= Solution
The overlap $\langle0|+\rangle=1/\sqrt2$ is nonzero, whereas <perfect discrimination of pure states requires orthogonality>. Consequently no single-copy <quantum measurement> can identify both preparations with certainty. After a nondegenerate <projective measurement>, the <loss of preparation information after a nondegenerate projective measurement> prevents further measurements from recovering the missing label: conditional on the first outcome, the post-measurement state is the same for every compatible initial hypothesis. Repeating the same basis gives that outcome again; changing basis probes the collapsed state. Fresh, independently prepared copies would be additional resources.
In the <computational basis>, input $|0\rangle$ always gives zero and remains $|0\rangle$. Input $|+\rangle$ gives zero or one with probability $1/2$, leaving the corresponding basis state. Combining these likelihoods with equal prior <probabilities>, the <Bayesian update after a projective measurement> gives
$$
\begin{array}{c|c|cc}
\text{outcome}&P(\text{outcome})&P(\text{input }0\mid\text{outcome})&P(\text{input }+\mid\text{outcome})\\\hline
0&3/4&2/3&1/3\\
1&1/4&0&1
\end{array}.
$$
For example, <Bayes' theorem> gives $P(0\text{ input}\mid0\text{ outcome})=(1/2)/(3/4)=2/3$. \b[Outcome one certifies input $|+\rangle$; outcome zero is ambiguous.] Both inputs that give outcome zero leave exactly the same post-measurement state $|0\rangle$.
In the <Hadamard basis>, input $|+\rangle$ always gives $+$, while input $|0\rangle$ gives $+$ or $-$ equally. Therefore
$$
\begin{array}{c|c|cc}
\text{outcome}&P(\text{outcome})&P(\text{input }0\mid\text{outcome})&P(\text{input }+\mid\text{outcome})\\\hline
+&3/4&1/3&2/3\\
-&1/4&1&0
\end{array}.
$$
\b[Outcome minus certifies input $|0\rangle$; outcome plus is ambiguous.] Guessing the more likely input after the ambiguous outcome gives success probability \b[$3/4$] for either of these two fixed-basis measurements.
Back to article page
