Helstrom-Holevo bound

Equiprobable pure states of overlap magnitude $c$ have optimal discrimination probability $\frac{1}{2}(1+\sqrt{1-c^2})$.

For ${\rho }_j=|{\alpha }_j⟩⟨{\alpha }_j|$, projecting onto the positive and negative eigenspaces of ${\rho }_0-{\rho }_1$ attains

Suppose one qubit is prepared in either of the known pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, with equal prior probabilities. Measuring in the computational basis seems natural, but a tilted projective measurement gives better quantum state discrimination. The two density operators and their difference areThe Pauli matrices give ${\Delta }^2=I/2$, and the eigenvalues of $\Delta$ are ${\lambda }_{\pm }=\pm 1/\sqrt{2}$. Their absolute sum gives trace distance $D=1/\sqrt{2}$, agreeing with trace distance between qubit states.

The positive spectral projector points along the unit Bloch vector $\mathbf{n}=(-1,0,1)/\sqrt{2}$:The outcome labels $+$ and $-$ here refer to the sign of the eigenvalue of $\Delta$; they do not denote a Hadamard basis measurement. Guess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. This sign choice is the equal-prior Helstrom measurement for two pure states, a case of the Holevo–Helstrom theorem. ^[ref]

An explicit orthonormal basis for this projective measurement iswith $P_{\pm }=|h_{\pm }⟩⟨h_{\pm }|$. Thus the real coefficient rotation angle is $\beta =-\pi /8$ relative to the computational basis, while the Bloch vector of $P_{+}$ is tilted by $\pi /4$ from the positive $z$ axis toward negative $x$. State-vector and Bloch sphere angles differ by a factor of two.

Write $a=(1+1/\sqrt{2})/2$ and $b=(1-1/\sqrt{2})/2$, so $a+b=1$. The Born rule gives the conditional outcome probabilitiesThe event $P_{+}$ has probability gap $a-b=1/\sqrt{2}=D$, attaining the variational characterization of trace distance. Both preparations are identified correctly with conditional probability $a$, so

For comparison, a $Z$ projective measurement in the computational basis has conditional outcome probabilities $(1,0)$ on ${\rho }_0$ and $(1/2,1/2)$ on ${\rho }_{+}$. Guess zero on outcome zero and plus on outcome one. The average success isand the measured total variation distance is only $1/2$. The optimal tilted projective measurement gains about $0.103553$ in absolute success probability by balancing the two conditional error rates instead of identifying $|0⟩$ perfectly at the expense of $|+⟩$.

These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted projective measurement is a different experiment, because the first measurement in quantum mechanics changes the density operator.

Two pure states can be distinguished with certainty in one measurement exactly when their inner product is zero.

Unambiguous quantum state discrimination allows an inconclusive outcome but forbids a wrong conclusive identification. Two distinct pure states admit such a measurement, whereas two full-rank density operators on the same support do not.

For normalized linearly independent pure states, their Gram matrix $G_{ij}=⟨{\psi }_i|{\psi }_j⟩$ is positive definite. The displayed reciprocal vectors satisfy $⟨{\chi }_i|{\psi }_j⟩={\delta }_{ij}$. The operator $R={\sum }_i|{\chi }_i⟩⟨{\chi }_i|$ has nonzero eigenvalues equal to those of $G^{-1}$: use the polar factorization of the matrix whose columns are ${\psi }_i$. Choosing $0<ϵ\le {\lambda }_{\min }(G)$ gives the POVM effects $E_i$ and $E_{?}=I-ϵR$. Each conclusive outcome has probability $ϵ$ on its own state and zero on every other input. The inconclusive effect preserves positivity, so the construction is a physical measurement in quantum mechanics.