= Solution
Consider any <POVM>, not necessarily rank one, with two distinct reported outcomes intended to identify the two states. Its effects satisfy $E_1,E_2\geq0$ and $E_1+E_2\leq I$; equality holds for a two-outcome measurement. If $\langle\psi_1|E_1|\psi_1\rangle=1$, then positivity and completeness force $\langle\psi_1|E_2|\psi_1\rangle=0$. Taking a positive square root gives $E_2|\psi_1\rangle=0$. Likewise, $\langle\psi_2|E_2|\psi_2\rangle=1$ and $0\leq E_2\leq I$ imply $(I-E_2)|\psi_2\rangle=0$, or $E_2|\psi_2\rangle=|\psi_2\rangle$.
Using Hermiticity,
$$
\langle\psi_1|\psi_2\rangle
=\langle\psi_1|E_2|\psi_2\rangle
=\langle E_2\psi_1|\psi_2\rangle=0.
$$
Thus \b[perfect discrimination of pure states requires orthogonality], contradicting the proposed nonzero overlap. This proves <perfect discrimination of pure states requires orthogonality> for arbitrary generalized measurements. Measurements with an inconclusive outcome can sometimes identify nonorthogonal inputs unambiguously, but <unambiguous quantum state discrimination> has a nonzero failure probability and is not perfect deterministic discrimination.
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