Interior finite-temperature stationarity gives Fermi-Dirac occupations

Let $H$ be any traceless Hermitian matrix. Since $D$ and $I-D$ are positive definite, their least eigenvalues have a positive minimum $\delta$. For $\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }H{\mathrm{\parallel }}_2<\delta$, both $D+tH$ and $I-D-tH$ remain positive definite, and the trace remains $r$. Thus $D+tH$ is feasible for both signs of small $t$. At a minimizer its directional derivative is zero; [F3] gives $\mathrm{tr}\text{\hspace{0.17em}⁣}\left([F_{\mathcal{E}}(D)+{\beta }^{-1}(\log D-\log (I-D))]H\right)=0.$

Put $A=F_{\mathcal{E}}(D)+{\beta }^{-1}(\log D-\log (I-D))$. This is Hermitian. Choose $H=A-(\mathrm{tr}A\mathrm{/}n)I$, which is Hermitian and traceless. Step 1.1 gives $0=\mathrm{tr}(AH)=\mathrm{tr}(H^2)=\mathrm{\parallel }H{\mathrm{\parallel }}_F^2$, hence $H=0$ and $A=\mu I$ with real $\mu =\mathrm{tr}A\mathrm{/}n$.

Rearranging step 2.1 gives $F_{\mathcal{E}}(D)=\mu I-{\beta }^{-1}(\log D-\log (I-D))$, so $F_{\mathcal{E}}(D)$ is a scalar function of $D$. In the spectral basis of $D$, for each eigenvalue $d\in (0,1)$ the scalar equation is $\lambda =\mu -{\beta }^{-1}\log (d\mathrm{/}(1-d))$, whose unique solution is $d=(1+\exp (\beta (\lambda -\mu )){)}^{-1}$. Applying this scalar identity to the spectral functional calculus [F4] proves the matrix formula.

For a fixed Hermitian $F$, diagonalize it by [F4] with real eigenvalues ${\lambda }_j(F)$, and define $g(\mu ):={\sum }_{j=1}^n(1+\exp (\beta ({\lambda }_j(F)-\mu )){)}^{-1}$. By [F6] and $\beta >0$, each summand is continuous and strictly increasing in $\mu$. Its limits are $0$ as $\mu \to -\mathrm{\infty }$ and $1$ as $\mu \to +\mathrm{\infty }$, so $g$ is continuous and strictly increasing with limits $0$ and $n$. Since $0<r<n$, these limits give finite real numbers ${\mu }_{-}$ and ${\mu }_{+}$ with $g({\mu }_{-})<r<g({\mu }_{+})$. Strict increase implies ${\mu }_{-}<{\mu }_{+}$. Applying [F5] to $g$ on the closed interval $[{\mu }_{-},{\mu }_{+}]$ gives a solution with $g(\mu )=r$, and strict increase makes it unique. This scalar argument proves uniqueness of the multiplier for the fixed field only.