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Is there a general pure entangled bipartite system that all other entangled bipartite systems are a special case of it?

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  • $\begingroup$ It's much easier the other way: there is a general separable bipartite state $\sum_i p_i \hat{\rho}_i\otimes\hat{\sigma}_i$ such that all other separable bipartite states are a special case of it. Then entangled states are all of the states not belonging to this. $\endgroup$ Commented Sep 7, 2023 at 13:56
  • $\begingroup$ I want a same formula for entanglement, if it is exist. $\endgroup$
    – reza
    Commented Sep 7, 2023 at 14:07
  • $\begingroup$ What does a state being a "special case" of another state mean? $\endgroup$ Commented Sep 7, 2023 at 20:17
  • $\begingroup$ It means u can make any pure entangled bipartite state via it. $\endgroup$
    – reza
    Commented Sep 7, 2023 at 20:21

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I didn't see the question asked about pure states - then yes! A general entangled pure state takes the following form, via the Schmidt decomposition: $$|\Psi\rangle=\sum_i \psi_i |a_i\rangle\otimes |b_i\rangle$$ for some non-negative coefficients $\psi_i$ and some pure states $|a_i\rangle$ and $|b_i\rangle$. To be entangled, more than one of the coefficients $\psi_i$ must be nonzero. This is a bit of a strange way of writing the set, but we can be clever and say that any state of the form of $|\Psi\rangle$ with all Schmidt coefficients $|\psi_i|<1$ is entangled. Unentangled states are those with the same form but with one of the coefficients having magnitude equal to unity.

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