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Write $n=n_0+y\mathcal{L}{n}_0+O(y^2)$, with $n_0=(e^x-1{)}^{-1}$ and $\mathcal{L}=x^{-2}{\partial }_x(x^4{\partial }_x)$. Differentiating the Planck photon distribution givesDividing by $n_0$ proves the first-order thermal Sunyaev-Zeldovich effect:In the Rayleigh-Jeans law limit, $x{e}^x/(e^x-1)\to 1$ and $x\coth (x/2)\to 2$, so $\Delta n/n_0\to -2y$. In the Wien approximation, the prefactor approaches $x$ and the bracket approaches $x-4$, so $\Delta n/n_0\sim yx(x-4)\sim y{x}^2$. The last limit is an asymptotic description of the first-order coefficient: a perturbative prediction at large $x$ still requires $y{x}^2\ll 1$.

Hot electrons transfer energy to the cosmic microwave background, removing low-frequency photons and enhancing the high-frequency tail. The first-order distortion crosses zero where $x\coth (x/2)=4$, at $x\simeq 3.830$; for today's CMB this is near $217\mathrm{GHz}$. Photon conservation follows directly from the boundary termIn contrast, integration by parts gives $\int x^3\Delta ndx=4y\int x^3{n}_{0}dx$, so radiation gains a fraction $4y$ of its original energy. The thermal Sunyaev-Zeldovich effect is thus a spectral distortion, not simply a new blackbody temperature.

A galaxy cluster with bulk peculiar velocity along the line of sight additionally produces the kinetic Sunyaev-Zeldovich effect. Define $v_{\parallel }>0$ for recession and $\tau =\int n_e{\sigma }_{T}d\ell$. To first order in velocity and optical depth, $\Delta T/T=-\tau v_{\parallel }/c$: receding gas gives a decrement and approaching gas an increment. Its occupation number perturbation is $(\Delta T/T)x{e}^x/(e^x-1{)}^2$, the spectrum of a small blackbody temperature change. It remains nonzero at the thermal null and therefore complicates measurements near that frequency.

An adiabatic change of box size preserves the mode labels, whose momenta scale as $p\propto L^{-1}$. Since $V=L^3$, differentiation gives $dp/dV=-p/(3V)$. Occupation numbers are fixed under this slow mechanical change, so summing the changes in the individual particle energies givesComparison with $dE=-PdV$ at fixed entropy and particle number yieldsThe integration notation means summing over the original momentum-mode labels; it does not assume that the density in momentum space is unchanged after rescaling. In the ultrarelativistic limit $E(p)=cp$, so $p{E}^{\prime }(p)=E(p)$ and $P=E/(3V)=\rho c^2/3$.

Adiabatic radiation obeys $d(E{a}^3)=-Pd(a^3)$, giving energy density proportional to $a^{-4}$. Equilibrium photon energy density is proportional to $T^4$, so $T\propto a^{-1}$. Therefore the photon number in a comoving volume, $n_{\gamma }{a}^3\propto T^3{a}^3$, is constant. Its entropy is also constant: photon entropy density is $({ϵ}_{\gamma }+P)/T\propto T^3$. Photon number is not a microscopic conserved charge, but equilibrium processes maintaining the spectrum do not alter this comoving-number scaling in adiabatic expansion; entropy-producing processes would invalidate the assumptions.

Use the Holstein–Primakoff transformation on the physical occupation number subspace $0\le n\le 2S$. With the canonical commutation relation $[a,a^{†}]=1$, the ordered square root givesBoth endpoints are respected: $S^{+}|0⟩=0$ and $S^{-}|2S⟩=0$. Consequently,Since these Fock states form a basis of the physical spin space, $[S^{+},S^{-}]=2S^z$ there. Similarly $[S^z,S^{\pm }]=\pm S^{\pm }$. The Holstein–Primakoff occupation constraint is essential: unrestricted bosonic occupation would not represent a spin-$S$ Hilbert space.

