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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.

Perfect identification on every trial is impossible, because the family contains a nonorthogonal pair. Any physical identification device induces a POVM whose effect $E_i$ corresponds to its declaration of label $i$. Certainty of correct identification would require $⟨{\psi }_i|E_i|{\psi }_i⟩=1$ and $⟨{\psi }_j|E_i|{\psi }_j⟩=0$ for every $j\ne i$.

Since $0\le E_i\le I$, the first identity gives $(I-E_i)|{\psi }_i⟩=0$, while positivity and the second identity give $E_i|{\psi }_j⟩=0$. For example these conclusions follow by writing the zero expectations as squared norms of $(I-E_i{)}^{1/2}{\psi }_i$ and $E_i^{1/2}{\psi }_j$. ThusPerfect identification would force the entire family to be orthogonal, contrary to the hypothesis. This argument permits ancillas, general measurements and disturbance of the input; it does not assume that the device must preserve the state.

Unambiguous identification with a nonzero success probability for every input is possible. Normalize the given kets and let $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ be their Gram matrix. Their linear independence makes $G$ positive definite. Define reciprocal vectors within their span byThen $⟨{\psi }_k|{\chi }_i⟩={\sum }_j{G}_{kj}(G^{-1}{)}_{ji}={\delta }_{ki}$, and hence also $⟨{\chi }_i|{\psi }_k⟩={\delta }_{ik}$. Put $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$. It is positive and has finite operator norm. Choosing $ϵ=1/(2\Vert R\Vert )>0$, setAll these effects are positive, and $E_{?}\ge I/2$, so they sum to the identity and define a POVM. The Born rule givesThus every label is correct whenever it occurs and every input has the strictly positive success probability $p_k=ϵ$. This is the reciprocal-state construction of unambiguous discrimination.

For an explicit success constant, collect the input kets into a matrix $\Psi$, so ${\Psi }^{†}\Psi =G$. Its polar factorization is $\Psi =U{G}^{1/2}$ with $U^{†}U=I_N$. The reciprocal columns are $\Psi G^{-1}=U{G}^{-1/2}$, and therefore $R=U{G}^{-1}{U}^{†}$. Its largest eigenvalue is $1/{\lambda }_{\min }(G)$, so one can take $ϵ={\lambda }_{\min }(G)/2$. Since the normalized Gram matrix has trace $N$, this is at most $1/2$.

To realize the device, use measurement operators $M_i=\sqrt{E_i}$ and $M_{?}=\sqrt{E_{?}}$. The map $|v⟩\mapsto {\sum }_i{M}_i|v⟩|i⟩+M_{?}|v⟩|?⟩$ preserves inner products because the effects sum to $I$. Extend this linear isometry to a unitary on the input and an initially blank ancilla, then projectively measure the ancilla's orthogonal outcome labels. Its outcome probabilities are precisely those above. This gives a physical unambiguous quantum state discrimination device rather than only a formal set of effects.

The Bloch vector representation of a qubit density matrix is $\rho =\frac{1}{2}(I+r_{x}X+r_{y}Y+r_{z}Z)$, where $r_j=\mathrm{Tr}(\rho {\sigma }_j)$ and $|\mathbf{r}|\le 1$. Pure states have $|\mathbf{r}|=1$, while the origin represents the maximally mixed state $I/2$, not a pure state.

For the two preparations, direct evaluation of the Pauli matrices givesTheir dot product is $-1/2$, so their Bloch vectors meet at angle $\boxed{2\pi /3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.

Completeness of the POVM requires $E_3=I-E_1-E_2$. In the computational basis, subtraction givesAn overall phase of ${\varphi }_3$ is immaterial. The matrix has eigenvalues $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid trine qubit POVM.

The Born rule gives $\mathrm{Pr}(j\mid {\rho }_i)=\mathrm{Tr}(E_j{\rho }_i)=\frac{2}{3}|⟨{\varphi }_j|{\psi }_i⟩{|}^2$. The conditional probabilities areOutcome one identifies ${\rho }_2$, since ${\varphi }_1$ is orthogonal to ${\psi }_1$; outcome two identifies ${\rho }_1$, since ${\varphi }_2$ is orthogonal to ${\psi }_2$. Outcome three is inconclusive. If Alice's prior probability for ${\rho }_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials. This is unambiguous quantum state discrimination.

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.

LetThe three effectsform a POVM, because the largest eigenvalue of the sum of the two rank-one projectors is $1+\cos \theta$. Outcome 1 never occurs on $|\theta ⟩$, while outcome 2 never occurs on $|0⟩$. Thus conclusive outcomes are never wrong; $E_{?}$ records failure. This is unambiguous quantum state discrimination.

Choose three pure qubit states whose unit Bloch vectors are coplanar, sum to zero and have pairwise angle $120$ degrees. The three effects $E_j=\frac{2}{3}|{\varphi }_j⟩⟨{\varphi }_j|$ sum to the identity, giving a POVM. If an input is the state orthogonal to ${\varphi }_1$ or to ${\varphi }_2$, outcome one excludes the first input and outcome two excludes the second. Each conclusive outcome has probability one half on its corresponding input; the third outcome is inconclusive with probability one half on either input. This gives unambiguous quantum state discrimination, rather than identification on every trial.