2.3 Further Removal of the Degeneracy of the N-electron States
53
Fig. 2.3 Energies of the three components of a triplet state in an external field along the z-axis
(left) and perpendicular to it (right)
not change, and hence, we can concentrate on the Zeeman interaction. The action of
ˆ
S x on |1, 0 is easiest obtained by using the expression of ˆ
S x in terms of the ladder
operators ˆ
S + and ˆ
S − .
1, 1|μ B g x H x
1
2
( ˆ
S
+ + ˆ
S
− )|1, 0
= μ B g x H x
1
2
1, 1|
√
2|1, 1++1, 1|
√
2|1 − 1
=
μ B g x H x
√
2
(2.70)
The other off-diagonal elements are the same except the interaction between |1, 1
and |1 − 1, which is zero. The full Hamiltonian takes this form
|1, 1| 1, 0| 1, −1
1, 1|
1
3 D
1
√
2
μ B g x H x
0
1, 0|
1
√
2
μ B g x H x
−
2
3 D
1
√
2
μ B g x H x
1, −1|
0
1
√
2
μ B g x H x
1
3 D
(2.71)
After shifting the diagonal by
2
3 D, the energy eigenvalues can be determined as
E 1 = D ; E 2,3 =
1
2
D ±
D 2 + 4μ 2
B g 2
x H 2
x
(2.72)
2.11 Confirm that the only effect of uniformly shifting the diagonal elements
is the same shift of the energy eigenvalues.
The expressions for E 2,3 can be simplified by the Taylor expansion
√
p + q =
√
p+
1
2 q/
√ p+···. Assuming that D 2 is (much) larger than 4μ 2
B g 2
x H 2
x , the expansion