The linear spin-wave approximation expands about the rotated all-up state, with ${\hat{n}}_n=a_n^{†}{a}_n$:In the longitudinal product, $S_n^z{S}_{n+1}^z=S^2-S({\hat{n}}_n+{\hat{n}}_{n+1})+O(S^0)$. Each site belongs to two bonds, so the resulting quadratic Hamiltonian isThe omitted terms are of order $S^0$ at fixed small occupation number. The pair terms describe the quantum fluctuations that were missing from the classical Néel state.

For the harmonic-oscillator number-state second moments, the number operator gives $⟨a^{†}a{⟩}_n=n$, while the canonical commutation relation gives $⟨a{a}^{†}{⟩}_n=n+1$. Terms with $a^2$ or $(a^{†}{)}^2$ have zero diagonal matrix elements, since they change the occupation number by two. ThereforeConsequently the potential energy and kinetic energy means areTheir sum is the oscillator energy eigenvalue $E_n=\hslash \omega (n+1/2)$. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

Add the one-particle energy term $H_0={ϵ}_1{n}_1+{ϵ}_2{n}_2$. It is diagonal in the same Fock state basis as the contact interaction between two ring modes. For $N=2$, the ordered basis $(|2,0⟩,|1,1⟩,|0,2⟩)$ givesFor $g>0$, the middle eigenvalue exceeds the mean of the two endpoint eigenvalues by $g/L$, so it cannot be the smallest. The ground state is therefore $|2,0⟩$ if ${ϵ}_1<{ϵ}_2$, and $|0,2⟩$ if ${ϵ}_2<{ϵ}_1$, withIf ${ϵ}_1={ϵ}_2$, every normalized linear combination of the two endpoint Fock states is a ground state. The same endpoint argument works for general fixed $N$, since the interaction contribution is a concave quadratic function of the mode occupation number.

For distinct positive integers $p,q$, use normalized box modes ${\phi }_1=\sqrt{2/L}\sin (p\pi x/L)$ and ${\phi }_2=\sqrt{2/L}\sin (q\pi x/L)$. Their contact interaction overlaps obeyUnlike traveling ring modes, these standing modes also allow pair conversion. For the first two box modes, or any distinct pair without a third-harmonic resonance, the contact interaction between two box modes isThe last two terms change an individual mode occupation number by two while conserving total particle number. Thus the Hamiltonian operator is not diagonal in the Fock state basis, and minimizing diagonal entries alone does not find its ground state. For example, at $N=2$ with no dispersion, $\left(|2,0⟩-|0,2⟩\right)/\sqrt{2}$ has the lowest energy $g/(2L)$, below either unmixed endpoint's $3g/(2L)$.

The wording permits modes beyond the first two. For completeness, the identity ${\sin }^{3}u=(3\sin u-\sin 3u)/4$ givesWhen $q=3p$, add $-g(a_1^{†2}{a}_1{a}_2+a_1^{†}{a}_2^{†}{a}_1^2)/(2L)$; when $p=3q$, add $-g(a_1^{†}{a}_2^{†}{a}_2^2+a_2^{†2}{a}_1{a}_2)/(2L)$. These extra terms change an individual occupation number by one and require diagonalizing the corresponding fixed-$N$ matrix as well.

For independent bosonic modes, the canonical commutation relation isFor independent fermionic modes, the canonical anticommutation relations arewhere $[A,B]=AB-BA$ and $\{A,B\}=AB+BA$. The creation operator is the adjoint of the annihilation operator. In either case the mode's number operator iswith $a_i=b_i$ or $c_i$ as appropriate. It measures the mode's occupation number. Its possible eigenvalues areFor fermions, $c_i^2=0$ implies $n_i^2=c_i^{†}(1-c_i^{†}{c}_i)c_i=n_i$, so the number operator is a projection. This expresses the exclusion of double occupation of one fermionic mode. The total particle number is $N={\sum }_i{n}_i$.

