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Exact finite-ring correlations and limiting correlation length at a chosen coupling

Physical scope

Zero-field finite periodic nearest-neighbor Ising ring at the exact parameter choice beta J=(log 2)/2; no observed magnetic data.

Example

For any finite $N \geq 2$ at zero field, choose coupling $J > 0$ in joules and inverse temperature $β = (\log 2) / (2 J)$ in J $^{- 1}$ . Then $t = \tanh (β J) = 1 / 3$ and the exact ring correlation at separation $1 \leq r < N$ is $E_{N} [s_{i} s_{i + r}] = \frac{3^{- r} + 3^{- (N - r)}}{1 + 3^{- N}} .$ At fixed $r$ , the limit is $3^{- r} = e^{- r \log 3}$ , so the correlation length is $1 / \log 3$ lattice spacings.

Facts & Assumptions

Given: The finite zero-field ring theorem Finite periodic Ising-ring correlations and exact decay with $J > 0$ , $β > 0$ , and its indexed-bond convention.

[F1]

Finite periodic Ising-ring correlations and exact decay: with $t = \tanh (β J)$ , $C_{N} (r) = (t^{r} + t^{N - r}) / (1 + t^{N})$ and, for $0 < t < 1$ , $ξ = - 1 / \log t$ .

[A1]

Set $β J = (\log 2) / 2$ in the finite model.

Verification

1.1A1algebra

Since $\tanh (x) = (e^{2 x} - 1) / (e^{2 x} + 1)$ , the choice [A1] gives $t = (2 - 1) / (2 + 1) = 1 / 3$ .

2.1F1step 1.1algebra∎

Substituting $t = 1 / 3$ in [F1] gives the displayed finite- $N$ correlation. Its fixed- $r$ limit is $3^{- r}$ by the theorem's error bound, and $3^{- r} = e^{- r \log 3}$ gives $ξ = 1 / \log 3$ .

Remarks

This is a finite-model calculation. The displayed coupling is a chosen parameter and is not an experimentally fitted value.

Exact finite-ring correlations and limiting correlation length at a chosen coupling

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources