Finite-regulator Gibbs trace and bounded expectations

Each ${ϵ}_{\mathbf{n}}$ is finite, and only finitely many multi-indices have ${ϵ}_{\mathbf{n}}\le M$ for any real $M$: for $M\ge 0$, that inequality bounds every nonnegative integer $n_j$ by $M\mathrm{/}(2{\alpha }_j)$, while for $M<0$ there are none. Fix the explicit degree-then-lexicographic enumeration of ${\mathbb{N}}_0^N$. Recursively, among the remaining tuples choose one of least energy, breaking a tie by that fixed enumeration; such a least energy exists because below the energy of the first remaining tuple there are only finitely many candidates. Every tuple is eventually listed, since only finitely many tuples have energy at most its energy. This gives the nondecreasing list $({ϵ}_k{)}_{k\ge 0}$ with multiplicity.

On $C_c^{\mathrm{\infty }}$, the shifted interacting form has weight $V+C_0+2\ge c\mathrm{\mid }q{\mathrm{\mid }}^2+2$, so it dominates the shifted oscillator form from [F2]. Therefore a Cauchy sequence in $Q$ is Cauchy in the oscillator form norm; the identity map extends to a continuous map $Q\to Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$, with the same $L^2$ representative because both completions inject into $H_N$ by [F3]. Subtracting the shifts in the two form norms gives $q_h(u)=a(u,u)-(C_0+2)\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\ge q_{\mathrm{o}\mathrm{s}\mathrm{c}}(u)-C_0\mathrm{\parallel }u{\mathrm{\parallel }}_2^2(u\in Q)\mathrm{.}$

Fix $k\ge 0$ and let $L_k=\mathrm{span}\{{\psi }_0,\dots ,{\psi }_k\}$. The linear map from $L_k$ to the span of the first $k$ oscillator eigenvectors, sending $u$ to its $k$ oscillator coefficients, has a nonzero kernel because $\dim L_k=k+1>k$. Choose $u\mathrm{\ne }0$ in that kernel. Since $u$ lies in the span of the first $k+1$ interacting eigenvectors, [F1] gives $q_h(u)\le {\lambda }_k\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. Its oscillator coefficients at indices below $k$ vanish, so [F2] gives $q_{\mathrm{o}\mathrm{s}\mathrm{c}}(u)\ge {ϵ}_k\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. Step 1.2 now yields ${\lambda }_k\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\ge q_h(u)\ge ({ϵ}_k-C_0)\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$, hence ${\lambda }_k\ge {ϵ}_k-C_0\mathrm{.}$ This argument also covers $k=0$, when the span of zero oscillator vectors is $\{0\}$.

By step 2.1 and $\beta >0$, ${\sum }_{k\ge 0}{e}^{-\beta {\lambda }_k}\le e^{\beta C_0}{\sum }_{\mathbf{n}\in {\mathbb{N}}_0^N}{e}^{-\beta {ϵ}_{\mathbf{n}}}=e^{\beta C_0}{\prod }_{j=1}^N\left({\sum }_{n_j\ge 0}{e}^{-\beta (2n_j+1){\alpha }_j}\right)=e^{\beta C_0}{\prod }_{j=1}^N\frac{e^{-\beta {\alpha }_j}}{1-e^{-2\beta {\alpha }_j}}\mathrm{.}$ The factorization follows first for finite rectangular sums by distributivity and then by increasing limits of nonnegative sums. Each denominator is positive because $0<e^{-2\beta {\alpha }_j}<1$. The lower bound $Z_{\beta }>0$ follows from its positive ground term.

The coefficients satisfy $0<e^{-\beta {\lambda }_k}\le e^{\beta C_0}$ by [F1]. For each $f$, Parseval makes the finite partial sums in the defining series for ${\rho }_{\beta }f$ Cauchy in $H_N$, so the series converges; its coefficient formula gives linearity and $\mathrm{\parallel }{\rho }_{\beta }f{\mathrm{\parallel }}_2\le e^{\beta C_0}\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Moreover Parseval gives $\mathrm{\parallel }{\rho }_{\beta }-{\rho }_{\beta ,M}\mathrm{\parallel }\le {\sup }_{k>M}{e}^{-\beta {\lambda }_k}⟶0,$ since ${\lambda }_k\to +\mathrm{\infty }$ and $\beta >0$. Each truncation is finite rank and hence compact by [F10], so the operator-norm limit ${\rho }_{\beta }$ is compact by [F11]. The same convergence makes its rank-one partial sums converge in operator norm, and their coefficient sum is ${\sum }_k{e}^{-\beta {\lambda }_k}=Z_{\beta }<\mathrm{\infty }$; thus ${\rho }_{\beta }$ is trace class by [F4].

