Past exam of the quantum, atomic and molecular physics course of the University of Oxford / 2016 / 4 / b / Solution

At first order in time-dependent perturbation theory set $c_1\simeq 1$ and use the parity result $V_{22}=0$. Decomposing the cosine into its two exponentials givesBoth exponentials contribute to the exact first-order result:Each apparent zero denominator has a continuous limit. In particular the near-resonant fraction tends to $it$ as $\omega \to {\omega }_0$, while the counter-rotating fraction remains bounded on the optical timescale. One may equivalently write the expression asThis form makes $c_2(0)=0$ explicit and avoids cancellation at resonance. At extremely short time both sinc functions tend to one, so $c_2\simeq -i\Omega t$, agreeing with the initial derivative of the original cosine-driven equation. No rotating-wave approximation has yet been made. The separate assumption of weak excitation remains necessary: the first-order expression must not be extended until it predicts probabilities comparable to or larger than one.