Thermal product state has no normal vacuum extension

For $s\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$ and finite $F\subseteq \mathrm{\Gamma }$, let $P_F^s\in {\mathcal{A}}_F$ be the projection onto the one-dimensional local basis vector with configuration $s{\mathrm{\mid }}_F$. If $t\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$, then $\pi (P_F^s)e_t=e_t$ exactly when $t{\mathrm{\mid }}_F=s{\mathrm{\mid }}_F$, and otherwise it is zero. Thus, as $F$ increases over finite subsets ordered by inclusion, $\pi (P_F^s)$ decreases strongly to the rank-one projection $E_s:=\mathrm{\mid }{e}_s⟩⟨e_s\mathrm{\mid }$: each basis vector $e_t$ with $t\mathrm{\ne }s$ eventually fails at a site where it differs from $s$, and $e_s$ remains fixed. Since all these projections have norm at most one and the finite span of the basis is dense, convergence on basis vectors extends to strong convergence on every vector.

Let $k_s:=\mathrm{\mid }\{x:s_x\mathrm{\ne }0\}\mathrm{\mid }$. Once $F$ contains this finite support, the product-state expectation is ${\omega }_r(P_F^s)={\prod }_{x\in F}⟨s_x,r{s}_x⟩=({\prod }_{x:s_x\mathrm{\ne }0}⟨s_x,r{s}_x⟩)p_0^{\mathrm{\mid }F\mathrm{\mid }-k_s}\mathrm{.}$ Because $0<p_0<1$ and $\mathrm{\Gamma }$ is infinite, this tends to zero along the directed set of finite regions: for every $N$ all sufficiently large $F$ have $\mathrm{\mid }F\mathrm{\mid }-k_s\ge N$.

Suppose a normal state $\tilde{\omega }$ on $\mathcal{B}({\mathcal{H}}_{\mathrm{v}\mathrm{a}\mathrm{c}})$ extended ${\omega }_r$. The complements $Q_F:=I-\pi (P_F^s)$ are increasing projections. The projection onto the closure of the union of their increasing ranges is their least upper bound: it dominates each $Q_F$, and any upper bound dominates that closed span. Strong convergence in step 1.1 identifies this supremum as $I-E_s$. Normality and step 1.2 give $\tilde{\omega }(I-E_s)={\sup }_F\tilde{\omega }(I-\pi (P_F^s))={\sup }_F(1-{\omega }_r(P_F^s))=1,$ so $\tilde{\omega }(E_s)=0$ for every $s\in S_{\mathrm{f}\mathrm{i}\mathrm{n}}$. For finite $J\subseteq S_{\mathrm{f}\mathrm{i}\mathrm{n}}$, the projections $R_J:={\sum }_{s\in J}{E}_s$ increase to $I$ as $J$ ranges over finite subsets, because $(e_s)$ is a complete orthonormal basis. Applying normality again gives $1=\tilde{\omega }(I)={\sup }_J\tilde{\omega }(R_J)=0,$ a contradiction. Therefore no normal extension exists.

Suppose instead a positive trace-class density $\rho$ represented an extension. By [F4], for each $s$ and finite $F$, $\mathrm{Tr}(\rho \pi (P_F^s))={\sum }_j{\lambda }_j⟨\pi (P_F^s)u_j,u_j⟩\mathrm{.}$ For every fixed $N$, strong convergence in step 1.1 makes the first $N$ terms converge to ${\sum }_{j\le N}{\lambda }_j⟨E_s{u}_j,u_j⟩$. The tails of both the $\pi (P_F^s)$ sum and its proposed $E_s$ limit are bounded by ${\sum }_{j>N}{\lambda }_j$, because each operator is a projection. Hence their tail difference is at most $2{\sum }_{j>N}{\lambda }_j$, which tends to zero; therefore $\mathrm{Tr}(\rho \pi (P_F^s))\to \mathrm{Tr}(\rho E_s)$. Step 1.2 then gives $\mathrm{Tr}(\rho E_s)=0$ for each $s$. The same finite-tail argument applied to $R_J\uparrow I$ gives $\mathrm{Tr}(\rho I)={\sup }_J\mathrm{Tr}(\rho R_J)=0$, contradicting $\mathrm{Tr}\rho =1$. Thus no such trace-class density exists.