(c)
The effect of a virtual transition is to lower the ground state energy of the system(the electrons have
a bigger space to move in), with the maximum effect being to arrange the
M
2+
n
antiferromagnetally.
3.7
Ferromagnetic GroundState and Excitations of the Heisenberg Hamiltonian
In this section we will consider only a 1D chain of ions.
The Heisenberg Hamiltonian is
ˆ
H
=


J

~
2
X
i
ˆ
S
i
·
ˆ
S
i
+1
(3.46)
Suppose all ions have spin 1/2, and the chain forms a closed loop so that
ˆ
S
N
+1
=
ˆ
S
1
. If the spins are aligned,
the energy takes the maximum negative value.
Ground States
Consider
χ
0
=
↑
1
↑
2
↑
3
...
↑
N
(3.47)
45
PHYS40352 Spring 2016
3
MAGNETISM
We will focus on
↑
1
↑
2
. We will start by applying
ˆ
S
1
·
ˆ
S
2
on
↑
1
↑
2
, where
ˆ
S
1
·
ˆ
S
2
=
1
2
ˆ
S
1
+
ˆ
S
2
2

ˆ
S
2
1

ˆ
S
2
2
(3.48)
Note that
↑
1
↑
2
is one of the triplet states with total spin of ions as
S
= 1
. Therefore
ˆ
S
1
·
ˆ
S
2
↑
1
↑
2
=
1
2
ˆ
S
1
+
ˆ
S
2
2

3
2
~
2
↑
1
↑
2
=
1
2
1(1 + 1)
~
2

3
2
~
2
↑
1
↑
2
=
1
4
~
2
↑
1
↑
2
Note that as
ˆ
S
1
,
ˆ
S
2
only apply to
↑
1
,
↑
2
, we could multiply this by
↑
3
, ...,
↑
N
without changing anything.
Hence
ˆ
S
1
·
ˆ
S
2
χ
0
=
1
4
~
2
χ
0
(3.49)
So in general for neighbouring spins
ˆ
S
m
·
ˆ
S
m
+1
χ
0
=
1
4
~
2
χ
0
(3.50)
Therefore
ˆ
Hχ
0
=


J

~
2
X
m
ˆ
S
m
·
ˆ
S
m
+1
χ
0
=


J

~
2
N
~
2
4
χ
0
=

1
4

J

N χ
0
hence
χ
0
is an energy eigenstate with energy eigenvalue
E
(0)
=

1
4

J

N
(3.51)
where
E
(0)
is the ground state energy. Now, in this state
S
z
=
N
2
~
, the maximum possible, with
S
=
N
2
. The
degeneracy is
2
S
+ 1
, giving a degeneracy of
N
+ 1
which is huge (the ground state is not unique  there are
many different possibilities).
This high degeneracy is closely connected with the broken symmetry of
χ
0
.
ˆ
H
has full rotational symmetry
(in spin space), but
χ
0
has much lower symmetry  only rotation about the zaxis.
Excited States
Consider a state with 1 spin flipped:
χ
n
=
↑
1
↑
2
...
↑
n

1
↓
n
↑
n
+1
Note that
ˆ
S
m
·
ˆ
S
m
+1
χ
n
=

1
4
~
2
χ
n
provided that
m
6
=
n
or
m
6
=
n

1
.
46
PHYS40352 Spring 2016
3
MAGNETISM
Let’s consider the effect of
ˆ
S
n
·
ˆ
S
n
+1
on
χ
n
ˆ
S
n
·
ˆ
S
n
+1
↓
n
↑
n
+1
=
ˆ
S
n
·
ˆ
S
n
+1
1
2
(
↓
n
↑
n
+1
 ↑
n
↓
n
+1
) +
1
2
(
↓
n
↑
n
+1
+
↑
n
↓
n
+1
)
=
1
2
ˆ
S
n
+
ˆ
S
n
+1
2

3
2
~
2
1
2
(
↓
n
↑
n
+1
 ↑
n
↓
n
+1
) +
1
2
(
↓
n
↑
n
+1
+
↑
n
↓
n
+1
)
=
1
2
1
2
0(0 + 1)
~
2

3
2
~
2
(
↓
n
↑
n
+1
 ↑
n
↓
n
+1
) +
1
2
1(1 + 1)
~
2

3
2
~
2
(
↓
n
↑
n
+1
+
↑
n
↓
n
+1
)
=
1
4

3
2
+
1
2
(
↓
n
↑
n
+1
) +
3
2
+
1
2
↑
n
↓
n
+1
~
2
=
1
4
~
2
(
 ↓
n
↑
n
+1
+2
↑
n
↓
n
+1
)
ˆ
S
n
·
ˆ
S
n
+1
χ
n
=
1
4
~
2
(

χ
n
+ 2
χ
n
+1
)
Hence the effect is to either leave the flipped spin where it is, or to transfer it to the adjacent ion. So far we
have the spinflip on site
n
,
ˆ
S
n
·
ˆ
S
n
+1
χ
n
=
1
4
~
2
(

χ
n
+ 2
χ
n
+1
)
(3.52)
or for
m
6
=
n
, with
N

2
states,
ˆ
S
m
·
ˆ
S
m
+1
χ
n
=

~
2
4
χ
n
(3.53)
and lastly we can apply
ˆ
S
n

1
·
ˆ
S
n
χ
n
=
1
4
~
2
(

χ
n
+ 2
χ
n

1
)
.
(3.54)
We can combine these three equations to find the total spins,
ˆ
Hχ
n
=


J

~
2
N
X
m
=1
ˆ
S
m
·
ˆ
S
m
+1
χ
n
=


J

~
2
(
N

2)
1
4
~
2
χ
n
+
~
2
4
(

χ
n
+ 2
χ
n
+1
) +
~
2
4
(

χ
n
+ 2
χ
n

1
)
=

1
4
N

J

χ
n
 
J


2
4

1
4

1
4
χ
n
+
2
4
χ
n
+1
+
2
4
χ
n

1
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 Summer '19
 Electron, Solid State Physics, Cubic crystal system, Crystal system, Reciprocal lattice, square lattice