Molecular Symmetry and Group Theory

Workshop 3 — Character tables


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  • Exercise 3

    Use the character tables provided to reduce the following representations (Γ1, Γ2 .... Γ6):

    C3v E 2C3 v
    Order =
     
    Γ1 4 1 0    
    A1 1 1 1
    A2 1 1 -1
    E 2 -1 0
    N×χR×χI(A1)
    Σ(N×χR×χI(A1)) =
    Σ(N×χR×χI(A1))/Order =
    N×χR×χI(A2)
    Σ(N×χR×χI(A2)) =
    Σ(N×χR×χI(A2))/Order =
    N×χR×χI(E)
    Σ(N×χR×χI(E)) =
    Σ(N×χR×χI(E))/Order =

    Therefore, the composition of Γ1 is:   Γ1 = A1 + A2 + E


    C3v E 2C3 v
    Order =
     
    Γ2 6 0 -2    
    A1 1 1 1
    A2 1 1 -1
    E 2 -1 0
    N×χR×χI(A1)
    Σ(N×χR×χI(A1)) =
    Σ(N×χR×χI(A1))/Order =
    N×χR×χI(A2)
    Σ(N×χR×χI(A2)) =
    Σ(N×χR×χI(A2))/Order =
    N×χR×χI(E)
    Σ(N×χR×χI(E)) =
    Σ(N×χR×χI(E))/Order =

    Therefore, the composition of Γ2 is:   Γ2 = A1 + A2 + E


    C2v E C2 σ(xz) σ(yz)
    Order =
     
    Γ3 4 0 0 -4    
    A1 1 1 1 1
    A2 1 1 -1 -1
    B1 1 -1 1 -1
    B2 1 -1 -1 1
    N×χR×χI(A1)
    Σ(N×χR×χI(A1)) =
    Σ(N×χR×χI(A1))/Order =
    N×χR×χI(A2)
    Σ(N×χR×χI(A2)) =
    Σ(N×χR×χI(A2))/Order =
    N×χR×χI(B1)
    Σ(N×χR×χI(B1)) =
    Σ(N×χR×χI(B1))/Order =
    N×χR×χI(B2)
    Σ(N×χR×χI(B2)) =
    Σ(N×χR×χI(B2))/Order =

    Therefore, the composition of Γ3 is:   Γ3 = A1 + A2 + B1 + B2


    C2v E C2 σ(xz) σ(yz)
    Order =
     
    Γ4 3 -1 -3 1    
    A1 1 1 1 1
    A2 1 1 -1 -1
    B1 1 -1 1 -1
    B2 1 -1 -1 1
    N×χR×χI(A1)
    Σ(N×χR×χI(A1)) =
    Σ(N×χR×χI(A1))/Order =
    N×χR×χI(A2)
    Σ(N×χR×χI(A2)) =
    Σ(N×χR×χI(A2))/Order =
    N×χR×χI(B1)
    Σ(N×χR×χI(B1)) =
    Σ(N×χR×χI(B1))/Order =
    N×χR×χI(B2)
    Σ(N×χR×χI(B2)) =
    Σ(N×χR×χI(B2))/Order =

    Therefore, the composition of Γ4 is:   Γ4 = A1 + A2 + B1 + B2


    C2h E C2 i σh
    Order =
     
    Γ5 8 0 6 2    
    Ag 1 1 1 1
    Bg 1 -1 1 -1
    Au 1 1 -1 -1
    Bu 1 -1 -1 1
    N×χR×χI(Ag)
    Σ(N×χR×χI(Ag)) =
    Σ(N×χR×χI(Ag))/Order =
    N×χR×χI(Bg)
    Σ(N×χR×χI(Bg)) =
    Σ(N×χR×χI(Bg))/Order =
    N×χR×χI(Au)
    Σ(N×χR×χI(Au)) =
    Σ(N×χR×χI(Au))/Order =
    N×χR×χI(Bu)
    Σ(N×χR×χI(Bu)) =
    Σ(N×χR×χI(Bu))/Order =

    Therefore, the composition of Γ5 is:   Γ5 = Ag + Bg + Au + Bu


    C2h E C2 i σh
    Order =
     
    Γ6 3 1 -3 -1    
    Ag 1 1 1 1
    Bg 1 -1 1 -1
    Au 1 1 -1 -1
    Bu 1 -1 -1 1
    N×χR×χI(Ag)
    Σ(N×χR×χI(Ag)) =
    Σ(N×χR×χI(Ag))/Order =
    N×χR×χI(Bg)
    Σ(N×χR×χI(Bg)) =
    Σ(N×χR×χI(Bg))/Order =
    N×χR×χI(Au)
    Σ(N×χR×χI(Au)) =
    Σ(N×χR×χI(Au))/Order =
    N×χR×χI(Bu)
    Σ(N×χR×χI(Bu)) =
    Σ(N×χR×χI(Bu))/Order =

    Therefore, the composition of Γ6 is:   Γ6 = Ag + Bg + Au + Bu

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