= Mixed-state concurrence of two qubits
{title2=$C(\rho)=\inf_{\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|}\sum_jp_j\,2|\det M_j|$}
For a normalized two-<qubit> <pure state> with coefficient matrix $M$, <concurrence> is $2|\det M|$. The displayed <convex roof extension> defines its mixed-state version. Combining decompositions proves convexity: $C(\sum_jr_j\rho_j)\leq\sum_jr_jC(\rho_j)$. Every <separable quantum state> has a decomposition into product vectors, each with zero determinant, so its concurrence is zero.
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