In cold atom experiments, one very basic question one can ask is, given some atom cloud, what is the number of atoms \(N\) in the cloud? One way is to basically shine some light on the atom cloud and see how much is absorbed. This absorption effect is quantified by the Beer-Lambert law.
\[I(z)=I(0)e^{-n\sigma z}\]
where \(n=N/V\) is the number density of atoms in the cloud of volume \(V\) and \(\sigma=\sigma(\omega_{\text{ext}})\) is the optical absorption cross-section presented by each atom in the cloud to incident monochromatic light of frequency \(\omega_{\text{ext}}\).
It is instructive to derive the Beer-Lambert law from first principles. In particular, the derivation is meant to emphasize that, for the most part, one can basically just think of the Beer-Lambert law as a mathematical theorem about probabilities, with some quantum mechanical asterisks to that statement. To get a sense of this, consider first a \(2\)D version of the Beer-Lambert law, in which one has an atom cloud confined to a plane, along with an incident beam of photons of frequency \(\omega_{\text{ext}}\) travelling along the (arbitrarily defined) \(z\)-direction.
The (average) number density of atoms is \(n\) (units: \(\text{atoms}/\text m^2\)) and each atom can be thought of as a “hard circle” with diameter \(\sigma\) (units: \(\text m/\text{atom}\)). In that case, in a small strip of width \(dz\), there will be \(ndz\) atoms per unit length along the strip, or equivalently the average interatomic spacing is \(1/ndz\) along the strip (see the picture). The probability that a given photon “collides” with such an atom is therefore \(\sigma/(1/ndz)=n\sigma dz\); such photons are depicted red on the diagram, while those that make it through the first layer \(dz\) are depicted green. Over many photons, this manifests as a loss \(dI<0\) in their collective intensity \(I\) across the layer \(dz\), so one may equate the fractional loss of intensity with the absorption probability:
\[\frac{dI}{I}=-n\sigma dz\]
for which the solution of this ODE yields the Beer-Lambert law:
\[I(z)=I(0)e^{-n\sigma z}\]
where \(1/n\sigma\) is the length scale of this exponential attenuation in the beam intensity. Of course, this argument generalizes readily to the \(3\)D case where now \(n\) (units: \(\text{atoms}/\text m^3\)) is the number density of atoms in \(\textbf R^3\) and \(\sigma\) (units: \(\text m^2/\text{atom}\)) is now the optical cross-section presented by each atom. As stressed earlier, there isn’t really much physics going on here, it’s just a statement about the statistics of a \(3\)D Galton board.
At this point however, one would like to introduce some quantum mechanical modifications to this simple Beer-Lambert law. As usual, suppose the laser light \(\omega_{\text{ext}}\) is not too detuned from a particular atomic transition \(\omega_{01}\) between some ground state \(|0\rangle\) and some excited state \(|1\rangle\) in each of the atoms in the cloud (also assume for simplicity that both \(|0\rangle\) and \(|1\rangle\) are non-degenerate). In that case, it makes sense to distinguish \(n=n_0+n_1\) between the number density \(n_0\) of atoms in the ground state \(|0\rangle\) vs. the number density \(n_1\) of atoms in the excited state \(|1\rangle\) since only the atoms in the ground state \(|0\rangle\) can absorb the incident photons, after which they go into the excited state \(|1\rangle\) and so are no longer able to absorb any more photons. Thus, one might think that the correct form of the Beer-Lambert law should be:
\[\frac{dI}{I}=-n_0\sigma dz\]
But this is forgetting that atoms in the excited state \(|1\rangle\) can undergo stimulated emission too back down to the ground state \(|0\rangle\) (and in the steady state, recall from Einstein’s statistical argument that the rates of stimulated absorption and emission are equal). In contrast to absorption, this would have the effect of actually increasing the intensity \(I\) because the atom emits a photon back into the beam. Thus, the correct form of the Beer-Lambert law is actually:
\[\frac{dI}{I}=-n_0\sigma dz+n_1\sigma dz=(n_1-n_0)\sigma dz\]
