Suppose we're given a 2-qubit density matrix($\rho_{4\times4}$). we can apply two local maps on each of these qubits seperatly. So the output is density matrix($\rho^{\prime}_{4\times4}$).
I'm wondering if we can find two sets of Kraus operators (two local maps and specific for the given $\rho$), transforming $\rho$ into the desired density matrix $\rho^{\prime}$ which has the form:
$$\rho^{\prime}=p|uv\rangle\langle uv|+(1-p)|\phi\rangle\langle\phi|$$
where $|uv\rangle$ is seperable state and $|\phi\rangle$ is completely entangled state. also $\langle\phi|uv\rangle=0$
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1$\begingroup$ Is it even clear that the answer does not depend on how you define "lower"? $\endgroup$– Norbert SchuchCommented May 23 at 21:21
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$\begingroup$ It doesn't matter how low the entanglement of the output will be (but should not be zero!); what matters is the structure of the output density matrix. For instance, the output density matrix should be a mixture of only two Bell states. $\endgroup$– xhianCommented May 23 at 22:29
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$\begingroup$ I think your question would profit from more detail. $\endgroup$– Norbert SchuchCommented May 24 at 8:55
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$\begingroup$ I have revised my question. Thank you for pointing that out. $\endgroup$– xhianCommented May 24 at 10:21
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$\begingroup$ Thanks. Some arbitrary |u>, |v>, |phi> (which I can choose), or some given ones (which you choose beforehand)? -- Also, why the orthogonality condition? $\endgroup$– Norbert SchuchCommented May 24 at 10:38
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