# Does the definition of separability of pure states require the components of the summands to be pure?

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?)

• – glS
May 30 at 16:51

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.

• Thanks! It is pretty clear and is what I want to know!
– Eric
Jun 1 at 8:54

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.

• Nevertheless, any separable state can be decomposed as a convex combination of pure product states. May 29 at 17:23
• Yes, definitely May 30 at 19:48