 Solution

With no access to the discarded qubit, the two preparations are exactly the reduced density matrices in c, with equal prior probabilities. Their difference isSince $\mathrm{Tr}\Delta =0$, its eigenvalues are $\pm \sqrt{2}/6$. The trace norm and trace distance are consequentlyThe Holevo–Helstrom theorem gives the optimumTo establish attainability, rather than only an upper bound, measure the Hermitian operatorwhere $X,Z$ are Pauli matrices. Since $\Delta =(\sqrt{2}/6)B$, report first qubit for the positive outcome; second qubit for the negative outcome. The corresponding orthogonal projections are $E_{\pm }=(I\pm B)/2$. Their success probability iswhich reaches the bound. Identical eigenvalues of the two density matrices do not make them indistinguishable: their eigenvectors differ, and this quantum measurement detects that difference.