prepared atoms for a cloverleaf apparatus for magnetic trapping of neutral atoms. Microwave evaporation removed energetic lithium-7 bosons; elastic collisions cooled lithium-6 fermions by sympathetic cooling.^[ref]

Sequential absorption imaging recorded both trapped clouds on a camera. Fits to axial profiles used the boson cloud as a thermometer. At common temperature, the nearly equal magnetic trapping potentials provided a comparison: the clouds had similar sizes near $T/T_F=1$, but the fermion cloud remained broader as cooling reached $T/T_F=0.25$. Figure 3 compared its radius with ideal Fermi gas and classical predictions.^[ref]

Varying the wait before imaging tested thermal equilibrium. Finite imaging resolution restricted the analysis to the axial direction; fits neglected interaction-induced density shifts. Cooling eventually suffered from the dwindling boson heat capacity. The result supports quantum occupation effects, without directly resolving individual occupied states or reaching zero temperature.^[ref]

These were neutral alkali atoms in a spatially varying trap. The Fermi energy of a model electron gas instead uses uniform electrons, two equally populated spin states, and the electron mass. The measured trapped-atom radii are not measurements of that electron model's $p_F$ or $E_F$; comparing their numerical Fermi energies requires the appropriate particle mass, internal-state populations, and confining potential.

Use the supplied effective electrostatic term and enforce charge neutrality while varying the chain length: $R=Z/(2n_e)$, with $N$ and $Z$ fixed. The total energy per positive ion becomesThe positive quadratic term supplies degeneracy pressure against compression. Differentiating gives the neutral one-dimensional metal equilibrium density,where $a_B=4\pi {ϵ}_0{\hslash }^2/(m_e{e}^2)$ is the Bohr radius. The second derivative $Z{\pi }^2{\hslash }^2/(12m_e)>0$ verifies a minimum. Within this specified toy energy, the minimizing electron density is independent of $Z$; instead the equilibrium ion spacing is $2R^{\ast }=Z/n_e^{\ast }$.

At the minimum the electrostatic contribution is minus twice the kinetic contribution, so ${\epsilon }_{\ast }=-Z{E}_F^{\ast }/3$. These follow from the given electrostatic approximation, not from a full self-consistent calculation of a real one-dimensional metal. In particular, additional Coulomb terms cannot be appended without the corresponding neutral-background subtraction that defines the supplied finite energy.

Within the prescribed uniform-density approximation, $n=3M/(4\pi m_p{R}^3)$. Treating the given approximate pressure coefficient as one, the degeneracy pressure and specified kinetic-energy relation giveFor a uniform sphere, $m(r)=M(r/R{)}^3$, so shell assembly givesThusAt fixed mass, this energy has its minimum at $R_{\ast }=2\alpha /\beta$, with $E^{\prime \prime }(R_{\ast })=2\alpha /R_{\ast }^4>0$. Expansion beyond this radius raises the energy, while strong compression costs positive energy growing as $R^{-2}$, faster than the attractive $R^{-1}$ term. This explains radial stability of the model white dwarf. The numerical coefficient in $\alpha$ inherits the approximate pressure law; a relativistic degeneracy law would change this stability argument.

Use the uniform-density configuration as a trial configuration for the white dwarf. Write $P=C_F{h}^2{n}^{5/3}/m_e$, with $C_F>0$ the numerical coefficient suppressed by the approximate equation of state. Its particle density is $n=3M/(4\pi m_p{R}^3)$, and the nonrelativistic degeneracy pressure has kinetic-energy density $3P/2$. HenceFor uniform density $m(r)=M(r/R{)}^3$, so integration of the shell energy givesBoth constants are positive and independent of $R$. Differentiating $E=\alpha /R^2-\beta /R$ gives its unique stationary radius $R=2\alpha /\beta$. The second derivative there is $2\alpha /R^4>0$, and this is the global minimum since $E\to +\infty$ at zero and $E\to 0^{-}$ at infinity. ThusThe uniform-density approximation is used for this energy estimate; it does not require a spatially constant pressure to satisfy the exact hydrostatic equilibrium equation.

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.

The nuclear Fermi gas model treats protons and neutrons as two species of spin one-half fermions moving approximately independently in a confining nuclear mean field. At a temperature small compared with their Fermi temperature, the Pauli exclusion principle fills the lowest available momentum states. It does not allow all nucleons to sit at zero momentum.

For species $i=p,n$, two spin states occupy each momentum-space cell $(2\pi \hslash {)}^3/V$. Counting a filled sphere giveswhere $n_i=N_i/V$. The Fermi energy is measured from the mean-field well bottom; it need not equal the energy needed to remove a nucleon into vacuum. Occupied-state averaging givesEven at zero temperature there is kinetic energy and degeneracy pressure. As an explicitly illustrative input, total density $n=0.16{\mathrm{fm}}^{-3}$ and $N=Z$ give $p_{F}c\simeq 263\mathrm{MeV}$ and $E_F\simeq 37\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.

