Appendix E: Solutions
239
Exercise 6.3 The remaining zeros correspond to Φ 1 | ˆ
H|Φ 2 , Φ 2 | ˆ
H|Φ 5 ,
Φ 3 | ˆ
H|Φ 6 and Φ 4 | ˆ
H|Φ 5 . In all cases the number of different columns of the
bra-determinant and the ket-determinant is larger than two. These matrix elements
are zero always as reflected in the Slater-Condon rules. Because of the centrosymmetric nature of the model, a 2 can be replaced by b 2 (and vice versa) in the integrals
of the Hamiltonian.
Exercise 6.4 Contributions from the one-electron integrals and Coulomb integrals
are the same in the three cases (h 11 + h 22 + h 33 + J 12 + J 13 + J 23 ) and will
be omitted. ϕ 1 ϕ 2 ϕ 3 | ˆ
H|ϕ 1 ϕ 2 ϕ 3 =− K 12 − K 13 − K 23 . ϕ 1 ϕ 2 ϕ 3 | ˆ
H|ϕ 1 ϕ 2 ϕ 3 =
−K 12 . With K 13 = K 23 = K, the energy difference becomes 2K, in line with
Eq. 6.29. The doublet function with triplet coupling for ϕ 1 and ϕ 2 is given in
Eq. 1.46. (1/6)2ϕ 1 ϕ 2 ϕ 3 − ϕ 1 ϕ 2 ϕ 3 − ϕ 1 ϕ 2 ϕ 3 | ˆ
H|2ϕ 1 ϕ 2 ϕ 3 − ϕ 1 ϕ 2 ϕ 3 − ϕ 1 ϕ 2 ϕ 3 =
(1/6)[−4K 12 + 2K 23 + 2K 13 + 2K 23 − K 13 − K 12 + 2K 13 − K 12 − K 23 ]=
−K 12 + (1/2)(K 23 + K 23 ). The energy difference with |ϕ 1 ϕ 2 ϕ 3 | is 3K as expected
from Eq. 6.30.
Exercise 6.5 The expression for the second-order correction is
a Φ I | ˆ
H|
Φ a Φ a | ˆ
H|Φ I /(E I − E a ), where I = S, T and a is one of the determinants other
than Ψ 1 or Ψ 2 in the matrices. Only Ψ 7 and Ψ 9 have non-zero matrix elements with
Ψ 1 leading to the expression given in Eq. 6.47 for the triplet. Ψ 2 interacts directly
with Ψ 8 (−t pd ), Ψ 10 (−t pd ) and Ψ 13 (2t ab ). The application of the formula gives the
second order corrected energy for the singlet.
Exercise 6.6 Ψ ± =
1
2
αα(αβ±βα) + (αβ±βα)αα
. −
1
2 J(ˆ s + (1) +ˆ s + (2))(ˆ s − (3) +
ˆ
s − (4))(αααβ ±ααβα) = 0; −
1
2 (ˆ s + (1) +ˆ s + (2))(ˆ s − (3)+ˆ s − (4))(αβαα ±βααα) =
−
1
2 J((ˆ s + (1) +ˆ s + (2))(αββα + βαβα ± (αβαβ + βααβ)) =∓J(αααβ + ααβα);
−
1
2 J(ˆ s − (1) +ˆ s − (2))(ˆ s + (3)+ˆ s + (4))(αααβ ±ααβα) =∓J(ˆ s − (1)+ˆ s − (2))αααα =
∓J(αααβ + ααβα); −
1
2 (ˆ s − (1) +ˆ s − (2))(ˆ s + (3) +ˆ s + (4))(αβαα ± βααα) = 0;
(ˆ s z (1) +ˆ s z (2))(ˆ s z (3) +ˆ s z (4))(αααβ ± ααβα) = (
1
2 +
1
2 )(
1
2 −
1
2 ) + (
1
2 +
1
2 )(−
1
2 +
1
2 )(αααβ ± ααβα) = 0 and similar for the other term of Ψ ± . From this: ˆ
HΨ ± =
∓JΨ ± . The eigenvalues of ˆ
S 2 can be determined in a similar way: (ˆ s + (1) +
ˆ
s + (2) +ˆ s + (3) +ˆ s + (4))(ˆ s − (1) +ˆ s − (2) +ˆ s − (3) +ˆ s − (4))ααβα = (ˆ s + (1) +ˆ s + (2) +
ˆ
s + (3) +ˆ s + (4))(βαβα + αββα + 0 + ααββ) = 3ααβα + βααα + αβαα + αααβ;
(ˆ s + (1) +ˆ s + (2) +ˆ s + (3) +ˆ s + (4))(ˆ s − (1) +ˆ s − (2) +ˆ s − (3) +ˆ s − (4))αααβ = 3αααβ +
βααα + αβαα + ααβα; (ˆ s + (1) +ˆ s + (2) +ˆ s + (3) +ˆ s + (4))(ˆ s − (1) +ˆ s − (2) +ˆ s − (3) +
ˆ
s − (4))αβαα = 3αβαα + βααα + ααβα + αααβ; (ˆ s + (1) +ˆ s + (2) +ˆ s + (3) +
ˆ
s + (4))(ˆ s − (1) +ˆ s − (2) +ˆ s − (3) +ˆ s − (4))βααα = 3βααα + αβαα + ααβα + αααβ;
(ˆ s z (1) +ˆ s z (2) +ˆ s z (3) +ˆ s z (4))ααβα = (
1
2 +
1
2 −
1
2 +
1
2 )ααβα = ααβα and similar
for the other terms. The ˆ
S 2
z operator has the same eigenvalues and cancels the effect
of ˆ
S z because they appear with opposite signs in the expression of ˆ
S 2 .F r o mt h i s
ˆ
S 2 Ψ + = 6Ψ + (quintet) and ˆ
S 2 Ψ − = 2Ψ − (triplet).
Exercise 6.7 (a) α(1)β(2)α(3)β(4)α(5)β(6)α(7)β(8) (b) In a step-by-step procedure, the expectation value of Φ 0 is determined. The products of spin-up and spindown operators change the wave function and hence give a zero contribution to