Consider a quantum mechanical system with Hamiltonian $\hat{H}$ and assume the eigenvalue problem $\hat{H}|n\rangle=E_n|n\rangle$ to be solved such that we have an orthonormal basis $\{|0\rangle, |1\rangle,\ldots\}$. Disturb the system now with a potential $\hat{V}(t)=(\hat{A}e^{-i\omega t}+\hat{A}^{\dagger}e^{i\omega t})\mathbb{1}_{t\ge 0}$ where $\hat{A}$ is an operator which does not explicitly depend on time.
Assignment: Compute $P_{|n\rangle\to|m\rangle}(t)$, the transition probability from state $|n\rangle$ to state $|m\rangle$.
Attempt: From first order perturbation theory, we know that $P_{|n\rangle\to|m\rangle}(t)=\left| \frac{1}{\hbar}\int_0^{t}dt e^{i(E_m-E_n)t/\hbar}\cdot\langle m|\hat{V}(t|n\rangle) \right|^2=\frac{1}{\hbar^2}\left| \frac{e^{i(E_m-E_n-\omega\hbar)t/\hbar}-1}{i(E_m-E_n-\omega\hbar)/\hbar}\langle m|\hat{A}|n\rangle + \frac{e^{i(E_m-E_n+\omega\hbar)t/\hbar}-1}{i(E_m-E_n+\omega\hbar)/\hbar}\langle m|\hat{A}^{\dagger}|n\rangle \right|^2=\frac{1}{\hbar}^2\left| e^{i(E_m-E_n-\omega\hbar)t/(2\hbar)}\cdot\frac{\sin\left(\frac{(E_m-E_n-\omega\hbar)t}{2\hbar}\right)}{\left(\frac{E_m-E_n-\omega\hbar}{2\hbar}\right)}\langle m|\hat{A}|n\rangle + e^{i(E_m-E_n+\omega\hbar)t/(2\hbar)}\cdot\frac{\sin\left(\frac{(E_m-E_n+\omega\hbar)t}{2\hbar}\right)}{\left(\frac{E_m-E_n+\omega\hbar}{2\hbar}\right)}\langle m|\hat{A}^{\dagger}|n\rangle\right|^2$.
Define now $\Delta_t(x)=\frac{\sin^2(xt)}{\pi x^2t}$ for $t>0$ then $\int_{-\infty}^{\infty}dx \Delta_t(x)=1$ and $\lim_{t\to\infty} \Delta_t(x)=\delta(x)$, the Dirac $\delta$-distribution. According to my lecture notes, the following now holds: $P_{|n\rangle\to|m\rangle}(t)=\frac{\pi t}{\hbar^2}\delta\left( \frac{E_m-E_n-\omega\hbar}{2\hbar} \right)|\langle m|\hat{A}|n\rangle|^2 +\frac{\pi t}{\hbar^2}\delta\left( \frac{E_m-E_n+\omega\hbar}{2\hbar} \right)|\langle m|\hat{A}^{\dagger}|n\rangle|^2 $.
Why does the last equality hold? If we evaluate the product $|.|^2$ from above, we should obtain mixture terms containing both$\langle m|\hat{A}|n\rangle$ and $\langle m| \hat{A}^{\dagger}|n\rangle$ and I cannot see why they should vanish unless $\langle m|\hat{A}|n\rangle\ \cdot\langle m| \hat{A}^{\dagger}|n\rangle$ is purely imaginary, which shouldn't hold for an arbitrary operator.