Explicit diagonal Fock domains and occupation trace

The tensor factors and product pairing in the Definition are supplied by [F1]. For every $n\ge 1$, [F2] makes $P_{\pm }^{(n)}$ an orthogonal projection. The range of an orthogonal projection is closed: if $P{x}_k=x_k$ and $x_k\to x$, boundedness of $P$ gives $Px={\lim }_{k}P{x}_k={\lim }_k{x}_k=x$. Therefore each ${\mathcal{H}}_{\pm }^{(n)}$ is a closed subspace of the complete tensor Hilbert space and is itself complete.

For $\psi ,\phi \in {\mathcal{F}}_{\pm }(\mathfrak{h})$, Cauchy–Schwarz for the square-summable nonnegative sequences $(\mathrm{\parallel }{\psi }_n\mathrm{\parallel }{)}_n$ and $(\mathrm{\parallel }{\phi }_n\mathrm{\parallel }{)}_n$ gives ${\sum }_n\mathrm{\mid }⟨{\psi }_n,{\phi }_n⟩\mathrm{\mid }\le ({\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}({\sum }_n\mathrm{\parallel }{\phi }_n{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Hence the displayed pairing converges absolutely, is sesquilinear, and is positive definite because $⟨\psi ,\psi ⟩={\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2=0$ forces every coordinate to vanish.

Let $({\psi }^{(k)}{)}_k$ be Cauchy in this norm. For each fixed $n$, $({\psi }_n^{(k)}{)}_k$ is Cauchy in the complete sector ${\mathcal{H}}_{\pm }^{(n)}$, so it converges to some ${\psi }_n\in {\mathcal{H}}_{\pm }^{(n)}$. If $k\ge K$ and $\mathrm{\parallel }{\psi }^{(k)}-{\psi }^{(m)}\mathrm{\parallel }\le \epsilon$ for all $m\ge K$, then for every finite $N$, passing $m\to \mathrm{\infty }$ in the finite sum yields ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n-{\psi }_n^{(k)}{\mathrm{\parallel }}^2\le {\epsilon }^2$. Also ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }{\psi }^{(k)}{\mathrm{\parallel }}^2+2{\epsilon }^2$. Taking suprema over finite $N$ proves ${\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $\psi \in {\mathcal{F}}_{\pm }(\mathfrak{h})$, and then $\mathrm{\parallel }{\psi }^{(k)}-\psi \mathrm{\parallel }\le \epsilon$ for all $k\ge K$. Thus the direct sum is complete.

For $\psi \in {\mathcal{F}}_{\pm }(\mathfrak{h})$, truncating after sector $N$ changes its squared norm by ${\sum }_{n>N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\to 0$. Therefore the finite-particle subspace is dense and $\mathrm{\Omega }$ is a unit vector.

Let $n=\mathrm{\mid }m\mathrm{\mid }$. In the bosonic permutation average $P_{+}^{(n)}{x}_m$, each distinct product word with occupation profile $m$ occurs ${\prod }_j{m}_j!$ times out of $n!$ permutations. Therefore $P_{+}^{(n)}{x}_m=\frac{{\prod }_j{m}_j!}{n!}{\sum }_{w\in \mathcal{W}(m)}w,\mathrm{\mid }\mathcal{W}(m)\mathrm{\mid }=\frac{n!}{{\prod }_j{m}_j!}\mathrm{.}$ The product words are orthonormal by [F1], so the squared norm of this average is ${\prod }_j{m}_j!\mathrm{/}n!$; the prefactor in $\mathrm{\mid }m{⟩}_B$ makes its norm one. Distinct occupation profiles have disjoint word supports, hence their vectors are orthogonal.

If a fermionic list repeats a mode, [F2] makes its antisymmetrization zero; ${\mathrm{\Omega }}_F$ excludes these profiles. For distinct occupied modes, the $n!$ permuted product words are distinct orthonormal vectors with coefficients $\pm 1\mathrm{/}n!$, so $\mathrm{\parallel }{P}_{-}^{(n)}{x}_m{\mathrm{\parallel }}^2=1\mathrm{/}n!$. The prefactor $\sqrt{n!}$ normalizes the vector. Distinct occupied subsets have disjoint product-word support and are orthogonal.

