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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.

Keep $N$ fixed and put $q=\tanh K$, so $q\to 1^{-}$ as temperature tends to zero. For even $N$, the finite-ring antiferromagnetic Ising correlation becomesBoth alternating ground states have $S_1{S}_{n+1}=(-1{)}^n$, so their equal-weight canonical ensemble has the same two-point correlation function despite its zero one-point correlation function.

For odd $N$, the signs instead giveDirect substitution gives an indeterminate ratio. Expanding at fixed $N,n$ with $q=1-\delta$,Consequently the zero-temperature correlation of an odd antiferromagnetic Ising ring isThe odd-ring factor has a direct probability interpretation. At zero temperature, the canonical ensemble gives equal weight to its $2N$ ground states. The single frustrated bond in an antiferromagnetic Ising ring is uniformly located, with probability $n/N$ of lying on the chosen $n$-bond path. If it is absent from that path, $S_1{S}_{n+1}=(-1{)}^n$; if it is present, this product is $-(-1{)}^n$. Averaging gives $(-1{)}^n[(1-n/N)-n/N]$.

Thus the decay with separation here comes from averaging a movable, compulsory frustrated bond in an antiferromagnetic Ising ring, even after additional thermally excited frustrated bonds in an antiferromagnetic Ising ring disappear. In particular $C_1=-1+2/N$ for odd $N$, and $JN{C}_1=-J(N-2)$ reproduces the ground-state energy. The formula also respects $C_n=C_{N-n}$, although its factor $1-2n/N$ changes sign halfway around the ring.

The order and scale of limits matter. Taking the thermodynamic limit first at positive temperature gives the exponential function in part (c). Cooling an odd finite ring first leaves the linear factor $1-2n/N$. Both orders subsequently give $(-1{)}^n$ at fixed $n$, but this agreement is not uniform in separation: when $n/N$ stays nonzero, the odd-ring linear factor survives in the zero-temperature expression.

Set $K=\beta J$. The transfer matrix for the one-dimensional Ising model assigns a Boltzmann factor to each bond. In the Ising spin order $(+1,-1)$, it isMultiplying these bond factors and summing every Ising spin implements the canonical ensemble. The periodic boundary conditions close the product into a trace:The eigenvectors $|+⟩=(1,1{)}^T/\sqrt{2}$ and $|-⟩=(1,-1{)}^T/\sqrt{2}$ form an orthonormal basis, with eigenvaluesThe negative second eigenvalue encodes alternating correlations. Defining $q={\lambda }_{-}/{\lambda }_{+}=-\tanh K$, the exact partition function and total Helmholtz free energy areThe parity-dependent term must be retained for finite $N$; in particular it nearly cancels the leading term of $Z_N$ at low temperature when $N$ is odd.

Represent insertion of one Ising spin by $\Sigma =\mathrm{diag}(1,-1)$. In the eigenbasis of $T$, this matrix swaps $|+⟩$ and $|-⟩$. Its diagonal entries in that eigenbasis vanish, giving the exact one-point correlation functionEquivalently, configurations pair with their opposites under spin inversion symmetry, with the same Boltzmann factor and opposite $S_1$.

Two insertions separated by $n$ bonds give the two-point correlation functionThus the exact finite antiferromagnetic Ising-ring correlation isThe two numerator terms represent the two arcs between the sites. The formula has $C_N(0)=1$ and $C_N(n)=C_N(N-n)$, as required by the periodic boundary conditions. Since the one-point correlation function is zero, this is also the connected correlation function. At $K=0$, every Ising spin is independent, so $C_N(n)=0$ for distinct sites and $C_N(0)=1$; powers with exponent zero are understood as one.

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.