I am trying to work out the frequency shifts to the hyperfine energy levels in $^{39}$K $\,$ S$_{1/2}$ (the ground state).

I diagonalise the Hamiltonian for different values of the $B_z$ field, with a basis that is an eigenstate of $$ \hat{\mathbf{L}}\,\otimes \hat{\mathbf{S}}\,\otimes\hat{\mathbf{I}},$$ these being the orbital, spin and nuclear angular momenta respectively.

I get something like this:enter image description here

Which looks qualitatevely correct.

Question: how would I label these states?

The eigenvectors corresposing to the eigenvalues in the graph are $B_z$ dependent, so I guess I can't really use $m_F$ as a good quantum number. But that's what is usually done in textbooks, along with a $m_J$ number as well. How would I get these from my eigenstates?

Also, 2 of the e-states are $B_z$ independent (the orange and the dark blue one)... is there a physical interpretation for this?


First of all, I think you have a sign error and your manifolds should actually be flipped around. Oops! This is easy to do if you use the wrong sign for $g_I$ and $g_J,$ and different authors use different conventions about where these signs go.

The convention is to label the eigenstates by the state they are adiabatically connected to at zero field. This means labelling them by $(F,m_F)$, where $F$ and $m_F$ are the total angular momentum and the projection of the total angular momentum along the magnetic field.

The $F$ label corresponds to the hyperfine manifold, which is $F=2$ for the five states that go to the same energy at zero field, and $F=1$ for the other three. Then, for each manifold, you can find the $m_F$ number by looking at the low field limit where they are all linear. In this regime, $$U=g_F m_F B$$ , so for a given $F$ and $B$ the $m_F$ states go in order from $m_F=-F$ to $m_F=F$.

$g_F$ is the total angular momentum gyromagnetic ratio, which can be calculated from $g_I$ and $g_J$ as:

$g_F\sim g_J \frac{F(F+1)-I(I+1)+J(J+1)}{2F(F+1)} $$

again being careful about the sign convention used (1).

Finally, if you feel like cheating, you can find the correct diagram here.

  • $\begingroup$ So if I want to get the $mF$ quantum number (say, $1/2$) from acting onto the eigenstate with the $F_z$ operator, how would I do it? $\endgroup$ – SuperCiocia Sep 9 '16 at 20:45
  • $\begingroup$ @SuperCiocia $F$ and $mF$ are just the total angular momentum and total projection along the quantization direction, so $F_z=I_z+J_z$. You can find more details in any atomic physics textbook (such as this one: amazon.com/Atomic-Physics-Oxford-Master-Optical/dp/0198506961/… ) $\endgroup$ – Rococo Sep 9 '16 at 21:21
  • $\begingroup$ I know. I want to be able to identify which line in the plot corresponds to which mF state. I can't just apply Fz because the eigenstate with a magnetic field isn't an eigenstate of Fz and it wouldn't give me an integer. How can I overcome this? Is there a trick or something? Should I apply Fz to the eigenstate in zero field? $\endgroup$ – SuperCiocia Sep 10 '16 at 12:39
  • $\begingroup$ I think the most correct thing to do is not to go to zero field (where the states are degenerate so you can't adiabatically connect them to nonzero field), but to go to the low field limit. In this limit mF is still a good quantum number to first order, and the energies are given by the linear term shown in my answer. So the simplest way to identify the $m_F$ state is by the slope of the energy with respect to $B$, rather than by acting on it with an operator. $\endgroup$ – Rococo Sep 10 '16 at 16:00
  • $\begingroup$ That said, if you insist on using the form of the states themselves you should be able to act on the exact low-field states with $F_z$ and see which quantum number is "almost good" for each one, or more precisely which quantum number would become good in the zero-field limit. $\endgroup$ – Rococo Sep 10 '16 at 16:01

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