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Bose statistics is the symmetric exchange statistics of identical bosons. Commuting creation operators build the symmetric bosonic Fock space; a normalized one-particle mode has occupation states $(a^{†}{)}^n|0⟩/\sqrt{n!}$ for every nonnegative integer $n$. At thermal equilibrium these statistics lead to the Bose-Einstein distribution. Exchange symmetry itself does not require thermal equilibrium.

Commuting creation operators make multiparticle states invariant under exchange of their labels. With $[a,a^{†}]=1$, the normalized number states $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exist for all $r\ge 0$, giving unrestricted bosonic occupation numbers. The single-mode thermal sum is $Z={\sum }_{r\ge 0}{e}^{-\beta Er}=(1-e^{-\beta E}{)}^{-1}$ and the mean occupation is $(e^{\beta E}-1{)}^{-1}$. Thus the exchange symmetry and the Bose-Einstein distribution are two expressions of the same bosonic algebra.

For a dilute nonrelativistic gas with $x=m/T\gg 1$ and zero chemical potential, the Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution. Expanding $E=m+p^2/(2m)+\cdots$ in the number density integral gives $n\sim g(mT/(2\pi ){)}^{3/2}{e}^{-x}$. Since $\rho =mn+3nT/2+\cdots$ and $P=nT$, the entropy density at zero chemical potential gives $s=n(x+5/2+\cdots )$ and hence the displayed leading expression. This includes the internal multiplicity $g$; a single real scalar field has $g=1$.

Use natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose–Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.

Canonical mode mixing and its inverse. Treat annihilation operators as column vectors and keep complex conjugation, transpose and Hermitian adjoint distinct. Computing the canonical commutation relations for $a^{\prime }=\alpha a+\beta a^{†}$ givesConsequently the required conditions areFor complete invertible mode mixing, putThe canonical identities for a bosonic Bogoliubov transformation are $\mathcal{T}J{\mathcal{T}}^{†}=J$, giving ${\mathcal{T}}^{-1}=J{\mathcal{T}}^{†}J$. Its upper blocks yieldEquivalently, the column identities are ${\alpha }^{†}\alpha -{\beta }^T\overline{\beta }=I$ and ${\alpha }^{†}\beta ={\beta }^T\overline{\alpha }$. In finite dimension invertibility follows from the canonical matrix identity. In infinitely many modes, the row identities alone need not give a complete inverse: $a_i^{\prime }=a_{i+1}$ preserves them but discards one mode. Here completeness of the two field-mode expansions supplies the needed invertible transformation.

Obtaining coefficients from the modes. Use the Klein-Gordon inner product, antilinear in its first argument,The normalized positive-frequency solutions satisfy $(p_i,p_j)={\delta }_{ij}$, $({\overline{p}}_i,{\overline{p}}_j)=-{\delta }_{ij}$ and $(p_i,{\overline{p}}_j)=0$. Current conservation makes this product independent of the Cauchy hypersurface when boundary flux vanishes. Taking the product of the field with a mode gives $a_i=(p_i,\hat{\Phi })$ and henceWith this operator convention, the corresponding mode expansion isThe minus sign comes from the negative norm of conjugate modes. Using a different convention for mode coefficients can move this sign and the complex conjugates; the operator formula fixes them unambiguously here.

Particle count. In the in-vacuum, only the contraction $⟨0|a_j{a}_k^{†}|0⟩={\delta }_{jk}$ survives. ThereforeThis is particle number from Bogoliubov coefficients: the old vacuum contains that expected number of particles in the new mode $i$. Nonzero negative-frequency mixing is the source of particle production. For continuum modes one uses normalized wave packets and replaces the sum by the corresponding integral.

The squeezed-state relation. First use finitely many modes, or an implementable infinite-mode limit. SeekThe canonical commutation relations giveAll further nested commutators vanish since $\hat{F}$ contains only creation operators. ThusThe old annihilation conditions hold exactly whenThe inverse transformation's canonical identity $A{B}^T=B{A}^T$ shows $A^{-1}B=(A^{-1}B{)}^T$, so the required symmetry is automatic. Also $A{A}^{†}-B{B}^{†}=I$ givesHence the singular values of $M$ are less than one. The Autonne-Takagi factorization $M=U\mathrm{diag}(s_j)U^T$ changes to independent canonical oscillators. For each factor, the squared norm of $\exp (s_j{b}_j^{†2}/2)|0⟩$ is the even-occupation seriesMultiplying these norms gives the finite-mode normalizationThe phase of $C$ is arbitrary. The resulting multimode squeezed vacuum is annihilated by all old annihilation operators, so uniqueness of the normalized Fock vacuum identifies it with $|0⟩$.

