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In the homogeneous spin-field sector, choose $\mathbf{M}=M{\hat{\mathbf{e}}}_z$. The fermionic levels become ${ϵ}_{\mathbf{k}}-\mu -\sigma M$, where $\sigma =\pm 1$ is the Pauli eigenvalue. The fermionic one-level thermal determinant can be applied independently to every momentum and spin, while the field cost is $\beta N{M}^2/(2U)$. Hence the homogeneous spin-field effective action isThe remaining homogeneous-field integral is $\mathcal{Z}\simeq \int \mathcal{D}M{e}^{-{\mathcal{S}}_{\mathrm{eff}}}$ in this mean-field restriction. Orientation and fluctuation-independent measure factors do not alter the displayed extensive action or its stationary equation. The two bands are exchange split by $2M$; $M$ has energy units and is proportional to the spin polarization, rather than being an electron number itself.

Differentiate the homogeneous spin-field effective action:At a stationary field, $M=(U/N){\sum }_{\mathbf{k},\sigma }\sigma n_F({ϵ}_{\mathbf{k}}-\sigma M)$. Define the spin-summed density of states per lattice site byThen the requested form isThe factor one half fixes the spin convention. With a single-spin density of states ${\nu }_s=\nu /2$, the prefactor would instead be $U$. In terms of site occupations, the same equation is $M=U(n_{\uparrow }-n_{\downarrow })=2U⟨S^z⟩$. The Fermi distribution includes the chemical potential in its argument definition; it must not be subtracted a second time.