Take two pure bi-partite states $\psi$ and $\phi$ that have the same amount of entanglement in them as quantified by concurrence (does the measure make a difference?). Can any such states be transformed into each other using local unitaries?


Any two bipartite pure states $\psi$ and $\phi$ can be transformed into each other with local unitaries if and only if they have the same Schmidt coefficients. (To prove the 'only if' part, note that the reduced density matrix of either qubit has eigenvalues that are the squares of the Schmidt coefficients, and are unchanged by unitaries).

The concurrence is typically specified for two-qubit states. Any reasonable entanglement monotone on two qubits, when calculated for pure states, must be a decreasing function of the largest Schmidt coefficient of the state, and thus two states having the same entanglement would have the same Schmidt coefficients, and hence be unitarily equivalent.

If you generalise beyond qubits, there will be entanglement measures that do not contain as much fine-grained information as the Schmidt coefficients, and so equality of that entanglement measure will be necessary but not sufficient for unitary equivalence.

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    $\begingroup$ But see also quantum embezzlement. "in the presence of arbitrary catalysts, any pure bipartite entangled state can be converted into any other to unlimited accuracy without the use of any communication, quantum or classical" $\endgroup$ Aug 21 '18 at 22:56
  • $\begingroup$ @DaftWullie Suppose that the two states are $|\psi> = \sqrt{0.5} |00> + \sqrt{0.4} |01> + \sqrt{0.05} |10> + \sqrt{0.05} |11>$ and $|\phi> = \sqrt{0.05} |00> + \sqrt{0.4} |01> + \sqrt{0.05} |10> + \sqrt{0.5} |11>$. They have the same concurrence $C = 2|\alpha_{00} \alpha_{11} - \alpha_{01} \alpha_{10}| = 0.0333851$. But if we write the second state as a Schmidt decomposition by applying $X_1 \otimes X_2$ then $H_1 \otimes I_2$, we see that the Schmidt coefficients of the two states are different. They do not have the same Schmidt coefficients even though they have the same concurrence. $\endgroup$
    – user120404
    Aug 22 '18 at 10:11
  • $\begingroup$ As a Schmidt decomposition, $|\phi>$ is just $|\phi> = \frac{1}{\sqrt{2}} (\sqrt{0.4} + \sqrt{0.5}) |00> + \sqrt{0.1} |01> + \frac{1}{\sqrt{2}} (\sqrt{0.5} - \sqrt{0.4}|10>)$. So the two states are not interconvertible, at least not without a catalyst. $\endgroup$
    – user120404
    Aug 22 '18 at 10:23
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    $\begingroup$ @user120404 That's not the Schmidt decomposition! By my calculation they have the same Schmidt decomposition, with Schmidt coefficients $\frac12\pm\sqrt{\frac{33}{200}+\frac{1}{\sqrt{500}}}$ $\endgroup$
    – DaftWullie
    Aug 22 '18 at 11:25
  • $\begingroup$ Yup! Silly mistake on my part! $\endgroup$
    – user120404
    Aug 22 '18 at 13:37

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