= Solution
At the last step, $|\psi_{0,n}\rangle=|1\rangle$ and $|\psi_{1,n}\rangle=|0\rangle$. Ignoring the recorded outcome, the final <density matrix> is
$$
\rho_n=q_n|1\rangle\langle1|+(1-q_n)|0\rangle\langle0|.
$$
For large $n$, the <Taylor expansion> $\log\cos x=-x^2/2+O(x^4)$ gives
$$
n\log\cos(\pi/n)=-\frac{\pi^2}{2n}+O(n^{-3})\longrightarrow0.
$$
Thus $q_n\to1$, and \b[the final state tends to $|1\rangle\langle1|$], or $|1\rangle$ as a pure-state representative. In fact $1-q_n\sim\pi^2/(4n)$, and its <trace distance> from $|1\rangle\langle1|$ is exactly $1-q_n$. With records retained, the final conditional state is $|1\rangle$ with <probability> tending to one. These are compatible descriptions, not a coherent sum of the two outcome branches. This <quantum Zeno effect> guides a known preparation through slowly rotating measurement bases; it is not a unitary <Pauli X gate> acting coherently on an arbitrary unknown input.
\Image[/past-exam-of-the-computer-science-course-of-the-university-of-oxford/2019/quantum-information-measurement-guided-transfer.png]
{title=Total probability of the final target outcome versus the uninterrupted measurement record, and the asymptotic target-error probability in measurement-guided Zeno transfer}
Back to article page
