Chemistry 431Lecture 13

Group TheorySymmetry OperationsProperties of a Group

Point GroupsReducible Representations

NC State University

Symmetry operation: an operation that moves a moleculeinto a new orientation that appears equivalent to the original one.

Example: rotation by 2π/3

(In general, Cn is a rotation by 2π/n)

A

BC

Definition of a Symmetry Operation

C

AB

C3

Symmetry element: a point line or plane with respect towhich the symmetry operation is performed.

Example: the B-F1 bond can be defined as an element.

Note that the rotated molecule is indistinguishable.

Definition of a Symmetry Element

F1

F2F3

F3

F1F2

C3

The identity operation is given the symbol E. It simplymeans that there is no change in the position of the molecule or object.

Operations: The Identity Operation

F1

F2F3

F1

F2F3

E

The reflection operation is given the symbol σ.Note that for the example of BF3 given below the effect ofvertical mirror plane about the symmetry element B-F1 is to reverse the position of atoms F2 and F3.

Reflection through a Plane

F1

F2F3

σ F1

F3F2

In this case the reflection is perpendicular to anotherpossible reflection plane shown in this diagram.We can define two symmetry planes shown as:1. The horizontal reflection plane, which has symbol σh.2. The vertical reflection plane, which has symbol σv.

Two Types of Reflection Plane

F1

F2F3

σv

σh

The horizontal reflection plane does not change the position of any of the atoms in BF3 since all of the atomsare in the mirror plane.

The horizontal reflection plane

F1

F2F3 σh

F1

F2F3

σh

We can see a case where the horizontal plane of reflectionresults in a change in the two axial carbonyls, but not theequatorial carbonyls.

The horizontal reflection plane

σh

O4C Fe

CO3

CO2

CO1

CO5

axial

equatorial

O4C Fe

CO3

CO2

CO5

CO1

To complete this introduction to reflection planes we canconsider the effect of a vertical reflection on Fe(CO)5. InThis case we have defined the symmetry element as thePlane containing CO1, CO3 and CO5.

Note that CO2 and CO4 switch positions as a result.

The vertical reflection plane

σv

O4C Fe

CO3

CO2

CO1

CO5

axial

equatorial

O2C Fe

CO3

CO4

CO1

CO5

Just as there are two types of reflection plane in Fe(CO)5We can also define two types of rotation shown below.

Rotations in Fe(CO)5

O4C Fe

CO3

CO2

CO1

CO5

axial

equatorial

C2

C3

Inversion, i involves passing each atom through the center of the molecule. If an atom has coordinate (x,y,x) then ithas coordinates (-x,-y,-z) after inversion. This only worksif the molecule has a center of symmetry. Fe(CO)5 does not have a center of symmetry.

Definition of Inversion

O4C Fe

CO3

CO2

CO1

CO5No inversion possible!

A molecule such as Mo(CO)6 does have a center of symmetry and can be inverted as shown below.

Example of Inversion

MoO4C

O5C CO2

CO3

CO1

CO6

MoO2C

O3C CO4

CO5

CO6

CO1

i

An improper rotation is a combined rotation followed byA reflection about a plane perpendicular to the axis ofrotation. It is given the symbol Sn, where n is order of the rotation.

Improper rotation

S4

C C C

H4

H3

H2

H1 C C C

H2

H1

H3

H4

A symmetry operation can be applied multiple times. For example a C3 rotation may be applied twice. Returning toBF3 we can consider C3

2 rotation, which consists of twoApplications of the C3 rotation.

First application from before.

Multiple applications of symmetry operations

F1

F2F3

F3

F1F2

C3

A symmetry operation can be applied multiple times. For example a C3 rotation may be applied twice. Returning toBF3 we can consider C3

2 rotation, which consists of twoApplications of the C3 rotation.

C32 = C3C3

Multiple applications of symmetry operations

F1

F2F3

C32

F2

F3F1

If we apply the C3 rotation one more time (for a total of three applications) the geometry returns to the starting structure. In other words we can write:

C33 = C3C3C3 = E

The product of multiple symmetryoperations can give the identity

F1

F2F3

C33 = E

F1

F2F3

In general the order of operation will affect the final result.In this case the operators do not commute.

