Quantum Chemistry: Group Theory

Prof Stuart Mackenzie

Atomic & Molecular Spectroscopy

Atomic Structure Quantum theory

atoms / molecules

Quantum Mechanics

Valence

Molecular Symmetry and Applications

Group Theory

Photochemistry

Recommended Books Books:

Molecular Quantum Mechanics (4th ed. Chapter 5) by Atkins and Friedman

Chemical Applications of Group Theory by F.A. Cotton

Symmetry and Structure by Sidney F.A. Kettle

Molecular Symmetry and Group Theory by Alan Vincent

This Course

Aims:

An introduction to Group Theory as the mathematical framework underlying molecular symmetry and structure

To reinforce and extend some of the ideas and concepts introduced in the Symmetry I course

To highlight the basis of important results which simplify the life of the chemist

Part I: Group theory in the abstract

A group is any set of elements {G}, together with a rule (or binary operation) for combining them, , which obey the group axioms:

1. Closure

2. Associativity

3. Identity

4. Inverse

Definition:

Which mean what, exactly?

Axiom 1: Closure (or group property)

For all elements A, B of the set {G}, the result of combining A and B, i.e., AB, is:

i) defined, and

ii) also a member of the set {G}

n.b.: Clearly this means that the “squares” of each element, AA, BB, etc., (and thus, all subsequent “powers”) are also elements of the group

It does not imply that AB = BA (see commutativity later)

Axiom 1: Closure

Examples: The set of permutations of objects is closed with respect to successive permutations: e.g., {XYZ}, with permutation defined by e.g., (312){XYZ} = {ZXY} (312){XYZ} = {ZXY} and (312){ZXY} = {YZX}, etc.,

Think! Is the set of integers () closed under:

i) addition, +? ii) subtraction, -? iii) multiplication, ? iv) division, ?

[A set with closed combination rule comprises a gruppoid but since we know of few useful properties for them they are uninteresting]

Axiom 2: Associativity

For all elements A, B, C of the set {G},

(AB) C = A(B C)

i.e., When more than one combination is involved, it doesn’t matter which order we do them in.

Examples: Successive translations or rotations are associative Permutations of objects are associative

Think! Is the addition of integers associative? What about their subtraction, multiplication and division?

[A set with combination rule which is closed and associative comprises a semigroup or monoid.]

Axiom 3: Identity

The set {G} contains an element E such that for any element, A, of the set

EA = AE = A

Examples: For the set of integers under addition the identity is 0 (i.e., zero) For permutation the identity is no permutation, i.e., (123)

Think! What is the identity for integers under multiplication? And for integers under division?

Axiom 4: Inverse

For every element A of the set {G} there exists an element denoted A-1 such that

A A-1 = A-1 A = E

Examples: For the set of integers under addition the inverse of A is –A For every permutation in the set of permutations there is an inverse which restores the original order

Think! There is no inverse for integers under multiplication, why? What about integers under division?

Commutativity and Abelian Groups

If any two elements A, B {G} are such that AB = BA, we say the elements A, B commute.

We do not require commutativity of a group but......

..... a group in which all elements commute is called an Abelian Group.

Examples: Integer addition is commutative as is integer multiplication Permutations of N objects are, in general, non commutative (except for N =2) Think! Is integer subtraction commutative?

Summary of Axioms

The elements of a set {G}, together with a rule (or binary operation) for combining them, , form a group G iff:

A, B {G}, AB {G} and BA {G} (closure)

A, B, C {G}, A(B C) = (AB)C (associativity)

E {G} such that AE = EA = A (identity)

A {G}, A-1 {G} such that A-1A = AA-1 = E (inverse)

If, A, B {G}, AB = BA, then the group is Abelian

There is only one identity per group

A few trivial theorems for groups

Proof: Assume E, F {G} are both identities: E = E F = F

Each element has a unique inverse

Proof: Assume Y, Z {G} are both inverses of the element A {G} Y = Y E (identity) = Y (A Z) (inverse) = (Y A) Z (associativity) = E Z = Z (identity)

The inverse of the combination of two or more elements of a group equals the combination of the inverses in reverse order:

i.e., (AB)-1 = B-1A-1

A few trivial theorems for groups

Proof: Consider A, B {G} Then (AB)(B-1A -1) = A(B B-1)A -1 (associativity) = AEA -1 (identity) = AA -1 = E (inverse) Hence, (B-1A -1) must be the inverse of (AB) i.e., (AB)-1 = B-1A -1 This clearly extends to ternary and greater products.

