= Perfect discrimination of pure states requires orthogonality
If a <positive operator-valued measure> distinguishes normalized <pure states> $|u\rangle,|v\rangle$ with certainty, the effect $E$ for the first answer satisfies $\langle u|E|u\rangle=1$ and $\langle v|E|v\rangle=0$. Positivity of $E$ and $I-E$ implies $E|v\rangle=0$ and $E|u\rangle=|u\rangle$, because a positive operator with zero expectation annihilates the vector. Consequently $\langle u|v\rangle=\langle u|E|v\rangle=0$. Conversely, orthogonal states can be measured with their rank-one <linear projections>. A common <unitary operator> preserves this <inner product> and cannot remove the obstruction.
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