The order of operation matters

σvC3

O4C Fe

CO3

CO2

CO1

CO5

C3σv

O4C Fe

CO3

CO2

CO1

CO5

O3C Fe

CO4

CO2

CO1

CO5

O3C Fe

CO2

CO4

CO1

CO5

Groups

1. There must exist an identity operator which commutes with all other operators.

2. The product of any two operators must also be a member of the group.

3. Multiplication is associative, but not necessarily commutative.

4. There must exist an inverse (or reciprocal) for each element in the group.

Corollaries: 1. The identity operator is its own inverse.2. A similarity transform is an operation:

Z-1XZ = Y

Properties of a group

We can assemble the operations of the group into amultiplication table. This group of operations satisifes all of the requirements of a mathematical group and is called a point group. Point groups get their name from thefact that at least one point in space remains unchangedfor all operations in the group.

C1 is a point group whose only symmetry operation is E,the identity. In other words there is no symmetry.

Cs is a point group whose symmetry operations are E andσ. The symmetry is restricted to a mirror plane.

Point groups

We can assemble the operations of the group into amultiplication table. This group of operations satisifes all of the requirements of a mathematical group and is called a point group. Point groups get their name from thefact that at least one point in space remains unchangedfor all operations in the group.

C1 is a point group whose only symmetry operation is E,the identity. In other words there is no symmetry.

Cs is a point group whose symmetry operations are E andσ. The symmetry is restricted to a mirror plane.

Point group examples C1 and Cs

C1 Cs

E only E and σ

Point group examples C1 and Cs

NCl F

HN

H FH

C1 Cs

E only E and σ

Point group examples C1 and Cs

NCl F

HN

H FH

NH H

H

Point group example: Ammonia C3v

The symmetry operation Eexists for all groups.

Point group example: Ammonia C3v

A vertical reflection planeσv is shown. There are threesuch planes in molecules in the C3v point group.

NH H

H

Point group example: Ammonia C3v

There are two possible Rotations about a 3-foldaxis. The first is a 120o

rotation and the second isa 240o rotation.

NH H

H

Point group example: Ammonia C3vThe group consists of thesethree symmetry operations.The order of the group is h=6.There are three irreducible representations in the point group C3v, which are given in the character table below.

NH H

H

Similarity TransformThe operations X and Y are said to be conjugate if they are related by a similarity transform Z-1XZ = Y where Zis at least one member of the group.

A class is a complete set of operations that are conjugate to each other.

For example, in the point group C3v one can show thatThe operations C3 and C3

2 are in the same class.We note that (σv)-1 = σv

(σv)-1 C3 σv = σv C3 σv = σv σv”= C32

1

23

1

32

2

13

2

31

C3

σv

σv

Point GroupsC1 : no symmetryCs : only a plane of symmetryCk : only a k rotational axisCi : only an inversion centerCkh : a k rotational axis and σhCkv : a k rotational axis and k σvDk : only Ck and k C2 rotational axesDkh : operations of Dk and σh which implies k σvDkd : operations of Dk and k σd which bisect the angles of C2Sk : only the improper rotation SkTd : tetrahedralOh : octahedral

Systematic assignment of a molecule to a point group

Symmetry properties are used to determine the molecular orbitals and spectral features of a molecule. It is importantto have a systematic approach to assignment of the point group.

The scheme gives a systematic series ofquestions that lead tothe point group assignment.

Determining the point group to which a molecule belongs will be the first step in a treatment of the molecular orbitals or spectra of a compound.

It is important that this be done somewhat systematically. The flow chart in the figure is offered as an aid, and a few examples should clarify the process.

Examples of point group assignment

What are the point groups for the following Pt(II) ions?

ClPt

Cl Cl

ClPt

Cl

Cl

2- ClPt

Br Cl

ClPt

Cl

Br

2- ClPt

Br Cl

ClPt

Br

Cl

2-

A B C

What are the point groups for the following Pt(II) ions?