The set of integers under algebraic addition

Simple examples of groups

Test against the group axioms: Closure: addition of two integers yields an integer Associativity: e.g., (9 + 4) + (-23) = -10 = 9 + (4+(-23)) Identity: 0 Inverse: the inverse of n is –n n.b., This set is also Abelian as addition is commutative (the order of addition is

unimportant)

Cyclic groups A cyclic group is one in which every element may be written as combinations of one particular element (the generator, g)

Simple examples of groups

All integers (an infinite group) can generated by successive application of +1 (or -1). Hence 1 (-1) are the generators.

The integers under addition is an example of a cyclic group

Cyclic groups have a number of important properties: They are Abelian Each finite sub-group of a cyclic group is also cyclic In finite cyclic groups the identity is g0 Infinite cyclic groups have exactly two generators

Multiplication Tables Consider three objects {XYZ} and permutations of these with operators (ijk). There are six such operators:

(123){XYZ} ={XYZ}

(312){XYZ} ={ZXY}

(231){XYZ} ={YZX}

(132){XYZ} ={XZY}

(321){XYZ} ={ZYX}

(213){XYZ} ={YXZ}

Label these operators, respectively as {E, A, B, C, D, F}

We can write down the combinations (products) of these operators and confirm they form a group (actually the symmetric group S (3)) by means of group multiplication table which summarises the group:

E

A

B

C

D

F

Multiplication Tables

E A B C D F

E

A

B

C

D

F

First

Seco

nd

CA

i.e., The combination CA appears in the row labelled C and the column labelled A. In this case, we note: CA{XYZ} = (132){ZXY} = {ZYX} = D{XYZ}, i.e., CA = D

Multiplication Tables

E A B C D F

E E A B C D F

A A

B B

C C D

D D

F F

First

Seco

nd

All combinations involving the E (the identity) are, of course, trivial. We can simply work the rest out:

Multiplication Tables

E A B C D F

E E A B C D F

A A B E F C D

B B E A D F C

C C D F E A B

D D F C B E A

F F C D A B E

First

Seco

nd

We have closure (no new elements are generated), E is the identity, We have associativity (convince yourself), Each element has an inverse (E appears in every row and column)

This is a (non Abelian) group of order 6 (it has 6 elements)

E A B C D F

E E A B C D F

A A B E F C D

B B E A D F C

C C D F E A B

D D F C B E A

F F C D A B E

Sub-groups

E A B C D F

E E A B C D F

A A B E F C D

B B E A D F C

C C D F E A B

D D F C B E A

F F C D A B E

First

Seco

nd

Some elements combine with themselves only, to form sub-groups.

Here, {E, A, B} form a proper sub-group of order 3; {E, C}, {E, D} and {E, F}, form proper sub-groups of order 2 which are isomorphic to one another (they share a common multiplication Table).

“The order of a sub-group must be a divisor of the group order” (Lagrange)

Group structures

We might reasonably ask ourselves how many unique group structures there are for a given order, h.

If h is a prime number, the answer is one, isomorphic to the cyclic group of order h (which, incidentally, is Abelian)

Hence for h = 1, 2, 3, respectively, there is only one group structure:

E

E E

E A

E E A

A A E

E A B

E E A B

A A B E

B B E A

However, for h>3, we start to see multiple structures. For h=4 there are two:

Groups of order 4 and higher

E A B C

E E A B C

A A B C E

B B C E A

C C E A B

E A B C

E E A B C

A A E C B

B B C E A

C C B A E

The LHS table is isomorphic to the cyclic group of order 4. The RHS is the Vierergruppe. Both forms are Abelian (check symmetry about the main diagonal).

