If I have a pure state vector: $$ \left|\psi_d\right\rangle=\frac{1}{\sqrt{d}}\sum^{d}_{i=1}\left|ii\right\rangle $$ then the distillable entanglement satisfies: $$ E_{D}(\left|\psi_d\right\rangle\langle\psi_d|)\geq\log d. $$ I am not seeing how to prove this as I thought if I have a pure state then the distillable Entanglement is equal to the entanglement entropy which is $\leq\log d$.

  • 1
    $\begingroup$ please add a link to where you encountered this statement (by editing the post, not in the comments) $\endgroup$
    – glS
    May 23, 2023 at 16:02
  • $\begingroup$ I have a lecture notes but it's not public according to my professor. I can screenshot the theorem if that helps. $\endgroup$
    – MZperX
    May 24, 2023 at 14:44

1 Answer 1


The distillable entanglement of this state is equal $\log d$ by definition. Thus, both $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\ge \log d$ and $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\le \log d$ holds true.

More generally, the distillable entanglement of a pure state is equal to the Shannon entropy $-\sum p_k\log p_k$ of its Schmidt coefficients $p_k$ (equivalently, the von Neumann entropy of either reduced density matrix) by definition. In the case of the state you give, this entropy equals $\log d$.


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