# Can two states with the same entanglement be transformed into each other using local unitaries?

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).
• @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. Aug 22 '18 at 10:11
• 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. Aug 22 '18 at 10:23
• @user120404 That's not the Schmidt decomposition! By my calculation they have the same Schmidt decomposition, with Schmidt coefficients $\frac12\pm\sqrt{\frac{33}{200}+\frac{1}{\sqrt{500}}}$ Aug 22 '18 at 11:25