6.1 Electron Hopping
181
and the hopping parameter can be directly determined from the off-diagonal matrix
element.
6.1 Show that the expression of t
+
ab for the non-centrosymmetric case reduces
to ∆E 12 /2 for a centrosymmetric system.
The determination of t 0
ab has already been discussed in Sect. 5.2. It requires the
construction of a 4 × 4 effective Hamiltonian with a basis of two neutral and two
ionic determinants. The hopping integral is defined as the matrix element between
neutral and ionic determinants. The calculation of t
−
ab is analogous to the procedure
for estimating t
+
ab . The two doublets that can be defined in a centrosymmetric complex
with three electrons in the two magnetic orbitals g and u (omitting hh for simplicity)
D 1 =|guu|
D 2 =|ggu|
(6.10)
are re-expressed in the orthogonal atomic-like orbitals a and b
D 1 =
1
2
√
2
|(a + b)(a − b)(a − b)|=
1
2
√
2
|−aba + abb + baa − bab|
=
1
√
2
(|abb|+|aab|)
(6.11a)
D 2 =
1
2
√
2
|(a + b)(a + b)(a − b)|=
1
2
√
2
|−aab − abb + baa + bba|
=
1
√
2
(|abb|−|aab|)
(6.11b)
The energies of the two states are
E 1 =
1
2
abb + aab| ˆ
H|abb + aab
=
1
2
abb| ˆ
H|abb+2abb| ˆ
H|aab++aab| ˆ
H|aab
(6.12a)
E 2 =
1
2
abb − aab| ˆ
H|abb − aab
=
1
2
abb| ˆ
H|abb−2abb| ˆ
H|aab++aab| ˆ
H|aab
(6.12b)
181
and the hopping parameter can be directly determined from the off-diagonal matrix
element.
6.1 Show that the expression of t
+
ab for the non-centrosymmetric case reduces
to ∆E 12 /2 for a centrosymmetric system.
The determination of t 0
ab has already been discussed in Sect. 5.2. It requires the
construction of a 4 × 4 effective Hamiltonian with a basis of two neutral and two
ionic determinants. The hopping integral is defined as the matrix element between
neutral and ionic determinants. The calculation of t
−
ab is analogous to the procedure
for estimating t
+
ab . The two doublets that can be defined in a centrosymmetric complex
with three electrons in the two magnetic orbitals g and u (omitting hh for simplicity)
D 1 =|guu|
D 2 =|ggu|
(6.10)
are re-expressed in the orthogonal atomic-like orbitals a and b
D 1 =
1
2
√
2
|(a + b)(a − b)(a − b)|=
1
2
√
2
|−aba + abb + baa − bab|
=
1
√
2
(|abb|+|aab|)
(6.11a)
D 2 =
1
2
√
2
|(a + b)(a + b)(a − b)|=
1
2
√
2
|−aab − abb + baa + bba|
=
1
√
2
(|abb|−|aab|)
(6.11b)
The energies of the two states are
E 1 =
1
2
abb + aab| ˆ
H|abb + aab
=
1
2
abb| ˆ
H|abb+2abb| ˆ
H|aab++aab| ˆ
H|aab
(6.12a)
E 2 =
1
2
abb − aab| ˆ
H|abb − aab
=
1
2
abb| ˆ
H|abb−2abb| ˆ
H|aab++aab| ˆ
H|aab
(6.12b)
