 Solution

Write $K=\beta J$, where $\beta =1/(k_{B}T)$ is the inverse temperature, $T$ is temperature and $k_B$ is the Boltzmann constant. In the canonical ensemble, each spin configuration has probability $e^{-\beta H}/Z_N$. The nearest-neighbor Boltzmann factors multiply, so the spin-chain transfer matrix, in the order $+1,-1$, isMultiplying its entries around the ring and summing over every Ising spin yields the partition function:The trace implements the periodic boundary conditions; replacing it by an endpoint sum would describe a different partition function. The normalized eigenvectors and corresponding eigenvalues areThe negative eigenvalue is essential for the antiferromagnetic Ising model. Introduce their ratioThe exact finite-$N$ partition function and Helmholtz free energy areFor finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of geometric frustration.

To insert an Ising spin into this trace, use the diagonal matrix $\mathsf{S}=\mathrm{diag}(1,-1)$. In the eigenbasis of $\mathsf{T}$, it exchanges $|+⟩$ and $|-⟩$, and hence has vanishing diagonal entries. The one-point correlation function is thereforeThe same conclusion follows directly from spin inversion symmetry: reversing every Ising spin preserves the energy and the Boltzmann factor while reversing $S_1$.

For the two-point correlation function, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. ThusIn particular, the exact finite-ring antiferromagnetic Ising correlation isBecause the one-point correlation functions vanish, this is also the connected correlation function. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct Ising spins are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.