quasi-free state and its $1$-pdm has finite trace.

Quasi-free states with finite expected particle number are called “generalized Hartree-Fock" states in Ref. [@BachLiebSolovej]. Accordingly, free states are gauge-invariant generalized Hartree-Fock states.

Gauge-invariant quasi-free states are discussed in Section \[Gauge-invariant quasi-free states\]. In Section \[Free states appendix\] we will prove that all free states are limits of substates of Slater determinant states:

\[free at last\] A density operator on a fermion Fock space is free if and only if (i) its $1$-pdm has finite trace, and (ii) it is a limit in trace norm of density operators that represent substates of Slater determinant states.

Free states whose natural occupation numbers are all strictly positive and less than $1$ have the form of (grand canonical) Gibbs states for noninteracting fermions [@GottliebMauserArchived]. Let $\hat{a}_1,\hat{a}_2,\ldots$ denote the fermionic annihilation operators associated to a complete orthonormal system of reference orbitals, so that $\hat{n}_i = \hat{a}^*_i \hat{a}_i$ represents the observable of “the number of fermions in the $i^{th}$ orbital" (either $0$ or $1$). Any density operator $\Gamma$ of the Gibbs form $$\Gamma \ \propto \  \exp\big( - \sum \lambda_i \hat{n}_i\big)
 \label{Gibbs form}$$ is free. Formula (\[Gibbs form\]) defines a density operator if and only if $\sum 