It is in the states of definitefandmfthat the hyperfine perturbation will be diagonal. In essence,
we are doing degenerate state perturbation theory. We could diagonalize the 4 by 4 matrix for the
perturbation to solve the problem or we can use what we know to pickthe right states to start with.
Again like the spin orbit interaction, thetotal angular momentum stateswill be the right states
because we can write the perturbation in terms of quantum numbers of those states.
~S·I~=^1
2
(
F^2 −S^2 −I^2
)
=
1
2
̄h^2
(
f(f+ 1)−
3
4
−
3
4
)
∆E=^23 (Zα)^4
(
m
MN
)
(mc^2 )gNn^13
(
f(f+ 1)−^32
)
≡A 2
(
f(f+ 1)−^32
)
For the hydrogen ground state we are just adding two spin^12 particles so the possible values are
f= 0,1.
- See Example 24.3.1:The Hyperfine Splitting of the Hydrogen Ground State.*
The transition between the two states gives rise to EM waves withλ= 21 cm.
24.2 Hyperfine Splitting in a B Field
If we apply a B-field the states will split further. As usual, we chooseour coordinates so that the
field is in ˆzdirection. The perturbation then is
Wz = −B~·(~μL+~μS+~μI)
=
μBB
̄h
(Lz+ 2Sz) +
gμN
̄h
BIz
where the magnetic moments from orbital motion, electron spin, and nuclear spin are considered
for now. Since we have already specialized to s states, we can drop the orbital term. For fields
achievable in the laboratory, we canneglect the nuclear magnetic momentin the perturbation.
Then we have
Wz= 2μBB
Sz
̄h
.
As an examples of perturbation theory, we will work this problem forweak fields, for strong fields, and
also work the general case for intermediate fields. Just as in the Zeeman effect,if one perturbation
is much bigger than another, we choose the set of states in which the larger perturbation
is diagonal. In this case, the hyperfine splitting is diagonal in states of definitefwhile the above
perturbation due to the B field is diagonal in states of definitems. For a weak field, the hyperfine
dominates and we use the states of definitef. For a strong field, we use thems,mfstates. If the
two perturbations are of the same order, we must diagonalize the full perturbation matrix. This
calculation will always be correct but more time consuming.
We can estimate the field at which the perturbations are the same size by comparingμBBto
2
3 α
4 me
mpmc
(^2) gN= 2. 9 × 10 − (^6). The weak field limit is achieved ifB≪500 gauss.