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6 Magnetism and Conduction
are essential to the basic description of the hopping process of the electron between
the a 2 and b 2 orbitals. The six determinants are
Φ 1 =|a 1 b 1 a 2 |
Φ 4 =|a 1 b 1 b 2 |
Φ 2 =|a 1 b 1 a 2 |
Φ 5 =|a 1 b 1 b 2 |
(6.14)
Φ 3 =|a 1 b 1 a 2 |
Φ 6 =|a 1 b 1 b 2 |
6.2 Find the four determinants (or linear combinations of determinants) that
represent a triplet spin coupling of the electrons on center a or b. What is the
spin coupling of the other two (linear combinations) of determinants? What
do you expect for the relative energies of the two groups?
The interaction matrix elements are readily written down using the Slater–Condon
rules given in Chap. 1. We will work out three examples and leave the others as
exercise to the reader.
Φ 1 | ˆ
H|Φ 1 ==a 1 b 1 a 2 | ˆ
H|a 1 b 1 a 2 ==a 1 | ˆ
h|a 1 ++b 1 | ˆ
h|b 1 ++a 2 | ˆ
h|a 2
++a 1 b 1 |
1 − ˆ
P 12
r 12
|a 1 b 1 ++a 1 a 2 |
1 − ˆ
P 12
r 12
|a 1 a 2 ++b 1 a 2 |
1 − ˆ
P 12
r 12
|b 1 a 2
(6.15)
The two-electron part becomes
a 1 b 1 |
1
r 12
|a 1 b 1 −−a 1 b 1 |
1
r 12
|b 1 a 1 ++a 1 a 2 |
1
r 12
|a 1 a 2 ++b 1 a 2 |
1
r 12
|b 1 a 2
= J a 1 b 1 + J a 1 a 2 + J b 1 a 2 − K a 1 b 1
(6.16)
The one-electron part and the Coulomb integrals J xy are common to all diagonal
matrix elements and the sum of these terms can be taken as the zero of energy. Then,
the matrix element Φ 1 | ˆ
H|Φ 1 reduces to the exchange integral −K a 1 b 1 . Likewise,
the diagonal elements involving Φ 2 and Φ 3 reduce to K a 1 a 2 and to 0, respectively.
The off-diagonal matrix element between Φ 2 and Φ 3 is relatively simple
Φ 2 | ˆ
H|Φ 3 ==a 1 b 1 a 2 | ˆ
H|a 1 b 1 a 2 ==a 1 b 1 |
1 − ˆ
P 12
r 12
|a 1 b 1
=−−a 1 b 1 |
1
r 12
|b 1 a 1 =−K a 1 b 1
(6.17)