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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 \begin{equation} \rho_{A \cup B} = \sum_{i,j} p_{ij} \rho_{A}^{(i)} \otimes \rho_{B}^{(j)} , \end{equation} 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 \begin{equation} \rho_{A \cup B} = \sum_{i} p_{i} \rho_{A}^{(i)} \otimes \rho_{B}^{(i)} , \end{equation} with $\sum_{i} p_{i}=1$, and $\rho_{A,B}^{(i)}$ are pure states.

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    $\begingroup$ 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. $\endgroup$ Sep 19, 2022 at 2:34
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    $\begingroup$ @Everiana This should be an answer. $\endgroup$ Sep 19, 2022 at 5:50
  • $\begingroup$ Indeed! Why did I not think about it?! Thanks @Everiana! $\endgroup$
    – Kaio
    Sep 19, 2022 at 15:59

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As suggested, I put here as an answer.

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.

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