Since each $e^{-\beta {\lambda }_k}$ is real, the Fourier expansion [F12] gives $⟨{\rho }_{\beta }f,g⟩={\sum }_k{e}^{-\beta {\lambda }_k}⟨f,{\psi }_k⟩⟨{\psi }_k,g⟩=⟨f,{\rho }_{\beta }g⟩$ for all $f,g\in H_N$; the series is absolutely convergent by Cauchy--Schwarz for the coefficient sequences and boundedness of the weights. Thus ${\rho }_{\beta }$ is self-adjoint. It is positive because $⟨{\rho }_{\beta }f,f⟩={\sum }_k{e}^{-\beta {\lambda }_k}\mathrm{\mid }⟨f,{\psi }_k⟩{\mathrm{\mid }}^2\ge 0.$ The defining series gives ${\rho }_{\beta }{\psi }_k=e^{-\beta {\lambda }_k}{\psi }_k$. If ${\rho }_{\beta }x=\mu x$, taking Fourier coefficients shows $(e^{-\beta {\lambda }_k}-\mu )⟨x,{\psi }_k⟩=0$ for every $k$. Completeness [F1, F12] then implies that every nonzero eigenspace is exactly the span of basis vectors with the corresponding repeated diagonal value; since every diagonal value is positive, $0$ is not an eigenvalue. Thus the eigenvalues with multiplicity are precisely $(e^{-\beta {\lambda }_k})$, and [F5] gives $\mathrm{Tr}({\rho }_{\beta })=Z_{\beta }$. For bounded $B$, each ${\rho }_{\beta ,M}B$ is finite rank and hence compact by [F10], while $\mathrm{\parallel }{\rho }_{\beta ,M}B-{\rho }_{\beta }B\mathrm{\parallel }\le \mathrm{\parallel }{\rho }_{\beta ,M}-{\rho }_{\beta }\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }\to 0$ by step 4.1. Therefore ${\rho }_{\beta }B$ is compact by [F11]. The functional $x\mapsto ⟨Bx,{\psi }_k⟩$ is linear and has norm at most $\mathrm{\parallel }B\mathrm{\parallel }$ by [F9] and [F8]. By [F6] it is represented by a vector $w_k$ with $\mathrm{\parallel }{w}_k\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }$ and $⟨Bx,{\psi }_k⟩=⟨x,w_k⟩$. Hence ${\rho }_{\beta }Bx={\sum }_k⟨x,w_k⟩e^{-\beta {\lambda }_k}{\psi }_k\mathrm{.}$ The partial sums converge in operator norm: the norm of the tail after $M$ is at most $\mathrm{\parallel }B\mathrm{\parallel }{\sum }_{k>M}{e}^{-\beta {\lambda }_k}\to 0$ by step 3.1. This nuclear representation has coefficient sum at most $\mathrm{\parallel }B\mathrm{\parallel }{Z}_{\beta }$. Since ${\rho }_{\beta }B$ is compact as shown above, [F4] gives that it is trace class. Its basis trace in $({\psi }_k)$ is absolutely convergent and equals the trace by [F7]. The $k$th diagonal entry is $⟨{\rho }_{\beta }B{\psi }_k,{\psi }_k⟩=e^{-\beta {\lambda }_k}⟨B{\psi }_k,{\psi }_k⟩,$ so this basis sum is the asserted numerator. Each term is bounded in modulus by $e^{-\beta {\lambda }_k}\mathrm{\parallel }B\mathrm{\parallel }$ by [F8] and [F9], giving the stated absolute bound. Division by $Z_{\beta }>0$ proves the expectation formula.