Optical pumping of Rubidium - degeneracy of energy levels I'm currently doing a lab experiment on the Optical pumping of rubidium and I'm fairly confused about the different levels of splitting. 
As far as I am aware, an atom has a fine structure and then a hyperfine structure. The hyperfine structure can then be split further through the Zeeman effect. 
I understand degeneracy as two or more quantum states occupying the same energy. However, what is the difference in 'degeneracy' for the fine, hyperfine and Zeeman splittings? Are these states equally degenerate?
We've always been taught in lectures as the Zeeman effect lifting the 'degeneracy' of the state. But is this the degeneracy of the fine state or the hyperfine state? 
I'm fairly confused, so any insight is very valuable. 
 A: Everything depends on the strength of the magnetic field relative to the other parts of our Hamiltonian. Let's start from the beginning. Our atom is described by the Hamiltonian
$$\hat{H}=\hat{H}_{0}+\hat{H}_{\rm FS}+\hat{H}_{\rm HFS}$$
Here $\hat{H}_{0}$ depicts the gross structure of the atom, i.e. the levels $\left|n,l,m_{l}\right>$; $\hat{H}_{\rm FS}$ the fine-structure and $\hat{H}_{\rm HFS}$ the hyperfine-structure. The order matters (at least aesthetically), since we have a strict order of magnitude relation $E_{0}\gg E_{\rm FS}\gg E_{\rm HFS}$. Now we add some magnetic field through an interaction Hamiltonian $\hat{H}_{\rm I}$. The question now is where are we going to place it?
Weak field
If the magnetic field is very weak such that $E_{\rm HFS}\gg E_{\rm I}$, we can treat our Hamiltonian as
$$\hat{H}=\underbrace{\hat{H}_{0}+\hat{H}_{\rm FS}+\hat{H}_{\rm HFS}}_{\rm main \: contribution}+\underbrace{\hat{H}_{\rm I}}_{\rm perturbation}$$
This means $F$ and $m_{F}$ are good quantum numbers and we treat $\hat{H}_{\rm I}$ as a perturbation to the $\left|F,m_{F}\right>$ states. Each such state has degeneracy $2F+1$ in $m_{F}$ and this degeneracy is lifted by the magnetic field $E_{\rm I}\propto m_{F}$. In other words the magnetic field splits the hyperfine-structure levels.
Strong field
If we increase the strength of our field such that $E_{\rm FS}\gg E_{\rm I}\gg E_{\rm HFS}$, the correct way to write our Hamiltonian is
$$\hat{H}=\underbrace{\hat{H}_{0}+\hat{H}_{\rm FS}}_{\rm main \: contribution}+\underbrace{\hat{H}_{\rm I}}_{\rm perturbation}+{\rm even \: weaker \: terms}$$
Now $\left|J,m_{J};I,m_{I}\right>$ is the good basis to work in and treating $\hat{H}_{\rm I}$ as perturbation leads to splitting of the fine-structure levels $E_{\rm I}\propto m_{J}+\left({\rm weaker\:}m_{I}{\rm \:terms}\right)$.
You can continue with this logic and find out what happens when the field term is even more dominant than the fine-structure interaction. This will lead to an $m_{l}$-dependent splitting.
The first two regimes for Rb-$87$ are depicted in the following figure, taken from Wikipedia

Rb-$87$ is an alkali metal, and thus have a single outer electron which, in the case of Rb, sits in the $5s$ orbital. All the inner shells are filled and don't contribute to the orbital angular momentum and/or the spin quantum numbers. Therefore $L=0$ and $S=\frac{1}{2}$ and consequently $J=\frac{1}{2}$. For this particular isotope of Rb the nuclear spin is $I=\frac{3}{2}$, leading to $F=1,2$, the first being the ground state. As you can see in the graph, for weak field each $F=1,2$ manifold is Zeeman-splitted into $2F+1=3,5$ different energy levels respectively. On the other hand, for stronger field all the energy levels are grouped into two manifold corresponding to $m_{J}=\pm\frac{1}{2}$, the lifting of the $2J+1=2$ degeneracy in the fine-structure levels. You still have $2I+1=4$ 'degeneracy' in each of these manifolds as a results of weaker nuclear spin correction to the Zeeman-splitting.
A: You're quite right that "degeneracy" refers to two (or more) states having the same energy.
In quantum mechanics classes, we often deal with hydrogen as the prime example. When you first solve Schrodinger's equation for the hydrogen atom Hamiltonian, you get an energy that depends solely on the principal quantum number $n$. Of course, all $n$ states have various angular momentum states associated with them, and the electron has spin $1/2$, so there is a set of ${j,m_j}$ states that are all degenerate for the hydrogen atom.
All of these results apply to rubidium, although the exact mathematics is a bit more difficult!
But there is a set of relativistic corrections to this first treatment of the hydrogen atom, which is fine structure. This "splits the degeneracy" of the $n$ states—that is, it makes some of the angular momentum states of any particular $n$ have different energies.
And then the electron interacts with the proton's spin, which is hyperfine structure, and which further "splits degeneracy," meaning now all of the angular momentum states of the atom's total angular momentum $F=L+S_e+I$, where $I$ is the nuclear (proton's) spin, are split: all $F$ values now have different energies. But this angular momentum $F$ still has different projections $m_F$, all of which are degenerate.
But you can add on another layer, which is an external magnetic field. Because we're thinking of $F$ states, the atom (electron + proton) is essentially a single magnetic dipole, all of whose $m_F$ states respond with different strengths to a magnetic field. An external magnetic field lifts this degeneracy, namely the degeneracy that remains after hyperfine splitting.