There is exactly one group structure for order 5, two group structures for order 6 (of which one is cyclic), one for order 7, three for order 8.

Conjugacy and Classes of Groups

Equivalently we say A is the similarity transformation of B by X.

The sub-group of elements of {G} which are all mutually conjugate is called a conjugacy class or simply a class. Note: If the group is Abelian then, by definition,

A = XAX-1 and so A = XBX-1 implies A = B

i.e., each element of an Abelian group is in a class of its own

If there is at least one element X {G} such that A = XBX-1 A, B {G}

then A is conjugate to B and B is conjugate to A (A and B are mutually conjugate)

Conjugacy and Classes of Groups: example of S (3) Consider S (3) again:

CAC-1 = B so A and B are mutually conjugate & AC A-1 = D; ADA-1 = F; AFA-1 = C; so C, D, F are mutually conjugate; E is, by necessity, always in a class of its own So S (3) comprises three classes: {A,B}, {C,D,F} and {E}

E A B C D F

E E A B C D F

A A B E F C D

B B E A D F C

C C D F E A B

D D F C B E A

F F C D A B E

All of which is absolutely fascinating, but so what?

Why should chemists be interested in group theory? Answer: a complete set of the symmetry operations generated by the symmetry

elements of a molecule constitute a mathematical group. So we can use everything known about groups and apply it to symmetry operation. You know by now that molecules can be ascribed to different molecular point groups depending on their symmetry with respect to various symmetry operations. The remainder of this course will be concerned with the application of group theory to molecular point groups. But first, let’s consider briefly why symmetry operations are of particular interest:

Part II. Symmetry and Quantum Mechanics

Now, consider the group of transformations G (defined as symmetry operations) whose elements R commute with the Hamiltonian:

ˆk k kH E

The energy levels of a quantum mechanical system are the solutions of the Schrödinger equation:

1ˆ ˆ ˆ ˆ(or, equivalently, ) HR RH RHR H

in which Ĥ is the Hamiltonian (the total energy) operator and k the wavefunction for the k th level.

i.e., Ĥ is “invariant under G ” or, equivalently, Ĥ is“totally symmetric with respect to the elements of G ”

Since Ĥk =Ekk , it follows that ĤRk =EkRk

i.e., Rk is also an eigenfunction of the Hamiltonian with the same eigenvalue, Ek

Two possibilities arise:

1) Rk is the same function as k (possibly multiplied by a constant)

For example, there may be functions for which Rk =k or Rk = –k allowing us to classify functions according to their symmetry.

Or

2) Rk and k are different functions with the same energy

We postulate that such degeneracy is a consequence of some kind of symmetry.

i.e., we don’t get degeneracy by accident

So, what sort of operations don’t change the energy of a system (don’t change the Hamiltonian)?

Think about the properties of Rk:

The Molecular Hamiltonian

The full Coulomb (non-relativistic) Hamiltonian of an isolated molecule is:

2 2 1 1 11

2,

1ˆ2

i A ij A iA A B ABi A i j i A A BA

H r Z r Z Z rM

nuclear KE

electron KE

e-- e- repulsion

(PE)

nuclear repulsion

(PE)

e- - nuclear attraction

(PE)

n.b.: The Hamiltonian contains only scalar distances and is thus

invariant to any transformation which preserves the distances between particles.

Transformations such as.......

1:Translations of all particles by the same distance, d

The PE terms clearly don’t change as the rn,m terms don’t change.

2If ' , then so the kinetic energy ( ) terms are similarly unchanged'

x x dx x

So, moving a molecule to a new location doesn’t change its energy.

This means the derivative of the energy with respect to position

(i.e., the net force on the molecule) is zero.

By Newton’s 2nd law, this means the momentum, too is unchanged by translation.

This symmetry implies conservation of linear momentum.

[This symmetry is important is describing crystalline materials]

This symmetry is referred to as the homogeneity of space

2: Overall rotation of the molecule by an angle q

The PE terms clearly don’t change as the rn,m terms don’t change.

