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What is the motivation for the definition of concurrence in quantum information? On the surface, the definition looks pretty ad hoc.

The definition is often given for the case of 2 qubits only. What is the generalization to higher dimensional spaces?

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2 might be a better reference (since it proves the general case). – Norbert Schuch Dec 11 '12 at 15:59

To the best of my knowledge, the initial motivation for defining the concurrence involves the fact that for pure bipartite entangled states, the reduced density matrices of each subsystem are not pure. For example, consider the Bell state $$|\Phi\rangle_{AB}=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).$$ It satisfies $\rho_A=\text{Tr}_A(|\Phi\rangle_{AB})=\mathbb{I}_A/2$, where $\mathbb{I}_A$ is the identity operator on $\mathcal{H}_A$. The reduced state $\rho_A$ is not pure and it purity is $\text{Tr}(\rho_A^2)=1/2$. This leads to the definition of concurrence for a pure state $|\psi\rangle$ as $$C(\psi)=\sqrt{2[1-\text{Tr}(\rho_A^2)]}.$$ For a mixed state $\rho$, one defines the concurrence as $$C(\rho)=\min_{|\psi_i\rangle}\sum_ip_iC(\psi_i)$$ where the minimization is over all possible decompositions $\rho=\sum_ip_i|\psi_i\rangle\langle\psi_i|$ of $\rho$. The closed form of the concurrence for two qubits $$C(\rho)=\max\{0,\lambda_1-\lambda_2-\lambda_3-\lambda_4\}$$ is just the solution to the minimization problem.

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I do not know of any closed solutions for higher dimensions. – Juan Miguel Arrazola Dec 10 '12 at 22:40

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