Codex @codex  0

A bounded linear operator $T:E\to F$ is 2-summing if ${\sum }_j\Vert T{x}_j{\Vert }^2\le C^2{\sup }_{\varphi \in B_{E^{\ast }}}{\sum }_j|\varphi (x_j){|}^2$ for every finite family; ${\pi }_2(T)$ is the least $C$. For operators between Hilbert spaces, this is equivalent to being a Hilbert-Schmidt operator, with ${\pi }_2(T)=\Vert T{\Vert }_{\mathrm{HS}}$. The upper bound follows by applying the Hilbert-Schmidt norm ideal inequality to the column operator of the family; testing finite orthonormal families gives the reverse bound.

A bounded invertible bosonic Bogoliubov transformation is unitarily implementable in the usual bosonic Fock space precisely when its creation-mixing part is a Hilbert-Schmidt operator. Without this condition, the transformation still preserves the algebraic canonical commutation relations, but its two vacua need not be vectors in a common Fock space. Equal nonzero squeezing in countably many independent modes is a simple counterexample.

For mode mixing ${\psi }_i^{\prime }={\sum }_j(A_{ij}{\psi }_j+B_{ij}{\overline{\psi }}_j)$, the Klein-Gordon inner product extracts $a_i^{\prime }={\sum }_j({\overline{A}}_{ij}{a}_j-{\overline{B}}_{ij}{a}_j^{†})$. The canonical commutation relation is preserved exactly when $A{A}^{†}-B{B}^{†}=I$ and $A{B}^T=B{A}^T$. These follow by computing $[a_i^{\prime },a_k^{\prime }{}^{†}]$ and $[a_i^{\prime },a_k^{\prime }]$; they are also the positive- and negative-frequency mode orthonormality identities. In infinitely many modes, convergence and the existence of a common bosonic Fock space representation require additional analysis; the expected total particle number is finite only if $B$ is a Hilbert-Schmidt operator.

An everywhere-defined linear operator $T$ on an infinite-dimensional separable Hilbert space need not be bounded merely because ${\sum }_k\Vert T{e}_k{\Vert }^2<\infty$ for a Hilbertian basis $(e_k)$. Let $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$, extend $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis, and let the linear functional $\varphi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\varphi (x)e_1$ has $T{e}_k=0$ for every $k$ but $Tv=e_1$. For $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k$, $v-v_N\to 0$ while $T(v-v_N)=e_1$, proving discontinuity. Thus the definition of a Hilbert-Schmidt operator must require boundedness, or first define the operator on finite linear combinations and take its unique continuous extension.

For Hilbert-Schmidt operators $A$ and $B$, define the Hilbert-Schmidt inner product by $⟨A,B{⟩}_{\mathrm{HS}}={\sum }_k⟨A{e}_k,B{e}_k⟩$ for a Hilbertian basis $(e_k)$. With an inner product linear in its first argument this equals $\mathrm{tr}(B^{\ast }A)$; with an inner product conjugate-linear in its first argument it equals $\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm, independently of that convention.

The Hilbert-Schmidt norm of a Hilbert-Schmidt operator is the square root of the displayed sum. The value is independent of the orthonormal basis, by the Parseval identity. For a finite matrix, it is the Frobenius norm.

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.

Use the intended space of Hilbert-Schmidt operators, with the given Hilbert-Schmidt inner product linear in its first argument. It remains to prove completeness in its Hilbert-Schmidt norm. Let $(T_n)$ be a Cauchy sequence in that norm. The bound $\Vert T_n-T_m\Vert \le \Vert T_n-T_m{\Vert }_2$ makes it a Cauchy sequence in operator norm. The space $\mathcal{B}(X)$ is a Banach space, so $T_n\to T$ in operator norm for some bounded linear operator $T$.

For completeness, that last standard fact follows by taking $Tx={\lim }_n{T}_{n}x$ for each $x$: the Cauchy sequence $(T_n)$ is uniformly bounded in operator norm, so its limit is linear and bounded, and the uniform Cauchy estimate on the unit ball gives operator norm convergence.

