Why does optical pumping of Rubidium require presence of magnetic field?

Question

The optical pumping experiment of Rubidium requires the presence of magnetic field, but I don't understand why.

The basic principle of pumping is that the selection rule forbids transition from $m_F=2$ of the ground state of ${}^{87} \mathrm{Rb}$ to excited states, but not the other way around ($\vec{F}$ is the total angular momentum of electron and nucleus). After several round of absorption and spontaneous emission, all atoms will reach the state of $m_F=2$, hence the optical pumping effect.

But what does the Zeeman splitting have anything to do with optical pumping? Granted, the ground state, even after fine structure and hyperfine structure considered, is degenerate without Zeeman splitting, but the states with different $m_F$ still exists.

In addition, how is the strength of optical pumping related to the intensity of magnetic field applied?

Answer 1

There are two kinds of optical pumping that are possible. Hyperfine Optical pumping and Zeeman Optical Pumping. The latter is under zero magnetic field and former requires a magnetic field. The spin-wave (ground state coherence) you create depends mainly on the polarization of the light being used and its intensity. If you increase the strength of the magnetic field, the separation between the Zeeman levels increases. For large values, you get a "level-crossing" effect.

What do you mean by "strength of optical pumping?" If you mean the population redistribution, then the strength of the magnetic field alone does not play a role. You have to consider the nature of the light (intensity, polarization,detuning,coherence) and the decoherence of the atomic ensemble as well.

Other applications:

In many experiments that exploit atomic coherence, using the Zeeman levels (i.e a 3-level system that consists of two ground state Zeeman levels of the same hyperfine state and the excited state) alleviates the problem of requiring two identical lasers. You can use one laser (perfectly correlated with itself!) and shift its frequency by a small amount using an AOM (say 80Mhz) as opposed to 6.1 GHz between the ground state hyperfine levels.

Answer 2

In a hyperfine pumping you pump atoms to the other hyperfine level, let's say you apply your laser to the $F_g = 2 \rightarrow F_e = 2$ transition of the $^{87}$Rb D$_1$, then the atoms from the $F_e = 2$ level will decay to both ground state hyperfine levels $F_g=1,2$ and eventually will be pumped into the $F_g = 1$ level.

In a Zeeman pumping scheme the polarization of the exciting laser becomes important. If you apply let's say $\sigma^+$ polarized laser to the same transition $F_g = 2 \rightarrow F_e = 2$, then magnetic quantum number is changed by $\Delta m = 1$ and the $m_g = +2$ sublevel cannot absorb the exciting radiation as there is no $m_e = +3$ sublevel in the excited state, as atoms from the excited state sublevels $m_e = +1,+2$ can and do sponataneously decay to the sublevel $m_g = +2$ atoms that haven't been pumped to $F_g = 1$ level by the hyperfine pumping are pumped to the $m_g = +2$ by the Zeeman pumping.

What's the role of the magnetic field? I would say it's needed only to detect that you have pumped your atoms into $m_g = +2$ sublevel. As long as all the Zeeman sublevels are degenerate you cannot distiunguish the. Once the magnetic field is applied and the degeneracy is removed, you can use radiofrequency to detect the atoms, if you satisfy equality condition for the Zeeman shifts befor neighbouring magnetic sublevels and anergy of the radiofrequency quantum $\Delta E = \mu_BgB = h\nu_{rf}$, then the radio frequency induces transitions of $\Delta m = \pm 1$ within the ground state hyperfine level and atoms are brought back to the sublevels that can absorb light ($m_g \neq +2$) allowing to observe some fluorescence or changes in absorption.