2.3 Further Removal of the Degeneracy of the N-electron States
49
2.9 Show that the ˆ
L · ˆ
S matrix elements of ψ
(0)
0 and an excited state ψ
(0)
i are
equal to
1
2 ψ
(0)
i | ˆ
L z |ψ
(0)
0 when the spin part of ψ
(0)
0 and ψ
(0)
i
is identical and
equal to
1
2 ψ
(0)
i | ˆ
L + |ψ
(0)
0 when ψ
(0)
0 has α- and ψ
(0)
i has β-spin.
For a magnetic field along the z-axis (the quantization axis of the system),
H can be replaced by a scalar and the two expressions of the Zeeman Hamiltonian
(Eqs. 2.23 and 2.48) reduce to μ B H( ˆ
L z + g e ˆ
S z ) and μ B Hg zz ˆ
S z . The corresponding
matrix elements are
μ B Hg zz ψ
(1) | ˆ
S z |ψ
(1) =
1
2
μ B Hg zz = μ B Bψ
(1) | ˆ
L z + g e ˆ
S z |ψ
(1)
(2.51)
μ B Hg zz ψ
(1) | ˆ
S z |ψ
(1) =−
1
2
μ B Hg zz = μ B Hψ
(1) | ˆ
L z + g e ˆ
S z |ψ
(1)
(2.52)
The off-diagonal elements are zero in both Hamiltonians due to the spin-orthogonality.
Now an expression for g zz emerges from either equation as
g zz = 2
g e ψ
(1) | ˆ
S z |ψ
(1) ++ψ
(1) | ˆ
L z |ψ
(1)
= g e + 2ψ
(1) | ˆ
L z |ψ
(1)
(2.53)
The last term can be specified by substitution of the definitions given in Eq. 2.50
ψ
(1) | ˆ
L z |ψ
(1) ==ψ
(0) | ˆ
L z |ψ
(0) +
1
2
ζ
i =0
ψ
(0)
i | ˆ
L z |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
2
ζ
i =0
ψ
(0)
i | ˆ
L x + i ˆ
L y |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
4
ζ
2 ...
++ψ
(0) | ˆ
L z |ψ
(0) +
1
2
ζ
i =0
ψ
(0)
i | ˆ
L z |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
2
ζ
i =0
ψ
(0)
i | ˆ
L x − i ˆ
L y |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
4
ζ
2 ...
(2.54)
Here, ˆ
L ± is replaced by the expression in terms of ˆ
L x,y and the minus sign in front of
i ˆ
L y in the seventh term on the right arises from the fact that the complex conjugated
function Ψ
†(1)
1
was written. Moreover, the third and seventh terms are zero because of
the spin-orthogonality. The terms that are quadratic in ζ are neglected. This somewhat
awkward expression can be further simplified by taking into account that the zerothorder wave function of the ground state has no orbital moment, ψ (0) | ˆ
L z |ψ (0) =0.
49
2.9 Show that the ˆ
L · ˆ
S matrix elements of ψ
(0)
0 and an excited state ψ
(0)
i are
equal to
1
2 ψ
(0)
i | ˆ
L z |ψ
(0)
0 when the spin part of ψ
(0)
0 and ψ
(0)
i
is identical and
equal to
1
2 ψ
(0)
i | ˆ
L + |ψ
(0)
0 when ψ
(0)
0 has α- and ψ
(0)
i has β-spin.
For a magnetic field along the z-axis (the quantization axis of the system),
H can be replaced by a scalar and the two expressions of the Zeeman Hamiltonian
(Eqs. 2.23 and 2.48) reduce to μ B H( ˆ
L z + g e ˆ
S z ) and μ B Hg zz ˆ
S z . The corresponding
matrix elements are
μ B Hg zz ψ
(1) | ˆ
S z |ψ
(1) =
1
2
μ B Hg zz = μ B Bψ
(1) | ˆ
L z + g e ˆ
S z |ψ
(1)
(2.51)
μ B Hg zz ψ
(1) | ˆ
S z |ψ
(1) =−
1
2
μ B Hg zz = μ B Hψ
(1) | ˆ
L z + g e ˆ
S z |ψ
(1)
(2.52)
The off-diagonal elements are zero in both Hamiltonians due to the spin-orthogonality.
Now an expression for g zz emerges from either equation as
g zz = 2
g e ψ
(1) | ˆ
S z |ψ
(1) ++ψ
(1) | ˆ
L z |ψ
(1)
= g e + 2ψ
(1) | ˆ
L z |ψ
(1)
(2.53)
The last term can be specified by substitution of the definitions given in Eq. 2.50
ψ
(1) | ˆ
L z |ψ
(1) ==ψ
(0) | ˆ
L z |ψ
(0) +
1
2
ζ
i =0
ψ
(0)
i | ˆ
L z |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
2
ζ
i =0
ψ
(0)
i | ˆ
L x + i ˆ
L y |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
4
ζ
2 ...
++ψ
(0) | ˆ
L z |ψ
(0) +
1
2
ζ
i =0
ψ
(0)
i | ˆ
L z |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
2
ζ
i =0
ψ
(0)
i | ˆ
L x − i ˆ
L y |ψ
(0)
0 ψ
(0)
0 | ˆ
L z |ψ
(0)
i
E 0 − E i
+
1
4
ζ
2 ...
(2.54)
Here, ˆ
L ± is replaced by the expression in terms of ˆ
L x,y and the minus sign in front of
i ˆ
L y in the seventh term on the right arises from the fact that the complex conjugated
function Ψ
†(1)
1
was written. Moreover, the third and seventh terms are zero because of
the spin-orthogonality. The terms that are quadratic in ζ are neglected. This somewhat
awkward expression can be further simplified by taking into account that the zerothorder wave function of the ground state has no orbital moment, ψ (0) | ˆ
L z |ψ (0) =0.