Is the sum of two gaussian density matrices also gaussian?

Question

Gaussian density matricies are nice because they are fully characterised by its 2-point correlation function. Consider a free fermionic theory with creation/annihilation ops $c_i,c^{\dagger}_i$, the 2-point correlator is given by $$ C_{ij} =\langle c_i^{\dagger} c_j\rangle \equiv \text{Tr} (c_i^{\dagger}c_j \rho). $$ It admits the following representation (quadratic in fermionic ops) $$ \rho = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j), $$ where $\mathcal{K},A_{ij}$ are some constants.

Is the linear combination of gaussian matrices also gaussian? i.e. $$ \rho_1 + \rho_2 = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j) + \mathcal{Q}\exp(-\sum_{ij}B_{ij}c_i^{\dagger}c_j) =? \mathcal{W}\exp(-\sum_{ij}C_{ij}c_i^{\dagger}c_j), $$ for some suitable choice of $\mathcal{W},C_{ij}$. It might be useful to note that one can diagonalise the exponent, i.e. $$ \rho_1 = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j) = \mathcal{K}\exp(-\sum_{k}E_{k}d_k^{\dagger}d_k) $$

Answer 1

No.

Just take two states with opposite 2-point correlators (like the all-zero and the all-one state). ("Opposite" in a suitable normalization.)

Then, their sum has correlators zero (again, zero in a suitable normalization - but in any normalization, the correlation matrix of the maximally mixed state). Yet, it is not the maximally mixed state (unless you consider a single mode).