The fixed enumeration identifies a finite-support configuration with a finite sequence of natural indices and finitely many nonnegative integer labels (or bits); the union over all finite sequence lengths is countable. Thus ${\mathrm{\Omega }}_B$ and ${\mathrm{\Omega }}_F$ are countable and their ${\mathrm{\ell }}^2$ coordinate descriptions are well defined.

On the bosonic basis, shifting the coordinate index gives $\mathrm{\parallel }{a}_{j}c{\mathrm{\parallel }}^2={\sum }_m{m}_j\mathrm{\mid }{c}_m{\mathrm{\mid }}^2$ and $\mathrm{\parallel }{a}_j^{†}c{\mathrm{\parallel }}^2={\sum }_m(m_j+1)\mathrm{\mid }{c}_m{\mathrm{\mid }}^2$. Hence the stated weighted domains are exactly the domains on which the shifted coefficient sequences lie in ${\mathrm{\ell }}^2$. If $c^{(r)}\to c$ and either shifted sequence converges in ${\mathrm{\ell }}^2$, coordinate convergence shows the limiting output has the same weighted-shift coordinates; its square summability puts $c$ in the displayed domain. Thus both maximal shifts are closed. Testing the adjoint identity against each basis vector forces the reverse weighted shift [F3], while Cauchy--Schwarz gives the converse pairing bound [F4]. Truncating to finite initial sets of configurations makes the weighted tails tend to zero in graph norm, so ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is a core for both shifts under the graph definition in [F3].

Product words span a dense subspace of each tensor power by [F1]. Applying the bounded projections $P_{\pm }^{(n)}$ to that dense span gives a dense subspace of each projected sector. Every projected product word is either zero or a scalar multiple of the corresponding occupation vector from steps 1.1–1.2. Thus the occupation vectors are complete in each sector and, after taking the direct sum over $n$, form orthonormal bases of the Fock spaces.

For $i=j$, direct action on $\mathrm{\mid }m{⟩}_B$ gives $a_j{a}_j^{†}\mathrm{\mid }m{⟩}_B=(m_j+1)\mathrm{\mid }m{⟩}_B$ and $a_j^{†}{a}_j\mathrm{\mid }m{⟩}_B=m_j\mathrm{\mid }m{⟩}_B$. For $i\mathrm{\ne }j$, both orders of distinct coordinate shifts give the same coefficient and same occupation vector, so the shifts commute; the same direct coordinate calculation applies to two annihilators and two creators. This proves the CCR on ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$, without asserting that unbounded commutators are defined on all Fock space.

Each fermionic shift maps a subset of an orthonormal basis into another orthonormal subset with coefficients of modulus at most one, so it is a contraction on the finite span and extends boundedly. For distinct $i,j$, the two orders in any CAR anticommutator are either both zero or have equal magnitudes and opposite signs: toggling the earlier mode changes by one the number of occupied modes preceding the later index. For $i=j$, if $m_j=0$ then $c_j{c}_j^{†}\mathrm{\mid }m{⟩}_F=\mathrm{\mid }m{⟩}_F$ and $c_j^{†}{c}_j\mathrm{\mid }m{⟩}_F=0$; if $m_j=1$ the roles reverse. The other same-mode anticommutators vanish. Thus all CAR identities hold on the basis span and extend by boundedness to the whole Fock space.

The configuration set ${\mathrm{\Omega }}_{\pm }$ is countable by [F1]. Every $m$ has finite support, so the sums defining $E(m)$ and $\mathrm{\mid }m\mathrm{\mid }$ are finite real sums. The finite-support vectors form a dense subspace of ${\mathrm{\ell }}^2({\mathrm{\Omega }}_{\pm })$.

If $I$ is empty, its only occupation configuration is the vacuum: $E=N=K_{\mu }=0$, $G=I$ on $\mathbb{C}$, $\mathrm{\Xi }=1$ and $\rho =1$. The empty products are one, the total number has mean and variance zero, and every stated parameter derivative is zero; the convention $q_{\ast }=0$ makes the bosonic hypotheses valid. For a finite mode set $J$, summing the positive diagonal weights independently over each bosonic coordinate gives ${\prod }_{j\in J}{\sum }_{n\ge 0}{q}_j^n={\prod }_{j\in J}(1-q_j{)}^{-1}$. In the fermionic case each coordinate has only two values and contributes $1+q_j$. This proves the finite-mode trace products.

