I was trying to understand the concept of entanglement entropy and for that I was studying the density operator formalism in the context of a bipartite quantum system.

In some references, people define the maximally entangled state as the one whose reduced density matrix (for the lower dimensional subsystem) is maximally mixed. I believe I understand the concept of maximally mixed - it just means that the state is descibed by an ensemble with every possible quantum state having the same probability (in some sense it describes the most random ensemble one can have with a certain basis of quantum states). Now what I don't understand in a clear way is why we define a maximally entangled state in such a way - is there any more insightful reason or is it just a definition that is consistent?

Indeed, this question arised when I was thinking about entanglement entropy. Still in the bipartite system, one can show that for a pure state of the system, the Von Neumann entropy of the subsystem has some properties that lead us to define this entropy as a measure of entanglement, yet these same properties are dependent on our notion of maximally entangled that I don't understand (http://www.mpmueller.net/seminar/talk2.pdf see Theorem.6)

Summing up, my questions are:

  1. When one defines maximal entanglement, do we already have entropy in mind which, indeed, quantifies entanglement? If not, why can one talk about maximal entanglement in some state a priori?
  2. Is there any way to see that Von Neumann entropy for a subsystem of a pure bipartite state really is quantifying entanglement in a more concrete way? Well, this is my main question, because although I see that this is plausible I don't understand clearly why.
  • $\begingroup$ The notes you link don't seem to provide the correct explanation for using the von Neumann entropy. The point is that in an asymptotic scenario -- i.e., when you are given many copies $N_\psi$ of a state $\vert\psi\rangle$ and you want to convert them into a number of copies $N_\phi$ of a state $\vert\phi\rangle$, the protocol can be carried out reversibly as the number of copies goes to infinity, as long as $N_\psi S(\psi)=N_\phi S(\phi)$. This is, in this scenario (and only in this scenario!) the von Neumann entropy uniquely quantifies the entanglement contained in the state. $\endgroup$ – Norbert Schuch Jul 25 '16 at 17:35
  • $\begingroup$ Thank you, do you know any good reference (not too lengthy) that gives a good explanation on the bridge between Von Neumann entropy and entanglement entropy? Maybe some way of understanding it in a simple setting and not with full generality. $\endgroup$ – blackhole1511 Jul 25 '16 at 17:38
  • $\begingroup$ Well, you need to know about typical sequences etc.. If you know about these, it is indeed a quite simple argument. I would check the book by Nielsen and Chuang first. (EDIT: There is not much difference between the "simple setting" and "full generality" -- the tricky part is dealing with the typical states appearing in an asymptotic scenario.) $\endgroup$ – Norbert Schuch Jul 25 '16 at 17:40
  • $\begingroup$ Okay, thank you! I was trying to have a peak into this topic without the full machinery of quantum information, but I see some background is needed in order to have a good understanding $\endgroup$ – blackhole1511 Jul 25 '16 at 17:42
  • 2
    $\begingroup$ Well, if you disregard the epsilons it's not too bad ... just take $\sqrt{p}|00\rangle+\sqrt{1-p}|11\rangle$ and try to figure out what the typical number of 1's on Alice's side is if you take many copies, and how many such configurations there are. You should find that it is exactly given by the exponential of the entropy (=a binomial coefficient). $\endgroup$ – Norbert Schuch Jul 25 '16 at 17:44

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.