# Basic labeling of the hyperfine structure of $^{133}$Cs

I am practicing some questions in preparation for an upcoming exam and the following has me stumped:

The D2 line of $$^{133}$$Cs involves transitions between the ground state $$6S_{1/2}$$ and the second excited state $$6P_{3/2}$$. Knowing that the nuclear spin for this atomic species is $$I=7/2$$, sketch a diagram where the hyperfine structure is resolved for both the $$6S_{1/2}$$ and the $$6P_{3/2}$$ levels and clearly label each hyperfine level.

How do I actually work out how many hyperfine states there should be for a given energy level? I know $$F = I + J$$ so taking $$I$$ as $$7/2$$ and $$J$$ as $$1/2$$ and $$3/2$$ respectively then $$F$$ has a degeneracy of $$9$$ ($$-4$$ through $$4$$) and $$11$$ ($$-5$$ through $$5$$) respectively. This is obviously incorrect as I believe Cs should only have two hyperfine states for the ground level. Is it as simple as that for any fine state there are two hyperfine states? If so then do I just label them $$F=0$$ and $$F=1$$?

I know $$F = I + J$$

No, that's not correct. The vector addition $$\mathbf F = \mathbf I + \mathbf J$$ produces values of $$F$$ from $$|I-J|$$ going up through to $$I+J$$. This means that for the $$6S_{1/2}$$ manifold, you will have both $$F=3$$ and $$F=4$$ states available, each with their corresponding degeneracies.

The $$6P_{3/2}$$, on the other hand, has even more values of $$F$$ and even more states within those degenerate level manifolds. Does that sound like a lot of states? Absolutely! To check the precise number, count all of the states and then make sure that they match the product between the $$I$$ degeneracy and the $$J$$ degeneracy.

There is also another misconception in this statement:

This is obviously incorrect as I believe Cs should only have two hyperfine states for the ground level.

The caesium standard operates between the $$F=3$$ and $$F=4$$ hyperfine manifolds, which are degenerate for an isolated atom. The presence of degeneracy in either state is not a problem.

• Got it thank you! Using $F_{min} = I - J$ then $F_{max} = I + J$ gave me what I was expecting.
– Tech
Commented Apr 15, 2019 at 14:06