= Solution
For $0<K<\infty$, $|q|=\tanh K<1$. Taking the <thermodynamic limit> $N\to\infty$ at fixed separation $n$ removes $q^N$ and $q^{N-n}$ from the <finite antiferromagnetic Ising-ring correlation>. Therefore
$$
\boxed{C_\infty(n)=(-\tanh K)^n=(-1)^ne^{-n/\xi},
\qquad \xi=-\frac1{\ln\tanh(\beta J)}=\frac1{\ln\coth(\beta J)}.}
$$
The <correlation length> $\xi$ is measured in lattice spacings. If the lattice spacing is $a$, its physical value is $a\xi$. The factor $(-1)^n$ expresses <antiferromagnetism>, while $e^{-n/\xi}$ measures the decay of the staggered <two-point correlation function>.
At every positive <temperature>, $\xi$ is finite. At low <temperature>, $\tanh K=1-2e^{-2K}+O(e^{-4K})$, giving $\boxed{\xi\sim\tfrac12e^{2\beta J}}$. At infinite <temperature>, $K=0$, distinct <Ising spins> are uncorrelated; the limiting <correlation length> is zero.
Back to article page
