Environmental Engineering Reference
In-Depth Information
If the usual basis vectors in the Hilbert space created bytheBoseoper-
ators are denoted by
n )= n
|
n ), i.e. a
|
|
n
1) where n =0 , 1 , 2 ,
···
,
,
then by the definition (5.2.7),
|
n )=
|
J z = J
n> for n =0 , 1 , 2 ,
···
, 2 J ,
but there is no physical
n )when n> 2 J .
It is straightforward to see that the Bose representation (5.2.8) produces
the right matrix-elements of the angular momentum operators, as long
as
|
J z > -state corresponding to
|
2 J ,
but this representation presupposes the presence of an infinite number
of states. In the ferromagnetic case, the unphysical states are at high
energies, if J is large and T is low, and their influence on the thermal
averages is negligible. In this regime of J and T ,the Holstein-Primakoff
transformation is useful and the results derived from it are trustworthy.
In order to be able to treat the Bose operators under the square
roots in eqn (5.2.8), we shall utilize 1 /J as an expansion parameter.
This means that, instead of the J ± given by (5.2.8), we shall use
|
n ) is restricted to the physical part of the Hilbert space, n
2 J a
4 J a + aa .
J + =( J )
1
(5 . 2 . 9)
It is important here to realize that the expansion parameter is 1 /J and
not, for instance, 'the number of deviation operators'. If the latter
were the case, a well-ordered expansion of J + (Lindg ard and Danielsen
1974) would suggest instead J + = 2 J
1
1 / 2 J ) a + aa +
{
a
(1
···}
,
1
4 J
1
8 J
corresponding to a replacement of 4 J
). We
emphasize that we shall be expanding the reduced operators (1 /J ( l ) ) O l ,
leaving no ambiguities either in (5.2.9) or in the following. Using eqn
(5.2.9) and J z = J
in (5.2.9) by
(1 +
+
···
a + a , it is straightforward to express the Stevens
operators in terms of the Bose operators. For O 2 ,weget
O 2 =3 J z
a + a ) 2
J ( J +1)=3( J
J ( J +1)
2
2
) a + a +3 a + a + aa
=2 J ( J
)
6( J
(5 . 2 . 10)
=2 J (2) 1
(1 /J 3 ) .
3
3
J a + a +
2 J 2 a + a + aa +
O
Here we have used [ a, a + ] = 1 to arrange the operators in 'well-ordered'
products, with all the creation operators to the left, and in the last line
1 /J (2) has been replaced by 1 /J 2 in the term of second order in 1 /J .In
thesameway,weobtain
)= J (2) 1
O 2 = 1
( J + + J 2
J ( a + a + + aa )
2
(1 /J 3 ) .
(5 . 2 . 11)
1
4 J 2
( a + a + + aa
2 a + a + a + a
2 a + aaa )+
+
O
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