The PDF's squeezed-vacuum claim needs qualification for infinitely many modes. Bosonic mode mixing implementability requires $\beta$ to be a Hilbert-Schmidt operator, ${\sum }_{ij}|{\beta }_{ij}{|}^2<\infty$, for a common ordinary bosonic Fock space. For a counterexample, take $\alpha =\cosh sI$ and $\beta =\sinh sI$ on countably many modes with fixed $s>0$. The algebraic canonical conditions hold, but $M=\tanh sI$ and the finite-$N$ normalization is $C_N=(\cosh s{)}^{-N/2}\to 0$. Every mode has a fixed positive probability of nonzero occupation in the required product state; the probability that all but finitely many modes are empty is zero. Ordinary Fock space vectors instead have total occupation finite with probability one, even if their expected occupation is infinite. Thus there is no nonzero common-Fock-space vector of the prescribed form. The squeezed expression is a normalizable vacuum relation with a finite-mode regulator or the implementability condition, not solely from the commutator identities.

Black-hole radiation. In a collapse spacetime, choose early positive-frequency solutions with respect to affine incoming time $v$ on past null infinity, and late outgoing modes proportional to $e^{-i\omega u}$ on future null infinity. The early state is their in-vacuum. The Hawking exponential ray map hasfor late outgoing rays, with $\kappa >0$. Tracing a late mode backwards therefore gives a profile proportional toIt is not a pure positive-frequency incoming wave, so the Bogoliubov transformation has nonzero $\beta$.

To see the thermal factor, put $x=v_H-v>0$ and $a=\omega /\kappa$. The positive- and negative-frequency Fourier pieces have, apart from common normalization and phases, the regulated integralsAs $\epsilon \downarrow 0$, the positive-frequency piece has squared-modulus factor $e^{+\pi a}$ and the negative-frequency piece has $e^{-\pi a}$. Consequently the thermal ratio of Hawking Bogoliubov coefficients isFor a normalized narrow-frequency outgoing wave packet, combine this ratio with ${\sum }_j(|{\alpha }_{ij}{|}^2-|{\beta }_{ij}{|}^2)=1$. The occupation is the Bose-Einstein distributionThe Hawking temperature is thus the same one found by Euclidean regularity. For Schwarzschild, $\kappa =1/(4M)$ gives $T_H=1/(8\pi M)$ in natural units. Propagation through the exterior potential multiplies the asymptotic occupation by the greybody factor ${\Gamma }_{\ell }(\omega )$.

The outgoing radiation has partner-mode correlations across the horizon: the total state can be pure while the reduced outgoing state is thermal. A complete mode basis must include modes entering the horizon as well as those reaching infinity. The particle statement is made for normalized packets and finite observation intervals; idealized infinite-duration emission need not define the two vacua in one global Fock space. This explains both the squeezed-state structure and the physically measurable Hawking radiation.

Since the species decouples while relativistic, its comoving particle number is conserved. Once it becomes nonrelativistic, its energy density is ${\rho }_{\nu ,0}\simeq m_{\nu }{n}_{\nu ,0}$. With $R_n:=n_{\nu ,0}/n_{\gamma ,0}$,This is a density bound on a relativistically decoupled massive relic. Using literally the printed energy-density coefficient, the supplied number-density coefficient gives $⟨E_{\gamma }⟩=(3.3/0.12)T_0=27.5T_0$. For $g_{\nu }=2$ and $g_d=100$,The estimate scales as $(2/g_{\nu })(g_d/100)$. At the bound the particles are safely nonrelativistic today, so neglecting their kinetic energy is consistent.

