= Solution
In a fixed fermion-parity sector use lattice spacing one and the unitary <discrete Fourier transform>
$$
c_n=\frac1{\sqrt N}\sum_k e^{ikn}c_k,\qquad
c_k=\frac1{\sqrt N}\sum_n e^{-ikn}c_n.
$$
The 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 $\sum_k2\cos k\,c_k^\dagger c_k$ and $\sum_kc_k^\dagger c_k$. Thus
$$
\boxed{H=E+\sum_k\epsilon_k c_k^\dagger c_k,\qquad
E=-hN/2,\qquad\epsilon_k=h-J\cos k.}
$$
Here $E$ 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 limit>
$$
|\mathrm{gs}\rangle=\prod_{|k|<k_F}c_k^\dagger|0\rangle,
\qquad k_F=\arccos(h/J),\quad 0<h<J.
$$
This 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 $|\epsilon_k|$. Compare the two sector energies. A zero-energy mode allows degenerate occupations. This restores the physical <spin> <Hilbert space> without changing the stated band.
\Image[/past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2015/paper-1-xx-band.png]
{title=Original XX-chain dispersion, occupied interval and thermodynamic magnetization; illustrative field ratios}
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