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I'll be using the diagrams here to explain a bit:

https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=5393

In dichroic atomic vapor spectroscopy you have a constant longitudinal field through a vapor cell, you send linearly polarized light through the cell, and then measure the output on a photodetector. Because of the Zeeman effect, the energy levels split and you can find different transition frequencies through spectroscopy. So if you probe the cell with linearly polarized light with frequency $f_0$ (say, corresponding to the D1-line of rubidium), then you actually see two red-shifted and blue-shifted absorption peaks; this is because the linearly polarized light can be seen as a superposition of left-handed and right-handed circularly polarized light.

I'm confused because it doesn't seem (or at least, no one has thought it useful to mention) that it's possible to still see the broad central peak at $f_0$ due to linearly polarized light still interacting with the atoms as linearly polarized. Those transitions are still available in terms of frequency differences (since the Zeeman splitting doesn't offset the $m_l = 0$ line), and I don't believe there are selection rules that I'm overlooking, but it would be a convenient explanation if that were the case.

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Good question. If we take the quantization axis to be the magnetic field direction then linearly polarized light propagating along the magnetic field direction is a superposition of $\sigma^+$ and $\sigma^-$. It has no $\pi$ component so it can’t drive the $0\rightarrow 0$ transition you are talking about. However, light which is propagating perpendicular to the magnetic field direction and which is polarized such that the electric field is parallel to the magnetic field is entirely composed of $\pi$ polarized light so it would drive the transition you are considering.

It is a common confusion that circularly polarized light is always $\sigma$ light and linearly polarized light is always $\pi$ Light but this is not the case. The definition of $\sigma^{\pm}$ and $\pi$ light depends on the choice of quantization axis. When I’m at a computer I will link two answers that explain this in more detail.

TLDR: geometry matters when it comes to polarization and selection rules.

Some other relevant questions and answers:

$\pi, ~\sigma$ - atomic transitions with respect to a quantization axis

$\pi$, $\sigma$ - atomic transitions with respect to the magnetic field axis

$\pi$, $\sigma^\pm$ components with no magnetic field?

The definition of a $\pi$ polarized photon?

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Single photons have spin of $\pm 1$ along the direction of motion. So fundamentally a photon is circularly polarised. This is why the spectrum has selection rules. Based on the conservation of angular momentum.

This would mean that $m_l= 0\to m_l=0$ transitions are forbidden because they don’t conserve angular momentum. So each transition to an $n,l$ level will have slightly different energies for different $m_l$ values.

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