236
Appendix E: Solutions
When λ ≈ µ, the term with the ionic determinants tends to zero, demonstrating that
S g is dominated by neutral determinants. In case of λ ≫ µ, an approximately 50 %
mixture of neutral (or covalent) and ionic determinants appears, typical of a covalent
bond in closed shell molecules such as H 2 .
Exercise 5.2 aa| ˆ
H|aa== a| ˆ
h|a++ a| ˆ
h|a++ aa|
1− ˆ
P 12
r 12
|aa=h aa + h aa +
J aa .bb| ˆ
H|bb=h bb + h bb + J bb . Only under the assumption that h aa = h bb and
J aa = J bb one can write the relative energy of the ionic determinants as U = J aa −J ab .
This is the case for centro-symmetric systems.
Exercise 5.3 Substitution of
U 2 + 16t 2
ab ≈
√
U 2 +
1
2 16t 2
ab /U = U + 8t 2
ab /U in
Eq. 5.7 gives (U − U + 8t 2
ab /U)/2 = 4t 2
ab /U.
Exercise 5.4 |ab|
t ba
− →|aa|
t ab
− →|ba|.
Exercise 5.5 |hhab|
t ha
− →| haab|
t hb
− →| aabb|
t bh
− →| ahba|
t ah
− →| hhba|, with the intermediate determinants at ∆E CT , ∆E 2CT , and ∆E CT , respectively. Using the expression
in Eq. 5.12, we arrive at (t h a · t hb · t bh · t ah )/(∆E CT · ∆E 2CT · ∆E CT ). Since there are
four different pathways, the final perturbative estimate of the contribution to J reads
−8(t
eff
ab ) 2 /(2∆E CT ∆E 2CT ), introducing an effective hopping parameter between the
magnetic centers t
eff
ab = t ha t hb .
Exercise 5.6 The two electron pairs have S 1 = 1 and S 2 = 1. The total spin of
these two electron pairs can in principle take the values S 1 + S 2 = 2 (quintet),
S 1 + S 2 − 1 = 1 (triplet), and S 1 − S 2 = 0 (singlet). For a binuclear Cu 2+ complex
(and all other systems with two S = 1/2 spin moments), only the triplet and singlet
couplings are relevant.
Exercise 5.7 Substituting t eff =−2218 cm −1 and ∆E ST =−362 cm −1 in U eff =
4t eff /∆E ST gives a value of 13590 cm −1 (6.74 eV) for the effective on-site repulsion,
aloweringof∼19 eV with respect to the bare valence-only value.
Exercise 5.8
Ψ i are the projections of Ψ on the model space,
Ψ ′
i are the normalized projections, and
Ψ
′†
i the biorthonormal projections. |
Ψ 1 | 2 = (−0.9224) 2 +
(−0.1223) 2 = 0.8658, |
Ψ 2 | 2 = (−0.6626) 2 = 0.4390, |
Ψ 7 | 2 = 0.4159 2 =
0.1730, |
Ψ 8 | 2 = 0.1704 2 + (−0.5324) 2 = 0.3125;
Ψ ′
1 =− 0.9913
|ab|+
|ba|
− 0.1315
|aa|+| bb|
;
Ψ ′
2 =
|ab|−| ba|
/
√
2;
Ψ ′
7 =
|aa|−| bb|
/
√
2;
Ψ ′
8 = 0.3048
|ab|+|ba|
− 0.9524
|aa|+|bb|
.
Ψ ′
1 |
Ψ ′
8 =(−0.9913 × 0.3048 +
−0.1315 ×− 0.9524) =− 0.1769, the other overlaps are zero.
Ψ ′
1 |
Ψ
′†
1 =
−0.9913 ×−0.9524 +−0.1315 ×−0.3048 = 0.9842 ==
Ψ ′
2 |
Ψ
′†
2 ;
Ψ ′
1 |
Ψ
′†
2 =
−0.9913 × 0.1315 +− 0.1315 ×− 0.9913 = 0 ==
Ψ ′
2 |
Ψ
′†
1 ;
Ψ ′
1 †|
Ψ
′†
2 =
−0.9524 × 0.1315 +−0.3048 ×−0.9913 = 0.1769 ==
Ψ ′
1 |
Ψ ′
2 .
Exercise 5.9 Remember that a and b are normalized, orthogonal orbitals: a ′ |b ′ =
2sinα cos α = sin(2α)/2. Overlaps for the listed values of α are 0, 0.052, 0.155,
0.5 and 0.
