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Bipartite operators close enough to the identity are separable. But how does one compute the product operator terms of the separable expansion? In particular, if $\left| \Phi \right> = m^{-1/2} \sum_{i=1}^n \left| i \right> \otimes \left| i \right>$ is the $n$-dimensional Bell state then $I \otimes I - \left| \Phi \right>\left< \Phi \right|$ is separable. But how do I write this as a sum of positive product operators $\sum A_i \otimes B_i$?

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The operator I gave is proportional to $\int_{\psi} \left|\psi\right>\left<\psi\right| \otimes (I-\left|\psi\right>\left<\psi\right|)$. This can be written as a sum instead of an integral by using a complex projective 2-design. I am curious if there are any simpler solutions.

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