Fixed-particle-number occupation bases and the countable-mode partition criterion

[Product basis and finite-prefix exhaustion] The tensors $e_{i_1}\otimes \cdots \otimes e_{i_N}$ are orthonormal by [F1]. Their span is dense: approximate each factor of an elementary tensor by its finite basis truncation using [F3]; the tensor norm formula and the telescoping difference of two products show convergence of these approximations, and finite sums of elementary tensors are dense by [F1]. Every finite collection of words is contained in some $H_m^{\otimes N}$. Since each $P_{\sigma }^{(N)}$ is bounded, its images of this dense word span are dense in its range. In the finite prefix, these images span precisely the finite Slater or orbit-sum basis of [F2]. A word in $\mathrm{orb}(n)$ uses only finitely many modes, and the orbit consists of distinct orthonormal product vectors, so its sum has norm squared $\mathrm{\mid }\mathrm{orb}(n)\mathrm{\mid }>0$. Thus its displayed normalization agrees in every larger prefix. The finite-prefix Slater vectors likewise agree in every larger prefix. Any pair of proposed occupation vectors lies in a common prefix, where [F2] gives orthogonality; the normalized vectors have norm one. Their span is dense in the sector by the projection argument, proving (a). The index sets are countable because they are countable unions of finite prefix index sets.

[Diagonal domain and self-adjointness] Write either sector in the orthonormal coordinates of step 1.1. The specified domain contains every finite coordinate vector, hence is dense. For a vector in that domain, Parseval makes the defining energy series converge. If $y$ belongs to the adjoint domain, with adjoint value $z$, testing against each $b_c^{\sigma }$ gives $z_c=E_c{y}_c$ because $E_c$ is real. Since $z$ is square summable, ${\sum }_c\mathrm{\mid }{E}_c{y}_c{\mathrm{\mid }}^2<\mathrm{\infty }$, so $y$ belongs to the specified domain. Conversely, if that sum is finite, Cauchy-Schwarz applied to the square-summable coefficient families gives $⟨H_N^{\sigma }x,y⟩=⟨x,H_N^{\sigma }y⟩$ for every domain vector $x$, so $y$ belongs to the adjoint domain with the stated value. The domains and values agree, proving self-adjointness by [F3]. Since every $E_c\ge Nb$, the absolutely convergent energy quadratic sum is at least $Nb\mathrm{\parallel }x{\mathrm{\parallel }}^2$. Finite coordinate truncations converge in both norm and graph norm since both square-sum tails vanish, so finite occupation sums are a core and the operator is their diagonal closure.

[Labelled energy and spectral measure] The same coordinate-domain construction on the product-word basis defines a self-adjoint labelled operator with energies ${\sum }_k{\epsilon }_{i_k}$, agreeing with the free slot-energy sum on finite mode words. Every permutation of a word has the same energy, so its domain and action commute with each permutation unitary and the projections [F1]. Their sector restrictions have exactly the occupation energies and domain of step 2.1. For the displayed $E_N^{\sigma }(B)$, the projection and intersection identities follow coordinatewise; for disjoint $B_l$, the tails of the square-summable coordinates give strong countable additivity and total mass $I$, so it is a PVM. The scalar measure of a vector $\psi ={\sum }_c{a}_c{b}_c^{\sigma }$ is ${\sum }_c\mathrm{\mid }{a}_c{\mathrm{\mid }}^2{\delta }_{E_c}$, so [F4] gives the identity-integral domain exactly ${\sum }_c{E}_c^2\mathrm{\mid }{a}_c{\mathrm{\mid }}^2<\mathrm{\infty }$; bounded spectral truncations converge to the displayed coordinate action there. This PVM is also uniquely determined by that operator: if another PVM represents it, the zero norm of $(H_N^{\sigma }-E_c)b_c^{\sigma }$ forces its scalar measure at $b_c^{\sigma }$ to be supported at $E_c$. Its projections therefore act as ${\mathbf{1}}_B(E_c)$ on every $b_c^{\sigma }$, and density gives equality on the whole sector. For the exponential, use the bounded Borel function $g(a)=e^{-\beta \max \{a,Nb\}}$ on $\mathbb{R}$; it agrees with $e^{-\beta a}$ on the support of this PVM. [F4] therefore gives the bounded positive operator with eigenvalues $e^{-\beta E_c}\le e^{-\beta Nb}$, namely $e^{-\beta H_N^{\sigma }}$. This proves (b) and identifies the exponential used in (c).

[Extended trace and occupation sums] Put $r_c:=e^{-\beta E_c}$, a bounded positive family. In the supplied occupation basis the extended positive trace is ${\sum }_c{r}_c$. This value is basis independent even when infinite: if $(v_l)$ is any other orthonormal basis, then ${\sum }_l⟨e^{-\beta H_N^{\sigma }}{v}_l,v_l⟩={\sum }_l{\sum }_c{r}_c\mathrm{\mid }⟨v_l,b_c^{\sigma }⟩{\mathrm{\mid }}^2={\sum }_c{r}_c$, since nonnegative-series sums can be interchanged as suprema over finite subsums, and Parseval gives ${\sum }_l\mathrm{\mid }⟨v_l,b_c^{\sigma }⟩{\mathrm{\mid }}^2=1$. A Bose occupation sequence has a unique sorted word $i_1\le \cdots \le i_N$, and a Fermi subset has a unique increasing word; $e^{-\beta E_c}$ is the product of the corresponding $q_j$. Thus the trace equals $h_N(q)$ or $e_N(q)$ as stated. If the sum is finite, the series of rank-one operators ${\sum }_c{r}_c⟨\cdot ,b_c^{\sigma }⟩b_c^{\sigma }$ is nuclear and converges in operator norm to the exponential: its norm tail is at most the summable scalar tail. Its finite-rank partial sums and [F5] first give compactness, and the nuclear criterion then gives trace class and the ordinary trace formula. Conversely trace class forces the diagonal sum to be finite by [F5].

[Upper and lower bounds and finiteness] Let $Q:={\sum }_j{q}_j\in (0,\mathrm{\infty }]$. The sum over all ordered words of length $N$ is $Q^N$: on the first $m$ modes it is $({\sum }_{j\le m}{q}_j{)}^N$, and every finite collection of words is contained in one such prefix. The nondecreasing and strictly increasing index sets are subsets of the ordered-word set, giving $Z_N^{\sigma }\le Q^N$. For $N\ge 2$, configurations selecting modes $1,\dots ,N-1$ once each and one additional mode $j\ge N$ are valid for both statistics, distinct as $j$ varies, and have weight $({\prod }_{k=1}^{N-1}{q}_k)q_j$. Their sum gives the lower bound. For $N=1$, every configuration is the single mode $j$, so both traces equal $Q$; the empty product is $1$ and the tail starts at $1$. For any $N$, the lower-bound coefficient is strictly positive and finite, and the omitted initial segment of $Q$ is finite. Hence $Q=\mathrm{\infty }$ forces $Z_N^{\sigma }=\mathrm{\infty }$, whereas $Q<\mathrm{\infty }$ gives $Z_N^{\sigma }<\mathrm{\infty }$ by the upper bound. The single configuration $\{1,\dots ,N\}$ has positive weight for either sign, so both traces are strictly positive. This proves all of (c). No sum over particle numbers or full-Fock product formula is used.