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 $|\sqrt\rho\sqrt{\tilde\rho}|$, using the notation $|A|\equiv\sqrt{AA^\dagger}$, and $\tilde\rho\equiv(\sigma_y\otimes \sigma_y)\bar\rho(\sigma_y\otimes \sigma_y)$.
This formula comes, as far as I understand, from defining the concurrence as $\sqrt{1-\operatorname{Tr}(\rho_A^2)}$ for pure states, and then defining it as the minimum of this quantity over all possible pure state decompositions, for general states. Solving the minimisation problem leads to the above formula.
Is there a good way to see why the result turns out to involve an operator such as $\tilde\rho$, and in particular seems to place a special role into the $\sigma_y$ operators?