For a fixed $\mu$, the countable configuration model has weights $e^{-\beta (E(m)-\mu \mathrm{\mid }m\mathrm{\mid })}$ and degeneracies one. On the nonempty open interval $J$, the compact $\beta$-envelope in the Statement ensures that [F2]–[F3] apply to this exact countable index set and energy $K_{\mu }(m)$, giving the displayed beta-derivative identities and finite moments. The envelope is stronger than trace finiteness and is stated separately.

For joint $(\beta ,\mu )$ response, put $X_1(m)=E(m)$, $X_2(m)=\mathrm{\mid }m\mathrm{\mid }$, and natural parameter $\theta =(-\beta ,\beta \mu )$. The map $(\beta ,\mu )\mapsto \theta$ is a smooth bijection onto its open image for $\beta >0$, with nonzero Jacobian. The stated joint envelope is exactly the required second-moment envelope for [F5]; differentiating at fixed $\beta$ gives ${\mathrm{\partial }}_{\mu }\mathbb{E}(N)=\beta \mathrm{Var}(N)$ and ${\mathrm{\partial }}_{\mu }\mathbb{E}(E)=\beta \mathrm{Cov}(E,N)$.

Let $D_{\lambda }$ multiply the coordinate $c_m$ by a real number ${\lambda }_m$, on the domain ${\sum }_m{\lambda }_m^2\mathrm{\mid }{c}_m{\mathrm{\mid }}^2<\mathrm{\infty }$. Testing the adjoint identity on each coordinate vector forces the adjoint coordinates to be ${\lambda }_m{c}_m$, so $D(D_{\lambda }^{\ast })$ is exactly this weighted ${\mathrm{\ell }}^2$ domain by [F8]. Conversely, on that domain Cauchy–Schwarz makes the coordinate pairing bounded and identifies the adjoint action with the same multiplier. Apply this argument to ${\lambda }_m=E(m)$, $\mathrm{\mid }m\mathrm{\mid }$, and $E(m)-\mu \mathrm{\mid }m\mathrm{\mid }$; all three displayed operators are self-adjoint.

Enumerate a countable mode set by increasing finite initial sets $J_M$. Their finite-support configuration sets increase to ${\mathrm{\Omega }}_{\pm }$, so the full nonnegative trace sum is the increasing limit of the finite-mode sums. For bosons, $-\log (1-q_j)={\int }_0^{q_j}(1-u{)}^{-1}du\le q_j\mathrm{/}(1-q_{\ast })$, hence ${\sum }_j-\log (1-q_j)<\mathrm{\infty }$. For fermions, $\log (1+q_j)\le q_j$, hence ${\sum }_j\log (1+q_j)<\mathrm{\infty }$. The products therefore have finite positive limits and equal the full diagonal trace sums; the negative fermionic gaps are finite by hypothesis, and each has finite weight.

If ${\mathrm{\Omega }}_{\pm }$ is infinite, enumerate it as $m_1,m_2,\dots$ and let $G_M$ retain the first $M$ diagonal entries of $G_{\beta ,\mu }$; in the finite case take the full finite sum. Each $G_M$ has finite rank and hence is compact. Since the omitted entries are nonnegative and summable, their supremum is bounded by their sum, so $\mathrm{\parallel }{G}_{\beta ,\mu }-G_M\mathrm{\parallel }\le {\sum }_{r>M}{e}^{-\beta (E(m_r)-\mu \mathrm{\mid }{m}_r\mathrm{\mid })}\to 0$. The norm-limit theorem therefore makes $G_{\beta ,\mu }$ compact. It is positive diagonal and satisfies $G_{\beta ,\mu }^{\ast }{G}_{\beta ,\mu }=G_{\beta ,\mu }^2$; uniqueness of its compact positive square root gives $\mathrm{\mid }{G}_{\beta ,\mu }\mathrm{\mid }=G_{\beta ,\mu }$. Its singular values are the nonnegative diagonal entries with multiplicity, whose sum is $\mathrm{\Xi }<\mathrm{\infty }$. Thus [F7] and the trace-class definition make it trace class and give $\mathrm{Tr}{G}_{\beta ,\mu }=\mathrm{\Xi }$. Dividing by $\mathrm{\Xi }>0$ gives a positive trace-class operator of trace one.

