Spin decimation

Spin decimation sums over spins at selected lattice sites while retaining the others. For alternate-site elimination in a nearest-neighbour periodic chain of even length, the coarse spin-chain transfer matrix is $W^2$, because its $(s,t)$ entry sums $W_{sa}{W}_{at}$ over the eliminated middle spin. Exact preservation $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$ requires keeping the scalar normalization as well as the entry ratios.

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.

Checkerboard spin decimation retains one parity sublattice of a square-lattice Ising model and sums over the other. The retained primitive vectors are $a(1,1)$ and $a(1,-1)$, so the coarse lattice is again square with spacing $\sqrt{2}a$. A rotation and length rescaling restore the original geometry. Every original nearest-neighbour bond joins the two parity classes, whereas a diagonal next-nearest-neighbour bond stays within one class.

For dimensionless nearest-neighbour coupling $K$ and diagonal coupling $L$, the second cumulant in checkerboard decimation of the square-lattice Ising model generates $K^2$ for each pair of retained neighbours of an eliminated centre. A retained diagonal pair shares two centres, and an axial pair at distance $2a$ shares one. The displayed two-coupling recursion neglects $K^4$, $K^{2}L$, and $L^2$, treating $L$ as order $K^2$. Its finite nontrivial renormalization-group fixed point is $(K,L)=(1/3,1/9)$. Higher orders generate additional operators, so the closure is a property of the truncation.

For the spin inversion symmetry parameterization $W=c\left(\begin{array}{ccc}1& x& y\\ x& z& x\\ y& x& 1\end{array}\right)$ with positive entries, spin decimation gives the same form with $x^{\prime }=x(1+y+z)/(1+x^2+y^2)$, $y^{\prime }=(x^2+2y)/(1+x^2+y^2)$ and $z^{\prime }=(z^2+2x^2)/(1+x^2+y^2)$. Multiply $W$ by itself and divide by its $(+,+)$ entry to prove these ratios. The leftover scalar is an additive free-energy coupling; dropping it leaves normalized expectations unchanged but loses the full free energy.