Rotating mode Gibbs obstruction

By [F3], $\mathcal{H}$ with the displayed inner product is complete, so it is a Hilbert space. For $x\in \mathcal{H}$, put $S:={\sum }_{n\ge 0}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and $S_M:={\sum }_{n=0}^M\mathrm{\mid }{x}_n{\mathrm{\mid }}^2$. Every finite subset of ${\mathbb{N}}_0$ is contained in some $\{0,\dots ,M\}$, so $S_M\uparrow S$. The finite-subsum splitting identity in [F1], applied to $\{0,\dots ,M\}$, gives $\mathrm{\parallel }x-{\sum }_{n=0}^M{x}_n{e}_n{\mathrm{\parallel }}_2^2={\sum }_{n>M}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=S-S_M\to 0$. Thus finite-support sequences are dense and lie in $D(\mathcal{N})$, so $\mathcal{N}$ is densely defined. For $x,y\in D(\mathcal{N})$, Cauchy–Schwarz [F2] gives convergence of both inner-product sums and $⟨\mathcal{N}x,y⟩={\sum }_{n}n{x}_n\overline{y_n}={\sum }_n{x}_n\overline{n{y}_n}=⟨x,\mathcal{N}y⟩,$ so the operator is symmetric.

On the basis vector $e_n$, the Gibbs weight for $K$ is $e^{-{\beta }_{K}an}$, so the partition sum is the displayed nonnegative series. If $a>0$, put $r=e^{-{\beta }_{K}a}\in (0,1)$; its partial sums are $(1-r^{M+1})\mathrm{/}(1-r)$ and tend to $(1-r{)}^{-1}$. If $a=0$, every term is $1$ and the partial sums tend to $+\mathrm{\infty }$. If $a<0$, the terms $e^{{\beta }_K\mathrm{\mid }a\mathrm{\mid }n}$ do not tend to zero, so the nonnegative series diverges.

If $y\in D(\mathcal{N})$, then $z:=\mathcal{N}y\in \mathcal{H}$, and step 1.1 shows $⟨\mathcal{N}x,y⟩=⟨x,z⟩$ for every $x\in D(\mathcal{N})$; thus $y\in D({\mathcal{N}}^{†})$ and ${\mathcal{N}}^{†}y=\mathcal{N}y$. Conversely, if $y\in D({\mathcal{N}}^{†})$ with representing vector $z\in \mathcal{H}$, test the defining identity on each $x=e_n$ to obtain $z_n=n{y}_n$. Since $z$ is square-summable, ${\sum }_n{n}^2\mathrm{\mid }{y}_n{\mathrm{\mid }}^2<\mathrm{\infty }$, so $y\in D(\mathcal{N})$ and $z=\mathcal{N}y$. Therefore $D({\mathcal{N}}^{†})=D(\mathcal{N})$ and ${\mathcal{N}}^{†}=\mathcal{N}$.

Multiplication by a real nonzero scalar $c$ leaves the adjoint domain $D(\mathcal{N})$ unchanged and multiplies the adjoint action by $c$: the same coordinate test as step 2.1 forces $z_n=cn{y}_n$, and $(cn{y}_n{)}_n$ is square-summable exactly when $(n{y}_n{)}_n$ is. Thus $c\mathcal{N}$ is self-adjoint on $D(\mathcal{N})$. The zero operator on all of $\mathcal{H}$ is self-adjoint: its adjoint relation forces $⟨x,z⟩=0$ for every $x\in \mathcal{H}$, in particular $x=z$, so $z=0$. Taking $c=ϵ$ and $c=\mathrm{\hslash }m$ when $m\mathrm{\ne }0$ proves the assertions for $H$ and $J$; when $m=0$, $J=0$ on all of $\mathcal{H}$. On the common domain, $K_0=a\mathcal{N}$. If $a\mathrm{\ne }0$, this operator is self-adjoint and closed by the scalar argument, so its closure is itself. If $a=0$, $K_0$ is zero on the dense finite-support domain $D(\mathcal{N})$; its graph closure is $\mathcal{H}\times \{0\}$ and hence $K$ is zero on all of $\mathcal{H}$, which is self-adjoint.

In the finite case $a>0$, $p_n=(1-r)r^n$ are positive and sum to one. For each $B\in \mathcal{B}(\mathcal{H})$, Cauchy–Schwarz [F2] and the operator norm give $\mathrm{\mid }⟨B{e}_n,e_n⟩\mathrm{\mid }\le \mathrm{\parallel }B{e}_n{\mathrm{\parallel }}_2\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_2\le \mathrm{\parallel }B\mathrm{\parallel }$, hence ${\sum }_n{p}_n\mathrm{\mid }⟨B{e}_n,e_n⟩\mathrm{\mid }\le \mathrm{\parallel }B\mathrm{\parallel }$; the defining series for ${\omega }_a(B)$ is absolutely convergent and $\mathrm{\mid }{\omega }_a(B)\mathrm{\mid }\le \mathrm{\parallel }B\mathrm{\parallel }$. Absolute convergence gives complex linearity, and ${\omega }_a(I)={\sum }_n{p}_n=1$. For every $C\in \mathcal{B}(\mathcal{H})$, ${\omega }_a(C^{\ast }C)={\sum }_n{p}_n⟨C^{\ast }C{e}_n,e_n⟩={\sum }_n{p}_n\mathrm{\parallel }C{e}_n{\mathrm{\parallel }}_2^2\ge 0$. Thus ${\omega }_a$ is a bounded positive normalized linear functional, hence a state. If $a\le 0$, step 1.2 gives no finite normalizer for these same occupation weights. This proves exactly the asserted trace-Gibbs obstruction without any field-theoretic or spacetime extension.