Fix one mode $j$, and take increasing finite mode sets $J_M$ containing it. For bosons, the trace weight of configurations in $J_M$ with occupation $n$ at $j$ is $q_j^n{\prod }_{i\in J_M\setminus \{j\}}(1-q_i{)}^{-1}$; division by the total finite trace gives $(1-q_j)q_j^n$. For fermions the same calculation gives probability $q_j^n\mathrm{/}(1+q_j)$, for $n\in \{0,1\}$. Taking increasing limits of the nonnegative configuration sums preserves these factorizations, because every finite-support configuration is eventually included and the finite products converge by step 2.2. Factoring over any fixed finite list of modes gives the product joint law, hence independence of finite collections in this diagonal grand state.

For finite $J$, multiplying the geometric series in each bosonic coordinate gives the coefficient identity for $Z_{B,N}(J)$; multiplying the two-term polynomials in each fermionic coordinate gives the identity for $Z_{F,N}(J)$. Each coefficient sums exactly the finite canonical Boltzmann weights with total occupation $N$. For countable $I$, increasing finite mode sets increase each coefficient to its sum over all finite-support configurations; for any admissible fugacity $z=e^{\beta \mu }>0$, nonnegativity gives $z^N{Z}_{\pm ,N}\le \mathrm{\Xi }$, hence $Z_{\pm ,N}\le z^{-N}\mathrm{\Xi }<\mathrm{\infty }$, and a canonical density exists exactly when the coefficient is positive. Conditioning the grand law on $\mathrm{\mid }m\mathrm{\mid }=N$ multiplies by the indicator of that constraint and generally couples the mode occupations.

If $0<q_j<1$, [F4] gives the stated Bose and Fermi means and variances. For a fermionic mode with $q_j\ge 1$, direct summation of its two probabilities gives $\mathbb{E}{N}_j=q_j\mathrm{/}(1+q_j)$ and $\mathrm{Var}(N_j)=q_j\mathrm{/}(1+q_j{)}^2$; no geometric-series claim is used there. The bosonic bounds $\mathbb{E}{N}_j\le q_j\mathrm{/}(1-q_{\ast })$ and $\mathrm{Var}(N_j)\le q_j\mathrm{/}(1-q_{\ast }{)}^2$, and the fermionic bounds $\mathbb{E}{N}_j\le q_j$ and $\mathrm{Var}(N_j)\le q_j$, are summable. For finite initial mode sets, independence from step 3.2 gives $\mathbb{E}{N}_M={\sum }_{j\in J_M}\mathbb{E}{N}_j$ and $\mathbb{E}{N}_M^2={\sum }_{j\in J_M}\mathrm{Var}(N_j)+({\sum }_{j\in J_M}\mathbb{E}{N}_j{)}^2$. Each configuration has finite support, so $N_M\uparrow N$; passing to the increasing limits of the defining nonnegative countable sums yields finite $\mathbb{E}N$ and $\mathbb{E}{N}^2$, with $\mathrm{Var}(N)={\sum }_j\mathrm{Var}(N_j)<\mathrm{\infty }$.

In the one-dimensional oscillator specialization, [F6] supplies the Hermite eigenbasis, energy list and exact diagonal domain; substitution of those eigenvalues into the Fock multiplier gives the stated occupation construction. This uses the actual self-adjoint oscillator carrier rather than a formal differential eigenfunction list.

For $\mu <0$, the Bose denominator is bounded below by $e^{\beta (-\mu )}-1>0$ on compact sets; for finite $\mu$, the Fermi denominator is at least $1$. At large $\mathrm{\mid }p\mathrm{\mid }$, each occupation is bounded by $C{e}^{-c\mathrm{\mid }p{\mathrm{\mid }}^2}$ for positive constants $C,c$ depending on a compact parameter neighborhood. A polynomial factor $\mathrm{\mid }p{\mathrm{\mid }}^s$ is absorbed by a slower Gaussian because $r^s{e}^{-c{r}^2\mathrm{/}2}$ is bounded on $[0,\mathrm{\infty })$. Thus the continuum integrands and their polynomially weighted tails are integrable by comparison with products of one dimensional Gaussians and Tonelli [F3,F1]. For each fixed $h>0$, the max-norm shell $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}=n$ has at most $C_d{n}^{d-1}$ points, and its summand is bounded by a fixed polynomial in $n$ times $e^{-c{h}^2{n}^2}$; the ratio of successive polynomial-Gaussian terms is eventually below a constant less than one, proving every lattice sum is finite.

