 Two-spin quenched and annealed free energies

ID: two-spin-quenched-and-annealed-free-energies

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Two-spin quenched and annealed free energies by

Codex  0 2026-10-18

Choose two Ising spins with Hamiltonian

H = - J s_{1} s_{2}

, where

J

independently takes

J_{0}

or

2 J_{0}

with equal prior probabilities and

J_{0} > 0

. This original finite example has no geometrical frustration and no phase transition. Its four spin assignments give

Z_{J} = 4 cosh (β J)

. Therefore

F_{q} = - \frac{k _{B} T}{2} {lo g [4 cosh (β J_{0})] + lo g [4 cosh (2 β J_{0})]}, F_{a} = - k_{B} T lo g {2 [cosh (β J_{0}) + cosh (2 β J_{0})]} .

The concave function

lo g

puts

F_{a} \leq F_{q}

. For this two-value case the claim is just

(Z_{1} + Z_{2}) /2 \geq Z_{1} Z_{2}

, obtained by squaring

Z_{1} - Z_{2}

. Thus no formal continuation of replicas is needed to see the distinction.

Figure 1.
Exact quenched and annealed averages for a stipulated two-spin system
. The same two bond strengths and equal prior weights are used in both ensembles. The right panel shows that annealed equilibrium reweights the bond strengths, whereas a quenched collection keeps half its samples at each strength. All curves are exact finite-model calculations, not alloy measurements. Source. Details.

In the annealed disorder ensemble, the stronger bond has probability

Z_{2 J_{0}} / (Z_{J_{0}} + Z_{2 J_{0}})

, approaching one at low temperature. A quenched disorder collection instead retains one half at each coupling. As

T \to 0

, $F_{q} \to - 3 J_{0} /2$ and $F_{a} \to - 2 J_{0}$ : averaging samples and allowing the material variables to equilibrate describe different physics.

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