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6.3 Zeeman effect and hyperfine structure 109

sum of the electronic and nuclear moments (cf. eqn 5.9):

μatom=−gJμBJ+gIμNI−gJμBJ. (6.27)

SinceμNμBwe can neglect the nuclear contribution (for all but the
most precise measurements), so that the Hamiltonian for the interaction
with an external fieldBis just that for the electronic magnetic moment:

H=gJμBJ·B. (6.28)

This interaction does not depend on the nuclear spin. However, its
expectation value does depend on the hyperfine structure. We consider
first the weak-field regime where the interaction with the external field
is weaker thanAI·Jso that it can be treated as a perturbation to the
hyperfine structure. We then treat the strong-field regime, and also the
intermediate situation.

6.3.1 Zeeman effect of a weak field,μBB<A

I

J

F

Fig. 6.8TheIJ-coupling scheme.

If the interaction with the external field in eqn 6.28 is weaker than the
hyperfine interactionAI·J,theninthevectormodelJandImove
rapidly about their resultantF, as illustrated in Fig. 6.8, whilstFitself
precesses more slowly about the magnetic field (z-axis). In this regime
FandMFare good quantum numbers, butMIandMJare not. Tak-
ing the projection of the magnetic moments alongFgives the effective
Hamiltonian

H=gJμB

〈J·F〉

F(F+1)

F·B=gFμBF·B=gFμBBFz, (6.29)

where

gF=

F(F+1)+J(J+1)−I(I+1)

2 F(F+1)

gJ. (6.30)

Here the factorgFarises from the projection ofJontoF, as illustrated
in Fig. 6.9, in the just same way asgJis given by the projection ofL
andSontoJin Section 5.5. The Zeeman energy is

E=gFμBBMF. (6.31)

As an example, consider the ground-state hyperfine levels in hydrogen
(I=J=1/2andgJ=gs2). ForF=1wefindgF=1sothethree
statesMF=−1, 0 and 1 are spaced byμBB.TheF=0,MF=0state
has no first-order Zeeman shift (see Fig. 6.10).
In summary, the calculation of the Zeeman effect of a weak magnetic
field on the hyperfine structure is simple because only the magnetic mo-
ment of the electron(s) alongJcontributes, whereas in theLS-coupling
scheme there are components along bothLandS. However,IaffectsgF
because the nuclear angular momentumIis not small, and has a major
effect on theIJF-triangle (thinking in terms of vectors as in Fig. 6.9),
even though the nuclear magnetic moment is negligible.