= Solution
{c}
Continue with the initial <quantum state> $\psi_1$ from part (ii), and let $\Delta t=T/N$. Conditional on every preceding outcome being $+\hbar/2$, 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 probability>
$$
\boxed{\mathbb P_N(T)=\left[\cos^2\left(\frac{\omega T}{N}\right)\right]^N.}
$$
This 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> $\cos^2u=1-u^2+O(u^4)$ gives the <quadratic short-time quantum survival probability>. For sufficiently large $N$, taking logarithms gives
$$
\log\mathbb P_N(T)=N\log\left(1-\frac{z^2}{N^2}+O(N^{-4})\right)
=-\frac{z^2}{N}+O(N^{-3})\longrightarrow0.
$$
Here the remainder uses <Big O notation>. Exponentiating proves
$$
\boxed{\lim_{N\to\infty}\mathbb P_N(T)=1,\qquad
\mathbb P_N(T)=1-\frac{(\omega T)^2}{N}+O(N^{-2}).}
$$
This 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 $+\hbar/2$ at every measurement in the ideal limit of instantaneous measurements with <unitary time evolution> between them.
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