The occupation formula from part (b) givesConsequentlyThe independent occupation numbers $n_1,n_2$ are each zero or one. The Hamiltonian is independent of $n_1$, so its four states haveEach energy admits $n_1=0$ or $n_1=1$; these states differ by occupancy of the zero-energy mode. The ground energy is $-|t|$ for either sign of $t$. These degeneracies refer to the full two-mode fermionic Fock space. If one separately restricts to a fixed global fermion parity, only one state at each energy remains in that parity sector.

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

In the grand canonical ensemble, the system exchanges both energy and particles with a reservoir at fixed temperature $T$ and chemical potential $\mu$. The probability of a particular many-body microstate $i$, with energy $E_i$ and particle number $N_i$, is its Boltzmann weight $e^{-\beta (E_i-\mu N_i)}$ divided by the grand canonical partition function $\Xi$, where $\beta =1/(k_{B}T)$. The denominator normalizes ${\sum }_i{p}_i=1$; a degeneracy is accounted for by counting distinct microstates, rather than changing an individual state's probability.

For noninteracting particles, specify a microstate by the occupation numbers $n_r$ of one-particle levels of energy ${ϵ}_r$. Then $E-\mu N={\sum }_r({ϵ}_r-\mu )n_r$, so the Gibbs probability factorizes between levels. Put $x_r=e^{-\beta ({ϵ}_r-\mu )}$.

For fermions, the Pauli exclusion principle permits $n_r=0,1$ in each complete one-particle state, giving ${\Xi }_r=1+x_r$. Its mean occupancy isthe Fermi-Dirac distribution.

For bosons, $n_r=0,1,2,\dots$, so the geometric sum is ${\Xi }_r=1/(1-x_r)$ when $x_r<1$. Differentiating that sum givesthe Bose-Einstein distribution. Finite-system convergence requires $\mu <{ϵ}_{\min }$; condensation is the limiting approach to the lowest energy with macroscopic ground-state occupation. Both quantum distributions reduce to $e^{-\beta ({ϵ}_r-\mu )}$ when occupation is small. The quantities derived here are mean occupancies of states, rather than normalized probabilities over the set of one-particle energies.

The Pauli exclusion principle says that each complete one-particle state can have fermionic occupation number zero or one. “Complete” includes spin and any other internal labels, as well as the spatial wavefunction. Thus two electrons can share a spatial orbital with opposite spin projections: these are two different spin orbitals. Equal energy alone does not mean equal quantum state, because degenerate energy levels contain several independent one-particle states.

This distinction explains how an electron shell can accommodate several electrons while exclusion still allows only one per complete one-particle state. It also makes a finite model of fermionic state counting surprisingly simple: two spatial orbitals with two spin choices each give four complete one-particle states, and exactly six two-electron basis states. The following diagram enumerates them; its boxes represent occupation numbers, not locations in space.

Pauli's Exclusion principle and quantum mechanics by Wolfgang Pauli is his retrospective account, delivered on 13 December 1946 for the 1945 Nobel Prize. On printed pp. 29–30, he describes proposing an extra two-valued electron property in autumn 1924 and formulating exclusion in 1925. The crucial clue linked counts of resolved energy eigenstates in alkali spectra under a magnetic field with the populations of closed electron shells. Resolving every label turned a complicated shell capacity into one electron per complete one-particle state. This historical route preceded the modern spin interpretation and the antisymmetric wavefunction formulation; it was not originally a deduction from a Slater determinant.

