Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2023 / 1 / a / ii / Solution

Let $w$ count the unequal neighbouring pairs, or domain walls. The energy is $E=-JL+Jw$. A single domain wall is impossible with periodic boundary conditions: following all the other, equal-state bonds around the ring would force its endpoints to be equal too. Two domain walls are possible, splitting the ring into two nonempty uniform arcs carrying different states. Thus the first excited state hasChoose the two wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways. For that choice, identify one of the resulting arcs by its smallest site label. Its state has $Q$ choices and the other arc has $Q-1$. This counts each configuration exactly once, giving the two-domain-wall degeneracy of a Potts ringFor example, $Q=2,L=4$ gives $d_0=2$ and $d_1=12$; the remaining two configurations are alternating and have four domain walls. The formula also covers $L=2$, where the specified ring energy counts the two directed neighbouring bonds, and an unequal pair breaks both. If $Q=1$ or $L=1$, every allowed configuration is a ground state and there is no first excited state. For $Q\ge 3$, the number of domain walls need not be even: three distinct states on a three-site ring give three, although one remains impossible.