where by time-reversal symmetry the optical cross-section \(\sigma\) is the same for both stimulated absorption and emission. In the steady state (i.e. when \(\dot n_1=\dot n_2=0\) reach an equilibrium), it is clear that one must also have \((n_0-n_1)\sigma I=n_1\Gamma\hbar\omega_{\text{ext}}\) where \(\Gamma=A_{10}\) is the rate of spontaneous emission/decay from the excited state \(|1\rangle\) back down to the ground state \(|0\rangle\) (note that it really is \(\hbar\omega_{\text{ext}}\) and not \(\hbar\omega_{01}\) in the formula; whatever frequency an atom absorbs must also be what it emits by energy conservation). On the other hand, also in the steady state, the optical Bloch equations assert that:
\[\rho_{11}=\frac{n_1}{n}=\frac{1}{2}\frac{s}{1+s+(2\delta/\Gamma)^2}\]
where \(s=I/I_{\text{sat}}=2(\Omega/\Gamma)^2\) is the saturation. Combining these two expressions allows one to obtain an explicit formula for how the optical cross-section \(\sigma\) depends on the “driving frequency” \(\omega_{\text{ext}}\) of the incident photons in e.g. a laser:
\[\sigma(\omega_{\text{ext}})=\frac{1}{1+(2\delta/\Gamma)^2}\frac{\hbar\omega_{\text{ext}}\Omega^2}{\Gamma I}\]
where there is also an \(\omega_{\text{ext}}\)-dependence hiding in the detuning \(\delta=\omega_{\text{ext}}-\omega_{01}\). At first glance, this seems to suggest that the optical cross-section \(\sigma\), in addition to depending on \(\omega_{\text{ext}}\) also depends on the intensity \(I\) of the incident photons, but actually this is an illusion, because the Rabi frequency \(\Omega\) also depends on \(I\) in such a way that the two effects cancel out so as to actually make \(\sigma\) independent of \(I\). To see this, recall that the time-average of the Poynting vector over a period \(2\pi/\omega_{\text{ext}}\) is \(I=\varepsilon_0 c|\textbf E_0|^2/2\) and that the Rabi frequency is \(\hbar\Omega=e\textbf E_0\cdot \langle 1|\textbf X|0\rangle\). The unsightly presence of the matrix element can be further removed by recalling that (in the dipole approximation) one has \(\Gamma=4\alpha\omega_{01}^3|\langle 1|\textbf X|0\rangle|^2/3c^2\). Therefore, in the best case where the incident light is polarized along the dipole moments of the atoms, then \(\Omega^2=e^2|\textbf E_0|^2|\langle 1|\textbf X|0\rangle|^2/\hbar^2\). If on the other hand the incident light were unpolarized or the atoms in the cloud were randomly oriented, then isotropic averaging would contribute an additional factor of \(1/3\):
\[\langle\cos^2\theta\rangle_{S^2}=\frac{1}{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi}d\theta\cos^2\theta\sin\theta=\frac{1}{3}\]
Sticking to the best case scenario (which can be thought of as an upper bound if one likes though it is experimentally the typical situation since one often tries to maximize \(\sigma\) anyways), this leads to the explicitly \(I\)-independent form of the optical cross-section:
\[\sigma(\omega_{\text{ext}})=\frac{1}{1+(2\delta/\Gamma)^2}\frac{6\pi\omega_{\text{ext}}c^2}{\omega_{01}^3}\]
so the optical cross-section takes its maximum value at \(\omega_{\text{ext}}=\sqrt{\omega_{01}^2+(\Gamma/2)^2}\) but because the line width \(\Gamma\ll\omega_{01}\) is typically much less than the transition frequency itself, this is basically just \(\omega_{\text{ext}}\approx \omega_{01}\) so the maximum cross-section \(\sigma_{01}\) occurs on resonance and is given by:
\[\sigma_{01}=\sigma(\omega_{01})=\frac{6\pi c^2}{\omega_{01}^2}=\frac{3\lambda_{01}^2}{2\pi}\]
This also allows one to approximate the spectrum of the optical cross-section \(\sigma\) as just a Lorentzian profile centered at \(\omega_{\text{ext}}\approx \omega_{01}\) with \(\Gamma\) being its FWHM:
\[\sigma(\omega_{\text{ext}})\approx\frac{\sigma_{01}}{1+(2\delta/\Gamma)^2}\]
Typical transition wavelengths (e.g. visible light) might be around \(\lambda_{01}\sim 10^{-7}\text{ m}\) which far exceeds the length scale \(\sim a_0\sim 10^{-11}\text{ m}\) of the individual atoms themselves. The corresponding optical cross-section \(\sigma_{01}\sim\lambda_{01}^2\) is thus much larger than the actual “size” of the atoms themselves, so this emphasizes another quantum mechanical discrepancy to the classically-minded picture where \(\sigma\) would have just been interpreted as the size of individual “hard sphere” atoms (and in that case it wouldn’t have any \(\omega_{\text{ext}}\)-dependence in the first place). Moreover, the fact that near resonance \(\sigma\) is much larger than the atoms themselves also helps to ensure laser cooling actually works since it gives each photon more “leeway” in that it doesn’t need to hit an atom “head-on” to be absorbed, but merely has to pass within the cross-section \(\sigma\).