At $\mu =0$ and $p\mathrm{\ne }0$, $e^x-1\ge x$ gives $F_B(p)\le (\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2{)}^{-1}$. If $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}=n$, the number of integer lattice points in that shell is $(2n+1{)}^d-(2n-1{)}^d\le C_d{n}^{d-1}$, while $\mathrm{\mid }k\mathrm{\mid }\ge n$. Hence for $0<h\le \delta$, $h^d{\sum }_{0<\mathrm{\mid }hk\mathrm{\mid }\le \delta }{F}_B(hk)\le C{h}^{d-2}{\sum }_{n\le \delta \mathrm{/}h}{n}^{d-3}\le C^{\mathrm{\prime }}{\delta }^{d-2}$ when $d>2$; the last estimate follows by comparison on each unit interval with the integral of $x^{d-3}$ (and for $d=3$ the sum is bounded by a constant times $\delta \mathrm{/}h$). The continuum small-ball integral has the same bound by decomposing into dyadic annuli, each with volume at most a constant times its radius to power $d$.

On a fixed compact cube, uniform continuity of $\mathrm{\mid }p{\mathrm{\mid }}^{s}F(p)$ implies that replacing its value on each mesh cube by its value at the lattice corner changes the integral by at most the cube volume times the common modulus of continuity; summing finitely many cubes makes this error tend to zero. The measure of each mesh cube is $h^d$ by [F5]. This proves convergence on compact cubes.

To control tails uniformly for $0<h\le 1$, let $Q_{k,h}=hk+[-h\mathrm{/}2,h\mathrm{/}2{]}^d$. For $p\in Q_{k,h}$, $\mathrm{\mid }hk{\mathrm{\mid }}^2\ge \mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}2-d{h}^2\mathrm{/}4$. The Gaussian bound from step 1.1 therefore bounds each weighted lattice summand times $h^d$ by a fixed constant times the integral of a slower Gaussian over $Q_{k,h}$; the polynomial weight is absorbed by reducing its exponent once more. Cubes whose centers satisfy $\mathrm{\mid }hk\mathrm{\mid }>R$ lie outside the ball of radius $R-\sqrt{d}\mathrm{/}2$. The integral of the slower Gaussian on that exterior tends to zero as $R\to \mathrm{\infty }$, by [F1,F3]. Compact convergence from step 2.1 and this uniform tail bound prove the stated full-space limit and finiteness of every lattice sum.

Outside any fixed ball around zero, $F_B$ is continuous and has Gaussian tails, so the compact-annulus and tail argument in steps 2.1 and 3.1 applies. Split both the sum and integral into $0<\mathrm{\mid }p\mathrm{\mid }\le \delta$, $\delta <\mathrm{\mid }p\mathrm{\mid }\le R$, and $\mathrm{\mid }p\mathrm{\mid }>R$. First take $h\downarrow 0$ on the annulus, then $R\to \mathrm{\infty }$ using Gaussian tails, and finally $\delta \downarrow 0$ using step 1.2. This proves convergence of the missing-origin Bose sums without assigning an origin value.

For $p\mathrm{\ne }0$, the positive geometric series gives $F_B(p;0)={\sum }_{\mathrm{\ell }\ge 1}{e}^{-\mathrm{\ell }\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2}$. Its partial sums increase pointwise, so [F4] and Tonelli [F3] give the equality as extended nonnegative integrals $\int F_B={\sum }_{\mathrm{\ell }\ge 1}\int e^{-\mathrm{\ell }\beta \alpha \mathrm{\mid }p{\mathrm{\mid }}^2}dp$. The product Gaussian integral [F1] evaluates each term as $(\pi \mathrm{/}(\mathrm{\ell }\beta \alpha ){)}^{d\mathrm{/}2}$. The resulting positive series converges exactly when $d\mathrm{/}2>1$, by comparison of $\sum {\mathrm{\ell }}^{-d\mathrm{/}2}$ with ${\int }_1^{\mathrm{\infty }}{x}^{-d\mathrm{/}2}dx$, and diverges otherwise. For $\mu <0$, $F_B(p;\mu )$ increases pointwise to $F_B(p;0)$ as $\mu \uparrow 0$, so [F4] also gives convergence of the corresponding integrals to this endpoint value. This proves the extended endpoint formula, its divergence for $d\le 2$, and the integral limit used by the Bose condensation theorem.