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Share each on-site field energy equally between its two neighboring bonds. In the ordered basis $\sigma =+1,-1$, the symmetric transfer matrix for a classical spin chain isThe product ${\prod }_j{T}_{{\sigma }_j{\sigma }_{j+1}}$ recovers each bond and each field term once. Summing the periodic indices closes a trace, giving $Z_N=\mathrm{Tr}{T}^N$. The two eigenvalues areThe square-root argument is positive. Moreover $\det T=e^{-2K}-e^{2K}<0$, so ${\lambda }_{-}<0$ even at nonzero field. Its sign must not be discarded merely because all matrix entries are positive. The Perron–Frobenius theorem ensures that the positive eigenvalue ${\lambda }_{+}$ has the larger modulus, and the infinite-chain free energy is $f_{\infty }=-{\beta }^{-1}\log {\lambda }_{+}$. For even $N$, both terms in the finite-size partition function are positive.

Insert the two observed spins into the configuration sum for the transfer matrix for a classical spin chain trace. Since the diagonal matrix ${\sigma }^z$ has entry $s$ on state $s$, multiplication by it at positions $j$ and $j+r$ produces the factor ${\sigma }_j{\sigma }_{j+r}$. The trace representation is consequentlyThere are exactly $N$ transfer factors: $(j-1)+r+(N+1-j-r)=N$. The distance between the two insertions is $r$ bonds, not $r-1$. Cyclicity of the trace combines the outside factors and gives the equivalent expressionwhich makes independence of the starting site manifest. No assumption that $T$ commutes with ${\sigma }^z$ is used; in general it does not. For $r=0$ the two insertions multiply to the identity and $g(0)=1$, a useful normalization check.

At zero field the normalized transfer matrix for a classical spin chain eigenvectors and eigenvalues areThe spin insertion exchanges the two eigenvectors: ${\sigma }^z|+⟩=|-⟩$ and conversely. Evaluating the trace in c therefore gives, with $x={\lambda }_{-}/{\lambda }_{+}=-\tanh K$,This is the finite-ring result. In the thermodynamic limit, or for $1\ll r\ll N$ away from the wrap-around image, it becomesHere $\xi$ is in lattice-site units. The alternating sign is antiferromagnetic order at short distances; the exponential envelope shows the absence of long-range order at every positive temperature in this one-dimensional finite-range model. As $T\to 0$, $\xi$ diverges, asymptotically $\xi \sim \frac{1}{2}{e}^{2\beta J}$. For the even ring the zero-temperature result is $g(r)=(-1{)}^r$, consistent with the two alternating ground configurations. At fixed finite $N$, “large distance” must still respect periodicity: $g_N(r)=g_N(N-r)$ rather than decaying indefinitely beyond the ring.

For each pair of adjacent spins, define the transfer matrix for a classical spin chain by $T_{s{s}^{\prime }}=e^{-\beta E_0(s,s^{\prime })}$. Multiplying its entries gives the Boltzmann factor of a whole configuration. The periodic boundary condition identifies the last index with the first, soThe assumed interchange symmetry of $E_0$ makes $T$ a real symmetric matrix, hence it has an orthonormal eigenbasis. At finite positive temperature, every matrix entry is strictly positive, so the Perron–Frobenius theorem supplies a unique positive largest eigenvalue ${\lambda }_{+}$, larger in modulus than every other one. Therefore $L^{-1}\log Z_L\to \log {\lambda }_{+}$.

Local observables become matrix insertions. For example, with $S_{s{s}^{\prime }}=s{\delta }_{s{s}^{\prime }}$,Thus partition functions and correlation functions reduce to a fixed-dimensional eigenvalue problem, rather than a sum over exponentially many microstates.

A homogeneous Zimm-Bragg chain assigns propagation weight $s>0$ to each helical unit and initiation factor $\sigma >0$ to each helical run. With states ordered as coil and helix, one possible transfer matrix for a classical spin chain isIts largest eigenvalue is ${\lambda }_{+}=(1+s+\sqrt{(s-1{)}^2+4s\sigma })/2$. In the long-chain limit the partition function grows as ${\lambda }_{+}^N$, so logarithmic differentiation gives the helical fraction $f_h=s\partial \ln {\lambda }_{+}/\partial s$, equal to the displayed expression. If $Q=(s-1{)}^2+4s\sigma$, thenFor $\sigma =1$, $f_h=s/(1+s)$ and units are noncooperative. A small initiation factor suppresses boundaries and sharpens the crossover around $s=1$, with width of order $\sqrt{\sigma }$. For every fixed $\sigma >0$ the long-chain expression is smooth; a discontinuous step appears only in the singular $\sigma \downarrow 0$ limit. ^[ref]