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Represent any universally quantum device, including its auxiliaries and arbitrary final readout, by a positive operator-valued measure $\{E_i\}$ on the input. Perfect identification would requireSince a positive operator has a positive square root, zero expectation implies $E_i|{\varphi }_j⟩=0$ for $j\ne i$. Also $I-E_i\ge 0$ and its expectation in ${\varphi }_i$ is zero, so $E_i|{\varphi }_i⟩=|{\varphi }_i⟩$. ConsequentlyThis proves that perfect discrimination of pure states requires orthogonality. Any nonzero overlap contradicts perfect identification, independently of whether the device disturbs its input.

Equivalently, a unitary operator coupling the input to a readout apparatus must preserve inner products. Distinct perfectly readable records have zero inner product, so their corresponding input vectors must already be orthogonal. Nonorthogonal candidate states cannot be identified with certainty in one use. Probabilistic unambiguous quantum state discrimination can have an inconclusive outcome, which is excluded by the guaranteed-identification requirement.

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.

In the ordered computational basis $00,01,10,11$, the first two qubits of $|{\psi }_{0000}⟩$ have amplitudes $(1,1,1,1)/2$, while those of $|{\psi }_{1111}⟩$ have amplitudes $(-1,1,1,1)/2$. Both have the same normalized answer qubit $|-⟩$. Their inner product is thereforeA common unitary operator $U\otimes I$ preserves this nonzero inner product. To distinguish the alternatives with certainty using the prescribed quantum measurement in the computational basis, their first-register outcome supports would have to be disjoint: every outcome possible under one alternative must be impossible under the other. Vectors with disjoint computational basis supports have zero inner product, contradicting its preservation. Thus no such $U$ exists. This is also an instance of perfect discrimination of pure states requires orthogonality; allowing a more general measurement in quantum mechanics would not rescue this pair of states. The obstruction concerns these prepared oracle-output states, rather than every conceivable oracle-query procedure.