Stationarity alone does not give quantum thermal equilibrium

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.