232
Appendix E: Solutions
=
⎛
⎜
⎜
⎜
⎜
⎝
H 11
H 12 +H 13
√
2
H 14
H 12 −H 13
√
2
H 21 +H 31
√
2
H 22 +H 32 +H 23 +H 33
2
H 24 +H 34
√
2
H 22 +H 32 −H 23 −H 33
2
H 41
H 42 +H 43
√
2
H 44
H 42 −H 43
√
2
H 21 −H 31
√
2
H 22 −H 32 +H 23 −H 33
2
H 24 −H 34
√
2
H 22 −H 32 −H 23 +H 33
2
⎞
⎟
⎟
⎟
⎟
⎠
Substituting the definition of H ij of the matrix in the uncoupled basis gives the
representation in the coupled basis. As example:
H 12 +H 13
√
2
=
1
√
2
(
1
4 D xz −
1
4 iD yz +
1
4 D xz −
1
4 iD yz ) =
1
2
√
2
(D xz − iD yz ) ==T + | ˆ
H|T 0 .
Exercise 3.14 From
1
2 A ij = D ij −
1
2 A ji and −
1
2 A ji = d ij −
1
2 A ij follows
1
2 A ij =
D ij + d ij −
1
2 A ij ⇒ A ij = D ij + d ij . Substituting this in the expression for A ji gives
−
1
2 A ji = d ij −
1
2 D ij −
1
2 d ij ⇒ A ji = D ij − d ij = D ji + d ji .
Problem 3.1 Ψ(0, 0) = N ′
|φ a φ b |+| φ b φ a |
with φ a = N(ψ a + νψ b ), φ b =
N(ψ b + νψ a ), N = 1/
√
1 + ν 2 and ψ a |ψ b =0. So S ab = 2ν/(1 + ν 2 ),
s e ea l s oE q .3.18. Substitution gives Ψ(0, 0) = N ′
|ψ a ψ b |+|ψ b ψ a |+(2ν/(1 +
ν 2 ))
|ψ a ψ a |+|ψ b ψ b |
. Since ψ a = (1/
√
2)(φ 1 + φ 2 ) and ψ b = (1/
√
2)(φ 1 − φ 2 )
(see Eq. 3.10a), we get |ψ a ψ b |+|ψ b ψ a |=|φ 1 φ 1 |−|φ 2 φ 2 | and |ψ a ψ a |+|ψ b ψ b |=
|φ 1 φ 1 |+|φ 2 φ 2 |.Now ,Ψ(0, 0) = N ′
|φ 1 φ 1 |−|φ 2 φ 2 |+S ab
|φ 1 φ 1 |+|φ 2 φ 2 |
=
N ′
(S ab + 1)|φ 1 φ 1 |+(S ab − 1)|φ 2 φ 2 |
. Hence, c 2 /c 1 = (S ab − 1)/(S ab + 1).
Problem 3.2 (a) |g 1 g 1 |=
1
2 |(a 1 + b 1 )(a 1 + b 2 )|=
1
2
|a 1 a 1 |+|a 1 b 1 |+|b 1 a 1 |+
|b 1 b 1 |
, 50 % neutral, 50 % ionic, eigenfunction of ˆ
S 2 (singlet); |g 1 g 2 |=
1
2 |(a 1 + b 1 )(a 2 + b 2 )|=
1
2
|a 1 a 2 |+|a 1 b 2 |+|b 1 a 2 |+|b 1 b 2 |
, 50 % neutral, 50 %
ionic, eigenfunction of ˆ
S 2 (triplet); |g 1 u 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )|=
1
2
|a 1 a 1 |−
|a 1 b 1 |−|b 1 a 1 |+|b 1 b 1 |
, 50 % neutral, 50 % ionic, not an eigenfunction of ˆ
S 2 .(b)
1
√
2
(|g 1 g 1 |+|u 1 u 1 |) =
1
2
|(a 1 + b 1 )(a 1 + b 1 )|+|(a 1 − b 1 )(a 1 − b 1 |
=
1
2
√
2
