 Solution

Let $S=\mathrm{diag}(1,-1)$ be the base-spin insertion matrix. In the eigenbasis from solution, its diagonal matrix elements vanish: $⟨+|S|+⟩=⟨-|S|-⟩=0$. Hencenot only at large $L$ but for every finite ring at zero field.

This vanishing is not merely an averaging artifact hiding a finite-temperature ordered phase. The two-point correlation function is $⟨{\sigma }_1{\sigma }_{1+r}⟩=({\lambda }_{-}^r{\lambda }_{+}^{L-r}+{\lambda }_{+}^r{\lambda }_{-}^{L-r})/Z_L$. In the thermodynamic limit it becomes $({\lambda }_{-}/{\lambda }_{+}{)}^r$, with finite correlation length ${\xi }^{-1}=\log ({\lambda }_{+}/{\lambda }_{-})$ for every $T>0$ and $J>0$. Thermal domain walls destroy long-range alignment in this one-dimensional short-range system. At zero temperature, the two fully aligned ground states are degenerate; that limiting case must be distinguished from any fixed positive temperature.