 Solution

In a fixed fermion-parity sector use lattice spacing one and the unitary discrete Fourier transformThe allowed momenta satisfy $e^{ikN}=-P$: they are $k=2\pi (\ell +1/2)/N$ for $P=+1$ and $k=2\pi \ell /N$ for $P=-1$, with $N$ distinct values modulo $2\pi$. The hopping and number sums become ${\sum }_{k}2\cos k{c}_k^{†}{c}_k$ and ${\sum }_k{c}_k^{†}{c}_k$. ThusHere $E$ is the energy of the all-up spin configuration, not generally the ground-state energy. The dispersion has minimum $h-J$ at $k=0$, maximum $h+J$ at $k=\pi$, and positive curvature at its minimum. Since the fermions do not interact, each mode can be occupied zero or one time. A negative-energy occupation lowers the many-body energy, so in the thermodynamic limitThis is a filled Fermi sea containing many fermions; it is not obtained by putting a single particle into the band minimum. For $h>J$ every mode has positive energy and the vacuum is the ground state. Its spin interpretation is complete longitudinal polarization. At $h=J$ the zero-momentum one-particle excitation first becomes gapless.

For an exact finite ring use parity-consistent filling of a free-fermion band. In each momentum grid fill negative-energy levels, check that their particle number has the chosen parity, and if necessary toggle a mode with the smallest $|{ϵ}_k|$. Compare the two sector energies. A zero-energy mode allows degenerate occupations. This restores the physical spin Hilbert space without changing the stated band.