Appendix E: Solutions
When λ ≈ µ, the term with the ionic determinants tends to zero, demonstrating that
S g is dominated by neutral determinants. In case of λ ≫ µ, an approximately 50 %
mixture of neutral (or covalent) and ionic determinants appears, typical of a covalent
bond in closed shell molecules such as H 2 .
Exercise 5.2 aa| ˆ
H|aa== a| ˆ
h|a++ a| ˆ
h|a++ aa|
1− ˆ
P 12
r 12
|aa=h aa + h aa +
J aa .bb| ˆ
H|bb=h bb + h bb + J bb . Only under the assumption that h aa = h bb and
J aa = J bb one can write the relative energy of the ionic determinants as U = J aa −J ab .
This is the case for centro-symmetric systems.
Exercise 5.3 Substitution of
U 2 + 16t 2
ab ≈
√
U 2 +
1
2 16t 2
ab /U = U + 8t 2
ab /U in
Eq. 5.7 gives (U − U + 8t 2
ab /U)/2 = 4t 2
ab /U.
Exercise 5.4 |ab|
t ba
− →|aa|
t ab
− →|ba|.
Exercise 5.5 |hhab|
t ha
− →| haab|
t hb
− →| aabb|
t bh
− →| ahba|
t ah
− →| hhba|, with the intermediate determinants at ∆E CT , ∆E 2CT , and ∆E CT , respectively. Using the expression
in Eq. 5.12, we arrive at (t h a · t hb · t bh · t ah )/(∆E CT · ∆E 2CT · ∆E CT ). Since there are
four different pathways, the final perturbative estimate of the contribution to J reads
−8(t
eff
ab ) 2 /(2∆E CT ∆E 2CT ), introducing an effective hopping parameter between the
magnetic centers t
eff
ab = t ha t hb .
Exercise 5.6 The two electron pairs have S 1 = 1 and S 2 = 1. The total spin of
these two electron pairs can in principle take the values S 1 + S 2 = 2 (quintet),
S 1 + S 2 − 1 = 1 (triplet), and S 1 − S 2 = 0 (singlet). For a binuclear Cu 2+ complex
(and all other systems with two S = 1/2 spin moments), only the triplet and singlet
couplings are relevant.
Exercise 5.7 Substituting t eff =−2218 cm −1 and ∆E ST =−362 cm −1 in U eff =
4t eff /∆E ST gives a value of 13590 cm −1 (6.74 eV) for the effective on-site repulsion,
aloweringof∼19 eV with respect to the bare valence-only value.
Exercise 5.8
Ψ i are the projections of Ψ on the model space,
Ψ ′
i are the normalized projections, and
Ψ
′†
i the biorthonormal projections. |
Ψ 1 | 2 = (−0.9224) 2 +
(−0.1223) 2 = 0.8658, |
Ψ 2 | 2 = (−0.6626) 2 = 0.4390, |
Ψ 7 | 2 = 0.4159 2 =
0.1730, |
Ψ 8 | 2 = 0.1704 2 + (−0.5324) 2 = 0.3125;
Ψ ′
1 =− 0.9913
|ab|+
|ba|
− 0.1315
|aa|+| bb|
;
Ψ ′
2 =
|ab|−| ba|
/
√
2;
Ψ ′
7 =
|aa|−| bb|
/
√
2;
Ψ ′
8 = 0.3048
|ab|+|ba|
− 0.9524
|aa|+|bb|
.
Ψ ′
1 |
Ψ ′
8 =(−0.9913 × 0.3048 +
−0.1315 ×− 0.9524) =− 0.1769, the other overlaps are zero.
Ψ ′
1 |
Ψ
′†
1 =
−0.9913 ×−0.9524 +−0.1315 ×−0.3048 = 0.9842 ==
Ψ ′
2 |
Ψ
′†
2 ;
Ψ ′
1 |
Ψ
′†
2 =
−0.9913 × 0.1315 +− 0.1315 ×− 0.9913 = 0 ==
Ψ ′
2 |
Ψ
′†
1 ;
Ψ ′
1 †|
Ψ
′†
2 =
−0.9524 × 0.1315 +−0.3048 ×−0.9913 = 0.1769 ==
Ψ ′
1 |
Ψ ′
2 .
Exercise 5.9 Remember that a and b are normalized, orthogonal orbitals: a ′ |b ′ =
2sinα cos α = sin(2α)/2. Overlaps for the listed values of α are 0, 0.052, 0.155,
0.5 and 0.