|a 1 a 1 |+
|a 1 b 1 |+|b 1 a 1 |+|b 1 b 1 |+|a 1 a 1 |−|a 1 b 1 |−|b 1 a 1 |+|b 1 b 1 |
=
1
√
2
|a 1 a 1 |+|b 1 b 1 |
,
100 % ionic, eigenfunction of ˆ
S 2 (singlet);
1
√
2
(|g 1 g 1 |−|u 1 u 1 |) =
1
2
|(a 1 + b 1 )(a 1 +
b 1 )|−|(a 1 − b 1 )(a 1 − b 1 |
=
1
2
√
2
|a 1 a 1 |+|a 1 b 1 |+|b 1 a 1 |+|b 1 b 1 |−|a 1 a 1 |+
|a 1 b 1 |+|b 1 a 1 |−|b 1 b 1 |
=
1
√
2
|a 1 b 1 |+|b 1 a 1 |
, 100 % covalent, eigenfunction
of ˆ
S 2 (singlet). (c) |g 1 u 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )|=
1
2 (−|a 1 b 1 |+| b 1 a 1 |) =
1
2 (|b 1 a a |+|b 1 a 1 |) =|b 1 a 1 |, 100 % covalent, eigenfunction of ˆ
S 2 (triplet). |g 1 u 1 v 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )c 1 |=
1
2
−|a 1 b 1 c 1 |+|b 1 a 1 c 1 |
=
1
2
−|a 1 b 1 c 1 |−|a 1 b 1 c 1 |
=
−|a 1 b 1 c 1 |, 100 % covalent, eigenfunction of ˆ
S 2 (quartet). (d) For simplicity, we drop
the subscript and multiply with the normalization constant at the end. 2|guv|−|guv|−
|guv|=|(a + b)(a − b)c|−
1
2 |(a + b)(a − b)c|−
1
2 |(a + b)(a − b)c|=−|abc|+
|bac|−
1
2
|aac|−|abc|+|bac|−|bbc|
−
1
2
|aac|−|abc|+|bac|−|bbc|
=−2|abc|+
|abc|+|abc|. After multiplying with 1/
√
6, the doublet spin eigenfunction appears,
with 100 % covalent character.
Appendix E: Solutions
=
⎛
⎜
⎜
⎜
⎜
⎝
H 11
H 12 +H 13
√
2
H 14
H 12 −H 13
√
2
H 21 +H 31
√
2
H 22 +H 32 +H 23 +H 33
2
H 24 +H 34
√
2
H 22 +H 32 −H 23 −H 33
2
H 41
H 42 +H 43
√
2
H 44
H 42 −H 43
√
2
H 21 −H 31
√
2
H 22 −H 32 +H 23 −H 33
2
H 24 −H 34
√
2
H 22 −H 32 −H 23 +H 33
2
⎞
⎟
⎟
⎟
⎟
⎠
Substituting the definition of H ij of the matrix in the uncoupled basis gives the
representation in the coupled basis. As example:
H 12 +H 13
√
2
=
1
√
2
(
1
4 D xz −
1
4 iD yz +
1
4 D xz −
1
4 iD yz ) =
1
2
√
2
(D xz − iD yz ) ==T + | ˆ
H|T 0 .
Exercise 3.14 From
1
2 A ij = D ij −
1
2 A ji and −
1
2 A ji = d ij −
1
2 A ij follows
1
2 A ij =
D ij + d ij −
1
2 A ij ⇒ A ij = D ij + d ij . Substituting this in the expression for A ji gives
−
1
2 A ji = d ij −
1
2 D ij −
1
2 d ij ⇒ A ji = D ij − d ij = D ji + d ji .