Intensity Saturation & Broadening
At low incident intensities \(s\ll 1\), spontaneous emission dominates stimulated absorption/emission \(\Gamma\gg\Omega\) and so any atom which is excited from the ground state \(|0\rangle\) into the excited state \(|1\rangle\) will quickly decay back down to the ground state \(|0\rangle\) by spontaneous emission. However, as one ramps up the laser intensity to saturation \(s\to 1\) and even \(s>1\), although there is a cap \(\rho_{11}<1/2\) on the excited state population, nevertheless the ground state population \(\rho_{00}\to 1/2\) will have depleted so much that there won’t be that many atoms left to absorb any more incident photons, so one would expect the sample to get worse and worse at absorbing incident photons. Recalling that \(s=I/I_{\text{sat}}=2(\Omega/\Gamma)^2\) (note that in the optimal case \(I_{\text{sat}}=\hbar\omega_{01}^3\Gamma/12\pi c^2\) but importantly is an intrinsic property of the atomic transition that scales with the transition frequency as \(I_{\text{sat}}\propto\omega_{01}^6\) due to the extra factor of \(\omega_{01}^3\) in \(\Gamma\)), it is clear that when \(s\to 1\), the Rabi frequency \(\Omega\) grows to the point of being comparable with the spontaneous decay rate \(\Gamma\), so now stimulated emission starts competing with spontaneous emission. In order to see this mathematically, it is useful to look at the absorption coefficient whose reciprocal directly governs the length scale of attenuation in the Beer-Lambert law:
\[(n_0-n_1)\sigma=\frac{n\sigma_{01}}{1+s}\frac{1}{1+(2\delta/\Gamma\sqrt{1+s})^2}\]
This is just another Lorentzian similar to the cross-section \(\sigma(\omega_{\text{ext}})\) itself. But there’s a crucial difference; whereas the FWHM of the Lorentzian for \(\sigma\) was fixed at \(\Gamma\), here it is \(\Gamma\sqrt{1+s}\); but this is now dependent on the laser intensity \(s\), causing the Lorentzian to broaden as \(s\) increases (this is exactly the same kind of broadening seen in \(\rho_{11}\); one difference though is that while \(\rho_{11}\to 1/2\) saturates, here the resonant absorption coefficient \(n\sigma_{01}/(1+s)\) just decreases monotonically as \(s\) is ramped up).
Finally, one can revisit the original Beer-Lambert law \(I(z)=I_0e^{-n\sigma z}\) and ask what becomes of it after all the modifications; from the expression for the absorption coefficient above, one has:
\[\frac{ds}{dz}=-\frac{n\sigma_{01}s}{1+s+(2\delta/\Gamma)^2}\]
In terms of the line-of-sight atomic column density \(n_c(z):=\int_0^zn(z’)dz’\), this ODE is trivial to integrate:
\[n_c\sigma_{01}=\ln\frac{I_0}{I}+\frac{I_0-I}{I_{\text{sat}}}\]
where \(I_0:=I(z=0)\) is the incident irradiance. The quantity \(\ln I_0/I\) is often called the optical density (OD) in AMO physics, or the absorbance in chemistry. In practice, this formula cannot just be used as is, but rather requires calibrating for the polarization, detuning fluctuations, optical pumping losses, etc. by sweeping over a range of incident intensities \(I_0\) and, using some known atom number \(N=n_cA\) obtained by other methods, choosing \(I_{\text{sat}}\) so that \(n_c\sigma_{01}\) is approximately invariant for all \(I_0\) and corresponding \(I\).