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For noninteracting particles in a three-dimensional isotropic quantum harmonic oscillator, measure the one-particle ground energy as zero. The single-particle partition function is $Z_1=(1-e^{-\beta \hslash \omega }{)}^{-3}$, from the product of three geometric series. In the dilute Maxwell–Boltzmann statistics regime, $\ln \Xi =e^{\beta \mu }{Z}_1$, so the grand potential is the displayed expression and $N=e^{\beta \mu }{Z}_1$. For $k_{B}T\gg \hslash \omega$,The same approximation follows from integrating the density of states $g(E)\simeq E^2/[2(\hslash \omega {)}^3]$ against $e^{-\beta E}$. Extrapolating the classical formula until the fugacity is one gives the scale of Bose condensation in a three-dimensional harmonic trap, but the quantitative transition temperature requires the Bose occupation law rather than Maxwell-Boltzmann statistics.

The Maxwell-Boltzmann distribution describes the random velocities and speeds of particles in a classical ideal gas at thermal equilibrium. Individual particles keep changing velocity through collisions, while the equilibrium probability distribution stays fixed. For one species of particle mass $m$ at positive temperature $T$, the characteristic scale iswhere $k_B$ is the Boltzmann constant. A signed velocity component follows a normal distribution; the nonnegative magnitude follows the Maxwell speed distribution. Their peaks differ because a speed measurement groups together all directions of motion.

Use the gas rest frame, three spatial dimensions, and nonrelativistic particles, with $k_{B}T\ll m{c}^2$ where $c$ is the speed of light. Assume a dilute gas whose interactions between encounters can be neglected in the ideal gas approximation and whose translational motion is classical. A sufficient condition for negligible quantum exchange effects is $n{\lambda }^3\ll 1$, where $n$ is the number density and $\lambda$ the thermal de Broglie wavelength. Diluteness alone does not guarantee thermal equilibrium; the gas must have equilibrated. These are the assumptions behind the classical Maxwell-Boltzmann velocity distribution. ^{[ref] [ref]}

The connection to Maxwell–Boltzmann statistics is the Boltzmann factor for translational kinetic energy, $E_{\mathrm{kin}}=m|\mathbf{v}{|}^2/2$. Normalizing $e^{-E_{\mathrm{kin}}/(k_{B}T)}$ over velocity gives a probability density; an occupation factor $e^{(\mu -E)/(k_{B}T)}$ instead specifies a mean occupation, with chemical potential $\mu$. Their normalizations serve different purposes. ^[ref]

This connects microscopic motion to measurements: a gas thermometer uses the ideal gas law, collisions and the mean free path govern transport, and measurements of spectral lines use thermal Doppler broadening to infer temperature. Each probes a different aspect of the same thermal motion.

In 1860, James Clerk Maxwell presented the component and speed laws in Part I, Proposition IV of Illustrations of the Dynamical Theory of Gases, pp. 22–23. His argument combined independence of the three velocity components with isotropy, and he explicitly compared the component law with the normal distribution of observational errors. Independence is an assumption of that argument; perpendicular coordinate axes alone do not establish it. ^[ref]

Maxwell–Boltzmann statistics describes a dilute classical gas whose single-particle occupation factors have the displayed exponential form. It is the low-occupation classical limit of Bose--Einstein and Fermi--Dirac statistics. Integrating this factor over momentum gives the nonrelativistic Maxwell--Boltzmann number density when the particle energy is $m{c}^2+p^2/(2m)$.

In the dilute Maxwell--Boltzmann regime, a species has number density proportional to $g(m{k}_{B}T{)}^{3/2}{e}^{(\mu -m{c}^2)/(k_{B}T)}$.

Integrating Maxwell–Boltzmann statistics over momentum for nonrelativistic particles of mass $m$, spin-state count $g_s$, and energy $m{c}^2+p^2/(2m)$ gives

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.

