= Solution
\b[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 $a'=\alpha a+\beta a^\dagger$ gives
$$
[a'_i,a_j'{}^\dagger]
=\sum_k(\alpha_{ik}\overline{\alpha_{jk}}-\beta_{ik}\overline{\beta_{jk}}),
\qquad
[a'_i,a'_j]=\sum_k(\alpha_{ik}\beta_{jk}-\beta_{ik}\alpha_{jk}).
$$
Consequently the required conditions are
$$
\boxed{\alpha\alpha^\dagger-\beta\beta^\dagger=I,\qquad
\alpha\beta^T=\beta\alpha^T.}
$$
For complete invertible mode mixing, put
$$
\mathcal T=
\begin{pmatrix}\alpha&\beta\\\overline\beta&\overline\alpha\end{pmatrix},
\qquad J=\begin{pmatrix}I&0\\0&-I\end{pmatrix}.
$$
The <canonical identities for a bosonic Bogoliubov transformation> are $\mathcal T J\mathcal T^\dagger=J$, giving $\mathcal T^{-1}=J\mathcal T^\dagger J$. Its upper blocks yield
$$
\boxed{A=\alpha^\dagger,\qquad B=-\beta^T,\qquad
A_{ij}=\overline{\alpha_{ji}},\quad B_{ij}=-\beta_{ji}.}
$$
Equivalently, the column identities are $\alpha^\dagger\alpha-\beta^T\overline\beta=I$ and $\alpha^\dagger\beta=\beta^T\overline\alpha$. In finite dimension invertibility follows from the canonical matrix identity. In infinitely many modes, the row identities alone need not give a complete inverse: $a'_i=a_{i+1}$ preserves them but discards one mode. Here completeness of the two field-mode expansions supplies the needed invertible transformation.
\b[Obtaining coefficients from the modes.] Use the <Klein-Gordon inner product>, antilinear in its first argument,
$$
(f,g)_{\rm KG}
=i\int_\Sigma d\Sigma^\mu
\left(\overline f\,\nabla_\mu g-g\nabla_\mu\overline f\right).
$$
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 $a_i=(p_i,\widehat\Phi)$ and hence
$$
\boxed{\alpha_{ij}=(p'_i,p_j)_{\rm KG},\qquad
\beta_{ij}=(p'_i,\overline p_j)_{\rm KG}.}
$$
With this operator convention, the corresponding mode expansion is
$$
\boxed{p'_i=\sum_j\left(\overline{\alpha_{ij}}p_j-
\overline{\beta_{ij}}\,\overline p_j\right).}
$$
The 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.
\b[Particle count.] In the <in-vacuum>, only the contraction $\langle0|a_j a_k^\dagger|0\rangle=\delta_{jk}$ survives. Therefore
$$
\boxed{\langle0|a_i'{}^\dagger a'_i|0\rangle=\sum_j|\beta_{ij}|^2.}
$$
This 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.
\b[The squeezed-state relation.] First use finitely many modes, or an implementable infinite-mode limit. Seek
$$
|\psi\rangle=C e^{\widehat F}|0'\rangle,\qquad
\widehat F=\frac12\sum_{ij}M_{ij}a_i'{}^\dagger a_j'{}^\dagger,
\qquad M=M^T.
$$
The <canonical commutation relations> give
$$
[a'_i,\widehat F]=\sum_jM_{ij}a_j'{}^\dagger.
$$
All further nested commutators vanish since $\widehat F$ contains only <creation operators>. Thus
$$
a_i e^{\widehat F}|0'\rangle
=e^{\widehat F}\sum_j(AM+B)_{ij}a_j'{}^\dagger|0'\rangle.
$$
The old annihilation conditions hold exactly when
$$
\boxed{M=-A^{-1}B=(\alpha^\dagger)^{-1}\beta^T.}
$$
The inverse transformation's canonical identity $AB^T=BA^T$ shows $A^{-1}B=(A^{-1}B)^T$, so the required symmetry is automatic. Also $AA^\dagger-BB^\dagger=I$ gives
$$
I-MM^\dagger=A^{-1}(A^{-1})^\dagger>0.
$$
Hence the singular values of $M$ are less than one. The <Autonne-Takagi factorization> $M=U\operatorname{diag}(s_j)U^T$ changes to independent canonical oscillators. For each factor, the squared norm of $\exp(s_j b_j^{\dagger2}/2)|0\rangle$ is the even-occupation series
$$
\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}(n!)^2}s_j^{2n}=(1-s_j^2)^{-1/2}.
$$
Multiplying these norms gives the finite-mode normalization
$$
\boxed{|C|=\det(I-MM^\dagger)^{1/4}.}
$$
The 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 $|0\rangle$.
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 $\alpha=\cosh s\,I$ and $\beta=\sinh s\,I$ on countably many modes with fixed $s>0$. The algebraic canonical conditions hold, but $M=\tanh s\,I$ and the finite-$N$ normalization is $C_N=(\cosh s)^{-N/2}\to0$. 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. \b[The squeezed expression is a normalizable vacuum relation with a finite-mode regulator or the implementability condition, not solely from the commutator identities.]
\b[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> has
$$
v_H-v=C_0e^{-\kappa u}
$$
for late outgoing rays, with $\kappa>0$. Tracing a late mode backwards therefore gives a profile proportional to
$$
\mathbf1_{\{v<v_H\}}(v_H-v)^{i\omega/\kappa}.
$$
It 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 integrals
$$
I_\pm=\int_0^\infty x^{ia}e^{-(\varepsilon\pm i\omega')x}\,dx
=\Gamma(1+ia)(\varepsilon\pm i\omega')^{-1-ia},
\qquad \varepsilon>0.
$$
As $\varepsilon\downarrow0$, 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> is
$$
\boxed{|\beta_{\omega\omega'}|^2
=e^{-2\pi\omega/\kappa}|\alpha_{\omega\omega'}|^2.}
$$
For 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 distribution>
$$
\boxed{\langle N_\omega\rangle
=\frac1{e^{2\pi\omega/\kappa}-1},\qquad
T_H=\frac{\kappa}{2\pi}.}
$$
The <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>.
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