6.3 Symmetry Properties of the Coupling Coefficients
119
If the coupled electrons are equivalent, i.e. belong to the same shell, the situation
is different. In this case, Γ a = Γ b , and the direct product becomes a direct square.
For non-equivalent irreps, exchange of the components γ a and γ b was not possible,
because they refer to different irreps. However, when they are components of the
same irrep, this exchange is an important symmetry of the product space. Indeed,
the squared space can be split into two separate blocks, one block which contains
product functions that are symmetric under exchange of the component labels and
one block which is antisymmetric. This implies that we can define two separate sets
of direct square coupling coefficients, which are either symmetric or antisymmetric under exchange of the labels, i.e.: Γ a γ a Γ a γ b |Γγ a γ b Γ a γ a |Γγ.T h e
symmetrized part of the direct square is denoted as [Γ a ] 2 .F o rn = dim(Γ a ),t h e
dimension of this subspace is equal to the number of symmetric combinations:
dim
[Γ a ]
2
=
γ a
1 +
γ a <γ b
1 = n + n(n − 1)/2 = n(n + 1)/2
(6.19)
On the other hand, if the coupling coefficients are antisymmetric under exchange of
the labels, the coupled state belongs to the antisymmetrized direct square, denoted as
{Γ a } 2 . This product space is restricted to combinations with γ a = γ b ; its dimension
is equal to n(n − 1)/2. The characters for either part of the square can be determined
separately. For the character of the {Γ a } 2 part the derivation runs as follows: one first
applies a symmetry operator to an arbitrary antisymmetric function. The ket product
|Γ a γ a (1)|Γ a γ b (2) will be abbreviated here as: γ a (1)γ b (2).
ˆ
R
γ a (1)γ b (2) − γ b (1)γ a (2)
=
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2) − γ
′
b (1)γ
′
a (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R)
=
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R) − D
Γ a
γ ′
a γ b
(R)D
Γ a
γ ′
b γ a
(R)
=
1
2
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2) − γ
′
b (1)γ
′
a (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R) − D
Γ a
γ ′
a γ b
(R)D
Γ a
γ ′
b γ a
(R)
(6.20)
Taking the trace then yields:
χ
{Γ a } 2
(R) =
1
2
γ a γ b
D
Γ a
γ a γ a
(R)D
Γ a
γ b γ b
(R) − D
Γ a
γ a γ b
(R)D
Γ a
γ b γ a
(R)
=
1
2
χ
Γ a (R)
2 −
γ a
D
Γ a
γ a γ a
R
2
=
1
2
χ
Γ a (R)
2 − χ
Γ a
R
2
(6.21)
119
If the coupled electrons are equivalent, i.e. belong to the same shell, the situation
is different. In this case, Γ a = Γ b , and the direct product becomes a direct square.
For non-equivalent irreps, exchange of the components γ a and γ b was not possible,
because they refer to different irreps. However, when they are components of the
same irrep, this exchange is an important symmetry of the product space. Indeed,
the squared space can be split into two separate blocks, one block which contains
product functions that are symmetric under exchange of the component labels and
one block which is antisymmetric. This implies that we can define two separate sets
of direct square coupling coefficients, which are either symmetric or antisymmetric under exchange of the labels, i.e.: Γ a γ a Γ a γ b |Γγ a γ b Γ a γ a |Γγ.T h e
symmetrized part of the direct square is denoted as [Γ a ] 2 .F o rn = dim(Γ a ),t h e
dimension of this subspace is equal to the number of symmetric combinations:
dim
[Γ a ]
2
=
γ a
1 +
γ a <γ b
1 = n + n(n − 1)/2 = n(n + 1)/2
(6.19)
On the other hand, if the coupling coefficients are antisymmetric under exchange of
the labels, the coupled state belongs to the antisymmetrized direct square, denoted as
{Γ a } 2 . This product space is restricted to combinations with γ a = γ b ; its dimension
is equal to n(n − 1)/2. The characters for either part of the square can be determined
separately. For the character of the {Γ a } 2 part the derivation runs as follows: one first
applies a symmetry operator to an arbitrary antisymmetric function. The ket product
|Γ a γ a (1)|Γ a γ b (2) will be abbreviated here as: γ a (1)γ b (2).
ˆ
R
γ a (1)γ b (2) − γ b (1)γ a (2)
=
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2) − γ
′
b (1)γ
′
a (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R)
=
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R) − D
Γ a
γ ′
a γ b
(R)D
Γ a
γ ′
b γ a
(R)
=
1
2
γ ′
a γ ′
b
γ
′
a (1)γ
′
b (2) − γ
′
b (1)γ
′
a (2)
D
Γ a
γ ′
a γ a
(R)D
Γ a
γ ′
b γ b
(R) − D
Γ a
γ ′
a γ b
(R)D
Γ a
γ ′
b γ a
(R)
(6.20)
Taking the trace then yields:
χ
{Γ a } 2
(R) =
1
2
γ a γ b
D
Γ a
γ a γ a
(R)D
Γ a
γ b γ b
(R) − D
Γ a
γ a γ b
(R)D
Γ a
γ b γ a
(R)
=
1
2
χ
Γ a (R)
2 −
γ a
D
Γ a
γ a γ a
R
2
=
1
2
χ
Γ a (R)
2 − χ
Γ a
R
2
(6.21)