0198506961.pdf

112 Hyperfine structure and isotope shift

The energies are measured from the point midway between the hyperfine

(^30) Any choice for the point at which levels to streamline the algebra. (^30) The energy eigenvalues are
E= 0 leads to the same result, e.g.
taking the unperturbed energies asA/ 4
and− 3 A/4 as in Fig. 6.3. E=±

√

(A/2)^2 +(ζμBB)^2. (6.35)

This exact solution for all fields is plotted in Fig. 6.10. The approximate

solution for weak fields is

Eweak±

{

A

2

+

(ζμBB)^2

A

}

. (6.36)

WhenB = 0 the two unperturbed levels have energies±A/2. The

term proportional toB^2 is the usual second-order perturbation theory

expression that causes the levels toavoidone another (hence the rule

(^31) The avoided crossing of states that that states of the sameMdo not cross). 31
mix is a general feature of perturbation
theory.
For strong fields, whereμBBA, eqn 6.35 gives the energy of the
M= 0 states as
E(F=± 1 ,MF=0)±ζμBB. (6.37)
In a strong field the energy levels of the system are given bygJμBBMJ
and the twoMJ=± 1 /2 states have Zeeman energies ofgμBBMJ =
±μBB. Comparison with eqn 6.37 shows thatζ= 1, and so we have
found the energies for all field strengths. The other two states have
energiesE(M=±1) =^12 A±μBBfor all values ofB.
A similar approach can be used whenJ=1/2 for arbitrary values
ofI, which applies to the ground states of the alkalis. ForI> 1 / 2
there are more states to consider than in hydrogen (whereI=1/2), so
the Hamiltonian will have larger dimensions than in eqn 6.34. Actually
there were four basis states to consider in hydrogen, but because the
perturbation mixes only two of them it was only necessary to diagonalise
a2×2matrix.

6.4 Measurement of hyperfine structure

An apparatus similar to that for measuring the Zeeman effect, shown

in Fig. 1.7(a), can be used to observe hyperfine structure—a magnet is

not required because hyperfine structure arises from an internal mag-

netic field of the atom. Figure 6.11 shows a typical experimental trace

obtained from such an experiment with a pressure-scanned Fabry–Perot

́etalon for the 5s5p^3 P 0 −5s6s^3 S 1 line in cadmium. The^3 P 0 level has no

hyperfine structure (becauseJ= 0) and the observed splitting comes en-

tirely from the 5s6s^3 S 1 level. Both s-electrons contribute to the field at

(^32) The two spins are aligned in a triplet the nucleus (^32) so this level has an exceptionally large hyperfine splitting
state so the 5s- and 6s-electrons pro-
duce fields in the same direction.
that is greater than the Doppler broadening.
ThefactthattheJ= 1 level only gives two hyperfine levels implies
thatI=1/2. IfI1 then there would be three levels (by the rules for
addition of angular momenta). One of these peaks might, however, be
hidden underneath the large central peak? That this is not the case here