Exact zero-field Ising-chain decimation ($\tanh K^{\prime }=(\tanh K{)}^b$)

Keep every $b$th spin in a ferromagnetic nearest-neighbour Ising model at zero field, with coupling $K=J/(k_{B}T)$. Multiplying its two-state transfer matrix $b$ times yields the displayed recursion. For $b=2$, direct summation of one intermediate spin gives $2\cosh [K(s+s^{\prime })]=A{e}^{K^{\prime }s{s}^{\prime }}$, with $K^{\prime }=\frac{1}{2}\log \cosh 2K$. The new spacing is $ba$ before coordinate rescaling. All finite positive couplings flow toward the disordered fixed point $K=0$; the ordered zero-temperature endpoint is unstable to a nonzero domain-wall number per bond.

Use the symmetric transfer matrix $T_{s{s}^{\prime }}=\exp [Ks{s}^{\prime }+h(s+s^{\prime })/2]$ with dimensionless field $h$. For blocking length two set $W=T^2$, then write $W_{s{s}^{\prime }}=A\exp [K^{\prime }s{s}^{\prime }+h^{\prime }(s+s^{\prime })/2]$. Matching its three independent entries proves the title relations and gives $A=(W_{++}{W}_{--}{W}_{+-}^2{)}^{1/4}$. Explicitly $W_{\pm \pm }=e^{2K\pm 2h}+e^{-2K}$ and $W_{+-}=2\cosh h$. At zero field the exact chain recursion is recovered; at $K=0$ the independent-spin field stays unchanged. Splitting the field equally between neighboring bonds is essential for this normalization.

A reversed bond costs $2J$, so its Boltzmann factor $w=e^{-2J/(k_{B}T)}$ is a natural thermal coordinate near the zero-temperature chain fixed point. Exact length-two decimation gives $w^{\prime }\simeq 2w$, and the One-dimensional Ising correlation function gives $\xi /a=-1/\log \tanh K\sim 1/(2w)$. This is a length exponent one with respect to $w$. The correlation length instead diverges exponentially in inverse physical temperature; claiming a conventional finite power $T^{-1}$ would confuse scaling coordinates. Choosing $e^{-K}$ gives multiplier $\sqrt{2}$ and power two in that alternative coordinate, the same physical length.

For a periodic even chain, eliminating alternate spins gives $Z_N(K)=A^{N/2}{Z}_{N/2}(K^{\prime })$. Thus dimensionless free energy per original spin obeys $f(K)=-\frac{1}{2}\log A+\frac{1}{2}f(K^{\prime })$. The exact solution $f(K)=-\log (2\cosh K)$ satisfies this relation. The multiplicative constant cancels from normalized retained-spin probabilities but records eliminated-spin entropy; discarding it loses thermodynamics even if the interaction recursion is correct.