This link takes you to a short video description of optical pumping. The following text covers everything in the video more thoroughly, and the video is intended to supplement the information in the rest of this section.
Simply stated, optical pumping is a process by which an atom is caused to repeatedly absorb and emit photons so that the atom eventually ends up in an energy state which does not allow it to continue absorbing photons.
In this lab you will optically pump an ensemble of Rb atoms by placing them in a precisely controlled magnetic field and illuminating them with 794.8 nm photons which have been right circularly-polarized. Absorption of the right circularly-polarized photons drives the Rb atoms into a pumped state which cannot then absorb a right circularly-polarized photon. A photodetector is used to record the intensity of the light which passes through the Rb vapor. As more of the atoms are driven into the pumped state, fewer photons are absorbed by the gas and thus more light reaches the photodetector. Changes in the opacity of the vapor are correlated with the degree of pumping which has occurred.
Understanding the details of how optical pumping works in this case is analogous to the quantum description of the hydrogen atom and requires a conceptual understanding of the following:
In the next section we describe the optical pumping process in more detail.
Rubidium is a hydrogen-like atom in the sense that its ground state has just a single electron in the outermost shell. Thus. the structure of the energy states (outside the full inner shell) is very similar to that of hydrogen which also has only one electron in its outermost shell. Figure 1 below shows the energy level structure for the ground and first two excited states of 87Rb.
This diagram uses spectroscopic notation to denote the values of the quantum numbers of the various energy states. (See here for more details on how to read spectroscopic notation). As a refresher, the relevant quantum numbers are:
As indicated on this diagram, transitions between the first excited state $^{2}P_{1/2}$ and the ground state $^{2}S_{1/2}$ absorb (in the case of excitation) or emit (in the case of de-excitation) a photon with a wavelength $\lambda$ = 794.8 nm. Likewise transitions between the ground and second excited states involve photon wavelengths of wavelength 780.0 nm.
For an ensemble of Rb atoms at 50$^{\circ}$ C, the number of atoms in the ground and first excited states at any given moment is determined by the Boltzmann distribution, such that there will be more atoms in the ground state than are in the first excited state; this is the equilibrium state of the ensemble. This equilibrium state is the result of a balance between physical processes which drive atoms from the ground state into an excited state (such as the thermal motion of the atoms in the gas) and processes which cause excited atoms to de-excite (such as the natural lifetime of the excited state and thermal interactions).
Referring back to Fig 1, we see that both the ground and first excited states split into additional energy sub-states. Rubidium nuclei have an intrinsic spin (and corresponding magnetic moment) which interacts with the electron's magnetic moment resulting in two hyperfine states. If the atom is in an external magnetic field, (such as the Earth's magnetic field, for example), there will be additional splitting of the hyperfine states due to the interaction of the electron magnetic moment with the external field. This is known as the Zeeman effect.
Thus, in the presence of an external magnetic field, electronic transitions between the ground and first excited states are in fact transitions from the different Zeeman levels of the excited and ground states. Although there are a lot of possible combinations of transitions between ground and first excited state Zeeman levels, many of these combinations are “forbidden”. In this context, saying that a particular transition is “forbidden” simply means that the probability of that transition occurring is 0 or very near 0. You have probably encountered these quantum selection rules for electromagnetic transitions, summarized below:
| $\Delta L = 0, \pm 1$ (but not $0\rightarrow 0$) |
| $\Delta S = 0$ |
| $\Delta F = 0, \pm 1$ |
| $m_{F} = 0, \pm 1$ |
As an example, let's walk through one possible sequence of transitions for the case where the light being absorbed is right circularly-polarized. In Fig. 2 above, we show a sequence of five hypothetical electronic transitions events – (a), (b), (c), (d), and (f) – and discuss each in detail below.
(a): For purposes of this example, we pick an arbitrary starting point where the atom is in the $^{2}S_{1/2}(f = 1, m_{f}=0)$ ground state. This atom absorbs a right circularly-polarized photon. (Right circularly-polarized is another way of saying the photon carries +1 unit of angular momentum, so upon absorption the $m_{f}$ quantum number must increase by +1.) In this example then, the $F$ quantum number increases by 1, and since $m_{f}$ must increase by 1, this puts the atom in the $^{2}P_{1/2}(f = 2, m_{f}=1)$ state.
(b): When the atom de-excites, it is not required to lose +1 unit of angular momentum; that requirement only exists for absorption because of the circular polarization of the incident light. Instead, the atom can lose 0, +1 or -1 unit of angular momentum (which corresponds to the possible amounts of angular momentum that a photon can carry with it.) Therefore, the atom could potentially decay to any one of three possible states: $^{2}S_{1/2}(f = 2, m_{f}=0, 1, \textrm{ or } 2)$. As an example, let's suppose the atom decays to the $m_{f} = 1$ state.
(c): If the atom now absorbs another right circularly-polarized photon, it would only be able to be excited to the $^{2}P_{1/2}(f = 2, m_{f}=2)$ state because of the requirement that $\Delta m_{f} = +1$.
(d): At this point the atom can de-excite to more than one possible ground state. One possibility is that it ends up in the $^{2}S_{1/2}(f = 2, m_{f} = 2)$ state.
(f): The atom is now pumped because it cannot absorb another right circularly-polarized photon since that would require it to go to a $m_{f} = 3$ state (of which there are none). Therefore, unless some other process drives the atom into some other ground state (such as thermal agitation or stimulated emission by an RF frequency photon), the atom is stuck in the pumped state.
The sequence of excitations and de-excitations described above is just one of many possible permutations of allowed transitions. Not all atoms in the ensemble will end up in the pumped state, but some will end up there. The rate at which atoms are driven into the pump state is in part determined by the intensity of the light source. At the same time energy exchanges due to random thermal collisions between atoms will tend to de-pump atoms out of the pumped state. Equilibrium is reached when the rate of pumping equals the rate of depumping.
For ${}^{87}$Rb, the relation between the applied magnetic field and energy splitting is
| $E = -\vec{\mu}\cdot\vec{B} = g_f \left(\frac{e}{2m_e}\right)\vec{F}\cdot\vec{B} = g_f \left(\frac{e\hbar}{2m_e}\right)Bm_F$, |
where $\mu_B = e\hbar/2m_e \approx 5.7883 \times 10^{-9}$ eV/G is the Bohr magneton. The Landé g-factor, $g_f$, relates the atom's magnetic dipole moment to its quantum numbers, and is given by
| $g_f = g_j\dfrac{f(f+1) + j(j+1) - i(i+1)}{2f(f+1)}$. |
Here, $g_j$ relates the electron's contribution to the total magnetic moment and is given by
| $g_j = 1+\dfrac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}$. |
where $\ell$ is the orbital angular momentum of the electron (0 for the S shell, 1 for the P shell, etc) $s$ is the spin of the electron, and $j$ is the sum of the two.
The energy level shifts are therefore given by
| $\Delta E = \mu_B g_f \Delta m_f B \approx (5.7883 \times 10^{-9} \textrm{ eV/G })g_f \Delta m_f B$. |
Note that this energy difference represents the shift up or down from the un-split (hyperfine) level characterized by quantum number $g_f$.