M2L9m.txt

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I have discussed for you the two limiting cases.
The weak field and the strong field case.
But you can solve it for also intermediate fields.
You simply have to do an exact diagonalization
of the Hamiltonian, which involves
the hyperfine coupling.
And the hyperfine coupling, if you want,
can be diagonalized as eigenfunctions
where the quantum numbers are J, I,
couple to F, and the projection of F,
the magnetic quantum number is mF.
But now we have the Zeeman Hamiltonian where everything
is projected on the z-axis.
So we have mJ and mI.
And this Zeeman term can be diagonalized
in a different basis, which is the basis of J, I, mJ, and mI.
So I've shown you the weak field limit, where we simply
assumed those quantum numbers and calculated this term
perturbatively.
And I've shown you the high field limit where
we use those quantum numbers and calculated this field
perturbatively.
But in general, you just have to write down
the matrix element of this Hamiltonian in whatever basis
you choose.
You can use the weak field basis.
This term is diagonal, this is off diagonal.
Or you can use the strong field basis,
where this is the diagonal, this is off diagonal,
and simply diagonalize your Hamiltonian.
Find the wave function, find the eigenenergies.
And since for cases where S is equals 1/2, it's only a 2
by 2 matrix, which has to be diagonalized.
You can do it analytically, and this
leads to the famous Breit-Rabi formula.
So the solution is analytic for J equals 1/2.
And it's a beautiful example which you should solve
in your homework assignment.
Let me just sketch the solution.
When you go from the weak field from the strong field limit,
the z component of the total angular momentum in one case
is mF.
In the other case, it is mI plus mJ.
So when you go from one limit to the next,
you connect only states where mI plus mJ equals mF.

So the structure of the general solution
can be explained by repulsion and anti-crossings of states with the same mF quantum number.