A photon gas is a collection of photons described statistically as radiation. Here the equilibrium model is homogeneous, isotropic electromagnetic radiation in vacuum at temperature $T>0$: each photon has energy $E=c|\mathbf{p}|=h\nu$, with momentum $\mathbf{p}$ and frequency $\nu$, and there are two independent transverse states of electromagnetic polarization. An absorbing cavity with walls at a common temperature provides a way to establish this thermal equilibrium. The bulk formulas assume cavity dimensions much larger than the typical wavelength $hc/(k_{B}T)$, so the electromagnetic mode density in a box can replace individual cavity modes by a continuum. ^[ref]

The crucial constraint is whether photon emission and photon absorption, or other number-changing processes, can equilibrate the photon population. When they can, photon number is not an independently conserved quantity and the equilibrium photon chemical potential is ${\mu }_{\gamma }=0$. For each polarization and frequency, the mean occupation number then follows the Bose-Einstein distribution:If number-changing processes are ineffective but number-conserving interactions redistribute energy, a constrained Bose-Einstein distribution can instead have $f=[\exp ((h\nu -{\mu }_{\gamma })/(k_{B}T))-1{]}^{-1}$ with chemical potential ${\mu }_{\gamma }\le 0$ fixed by the number constraint. Thus thermal equilibrium with respect to energy exchange alone does not establish ${\mu }_{\gamma }=0$. The zero value can also persist in an already prepared Planck photon distribution during suitable subsequent evolution; continual photon emission is not required to preserve it. ^[ref]

Combining $f$ with the two-polarization density of states, $g_{\nu }/V=8\pi {\nu }^2/c^3$, gives the Planck photon distribution and the spectral photon energy density:Here $B_{\nu }$ is the spectral radiance in Planck's law; $u_{\nu }$ is stored energy density per unit frequency, not outgoing power per area. Their different units are connected by the full-sphere solid angle and the speed of light $c$. The Boltzmann constant is $k_B$, and $h$ is the Planck constant. ^[ref]

The equilibrium equation of state is particularly simple:The radiation constant $a_R$ is the same constant called $a_{\mathrm{rad}}$ in black-body radiation; $\sigma$ is the Stefan-Boltzmann constant. An outline of the energy density calculation substitutes $x=h\nu /(k_{B}T)$ and uses ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation pressure follows because the angular average of the squared directional component of photon momentum is $1/3$ for isotropy; the thermodynamic derivation of blackbody radiation pressure supplies an independent route. These formulas describe the radiation contribution, not the cavity walls. The isotropic massless relation $p=u/3$ survives a different isotropic spectrum, whereas the coefficient in $u=a_R{T}^4$ requires the zero-chemical potential Planck photon distribution.

For a computed example with stipulated $T=2.725\mathrm{K}$, the photon energy density is $u\simeq 4.17174\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$ and the radiation pressure is $p\simeq 1.39058\times 10^{-14}\mathrm{Pa}$. The photon number density in the International System of Units isHere $\zeta$ is the Riemann zeta function. The mean photon energy differs from the energy at the peak of a spectral curve: counting photons and weighting them by energy emphasize different frequencies. The calculations use the exact International System of Units values of the Planck constant, Boltzmann constant and speed of light. ^[ref]

A motivating observation is the cosmic microwave background spectrum measured by the Far Infrared Absolute Spectrophotometer (FIRAS) aboard COBE. Its apparatus used a polarizing Michelson interferometer, with one input viewing a roughly seven-degree sky patch and the other an internal blackbody. Scanning mirrors changed the optical path length difference; heat-sensitive detectors recorded the varying power, and Fourier transform spectroscopy recovered the spectral difference. Operation outside the terrestrial atmosphere removed atmospheric emission and absorption, while shielding and cooling the optics in a helium cryostat suppressed the instrument's own radiation. ^[ref]

Absolute calibration came from moving an external blackbody into the sky aperture, filling the same beam used for the sky. This was an iron-loaded epoxy absorber shaped as a re-entrant cone. Repeated reflections made it a close approximation to a blackbody, while thermometers supplied its temperature. Mather and colleagues explicitly evaluated reflections, temperature gradients, radiation leaking through the material and leakage around its edge. These are limits on how faithfully the calibrator realizes thermal equilibrium radiation, rather than details that can be eliminated merely by calling it black. ^[ref]