A contains three C2 axes, i.e., [Ck?] is yes with k=2. It contains a plane of symmetry so [σ?] is yes. The three C2 axes are perpendicular, i.e, there is a C2 axis and two perpendicular C2's which means that [⊥C2?] is yes. There is a plane of symmetry perpendicular to the C2 so [σh?] is yes and we arrive at the D2hpoint group.

ClPt

Cl Cl

ClPt

Cl

Cl

2- ClPt

Br Cl

ClPt

Cl

Br

2- ClPt

Br Cl

ClPt

Br

Cl

2-

A B C

What are the point groups for the following Pt(II) ions?

B contains only one C2 axis, no ⊥C2's, no σh, but it does have two σv's and is therefore a C2v ion.

ClPt

Cl Cl

ClPt

Cl

Cl

2- ClPt

Br Cl

ClPt

Cl

Br

2- ClPt

Br Cl

ClPt

Br

Cl

2-

A B C

What are the point groups for the following Pt(II) ions?

C contains a single C2 axis and a horizontal plane (the plane of the ion) and therefore has C2h symmetry.

ClPt

Cl Cl

ClPt

Cl

Cl

2- ClPt

Br Cl

ClPt

Cl

Br

2- ClPt

Br Cl

ClPt

Br

Cl

2-

A B C

Matrix representation of groupsThe basis is comprised by the labels attached to objects.For molecules the objects can be:1. Atoms2. Coordinates3. Orbitals4. Bonds5. AnglesThe number of basis functions or labels is called the dimension.

For example, when considering molecular motions we Can assign coordinates x, y and z to each atom. There are three coordinates for N atoms to give a total dimension of 3N.

C2(z) gives (xi→ -xj), (yi→ -yj), and (zi→ +zj) where i = j for O coordinates since the O lies along C2 but i ≠ j for the H since they do not lie along C2 and are therefore rotated into one another, e.g., x2 → -x3.

We can represent this transformation in matrix notation where each atom will have a 3x3 matrix,

The trace of the matrix is the sum of the diagonal elements. In this case it is -1.

(3r x 3c)(3r x 1c) gives a (3r x 1c)

xi

yi

zi

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=−1 0 00 −1 00 0 +1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

xj

yj

zj

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

=−x j

−y j

+ zj

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

C2(z)

H

O

H

z1y1

z2y2

z3y3

x3

x1

x2 H

O

H

z1y1

z3y3

z2y2x2

x1

x3

Motions of H2OSince there are atoms the full matrix for H2O is a 9x9:

The hydrogen atoms are not on the diagonal since theatoms themselves are moved as a result of the C2rotation.

The trace of this matrix is –1. This is also called the character.

Motions of H2OWe can consider also the result of the σv mirror plane, which is also a 9x9:

None of the atoms are moved by the symmetry operation so all of the submatrices representing the vectors lie along the diagonal. For this operation the trace (also known as the character) is +3.

Conclusions for symmetry operationsWe can conclude with 2 general rules:

1. Only those atoms, which remain in the place following an operation can contribute to the trace.

2. Each atom contributes the same amount to the trace since all of the atoms have the same 3x3 matrix.

Using these principles we can see that σv’ has a character of +1.The identity always has a character equal to the number of basis functions. Here E = 9.

Using the character of the 4 symmetry operations of the C2v point group we can construct a representation Γ.

C2

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

−1 0 0 0 0 0 0 0 00 −1 0 0 0 0 0 0 00 0 +1 0 0 0 0 0 00 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 −1 00 0 0 0 0 0 0 0 +10 0 0 −1 0 0 0 0 00 0 0 0 −1 0 0 0 00 0 0 0 0 +1 0 0 0

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

9x9 matrix for C2 rotation in H2OH

O

H

z1y1

z2y2

z3y3

x3

x1

x2

For σv(yz), no atoms are moved so the matrix representation is

σv

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

−1 0 0 0 0 0 0 0 00 +1 0 0 0 0 0 0 00 0 +1 0 0 0 0 0 00 0 0 −1 0 0 0 0 00 0 0 0 +1 0 0 0 00 0 0 0 0 +1 0 0 00 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 +1 00 0 0 0 0 0 0 0 +1

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

9x9 matrices (!) like those above could serve as the representations of the operations for the water molecule in this basis!

