It looks like there are two eigenstates $\chi_{\pm}$ of $Q$, and a measurement yielding $Q=\pm1$ leaves the system in the eigenstate $\chi_+$ for $Q=+1$. The time evolution operator being
$e^{-iHt/\hbar}$, after time $T/n$ where $n$ is large this state becomes
$e^{-iHT/(n \hbar)} \chi_+ = \left(I - \frac{i H T}{n\hbar}\right) \chi_+ + {\cal O}(1/n^2)$. So apparently (and this will be the result of whatever "the previous example" was), we must have
$$ \langle \chi_+, H \chi_+ \rangle= \dfrac{E_1 + E_2}{2}$$
If $A_n$ is as given, show that $|A_n|^2 = 1 + {\cal O}(1/n^2)$. What does this say about
$\log(|A_n|^2)$ and then about $\log(|A_n|^{2n}) = n \log(|A_n|^2)$ and $|A_n|^{2n}$?
It looks like there are two eigenstates $\chi_{\pm}$ of $Q$, and a measurement yielding $Q=\pm1$ leaves the system in the eigenstate $\chi_+$ for $Q=+1$. The time evolution operator being $e^{-iHt/\hbar}$, after time $T/n$ where $n$ is large this state becomes $e^{-iHT/(n \hbar)} \chi_+ = \left(I - \frac{i H T}{n\hbar}\right) \chi_+ + {\cal O}(1/n^2)$. So apparently (and this will be the result of whatever "the previous example" was), we must have $$ \langle \chi_+, H \chi_+ \rangle= \dfrac{E_1 + E_2}{2}$$
If $A_n$ is as given, show that $|A_n|^2 = 1 + {\cal O}(1/n^2)$. What does this say about $\log(|A_n|^2)$ and then about $\log(|A_n|^{2n}) = n \log(|A_n|^2)$ and $|A_n|^{2n}$?