There is a physical normalization defect in the printed coefficient. Direct integration of the Bose-Einstein distribution giveswhich is also the coefficient consistent with the supplied $1.66$ expansion-rate formula. It implies $⟨E_{\gamma }⟩={\pi }^4{T}_0/(30\zeta (3))\simeq 2.70T_0$, rather than $27.5T_0$. With this physically normalized photon gas, the same density bound on a relativistically decoupled massive relic becomes $m_{\nu }\lesssim 3.7\times 10^1\mathrm{eV}$, or about $4\times 10^1\mathrm{eV}$ using the rounded coefficients with $0.33$ in place of $3.3$. The two estimates differ by the printed factor of ten; the entropy-dilution result is unaffected.

The Bose-Einstein distribution gives $n_i=[\exp ((E_i-\mu )/(k_{B}T))-1{]}^{-1}$, with chemical potential below the lowest energy when its discrete occupation is finite. In the continuum approximation,If the zero-energy ground state is macroscopically occupied, its discrete occupation $N_0$ must be added to the first integral; it contributes no energy to the second. This distinction is essential when applying the formulas to Bose-Einstein condensation.

Use the dark-sector temperature in every dimensionless integration variable: $x=m_{\varphi }/T_{\varphi }$ and $y=p/T_{\varphi }$. Assume a single real scalar field, $g=1$, and fast number-changing reactions keeping ${\mu }_{\varphi }=0$. For $x\gg 1$, the populated momenta satisfy $y=O(\sqrt{x})$, soThis is the nonrelativistic Maxwell-Boltzmann distribution limit of the Bose-Einstein distribution. The given Gaussian Gamma function integral givesSince ${\rho }_{\varphi }=m_{\varphi }n+3n{T}_{\varphi }/2+\cdots$ and $P_{\varphi }=n{T}_{\varphi }$, the entropy density at zero chemical potential is $s_{\varphi }=n(x+5/2+\cdots )$. Its leading nonrelativistic entropy at zero chemical potential is thereforeAn internal multiplicity $g$ would multiply this expression; the absence of $g$ in the printed result assumes a real one-state scalar.

Now use the constant separately conserved visible and dark entropy ratio and the relativistic visible-bath entropy law, $s_{\mathrm{SM}}=(2{\pi }^2/45)g_{\ast s}^{\mathrm{SM}}{T}_{\gamma }^3=\xi s_{\varphi }$. Since $T_{\varphi }=m_{\varphi }/x$,Thus the cannibal-sector temperature ratio hasThe numerical coefficient is approximately $0.525$. The visible-bath temperature is the photon temperature while that bath is internally thermalized; after distinct visible species acquire different temperatures, its total entropy must instead use the appropriate temperature-weighted $g_{\ast s}^{\mathrm{SM}}$.

The displayed relativistic scalar entropy formula must not be continued into this regime with $g_{\ast s}^{\varphi }=1$ held constant. If it is used as an effective definition at arbitrary temperature, thenand $T_{\gamma }/T_{\varphi }=(\xi g_{\ast s}^{\varphi }/g_{\ast s}^{\mathrm{SM}}{)}^{1/3}$ gives the same result. This specifies how the relativistic expression can legitimately be combined with the nonrelativistic one.

The Bose-Einstein distribution gives one occupation for the nondegenerate ground level and $M$ identical contributions from the excited levels:

In the very early universe, assume species $i$ is in thermal equilibrium, is ultrarelativistic so that $kT\gg m_i{c}^2$ and $E(p)\simeq pc$, and has negligible chemical potential, ${\mu }_i\simeq 0$. Its occupation factor is then the Bose-Einstein distribution for a boson or the Fermi-Dirac distribution for a fermion.

Substitute $E=pc$ and set $y=pc/(kT)$. Since $p=(kT/c)y$ and $dp=(kT/c)dy$,

For equal internal degeneracy, the number density of an isotropic ultrarelativistic species is proportional toWith $x=\beta pc$, compare the Fermi-Dirac distribution and Bose-Einstein distribution usingFor $s>0$, a change of variables in the second term gives the fermion-to-boson thermal integral ratioParticle number has $s=2$, so $n_f=(1-2^{-2})n_b=3n_b/4$. The energy density contains one further factor $E=pc$, so it has $s=3$ and

Write the total derivative of the phase-space distribution function asSince $a\bar{T}$ is constant for the background photon gas, expand the Bose-Einstein distribution asAt first order the four terms are respectivelyThe angular-deflection velocity is already first order and multiplies the first-order angular dependence of $f$, so its contribution is second order. Thus the collisionless left-hand side of the Free-streaming photon Boltzmann equation is