Mixed-state concurrence of two qubits ($C(\rho )={\inf }_{\rho ={\sum }_j{p}_j|{\psi }_j⟩⟨{\psi }_j|}{\sum }_j{p}_{j}2|\det M_j|$)

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: $C({\sum }_j{r}_j{\rho }_j)\le {\sum }_j{r}_{j}C({\rho }_j)$. Every separable quantum state has a decomposition into product vectors, each with zero determinant, so its concurrence is zero.