Interpretation of the photon scattering rate? The photon scattering rate $\Gamma$ describes the rate at which photons scatter off an atom$^1$. In a two-level system, the ansatz for the photon scattering rate often is given by
\begin{equation}
\Gamma = \rho_{22}\gamma
\end{equation}
where $\rho_{22}$ is the probability to find the atom in the excited state and $\gamma$ is the rate of spontaneous decay. However, I don't see the connection between the ansatz above and what the photon scattering rate is physically meant to be. 

$^1$In my imagination, the photon scattering rate is the absorption rate for photons at a certain frequency $\omega$. Hence $\Gamma(\omega)$ shows the saturation broadened Lorentzian absorption line of the atom, centered around a resonance frequency.
 A: Why do you imagine the photon scattering rate to the be absorption rate? The scattering rate is the rate at which an atom absorbs AND re-emits incident photons. $\rho_{22}$ captures how excited an atom becomes for a particular incident field and $\gamma$ captures how quickly the atom decays and re-emits the excitation. Thus the scattering rate is $\Gamma = \rho_{22} \gamma$.
Note that $\rho_{22}$ depends on the intensity and frequency of the incident light. This is what gives the amplitude and frequency dependence of the scattering rate. If the light you're driving the atom with doesn't excite the atom at all we don't expect the atom to scatter any photons.
A: If $\rho_{22}$ is the population of the excited state then for sure $\rho_{22} \gamma$ is the rate at which the atom is emitting photons. I guess the question arises from not realising that $\rho_{22}$ is itself dependent on the conditions the atom is under. For example, if the atom has been left alone for a while then you will have $\rho_{22} = 0$. If the atom is in an electric discharge then you will have $\rho_{22} > 0$. If the atom is located in a beam of light then $\rho_{22}$ will have some value which depends on the intensity and frequency of the light. It seems that the question has the latter scenario in mind. In that case you will have
$$
\rho_{22} = \rho_{22}(I, \omega)
$$
and a typical dependence on intensity and frequency (for a two-level model of the atom, in steady state) is
$$
\rho_{22} = \frac{(1/2)\gamma^2 I/I_s}{(\omega-\omega_0)^2 + \gamma^2/4 + \gamma^2 I/I_s}
$$
where $I_s$ is the saturation intensity. This leads to the Lorentzian function for $\Gamma(\omega)$ which the question asks about.
A: Considering light as a stream of photons at energy hω, photon scattering is usually defined as cycles of absorption and subsequent spontaneous emission.
$$Γsc(r) = Pabs
/hω = 1
/he0c*Im(α) I(r).$$
http://cds.cern.ch/record/380296/files/9902072.pdf
The photon scattering rate is the radiated power divided by the photon energy hω.
$$Rsc =
Prad
/ω$$
http://atomoptics-nas.uoregon.edu/~dsteck/teaching/quantum-optics/quantum-optics-notes.pdf
A: The scattering rate is given by $F\cdot\sigma$ where $F$ is the incident flux and $\sigma$ the scattering cross section, integrated over the Lorentzian line profile. You don't need to know the excitation rate for this. In fact, bringing in the spontaneous decay constant $\gamma$ here is in general incorrect. Coherent scattering (which you would have in the absence of any disturbances) is a one-quantum process and can not be described by an absorption followed by a spontaneous emission.
