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Does the definition of separability of pure states require the components of the summands to be pure? More precisely, let $\rho$ be a pure state (i.e., $\rho=|\phi\rangle\langle\phi|$) on the space $H_A\otimes H_B$. If it is possible to write $\rho=\sum_{i}p_i(\rho_A^{i}\otimes\rho_B^{i})$, then we say that $\rho$ is separable. I want to ask whether here we require that all $\rho_A^{i}$'s and $\rho_B^{i}$'s are themselves being pure. (Or maybe we can prove that whether they are pure or mixed does not matter?)

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The definition of separability does not require any restrictions on the purity of the state. However, if $\rho$ is pure then it is separable on some bipartition $AB$ iff it can be written as $$ \rho = \rho_A \otimes \rho_B $$ where both $\rho_A$ and $\rho_B$ are pure states.

Here's a sketch of why that's true. Note that if $\rho$ is separable we have $$ \rho = \sum_i p_i \rho_A^i \otimes \rho_B^i. $$ Now as $\rho$ is pure the RHS must be rank one. Now for positive semidefinite matrices $X,Y$ we have that $\mathrm{rank}(X+Y) \geq \max\{\mathrm{rank}(X), \mathrm{rank}(Y)\}$ and so the RHS is rank one only if $\rho_A^i \otimes \rho_B^i$ is rank one for all $i$. In fact, that sum can only be rank one if each of the terms are scalar multiples of each other. Thus we end up with $\rho=\rho_A \otimes \rho_B$ which is rank one iff $\rho_A$ and $\rho_B$ both have rank one.

Note also that any separable state $\rho$ (not necessarily pure) can always be written as $$ \rho = \sum_i p_i \rho_A^i \otimes \rho_B^i $$ with $\rho_A^i$ and $\rho_B^i$ pure. This can be shown by taking any separable decomposition and using the spectral decomposition of the $\rho_A^i$ and $\rho_B^i$ to write them as a sum of pure states.

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  • $\begingroup$ Thanks! It is pretty clear and is what I want to know! $\endgroup$ – Eric Jun 1 at 8:54
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No, we do not require $\rho_A^i$ and $\rho_B^i$ to be pure. The state is separable any time it can be written in this form with positive weights $p_i$. As a basic example, we can choose a single $p_i$ to be nonzero; this is equivalent to saying that $$\rho=\rho_A^i\otimes \rho_B^i$$ is separable. This is true regardless of the purities of $\rho_A^i$ and $\rho_B^i$, by definition.

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    $\begingroup$ Nevertheless, any separable state can be decomposed as a convex combination of pure product states. $\endgroup$ – Rammus May 29 at 17:23
  • $\begingroup$ Yes, definitely $\endgroup$ – Quantum Mechanic May 30 at 19:48

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