= Frustrated Ising triangle
{title2=$H=J(s_1s_2+s_2s_3+s_3s_1),\quad J>0$}
Take three <Ising spins> $s_i\in\{-1,1\}$, with an <antiferromagnetic> interaction of <energy> scale $J$ on each <edge> of a triangle. An individually satisfied bond has $s_is_j=-1$. But multiplying all three bond products gives $(s_1s_2)(s_2s_3)(s_3s_1)=1$, whereas three satisfied bonds would give $-1$. At least one bond must therefore remain unsatisfied. This is an exactly finite example of <geometrical frustration>.
There are six <ground states> with one unsatisfied bond and \b[ground-state energy $-J$]. The two all-aligned <microstates> have <energy> $3J$. No interaction is switched off in a <ground state>: the system minimizes their sum by accepting one local disappointment. Allowing continuous three-component <spins> instead of <Ising spins> changes this optimization problem and permits the $120$-degree arrangement described under <geometrical frustration>.
\Image[/frustrated-ising-triangle-states.png]
{title=An antiferromagnetic Ising triangle and its complete finite energy spectrum}
{description=The blue solid bonds are satisfied and the orange dashed bonds are unsatisfied. The histogram counts all eight configurations of the stipulated three-spin model. It contains no experimental measurements.}
{source=frustrated-ising-triangle-states.py}
Writing $\beta=1/(k_BT)$ for <inverse temperature>, the exact <partition function> and mean <internal energy> are
$$
Z=6e^{\beta J}+2e^{-3\beta J},\qquad
\langle H\rangle=\frac{-6Je^{\beta J}+6Je^{-3\beta J}}{6e^{\beta J}+2e^{-3\beta J}}.
$$
At $T>0$ these expressions are smooth: \b[a finite triangle has no thermodynamic phase transition]. In zero <magnetic field>, <spin inversion symmetry> also gives $\langle s_i\rangle=0$ at every site, even as the <Boltzmann distribution> concentrates on the six frustrated <ground states>.
Back to article page