H

O

H

z1y1

z2y2

z3y3

x3

x1

x2

Fortunately, only the trace of this matrix is required. This sum is called the character, χ(R). Here, χ(σv) = 3

σv

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

−1 0 0 0 0 0 0 0 00 +1 0 0 0 0 0 0 00 0 +1 0 0 0 0 0 00 0 0 −1 0 0 0 0 00 0 0 0 +1 0 0 0 00 0 0 0 0 +1 0 0 00 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 +1 00 0 0 0 0 0 0 0 +1

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

3

For χ(C2) = -1

-1

C2

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

=

−1 0 0 0 0 0 0 0 00 −1 0 0 0 0 0 0 00 0 +1 0 0 0 0 0 00 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 −1 00 0 0 0 0 0 0 0 +10 0 0 −1 0 0 0 0 00 0 0 0 −1 0 0 0 00 0 0 0 0 +1 0 0 0

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

x1

y1

z1

x2

y2

z2

x3

y3

z3

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

For a reflection through the plane bisecting the H-O-H bond angle, χ(σv') = +1 since only the O is unshifted and a plane contributes +1 for each unshifted atom.

The character for the identity element will always be the dimension of the basis since all labels are unchanged. For water then, χ(E) = 9.

The representation (Γ) for water in this Cartesian basis is:

E C2 σv(yz) σv'(xz)

Γ 9 -1 3 1

Reducible and irreducible representationsWe call the representation Γ a reducible representation.Here we write the reducible representation as:

The irreducible representations form the basis of the point group in the same way that the vectors along x, y and z form the basis for three dimensional space.

H2O belongs the point group C2v. In this point groupThere are 4 irreducible representations, A1, B1, A2, B2.The decomposition of the reducible representation is aUnique determination of the irreducible reps (or irreps)zspanned by Γ.

E C2 σv(yz) σv'(xz)

Γ 9 -1 3 1

E each atom has 1 and not 3 labels so each operation is a 3x3 matrix as opposed to the 9x9 matrices of the previous basis. In addition, there can be no sign change for an s-orbital. The resulting representation is,

E =1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

; C2 =1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ; σv =

1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ; σv

' =1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

s-Orbitals as a basisI

IIIII

E each atom has 1 and not 3 labels so each operation is a 3x3 matrix as opposed to the 9x9 matrices of the previous basis. In addition, there can be no sign change for an s-orbital. The resulting representation is,

E =1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

; C2 =1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ; σv =

1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ; σv

' =1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Γ 3 1 3 1

s-Orbitals as a basis

IIIIII

If one basis (f ') is a linear combination of another basis (f) or f ' = Cf, then the representation in one basis should be similar to that in the other.

It can be shown that the matrix representations of operator R in these two basis sets (D(R) and D'(R)) are related by a similarity transformations:

D'(R) = C-1D(R)Cand

D(R) = CD'(R)C-1

i.e., the matrices D(R) and D'(R) are conjugate.

Transformation of a basis

For example, the linear combination of s-orbitals: X = I + IIY = I + IIIZ = II + III

which can be expressed in matrix form as:

f' = C fXYZ

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=1 1 01 0 10 1 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

IIIIII

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and inverse of C is C−1 =+0.5 +0.5 −0.5+0.5 −0.5 +0.5−0.5 +0.5 +0.5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

IIIIII

The matrix representation of C2 (D'(C2)) in the new basis then is given by D'(C2) = C-1D(C2)C

or D'(C2) = +0.5 +0.5 −0.5+0.5 −0.5 +0.5−0.5 +0.5 +0.5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

1 1 01 0 10 1 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=0 1 01 0 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Γ =1

C-1 C2 C

D'(C2) = D'(σv') = D'(E) = D'(σv) =0 1 01 0 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Note that the representation for C2 and σv' have changed, but in all cases the character is invariant with the similarity

transformation.

