 Solution

Continue with the initial quantum state ${\psi }_1$ from part (ii), and let $\Delta t=T/N$. Conditional on every preceding outcome being $+\hslash /2$, the post-measurement state after each ideal projective measurement is again ${\psi }_1$, up to global phase. Part (ii) therefore gives the same conditional probability ${\cos }^2(\omega \Delta t)$ for each successive outcome. Multiplying these conditional probabilities yields the repeated projective-measurement survival probabilityThis multiplication uses the quantum state reset at each successful projective measurement; the outcomes need not be assumed statistically independent.

For fixed $T$, put $z=\omega T$. The Taylor expansion ${\cos }^{2}u=1-u^2+O(u^4)$ gives the quadratic short-time quantum survival probability. For sufficiently large $N$, taking logarithms givesHere the remainder uses Big O notation. Exponentiating provesThis is the quantum Zeno effect: increasingly frequent ideal projective measurements inhibit the transition away from the initial quantum state. The departure probability in a short interval is quadratic in its duration, so $N$ intervals of duration $T/N$ produce a total leading departure probability proportional to $T^2/N$. Without intervening measurements, the return probability at time $T$ would be ${\cos }^2(\omega T)$. The result concerns obtaining $+\hslash /2$ at every measurement in the ideal limit of instantaneous measurements with unitary time evolution between them.