Appendix E: Solutions
229
Problem 2.3 (1) Ti III :[ 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 1 ], there is no ZFS for S =
1
2 .( 2 )
In the first place, the projected wave functions have to be expressed in terms of
the 3d orbitals by substituting the expressions of φ i in the multideterminantal wave
functions
Ψ ′
i . This gives
Ψ ′
1
Ψ ′
2
Ψ ′
3
Ψ ′
4
Ψ ′
5
3d z 2
0.0024 0.0166 −0.3142 0.1158 −0.9433
3d x 2 −y 2 0.2569 0.1497 −0.2999 0.8799 0.2101
3d xy
0.2249 −0.2178 −0.8409 −0.3731 0.2300
3d yz
0.3072 −0.9071 0.2313 0.1587 −0.0679
3d xz
−0.8883 −0.3271 −0.2250 0.2190 0.0923
The next step (cf. Eqs. 2.55 and 2.59) is the calculation of
Ψ ′
i | ˆ
L z |
Ψ ′
1 and
Ψ ′
1 | ˆ
L z |
Ψ ′
i
with i = 2, 3, 4, 5, i = 1 is the ground state. The same has to be done for ˆ
L x .
Ψ ′
2
Ψ ′
3
Ψ ′
4
Ψ ′
5
i| ˆ
L z |1 0.7270184 −0.4334652 −0.7957402 0.0556362
1| ˆ
L z |i− 0.7270184 0.4334652 0.7957402 −0.0556362
i| ˆ
L x |1 0.3192372 0.5594345 0.1304799 −0.7317820
1| ˆ
L x |i− 0.3192372 −0.5594345 −0.1304799 0.7317820
From this, g zz = g e − 0.150 and g xx = g e + 0.088. The deviation in z is significantly
larger than in x confirming the axial anisotropy.
Exercises and Problems of Chap. 3
Exercise 3.1 Φ 11 (0, 0)| ˆ
H|Φ 22 (0, 0)== φ 1 φ 1 | ˆ
H|φ 2 φ 2 == φ 1 φ 1 |1/r 12 |φ 2 φ 2 −
φ 1 φ 1 |1/r 12 |φ 2 φ 2 ==φ 1 φ 1 |1/r 12 |φ 2 φ 2 −0 = K 12
Exercise 3.2 ψ a = (φ a + φ b )/
√
2 = (1/
√
2)
(χ a + χ b )/
√
2(1 + S) + (χ a −
χ b )/
√
2(1 − S)
= (1/
√
2)
χ a [1/
√
2(1 + S)+1/
√
2(1 − S)]+χ b [1/
√
2(1 + S)−
1/
√
2(1 − S)]
.F o rS = 0.2, the coefficient of χ b is −0.1026 and for S = 0.003,
the coefficient reduces to −0.0015.
Exercise 3.3
|ψ a ψ b |+|ψ b ψ a |
?
=
|φ 1 φ 1 |−|φ 2 φ 2 |
: Substitute ψ a,b = (φ 1 ±
φ 2 )/
√
2:
1
2 {|(φ 1 + φ 2 )(φ 1 − φ 2 )|+|(φ 1 − φ 2 )(φ 1 + φ 2 )|} =
1
2 {|φ 1 φ 1 |−|φ 1 φ 2 |+
|φ 2 φ 1 |−| φ 2 φ 2 |+| φ 1 φ 1 |+| φ 1 φ 2 |−| φ 2 φ 1 |−| φ 2 φ 2 |} =
|φ 1 φ 1 |−| φ 2 φ 2 |
.
|ψ a ψ b |−| ψ b ψ a |
?
