Finite quasifree CAR states and positive quadratic Bose Hamiltonians

By [F3], write $C=V\mathrm{diag}({\nu }_1,\dots ,{\nu }_r)V^{\ast }$ with $V$ unitary and $0\le {\nu }_j\le 1$ . On the exterior-power decomposition define ${\mathrm{\Gamma }}_{-}(V):={⨁}_{k=0}^r{\bigwedge }^{k}V$ ; each summand is unitary, so ${\mathrm{\Gamma }}_{-}(V)$ is unitary: inner products of simple exterior products are determinants of their Gram matrices, and $V$ preserves each Gram matrix. In the occupation basis for the eigenmodes, define $w_j(0):=1-{\nu }_j$ and $w_j(1):={\nu }_j$ , and define the product density ${\rho }_D$ by ${\rho }_D\mathrm{\mid }m⟩=({\prod }_{j=1}^r{w}_j(m_j))\mathrm{\mid }m⟩,m\in \{0,1{\}}^r\mathrm{.}$ Each diagonal entry is nonnegative and the sum of the entries is ${\prod }_j((1-{\nu }_j)+{\nu }_j)=1$ , so ${\rho }_D$ is positive with trace one. Set ${\rho }_C:={\mathrm{\Gamma }}_{-}(V){\rho }_D{\mathrm{\Gamma }}_{-}(V{)}^{\ast }$ .

For the Gibbs assertion, diagonalize $h=W\mathrm{diag}({ϵ}_1,\dots ,{ϵ}_r)W^{\ast }$ by [F3], and set $d=W^{\ast }c$ in column notation. Exterior multiplication by $e_j$ is $c_j^{\ast }$, so the exterior-power definition in step 1.1 gives ${\mathrm{\Gamma }}_{-}(W)c_j^{\ast }{\mathrm{\Gamma }}_{-}(W{)}^{\ast }={\sum }_i{W}_{ij}{c}_i^{\ast }=d_j^{\ast }$; taking adjoints gives ${\mathrm{\Gamma }}_{-}(W)c_j{\mathrm{\Gamma }}_{-}(W{)}^{\ast }=d_j$. Thus the eigenmode occupation basis is $\mathrm{\mid }m{⟩}_W:={\mathrm{\Gamma }}_{-}(W)\mathrm{\mid }m⟩$, and $d_j^{\ast }{d}_j\mathrm{\mid }m{⟩}_W=m_j\mathrm{\mid }m{⟩}_W$ by [F2]. Substitution of $c=Wd$ gives $d\mathrm{\Gamma }(h-\mu I)={\sum }_j({ϵ}_j-\mu )d_j^{\ast }{d}_j$, hence $d\mathrm{\Gamma }(h-\mu I)\mathrm{\mid }m{⟩}_W=({\sum }_{j=1}^r({ϵ}_j-\mu )m_j)\mathrm{\mid }m{⟩}_W\mathrm{.}$ Therefore the positive exponential has eigenvalues ${\prod }_j{q}_j^{m_j}$, where $q_j=e^{-\beta ({ϵ}_j-\mu )}>0$, and its finite trace is $Z={\prod }_{j=1}^r(1+q_j)\in (0,\mathrm{\infty })\mathrm{.}$ In this basis the normalized density is a product Bernoulli density with occupation probabilities $p_j=q_j\mathrm{/}(1+q_j)=(1+e^{\beta ({ϵ}_j-\mu )}{)}^{-1}$; in the original basis it is ${\mathrm{\Gamma }}_{-}(W){\rho }_{\mathrm{diag}(p)}{\mathrm{\Gamma }}_{-}(W{)}^{\ast }$, with the diagonal product density defined as in step 1.1.

The density ${\rho }_D$ is diagonal in the occupation basis, so its state is invariant under the global gauge rotation. For a normally ordered monomial, repeated creation or annihilation indices give zero by the CAR. If the distinct creator and annihilator index sets differ, the monomial changes an occupation vector and has zero diagonal expectation. If the sets agree at $S$ , the CAR relations reorder the product to its permutation sign times ${\prod }_{j\in S}{c}_j^{\ast }{c}_j$ ; its product-state expectation is that sign times ${\prod }_{j\in S}{\nu }_j$ . The determinant of the diagonal two-point matrix has exactly the same value: it is zero for unequal index sets and, for equal sets $S$ , is the permutation sign times ${\prod }_{j\in S}{\nu }_j$ . Thus ${\rho }_D$ obeys the quasifree moment rule with diagonal covariance $\mathrm{diag}({\nu }_1,\dots ,{\nu }_r)$ . Moreover, every $c_i$ is a linear combination of the eigenmode annihilators and every $c_i^{\ast }$ is the corresponding conjugate linear combination. Expanding a normal-ordered product and using multilinearity of the determinant carries the Wick rule to covariance $V\mathrm{diag}({\nu }_1,\dots ,{\nu }_r)V^{\ast }$ . The unitary ${\mathrm{\Gamma }}_{-}(V)$ commutes with total particle number, so it preserves gauge invariance. Hence ${\rho }_C$ has the required covariance and moments.

