0
$\begingroup$

To illustrate discord and its use, Zurek in his paper on discord (NB pdf) gives example of a partially decohered bell state i.e. $$\rho_{AB}=\frac{1}{2}(|00\rangle\langle 00|+|11\rangle\langle 11|) + \frac{z}{2}(|00\rangle \langle 11|+|11\rangle \langle 00|)$$ using the measurement basis for Alice's side. $$\{\cos\theta |0\rangle + e^{i\phi} \sin\theta |1\rangle,\ e^{-i\phi} \sin\theta |0\rangle + \cos\theta |1\rangle\}$$ He plotted it in the Fig.1.

Now my problem is that I have not been able to reproduce this result. Here is what I got. I can not seem to locate what could be the source problem. The Mathematica code which I used is here. Can anyone see what might be going wrong here.

Improve this question
$\endgroup$
4
  • $\begingroup$ Your state $\psi_{AB}$ is not the same as Zurek's partially decohered Bell state in Eq. (17). (In particular, the latter is a mixed state.) $\endgroup$
    – Norbert Schuch
    Commented Nov 24, 2013 at 15:23
  • $\begingroup$ @Norbert Sorry I wrote it wrong, but you can see in mathematica code that I used it correctly. Thanks I'll correct it now. $\endgroup$
    – The Imp
    Commented Nov 24, 2013 at 22:54
  • 1
    $\begingroup$ The Mathematica code cannot be viewed without requesting some kind of permission. $\endgroup$
    – Norbert Schuch
    Commented Nov 25, 2013 at 9:40
  • $\begingroup$ @NorbertSchuch I have changed it's privacy please. this works now. I checked it by signing out. please try it now. $\endgroup$
    – The Imp
    Commented Nov 25, 2013 at 12:01

1 Answer 1

Reset to default
1
$\begingroup$

I have confirmed with Zurek, he told me that it was wrong (at least the period-wise) and it has been pointed out many times by other people including Animesh Dutta.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Could you provide references to, at least, the other rebuttals? $\endgroup$
    – Emilio Pisanty
    Commented Apr 4, 2014 at 10:57

Your Answer

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

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