Countable-mode product criterion

[Orthonormality] Let $I=(j_1\le \cdots \le j_n)$, $I^{\mathrm{\prime }}=(j_1^{\mathrm{\prime }}\le \cdots \le j_m^{\mathrm{\prime }})$ and first suppose $n=m\ge 1$. Using $P_{+}^{(n)}=P_{+}^{(n)\ast }$ and $P_{+}^{(n)}\chi =\chi$ for $\chi \in \mathrm{ran}{P}_{+}^{(n)}$, [F2] gives $⟨v_I,v_{I^{\mathrm{\prime }}}⟩=\frac{1}{n!}{\sum }_{\pi \in S_n}⟨e_{j_1}\otimes \cdots \otimes e_{j_n},U_{\pi }(e_{j_1^{\mathrm{\prime }}}\otimes \cdots \otimes e_{j_n^{\mathrm{\prime }}})⟩$. Each summand is $1$ exactly when $\pi$ maps the tuple $(j_1^{\mathrm{\prime }},\dots ,j_n^{\mathrm{\prime }})$ to $(j_1,\dots ,j_n)$ and $0$ otherwise, by orthonormality of the $e_j$; since both tuples are nondecreasing, such $\pi$ exists only if $I=I^{\mathrm{\prime }}$, and then the number of such $\pi$ is the product of the factorials of the multiplicities of $I$. Hence $⟨v_I,v_{I^{\mathrm{\prime }}}⟩=0$ for $I\mathrm{\ne }{I}^{\mathrm{\prime }}$ and $\mathrm{\parallel }{v}_I{\mathrm{\parallel }}^2\ge 1\mathrm{/}n!>0$, so $(u_I)$ is an orthonormal family; tensors from different $n$ are orthogonal by [F1].

[Factoring the finite sums] Fix $N\ge 1$. The map $I\mapsto (n_1,\dots ,n_N)$ assigning to a multiset $I\subseteq \{1,\dots ,N\}$ the number $n_j$ of occurrences of $j$ is a bijection onto ${\mathbb{N}}^N$, and ${\lambda }_I={\sum }_j{n}_j{\epsilon }_j$; by [F5] and the geometric series with $q=e^{-\beta {\epsilon }_j}\in (0,1)$, ${\sum }_{I\subseteq \{1,\dots ,N\}}{e}^{-\beta {\lambda }_I}={\sum }_{(n_1,\dots ,n_N)\in {\mathbb{N}}^N}{\prod }_{j=1}^N{e}^{-\beta n_j{\epsilon }_j}={\prod }_{j=1}^N{\sum }_{n\ge 0}{e}^{-\beta n{\epsilon }_j}={\prod }_{j=1}^N\frac{1}{1-e^{-\beta {\epsilon }_j}}$, the interchange of the finite product with the sums following by induction from [F5] (Tonelli for nonnegative series).

[Completeness] The ordered elementary tensors $e_{i_1}\otimes \cdots \otimes e_{i_n}$ have dense span in $H^{\otimes n}$ ([F2]); applying the bounded projection $P_{+}^{(n)}$ gives a dense subset of $\mathrm{ran}{P}_{+}^{(n)}$, and the vectors $P_{+}^{(n)}(e_{i_1}\otimes \cdots \otimes e_{i_n})$ over ordered tuples coincide with those over nondecreasing tuples because $P_{+}^{(n)}{U}_{\pi }=P_{+}^{(n)}$ ([F2]). Hence the span of $\{u_I:\mathrm{\mid }I\mathrm{\mid }=n\}$ is dense in $\mathrm{ran}{P}_{+}^{(n)}$; combining the sectors and using that every $\psi \in F(H)$ is the limit of its truncations ([F1]) shows that the span of all $u_I$ is dense in $F(H)$. With step 1.1 this proves (a).