=
|φ 1 φ 2 |−| φ 2 φ 1 |
: Substitute ψ a,b = (φ 1 ± φ 2 )/
√
2:
1
2 {|(φ 1 + φ 2 )(φ 1 − φ 2 )|−|(φ 1 − φ 2 )(φ 1 + φ 2 )|} =
1
2 {|φ 1 φ 1 |−|φ 2 φ 2 |−|φ 1 φ 2 |+
|φ 2 φ 1 |−|φ 1 φ 1 |+|φ 2 φ 2 |−|φ 1 φ 2 |+|φ 2 φ 1 |} = −
|φ 1 φ 2 |−|φ 2 φ 1 |
(the sign is not
relevant)
229
Problem 2.3 (1) Ti III :[ 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 1 ], there is no ZFS for S =
1
2 .( 2 )
In the first place, the projected wave functions have to be expressed in terms of
the 3d orbitals by substituting the expressions of φ i in the multideterminantal wave
functions
Ψ ′
i . This gives
Ψ ′
1
Ψ ′
2
Ψ ′
3
Ψ ′
4
Ψ ′
5
3d z 2
0.0024 0.0166 −0.3142 0.1158 −0.9433
3d x 2 −y 2 0.2569 0.1497 −0.2999 0.8799 0.2101
3d xy
0.2249 −0.2178 −0.8409 −0.3731 0.2300
3d yz
0.3072 −0.9071 0.2313 0.1587 −0.0679
3d xz
−0.8883 −0.3271 −0.2250 0.2190 0.0923
The next step (cf. Eqs. 2.55 and 2.59) is the calculation of
Ψ ′
i | ˆ
L z |
Ψ ′
1 and
Ψ ′
1 | ˆ
L z |
Ψ ′
i
with i = 2, 3, 4, 5, i = 1 is the ground state. The same has to be done for ˆ
L x .
Ψ ′
2
Ψ ′
3
Ψ ′
4
Ψ ′
5
i| ˆ
L z |1 0.7270184 −0.4334652 −0.7957402 0.0556362
1| ˆ
L z |i− 0.7270184 0.4334652 0.7957402 −0.0556362
i| ˆ
L x |1 0.3192372 0.5594345 0.1304799 −0.7317820
1| ˆ
L x |i− 0.3192372 −0.5594345 −0.1304799 0.7317820
From this, g zz = g e − 0.150 and g xx = g e + 0.088. The deviation in z is significantly
larger than in x confirming the axial anisotropy.
Exercises and Problems of Chap. 3
Exercise 3.1 Φ 11 (0, 0)| ˆ
H|Φ 22 (0, 0)== φ 1 φ 1 | ˆ
H|φ 2 φ 2 == φ 1 φ 1 |1/r 12 |φ 2 φ 2 −
φ 1 φ 1 |1/r 12 |φ 2 φ 2 ==φ 1 φ 1 |1/r 12 |φ 2 φ 2 −0 = K 12
Exercise 3.2 ψ a = (φ a + φ b )/
√
2 = (1/
√
2)
(χ a + χ b )/
√
2(1 + S) + (χ a −
χ b )/
√
2(1 − S)
= (1/
√
2)
χ a [1/
√
2(1 + S)+1/
√
2(1 − S)]+χ b [1/
√
2(1 + S)−
1/
√
2(1 − S)]
.F o rS = 0.2, the coefficient of χ b is −0.1026 and for S = 0.003,
the coefficient reduces to −0.0015.
Exercise 3.3
|ψ a ψ b |+|ψ b ψ a |
?
=
|φ 1 φ 1 |−|φ 2 φ 2 |
: Substitute ψ a,b = (φ 1 ±
φ 2 )/
√
2:
1
2 {|(φ 1 + φ 2 )(φ 1 − φ 2 )|+|(φ 1 − φ 2 )(φ 1 + φ 2 )|} =
1
2 {|φ 1 φ 1 |−|φ 1 φ 2 |+
|φ 2 φ 1 |−| φ 2 φ 2 |+| φ 1 φ 1 |+| φ 1 φ 2 |−| φ 2 φ 1 |−| φ 2 φ 2 |} =
|φ 1 φ 1 |−| φ 2 φ 2 |
.
|ψ a ψ b |−| ψ b ψ a |
?
=
|φ 1 φ 2 |−| φ 2 φ 1 |
: Substitute ψ a,b = (φ 1 ± φ 2 )/
√
2:
1
2 {|(φ 1 + φ 2 )(φ 1 − φ 2 )|−|(φ 1 − φ 2 )(φ 1 + φ 2 )|} =
1
2 {|φ 1 φ 1 |−|φ 2 φ 2 |−|φ 1 φ 2 |+
|φ 2 φ 1 |−|φ 1 φ 1 |+|φ 2 φ 2 |−|φ 1 φ 2 |+|φ 2 φ 1 |} = −
|φ 1 φ 2 |−|φ 2 φ 1 |
(the sign is not
relevant)