Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2025 / 1 / c / Solution

Take the thermodynamic limit $N\to \infty$ at fixed separation $n$ and fixed finite $K=\beta J>0$. Since $|t|=\tanh K<1$, the terms $t^{N-n}$ and $t^N$ in the finite-ring antiferromagnetic Ising correlation vanish, through either even or odd $N$. ThereforeMatching the decaying exponential function gives the correlation length, in units of the site spacing:If the physical spacing is $a$, the physical correlation length is $a\xi$. The alternating sign records the preferred opposite orientation of neighboring Ising spins; the finite correlation length shows that this preference does not maintain long-range Néel order at positive temperature.

For large $K$, $\tanh K=1-2e^{-2K}+O(e^{-4K})$, so $\xi \sim e^{2K}/2$: the correlation length becomes large on cooling, but diverges only at zero temperature. At infinite temperature, $K\to 0^{+}$ and $\xi \to 0$; the separate equal-site two-point correlation function remains $C_0=1$.