So, rotating a molecule through an angle q doesn’t change its energy.

This means the derivative of the energy with respect to angle (and hence the net torque on the molecule) is zero.

In other words, this symmetry implies conservation of angular momentum.

Similarly, rotation doesn’t change the 2 terms and thus the KE is unchanged.

This symmetry is referred to as the isotropy of space

[It follows that the total angular momentum in atoms is characterised by a “good quantum number” as is the rotational angular momentum in molecules.]

3: Permutation of the electrons

If we make the Born-Oppenheimer approximation, Ĥ = Ĥel + Ĥnuc

2 1 1 11

2,

ˆel i ij A iA A B AB

i i j i A A B

H r Z r Z Z R

where the nuclear coordinates {RA} are now fixed.

All electrons are equivalent (identical KE and PE terms for each one) and so permuting electrons cannot change the Hamiltonian.

Electron permutation is thus a symmetry operation.

n.b., The Pauli Principle requires that the total wavefunction be antisymmetric with respect to exchange of electrons

Generally: must be symmetric with respect to exchange of identical bosons (integer spin particles) and antisymmetric to exchange of identical fermions (half-integer spin).

4: Permutation of identical nuclei

Similarly,

where Eel,k(R{A}) is the electronic energy of the k th level (a function of the nuclear coordinates).

The nuclear Hamiltonian is invariant to nuclear permutations.

2

,

1ˆ ( { })2

nuc A el kA A

H E R AM

5: Parity (inversion of each particle through the origin)

Denoted E* (to distinguish it from the inversion operator i in finite symmetry groups), inversion leaves all relative distances and 2 terms unchanged (and hence the Hamiltonian invariant)

So...

.....the operations which make up the elements of the ordinary molecular symmetry group are:

translation of the molecule

rotation of the molecule

any permutation of the electrons

any permutation of identical nuclei

parity

These are “true” symmetries.

“Hang on, this is not what Dr Kukura just spent three weeks telling us......”

“..he was going on about rotations, reflections, etc.:”

2) Cn, a rotation through 2p/n radians about an axis

A C

B C

C3

Rot. by 120 A B

n.b. i) Move the body not the axis ii) RH screw gives sense of +ve rotn

1) E Identity – no change

Reminder: Symmetry operations are physical movements which bring a body into

coincidence with itself (or into an equivalent configuration) Symmetry elements are geometrical entities such as a point, a line (axis)

or a plane with respect to which symmetry operations may be performed

Likewise C32 is two successive C3 rotations:

A

B C Rot. by 2x120

B

C A

23C

33C gets us back to where we started, hence 3

3C E

3) s, a reflection in a plane

C C

H H

H H

s2

s1

The plane of the paper comprises a third plane of symmetry, s3

n.b., sh refers to a horizontal plane to the principal axis (s2) sv refers to a plane including (vertical to) the principal axis

sd refers to a dihedral plane which bisects the angle between two rotation axes. e.g., CH4 4) Sn, an improper rotation (rotation-reflection)

C

H1

H2 H4

H4

H3

C4 C H3

H2

H1

H2

C

H4

H1 H3 sh S4

n.b. If n even, then Sn

n E If n odd, then Sn

n s

5) i, inversion

Inversion of all coordinates in the origin. i.e., (x, y, z) (-x, -y, -z). This symmetry is possessed only by centro-symmetric molecules (e.g., CO2, benzene)

Note: Chemists use rotations, reflections and inversions simply because they are easy to visualise. Strictly, they are only applicable to rigid objects (i.e., not to molecules!). Quantum mechanically, however, we can only talk of the symmetry of i) the Hamiltonian and ii) wavefunctions which leads to the “true” symmetries we met earlier. That said, permutations often correspond to rotations and reflections which we can see by a simple example:

IMPORTANT: A complete set of the corresponding symmetry operations comprise a mathematical group allowing us to use the known properties of groups to infer properties of our molecule.

