Finite-mode bosonic Gibbs occupations and moments

For $\mathbf{n}\in {\mathbb{N}}^m$ put $N={\sum }_j{n}_j$, $t_{\mathbf{n}}=e_1^{\otimes n_1}\otimes \cdots \otimes e_m^{\otimes n_m}$, $C_{\mathbf{n}}=N!\mathrm{/}{\prod }_j{n}_j!$, and $\mathrm{\mid }\mathbf{n}⟩=\sqrt{C_{\mathbf{n}}}{P}_{s,N}{t}_{\mathbf{n}}$, with the vacuum at $N=0$. There are $C_{\mathbf{n}}$ distinct tensor words with these occupations, each occurring ${\prod }_j{n}_j!$ times in the permutation average, so $P_{s,N}{t}_{\mathbf{n}}$ is their average. These words are orthonormal by the tensor inner product; hence $\mathrm{\parallel }{P}_{s,N}{t}_{\mathbf{n}}{\mathrm{\parallel }}^2=C_{\mathbf{n}}^{-1}$. Distinct occupations give orthogonal vectors, and all symmetrized elementary basis tensors occur this way, so these vectors span the symmetric $N$-particle sector. Thus the occupation vectors form an orthonormal basis of Fock space, and their finite span is the finite-particle subspace, because every fixed-particle sector is finite-dimensional. These are also the vectors obtained from successive applications of the explicit maps in [F2], whose binomial normalization factors multiply to $\sqrt{C_{\mathbf{n}}}$. This identification uses only the splitting maps, without assuming any intertwining for complementary modes.

Put $q_j=e^{-\beta {\omega }_j}\in (0,1)$ and $w_{\mathbf{n}}={\prod }_j{q}_j^{n_j}$. For $0<q<1$, finite summation gives $(1-q){\sum }_{n=0}^{M}n{q}^n={\sum }_{n=1}^M{q}^n-M{q}^{M+1}$. To see $M{q}^{M+1}\to 0$, choose $q<r<1$; the ratio of successive positive terms is $q(1+1\mathrm{/}M)\le r$ for sufficiently large $M$, and a geometric bound applies. Therefore [F6] gives ${\sum }_{n\ge 0}n{q}^n=q\mathrm{/}(1-q{)}^2$ and ${\sum }_{n\ge 0}(n+1)q^n=(1-q{)}^{-2}$. Taking finite rectangular sums and then their increasing limits shows ${\sum }_{\mathbf{n}}{w}_{\mathbf{n}}(2+{\sum }_j{n}_j)<\mathrm{\infty }$.

The sector creation formula in [F1] and invariance of the full permutation average under prior symmetrization give $a_j^{†}\mathrm{\mid }\mathbf{n}⟩=\sqrt{N+1}\sqrt{C_{\mathbf{n}}\mathrm{/}{C}_{\mathbf{n}+{\mathbf{e}}_j}}\mathrm{\mid }\mathbf{n}+{\mathbf{e}}_j⟩=\sqrt{n_j+1}\mathrm{\mid }\mathbf{n}+{\mathbf{e}}_j⟩$, where ${\mathbf{e}}_j$ is the unit occupation vector. For $N>0$ and $n_j>0$, exactly the fraction $n_j\mathrm{/}N$ of the words in step 1.1 have first factor $e_j$; contraction by $e_j$ leaves the uniform average of the words with occupation $\mathbf{n}-{\mathbf{e}}_j$. Thus $a_j\mathrm{\mid }\mathbf{n}⟩=\sqrt{N}(n_j\mathrm{/}N)\sqrt{C_{\mathbf{n}}\mathrm{/}{C}_{\mathbf{n}-{\mathbf{e}}_j}}\mathrm{\mid }\mathbf{n}-{\mathbf{e}}_j⟩=\sqrt{n_j}\mathrm{\mid }\mathbf{n}-{\mathbf{e}}_j⟩$. It is zero if $n_j=0$, including the vacuum. This proves both ladder formulas on the same basis for every $j$ directly from the sector definitions. Consequently ${\sum }_j{\omega }_j{a}_j^{†}{a}_j$ has eigenvalue $E_{\mathbf{n}}={\sum }_j{\omega }_j{n}_j$ on that finite span.

For a Borel set $B\subseteq \mathbb{R}$, define $P(B)\psi ={\sum }_{E_{\mathbf{n}}\in B}{c}_{\mathbf{n}}\mathrm{\mid }\mathbf{n}⟩$, where $c_{\mathbf{n}}=⟨\mathbf{n}\mathrm{\mid }\psi ⟩$. These are coordinate orthogonal projections, $P(\mathbb{R})=I$, $P(\mathrm{\varnothing })=0$, and $P(B)P(C)=P(B\cap C)$. For pairwise disjoint Borel sets, the corresponding coordinate selections are orthogonal and their squared norms sum to at most $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$, proving strong countable additivity. The scalar measure is $P_{\psi }(B)={\sum }_{E_{\mathbf{n}}\in B}\mathrm{\mid }{c}_{\mathbf{n}}{\mathrm{\mid }}^2$. It is finite and regular: given $\epsilon >0$, choose a finite set of occupation indices whose omitted squared coefficients sum to less than $\epsilon$. For any Borel $B$, the energies in $B$ of these selected indices form a finite compact subset $K\subseteq B$ with $P_{\psi }(B\setminus K)<\epsilon$. The open set obtained by deleting their finitely many energies outside $B$ contains $B$ and has excess measure less than $\epsilon$. This proves compact inner and open outer approximation for every Borel set; finiteness gives local finiteness as well. Hence $P$ is a regular PVM on $\mathbb{R}$, supported on $[0,\mathrm{\infty })$.