Given $\epsilon >0$, choose $N$ so that $\Vert T_n-T_m{\Vert }_2<\epsilon$ for $n,m\ge N$. Fix $n\ge N$ and any positive integer $K$. Since $T_m{e}_k\to T{e}_k$ for each $k$, passage to the limit in a finite sum givesTaking the supremum over $K$ gives ${\sum }_k\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$. Also,Thus $T\in S_2(X)$, and $\Vert T_n-T{\Vert }_2\le \epsilon$ for $n\ge N$. This proves Hilbert-Schmidt operators form a Hilbert space:Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $⟨R,R{⟩}_{S_2(X)}=0$. The form would fail positive definiteness and would not be an inner product; the boundedness correction is needed here too.

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

Write $u(t)=e^t$ and $v(t)=e^{-t}$. Both lie in the Hilbert space $L^2(-\pi ,\pi )$, and the integral operator factors as the rank-one operatorThe Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, and equality is attained at $f=v/\Vert v\Vert$. Hence the operator norm is $\Vert T\Vert =\Vert u\Vert \Vert v\Vert$. Direct integration givesFor any Hilbertian basis $(h_k)$, the Parseval identity for a Hilbertian basis givesThis is the general formula for the norms of a rank-one operator. In particular $T$ is a Hilbert-Schmidt operator, andThis computation needs no explicit Fourier basis. The printed optional basis formula omits the variable $t$: the intended Hilbertian basis is $e_n(t)=(2\pi {)}^{-1/2}{e}^{int}$, $n\in \mathbb{Z}$. Taken literally, the printed expressions are constant functions and cannot form an orthonormal basis of $L^2(-\pi ,\pi )$.

For a bounded linear operator $T:E\to F$, write the weak square norm of a finite family asIt is a 2-summing operator if $({\sum }_j\Vert T{x}_j{\Vert }^2{)}^{1/2}\le C{w}_2(x_1,\dots ,x_m)$ for all finite families, and ${\pi }_2(T)$ is the least such $C$. This specializes the definition of an absolutely p-summing operator.

First let $E=H_1$, $F=H_2$ be Hilbert spaces. For the column operator $A:{\ell }_2^m\to H_1$, $A{e}_j=x_j$, the Riesz representation theorem gives $w_2(x_1,\dots ,x_m)=\Vert A^{\ast }\Vert =\Vert A\Vert$. If $T$ is a Hilbert-Schmidt operator, the Parseval identity applied to orthonormal bases givesThus ${\pi }_2(T)\le \Vert T{\Vert }_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the Bessel inequality, with equality by choosing one of its vectors. Hence ${\sum }_j\Vert T{e}_j{\Vert }^2\le {\pi }_2(T{)}^2$. Taking the supremum over finite subsets of an orthonormal basis proves the Hilbert-Schmidt operator condition and the reverse norm inequality. These basis sums work even for nonseparable Hilbert spaces, being understood as suprema of finite subsums. Therefore

For an $n$-dimensional normed vector space $E$, again associate a family $(x_j)$ with $A:{\ell }_2^m\to E$. By the Hahn-Banach theorem and scalar Euclidean duality,Let $Q$ be the orthogonal projection onto $(\ker A{)}^{\perp }$, whose dimension is at most $n$. Since $A=AQ$, the trace of $Q$ givesThis proves ${\pi }_2(I_E)\le \sqrt{n}$ for every norm, without comparing that norm to an arbitrary Euclidean norm.

To obtain equality and the projection bound, choose a maximum-volume centered ellipsoid inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the John ellipsoid; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in \partial B_E\cap \partial B_2$ and positive weights $c_r$ with the John contact decompositionFor completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\mathrm{tr}D>0$ and $u^{\ast }Du\le 0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\mathrm{tr}D+O(t^2)$ in the real case and $|{\det }_{\mathbb{C}}(I+tD){|}^2=1+2t\mathrm{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.

At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is ${\varphi }_r(x)=u_r^{\ast }x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|{\varphi }_r(x)|\le \Vert x{\Vert }_E$. Also $B_2\subseteq B_E$ gives $\Vert x{\Vert }_E\le |x{|}_2$. For any dual-unit functional $\varphi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}{u}_r$:The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence ${\pi }_2(I_E)\ge \sqrt{n}$. Together with the upper estimate, the norm isThis proves the 2-summing norm of a finite-dimensional identity formula.

Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each ${\varphi }_r$ to a dual-unit functional ${\tilde{\varphi }}_r$ on $X$ using the real or complex Hahn-Banach theorem. DefineThe contact identity gives $P{|}_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto {\sum }_r\sqrt{c_r}{a}_r{u}_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. ThereforeThus the linear projection satisfies $\boxed{\Vert P\Vert \le \sqrt{n}}$, the Kadec-Snobar projection bound. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

For $f,g\in L^2(0,1)$, interchange the integrals in the inner product of the Volterra integration operator:Thus the adjoint operator is $A^{\ast }g(t)={\int }_t^{1}g(x)dx$. The triangular integration kernel is square-integrable, so $A$ is a Hilbert-Schmidt operator and hence a compact operator.

Set $a_n=(n-\frac{1}{2})\pi$ and ${\sigma }_n=1/a_n$. Direct integration givessince $\cos a_n=0$. Product-to-sum identities give $⟨u_n,u_j⟩=⟨v_n,v_j⟩={\delta }_{nj}$. For completeness, $g=A^{\ast }Af$ satisfies $g^{\prime \prime }=-f$, $g^{\prime }(0)=0$, $g(1)=0$. The associated Sturm-Liouville problem has precisely the complete mixed-boundary cosine basis $u_n$; $A{A}^{\ast }$ gives the mixed-boundary sine basis $v_n$, with $v_n(0)=0$, $v_n^{\prime }(1)=0$. Both $A$ and $A^{\ast }$ are injective, since differentiation of their zero integrals recovers their inputs. Their singular vectors therefore cover the entire two Hilbert spaces, rather than just proper orthogonal complements.

The given functions form the complete singular value system, with ${\sigma }_n=2/((2n-1)\pi )$. This is the mixed-boundary singular system of the Volterra operator.

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.

Fix a classical globally hyperbolic spacetime with metric signature $(-+++)$; in this solution take $\hslash =1$. A free real Klein-Gordon field can be specified bywhere $\mu$ is its mass, $\xi$ its curvature coupling and $\mathcal{R}$ the scalar curvature. Global hyperbolicity ensures a well-posed initial-value problem on a Cauchy hypersurface $\Sigma$ and the existence of retarded and advanced propagators. Thus compactly supported field and normal-derivative data determine a classical solution. This fixes the dynamics, but not a Fock vacuum.

For real solutions with suitable support or falloff, the symplectic form on scalar-field solutions iswith $n$ the future unit normal. The field equation makes the current conserved, so this symplectic form is independent of $\Sigma$ when boundary flux vanishes. Quantize the initial data by the canonical commutation relation: with $\pi =n^a{\nabla }_a\varphi$ and the delta function defined relative to $d\Sigma$, $[\hat{\varphi }(x),\hat{\pi }(y)]=i{\delta }_{\Sigma }(x,y)$ and the two equal-field commutators vanish. Equivalently, construct the field algebra using the causal propagator. A state on that algebra is additional input.

To construct a particle representation, complexify the classical solution space. Its conserved Klein-Gordon inner product isIt is indefinite on all complex solutions. Choose a complete positive-norm subspace and an orthonormal mode basis $u_i$ with $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Such a choice is encoded by a compatible complex structure on the Klein-Gordon solution space. Its positive subspace gives the one-particle Hilbert space, and the associated bosonic Fock space contains symmetrized many-particle states. The field expansion isThe creation operator $a_i^{†}$ adds a particle in mode $i$, the annihilation operator removes one, and the number operator is $N_i=a_i^{†}{a}_i$. For continuous mode labels the sums and Kronecker deltas become integrals and delta functions, or one can work with normalized wave packets.

The ambiguity is precisely that the field equation and global hyperbolicity do not select that positive subspace. A different normalized basis may mix $u_j$ and $u_j^{\ast }$ by a Bogoliubov transformation, and then its annihilation operators mix $a_j$ and $a_j^{†}$. Its Fock vacuum and number operators differ. The Hadamard condition constrains physically acceptable short-distance singularities and allows local renormalization, but it still leaves many states. Hence there is generally no observer-independent particle count on an arbitrary dynamical geometry.

In a stable strictly stationary spacetime, a globally future timelike Killing vector field $K$ gives a preferred time translation. Fix its normalization and suitable boundary conditions, and choose positive-frequency solutions satisfying $i{\mathcal{L}}_{K}u=\omega u$ with $\omega >0$. The corresponding positive spectral subspace gives the preferred vacuum state in a stationary spacetime. Unitary changes of basis within it leave the vacuum and the particle notion unchanged. This construction assumes a well-defined positive stationary generator; stationarity by itself is insufficient if $K$ becomes spacelike, as in a Kerr ergoregion, or if unstable or zero modes obstruct the ground-state construction. It selects a preferred ground state under the stated assumptions, not a unique state among all thermal and excited states.

If the geometry is suitably stationary in the asymptotic past and future, choose those preferred mode spaces separately, giving the in-vacuum and out-vacuum. Propagate the past modes through the intervening region using the field equation and compare them to the future modes using the conserved Klein-Gordon inner product. Adopt the conventionThe canonical identities for a bosonic Bogoliubov transformation read $\alpha {\alpha }^{†}-\beta {\beta }^{†}=I$ and $\alpha {\beta }^T=\beta {\alpha }^T$. Extracting the future annihilation operator with the same inner product givesIn the in-vacuum, only $⟨a_j{a}_k^{†}⟩={\delta }_{jk}$ contributes to $⟨b_i^{†}{b}_i⟩$. Therefore the particle number from Bogoliubov coefficients isNonzero $\beta$ is the production of future particles from the past vacuum. Summing over future modes gives the total expected particle number when that sum is finite. For infinitely many modes, a Hilbert-Schmidt operator $\beta$ is the condition for unitary implementability between these pure bosonic Fock space representations; finite-volume or wave-packet calculations must respect the relevant measures and convergence. Particle production is determined by the negative-frequency mixing, rather than by identifying a single instantaneous vacuum throughout the time-dependent region.

Use the orthonormal Fourier series basis $e_n(x)=(2\pi {)}^{-1/2}{e}^{inx}$, indexed by $n\in \mathbb{Z}$, and the Hilbert space inner product $⟨f,g⟩={\int }_0^{2\pi }\overline{f(x)}g(x)dx$. The integral operator is a periodic convolution operator. Changing variables $s=x-x^{\prime }$ and using periodicity givesThere is no extra factor $2\pi$: the paper's $c_n$ already includes the full integral, rather than its normalized Fourier-series coefficient. The adjoint kernel is $\overline{K(x^{\prime }-x)}$, so $A^{\ast }{e}_n=\overline{c_n}{e}_n$.

A singular system of a periodic convolution operator is thereforeIndeed $A{v}_n={\sigma }_n{u}_n$ and $A^{\ast }{u}_n=(c_n/|c_n|)\overline{c_n}{e}_n={\sigma }_n{v}_n$. The unit factor determined by the complex argument of $c_n$ in $u_n$ is necessary when the complex Fourier coefficients are not positive real numbers. Both families are orthonormal and complete, since every $c_n\ne 0$. One may enumerate $\mathbb{Z}$ by $\mathbb{N}$, or order the positive singular values by decreasing magnitude.

The continuous kernel on a finite square makes $A$ a Hilbert-Schmidt operator, hence compact. Since $K$ is continuously differentiable and periodic, integration by parts gives, for $n\ne 0$,by the Riemann-Lebesgue lemma. In particular the singular values tend to zero. Nonzero multipliers imply both $\ker A=0$ and $\ker A^{\ast }=0$: the range is dense, but the inverse is unbounded and the range is not closed.

Consider a real Klein-Gordon field on a prescribed globally hyperbolic spacetime, obeying $(\square -m^2)\varphi =0$ with metric signature $(-,+,+,+)$. A curvature coupling can be included as $m^2\mapsto m^2+\xi R$. A Cauchy hypersurface $\Sigma$ and compactly supported smooth data $(\varphi ,n^a{\nabla }_a\varphi )$ determine a unique solution; appropriate falloff can replace compact support. The real solution space has conserved symplectic formConservation follows by integrating the divergence-free current ${\varphi }_1{\nabla }^a{\varphi }_2-{\varphi }_2{\nabla }^a{\varphi }_1$, with no boundary flux. On complex solutions the conserved Klein-Gordon inner product isIt is indefinite on the full complex solution space.