Finite-excitation vacuum spin representation and normality — Examples

By the specified interaction, $H_{\mathrm{\Lambda }}=0$ for every finite $\mathrm{\Lambda }$, so each finite-volume dynamics is the identity and its thermodynamic limit is ${\alpha }_t=\mathrm{i}\mathrm{d}$. Hence ${\omega }_0({\alpha }_t(A))={\omega }_0(A)$ for every $A$ and $t$, so ${\omega }_0$ is stationary. The vector formula is a normalized positive state because $\pi$ is a unital $\ast$-representation and $\mathrm{\parallel }\mathrm{\Omega }\mathrm{\parallel }=1$.

At any site $x$, set $A=E_x^{01}$ and $B=E_x^{10}$. Then $AB=E_x^{00}$ and $BA=E_x^{11}$, so ${\omega }_0(AB)=1$ and ${\omega }_0(BA)=0$. Therefore ${\omega }_0$ is not tracial.

If ${\omega }_0$ were $\beta$-KMS, its correlation function for this pair would have the constant lower boundary $F(t)=1$. The constant function $G(z)=1$ is bounded, continuous and holomorphic on the strip with the same lower edge; [F4] forces $F=G$ throughout, so its upper edge at $t=0$ would also be $1$. But the KMS upper edge equals ${\omega }_0(BA)=0$, a contradiction. Hence this stationary state is not $\beta$-KMS for any $\beta >0$.

By [F5], the unique $\beta$-KMS state of this zero-Hamiltonian model is the tracial product state with one-site density $q^{-1}I$, which differs from ${\omega }_0$. The counterexample distinguishes stationarity from thermal equilibrium without asserting that every stationary state fails KMS.

The finite linear span of $(e_s)$ is contained in $D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$ and is dense, so $H_{\mathrm{v}\mathrm{a}\mathrm{c}}$ is densely defined. Its diagonal eigenvalues $ϵN(s)$ are real, hence $⟨H_{\mathrm{v}\mathrm{a}\mathrm{c}}\psi ,\phi ⟩=⟨\psi ,H_{\mathrm{v}\mathrm{a}\mathrm{c}}\phi ⟩$ for $\psi ,\phi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$ by Cauchy–Schwarz. Thus it is symmetric.

Define $U_t{e}_s=e^{itϵN(s)\mathrm{/}\mathrm{\hslash }}{e}_s$; this is unitary because every diagonal coefficient has modulus one. A local matrix coefficient between $e_s$ and $e_t$ can be nonzero only if $s$ and $t$ agree outside its finite support $\mathrm{\Lambda }$. In that case $N(t)-N(s)={\sum }_{x\in \mathrm{\Lambda }}({\mathbf{1}}_{\{t_x=1\}}-{\mathbf{1}}_{\{s_x=1\}})$, so conjugation by $U_t$ multiplies the coefficient by the same phase as conjugation by the finite onsite Hamiltonian $H_{\mathrm{\Lambda }}=ϵ{\sum }_{x\in \mathrm{\Lambda }}\mathrm{\mid }1⟩⟨1{\mathrm{\mid }}_x$. Thus $U_t\pi (A)U_t^{\ast }=\pi ({\alpha }_t(A))$ for every local $A$; this operator identity extends to $\mathcal{A}$ by norm continuity.

For each $x\in \mathbb{Z}$, the one-excitation configuration $s^{(x)}$ with $s_x^{(x)}=1$ and all other labels $0$ contributes $e^{-\beta ϵ}$ to $Z_{\mathrm{v}\mathrm{a}\mathrm{c}}(\beta )$. These configurations are distinct, so for any $m$ distinct sites the finite subsum is $m{e}^{-\beta ϵ}$. Since $ϵ,\beta >0$, this grows without bound as $m\to \mathrm{\infty }$, and $Z_{\mathrm{v}\mathrm{a}\mathrm{c}}(\beta )=+\mathrm{\infty }$.

