# On the separability definition of mixed states

Consider a mixed quantum state $$\rho_{A \cup B}$$ acting on the Hilbert space $$H=H_A\otimes H_B$$. This state is separable if it can be written as a finite convex combination of pure product states (see Horodecki arXiv:quant-ph/9703004), that is if it can be written as $$$$\rho_{A \cup B} = \sum_{i,j} p_{ij} \rho_{A}^{(i)} \otimes \rho_{B}^{(j)} ,$$$$ with $$\sum_{ij} p_{ij}=1$$. Here $$\rho_{A,B}^{(i)}$$ should be pure states (projectors).

I was wondering if the condition of $$\rho_{A,B}^{(i)}$$ being projectors is necessary in the above definition of a separable state? For example, if one can write $$\rho_{A \cup B}$$ as in the equation above, but with $$\rho_{A,B}^{(i)}$$ being normalised Hermitian positive semidefinite operators only (e.g. mixed states), is $$\rho_{A \cup B}$$ still called separable?

Note that the standard definition of a separable state may be written as $$$$\rho_{A \cup B} = \sum_{i} p_{i} \rho_{A}^{(i)} \otimes \rho_{B}^{(i)} ,$$$$ with $$\sum_{i} p_{i}=1$$, and $$\rho_{A,B}^{(i)}$$ are pure states.

• If the goal is checking separability, there is no loss of generality since by spectral decomposition any density matrix $\rho_{A}^{(i)}$ can be written as sums of projectors (similarly for $B$). Then you can just redefine the probabilities that appear in the sum. Sep 19, 2022 at 2:34
• @Everiana This should be an answer. Sep 19, 2022 at 5:50
• Indeed! Why did I not think about it?! Thanks @Everiana!
– Kaio
Sep 19, 2022 at 15:59

The idea is that if the goal is to check for separability, there is no loss of generality since by spectral decomposition any density matrix $$\rho^{(i)}_A$$ (respectively for $$B$$) can be written as sums of projectors (similarly for $$B$$). Then you can just redefine the probabilities that appear in the sum.