Solution (source code)

= Solution

Resolve the weak <broadband radiation> into narrow <frequency> intervals whose relative phases are uncorrelated. The transition \b[<probabilities>], rather than the coherent amplitudes, can then be added after ensemble or time averaging. For the x-polarized field, the <energy> density in one interval is $\rho(\omega)d\omega=\epsilon_0\mathcal E_\omega^2/2$, so its squared coupling is $|d_{21}|^2\mathcal E_\omega^2/\hbar^2$. Inserting this into the weak rotating-wave result gives
$$
P_2(t)=\frac{|d_{21}|^2}{2\epsilon_0\hbar^2}
\int_0^\infty\rho(\omega)
\left[\frac{\sin[(\omega-\omega_0)t/2]}{(\omega-\omega_0)/2}\right]^2d\omega.
$$
For times long compared with the <optical coherence time> of the field, the narrow resonant kernel samples a nearly constant $\rho(\omega_0)$. Provided the resonance is well away from the zero-frequency boundary, extend the <Rabi detuning> integral to the full real line. Its value is $2\pi t$, yielding
$$
\boxed{P_2(t)\simeq W_{1\to2}t,\qquad
W_{1\to2}=\frac{\pi|\langle2|ex|1\rangle|^2}{\epsilon_0\hbar^2}\rho(\omega_0).}
$$
This is the <stimulated absorption> rate for the specified <polarization> and spectral <radiation energy density> \b[per angular-frequency interval]. The averaging window should also satisfy $Wt\ll1$ while the lower state is treated as undepleted. A deterministic phase-locked broadband pulse would not in general permit adding these <probabilities> without its <interference> terms.

For isotropic unpolarized radiation with total spectral <radiation energy density> $\rho_{\rm tot}$, orientational averaging instead gives $W=\pi|\mathbf d_{21}|^2\rho_{\rm tot}(\omega_0)/(3\epsilon_0\hbar^2)$. The factor one third is not inserted into the original x-polarized formula. If ordinary <frequency> is used, $\rho_\nu=2\pi\rho_\omega$ and the isotropic coefficient becomes $|\mathbf d_{21}|^2/(6\epsilon_0\hbar^2)$ multiplying $\rho_\nu(\nu_0)$. The spectral convention and <polarization> must be stated together to avoid spurious factors of three or $2\pi$.