Past exam of the field theories and collective phenomena in condensed matter course of the University of Oxford / 2025 / 2 / d / Solution

Put $r=\tanh K$, so $r\to 1^{-}$ as the temperature tends to zero. For even $N$, the finite antiferromagnetic Ising-ring correlation becomesThe two perfectly alternating ground states give exactly the same product $S_1{S}_{n+1}=(-1{)}^n$; averaging them preserves its magnitude even though the one-point correlation function remains zero.

For odd $N$ the signs instead giveBoth numerator and denominator vanish. Set $r=1-\delta$ and expand at fixed $N,n$:HenceThere is a direct ground state interpretation. The mandatory parallel bond of an odd antiferromagnetic Ising ring is uniformly distributed among its $N$ bonds in the limiting canonical ensemble. Along the directed $n$-bond arc from site 1 to site $n+1$, the product is $(-1{)}^n$ if the defect lies outside that arc, and $-(-1{)}^n$ if it lies inside. The respective probabilities are $1-n/N$ and $n/N$, giving the displayed two-point correlation function. Global spin inversion symmetry doubles the number of ground states but leaves each product unchanged.

The reduction on an odd ring survives at separations comparable with $N$, despite the vanishing defect density $1/N$. For fixed $n$, however, taking $N\to \infty$ after the zero-temperature limit recovers $(-1{)}^n$ for either parity, consistently with the diverging correlation length. The linear envelope at finite odd $N$ comes from one freely located frustration defect; it is distinct from the exponential decay caused by thermal defects in the thermodynamic limit.