1D Ising Renormalization-Group Flow

Decimate every other spin of the 1D Ising chain and the remaining spins obey the same model with renormalized couplings. The transformation is exact: $K'=\tfrac14\ln\frac{\cosh(2K+h)\cosh(2K-h)}{\cosh^2 h}$ , $h'=h+\tfrac12\ln\frac{\cosh(2K+h)}{\cosh(2K-h)}$ , with $K=\beta J$ , $h=\beta H$ . Iterating this map is the renormalization group. The plane is $(u=\tanh K,\;h)$ , so $K=0$ (infinite temperature) is the left edge and $K\to\infty$ (zero temperature) the right. Every trajectory flows left into the disordered sink at the origin: there is no finite-temperature fixed point, the exact statement that the 1D Ising chain has no phase transition. The only critical point sits at $T=0$ ( $u=1$ ), and it is unstable. Summing the per-step free-energy constants reconstructs the exact transfer-matrix free energy.

Figure 1. Exact decimation RG of the 1D Ising chain in a field: the flow plane (tanh K, h), the spin-chain decimation cascade, or the zero-field cobweb; all flow to the disordered sink (no finite-T transition). Method: exact real-space decimation (b = 2); free energy reconstructed from the per-step constants; representative seeded chains at each renormalized coupling.

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start h0.00

RG steps9

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WHAT TO TRY

Drag the starting coupling K and watch the RG flow: decimating every other spin sends K toward the stable T = infinity sink, the 1D Ising chain has no finite-temperature ordered phase.
Switch to the spin-chain decimation view: every other spin is integrated out and the survivors obey the same model with a renormalized coupling K-prime, the exact recursion the flow plane plots.
Add a field h and follow the flow plane: the trajectories curve, and the only fixed points are the trivial sink and the unstable T = 0 point, which is why 1D never orders.