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Questions tagged [concurrence]

A quantification of quantum entanglement that also serves as a separability criterion. Concurrence equal to zero indicates an unentangled/separable state. A non-zero concurrence "quantifies" how far the states in question are from achieving separability.

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Why does $\sigma_y$ seem to have a special role in the two-qubit concurrence?

The concurrence of a two-qubit state $\rho$ can be written as $$\mathcal C(\rho) = \max(0, \lambda_1-\lambda_2-\lambda_3-\lambda_4),$$ where $\lambda_i$ are the eigenvalues of $|\sqrt\rho\sqrt{\tilde\...
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What is the relation between fidelity and concurrence for a two qubit maximally mixed state?

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 ...
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  • 105
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How to sample from the uniform distribution over the tensor product of two Bloch spheres?

For some context, I am trying to assess the capacity that certain two qubit gates have to create entanglement. To do this I am using the idea of "entangling power", where one takes their ...
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6 votes
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Why do we use complex-conjugate instead of complex-conjugate-transpose when calculating the concurrence?

When we use the formula to calculate two-qubit entanglement, like these: $$ C(\rho)=\max \left\{\sqrt{e_{1}}-\sqrt{e_{2}}-\sqrt{e_{3}}-\sqrt{e_{4}}, 0\right\}\tag{18} $$ with the quantities $...
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  • 181
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Connection between the definitions of concurrence for a two-qubit states

The concurrence for a state $\rho$ as defined here is \begin{equation} C(\rho) = {\rm max}\{0, \lambda_1-\lambda_2-\lambda_3-\lambda_4\}. \end{equation} Where $\lambda_i$ are the eigenvalues of matrix ...
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8 votes
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Can two states with the same entanglement be transformed into each other using local unitaries?

Take two pure bi-partite states $\psi$ and $\phi$ that have the same amount of entanglement in them as quantified by concurrence (does the measure make a difference?). Can any such states be ...
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