130_notes.dvi

24 Hyperfine Structure

The interaction between the magnetic moment, due to the spin of the nucleus, and the larger
magnetic moment, due to the electron’s spin, results in energy shifts which are much smaller than
those of the fine structure. They are of ordermmepα^2 Enand are hence called hyperfine.

The hyperfine corrections may be difficult to measure in transitions between states of differentn,
however, they are quite measurable and important because they split the ground state. The different
hyperfine levels of the ground state are populated thermally. Hyperfine transitions, which emit radio
frequency waves, can be used to detect interstellar gas.

This material is covered inGasiorowicz Chapter 17,inCohen-Tannoudji et al. Chapter
XII,and briefly inGriffiths 6.5.

24.1 Hyperfine Splitting

We can think of the nucleus as a single particle with spinI~. This particle is actually made up of
protons and neutrons which are both spin^12 particles. The protons and neutrons in turn are made
of spin^12 quarks. The magnetic dipole moment due to the nuclear spin is much smaller than that of
the electron because the mass appears in the denominator. The magnetic moment of the nucleus is

~μN=

ZegN

2 MNc

I~

where~Iis thenuclear spinvector. Because the nucleus has internal structure, the nuclear gy-
romagnetic ratio is not just 2. For the proton, it isgp≈ 5 .56. This is the nucleus of hydrogen
upon which we will concentrate. Even though the neutron is neutral, the gyromagnetic ratio is
about -3.83. (The quarks have gyromagnetic ratios of 2 (plus corrections) like the electron but the
problem is complicated by the strong interactions which make it hard to define a quark’s mass.) We
can compute (to some accuracy) the gyromagnetic ratio of nucleifrom that of protons and neutrons
as we can compute the proton’s gyromagnetic ratio from its quark constituents.

In any case, the nuclear dipole moment is about 1000 times smaller than that for e-spin orL~. We
will calculate ∆Eforℓ= 0 states (see Condon and Shortley for more details). This is particularly
important because it will break the degeneracy of the Hydrogen ground state.

To get the perturbation, we should findB~ from~μ(see Gasiorowicz page 287) then calculate the

energy change in first order perturbation theory ∆E=

〈

−~μe·B~

〉

.Calculating(see section 24.4.1)

the energy shift forℓ= 0 states.

∆E=

〈e

mc

S~·B~

〉

=

4

3

(Zα)^4

(

m

MN

)

(mc^2 )gN

1

n^3

S~·I~

̄h^2

Now, just as in the case of theL~·S~, spin-orbit interaction, we will define the total angular momentum

F~=S~+~I.