1
$\begingroup$

I am trying to understand the relation between Fidelity and Concurrence for a two qubit maximally mixed state. When I calculate the Fidelity and Concurrence, I observe that Concurrence is zero whereas Fidelity is one.

In some of the papers, I read that the relation between Fidelity and Concurrence is $F = (c+2/3)$. But this violates the above observation.

Please let me know if I miss anything.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ Fidelity is measured between two quantum states but I only see you mention one. Are you perhaps taking the fidelity between the state you mention and the maximally entangled state? It may also be nice if you could cite the paper you mentioned regarding the relation between the concurrence and the fidelity. $\endgroup$
    – Rammus
    Jan 27, 2021 at 15:42
  • $\begingroup$ I took the following density matrix 1/4(|00><00| + |01><01| + |10><10| + |11><11|). I calculated the concurrence and since it is a maximally mixed state concurrence is zero. I calculated partial traces for qubit1 and qubit2, and then I calculated the fidelity. Now both the partial traces are coming equal and hence fidelity is coming as 1. But when Iooked into the following paper, it is mentioned that the relation between C and F as F = ((C + 2)/3). Please find the link for the paper apps.dtic.mil/dtic/tr/fulltext/u2/1044329.pdf $\endgroup$
    – Ganesh M
    Jan 28, 2021 at 14:20
  • 2
    $\begingroup$ I had a look at the document. It seems that by "fidelity of a state" you are referring to the fidelity between the initial and final state of a teleportation protocol. And the bound that you mention is for a particular family of states that the authors study. I would suggest rereading through the document and trying to understand in exactly what the context is that the authors are studying. You can edit the question to include these additional details (like exactly what you mean by fidelity), this will help someone with answering. $\endgroup$
    – Rammus
    Jan 28, 2021 at 14:39
  • 1
    $\begingroup$ Agree with Rammus. The paper OP referenced is specifically using isotropic states and it also cites another paper (arxiv.org/pdf/1508.01417.pdf) which has the derivation on page 2. In addition, the fidelity (F) is a different quantity from average fidelity (f). The relationship derived is for average fidelity and concurrence(C) i.e. f = (C+2)/3. $\endgroup$
    – Purva Thakre
    Jan 29, 2021 at 19:13

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.