 Solution

At zero field the normalized transfer matrix for a classical spin chain eigenvectors and eigenvalues areThe spin insertion exchanges the two eigenvectors: ${\sigma }^z|+⟩=|-⟩$ and conversely. Evaluating the trace in c therefore gives, with $x={\lambda }_{-}/{\lambda }_{+}=-\tanh K$,This is the finite-ring result. In the thermodynamic limit, or for $1\ll r\ll N$ away from the wrap-around image, it becomesHere $\xi$ is in lattice-site units. The alternating sign is antiferromagnetic order at short distances; the exponential envelope shows the absence of long-range order at every positive temperature in this one-dimensional finite-range model. As $T\to 0$, $\xi$ diverges, asymptotically $\xi \sim \frac{1}{2}{e}^{2\beta J}$. For the even ring the zero-temperature result is $g(r)=(-1{)}^r$, consistent with the two alternating ground configurations. At fixed finite $N$, “large distance” must still respect periodicity: $g_N(r)=g_N(N-r)$ rather than decaying indefinitely beyond the ring.