If $\phi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}}^{\ast })$, then by [F2] there is $\eta \in {\mathrm{\ell }}^2(S_{\mathrm{f}\mathrm{i}\mathrm{n}})$ such that $⟨H_{\mathrm{v}\mathrm{a}\mathrm{c}}\psi ,\phi ⟩=⟨\psi ,\eta ⟩$ for every $\psi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$. Taking $\psi =e_s$ gives ${\eta }_s=ϵN(s){\phi }_s$ for every $s$, so $(ϵN(s){\phi }_s{)}_s\in {\mathrm{\ell }}^2(S_{\mathrm{f}\mathrm{i}\mathrm{n}})$ and $\phi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$. Conversely, if $\phi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$, the diagonal sums satisfy $⟨H_{\mathrm{v}\mathrm{a}\mathrm{c}}\psi ,\phi ⟩=⟨\psi ,H_{\mathrm{v}\mathrm{a}\mathrm{c}}\phi ⟩$ for all $\psi \in D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$, hence $H_{\mathrm{v}\mathrm{a}\mathrm{c}}^{\ast }\phi =H_{\mathrm{v}\mathrm{a}\mathrm{c}}\phi$. Therefore $D(H_{\mathrm{v}\mathrm{a}\mathrm{c}}^{\ast })=D(H_{\mathrm{v}\mathrm{a}\mathrm{c}})$ and the two operators agree: $H_{\mathrm{v}\mathrm{a}\mathrm{c}}$ is self-adjoint.

If $G_{\beta }$ were trace class, [F3] applied to the supplied basis $(e_s)$ would make the nonnegative diagonal sum defining $Z_{\mathrm{v}\mathrm{a}\mathrm{c}}(\beta )$ finite, contrary to step 1.3. Thus no normalized trace-class Gibbs density is obtained from $G_{\beta }$. Yet [F4] gives a unique quasi-local KMS state, and [F5] confirms that it has no normal vacuum-representation extension.

For every finite region, $q^{\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}\ge 1$, so the finite-matrix clauses in [F3] apply and discharge exactly the C*-identity and positive-functional/density hypotheses of [F1]. Together with Countable Choice [A1], $h=0=h^{\ast }$, $\beta >0$ and $\mathrm{\hslash }>0$, these meet its hypotheses. Since $e^{-\beta h}=I$ and $\mathrm{Tr}I=q$, the density is $r=q^{-1}I$, so $r^{\otimes \mathrm{\Lambda }}=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{I}_{\mathrm{\Lambda }}$, and [F1] identifies its extension as the unique $\beta$-KMS state.

If $A,B$ are local, embed them into a common finite region $\mathrm{\Lambda }$ using [F3]. Then $\omega (AB)=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(AB)=q^{-\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid }}{\mathrm{Tr}}_{\mathrm{\Lambda }}(BA)=\omega (BA),$ by finite-matrix trace cyclicity.

For arbitrary $A,B\in \mathcal{A}$, norm density from [F4] and Countable Choice [A1] give local sequences $A_n\to A$ and $B_n\to B$. Their norms are bounded, and $\mathrm{\parallel }{A}_n{B}_n-AB\mathrm{\parallel }\le \mathrm{\parallel }{A}_n-A\mathrm{\parallel }\mathrm{\parallel }{B}_n\mathrm{\parallel }+\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{B}_n-B\mathrm{\parallel }⟶0,$ with the same estimate after exchanging $A_n,B_n$ and $A,B$. Boundedness of the state then gives $\omega (A_n{B}_n)\to \omega (AB)$ and $\omega (B_n{A}_n)\to \omega (BA)$. Step 2.1 gives equality at each $n$, so $\omega (AB)=\omega (BA)$; the state is tracial on the full algebra.

The density $r=q^{-1}I$ has all eigenvalues $1\mathrm{/}q>0$, hence is full rank, and its reference diagonal entry is $p_0=1\mathrm{/}q\in (0,1)$ because $q\ge 2$. Countable Choice [A1] and the finite-matrix clauses [F3] were discharged in step 1.1 for this same onsite model. Thus every hypothesis retained in [F2] holds, and the unique tracial KMS state has no normal extension in the finite-excitation vacuum representation.