0198506961.pdf

6.1 Hyperfine structure 99

Fig. 6.2(a) The probability density|ψ(r)|^2 of an s-electron at small distances (r
a 0 ) is almost constant. The distribution of nuclear matterρN(r) gives an indication
of the nuclear radiusrN. To calculate the interaction of the nuclear magnetic moment
with an s-electron the region is divided into two parts by a boundary surface of radius
r=rb rN(as also shown in Fig. 6.1). The inner region corresponds to a sphere
of uniform magnetisation that produces a flux densityBeatr= 0. The nuclear
magnetic moment interacts with this field.

This is called theFermi contact interactionsince it depends on|ψns(0)|^2
being finite. It can also be expressed as

HHFS=AI·J (6.8)

becauseJ=sforl= 0. It is useful to write down this more general
form at an early stage since it turns out that an interaction proportional
toI·Jis also obtained whenl
=0.
We have already considered the effect of an interaction proportional
to a dot product of two angular momenta when looking at the spin–orbit
interactionβS·L(eqn 5.4). In the same way the hyperfine interaction
in eqn 6.8 causesIandJto change direction but the total angular
momentum of the atomF=I+Jremains constant. The quantities
I·BandJ·Bare not constant in this precession ofIandJaroundF.
ThereforeMIandMJ are not good quantum numbers; we useF and
MFinstead and the eigenstates ofHHFSare|IJFMF〉.^5 The expectation

(^5) This should be compared with theLS-
coupling scheme where combinations
of the eigenstates |LMLSMS〉 form
eigenstates of the spin–orbit interaction
|LSJMJ〉. The same warning issued for
LS-coupling and the spin–orbit inter-
actionβS·Lalso applies. It is impor-
tant not to confuseIJ-coupling and the
interactionAI·J.IntheIJ-coupling
schemeIandJare good quantum num-
bers and the states are|IJMIMJ〉or
|IJFMF〉. The latter are eigenstates
of the interactionAI·J.
value of eqn 6.8 gives
EHFS=A〈I·J〉=

A

2

{F(F+1)−I(I+1)−J(J+1)}. (6.9)

Example 6.1 Hyperfine structure of the1s^2 S 1 / 2 ground state of hy-
drogen
The lowest level of the hydrogen atom is 1s^2 S 1 / 2 , i.e.J=^12 and the
proton has spinI=1/2sothatF= 0 and 1 and these hyperfine levels
have energies

EHFS=

A

2

{F(F+1)−I(I+1)−J(J+1)}=

{

A/4forF=1,

− 3 A/4forF=0.

Fig. 6.3The splitting between the hy-

perfine levels in the ground state of hy-

drogen.

The splitting between the hyperfine levels is ∆EHFS=A(see Fig. 6.3).
Substituting for|ψns(0)|^2 from eqn 2.22 gives

A=

2

3

μ 0 gsμBgIμN

Z^3

πa^30 n^3

. (6.10)