4.6 Subduction and Induction
75
Knowing the diagonal elements of the induction matrix, we can now calculate
the frequency of a given Γ irrep of the main group, using the character theorem:
c Γ (γ H ↑ G) =
1
|G|
g∈G
¯
χ
Γ (g)Tr
D
H ↑G (g)
=
1
|G|
g∈G
¯
χ
Γ (g)
κ
P κκ (g)χ
γ
g
−1
κ gg κ
(4.78)
The only elements ˆ
g that are allowed in the summation over κ are the ones such
that ( ˆ
g −1
κ ˆ
g ˆ
g κ ) ∈ H A . For other elements, P κκ (g) are zero. Let us denote by ˆ
h the
subelement that allows us to express ˆ
g as
ˆ
g =ˆ g κ ˆ
h ˆ
g
−1
κ
(4.79)
Introducing this substitution in Eq. (4.78) yields
c Γ (γ H ↑ G) =
1
|G|
h∈H
κ
¯
χ
Γ
g κ hg
−1
κ
χ
γ (h)P κκ
g κ hg
−1
κ
(4.80)
The first character in this equation belongs to the full group and is the same for all
elements of a conjugacy class, and hence,
χ
Γ
g κ hg
−1
κ
= χ
Γ (h)
(4.81)
Substituting the result of Eq. (4.81) and the sum rule in Eq. (4.75) into the character
expression finally gives
c Γ (γ H ↑ G) =
1
|G|
h∈H
κ
¯
χ
Γ (h)χ
γ (h)P κκ
g κ hg
−1
κ
=
1
|G|
h∈H
¯
χ
Γ (h)χ
γ (h)
κ
P κκ
g κ hg
−1
κ
=
1
|H |
h∈H
¯
χ
Γ (h)χ
γ (h)
= c γ (Γ G ↓ H)
(4.82)
which concludes the proof. Armed with the subduction tables, we can now read
these at once in the opposite sense and obtain the corresponding induction frequencies. As a simple example, consider a hydrogen atom in ammonia. The site symmetry is C s , and the subduction from C 3v reads:
A 1 → a
A 2 → b
E → a + b
(4.83)
75
Knowing the diagonal elements of the induction matrix, we can now calculate
the frequency of a given Γ irrep of the main group, using the character theorem:
c Γ (γ H ↑ G) =
1
|G|
g∈G
¯
χ
Γ (g)Tr
D
H ↑G (g)
=
1
|G|
g∈G
¯
χ
Γ (g)
κ
P κκ (g)χ
γ
g
−1
κ gg κ
(4.78)
The only elements ˆ
g that are allowed in the summation over κ are the ones such
that ( ˆ
g −1
κ ˆ
g ˆ
g κ ) ∈ H A . For other elements, P κκ (g) are zero. Let us denote by ˆ
h the
subelement that allows us to express ˆ
g as
ˆ
g =ˆ g κ ˆ
h ˆ
g
−1
κ
(4.79)
Introducing this substitution in Eq. (4.78) yields
c Γ (γ H ↑ G) =
1
|G|
h∈H
κ
¯
χ
Γ
g κ hg
−1
κ
χ
γ (h)P κκ
g κ hg
−1
κ
(4.80)
The first character in this equation belongs to the full group and is the same for all
elements of a conjugacy class, and hence,
χ
Γ
g κ hg
−1
κ
= χ
Γ (h)
(4.81)
Substituting the result of Eq. (4.81) and the sum rule in Eq. (4.75) into the character
expression finally gives
c Γ (γ H ↑ G) =
1
|G|
h∈H
κ
¯
χ
Γ (h)χ
γ (h)P κκ
g κ hg
−1
κ
=
1
|G|
h∈H
¯
χ
Γ (h)χ
γ (h)
κ
P κκ
g κ hg
−1
κ
=
1
|H |
h∈H
¯
χ
Γ (h)χ
γ (h)
= c γ (Γ G ↓ H)
(4.82)
which concludes the proof. Armed with the subduction tables, we can now read
these at once in the opposite sense and obtain the corresponding induction frequencies. As a simple example, consider a hydrogen atom in ammonia. The site symmetry is C s , and the subduction from C 3v reads:
A 1 → a
A 2 → b
E → a + b
(4.83)