Environmental Engineering Reference
In-Depth Information
Initially we assume that
J
z
=
J
at
T
= 0, which implies that the
ground state is the product of
J
z
=
J>
-states of the single ions. In
this case, we find, consistently with eqn (2.2.14),
|
Q
2
Q
2
|
J>
=
J
(2)
(3 cos
2
θ
=
<J
|
−
1)
,
1
2
)
n−
1
2
where, as before,
J
(
n
)
=
J
(
J
−
···
(
J
−
),andwehaveusedthe
1
2
J
and
J
z
=
J
2
,
J
x
expectation values
= 0. Analogously,
though with considerably more labour, we can show that, for instance,
=
J
z
J
x
|
2
(
J
ξ
+
iJ
η
)
6
+
2
Q
6
iJ
η
)
6
J>
=
J
(6)
sin
6
θ
cos 6
φ.
(5
.
2
.
4)
=
<J
(
J
ξ
−
|
For simplicity, we neglect for the moment
B
4
and
B
6
, and specifying
the direction of the magnetic field by the polar angles (
θ
H
,φ
H
), we find
that the ground-state energy is, within this approximation,
N
B
2
J
(2)
(3 cos
2
θ
1) +
B
6
J
(6)
sin
6
θ
cos 6
φ
U
(
T
=0)
−
(
0
)
J
2
,
(5
.
2
.
5)
where
θ
and
φ
are determined so that they minimize this expression. In
zero magnetic field,
H
= 0, (5.2.5) only gives two possibilities for
θ
,viz.
θ
=0for
B
2
J
(2)
<
}−
2
J
−
gµ
B
JH
{
cos
θ
cos
θ
H
+sin
θ
sin
θ
H
cos (
φ
−
φ
H
)
π
2
1
B
6
|
J
(6)
or
θ
=
for
B
2
J
(2)
>
1
B
6
|
J
(6)
.We
−
3
|
−
3
|
π
2
shall here be concerned with the second case of
θ
=
, i.e. the basal-
plane ferromagnet. In this case, the angle
φ
is determined by the sign
of
B
6
. The magnetic moments will be along an
a
-ora
b
-axis (
φ
=0
or
φ
=
π
2
)if
B
6
is respectively negative or positive. Having specified
the (approximate) ground state, we turn to the excitations, i.e. the spin
waves.
Instead of utilizing the standard-basis operators, defined by (3.5.11),
we shall introduce a Bose operator
a
i
for the
i
th ion, satisfying
[
a
i
,a
j
]=
δ
ij
[
a
i
,a
j
]=[
a
i
,a
j
]=0
,
;
(5
.
2
.
6)
which acts on the
J
z
>
-state vector of this ion (the site index is sup-
pressed) in the following way:
|
a |J − m>
=
√
m |J − m
+1
>
a |J>
=0
;
(5
.
2
.
7)
Holstein and Primakoff (1940) introduced the following representation
of the angular momentum operators:
a
+
a
J
z
=
J
−
J
+
=
2
J
a
+
a
2
a
−
(5
.
2
.
8)
J
−
=
a
+
2
J
a
+
a
2
.
−
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