2
$\begingroup$

Let a tripartite system be given with pure state $\rho_{ABC}$. My adviser said that the entanglement of formation satisfies the identity

$${E_F}_{A(BC)}^2={E_F}_{AB}^2+{E_F}_{AC}^2$$

Where ${E_F}_{A(BC)}$ is the entanglement of formation considering the bipartite state with parts $A$ and $B+C$.

He further said that ${E_F}_{A(BC)}^2=S_A^2$ the Von-Neumman entropy, and that because of that if we know ${E_F}_{AB}$ we get ${E_F}_{AC}$ by means of

$${E_F}_{AC}^2=S_B^2-{E_F}_{AB}^2.$$

Now is this identity true? He couldn't point me the specific reference, and I've tried searching and found none.

That ${E_F}_{A(BC)}=S_A$ is obvious because the entanglement of formation reduces to the entanglement entropy for pure states. Since $\rho_{ABC}$ is pure, considered a bipartite state between $A$ and $B+C$ it is pure, and the entanglement of formation is the entropy.

My doubt is about the very first identity. If it is true, why is it the case?

$\endgroup$
2
$\begingroup$

No. A counterexample is the tripartite GHZ state $|000\rangle+|111\rangle$.

$\endgroup$
2
$\begingroup$

The first equality is actually an inequality for qubit systems. (See This) I'm not so sure if this inequality holds in general dimension.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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