The CAR relations reorder any word in the generators into a linear combination of normally ordered monomials. These monomials determine the functional on the full finite matrix algebra: the vacuum projection is $P_0={\prod }_{j=1}^r(I-c_j^{\ast }{c}_j)$ , and for any two occupation subsets $S,T$ , the operator obtained by creating the modes in $S$ , applying $P_0$ , and annihilating the modes in $T$ is a nonzero scalar multiple of $\mathrm{\mid }S⟩⟨T\mathrm{\mid }$ . Expanding $P_0$ and reordering with CAR expresses each such matrix unit as a linear combination of normally ordered monomials. Since these matrix units span the full algebra, no two states can have the same stated moments unless they are equal. This proves uniqueness of ${\rho }_C$ .

The product density in the eigenmode basis obeys the determinant moment rule proved in step 2.2. Its covariance is $W\mathrm{diag}(p_1,\dots ,p_r)W^{\ast }$ by the convention [F1], and the finite spectral calculus [F3] identifies this matrix with $(I+e^{\beta (h-\mu I)}{)}^{-1}$ . Uniqueness from step 3.1 identifies the Gibbs density with ${\rho }_C$ . All sums are finite-mode sums; no infinite-mode implementability or thermodynamic limit is involved.

By [F5], write $C=V\mathrm{diag}({\nu }_1,\dots ,{\nu }_r)V^{\ast }$ with $V$ unitary and ${\nu }_j\in [0,1]$ . By [F2], after conjugation by the induced exterior-power unitary, ${\rho }_C$ is diagonal on the occupation basis with probabilities $p_m={\prod }_j{w}_j(m_j)$ , where $w_j(0)=1-{\nu }_j$ and $w_j(1)={\nu }_j$ . These probabilities sum to one because ${\sum }_{m\in \{0,1{\}}^r}{p}_m={\prod }_j((1-{\nu }_j)+{\nu }_j)=1$ .

Choose the ordered orthonormal bases of $K$ and $K^{\perp }$ using [F6]. The map from the full occupation vector to the tensor product of its $K$ -occupation and $K^{\perp }$ -occupation vectors sends one orthonormal basis onto another, hence extends to a unitary factorization. Because the $K$ modes come first, their creation and annihilation operators act on the first factor as $c_j^K\otimes I$ ; the CAR parity signs from preceding modes are therefore entirely within that first factor. The partial trace over $K^{\perp }$ has the same expectation as ${\rho }_C$ for every word in the $K$ modes. Restricting the determinant rule [F1] to those words gives the compressed covariance $C_K=P_{K}C{\mathrm{\mid }}_K$ and the quasifree moment rule; gauge invariance is inherited under restriction. Thus ${\rho }_K$ is the unique quasifree density for $C_K$ by [F2].

Unitary conjugation preserves the eigenvalues, so the entropy is $-{\sum }_m{p}_m\ln p_m$ , with zero terms omitted under [F3]. When $0<{\nu }_j<1$ , expand $\ln p_m={\sum }_j[m_j\ln {\nu }_j+(1-m_j)\ln (1-{\nu }_j)]$ and sum the finite product distribution; its $j$ th occupation has mean ${\nu }_j$ . The result is $-{\sum }_j[{\nu }_j\ln {\nu }_j+(1-{\nu }_j)\ln (1-{\nu }_j)]$ . If ${\nu }_j=0$ or $1$ , that mode has a deterministic occupation and contributes zero by [F3], which is the same formula with $0\ln 0=0$ . Spectral calculus for $C$ and $I-C$ now identifies this sum with the displayed matrix trace.

Apply the calculation of step 2.1 to the finite quasifree density ${\rho }_K$ and its positive contraction $C_K$ . The same endpoint convention $0\ln 0=0$ handles pure or rank-deficient reduced states. A change of orthonormal basis in $K$ conjugates ${\rho }_K$ by the induced unitary and $C_K$ by the corresponding matrix unitary, so [F3] makes the trace entropy unchanged. A change of basis in $K^{\perp }$ conjugates only the traced factor; for a unitary matrix $U$ on that factor, the partial-trace entry sum in [F3] contains ${\sum }_j{U}_{ja}\overline{U_{jb}}={\delta }_{ab}$ , so ${\rho }_K$ is unchanged. This proves the restriction formula and its basis independence.

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.