D(C2) = D (σv') = D(E) = D(σv) =1 0 00 0 10 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

NEW

OLD

χ=1 χ=3

χ=3χ=1

Thus, all the members of a class of operations can be treated together since they are related by a similarity transformation and must have the same characters.

The two 3x3 bases used to this point can be viewed as consisting of a 1x1 matrix (one basis vector not rotated into any of the others by any R: I & Z) and a 2x2 submatrix (two basis vectors rotated into one another by at least one R).

Thus, the 3x3 representation has been reduced to a 1x1 and a 2x2.

Indeed, the 2x2 matrix can be reduced into two 1x1 matrices. In this process, a large reducible representation is decomposed into smaller (usually 1x1 but sometimes 2x2 and 3x3) irreducible representations.

Consider the symmetry adapted linear combinations (SALC's)

A—C for water s-orbitals

ABC

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=1 0 00 1 10 1 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

IIIIII

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=I

II + IIIII − III

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

IIIIII

A = I B = II + III C = II - III

IIIII

In this basis, no basis vector is changed into another by a symmetry operation, i.e., this basis is symmetry adapted.

Old 3x3 is now three 1x1 matrices and reducible representation ΓS is now three irreducible representations, ΓA, ΓB and ΓC.

i.e., ΓS = ΓA + ΓB + ΓC

E C2 σ σ'

ΓA 1 1 1 1ΓB 1 1 1 1ΓC 1 -1 1 -1ΓS 3 1 3 1

IIIIII

I A

IIIII B

IIIII

ΓS = 3 1 3 1

C

ΓA = 1 1 1 1

ΓB = 1 1 1 1

ΓC = 1 -1 1 -1

The term irreducible representation is used so frequently that it is often abbreviated as "irrep". We will also use this abbreviation throughout this lecture,

irrep ≡ irreducible representation

Decomposing a reducible representation into irreps is a very important process, and a procedure to accomplish the decomposition will be described shortly.

Point Group RepresentationsA point group representation is a basis set in which the irreducible representations are the basis vectors.

The irreps form the a complete orthonormal basis for an m-dimensional space, where m is the number of irreducible representations and is equal to the number of classes in the group.

These considerations are summarized by the following rules.

1. The number of basis vectors or irreps (m) equals the number of classes.

2. The sum of the squares of the dimensions of the m irreps equals the order,

expect at least one χ(E) > 1 for Cn (n > 2)di2

i =1

m

∑ = h

The character of the identity operation equals the dimension of the representation, χ(E) = di which is referred to as the degeneracy of the irrep.

The degeneracy of most irreducible representations is 1 (non-degenerate representations are 1x1 matrices) but can sometimes be 2 or 3.

No character in an irreducible representation can exceed the dimension of the representation. Thus, in non-degenenerate representations, all characters must be ±1.

3. The member irreps are orthonormal, i.e., the sum of the squares of the characters in any irrep is equal to the order (row normalization), while the sum of the product of the characters over all operations in two different irreps is zero (orthogonality).

g(R)χi(R)R

∑ χ j (R) = hδij

• δij is the Kroniker delta (0 when i≠j and 1 when i=j), • where the sum is over all of the classes of operations, • g(R) is the number of operations, R, in the class, • χi(R) and χj(R) are the characters of the operation R in the ith and jth irreps, • h is the order of the group.

4. The sum of the squares of the characters of any operation over all of the irreps times the number of operations in the class is equal to h, i.e., columns of the representation are also normalized.

where m is the number of irreducible representations

g(R)χi2

i = 1

m

∑ (R) = h

5. The sum of the products of the characters of any two different operations over all of the irreps is zero, i.e., columns of the representation are also orthogonal.

χi (R' )χ ii = 1

m

∑ (R) = 0

6. The first representation is always the totally symmetricrepresentation in which all characters are +1.

7. Any reducible representation in the point group can be expressed as a linear combination of the irreducible representations.