During sky observations the internal blackbody temperature was adjusted to nearly null the spectral difference, reducing gain and drift errors. Fixsen and colleagues' 1996 analysis used calibrated spectra over wavenumbers $2$ to $21\mathrm{c}{\mathrm{m}}^{-1}$, about $60$ to $630\mathrm{GHz}$. They separated the sky's spatially uniform component, its dipole and Galactic emission, and varied sky cuts and foreground templates as controls. The resulting cosmic microwave background spectrum agreed with a single Planck function, with reported RMS deviations below 50 parts per million of its peak brightness. This is a residual statistic over the analyzed spectrum, not a pointwise relative error bound at every frequency. ^[ref]

The observation motivates using a photon gas as a physical model rather than merely a formal counting exercise. Its interpretation has a clear boundary: agreement with Planck's law constrains departures from an equilibrium spectral shape within the calibrated band and foreground assumptions; it does not directly measure radiation pressure or entropy density of a photon gas. Nor does it establish that today's cosmic microwave background is continually re-equilibrating with matter. A previously thermal spectrum can retain its Planck function shape as expansion of the universe lowers every photon frequency in the same proportion. The usefulness of the model is that one measured spectral temperature, together with its equilibrium assumptions, predicts energy density, radiation pressure, photon number density and entropy density of a photon gas consistently.

For a linear spin-wave approximation, the ground state is empty of the diagonal quasiparticles but contains the original spin bosons. If $a_k=u_k{\alpha }_k-v_k{\alpha }_{-k}^{†}$, their occupation number is $v_k^2$. The staggered magnetization is reduced from $S$ to $S-N^{-1}{\sum }_k{v}_k^2$. For the one-dimensional nearest-neighbor Heisenberg antiferromagnet, $v_k^2=(|\sin k{|}^{-1}-1)/2$ and the integral diverges logarithmically. This invalidates finite ordered magnetization within the approximation; it is not itself a determination of the exact spectral gap.

For a two-species uniform boson number state, set $S_i^{+}=a_i^{†}{b}_i$. A single spin ladder operator changes the species occupation numbers, so its expectation value vanishes. At distinct sites, $⟨S_i^{+}{S}_j^{-}⟩=n_A{n}_B/L^2$, proving the displayed transverse two-point correlation function. For $n_A/L,n_B/L$ tending to nonzero constants, this correlation function has a nonzero large-distance limit despite the global U(1) symmetry. This diagnoses long-range order in the auxiliary state; its interpretation as a state of fixed physical spins still requires the Schwinger boson constraint.

The normalized vacuum state annihilated by $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$, for positive $u,v$ with $u^2+v^2=1$. Since $(a^{†}{b}^{†}{)}^2=0$, it can also be written $u\exp [(v/u)a^{†}{b}^{†}]|0⟩$. Acting with either annihilation operator makes the two terms cancel. Its original occupation numbers each have expected value $v^2$ and its pair amplitude is $⟨ab⟩=-uv$. It has even fermion parity although the original total number operator is not sharp. For the two-orbital fermionic pairing Hamiltonian, it is the unique ground state of energy $\epsilon -\sqrt{{\epsilon }^2+{\Delta }^2}$.

Two independent boson species occupy uniform single-particle modes on $L$ sites, with fixed occupation numbers $n_A,n_B$. This differs from a fixed-number two-mode condensate whose particles all occupy one superposition of the two modes. Here $⟨a_i^{†}{a}_j⟩=n_A/L$, $⟨b_i^{†}{b}_j⟩=n_B/L$, and the local total number operator has variance $(n_A+n_B)L^{-1}(1-L^{-1})$. The latter follows from two independent binomial distributions with success probability $1/L$. This state satisfies a prescribed Schwinger boson constraint only in the mean unless the local variance vanishes.