238
Appendix E: Solutions
2h ab ++ab|1/r 12 |aa++ab|1/r 12 |bb=2t ab . Taking T u | ˆ
H|T u as reference energy,
the CAS(2, 2) matrix becomes
|Ψ 1 | Ψ 2 | Ψ 3 | Ψ 4
Ψ 1 | 2K ab 02 t ab
0
Ψ 2 | 00
0
0
Ψ 3 | 2t ab 02 K ab + U 0
Ψ 4 | 00
0
U
Problem 5.3 (a) The third term (sin α cos α)
|aa|+| bb|
is singlet spin eigenfunction with eigenvalue S(S + 1) = 0. (b) Φ BS =
1
2 cos 2 α|ab|+
1
2 cos 2 α|ab|+
1
2 sin 2 α|ba|+
1
2 sin 2 α|ba|+
1
2 cos 2 α|ba|−
1
2 cos 2 α|ba|+
1
2 sin 2 α|ab|−
1
2 sin 2 α|ab|+
(sin α cos α)(|aa|+|bb|) =
1
2 (cos 2 α+sin 2 α)(|ab|+|ba|)+
1
2 (cos 2 α−sin 2 α)(|ab|−
|ba|) + (sin α cos α)(|aa|+|bb|) =
1
2
√
2|S 1 +
1
2 cos 2α ·
√
2|T +(sin α cos α) ·
√
2|S 2 .(c)Φ BS | ˆ
S 2 |Φ BS =
1
4 ·2··S 1 | ˆ
S 2 |S 1 +
1
4 cos 2 (2α)·2T | ˆ
S 2 |T +(sin α cos α) 2 ·
2S 2 | ˆ
S 2 |S 2 =0 +
1
2 cos 2 (2α) · 1(1 + 1) + 0 = cos 2 (2α).
Problem 5.4 Energies relative to E T : S| ˆ
H|S=2K ab ; I 1 | ˆ
H|I 1 =U + 2K ab ;
I 2 | ˆ
H|I 2 =U −2K ab (all the Coulomb interactions are absorbed in E ref ). Interaction
matrix elements: S| ˆ
H|I=
1
2 [[ab| ˆ
H|aa++ab| ˆ
H|bb++ba| ˆ
H|aa++ba| ˆ
H|bb] =
2t. S| ˆ
H|I 2 =0. Second-order energy of S :: S| ˆ
H|S++S| ˆ
H|I 1 I 1 | ˆ
H|S/(E S −
E I 1 ) = 2K ab + (2t 2 · 2t 2 )(2K ab − (U + 2K ab )) = 2K ab − 4t 2 /U.
Problem 5.5 λ = B 2 /3K − J (2)2 /4K =[ (4B 2 − 3J (2)2 )/12K = (4t 4
13 + 4t 4
24 −
8t 2
13 t 2
24 ) − (3t 4
13 + 3t 4
24 + 6t 2
13 t 2
24 )]/12KU 2 = (t 4
13 + t 4
24 − 2t 2
13 t 2
24 )/12KU 2 =
(t 2
13 −
t 2
24 ) 2
/12KU 2 . Biquadratic exchange is maximum for maximal difference between
the two t-values and approaches zero when they become equal.
Problem 5.6 Neglecting the direct exchange contribution, the perturbative estimate
of J r = 80t 4 /U 3 ;forJ ij = 4t 2 /U ⇒ J r /(J 12 J 23 ) = (80t 4 × U 2 )/(U 3 × 16t 4 ) =
5/U. From this immediately follows that J r = (5J 12 J 23 )/U = 5 ×− 25.1 ×
−39.5)/3100 = 1.6meV .
Exercises and Problems of Chap. 6
Exercise 6.1 For a centrosymmetric system c 1 = c 2 = 1/
√
2. Substitution in
Eq. 6.9 leads to t
+
ab =
∆E 12 − (1/2 − 1/2)(H aa − H bb )
/4 · 1/2 = ∆E/2.
Exercise 6.2 Φ 2 and Φ 5 have two α electrons on A and B, respectively. Φ 1 + Φ 3 and
Φ 4 + Φ 6 are the M S = 0 components of the on-site triplets. The minus combinations
of these correspond to singlet coupling on the magnetic centers. Note that these
functions are not directly spin eigenfunctions of the whole complex.