Let’s generate the C2v point group, {C2v⏐E, C2, σ, σ'}

C2v E C2

σv σv'

Γ1 d1 ? ? ?Γ2 d2 a b cΓ3 d3 d e fΓ4 d4 g h i

Rule 2 d12 + d2

2 + d32 + d4

2 = h = 4

d12 = 1

d22 + d3

2 + d42 = 3

d2 = d3 = d4 = 1

Rule 6 (χ(Γ1)=+1)

C2v E C2 σv σv'Γ1 1 1 1 1Γ2 1 a b cΓ3 1 d e fΓ4 1 g h i

•Rule 3 (row 1 x row 2)• 1 + a + b + c = 0• a + b + c = -1• a = 1; b = c = -1 (others possible)

Use orthogonality to construct the irreducible representations

C2v E C2 σv σv'

Γ1 1 1 1 1Γ2 1 1 -1 -1Γ3 1 d e fΓ4 1 g h i

•Rule 3 (row 2 x row 3)• 1 + d - e - f = 0• d - e - f = -1• d = f = -1; e = +1 (others possible)

Use orthogonality to construct the irreducible representations

C2v E C2 σv σv'Γ1 1 1 1 1Γ2 1 1 -1 -1Γ3 1 -1 1 -1Γ4 1 g h i

•Rule 3 (row 3 x row 4)• 1 - g + h - i = 0• - g + h - i = -1• g = h = -1; i = +1

Use orthogonality to construct the irreducible representations

C2v E C2 σv σv'Γ1 1 1 1 1Γ2 1 1 -1 -1Γ3 1 -1 1 -1Γ4 1 -1 -1 1

Mulliken symbols for irreps:• "A"—symmetric wrt rotation about principle axis

(χ[Cn(z)] =+1)• "B"— irrep is antisymmetric wrt rotation about

the principle axis (χ[Cn(z)] = -1)• "E" —doubly degenerate irrep (d = 2 ⇒ χ(E) = 2)• "T"—triply degenerate irrep (d = 3 ⇒ χ(E) = 3)• "G" and "H”—degeneracies of 4 and 5, respectively.

In many instances there are more than one A, B, E , etc. irreps present in the point group so subscripts and superscripts are used

g or u subscripts are used in point groups with centers of symmetry (i) to denote gerade (symmetric) and ungerade(antisymmetric) with respect to inversion

' and " are used to designate symmetric and antisymmetric with respect to inversion through a σh plane

numerical subscripts are used otherwise

C2v E C2 σv σv'Α1 1 1 1 1Α2 1 1 -1 -1Β1 1 -1 1 -1Β2 1 -1 -1 1

For C2v:

Determination of Reducible Representations

Oxygen s-orbitals in water,

C2v E C2 σv σv'ΓO(s) +1 +1 +1 +1

s-orbitals on central elements will always transform as the totally symmetric representation but are not included in

character tables

I

ΓO(s) = a1

Oxygen p-orbitals in water,

C2v E C2 σv(xz) σv'(yz)pz +1 +1 +1 +1 a1

px +1 -1 +1 -1 b1

py +1 -1 -1 +1 b2Γp 3 -1 1 1

Thus, Γp = a1 + b1 + b2. The px orbital is said to • form the basis for the b1 representation,• have b1 symmetry, or • transform as b1

px pypz

Translations along the x, y and z directions (x, y, z) transform in the same way as px, py and pz.

C2v E C2 σv(xz) σv'(yz)Tz +1 +1 +1 +1 a1

Tx +1 -1 -1 +1 b2

Ty +1 -1 +1 -1 b1Γtrans 3 -1 1 1

Thus, Γtrans = a1 + b1 + b2

ΓT(y) = 1 -1 1 -1H

O

HH

O

H

z

xy

Rotation of the water molecule,

C2v E C2 σv(xz) σv'(yz)

Rz +1 +1 -1 -1 a2

Rx +1 -1 -1 +1 b2

Ry +1 -1 +1 -1 b1

Γrot 3 -1 1 1

Thus, Γrot = a2 + b1 + b2

ΓR(z) = 1 1 -1 -1H

O

HH

O

H

z

xy

In most character tables, C2v has the following form:

C2v E C2 σv(xz) σv'(yz)Α1 1 1 1 1 z x2, y2, z2

Α2 1 1 -1 -1 Rz xy

Β1 1 -1 1 -1 x, Ry xz

Β2 1 -1 -1 1 y, Rx yz

The final column gives squares and binary products and represent the transformation properties of the d-orbitals.

