 Solution

Use first-order time-dependent perturbation theory, setting $c_1=1$ in the equation for $c_2$ and retaining the generally complex matrix element $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$. With $\delta ={\omega }_0-\omega$ and $\Sigma ={\omega }_0+\omega$, the probability amplitude isFor $t>0$, split the integral at the pulse maximum. For either $x=\delta$ or $x=\Sigma$,Thus the full first-order answer, without discarding the counter-rotating contribution, isAt long times it givesThis full expression is not exactly one squared Lorentzian line shape. To obtain the intended near-resonance spectral line profile, additionally assume $b,|\delta |\ll {\omega }_0$, so the pulse has a narrow positive-frequency lobe and $F(\Sigma ,t)$ is negligible. ThenThe weak-excitation condition near resonance is $|d|/(\hslash b)\ll 1$. The squared Lorentzian line shape occurs because the symmetric exponential pulse has a Lorentzian line shape amplitude Fourier transform, which is then squared to obtain probability.

At half the maximum, $[1+(\delta /b{)}^2{]}^2=2$, soThe full width at half maximum refers to varying the drive frequency through the positive-frequency resonance. For a concrete failure of the unqualified squared-Lorentzian line shape claim, set $b=\omega ={\omega }_0$ while keeping $|d|/(\hslash b)\ll 1$. The full first-order resonant probability is $36/25$ times the one-lobe prediction. Weak radiation alone therefore does not justify dropping the second Fourier transform lobe.