e^{-\lambda_i} < \infty$ because the trace of the operator on the right-hand side of the formula equals $\prod\big(1+e^{-\lambda_i} \big)$. The reference orbitals of the density operator $\Gamma$ defined by (\[Gibbs form\]) are its natural orbitals. Its natural occupation numbers, the average values of the observables $\hat{n}_i$, are $ p_i = e^{-\lambda_i}/(1+e^{-\lambda_i}) $. For later use we note here that $\log\Gamma$ is the “quadratic Hamiltonian" [@BachLiebSolovej] operator $$\log \Gamma  
      \equals 
      \sum_i \big( \log(p_i) \hat{a}^*_i \hat{a}_i +  \log(1-p_i) \hat{a}_i \hat{a}^*_i \big) \ . 
 \label{quadratic}$$

Free states are characterized by statistically independent occupation of their natural orbitals. In Section \[Free states appendix\] we show that a density operator $\Gamma$ on $\FFF(\HH)$ is free if and only if orthogonal natural orbitals are occupied independently of one another. For example, in the free Gibbs state (\[Gibbs form\]) the expected value of the occupation observables $\hat{n}_i$ and $\hat{n}_j$ are $p_i$ and $p_j$, respectively, while the expected value of $\hat{n}_i\hat{n}_j$, i.e., the probability that the $i^{th}$ and $j^{th}$ orbitals are both occupied, equals $p_ip_j$ (assuming $i \ne j$).

In order to define nonfreeness and related correlation functionals, we will require the following well-known fact [@BachLiebSolovej; @AlickiFannes], which we will also prove in Section \[Free states appendix\].

\[free 