 Solution

At the last step, $|{\psi }_{0,n}⟩=|1⟩$ and $|{\psi }_{1,n}⟩=|0⟩$. Ignoring the recorded outcome, the final density matrix isFor large $n$, the Taylor expansion $\log \cos x=-x^2/2+O(x^4)$ givesThus $q_n\to 1$, and the final state tends to $|1⟩⟨1|$, or $|1⟩$ as a pure-state representative. In fact $1-q_n\sim {\pi }^2/(4n)$, and its trace distance from $|1⟩⟨1|$ is exactly $1-q_n$. With records retained, the final conditional state is $|1⟩$ 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.