Local spectrum and noncommuting terms

By [F4], $P_1$ and $D_{12}$ are the odd and even adjacent local generators in the stated Potts Temperley-Lieb representation. Since $P:=\mathrm{\mid }+⟩⟨+\mathrm{\mid }$, $P\mathrm{\mid }a⟩=Q^{-1\mathrm{/}2}\mathrm{\mid }+⟩$. Therefore $P_1{D}_{12}{P}_1={\sum }_{a=0}^{Q-1}(P\mathrm{\mid }a⟩⟨a\mathrm{\mid }P)\otimes \mathrm{\mid }a⟩⟨a\mathrm{\mid }=Q^{-1}{P}_1\mathrm{.}$ Both $P_1$ and $D_{12}$ are orthogonal projections. Their ranks are $Q$ in the $Q^2$-dimensional two-site space; for $Q\ge 2$ each has spectrum $\{0,1\}$.

The matrix entry from $\mathrm{\mid }00⟩$ to $\mathrm{\mid }10⟩$ is $⟨10\mathrm{\mid }H\mathrm{\mid }00⟩=-\kappa \sqrt{Q}⟨10\mathrm{\mid }{P}_1\mathrm{\mid }00⟩=-\kappa \mathrm{/}\sqrt{Q}\mathrm{\ne }0,$ because $D_{12}$ is diagonal in this basis. Hence $H$ is not diagonal.

Acting on $\mathrm{\mid }00⟩$ gives $P_1{D}_{12}\mathrm{\mid }00⟩=Q^{-1\mathrm{/}2}\mathrm{\mid }+⟩\mathrm{\mid }0⟩$ and $D_{12}{P}_1\mathrm{\mid }00⟩=Q^{-1}\mathrm{\mid }00⟩$. These vectors differ for every $Q\ge 2$, so $P_1{D}_{12}\mathrm{\ne }{D}_{12}{P}_1$.

Each eigenvalue $\mu$ of $H$ has a unit eigenvector $v$. By [F2], $H$ is self-adjoint; using the linear-first inner product from [F1], $\mu =⟨Hv,v⟩=⟨v,Hv⟩=\bar{\mu }$, so $\mu \in \mathbb{R}$. Since $\beta \mathrm{\ne }0$, multiplication $\mu \mapsto -\beta \mu$ is injective, and [F3] makes $\mu \mapsto e^{-\beta \mu }$ injective; [F3] also gives positive weights. The projections in [F1] sum to the identity, so $B_{\beta }={\sum }_{\mu }{e}^{-\beta \mu }{\mathrm{\Pi }}_{\mu }$ is positive definite: for nonzero $w$, $⟨B_{\beta }w,w⟩={\sum }_{\mu }{e}^{-\beta \mu }\mathrm{\parallel }{\mathrm{\Pi }}_{\mu }w{\mathrm{\parallel }}^2>0$. Now suppose ${\rho }_{\beta }$ were diagonal in the computational basis. By [F2], $Z_{\beta }>0$ and $B_{\beta }=Z_{\beta }{\rho }_{\beta }$, so $B_{\beta }$ would be diagonal. On the $\mu$-eigenspace of $H$, $B_{\beta }$ acts by $e^{-\beta \mu }$; these scalars are distinct for distinct real $\mu$, hence the eigenspaces of $B_{\beta }$ are exactly those of $H$. Every computational basis vector is an eigenvector of a diagonal $B_{\beta }$, and therefore would be an $H$-eigenvector. This contradicts the nonzero off-diagonal entry in step 1.2, so ${\rho }_{\beta }$ is not diagonal.

At $\beta =0$, the spectral definition in [F1] gives $B_0={\sum }_{\mu }{\mathrm{\Pi }}_{\mu }=I$, so [F2] gives ${\rho }_0=I\mathrm{/}{Q}^2$, which is diagonal. If $\kappa =0$, [F1] gives $H=-\lambda \sqrt{Q}{D}_{12}$, a diagonal matrix, so its spectral Gibbs density is diagonal as well. These boundary cases show why the nonzero assumptions are required.