Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2025 / 1 / a

Each neighboring pair of Ising spins contributes at least $-J$ to the energy, with equality exactly when the two Ising spins are opposite. For even $N$, alternation closes consistently under the periodic boundary conditions. The ground states are thereforeHere $g_0$ is the ground-state degeneracy. Choosing $S_1$ fixes every other Ising spin, so these are all the ground states; translations do not give additional spin configurations. The two ground states are exchanged by spin inversion symmetry.

Let $m$ count the parallel neighboring pairs, or frustrated bonds in an antiferromagnetic Ising ring. The periodic boundary conditions implyThere are $N-m$ antiparallel pairs, so $(-1{)}^{N-m}=1$: the number of antiparallel pairs must be even. For odd $N$, $m$ must consequently be odd. Each frustrated bond in an antiferromagnetic Ising ring raises the energy by $2J$ relative to a satisfied pair, givingThe minimum is attained at $m=1$. Choose its position among the $N$ labeled bonds, choose either orientation of one Ising spin, then alternate across every other bond. This constructs exactly two spin configurations per position. ThusThis is geometric frustration: every local interaction prefers opposite Ising spins, but the closed odd ring prevents all preferences being satisfied at once. No particular location is preferred for the single frustrated bond in an antiferromagnetic Ising ring.