Appendix E: Solutions
233
Problem 3.3 ˆ
S + (1) ˆ
S − (2)αα = 0; ˆ
S − (1) ˆ
S + (2)αα = 0; ˆ
S z (1) ˆ
S z (2)αα =
1
4 αα ⇒
−J ˆ
S(1) · ˆ
S(2)Φ(T ) =−
1
4 JΦ(T ). ˆ
S + (1) ˆ
S − (2)(αβ − βα) = 0 − αβ;
ˆ
S − (1) ˆ
S + (2)(αβ − βα) = βα − 0; ˆ
S z (1) ˆ
S z (2)(αβ − βα) =−
1
4 αβ +
1
4 βα ⇒ ˆ
S(1) ·
ˆ
S(2)(αβ − βα) =−
1
2 (αβ − βα) −
1
4 (αβ − βα) ⇒−J ˆ
S(1) · ˆ
S(2)Φ(S) =
3
4 JΦ(S).
Problem 3.4 (a) All determinants have two electrons with α spin and one with β
spin, hence M S of all determinants is
1
2 . Ψ 3 =
1
√
3
|φ 1 φ 2 φ 3 |+|φ 1 φ 2 φ 3 |+|φ 1 φ 2 φ 3 |
.
Separating the spatial and spin part: Ψ 3 =
1
√
3
|φ 1 φ 2 φ 3 |(ααβ +αβα +βαα).Thespin
part is the M S =
1
2 component of the quartet spin eigenfunction, the application of ˆ
S +
gives ααα.(b)Ψ 1 corresponds to D 2 , then J 12 = J 23 =
2
3 (E(Ψ 1 )−E(Ψ 3 )) =−27.24
meV and J 13 =−27.24−(E(D 2 )−E(D 1 )) = 0.07 meV. (c) Model space: |φ 1 φ 2 φ 3 |,
|φ 1 φ 2 φ 3 |, |φ 1 φ 2 φ 3 |, |
Ψ i | 2 = 0.8973, 0.9623, 1.0002, 0.0349, 0.0922 (the third is
due to round-off errors). (d) Ψ 1 , Ψ 2 and Ψ 3 .O n l yΨ 1 |Ψ 2 = 0, Gram-Schmidt
orthogonalization gives c 1 =− 0.4672, c 2 = 0.8135, c 3 =− 0.3463 for Ψ 1 and
c ′
1 =− 0.6696, c ′
2 =− 0.0699, c ′
3 = 0.7394 for Ψ 2 .( e ) ˆ
H
eff
11 =− 27.9615094,
ˆ
H
eff
21 =−0.0001464, ˆ
H
eff
22 =−27.9608149, ˆ
H
eff
13 = 0.0000786, ˆ
H
eff
23 = 0.0010935,
ˆ
H
eff
33 =−27.9587333. J 12 = 7.97 meV, J 13 =−4.28 meV, J 23 =−59.51 meV.
Problem 3.5 (a) Dividing the eigenvalues given in Fig. 3.6 by −J, we obtain
ˆ
S 1 ˆ
S 2 Q = 1 · Q; ˆ
S 1 ˆ
S 2 T =− 1 · T; ˆ
S 1 ˆ
S 2 S =− 2 · S. Applying the operator for the
second time leads to ˆ
S 1 ˆ
S 2 (1 · Q) = 1 · Q; ˆ
S 1 ˆ
S 2 (−1 · T) = 1 · T; ˆ
S 1 ˆ
S 2 (−2 · S) = 4 · S.
Multiplying with λ gives exactly the same eigenvalues as listed in Eq. 3.75.
Problem 3.6 (E(T ) − E(Q))/2 =− 42.58 meV, E(S) − E(T ) =− 37.54 meV, no
regular spacing. From Eq. 3.75 follows E(T )−E(Q) = 2J and E(S)−E(T ) = J +3λ.
This gives J =−42.58 meV and λ =[−37.54 − (−42.58)]/3 = 1.68 meV.
Exercises and Problems of Chap. 4
Exercise 4.1 Ψ S |Ψ S =(1/(2 + 2S))ab + ba|ab + ba=(1/(2 + 2S))
ab|ab
++ab|ba++ba|ab++ba|ba
= (1/(2 + 2S))(1 + S + S + 1) = 1;;Ψ T |Ψ T =
(1/(2 − 2S))ab − ba|ab − ba=(1/(2 − 2S))
ab|ab−−ab|ba−−ba|ab+
ba|ba
= (1/(2 − 2S))(1 − S − S + 1) = 1.
Exercise 4.2 J 11 == φ 1 φ 1 |1/r 12 |φ 1 φ 1 =(1/4)(φ a + φ b )(φ a + φ b )1/r 12 (φ a +
φ b )(φ a +φ b )=(1/4)
φ a φ a |1/r 12 |φ a φ a ++φ a φ a |1/r 12 |φ a φ b ++φ a φ a |1/r 12 |φ b φ a
++φ a φ a |1/r 12 |φ b φ b ++φ a φ b |1/r 12 |φ a φ a +...(eleven more terms)
= (1/4)(2J aa +
2J ab + 4K ab + 8φ a φ a |1/r 12 |φ a φ b ), where we have used that the system is centrosymmetric: J aa = J bb , etc. This is equal to the expression given in Eq. 4.20.
