Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.