A conserved finite current obstructs an integrable dc correlation

Each $h_{j,j+1}$ acts only on the span of $\mathrm{\mid }j⟩,\mathrm{\mid }j+1⟩$, so it commutes with every $q_k$ outside those two sites and with $q_j+q_{j+1}$. The definition [F1] therefore applies. By [F2], each bond current is self-adjoint, so $J_3=J_3^{\ast }$. Direct multiplication gives $H_3=-g(S+S^{\ast })$ and $J_3=(i{q}_{\ast }g\mathrm{/}\mathrm{\hslash })(S-S^{\ast })$. Since $S$ is unitary, $S$ commutes with $S^{\ast }$; hence $[H_3,J_3]=0$, and $J_3\mathrm{\ne }0$.

In the Fourier basis of the three-cycle, $H_3$ has eigenvalues $-2g,g,g$, while $J_3$ has eigenvalues $0,\sqrt{3}{q}_{\ast }g\mathrm{/}\mathrm{\hslash },-\sqrt{3}{q}_{\ast }g\mathrm{/}\mathrm{\hslash }$ in the corresponding order. The two states with energy $g$ have equal Gibbs weight, so ${\omega }_{\beta ,3}(J_3)=0$. Positive definiteness of ${\rho }_{\beta ,3}$ and $J_3\mathrm{\ne }0$ give $c:={\omega }_{\beta ,3}(J_3^2)=6q_{\ast }^2{g}^2\mathrm{/}({\mathrm{\hslash }}^2(e^{3\beta g}+2))>0$, with unit ${\mathsf{Q}}^2{\mathrm{s}}^{-2}$.

Because $[H_3,J_3]=0$, Heisenberg evolution fixes $J_3$ for all $t$. Thus $J_3^{\circ }=J_3$ and $C_3(t)=c\mathrm{/}3$ is a positive constant. Therefore ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }{C}_3(t)\mathrm{\mid }dt=+\mathrm{\infty }$, refuting the statement.

In the finite Mazur bound [F3], choose the single conserved operator $Q_1=J_3$. Then $G_{11}=v_1=c$, so the lower bound on the Cesàro average of the unnormalized autocorrelation $\mathrm{Re}{\omega }_{\beta ,3}({\alpha }_t(J_3)J_3)$ is $c>0$. The per-site convention in the definition gives $C_3(t)=\frac{1}{3}\mathrm{Re}{\omega }_{\beta ,3}({\alpha }_t(J_3)J_3)$, so the Mazur contribution for the Cesàro average of $C_3$ is $c\mathrm{/}3$, equal to the constant correlation found in step 3.1. This remains a finite-volume obstruction only; no thermodynamic Drude weight is inferred.