Problem 3.1 Ψ(0, 0) = N ′
|φ a φ b |+| φ b φ a |
with φ a = N(ψ a + νψ b ), φ b =
N(ψ b + νψ a ), N = 1/
√
1 + ν 2 and ψ a |ψ b =0. So S ab = 2ν/(1 + ν 2 ),
s e ea l s oE q .3.18. Substitution gives Ψ(0, 0) = N ′
|ψ a ψ b |+|ψ b ψ a |+(2ν/(1 +
ν 2 ))
|ψ a ψ a |+|ψ b ψ b |
. Since ψ a = (1/
√
2)(φ 1 + φ 2 ) and ψ b = (1/
√
2)(φ 1 − φ 2 )
(see Eq. 3.10a), we get |ψ a ψ b |+|ψ b ψ a |=|φ 1 φ 1 |−|φ 2 φ 2 | and |ψ a ψ a |+|ψ b ψ b |=
|φ 1 φ 1 |+|φ 2 φ 2 |.Now ,Ψ(0, 0) = N ′
|φ 1 φ 1 |−|φ 2 φ 2 |+S ab
|φ 1 φ 1 |+|φ 2 φ 2 |
=
N ′
(S ab + 1)|φ 1 φ 1 |+(S ab − 1)|φ 2 φ 2 |
. Hence, c 2 /c 1 = (S ab − 1)/(S ab + 1).
Problem 3.2 (a) |g 1 g 1 |=
1
2 |(a 1 + b 1 )(a 1 + b 2 )|=
1
2
|a 1 a 1 |+|a 1 b 1 |+|b 1 a 1 |+
|b 1 b 1 |
, 50 % neutral, 50 % ionic, eigenfunction of ˆ
S 2 (singlet); |g 1 g 2 |=
1
2 |(a 1 + b 1 )(a 2 + b 2 )|=
1
2
|a 1 a 2 |+|a 1 b 2 |+|b 1 a 2 |+|b 1 b 2 |
, 50 % neutral, 50 %
ionic, eigenfunction of ˆ
S 2 (triplet); |g 1 u 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )|=
1
2
|a 1 a 1 |−
|a 1 b 1 |−|b 1 a 1 |+|b 1 b 1 |
, 50 % neutral, 50 % ionic, not an eigenfunction of ˆ
S 2 .(b)
1
√
2
(|g 1 g 1 |+|u 1 u 1 |) =
1
2
|(a 1 + b 1 )(a 1 + b 1 )|+|(a 1 − b 1 )(a 1 − b 1 |
=
1
2
√
2
|a 1 a 1 |+
|a 1 b 1 |+|b 1 a 1 |+|b 1 b 1 |+|a 1 a 1 |−|a 1 b 1 |−|b 1 a 1 |+|b 1 b 1 |
=
1
√
2
|a 1 a 1 |+|b 1 b 1 |
,
100 % ionic, eigenfunction of ˆ
S 2 (singlet);
1
√
2
(|g 1 g 1 |−|u 1 u 1 |) =
1
2
|(a 1 + b 1 )(a 1 +
b 1 )|−|(a 1 − b 1 )(a 1 − b 1 |
=
1
2
√
2
|a 1 a 1 |+|a 1 b 1 |+|b 1 a 1 |+|b 1 b 1 |−|a 1 a 1 |+
|a 1 b 1 |+|b 1 a 1 |−|b 1 b 1 |
=
1
√
2
|a 1 b 1 |+|b 1 a 1 |
, 100 % covalent, eigenfunction
of ˆ
S 2 (singlet). (c) |g 1 u 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )|=
1
2 (−|a 1 b 1 |+| b 1 a 1 |) =
1
2 (|b 1 a a |+|b 1 a 1 |) =|b 1 a 1 |, 100 % covalent, eigenfunction of ˆ
S 2 (triplet). |g 1 u 1 v 1 |=
1
2 |(a 1 + b 1 )(a 1 − b 1 )c 1 |=
1
2
−|a 1 b 1 c 1 |+|b 1 a 1 c 1 |
=
1
2
−|a 1 b 1 c 1 |−|a 1 b 1 c 1 |
=
−|a 1 b 1 c 1 |, 100 % covalent, eigenfunction of ˆ
S 2 (quartet). (d) For simplicity, we drop
the subscript and multiply with the normalization constant at the end. 2|guv|−|guv|−
|guv|=|(a + b)(a − b)c|−
1
2 |(a + b)(a − b)c|−
1
2 |(a + b)(a − b)c|=−|abc|+
|bac|−
1
2
|aac|−|abc|+|bac|−|bbc|
−
1
2
|aac|−|abc|+|bac|−|bbc|
=−2|abc|+
|abc|+|abc|. After multiplying with 1/
√
6, the doublet spin eigenfunction appears,
with 100 % covalent character.