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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.

Expand the Grassmann fields in fermionic Matsubara frequencies ${\omega }_l=(2l+1)\pi /\beta$. For $U=0$, the Grassmann Gaussian integral factorizes into thermal fermionic one-level thermal determinants:Each factor counts the empty and occupied states of one spin orbital. Equivalently, the regularized Matsubara sum gives the logarithm of that factor; the raw infinite determinant requires thermal normalization of a fermion determinant. ThereforeThe printed $F$ is the grand potential $\Omega$, because the trace contains $H-\mu \hat{N}$. The Helmholtz free energy at fixed mean particle number is $F_{\mathrm{H}}=\Omega +\mu ⟨\hat{N}⟩$.