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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?

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Please give more details of your experiment, escpecially your light source (linewidht, polarization), your magnetic field (parallel or perpendicular) and your detection scheme (optically detected double resonance, absorption, fluorescence). Without further infos there are to many ways to do opticaly pumping. –  Alex1167623 Apr 10 '12 at 20:32

4 Answers 4

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.

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But what does the magnetic field do? In our experiments we cannot see any optical pumping effect without the presence of magnetic field. –  Siyuan Ren Mar 10 '12 at 10:11
It would help if you explained how your experiment is setup? What kind of light source, detectors...there are many reasons why optical pumping is getting washed out. Pulsed light or CW light? How are you probing the system? –  Antillar Maximus Mar 10 '12 at 10:15
Maybe you could just explain the difference between the hyperfine type and Zeeman type. I could not see any difference between this two. –  Siyuan Ren Mar 10 '12 at 11:21
We use an alternating magnetic field parallel to the laser. The intensity of light passing through Rb varies accordingly. When the magnetic field is stable, Rb becomes transparent; without, opaque. –  Siyuan Ren Mar 11 '12 at 4:09

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.

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But if atoms can be pumped to $m=2$ sublevel without magnetic field, it will become transparent to $D_1 \ \sigma^+$ light, a rather obvious effect. In our experiment the intensity of magnetic field is directly related to how much light passes through. –  Siyuan Ren Mar 10 '12 at 17:12
Maybe you could give some more details (geometry) about "your experiment"? Though at first glance it seems that you might be observing the "magneto-optical" resonances with magnetic field and laser beem being mutually orthogonal which would give a transparent media at $B = 0$ and non-transparent at $B \neq 0$. –  Linards Kalvans Mar 11 '12 at 9:52

Though it's too late to answer, I was looking for answer and saw your question. I observed the same thing through my experiments today. Zeroing the imposed magnetic field leads to vanish the dark state due to a circularly polarized light.

My reasoning is related to the most basic concepts of quantum mechanics; based on my knowledge it's impossible to measure a system (say atomic population) without changing that system. In fact, in the absence of any external agency, there is no preferred direction and we have a perfect symmetry (off course prefect doesn't mean perfect actually but it's valid considering our experimental restrictions for resolving broken symmetry). So, there is no Z direction for atoms. Finally, you can observe optical pumping and dark states only if you align atoms in a preferred direction which is defined by the external magnetic field.

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This is a great question I too have been pondering about. I think theoretically it has to do with good quantum number. The optical pumping relies on defining circular polarization transition selection rules as delta m = +1 for a RCP polarization. Why is that? It comes from the fact that B defines the quantization axes. Imagine the experiment in which B and the light are perpendicular to each other. Would you get optical pumping? Now you have a RCP light propagating in the x direction while B field is in the z-direction. So what you have here is an RCP light whose selection rule does not follow delta m=+1 but will follow linear selection rules (even though the light is RCP). So now the real question is - what is the reality behind quantization axes. Why should a B-field define a quantization and not the light field? Let us then do this the other way. Imagine you have a light field defining the quantization axes i.e., x-beam is RCP and it gives selection rule of delta m=+1, and you have a x as quantization axes and B NOT along the quantization axis. Now to begin with how would you even define your Zeeman states? I cannot answer this question and I am badly stuck at this point. Maybe you have found the answer. If so please share.

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