46
S. J. Blundell
(a)
(b)
Mn
O
Mn
I
↑
↑ ↓
↓
II ↑↓
↑↓
III ↑↓
↑↓
IV ↑
↑ ↓
↑
Fig. 2.4 a The crystal structure of MnO. Nearest neighbour pairs of Mn 2+ (manganese) ions are
connected via O 2− (oxygen) ions. b A simple model of superexchange for a Mn–O–Mn bond.
I: the antiferromagnetic ground state with opposite spins on the two Mn ions and a pair of electrons
on the oxygen anion. II and III: two excited states of the antiferromagnetic ground state in which
the electrons from (I) hop back and forth. IV: the competing ferromagnetic ground state. This is
energetically more costly because the excited states analogous to (II) and (III) are not available
because of the Pauli exclusion principle
where the first sum is over nearest neighbours; thus energy is lowered by hopping
(the first term on the right) but there is an energy penalty for double occupancy (the
second term on the right) due to the Coulomb repulsion energy U . Let us now restrict
this model to a system with two possible sites for electrons (here we are ignoring the
intermediate oxygen to for simplicity). We can start by putting a single electron with
spin ↑ into the system. Using a basis | ↑, 0 and |0, ↑↑, the Hamiltonian is given by
ˆ
H =
0 −t
−t 0
,
(2.19)
because with one electron there is no possibility of a Coulomb penalty, and so the
only energy to worry about is the energy saving you get from hopping. This is the
same as the H 2 problem we considered earlier and the eigenvalues are ±t and so the
lowest energy state is the bonding state, just as before.
Now let us put a second electron into the system with opposite spin to the first.
Now using a basis such that a general state can be written as
|ψ = a| ↑↓, 0 + b| ↑, ↓↓ + c| ↓, ↑↑ + d|0, ↑↓↓ ,
(2.20)
and we can easily show that in this basis the Hubbard Hamiltonian is
ˆ
H =
⎛
⎜
⎜
⎝
U t −t 0
t 0 0 t
−t 0 0 −t
0 t −t U
⎞
⎟
⎟
⎠ ,
(2.21)
S. J. Blundell
(a)
(b)
Mn
O
Mn
I
↑
↑ ↓
↓
II ↑↓
↑↓
III ↑↓
↑↓
IV ↑
↑ ↓
↑
Fig. 2.4 a The crystal structure of MnO. Nearest neighbour pairs of Mn 2+ (manganese) ions are
connected via O 2− (oxygen) ions. b A simple model of superexchange for a Mn–O–Mn bond.
I: the antiferromagnetic ground state with opposite spins on the two Mn ions and a pair of electrons
on the oxygen anion. II and III: two excited states of the antiferromagnetic ground state in which
the electrons from (I) hop back and forth. IV: the competing ferromagnetic ground state. This is
energetically more costly because the excited states analogous to (II) and (III) are not available
because of the Pauli exclusion principle
where the first sum is over nearest neighbours; thus energy is lowered by hopping
(the first term on the right) but there is an energy penalty for double occupancy (the
second term on the right) due to the Coulomb repulsion energy U . Let us now restrict
this model to a system with two possible sites for electrons (here we are ignoring the
intermediate oxygen to for simplicity). We can start by putting a single electron with
spin ↑ into the system. Using a basis | ↑, 0 and |0, ↑↑, the Hamiltonian is given by
ˆ
H =
0 −t
−t 0
,
(2.19)
because with one electron there is no possibility of a Coulomb penalty, and so the
only energy to worry about is the energy saving you get from hopping. This is the
same as the H 2 problem we considered earlier and the eigenvalues are ±t and so the
lowest energy state is the bonding state, just as before.
Now let us put a second electron into the system with opposite spin to the first.
Now using a basis such that a general state can be written as
|ψ = a| ↑↓, 0 + b| ↑, ↓↓ + c| ↓, ↑↑ + d|0, ↑↓↓ ,
(2.20)
and we can easily show that in this basis the Hubbard Hamiltonian is
ˆ
H =
⎛
⎜
⎜
⎝
U t −t 0
t 0 0 t
−t 0 0 −t
0 t −t U
⎞
⎟
⎟
⎠ ,
(2.21)