Let’s generate the C3v point group.

The operations of C3v are E, 2C3, 3σv (h=6, m=3)

d12 + d2

2 + d32 = 6

d1 = d2 = 1 and d3 = 2.

Since the dimensions of the irreps are the χ(E) and every group contains the totally symmetric irrep,

C3v 1E 2C3 3σv

Γ1 1 1 1

Γ2 1 j k

Γ3 2 m n

Orthogonality of Γ1 with Γ2:(1)(1)(1) + (2)(1)(j) + (3)(1)(k) = 01 + 2j + 3k = 0j = +1 and k = -1

Normalization of Γ3 means (1)(2)2 + 2(m2) + 3(n2) = 6so, m = -1 and n =0.

C3v 1E 2C3 3σv

Γ1 1 1 1Γ2 1 1 -1Γ3 2 -1 0

For C3v,

C3v 1E 2C3 3σv

Α1 1 1 1Α2 1 1 -1Ε 2 -1 0

The ammonia molecule (C3v point group) and the coordinate system used in the following discussion is given below (one N-H bond in the XZ plane).

The N-pz orbital is not changed by any of the operations of the group, i.e., it is totally symmetric and transforms as a1

However, px and py are neither symmetric nor antisymmetric with respect to the C3 or σv operations, but rather go into linear combinations of one another and must therefore be considered together as components of a 2 dimensional representation.

The matrices in this irreducible representation will be 2x2 and not 1x1. The character of the identity operation will then be 2 (the trace of a 2x2 matrix with 1's on the diagonal), i.e., χ(E)=2.

NHH

H

z

yx HH

H HHH HH

H

px py pz

A rotation through an angle 2π/n can be represented by the following transformation:

x'y'

⎛

⎝ ⎜

⎞

⎠ ⎟ =

cos(2π / n) sin(2π / n)-sin(2π / n) cos(2π / n)

⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜

⎞

⎠ ⎟

the trace of the Cn rotation matrix is 2cos(2π/n) which for n=3 is 2cos(2π/3) = 2(-0.5) = -1, i.e.,

χ(C3) = -1

The character for sv can be determined by the effect of reflection through any one of the three svsince they are all in the same class. Use sv(xz) which results in px → px and py → -py or,

x'y'

⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 −1

⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜

⎞

⎠ ⎟ HH

H HHH HH

H

px py pz

χ(sv) = 0

The transformation properties of the px and pyorbitals are represented as,

C3v E 2C3 3σv

Γpx,py 2 -1 0

The px and py orbitals are degenerate in C3v symmetry and are taken together to form a basis for the two-dimensional irreducible representation, e.

Γpx,py = e

Treating rotations and binary products as before, we can represent the C3v point group as

C3v E 2C3 3σv

Α1 1 1 1 z x2+y2; z2

Α2 1 1 -1 Rz

Ε 2 -1 0 (x,y);(Rx,Ry) (x2-y2,xy);(xz,yz)

The x2-y2 and xy orbitals are also degenerate as are the xz and yz orbitals

Decomposing Reducible Representations

In the determination of molecular orbital or vibrational symmetries, a reducible representation is generated from an appropriate basis set and then decomposed into its constituent irreducible representations.

ai =1h

g R( )R

∑ χi R( )χ R( )

ai: the # of times that ith irrep appears in the reducible representationh: the order of the groupR: an operation of the groupg(R): the number of operations in the classχi(R): the character of the Rth operation in the ith irrepχ(R): the character of the Rth operation in the reducible representation

Γred = 7 1 1 of the C3v point group. The order of the point group is 6.