For any bounded Borel $f$, first evaluate simple integrals on an occupation vector, obtaining $f(E_{\mathbf{n}})\mathrm{\mid }\mathbf{n}⟩$; uniform approximation in [F8] proves the same formula for $f(P)$. Thus [F9] gives $D(\int \lambda dP)=\{\psi :{\sum }_{\mathbf{n}}{E}_{\mathbf{n}}^2\mathrm{\mid }{c}_{\mathbf{n}}{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and the coordinate action $c_{\mathbf{n}}\mapsto E_{\mathbf{n}}{c}_{\mathbf{n}}$. By [F3] this operator is self-adjoint with spectral PVM $P$. Truncating occupation coordinates, for example to ${\sum }_j{n}_j\le M$, converges in its graph norm, so by step 2.1 it is exactly the closure $H$ in the Statement. It is nonnegative, since its quadratic form on its operator domain is ${\sum }_{\mathbf{n}}{E}_{\mathbf{n}}\mathrm{\mid }{c}_{\mathbf{n}}{\mathrm{\mid }}^2\ge 0$. Because $P((-\mathrm{\infty },0))=0$, [F9] permits replacing the exponential function by the bounded Borel function $f_{\beta }(\lambda )=e^{-\beta \max (\lambda ,0)}$. Its integral therefore is $e^{-\beta H}$ and acts on $\mathrm{\mid }\mathbf{n}⟩$ by $w_{\mathbf{n}}$.

Repeated finite rectangular sums, their increasing limits, and [F6] give ${\sum }_{\mathbf{n}}{w}_{\mathbf{n}}={\prod }_j(1-q_j{)}^{-1}<\mathrm{\infty }$. The diagonal expansion of $e^{-\beta H}$ has rank-one norms $w_{\mathbf{n}}$, so its finite sums converge in operator norm. The limit is compact: for each positive tolerance, a finite-rank approximation maps the unit ball into a bounded finite-dimensional set with a finite net, and the norm error supplies a finite net for the image under the limit; its closure is complete in the Hilbert space. It is trace class by [F5], with trace the displayed positive sum. Step 1.1 supplies a countable orthonormal basis, hence a nonzero separable complex Hilbert space. The same argument holds for every $\beta >0$, so [F4] applies on the open interval $(0,\mathrm{\infty })$ and gives the density operator ${\rho }_{\beta }$.

Write a product of two ladder operators as $X=BC$, with $C$ acting first. Each maps an occupation vector to a single occupation vector or zero. If $N={\sum }_j{n}_j$, the first coefficient has modulus at most $\sqrt{N+1}$ and, after this operation, total occupation is at most $N+1$, so the second coefficient is at most $\sqrt{N+2}$. Thus $\mathrm{\parallel }X\mathrm{\mid }\mathbf{n}⟩\mathrm{\parallel }\le \sqrt{(N+1)(N+2)}\le N+2$. For $\psi =\sum c_{\mathbf{n}}\mathrm{\mid }\mathbf{n}⟩$, let $u_M=Z^{-1}{\sum }_{N\le M}{w}_{\mathbf{n}}{c}_{\mathbf{n}}\mathrm{\mid }\mathbf{n}⟩$. Then $u_M\to {\rho }_{\beta }\psi$. Both $C{u}_M$ and $BC{u}_M$ converge absolutely in norm: $\mathrm{\mid }{c}_{\mathbf{n}}\mathrm{\mid }\le \mathrm{\parallel }\psi \mathrm{\parallel }$, the first coefficient is at most $\sqrt{N+1}\le N+2$, and ${\sum }_{\mathbf{n}}{w}_{\mathbf{n}}(N+2)<\mathrm{\infty }$ by step 1.2. Closedness of $C$ first gives ${\rho }_{\beta }\psi \in D(C)$ and $C{u}_M\to C{\rho }_{\beta }\psi$; closedness of $B$ then gives $C{\rho }_{\beta }\psi \in D(B)$ and the required image. Hence $X{\rho }_{\beta }$ is defined on all of Fock space. Its occupation expansion has sum of rank-one norms at most $Z^{-1}{\sum }_{\mathbf{n}}{w}_{\mathbf{n}}(2+{\sum }_j{n}_j)<\mathrm{\infty }$, and thus converges in operator norm. It is compact by the finite-net argument of step 5.1, trace class by [F5], and its trace is the sum of its diagonal occupation matrix elements.

For $i\mathrm{\ne }j$, both normal and antinormal products change the occupation vector, so their diagonal matrix elements vanish. For $i=j$, they are respectively $n_j$ and $n_j+1$. Products of two annihilations or two creations always change total particle number by two and therefore have zero diagonal matrix elements. Using step 1.2 and the product normalization gives $⟨a_j^{†}{a}_j{⟩}_{\beta }=q_j\mathrm{/}(1-q_j)=(e^{\beta {\omega }_j}-1{)}^{-1}$ and $⟨a_j{a}_j^{†}{⟩}_{\beta }=(1-q_j{)}^{-1}=1+(e^{\beta {\omega }_j}-1{)}^{-1}$. Absolute convergence and the trace-class domains are supplied by step 6.1. These are all the claimed moments; no infinite-volume limit has been invoked.