Certified positive finite quadratic Bose Hamiltonian is normalizable

By [F2], the certificate's matrices give a unitary $M$ satisfying $M^{\ast }{R}_{g}M=S_g{R}_g$ on the common transported Schwartz core. By [F3], we may take this implementer to preserve Schwartz space: its normalized metaplectic implementer has that property, and multiplying by a scalar phase leaves $M^{\ast }{R}_{g}M$ unchanged.

Conjugating each symmetrically ordered term of $H_{G,c}^{(0)}$ by $M$ replaces $R_g$ by $S_g{R}_g$ . Collecting the finite sums gives $M^{\ast }{H}_{G,c}^{(0)}M=\frac{1}{4}{\sum }_{a,b}(S_g^{T}G{S}_g{)}_{ab}(R_{g,a}{R}_{g,b}+R_{g,b}{R}_{g,a})+cI\mathrm{.}$ By [A2] this is the diagonal quadratic with coefficient $D_{\omega }$ on the same core.

On the finite Hermite core put $a_j=(Q_j+i{P}_j)\mathrm{/}\sqrt{2\mathrm{\hslash }}$ and $a_j^{\ast }=(Q_j-i{P}_j)\mathrm{/}\sqrt{2\mathrm{\hslash }}$ . The relation $[Q_j,P_j]=i\mathrm{\hslash }I$ gives $Q_j^2+P_j^2=2\mathrm{\hslash }(a_j^{\ast }{a}_j+\frac{1}{2}I)$ by direct multiplication. Consequently the diagonal quadratic is ${\sum }_j\mathrm{\hslash }{\omega }_j(a_j^{\ast }{a}_j+\frac{1}{2}I)+cI$ . Write $g_{\mathbf{n}}\in L^2({\mathbb{R}}^r)$ for the product Hermite basis and set ${\varphi }_{\mathbf{n}}:={\mathcal{U}}^{-1}{g}_{\mathbf{n}}\in {\mathcal{F}}_s({\mathbb{C}}^r)$ . Then $\mathcal{U}{\varphi }_{\mathbf{n}}=g_{\mathbf{n}}$ , and the diagonal quadratic has eigenvalue $E_{\mathbf{n}}=c+{\sum }_j\mathrm{\hslash }{\omega }_j(n_j+\frac{1}{2})$ . Denote this diagonal differential expression on $\mathcal{S}$ by ${\tilde{T}}_{\omega }^{(0)}$ , and its Fock-space transport on $\mathcal{D}$ by $T_{\omega }^{(0)}:={\mathcal{U}}^{-1}{\tilde{T}}_{\omega }^{(0)}\mathcal{U}$ .

Put ${ϵ}_j=\mathrm{\hslash }{\omega }_j$ and let $A$ be the maximal diagonal multiplier in the basis $({\varphi }_{\mathbf{n}})$ with eigenvalues ${\sum }_j{ϵ}_j{n}_j$ . By [F5], $A$ is self-adjoint on the domain where ${\sum }_{\mathbf{n}}\mathrm{\mid }{\sum }_j{ϵ}_j{n}_j{\mathrm{\mid }}^2\mathrm{\mid }{f}_{\mathbf{n}}{\mathrm{\mid }}^2<\mathrm{\infty }$ . The diagonal operator with eigenvalues $E_{\mathbf{n}}$ is $A+c_{0}I$ , where $c_0=c+\frac{1}{2}{\sum }_j{ϵ}_j$ . Its domain is the same: for real $x,c_0$ , $\mathrm{\mid }x+c_0{\mathrm{\mid }}^2\le 2\mathrm{\mid }x{\mathrm{\mid }}^2+2\mathrm{\mid }{c}_0{\mathrm{\mid }}^2$ and $\mathrm{\mid }x{\mathrm{\mid }}^2\le 2\mathrm{\mid }x+c_0{\mathrm{\mid }}^2+2\mathrm{\mid }{c}_0{\mathrm{\mid }}^2$ , and each coefficient sequence $(f_{\mathbf{n}})$ is in ${\mathrm{\ell }}^2$ . For $g\in D((A+c_{0}I{)}^{\ast })$ , [F7] says that boundedness of $f\mapsto ⟨(A+c_{0}I)f,g⟩$ and boundedness of $f\mapsto c_0⟨f,g⟩$ imply boundedness of $f\mapsto ⟨Af,g⟩$ . Thus $g\in D(A^{\ast })=D(A)$ and $(A+c_{0}I{)}^{\ast }g=Ag+c_{0}g$ . Conversely, every $g\in D(A)$ satisfies this adjoint identity, so $T=A+c_{0}I$ is self-adjoint. Finite Hermite truncations converge in its graph norm, because the tail of ${\sum }_{\mathbf{n}}(1+\mathrm{\mid }{E}_{\mathbf{n}}{\mathrm{\mid }}^2)\mathrm{\mid }{f}_{\mathbf{n}}{\mathrm{\mid }}^2$ tends to zero. Thus finite Hermite sums are a core for $T$ in the graph norm of [F8].