C3v 1E 2C3

3σv C3v 1E 2C3 3σv

Α1 1 1 1 Α2 1 1 -1

Γred 7 1 1 Γred 7 1 1

a(a1) = 1/6{(1)(1)(7)+(2)(1)(1)+(3)(1)(1)} = 1/6{12} = 2

C3v 1E 2C3 3σv

E 2 -1 0

Γred 7 1 1

a(e) = 1/6{(1)(2)(7) + (2)(-1)(1) + (3)(0)(+1)} = 1/6{12} = 2

a(a2) = 1/6{(1)(1)(7)+(2)(1)(1)+(3)(-1)(1)} = 1/6{6} = 1

The reducible representation is decomposed as: Γred = 2a1 + a2 + 2e

The results can be verified by adding the characters of the irreps,

C3v 1E 2C3 3σv

2a1 2 2 2a2 1 1 -12e 4 -2 0

Γred 7 1 1

Problem I.7 Decompose the following reducible representations of the C4v point group.

C4v E 2C4 C2 2σv 2σd

Γ1 11 1 -1 5 1Γ2 6 0 2 0 0Γ3 5 1 -3 -1 -1Γ4 4 -4 4 0 0

The reducible representation of the Cartesian displacement vectors for water was determined earlier and is given in the following table as Γcart

Γcart(E) = 3NHere is a shortcut for generating Γcart for any system:

Γcart = ΓunshΓxyz = Γunsh[Γx + Γy + Γz]

C2v E C2 σv σ'vA1 1 1 1 1 zA2 1 1 -1 -1 Rz

B1 1 -1 1 -1 x, Ry

B2 1 -1 -1 1 y,Rx

Γcart 9 -1 3 1

C2v E C2 σv σ'vA1 1 1 1 1 zA2 1 1 -1 -1 Rz

B1 1 -1 1 -1 x, Ry

B2 1 -1 -1 1 y,Rx

Γcart 9 -1 3 1

Decomposition of Γcart yields,a(a1) = 1/4 {(1)(1)(9) + (1)( 1)(-1) + (1)( 1)(3) + (1)( 1)(1)} = 1/4 {12} = 3a(a2) = 1/4 {(1)(1)(9) + (1)( 1)(-1) + (1)(-1)(3) + (1)(-1)(1)} = 1/4 { 4} = 1a(b1) = 1/4 {(1)(1)(9) + (1)(-1)(-1) + (1)( 1)(3) + (1)(-1)(1)} = 1/4 {12} = 3a(a2) = 1/4 {(1)(1)(9) + (1)(-1)(-1) + (1)(-1)(3) + (1)( 1)(1)} = 1/4 { 8} = 2

Γcart = 3a1 + a2 + 3b1 + 2b2

Of these 3N degrees of freedom, three are translational, three are rotational and the remaining 3N-6 are the vibrational degrees of freedom.

Thus, to get the symmetries of the vibrations, the irreducible representations of translation and rotation need only be subtracted from Γcart, but the irreps of rotation and translation are available from the character table.

For the water molecule, Γvib = Γcart - Γtrans - Γrot

= {3a1 + a2 + 3b1 + 2b2} - {a1 + b1 + b2} - {a2 + b1 + b2} = 2a1 + b1

Problem I.8 Determine the symmetries of the vibrations of NH3, PtCl42- and SbF5.

Direct Products.Direct Products: The representation of the product of two representations is given by the product of the characters of

the two representations.Verify that under C2v symmetry A2 ⊗ B1 = B2

C2v E C2 σv σ'vA2 1 1 -1 -1B1 1 -1 1 -1

A2 ⊗ B1 1 -1 -1 1

As can be seen above, the characters of A2 ⊗ B1 are those of the B2 irrep.

Verify that A2 ⊗ B2 = B1, B2 ⊗ B1= A2

Also verify that • the product of any non degenerate representation with itself is totally symmetric and • the product of any representation with the totally symmetric representation yields the original representation

Note that,

•A x B = B; while A x A = B x B = A•"1" x "2" = "2”; while "1" x "1" = "2" x "2" = "1”•g x u = u; while g x g = u x u =g.