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Can the concurrence be calculated in terms of the entanglement of formation?

+ 4 like - 0 dislike

If I somehow know the entanglement of formation, $E_F$ for two mixed qubits, where

\begin{equation} E_F = -x \log x - (1-x) \log (1-x), \end{equation}

where $x = (1+\sqrt{1-\mathcal{C}^2})/2$ and $\mathcal{C}$ is the concurrence, can I then calculate the concurrence from it? (Rather than calculating the concurrence 'normally' using $\mathcal{C} (\rho) = \max \{ 0, \lambda_1 - \lambda_2 - \lambda_3 - \lambda_4 \}$ where $\lambda_i$ are the square roots of the eigenvalues of $\rho S \rho^* S$ and $S = \sigma_y \otimes \sigma_y$).

This post has been migrated from (A51.SE)
asked Feb 14, 2012 in Theoretical Physics by Calvin (60 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

Concurrence was introduced exactly in the effort to find an analytic formula for entanglement of formation. Since one is a monotonic function of the other, you can imagine inverting the relation to obtain concurrence from entanglement of formation. Unfortunately the inverse mapping from $E_F$ to $\mathcal{C}$ probably does not look as nice as the mapping in the other direction (form $\mathcal{C}$ to $E_F$ that you know and write above.

This post has been migrated from (A51.SE)
answered Feb 14, 2012 by Marco (260 points) [ no revision ]
Thanks. You're probably right that it won't look as nice, but is there a way of finding the inverse?

This post has been migrated from (A51.SE)
By "does not look as nice as" I actually meant that you cannot write it down as a combination of elementary functions. Even just inverting the binary entropy $-x\log x - (1-x)\log (1-x)$ is not possible in this sense, as far as I know. You can just show that the inverse exists, and maybe calculate some properties, or try to expand it in series, but I believe that's it.

This post has been migrated from (A51.SE)
I plotted $E_F$ against concurrence. The graph looks so well behaved that it seems like there should be an explicit answer! But I think you're right that there isn't. Thanks for your answer.

This post has been migrated from (A51.SE)

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