# Do the states forming an orthonormal basis have the same amount of entanglement?

If $\{|\psi_{i}\rangle\}$ is an orthonormal basis for a bipartite system, will $E(|\psi_i\rangle) = E(|\psi_j\rangle)$ for all $i, j$, where $E$ is some entanglement measure?

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The relation $a |00> + b |11> = (\frac{1}{\sqrt{2}}(a + b)) (\frac{1}{\sqrt{2}}( |00> + |11>)) + (\frac{1}{\sqrt{2}}(a - b)) (\frac{1}{\sqrt{2}}( |00> - |11>))$, shows that it is not true. –  Trimok Jun 26 '13 at 16:39

No. Here is a counter-example. I will consider a $H_4 \otimes H_4$ space, where $H_4$ is four dimensional. I will express the counter-example basis in terms of the standard separable basis states $|0\rangle, \ldots, |3\rangle$. To form the basis, first take all states $|kj\rangle$, where $k$ and $j$ are such that $k \neq j$ or $k \in \{0,1\}$ or $j \in \{0,1\}$. So basically all the usual basis states except $|22\rangle$ and $|33\rangle$. Now we have 14 of the 16 states in our basis. The final two are $2^{-1/2} (|22\rangle + |33\rangle)$ and $2^{-1/2} (|22\rangle - |33\rangle)$. The last two are entangled, the rest are not.