From irreducible representations to character tables Recapitulation - PowerPoint PPT Presentation

From irreducible representations to character tables

Recapitulation Recipe for generating a unitary representation • 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 , Λ 1/2 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) Λ 1/2 2

Recapitulation Reducible and Irreducible representations D1(R) [0] …… [0] [0] D2(R) …… [0] A-1 D(R) A = for all R … … … [0] … … Dn(R) Γ = Γ1 ⊕ Γ2 ⊕ … ⊕ Γ n General case: Γ = Σ ai Γ i All D(R) in identical block diagonal form by similarity transformation: Reducible representaion If not: Irreducible representations 3

Irreducible representations: Great Orthogonality theorem Σ h [ Γ i ( R ) mn ] [ Γ ϕ ( Ρ ) µ ν ’ ’ ]∗ = διϕ δµµ δνν ’ ’ √ λιλϕ R 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 ) Five important working rules 4

The relationship between li and h Σ h [ Γ i ( R ) mn ] [ Γ ϕ ( Ρ ) µ ν ’ ’ ]∗ = διϕ δµµ δνν ’ ’ √ λιλϕ R Σ li2 = h’ i li2 = Number of orthogonal vectors for the i th IR Σ li2 = Total number of orthogonal vectors ≤ h’ i Also, li = χ i ( E ) Σ [ χ i ( E )] 2 = h i 5

The relationship between characters and h Σ h [ Γ i ( R ) mn ] [ Γ ϕ ( Ρ ) µ ν ’ ’ ]∗ = διϕ δµµ δνν ’ ’ √ λιλϕ R 6

The relationship between characters and h Σ h [ Γ i ( R ) mm ] [ Γ ϕ ( Ρ ) µ µ ’ ’ ]∗ = διϕ δµµ δνν ’ ’ √ λιλϕ R Σ h [ Γ i ( R ) mm ] [ Γ ι ( Ρ ) µ µ ’ ’ ]∗ = δµµ ’ Σ Σ li R Σ R Σ Σ Σ Σ m’ m h [ Γ ι ( Ρ ) µµ ] [ Γ ι ( Ρ ) µ µ ’ ’ ]∗ = δµµ ’ li m m’ m’ m χ i ( R ) χ i ( R )* li Σ [ χ i ( R )] 2 = h R 7

Orthogonality of characters of different IRs Σ h [ Γ i ( R ) mn ] [ Γ ϕ ( Ρ ) µ ν ’ ’ ]∗ = διϕ δµµ δνν ’ ’ √ λιλϕ R Σ [ χ i ( R )][ χ j ( R )] = 0 R 8

Characters of matrices belonging to a class are identical If Q -1 A Q = B , then A and B have the same traces Lecture 6 9

Number of IRs = Number of classes Σ [ χ i ( R )][ χ j ( R )] = h διϕ R Let there be k classes Each has a characteristic χ value for each IR If gp be the number of operations in the n th class, then k Σ [ χ i ( Rp )][ χ j ( Rp )] gp = h διϕ P = 1 10

The five rules at a glance Σ li2 = h’ i Σ [ χ i ( R )] 2 = h R Σ [ χ i ( R )][ χ j ( R )] = 0 R If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes 11

Character tables Σ E C2 σ v σ v’ C2v li2 = h’ i Σ Γ 1 [ χ i ( R )] 2 = Γ 2 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 R Γ 4 If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes 4 12

Character tables h = 4 Σ E C2 σ v σ v’ C2v li2 = h’ i 1 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 1 1 1 1 1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 1 1 1 1 1 R Γ 1 1 1 1 1 If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes li = 1 for i = 1 to 4 13

Character tables Σ E C2 σ v σ v’ C2v li2 = h’ i 1 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 1 1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 1 1 1 1 R Γ 4 1 1 1 1 If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes li = 1 for i = 1 to 4 14

Character tables Σ E C2 σ v σ v’ C2v li2 = h’ i 1 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 1 1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 1 1 1 1 R Γ 4 1 1 1 1 If Q -1 A Q = B , then A and B have the same traces Number of IRs All characters are + 1 = Number of classes li = 1 for i = 1 to 4 Two + 1, two - 1 15

Character tables Σ E C2 σ v σ v’ C2v li2 = h’ i 1 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 -1 -1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 1 -1 1 -1 R Γ 4 1 -1 -1 1 If Q -1 A Q = B , then A and B have the same traces Number of IRs All characters are + 1 = Number of classes li = 1 for i = 1 to 4 Two + 1, two - 1 16

The bases E C2 σ v σ v’ C2v 1 1 1 1 z 2 … Γ 1 z Γ 2 xy 1 1 -1 -1 Γ 3 zx 1 -1 1 -1 x Γ 4 1 -1 -1 1 y yz Irreducible representations: Symmetry species 17

Mulliken symbols 1D: A or B E C2 σ v σ v’ C2v χ ( Cn ) = 1 or - 1 1 1 1 1 z 2 … Γ 1 z Γ 2 xy 1 1 -1 -1 χ ( C2 ) / χ ( σ v ) = 1 or - 1 Γ 3 zx 1 -1 1 -1 x X1 X2 Γ 4 1 -1 -1 1 y yz 18

Mulliken symbols 1D: A or B E C2 σ v σ v’ C2v χ ( Cn ) = 1 or - 1 1 1 1 1 z 2 … z A1 xy A2 1 1 -1 -1 χ ( C2 ) / χ ( σ v ) = 1 or - 1 zx B1 1 -1 1 -1 x X1 X2 B2 1 -1 -1 1 y yz χ ( σ h ) = 1 or - 1 2D: E 3D: T X’ X” χ ( i ) = 1 or - 1 Homework: Practise Mulliken symbols from character tables 19 Xu Xg

An example of a 2D IR Σ E 2 C3 3 σ v C3v li2 = h’ i Σ Γ 1 [ χ i ( R )] 2 = Γ 2 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 R If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes 3 20

Character table h = 6 Σ E 2 C3 3 σ v C3v li2 = h’ i 1 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 1 1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ3 1 1 1 1 R 1 1 1 1 If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes l1 2 + l2 2 + l3 2 = 6 l1 = l2 = 1; l3 = 2 21

An example of a 2D IR Σ E 2 C3 3 σ v C3v li2 = h’ i 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 1 1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 2 1 1 1 R If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of classes 22

The other characters Σ E 2 C3 3 σ v C3v li2 = h’ i 1 1 1 Σ Γ 1 [ χ i ( R )] 2 = Γ 2 1 1 -1 h R Σ [ χ i ( R )][ χ j ( R )] = 0 Γ 3 2 -1 0 R If Q -1 A Q = B , then A and B have the same traces Number of IRs = Number of What is the meaning of -1 and 0 classes in this table? 23

Mulliken symbols E 2 C3 3 σ v C3v 1 1 1 A1 A2 1 1 -1 E 2 -1 0 24

Bases? E 2 C3 3 σ v C3v x, y ? z 1 1 1 A1 A2 1 1 -1 E ( x, y ) 2 -1 0 What is the meaning of -1 and 0 in this table? 25

After midsem • Complex characters • Derivation of GOT • Decomposition of RR to a sum of Irs • Applications 26