 Solution

The Hubbard model combines hopping with a local repulsion,Hopping favours a broad itinerant band; the interaction penalizes opposite-spin double occupancy. For uniform Stoner decoupling of the Hubbard interaction, let $s=n_{\uparrow }-n_{\downarrow }$ be the spin imbalance per site. The interaction is $U(n^2-s^2)/4$, so polarization lowers it by $U{s}^2/4$, at the expense of kinetic energy.

If $D(E_F)$ is the noninteracting density of states per spin per site, small polarization costs $s^2/[4D(E_F)]$. Thereforeis the mean-field ferromagnetic instability criterion. The exchange splitting is $\Delta =Us$, and the paramagnetic response is enhanced as $\chi ={\chi }_0/[1-UD(E_F)]$. A density of states including both spins changes the written factor in the criterion.

The model explains how itinerant bands can polarize without imposing pre-existing localized atomic spins. It is not a claim that every large-$U$ metal becomes ferromagnetic: near half filling the Hubbard model can instead form a Mott insulator with antiferromagnetic superexchange of scale $4t^2/U$. Band filling, bandwidth and fluctuations matter.^[ref] The interaction-energy and kinetic-cost expansion above is the stated mean-field derivation of the enhancement.^[ref]