3
$\begingroup$

I've read that, given a bipartite pure state $|\Phi\rangle$, its entanglement (equivalently here, von Neumann) entropy $E(\Phi)$ gives the asymptotic number of singlets required to create $n$ copies of $|\Phi\rangle$. More specifically, that for all $\epsilon>0$ there is a large enough $n$ such that, given $m$ copies of a singlet state $|\Phi^-\rangle$, Alice and Bob can produce $n$ copies of $|\Phi\rangle$ with $m/n\le (1+\epsilon)E(\Phi)$. I think this roughly means that $E(\Phi)$ is the number of singlets required to generate each copy of $|\Phi\rangle$.

This is mentioned in this paper by Wooters: http://www.rintonpress.com/journals/qic-1-1/eof2.pdf, around equation 3.

To back up this statement, they cite this paper by Bennett and collaborators. I know that the above statement is shown in this paper, but I was wondering if there was a way to briefly sketch the reasoning, or more generally a way to understand the intuition behind it. More recent takes on this result would also be welcome, if there is any.

$\endgroup$
3
  • $\begingroup$ Please link to abstracts, not pdfs. $\endgroup$ Jun 9 at 0:00
  • $\begingroup$ @NorbertSchuch sorry. The second link was a mistake, but for the first one I couldn't find a corresponding non-pdf link $\endgroup$
    – user15342
    Jun 9 at 19:18
  • 2
    $\begingroup$ Regarding your question, it can be explained, but it takes (at least me) a significant part of a 90'-lecture, since one has to introduce some basic concepts about iid sources (typical sequences etc.). In essence, it is the same as classical Shannon compression of random sources - if you know about this, explaining the quantum case is rather simple. -- You can find the whole explanation (including typicality etc.) e.g. on pg 21-36 of Sec. III.4 of these lecture notes of mine. $\endgroup$ Jun 9 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.