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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.

Yes, access to the original qubits can supply information about the secret encoding basis once the bit is known. For bit zero the possibilities are $|0⟩$ and $|+⟩$, whose overlap is $1/\sqrt{2}$; for bit one they are $|1⟩$ and $|-⟩$ with the same overlap magnitude. A Helstrom measurement for two pure states can identify which candidate was supplied more reliably than a blind basis guess. Measuring and preparing a state based on that evidence can improve an additional counterfeit, although the discrimination is not perfect and the measurement can disturb the original.

The no-cloning theorem rules out a deterministic perfect extra copy of both nonorthogonal candidates; it does not imply that access gives no advantage. The task of producing an additional accepted note should also be distinguished from simply returning the unaltered original, which already passes Alice's test. Requiring both the original and the extra note to survive imposes a further information-disturbance tradeoff.

Continue with equal prior probabilities as in part (c). The two Bloch vectors areTheir density-matrix difference is $\Delta =\sin \theta \cos \phi {\sigma }_x$. If a binary measurement effect $E$ declares the first state, its success isDiagonalize $\Delta$. Because $0\le E\le I$, the trace is maximized by taking $E$ to be its positive-eigenvalue projector. Thus perform a Pauli measurement of ${\sigma }_x$, equivalently measure in $(|0⟩\pm |1⟩)/\sqrt{2}$. If $\cos \phi >0$, identify the first state on the positive outcome and the second on the negative; reverse the labels if $\cos \phi <0$. The optimum isFor $\cos \phi =0$ the states coincide, so the best probability is $1/2$. This explicit positive-eigenspace optimization proves the Helstrom measurement for two pure states result in the present case, rather than only quoting its name.

The no-cloning theorem for two pure states states that, for distinct nonorthogonal states $|c_0⟩$ and $|c_1⟩$, there is no unitary operator $U$ and fixed blank state $|0⟩$ such that

To derive this from state discrimination, letand suppose such a cloner existed. Applying it repeatedly would produce $N$ copies. The inner product of the two possible $N$-copy states has magnitude $s^N$, so their optimal success probability under the Helstrom-Holevo bound would beAlready for $N=2$,because $0<s<1$ implies $s^4<s^2$. But cloning followed by the two-copy Helstrom measurement for two pure states would itself be a quantum procedure applied to the original single state, contradicting the optimal one-copy bound $P_1$. Hence the assumed unitary cloner cannot exist.

Equivalently, clone-assisted asymptotic state discrimination would give $P_N\to 1$, although two nonorthogonal states are not perfectly distinguishable. Thus the Helstrom--Holevo theorem implies the no-cloning theorem.