# Is the sum of two gaussian density matrices also gaussian?

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. $$\rho_1 = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j) = \mathcal{K}\exp(-\sum_{k}E_{k}d_k^{\dagger}d_k)$$

• I'm pretty sure the corresponding statement isn't true for gaussian distributions in probability theory so would be very surprised if it were true here. Apr 20, 2021 at 17:06

• Regardless of the basis one chooses for representing $\rho$ I meant. Apr 20, 2021 at 21:29