From irreducible representations to character tables Recapitulation - - PowerPoint PPT Presentation

From irreducible representations to character tables

Recipe for generating a unitary representation

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• For a chosen basis, work out the transformation matrices, D(R) for all

the symmetry operations R in the point group

• Find their adjoints, D(R)†
• Work out
• Find the eigenvalues of H by solving its characteristic equation. Hence,

construct Λ, Λ1/2 and Λ-1/2

• Work out U, the matrix of eigenvectors of H
• Construct the matrices D’(R) and D’(R)†: D’(R) = U-1 D(R) U for each

symmetry operation, R

• Hence, work out the unitary matrices D”(R) = Λ−1/2 D’(R) Λ1/2

Recapitulation

Reducible and Irreducible representations

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A-1 D(R) A = for all R D1(R) [0] …… [0] [0] D2(R) …… [0] … … … [0] … … Dn(R) Γ = Γ1 ⊕ Γ2 ⊕ … ⊕ Γn General case: Γ = Σ ai Γi

Recapitulation

All D(R) in identical block diagonal form by similarity transformation: Reducible representaion If not: Irreducible representations

Irreducible representations: Great Orthogonality theorem

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i, j: Identifiers for irreducible representations li, lj: Respective dimensionalities m, n: Identifiers for rows and columns, respectively h: Order of the point group (Total number of symmetry OPERATIONS )

[Γi (R)mn] [Γϕ (Ρ)µ ν ’ ’]∗ = διϕ δµµ δνν ’ ’

Σ

R

h √ λιλϕ Five important working rules

The relationship between li and h

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[Γi (R)mn] [Γϕ (Ρ)µ ν ’ ’]∗ = διϕ δµµ δνν ’ ’

Σ

R

h √ λιλϕ li2 = h’

Σ

i

li2 = Number of orthogonal vectors for the ith IR li2 = Total number of orthogonal vectors ≤ h’

Σ

i

Also, li = χi(E)

[χi(E)]2 = h

Σ

i

The relationship between characters and h

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[Γi (R)mn] [Γϕ (Ρ)µ ν ’ ’]∗ = διϕ δµµ δνν ’ ’

Σ

R

h √ λιλϕ

The relationship between characters and h

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[Γi (R)mm] [Γϕ (Ρ)µ µ ’ ’]∗ = διϕ δµµ δνν ’ ’

Σ

R

h √ λιλϕ [χi(R)]2 = h

Σ

R

[Γi (R)mm] [Γι (Ρ)µ µ ’ ’]∗ = δµµ’

Σ

R

h li

Σ

m

Σ

m’

Σ

R Σ m

[Γι (Ρ)µµ]

Σ

m’

[Γι (Ρ)µ µ ’ ’]∗ χi(R) χi(R)* = δµµ’ h li

Σ

m

Σ

m’

li

Orthogonality of characters of different IRs

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[Γi (R)mn] [Γϕ (Ρ)µ ν ’ ’]∗ = διϕ δµµ δνν ’ ’

Σ

R

h √ λιλϕ [χi(R)][χj(R)] = 0

Σ

R

Characters of matrices belonging to a class are identical

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If Q-1 A Q = B, then A and B have the same traces Lecture 6

Number of IRs = Number of classes

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[χi(R)][χj(R)] = hδιϕ

Σ

R

Let there be k classes Each has a characteristic χ value for each IR [χi(Rp)][χj(Rp)]gp = hδιϕ

Σ

k P = 1

If gp be the number of operations in the nth class, then

The five rules at a glance

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

Character tables

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C2v

E C2 σv σv’

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Γ1 Γ2 Γ3 Γ4

Character tables

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C2v

E C2 σv σv’ Γ1 Γ1 Γ1 Γ1

li = 1 for i = 1 to 4

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

h = 4

Character tables

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C2v

E C2 σv σv’ Γ1 Γ2 Γ3 Γ4

li = 1 for i = 1 to 4

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Character tables

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C2v

E C2 σv σv’

li = 1 for i = 1 to 4

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

All characters are + 1 Two + 1, two - 1

Γ1 Γ2 Γ3 Γ4

Character tables

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C2v

E C2 σv σv’

li = 1 for i = 1 to 4

1 1 1 1 1 1 -1 -1 1 -1 1 -1 1 -1 -1 1

All characters are + 1 Two + 1, two - 1

Γ1 Γ2 Γ3 Γ4

The bases

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C2v

E C2 σv σv’ 1 1 1 1 1 1 -1 -1 1 -1 1 -1 1 -1 -1 1 z x y z2 … xy yz zx Γ1 Γ2 Γ3 Γ4

Irreducible representations: Symmetry species

Mulliken symbols

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C2v

E C2 σv σv’ Γ1 Γ2 Γ3 Γ4 1 1 1 1 1 1 -1 -1 1 -1 1 -1 1 -1 -1 1 z x y z2 … xy yz zx 1D: A or B χ (Cn) = 1 or - 1 χ (C2) / χ (σv) = 1 or - 1 X1 X2

Mulliken symbols

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C2v

E C2 σv σv’ A1 A2 B1 B2 1 1 1 1 1 1 -1 -1 1 -1 1 -1 1 -1 -1 1 z x y z2 … xy yz zx 1D: A or B χ (Cn) = 1 or - 1 χ (C2) / χ (σv) = 1 or - 1 X1 X2 χ (σh) = 1 or - 1 X’ X” χ (i) = 1 or - 1 Xg Xu 2D: E 3D: T

Homework: Practise Mulliken symbols from character tables

An example of a 2D IR

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C3v

E 2C3 3σv

3

Γ1 Γ2 Γ3

Character table

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C3v

Γ1 Γ2 Γ3

l12 + l22 + l32 = 6

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E 2C3 3σv

h = 6 l1 = l2 = 1; l3 = 2

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C3v

Γ1 Γ2 Γ3 1 1 1 1 1 1 1 2 1 1 1 E 2C3 3σv

An example of a 2D IR

The other characters

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li2 = h’

Σ

i

[χi(R)]2 = h

Σ

R

[χi(R)][χj(R)] = 0

Σ

R

If Q-1 A Q = B, then A and B have the same traces Number of IRs = Number of classes

C3v

Γ1 Γ2 Γ3 1 1 1 1 1 -1 2 -1 0 E 2C3 3σv

What is the meaning of -1 and 0 in this table?

Mulliken symbols

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C3v

A1 A2 E 1 1 1 1 1 -1 2 -1 0 E 2C3 3σv

Bases?

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C3v

A1 A2 E 1 1 1 1 1 -1 2 -1 0 E 2C3 3σv z x, y? (x, y)

What is the meaning of -1 and 0 in this table?

After midsem

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• Complex characters
• Derivation of GOT
• Decomposition of RR to a sum of Irs
• Applications