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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.