# How is the Breit-Rabi formula derived?

Does anyone know how the Breit-Rabi formula (energy dependence of the hyper-fine splitting in presence of any external magnetic field for alkali atoms when $F=I \pm 1/2$) is derived, or where to find a reference for it? It doesn't seem to be in the original paper from 1931, or in the textbooks I've looked at.

EDIT: The low- and high-field solutions I understand, but in which step in the derivation is $F=I \pm 1/2$ used, and how does this lead to the exact solution?

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The newer references [12], [19] cited in steck.us/alkalidata/rubidium87numbers.1.6.pdf could be more useful than the original 1931 paper [21]. – Luboš Motl May 8 '11 at 4:37
@Lubos Thank you! If you answer, I'll accept. In the meantime an arxiv search turned up this paper which also seems relevant, especially sections II and III.a. – yatima2975 May 12 '11 at 17:54
The Wikipedia page on the Zeeman effect contains a derivation of the Breit-Rabi formula. – user8113 Mar 12 '12 at 11:39

I believe it is in Rabi's paper "On the Process of Space Quantization" published in 1935 if you can't find the 1931 paper you might enjoy that.

I am going to try and paraphrase the derivation he did in that although I am not sure how much it relates to your question though so apologies if you already know this already. Do check the paper Rabi has lots of pretty pictures and also goes into a lot more depth than I am able to do here.

When an atom has no external field acting and spin $l$, its angular momentum can be expressed as $$F_1=\left(l+\frac{1}{2}\right)$$ $$F_2=\left(l-\frac{1}{2}\right)$$ with units $\hbar$. According to Rabi, the positive moments in such a nucleus will have higher energy $F_1=(l+\frac{1}{2})$ and the negative moments will have lower energy $F_2=(l-\frac{1}{2})$.

Note that the energy difference between states $\Delta E=h\Delta f$ is due to the interactions between protons and electrons in the atom

Now to your question of what happens when an external field is applied. Rabi states without any proof that magnetic field splits into $2(l+\frac{1}{2})$ levels when the field is applied.

If we imagine there is some state that is neither affected by the external magnetic field nor the nuclear fields then we have the The energy shif of the levels with respect to this state is given by $$E_M=-\frac{\Delta E }{2(l+\frac{1}{2})}$$. For for a positive moment we have $$E_M=\frac{\Delta E }{2(l+\frac{1}{2})}$$.

For each state the effective moment is given by $$\frac{\partial{E_M}}{\partial{H}}$$. It goes on quite a bit longer than this, but see how that makes sense so far.

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Landau's Vol 3 on non-relativistic QM (first edition -- at least I think that's what I got here in front of me..) has it in Ch. XVI (Motion in a magnetic field), sec. 128 (the current density in a magnetic field), pg. 486 (the last problem in the chapter). As usual the L&L stuff isn't super explicit but it's there.

If you look in 'Introduction to Elementary Particles' by Griffiths in sec 5.5 on Hyperfine structure you'll see the hyperfine splitting carried out in excruciating detail for hydrogen and positronium. That has the usage of F = I +/- 1/2 in there explicitly.

I believe the same argument carries through for all the alkalis. It has been a while tho so don't bet the farm on it just yet but take a look.

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