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

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.