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The state $$ \frac{1}{2}\left(| \phi^+ \rangle \langle \phi^+ | + | \psi^+ \rangle \langle \psi^+ | \right) $$

where $$ | \phi^+ \rangle = \frac{1}{\sqrt2} \left(|00 \rangle + | 11 \rangle \right) $$ $$ | \psi^+ \rangle = \frac{1}{\sqrt2} \left(|01 \rangle + | 10 \rangle \right) $$

By PPT criteria, we know this is a separable state. If I wanted to find what is the mixture of separable states that form this, how would I go about it?

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I would start by writing this as a matrix, and recognising how it can be written in terms of Pauli matrices: $$ \frac14\left(\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}\right)=\frac14(\mathbb{I}\otimes\mathbb{I}+X\otimes X) $$ From here, I don't have a completely formulaic approach for how you do it. But, in this instance, I wrote $$ =\frac{1}{2}\left(\frac{\mathbb{I}+X}{2}\otimes \frac{\mathbb{I}+X}{2}+\frac{\mathbb{I}-X}{2}\otimes \frac{\mathbb{I}-X}{2}\right). $$ Now you can see that each of the terms in the tensor product is a separable state. Specifically, $$ (|++\rangle\langle ++|+|--\rangle\langle --|)/2 $$

One approach that I suppose I might have taken is to recognise the separable, diagonal basis of $X\otimes X$, and decompose $\mathbb{I}\otimes\mathbb{I}$ in the same basis: $$ \frac{1}{4}(|++\rangle\langle ++|+|+-\rangle\langle +-|+|-+\rangle\langle -+|+|--\rangle\langle --|)+\frac{1}{4}(|++\rangle\langle ++|-|+-\rangle\langle +-|-|-+\rangle\langle -+|+|--\rangle\langle --|), $$ which inevitably leads to that result.

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  • $\begingroup$ Just want to confirm, unlike a 1 qubit system where I/2 is the only state that is diagonal in multiple bases, in a 2 qubit system a state can be diagonal in multiple bases? $\endgroup$ – Mahathi Vempati Feb 27 at 10:03
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    $\begingroup$ Yes, because that is a result of degeneracy/multiplicity. The larger the space, the more opportunity there is for that. $\endgroup$ – DaftWullie Feb 27 at 11:03
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    $\begingroup$ Given that deciding separability is a computationally hard problem, it is not surprising that you do not have an algorithmic way to do it. $\endgroup$ – Norbert Schuch Feb 28 at 21:22

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