# What are good quantum numbers for the Hyperfine Hamiltonian?

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.