 Solution

Because $S_{\mathrm{tot}}^z=N/2-N_f$, the exact finite-size answer is $m_N=1/2-N_f^{\mathrm{gs}}/N$, with $N_f^{\mathrm{gs}}$ determined by the parity-consistent filling above. For the thermodynamic magnetization of the quantum XX chain, the occupied fraction is the length of the negative-energy interval divided by the Brillouin-zone length:ThereforeThe zero-field limit has zero magnetization and the high-field limit has all spins up. The finite-ring curve consists of magnetization steps approaching this continuous limit. Strictly above $h_c$ the fully polarized ground state is unique; exactly at $h_c$ a zero-energy spin-flip state can be degenerate with it. This finite-size qualification does not alter the saturation field or limiting magnetization density.