 Solution

Consider any POVM, not necessarily rank one, with two distinct reported outcomes intended to identify the two states. Its effects satisfy $E_1,E_2\ge 0$ and $E_1+E_2\le I$; equality holds for a two-outcome measurement. If $⟨{\psi }_1|E_1|{\psi }_1⟩=1$, then positivity and completeness force $⟨{\psi }_1|E_2|{\psi }_1⟩=0$. Taking a positive square root gives $E_2|{\psi }_1⟩=0$. Likewise, $⟨{\psi }_2|E_2|{\psi }_2⟩=1$ and $0\le E_2\le I$ imply $(I-E_2)|{\psi }_2⟩=0$, or $E_2|{\psi }_2⟩=|{\psi }_2⟩$.

Using Hermiticity,Thus 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.