Two-spin quenched and annealed free energies ($F_{\mathrm{a}}\le F_{\mathrm{q}}$)

Choose two Ising spins with Hamiltonian $H=-J{s}_1{s}_2$, 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 $Z_J=4\cosh (\beta J)$. ThereforeThe concave function $\log$ puts $F_{\mathrm{a}}\le F_{\mathrm{q}}$. For this two-value case the claim is just $(Z_1+Z_2)/2\ge \sqrt{Z_1{Z}_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.

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 $T\to 0$, $F_{\mathrm{q}}\to -3J_0/2$ and $F_{\mathrm{a}}\to -2J_0$: averaging samples and allowing the material variables to equilibrate describe different physics.