One-dimensional periodic Ising-chain limit

For each configuration, the product ${\prod }_{i=1}^N{T}_{s_i,s_{i+1}}$ has bond contribution $e^{\beta J{\sum }_i{s}_i{s}_{i+1}}$ and field contribution $e^{\beta h{\sum }_i(s_i+s_{i+1})\mathrm{/}2}=e^{\beta h{\sum }_i{s}_i}$. These factors equal $e^{-\beta H_N(s)}$. Summing over the finite spin sequences gives $Z_N={\sum }_{s_1,\dots ,s_N}{T}_{s_1{s}_2}\cdots T_{s_N{s}_1}=\mathrm{tr}(T^N)$.

In the ordered basis $+1,-1$, $T=\left(\begin{array}{cc}{e}^{\beta J+\beta h}& e^{-\beta J}\\ e^{-\beta J}& e^{\beta J-\beta h}\end{array}\right)$. Put $A=e^{\beta J}\cosh (\beta h)$ and $B=\sqrt{e^{2\beta J}{\sinh }^2(\beta h)+e^{-2\beta J}}$. The diagonal entries give $\mathrm{tr}T=2A$; direct multiplication gives $\det T=e^{2\beta J}-e^{-2\beta J}$. Using ${\cosh }^{2}x-{\sinh }^{2}x=1$, which follows by expanding their definitions, $A^2-\det T=e^{2\beta J}{\sinh }^2(\beta h)+e^{-2\beta J}=B^2$. Thus the characteristic polynomial has roots $A\pm B={\lambda }_{\pm }$. Since $A,B>0$, $A+B>\mathrm{\mid }A-B\mathrm{\mid }$, hence ${\lambda }_{+}>\mathrm{\mid }{\lambda }_{-}\mathrm{\mid }$.

For a $2\times 2$ matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, direct multiplication gives $M^2-(a+d)M+(ad-bc)I=0$. Apply this identity to $T$ and multiply by $T^n$; taking traces gives $z_{n+2}=(\mathrm{tr}T)z_{n+1}-(\det T)z_n$, where $z_n=\mathrm{tr}(T^n)$, $z_0=2$, and $z_1={\lambda }_{+}+{\lambda }_{-}$. The sequence ${\lambda }_{+}^n+{\lambda }_{-}^n$ has the same recurrence and initial values, so induction gives $Z_N=z_N={\lambda }_{+}^N+{\lambda }_{-}^N$.

Set $r={\lambda }_{-}\mathrm{/}{\lambda }_{+}$. Then $Z_N={\lambda }_{+}^N(1+r^N)$ and $\mathrm{\mid }r\mathrm{\mid }<1$, so $1+r^N>0$. Therefore $N^{-1}\log Z_N=\log {\lambda }_{+}+N^{-1}\log (1+r^N)$, whose second term tends to zero. For a compact parameter set, continuity and $\mathrm{\mid }r\mathrm{\mid }<1$ give $0<\rho <1$ with $\mathrm{\mid }r\mathrm{\mid }\le \rho$. The positive radicand in $B$ is a smooth positive combination of exponentials; the real-power derivative in [F4] with exponent $1\mathrm{/}2$, followed inductively by chain and product rules, makes $B$ and ${\lambda }_{\pm }$ smooth. Since ${\lambda }_{+}>0$, its reciprocal and logarithm are smooth as well, by the real-power derivative with exponent $-1$ and the logarithm derivative in [F4]. Every fixed finite collection of derivatives of $r$ is therefore bounded on the compact set. For a multi-index $\alpha$ of order at most $k$ and $N\ge k$, repeated product differentiation of the $N$ factors gives at most $C_k{N}^k$ terms; each has at least $N-k$ undifferentiated factors $r$, so $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }(r^N)\mathrm{\mid }\le C_k{N}^k{\rho }^{N-k}$. Put $u=r^N$. Since $1+u\ge 1-\rho$, repeated differentiation of $\log (1+u)$ gives finite sums of products of derivatives of $u$ divided by powers of $1+u$. Each term has at least one derivative factor and the product bounds yield $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }\log (1+u)\mathrm{\mid }\le C_k^{\mathrm{\prime }}{N}^k{\rho }^{N-k}$ after enlarging the constant. Divide by $N$; [F5] implies $N^{k-1}{\rho }^{N-k}\to 0$. The zero-order logarithm term also tends uniformly to zero. Hence the correction and every fixed-order derivative vanish uniformly on the compact set.