Exercise 4.3 Meta: the shortest contacts are formed by the aligned carbon atoms of
the benzene ring. These have opposite spin density, and hence, ρ i ρ j < 0, indicating
233
Problem 3.3 ˆ
S + (1) ˆ
S − (2)αα = 0; ˆ
S − (1) ˆ
S + (2)αα = 0; ˆ
S z (1) ˆ
S z (2)αα =
1
4 αα ⇒
−J ˆ
S(1) · ˆ
S(2)Φ(T ) =−
1
4 JΦ(T ). ˆ
S + (1) ˆ
S − (2)(αβ − βα) = 0 − αβ;
ˆ
S − (1) ˆ
S + (2)(αβ − βα) = βα − 0; ˆ
S z (1) ˆ
S z (2)(αβ − βα) =−
1
4 αβ +
1
4 βα ⇒ ˆ
S(1) ·
ˆ
S(2)(αβ − βα) =−
1
2 (αβ − βα) −
1
4 (αβ − βα) ⇒−J ˆ
S(1) · ˆ
S(2)Φ(S) =
3
4 JΦ(S).
Problem 3.4 (a) All determinants have two electrons with α spin and one with β
spin, hence M S of all determinants is
1
2 . Ψ 3 =
1
√
3
|φ 1 φ 2 φ 3 |+|φ 1 φ 2 φ 3 |+|φ 1 φ 2 φ 3 |
.
Separating the spatial and spin part: Ψ 3 =
1
√
3
|φ 1 φ 2 φ 3 |(ααβ +αβα +βαα).Thespin
part is the M S =
1
2 component of the quartet spin eigenfunction, the application of ˆ
S +
gives ααα.(b)Ψ 1 corresponds to D 2 , then J 12 = J 23 =
2
3 (E(Ψ 1 )−E(Ψ 3 )) =−27.24
meV and J 13 =−27.24−(E(D 2 )−E(D 1 )) = 0.07 meV. (c) Model space: |φ 1 φ 2 φ 3 |,
|φ 1 φ 2 φ 3 |, |φ 1 φ 2 φ 3 |, |
Ψ i | 2 = 0.8973, 0.9623, 1.0002, 0.0349, 0.0922 (the third is
due to round-off errors). (d) Ψ 1 , Ψ 2 and Ψ 3 .O n l yΨ 1 |Ψ 2 = 0, Gram-Schmidt
orthogonalization gives c 1 =− 0.4672, c 2 = 0.8135, c 3 =− 0.3463 for Ψ 1 and
c ′
1 =− 0.6696, c ′
2 =− 0.0699, c ′
3 = 0.7394 for Ψ 2 .( e ) ˆ
H
eff
11 =− 27.9615094,
ˆ
H
eff
21 =−0.0001464, ˆ
H
eff
22 =−27.9608149, ˆ
H
eff
13 = 0.0000786, ˆ
H
eff
23 = 0.0010935,
ˆ
H
eff
33 =−27.9587333. J 12 = 7.97 meV, J 13 =−4.28 meV, J 23 =−59.51 meV.
Problem 3.5 (a) Dividing the eigenvalues given in Fig. 3.6 by −J, we obtain
ˆ
S 1 ˆ
S 2 Q = 1 · Q; ˆ
S 1 ˆ
S 2 T =− 1 · T; ˆ
S 1 ˆ
S 2 S =− 2 · S. Applying the operator for the
second time leads to ˆ
S 1 ˆ
S 2 (1 · Q) = 1 · Q; ˆ
S 1 ˆ
S 2 (−1 · T) = 1 · T; ˆ
S 1 ˆ
S 2 (−2 · S) = 4 · S.
Multiplying with λ gives exactly the same eigenvalues as listed in Eq. 3.75.
Problem 3.6 (E(T ) − E(Q))/2 =− 42.58 meV, E(S) − E(T ) =− 37.54 meV, no
regular spacing. From Eq. 3.75 follows E(T )−E(Q) = 2J and E(S)−E(T ) = J +3λ.
This gives J =−42.58 meV and λ =[−37.54 − (−42.58)]/3 = 1.68 meV.
Exercises and Problems of Chap. 4
Exercise 4.1 Ψ S |Ψ S =(1/(2 + 2S))ab + ba|ab + ba=(1/(2 + 2S))
ab|ab
++ab|ba++ba|ab++ba|ba
= (1/(2 + 2S))(1 + S + S + 1) = 1;;Ψ T |Ψ T =
(1/(2 − 2S))ab − ba|ab − ba=(1/(2 − 2S))
ab|ab−−ab|ba−−ba|ab+
ba|ba
= (1/(2 − 2S))(1 − S − S + 1) = 1.
Exercise 4.2 J 11 == φ 1 φ 1 |1/r 12 |φ 1 φ 1 =(1/4)(φ a + φ b )(φ a + φ b )1/r 12 (φ a +
φ b )(φ a +φ b )=(1/4)
φ a φ a |1/r 12 |φ a φ a ++φ a φ a |1/r 12 |φ a φ b ++φ a φ a |1/r 12 |φ b φ a
++φ a φ a |1/r 12 |φ b φ b ++φ a φ b |1/r 12 |φ a φ a +...(eleven more terms)
= (1/4)(2J aa +
2J ab + 4K ab + 8φ a φ a |1/r 12 |φ a φ b ), where we have used that the system is centrosymmetric: J aa = J bb , etc. This is equal to the expression given in Eq. 4.20.
Exercise 4.3 Meta: the shortest contacts are formed by the aligned carbon atoms of
the benzene ring. These have opposite spin density, and hence, ρ i ρ j < 0, indicating