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

Assume $Q\ge 2$ and $L\ge 2$; the trivial exceptions are described below. In this periodic Potts chain, every satisfied bond contributes $-J$ to the energy, and no bond can contribute less. All $L$ bonds can be satisfied simultaneously by choosing the same state at every site. Conversely, satisfying every bond forces this uniform configuration. Therefore the ground state energy and degeneracy factor areThe $Q$ choices are distinct configurations on the labelled lattice: a global relabelling of the states is a symmetry, not an identification of microstates.

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.

The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix byHere ${\delta }_{ab}$ is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition functionHence, writing $I$ for the identity matrix and $P$ for the all-ones matrix,Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. ThusMore explicitly, for $u=(1,\dots ,1{)}^T$, $Pu=Qu$. If ${\sum }_a{v}_a=0$, then $Pv=0$. These complementary eigenspaces have dimensions $1$ and $Q-1$. Equivalently, $P/Q$ is the orthogonal projector onto the linear span of $u$. The eigenvalues and their eigenvalue multiplicities are thereforeSince the two matrices share these eigenvectors, the exact finite-ring canonical partition function isBoth terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

Put $x=e^{-\beta J}$. Factoring out the ground state Boltzmann factor from the exact canonical partition function givesAt fixed finite $L$, the binomial theorem givesTaking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,The constant term is the ground state energy. The term $-{\beta }^{-1}\log Q$ is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:Each pair of domain walls costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of domain walls, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta )$.

This is an expansion for fixed $L$, not a uniform thermodynamic limit. In particular, the small correction requires $\left(\genfrac{}{}{0ex}{}{L}{2}\right)(Q-1)x^2\ll 1$, with higher terms also controlled as $x\to 0$ at fixed $L,Q$. Taking $L\to \infty$ first instead selects ${\lambda }_{+}$ and givesIts leading correction is of order $x$. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.