= Two-spin quenched and annealed free energies
{title2=$F_{\rm a}\leq F_{\rm q}$}
Choose two <Ising spins> with <Hamiltonian> $H=-Js_1s_2$, where $J$ independently takes $J_0$ or $2J_0$ 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 $Z_J=4\cosh(\beta J)$. Therefore
$$
F_{\rm q}=-\frac{k_BT}{2}\left\{\log[4\cosh(\beta J_0)]+\log[4\cosh(2\beta J_0)]\right\},\qquad
F_{\rm a}=-k_BT\log\left\{2[\cosh(\beta J_0)+\cosh(2\beta J_0)]\right\}.
$$
The <concave function> $\log$ puts $F_{\rm a}\leq F_{\rm q}$. For this two-value case the claim is just $(Z_1+Z_2)/2\geq\sqrt{Z_1Z_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.
\Image[/quenched-annealed-two-spin-free-energy.png]
{title=Exact quenched and annealed averages for a stipulated two-spin system}
{description=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=quenched-annealed-two-spin-free-energy.py}
In the <annealed disorder> ensemble, the stronger bond has probability $Z_{2J_0}/(Z_{J_0}+Z_{2J_0})$, approaching one at low <temperature>. A <quenched disorder> collection instead retains one half at each coupling. As $T\to0$, \b[$F_{\rm q}\to-3J_0/2$ and $F_{\rm a}\to-2J_0$]: averaging samples and allowing the material variables to equilibrate describe different physics.
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