For $\psi \in \mathcal{D}$ , put $f=\mathcal{U}\psi \in \mathcal{S}$ . The coordinate expression ${\tilde{T}}_{\omega }^{(0)}$ sends $f$ to an $L^2$ function. Integration by parts against each Schwartz Hermite function gives $⟨T_{\omega }^{(0)}\psi ,{\varphi }_{\mathbf{n}}⟩=E_{\mathbf{n}}⟨\psi ,{\varphi }_{\mathbf{n}}⟩$ ; equivalently, this is the $L^2$ pairing of the differential expression on $f$ with $g_{\mathbf{n}}=\mathcal{U}{\varphi }_{\mathbf{n}}$ . Boundary terms vanish because both functions are Schwartz. For every finite set $F$ of occupation indices, orthogonal projection therefore gives ${\sum }_{\mathbf{n}\in F}\mathrm{\mid }{E}_{\mathbf{n}}⟨\psi ,{\varphi }_{\mathbf{n}}⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }{T}_{\omega }^{(0)}\psi {\mathrm{\parallel }}^2$ . These coefficients are square-summable, and completeness of $({\varphi }_{\mathbf{n}})$ from [F4] identifies their vector with $T_{\omega }^{(0)}\psi$ . Hence $\mathcal{D}\subseteq D(T)$ and the diagonal expression is $T$ on $\mathcal{D}$ . The finite Hermite sums from step 4.1 lie in $\mathcal{D}$ because [F1] places each $g_{\mathbf{n}}$ in $\mathcal{S}$ , and are graph-norm dense in $D(T)$ ; so the graph closure of this restriction is $T$.

By step 1.1, $M$ and $M^{\ast }$ preserve Schwartz space, and by step 2.1 $H_{G,c}^{(0)}=M{T}_{\omega }^{(0)}{M}^{\ast }$ there. Step 5.1 gives graph closure $T$ for the diagonal restriction. Conjugation by the unitary $M$ carries the graph and its closure in ${\mathcal{F}}_s\oplus {\mathcal{F}}_s$ to the graph and closure of $H_{G,c}^{(0)}$ , so $H=\overline{H_{G,c}^{(0)}}=MT{M}^{\ast }$ . This is self-adjoint because $T$ is self-adjoint and $M$ is unitary.

Set $q_j=e^{-\beta {ϵ}_j}=e^{-\beta \mathrm{\hslash }{\omega }_j}$ . Since $\beta >0$ and every ${\omega }_j>0$ , $0<q_j<1$ ; the mode set is finite, so $q_{\ast }<1$ and ${\sum }_j{q}_j<\mathrm{\infty }$ . Apply [F5] to the diagonal number-energy operator $A$ to get $\mathrm{Tr}(e^{-\beta A})={\prod }_j(1-q_j{)}^{-1}<\mathrm{\infty }$ . On the occupation basis, $T$ has eigenvalues $E_{\mathbf{n}}=c_0+{\sum }_j{ϵ}_j{n}_j\ge c_0$. Its spectral measure is supported on these eigenvalues because this basis is complete. Hence $e^{-\beta x}\le e^{-\beta c_0}$ on that support, so its spectral exponential is bounded and everywhere defined, with positive eigenvalues $e^{-\beta E_{\mathbf{n}}}=e^{-\beta c_0}{\prod }_j{q}_j^{n_j}$ by [F9]. Thus $e^{-\beta T}=e^{-\beta c_0}{e}^{-\beta A}$ on the occupation basis; [F10] and the trace-class conclusion for $e^{-\beta A}$ from [F5] show that $e^{-\beta T}$ is positive trace class. Its eigenvalue sum is $e^{-\beta c_0}{\prod }_j(1-q_j{)}^{-1}<\mathrm{\infty }$ by [F5], so [F6] identifies that sum with its trace. By steps 6.1 and 5.1, the vectors $M{\varphi }_{\mathbf{n}}$ form a complete orthonormal eigenbasis for $H$ with the same eigenvalues; equivalently $e^{-\beta H}=M{e}^{-\beta T}{M}^{\ast }$ . The unitary invariance in [F10] makes $e^{-\beta H}$ trace class with the same eigenvalue sum, and [F6] gives the same trace. Finally, $e^{-\beta c_0}{\prod }_j(1-e^{-\beta \mathrm{\hslash }{\omega }_j}{)}^{-1}=e^{-\beta c}{\prod }_j\frac{e^{-\beta \mathrm{\hslash }{\omega }_j\mathrm{/}2}}{1-e^{-\beta \mathrm{\hslash }{\omega }_j}}=e^{-\beta c}{\prod }_j{[2\sinh \text{\hspace{0.17em}⁣}\left(\frac{\beta \mathrm{\hslash }{\omega }_j}{2}\right)]}^{-1}\mathrm{.}$ Every factor is strictly positive and finite, proving $Z_{\beta }\in (0,\mathrm{\infty })$ . The only infinite-dimensional Hilbert space used here is the bosonic Fock space for the stated finite number $r$ of modes.