 Single-copy entanglement concentration

For one pure rank-two entangled state with descending squared Schmidt coefficients $({\lambda }_0,{\lambda }_1)$, the success operator $F=\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ gives a Bell pair with probability $2{\lambda }_1$. The complementary operator $\sqrt{I-F^{†}F}$ makes a valid quantum instrument. Optimality follows from the Schmidt-tail entanglement monotone: a successful Bell branch has smallest weight $1/2$, so its probability cannot exceed $2{\lambda }_1$. Failure branches have nonnegative monotone and cannot loosen this bound.