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Square-lattice Ising critical coupling from the two-state Potts model

Example

On $Λ_{n} = [- n, n]^{2} \cap Z^{2}$ , let $E_{n}$ be the nearest-neighbor edges with both endpoints in $Λ_{n}$ and fix plus boundary spins $σ_{x} = + 1$ whenever $∥ x ∥_{\infty} = n$ . For $σ_{x} \in {- 1, + 1}$ , define the finite-volume Ising Hamiltonian and Gibbs law by $H_{I, n} (σ) = - J_{I} \sum_{{x, y} \in E_{n}} σ_{x} σ_{y}, P_{β, n}^{I, +} (σ) \propto e^{- β H_{I, n} (σ)}, J_{I} > 0.$ Define $m_{I}^{+} (β) : = \lim_{n \to \infty} P_{β, n}^{I, +} (σ_{0} = + 1) - 1 / 2$ when the limit exists, and $β_{c}^{I} : = \inf {β > 0 : m_{I}^{+} (β) > 0}$ . The limit exists and $β_{c}^{I} J_{I} = \frac{1}{2} \log (1 + \sqrt{2}) .$

Example references

Square-lattice Potts critical inverse temperature

Facts & Assumptions

Given: The Ising interaction above, $J_{I} > 0$ , and the square-lattice $Q = 2$ Potts critical-temperature theorem.

[F1]

For $σ_{x}, σ_{y} \in {- 1, + 1}$ , $1_{{σ_{x} = σ_{y}}} = (1 + σ_{x} σ_{y}) / 2$ by checking the two possible products $+ 1$ and $- 1$ .

[F2]

For the two-state Potts model with energy $- J_{P} \sum_{{x, y}} 1_{{τ_{x} = τ_{y}}}$ , its critical inverse temperature $β_{c}^{P}$ satisfies $β_{c}^{P} J_{P} = \log (1 + \sqrt{2})$ (Square-lattice Potts critical inverse temperature).

Verification

technique · direct coupling conversion

1.1F1algebra

For each finite box and edge, [F1] gives $- J_{P} 1_{{σ_{x} = σ_{y}}} = - (J_{P} / 2) σ_{x} σ_{y} - J_{P} / 2$ . Setting $J_{P} = 2 J_{I}$ makes the finite Potts Hamiltonian equal to $H_{I, n} (σ) - J_{I} ∣ E_{n} ∣$ . The difference is independent of $σ$ , so it cancels from the normalized finite Gibbs probabilities. The boundary color $1$ in the Potts model corresponds to the Ising plus spin $+ 1$ . Thus the two models have the same plus-boundary probabilities and the same critical inverse temperature.

2.1F2step 1.1algebra∎

Let $β_{c}^{P}$ and $β_{c}^{I}$ denote the critical inverse temperatures in the two definitions. Step 1.1 gives $β_{c}^{I} = β_{c}^{P}$ . By [F2] with $J_{P} = 2 J_{I}$ , $β_{c}^{P} (2 J_{I}) = \log (1 + \sqrt{2})$ . Divide by $2$ to obtain $β_{c}^{I} J_{I} = \frac{1}{2} \log (1 + \sqrt{2})$ . This is an exact coupling conversion; the constant energy shift does not affect probabilities.

Remarks

The factor of two comes from $1_{{σ_{x} = σ_{y}}} = (1 + σ_{x} σ_{y}) / 2$ ; no critical exponent or transition order is inferred.

Depends on

Square-lattice Potts critical inverse temperature