Past exam of the computer science course of the University of Oxford / 2019 / Quantum information / 1 / e

Put $\delta =\pi /(2n)$. Between successive measurement bases, the squared overlap is $c={\cos }^2\delta$ for retaining the same label and $s={\sin }^2\delta$ for changing label. The first step obeys the same rule if we regard the initial state as label 0 at angle zero. Thus the recorded labels form a Markov chain with transition matrix $\left(\begin{array}{cc}c& s\\ s& c\end{array}\right)$.

Let $q_k$ be the probability of label 0 after step $k$. Then $q_0=1$ andSince $c-s=\cos (\pi /n)$, iteration yieldsThis is the total final-outcome probability, including histories that change label an even number of times. The smaller quantity ${\cos }^{2n}(\pi /(2n))$ counts only the history with label 0 at every step and is not the requested total. For example, at $n=3$ the total is $9/16$, whereas the uninterrupted history has probability $27/64$. At $n=1$, the final basis is $\{|1⟩,|0⟩\}$ up to signs, so the total probability of label 0 is zero, as the formula also gives. This distinguishes nonselective measurement-guided Zeno transfer from postselection on one record.

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.