Put $z=e^{\beta \mu }$, with $\beta =1/(k_{B}T)$. For a single quantum state of energy $ϵ$, dilute Maxwell–Boltzmann statistics gives $\ln {\Xi }_{ϵ}\simeq z{e}^{-\beta ϵ}$. For bosons this is the leading expansion of $-\ln (1-z{e}^{-\beta ϵ})$; the occupation must be small for that expansion. Thus the grand potential isWith the ground-state energy set to zero, a dilute classical gas requires $e^{\beta \mu }\ll 1$, giving negative $\mu$. This sign is relative to the chosen energy zero. For an ideal Bose gas, $\mu$ approaches the ground energy from below at condensation, so it approaches zero with this convention.

The printed degeneracy formula requires $n_x,n_y,n_z=0,1,2,\dots$, rather than strictly positive integers. With zero-point energy omitted these are the appropriate oscillator quantum numbers. The stars and bars count of three nonnegative integers summing to $n$ givesThere is one ground state at $n=0$. If the integers were strictly positive, the count would instead be $\left(\genfrac{}{}{0ex}{}{n-1}{2}\right)$ for $n\ge 3$, inconsistent with both the printed formula and its energy reference.

For large $n$, $d_n\simeq n^2/2$. Consecutive energy shells are separated by $\hslash \omega$, giving the continuum density of statesUse this when $k_{B}T\gg \hslash \omega$, and integrate the single-state grand potential over energy:because the integral is $2/{\beta }^3$. Particle number follows by differentiating at fixed temperature and trap frequency:These are the high-temperature expressions for the classical grand potential in a harmonic trap.

Extrapolating to $\mu =0$ gives the requested transition scaleIt is an order-one estimate because Maxwell-Boltzmann occupations cease to be accurate near condensation. The Bose-Einstein distribution gives the more precise ideal-trap result $N_{\mathrm{ex}}^{\max }=\zeta (3)(k_{B}T/\hslash \omega {)}^3$, and hence $T_c=(\hslash \omega /k_B)[N/\zeta (3){]}^{1/3}$, about $0.94$ times the estimate. Excess particles populate the ground state after the excited-state capacity saturates: Bose condensation in a three-dimensional harmonic trap.

The observation of Bose-Einstein condensation in rubidium in 1995 implemented laser precooling followed by radio-frequency evaporation from a magnetic trap. Imaging the released cloud revealed a narrow anisotropic component within a broader thermal population. This provides an experimental connection to the population-counting argument; the measured expansion also reflects interactions and imaging resolution, not just the ideal-trap density of states.^[ref]

An intrinsic semiconductor has negligible electrically active impurity doping, and thermal excitation across its band gap produces equal electron and hole densities $n=p=n_i$. In an extrinsic sample, semiconductor doping supplies donor impurities or acceptor impurities, so one carrier species dominates over a range of temperatures. Doping does not remove the possibility of low-temperature dopant freeze-out or high-temperature intrinsic behavior.

Assume parabolic isotropic band edges, spin degeneracy two, a single valley, thermal equilibrium and nondegenerate carriers. Write $E_F$ for the Fermi level, and measure conduction-band energy by $E=E_c+ϵ$. Counting momentum states gives the density of states per unit volumeThe low-occupation Maxwell–Boltzmann statistics givesThe substitution $x=ϵ/(k_{B}T)$ and ${\int }_0^{\infty }\sqrt{x}{e}^{-x}dx=\sqrt{\pi }/2$ give the coefficient. Similarly $p=N_v{e}^{(E_v-E_F)/(k_{B}T)}$ and $N_v=2[m_h^{\ast }{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Multiplication eliminates $E_F$, proving the semiconductor mass-action lawHere the supplied effective masses are treated as the masses entering the effective density of states. A detailed multivalley silicon model would require the appropriate valley degeneracy and density-of-states masses consistently; adding an extra factor without changing the stated model would alter the requested expression.

A pentavalent arsenic impurity supplies a weakly bound excess electron. In the hydrogenic donor state approximation, dielectric screening reduces the Coulomb attraction and the band effective mass replaces the free mass. Hence the donor binding energy isThis is the effective-mass estimate, without a material-specific central-cell correction.

