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The hyperfine hamiltonian for an atom with one electron in valence shell is of the form:

$ H = H_0 + A_1\vec{S}\cdot\vec{L} + A_2\vec{L}\cdot\vec{I} + A_3\vec{S}\cdot\vec{I} + A_4(\vec{I}\cdot\hat{r})(\vec{S}\cdot\hat{r})$

where $H_0$ is the kinetic term plus a central potential; $\vec{L}$ is the electron orbital angular momentum, $\vec{S}$ is the electron spin angular momentum and $\vec{I}$ is the nuclear spin angular momentum; $A_1,A_2,A_3,A_4$ are constants or functions dependent on only the radial coordinate $r$ and thus commute with $\vec{L},\vec{S},\vec{I}$.

On my book it doesn't say explicitly if there is a simple set of good quantum numbers for this hamiltonian, but sometimes it uses $|n,l,j,s,F,m_F\rangle$ where $n,l,j,s$ are as usual and $F,m_F$ are the quantum numbers associated respectively with $F^2,F_z$ (with $\vec{F} = \vec{L} + \vec{S} + \vec{I}$).

By looking at the hamiltonian however it doesn't look like this is true and I can't manage to prove it (I've tried, but the calculations get too messy and I get nowhere). For one, $L^2$ doesn't seem to commute with the hamiltonian due to the presence of $\hat{r}$.\ I'm not sure about the $F^2,F_z$ either since there is no $\vec{J}\cdot\vec{I}$ term (which one would write as $1/2(F^2 - J^2 - I^2)$, which clearly commutes with $F^2,J^2,I^2,F_z$).

If there was no spin I would simply use the fact that the system is rotationally symmetric to say that $L^2,L_z$ commute with the hamiltonian and thus $n,l,m_l$ are good quantum number. The presence of both electron and nuclear spin however makes it seem more complicated and I can't figure it out. I think it should be easier than this but I'm just getting confused at this point. Can anyone help me figure this out or point me to a book/resource where this is explained clearly? Thanks.

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