$\DD(\FFF)$ denotes the set of density operators on the fermion Fock space $\FFF$ that represent states of finite average particle number.

\[good lemma\] Suppose $\Delta \in \DD(\FFF)$ and let $\Gamma_{\Delta}$ denote the unique free state that has the same $1$-pdm as $\Delta$. If $\Gamma$ is free then $$\label{conclusion}  -  \Tr(\Delta \log\Gamma )   \equals - \Tr(\Gamma_\Delta \log\Gamma )\ . \$$

We prove this here for the case where $\Gamma$ is a free Gibbs state, i.e., when all natural occupation numbers $p_i$ of $\Gamma$ are strictly between $0$ and $1$. The proof is simple in this case because the operator $\log \Gamma$ is then quadratic in the creators and annihilators, while $\Delta$ and $\Gamma_\Delta$, having the same $1$-pdm, assign the same expectations to all such operators. The general case where some of the $p_i$ may equal $0$ or $1$ requires some care and is handled in Section \[Proof of Lemma 1\].

Suppose $\log \Gamma$ is the quadratic Hamiltonian operator (\[quadratic\]). By the defining property (\[defining property of $1$-pdm\]) of the $1$-pdm $\gamma_\Delta$, $$\begin{aligned}
- \Tr (\Delta \log \Gamma ) & = & 
- \sum_i \big( \log(p_i) \Tr (\Delta \hat{a}^*_i \hat{a}_i) -  \log(1-p_i) \Tr (\Delta \hat{a}_i \hat{a}^*_i)  \big)
\\
& = & 
- 