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By the Rayleigh-Ritz variational principle, $\mathrm{tr}({\Pi }_{0}D)$ is maximized by taking ${\Pi }_0$ to project onto the positive-eigenvalue eigenspace of $D$. Put $|⟨{\alpha }_0|{\alpha }_1⟩|=\cos \theta$. On $S$, the characteristic polynomial of $D$ isso its positive eigenvalue isPart (b) then gives the Helstrom-Holevo bound

With equal priors, the probability of a correct result from this quantum measurement isThe two required amplitudes areThereforeIts maximum occurs at $\beta =-\pi /8$ and is

Here $|⟨a_0|a_1⟩|=1/\sqrt{2}$, so $\theta =\pi /4$ and $\sin \theta =1/\sqrt{2}$. The stated Helstrom-Holevo bound is consequently exactly the same value, and the basis above is the Helstrom measurement for the zero and plus states.

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.

Apply the hypothetical quantum cloning device repeatedly to obtain $N$ copies of the unknown state. If $c=|⟨\varphi |\psi ⟩|<1$, the two possibilities arewhose overlap has modulus $c^N$. For equal prior probabilities, the Helstrom-Holevo bound gives optimal success probabilityThus, for any $\epsilon >0$, sufficiently many clones permit success probability above $1-\epsilon$. Orthogonal states already have $c=0$ and can be distinguished perfectly. This is clone-assisted asymptotic state discrimination.

Here distinct states means distinct rays. Vectors differing only by a global phase represent the same quantum state and cannot be distinguished.

The overlap isThe Helstrom-Holevo bound for two equiprobable pure states givesThe Breidbart basis is the eigenbasis of the difference of the two weighted state projectors, so its decision rule is precisely the Helstrom measurement and attains this bound.

The Helstrom-Holevo bound for equal priors isPerfect cloning twice changes overlap $c$ to $c^2$, so discrimination would succeed with $\frac{1}{2}(1+\sqrt{1-c^4})$, strictly exceeding $\frac{1}{2}(1+\sqrt{1-c^2})$ for $0<c<1$. For $|0⟩,|+⟩$, $c=1/\sqrt{2}$, producing $\frac{1}{2}(1+\sqrt{3}/2)>\frac{1}{2}(1+1/\sqrt{2})$, a contradiction.