= Spin-1 chain decimation recursion
{title2=$W\longmapsto W^2$}
For the <spin inversion symmetry> parameterization $W=c\begin{pmatrix}1&x&y\\x&z&x\\y&x&1\end{pmatrix}$ with positive entries, <spin decimation> gives the same form with $x'=x(1+y+z)/(1+x^2+y^2)$, $y'=(x^2+2y)/(1+x^2+y^2)$ and $z'=(z^2+2x^2)/(1+x^2+y^2)$. 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>.
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