[The infinite product and the criterion] The finite sums of step 1.2 increase with $N$ because all terms are positive, and every finite multiset $I$ lies in an initial segment, so ${\sum }_{I\in \mathcal{M}}{e}^{-\beta {\lambda }_I}={\lim }_N{\sum }_{I\subseteq \{1,\dots ,N\}}{e}^{-\beta {\lambda }_I}={\lim }_N{\prod }_{j\le N}(1-e^{-\beta {\epsilon }_j}{)}^{-1}\in (0,\mathrm{\infty }]$. Put $q_j=e^{-\beta {\epsilon }_j}\in (0,1)$. For every $j$, [F6] gives $q_j\le -\log (1-q_j)\le q_j\mathrm{/}(1-q_j)$. If infinitely many $q_j>1\mathrm{/}2$, both ${\sum }_j{q}_j$ and ${\sum }_j-\log (1-q_j)$ diverge, the latter already bounded below by the former termwise. Otherwise only finitely many such terms occur, all finite, and on the remaining tail $q_j\le -\log (1-q_j)\le 2q_j$; therefore the logarithmic sum is finite exactly when ${\sum }_j{q}_j$ is finite. By the product and reciprocal laws in [F7], $\log {\prod }_{j\le N}(1-q_j{)}^{-1}={\sum }_{j\le N}-\log (1-q_j)$. If these logarithmic partial sums are bounded, the increasing reciprocal products are bounded as well, since $\log$ is increasing and $\exp$ is its increasing inverse; hence [F8] gives a finite limit. Conversely, if the reciprocal products have a finite limit, they are bounded and [F7] makes their logarithmic partial sums bounded. In the finite case, if $P_N$ denotes the reciprocal partial product and $P_N\to P$, continuity of $\log$ gives $\log P={\sum }_j-\log (1-q_j)$, so $P=\exp ({\sum }_j-\log (1-q_j))$. This proves (c) and (d) in both large-factor branches without assuming summability in advance.

[The diagonal operator and the $\mathcal{M}$-indexed trace criterion] The set $\mathcal{M}$ is countably infinite: group its finite nondecreasing sequences by the integer $n+{\sum }_{r=1}^n{j}_r$, list each finite group lexicographically (its sequences have length and entries bounded by the group integer), skip empty groups, and concatenate the groups (the singleton sequences ensure infinitely many entries). Fix the resulting enumeration $I_1,I_2,\dots$ and put ${\mu }_m:=e^{-\beta {\lambda }_{I_m}}$. Put ${\mu }_I:=e^{-\beta {\lambda }_I}$ for every $I$. By step 2.1, $(f_n{)}_{n\ge 0}:=(u_{I_{n+1}}{)}_{n\ge 0}$ is a complete orthonormal family, and [F3] applies with coefficients ${\mu }_{I_{n+1}}$. Reindexing its sums by this bijection shows that the diagonal operator $D_{\beta }$ is self-adjoint; Parseval from [F4] gives $\mathrm{\parallel }{D}_{\beta }\psi {\mathrm{\parallel }}^2={\sum }_I{\mu }_I^2\mathrm{\mid }⟨\psi ,u_I⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }\psi {\mathrm{\parallel }}^2$, so its domain is all of $F(H)$ and it is bounded with norm at most one. Its coefficients are positive, so $⟨D_{\beta }\psi ,\psi ⟩={\sum }_I{\mu }_I\mathrm{\mid }⟨\psi ,u_I⟩{\mathrm{\mid }}^2\ge 0$; hence it is positive. If ${\sum }_I{\mu }_I<\mathrm{\infty }$, define $R_N\psi :={\sum }_{m=1}^N{\mu }_m⟨\psi ,u_{I_m}⟩u_{I_m}$. Each $R_N$ is finite rank and compact by [F4], while Parseval gives $\mathrm{\parallel }{D}_{\beta }-R_N\mathrm{\parallel }\le {\sup }_{m>N}{\mu }_m\le {\sum }_{m>N}{\mu }_m\to 0$. Thus $D_{\beta }$ is compact by [F4]. Its coefficient expansion is the nuclear representation $D_{\beta }\psi ={\sum }_{m\ge 1}⟨\psi ,u_{I_m}⟩({\mu }_m{u}_{I_m})$, whose nuclear sum is ${\sum }_m{\mu }_m<\mathrm{\infty }$; [F4] therefore proves that $D_{\beta }$ is trace class. Conversely, if $D_{\beta }$ is trace class, [F9] applied to the supplied basis $(u_I)$ gives $\mathrm{tr}{D}_{\beta }={\sum }_I⟨D_{\beta }{u}_I,u_I⟩={\sum }_I{\mu }_I<\mathrm{\infty }$. Hence $D_{\beta }$ is trace class exactly when the $\mathcal{M}$-indexed sum is finite, and then its trace is that sum.

[Conclusion] Combining step 3.1 with step 2.2 gives (e): $D_{\beta }$ is trace class if and only if ${\sum }_j{e}^{-\beta {\epsilon }_j}<\mathrm{\infty }$, and then $\mathrm{tr}{D}_{\beta }={\sum }_I{e}^{-\beta {\lambda }_I}={\prod }_{j\ge 1}(1-e^{-\beta {\epsilon }_j}{)}^{-1}$. Parts (a)-(e) are proved, and no thermal or measurement interpretation is claimed.