For the temperature plot use donor density $N_D=10^{15}{\mathrm{cm}}^{-3}$ and no acceptors. The charge neutrality with partial donor ionization model puts $E_D=E_c-E_b$ and includes the twofold spin degeneracy of an occupied donor:These equations give three regimes. At low temperature, $p$ is negligible and strong dopant freeze-out gives $n\simeq \sqrt{N_D{N}_c/2}{e}^{-E_b/(2k_{B}T)}$. At intermediate temperature, the donors are almost ionized and intrinsic holes remain rare, so $n\simeq N_D$. At high temperature, $n\simeq p\simeq n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$. Thus the log-density plot against inverse temperature has a steep intrinsic segment on the left, a nearly horizontal extrinsic plateau, and a freeze-out decline on the right; the $T^{3/2}$ and $T^{3/4}$ factors give slight curvature.

To quantify “approximately constant”, require $0.9N_D\le n\le 1.1N_D$. At $300\mathrm{K}$, the specified model gives $N_c\simeq 4.12\times 10^{18}$, $N_v\simeq 8.61\times 10^{18}$ and $n_i\simeq 2.33\times 10^9{\mathrm{cm}}^{-3}$. The full neutrality equations give the ten-percent plateau intervalThe lower boundary is set by donor ionization and the upper boundary by intrinsic electron-hole production. The large ratio $N_c/N_D$ allows substantial donor ionization even when $k_{B}T<E_b$; using only $k_{B}T\gg E_b$ would incorrectly exclude much of the plateau. The quoted interval depends on the chosen tolerance and the constant-gap band model, rather than being an exact measured range for arsenic-doped silicon.

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

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.

An ideal quantum gas has mean occupationwith one sign for bosons and the other for fermions.

For a single one-particle state with $x_k\ll 1$, the empty and singly occupied contributions dominate. A fermion state gives ${\Xi }_k=1+x_k$, a boson state gives $(1-x_k{)}^{-1}$, and Maxwell–Boltzmann statistics give $e^{x_k}$. Thus all have the same leading dilute response, $\ln {\Xi }_k=x_k+O(x_k^2)$. This approximation does not impose an exclusion principle on classical particles.

Consider a planar surface with outward normal along $z$. For an auxiliary equilibrium electron gas just outside it, let $n_v$ be the total number density, counting both signs of $v_z$. Its Maxwell-Boltzmann velocity distribution has normal-component densityAn electron with $v_z>0$ crosses more area per unit time when it moves faster. The outgoing number flux therefore uses $v_z{f}_z$, rather than just the probability of $v_z>0$:This integral contributes only $T^{1/2}$. Holding a classical bulk number density fixed and imposing a normal kinetic energy threshold $W$ would instead give $\Gamma =n\sqrt{k_{B}T/(2\pi m_e)}{e}^{-W/(k_{B}T)}$; that argument alone does not produce $T^2$.

For the ideal surface model, measure external electron energies from the metal's chemical potential $\mu$. A freely propagating external state has $E-\mu =W+p^2/(2m_e)$. When $W\gg k_{B}T$, its Fermi-Dirac distribution reduces to the Maxwell–Boltzmann statistics tail. Counting two spin states per phase space cell of volume $h^3$ givesEach of the three Gaussian integrals supplies $T^{1/2}$. Combining this $T^{3/2}$ state count with the $T^{1/2}$ crossing factor givesThe auxiliary atmosphere supplies an equilibrium state-counting argument; an operating collector removes the outgoing electrons. It need not maintain a real equilibrium gas throughout the gap. Unit transmission and the ideal free electron state count are substantive assumptions, so this calculation does not determine an arbitrary real surface's measured $A$.

In particular, electrons inside a metal generally form a degenerate Fermi gas. A classical distribution for the escaping tail does not imply a classical distribution for the whole interior. Richardson himself emphasized this distinction in his lecture.^[ref]