Separate proton and neutron seas explain a kinetic contribution to nuclear symmetry energy. At fixed $n$, put $\delta =(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. ThenIncreasing one species' population raises its Fermi surface and costs energy, favoring similar populations before electrostatic and interaction terms are included.

pauli blocking suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a nucleon into an empty state and reveal its initial Fermi motion, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent nuclear mean field.

A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete nuclear shells, nuclear spin-orbit splitting, nuclear pairing, short-distance correlations and deformation. It cannot assign all ground-state spins from a momentum sphere. Its useful prediction is the scale and occupancy of nucleon motion, with a surrounding mean field supplied separately.

identical particles have the same intrinsic properties. Exchanging their bookkeeping labels cannot distinguish a physical preparation or change a measurable prediction. This is stronger than merely failing to track classical trajectories. For ordinary bosons and fermions, the many-particle wavefunction has respectively bosonic exchange symmetry or fermionic exchange symmetry under every particle exchange operator, acting on spatial and spin coordinates together. The Spin-statistics theorem connects integer spin to bosons and half-integer spin to fermions; this connection uses relativistic quantum-field assumptions, rather than following from the nonrelativistic Schrödinger equation alone.

For one-particle states $u,v$ satisfying orthonormality, the two alternatives form a quantum superposition:Its probability density contains an exchange interference term, absent from an incoherent mixture of labelled alternatives. The antisymmetric expression vanishes when both one-particle states are identical: one complete one-particle state can contain at most one fermion. This Pauli exclusion principle explains electron shell filling, the electronic Fermi sea and degeneracy pressure. Opposite spins are different one-particle states, so two electrons may share an atomic orbital. Spatially separated wave packets with negligible overlap can often be treated as distinguishable for local measurements; the underlying exchange symmetry remains.

In thermal equilibrium, occupation of a level of energy $ϵ$ obeysbosons can accumulate macroscopically in one state, giving Bose-Einstein condensation; fermions instead fill successive states up to the Fermi energy at zero temperature. When occupation is dilute, both expressions reduce to the Maxwell-Boltzmann distribution. The identical-particle classical partition function contains $1/N!$, avoiding spurious entropy from permutations and the Gibbs paradox. Exchange effects become conspicuous when the thermal de Broglie wavelength is comparable to the interparticle spacing.

An operational test is the Hong-Ou-Mandel effect. Down-converted photon pairs enter separate ports of a balanced beam splitter; indistinguishable alternatives for one photon per output cancel. Scanning the relative arrival delay restores coincidences away from overlap. Matching polarization, spectral filtering and spatial alignment controls distinguishability; measuring delayed accidental coincidences controls background. The observed dip tests two-photon interference, and its finite depth must not automatically be interpreted as a failure of bosonic symmetry. The original apparatus varied beam-splitter position and subtracted a separately measured accidental background.^[ref]

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

For a single-particle state of energy $ϵ$, put $z=e^{\mu /(k_{B}T)}$. The equilibrium mean occupancies areThe Maxwell-Boltzmann distribution is the dilute classical limit. The Fermi-Dirac distribution applies to identical fermions with at most one particle in each complete single-particle state by the Pauli exclusion principle. The Bose-Einstein distribution applies to identical bosons with no such upper occupation bound. The chemical potential fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal Bose gas, $\mu$ cannot exceed the lowest single-particle energy.

Take $\mathbf{k}$ to be a wavevector and count periodic-box states with cell volume $(2\pi {)}^3/V$ in momentum space, parameterized by $\mathbf{k}=\mathbf{p}/\hslash$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_{s}V4\pi k^{2}dk/(2\pi {)}^3$. Since $ϵ=\alpha k$,Thus the one-particle canonical partition function isHere $\beta =1/(k_{B}T)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses ${\int }_0^{\infty }{u}^2{e}^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, givingThe factor three in the energy arises from the linear dispersion relation and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3N{k}_{B}T/2$ does not apply.

For spin-half fermions at $T=0$, all states up to the Fermi energy are occupied. With $g_s=2$,Integrating the occupied energy states givesThese are the corresponding ultrarelativistic ideal Fermi gas formulae with general slope $\alpha$. The nonzero ground-state internal energy and degeneracy pressure reflect the Pauli exclusion principle: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to 0$, but Maxwell–Boltzmann statistics is not valid at fixed positive $n$ as $T\to 0$.

At large $T$, the number equation gives $z\simeq N/Z_1\ll 1$, so expanding the Fermi-Dirac distribution recovers the Maxwell-Boltzmann distribution. A convenient crossover scale is the Fermi temperature,For $T\gg T_F$, both gases have $p\simeq n{k}_{B}T$. At fixed $n$, the Fermi curve begins at $p_0=n{ϵ}_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n{ϵ}_F)=1/4+({\pi }^2/6)(T/T_F{)}^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.

If the source's symbol $\mathbf{k}$ is interpreted literally as momentum rather than a wavevector, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hslash \alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.