Example: H2O

To understand the relationship we need to consider different frames:

We are interested mainly in the internal degrees of freedom (electronic / vibrational) which are referred to a local coordinate frame (x,y,z) defined only in terms of the nuclear labels (if the labels change, the axes may change).

The rotational wavefunction depends on the orientation of this frame relative to the global frame (X, Y, Z).

Our symmetry operations are:

(ab) exchange of proton labels a and b E* inversion through the origin (ab)E* the product of the above and E the identity

Let the origin of the local frame lie at the centre of mass. The molecule lies in the y, z plane with the z axis bisecting the HOH angle. By convention, the x axis completes a right hand set:

x

y

z

a b

Right-handed frame

Symmetry operations in H2O: permutation (ab)

Under (ab) the y-axis changes direction. The x axis must also change direction to maintain the right hand coordinate frame:

Compare with C2 rotation:

(ab)

x

y

z

a b

1sa

2px

x y

z

b a

1sb

-2px

Symmetry operations in H2O: permutation (ab)

Under (ab) the y-axis changes direction. The x axis must also change direction to maintain the right hand coordinate frame:

(ab)

x

y

z

a b

2zC

1sb

The effect of (ab) on the internal coordinates is the same as C2. Remember: C2 rotates the body (functions and coordinates) but not nuclear labels / axes.

x

y

z

a b

1sa

2px

x y

z

b a

1sb

-2px

-2px

Symmetry operations in H2O: E*

x

y

z

a b

x

y

z

E*

Compare with syz:

2px

-2px

a b

1sa

1sa

Symmetry operations in H2O: E*

x

y

z

a b

x

y

z

E*

x

y

z

a b

syz

E* corresponds to reflection of the internal coordinates in the plane containing the nuclei (yz).

2px

-2px

-2px

a b

1sa

1sa

1sa

Symmetry operations in H2O: (ab)E* (or (ab)*)

x

y

z

a b

x y

z

(ab)*

x

y

z

a b

σxz

(ab) corresponds to C2, E* corresponds to syz, Hence, the product (ab)E* corresponds to the product C2s

yz which is sxz

2px

2px

2px

b a

1sa

1sb

1sb

A note on other symmetries

Charge conjugation, C: in the absence of fields, the Hamiltonian is invariant to reversal of the signs of all charges. This is reflected in the wavefunction.

[n.b. A full relativistic treatment, important in particle physics, leads to atoms / molecules of anti-particles, e.g., anti hydrogen]

Time reversal, T: The time-dependent Schrödinger equation is unaffected by a change in the sign of t (provided we take the complex conjugate):

H i

t

CPT: Particle physics has long since shown conclusively that the weak nuclear force can lead to breakdowns in P and C symmetries.

However, as far as we know the combined operation CPT (conjugation parity time reversal) is a true symmetry operation!

Part III. Molecular Point Groups (Schönflies system) You are, by now, familiar with the idea that molecules may only possess certain combinations of all possible symmetry elements and can be classified accordingly:

C1 only the identity, E

Ci E and inversion i, only

Cs E and a plane of reflection, s

Cn E and an n-fold rotation axis

Cnh E, Cn axis and a sh plane

Cnv E, Cn axis and a sv plane

Dn E, Cn axis and n perpendicular C2 axes

Dnh E, Cn axis, n perpendicular C2 axes and a sh plane

Dnd E, Cn axis, n perpendicular C2 axes and n dihedral mirror planes

Sn E, Sn axis

D3d

D3h

C3h C3v

C3

D3

Molecular Point Groups (cont.)

More than one principal axis:

Regular polyhedra:

Td tetrahedral Oh octahedral Ih icosahedral

Linear molecules: Cv: linear asymmetrical molecules e.g., HCN, heteronuclear diatomics, etc. Dh: linear symmetrical molecules e.g., CO2, homonuclear diatomics, HCCH, etc.

Flow charts for determining symmetry groups:

Atkins and de Paula, Physical Chemistry, 8th ed.

There are several:

Example is for square planar [AuCl4]- (D4h)

e.g., H2O

C2v

Multiplication tables for point groups

Consider C3v (symmetry operations E, C3, C32, sa, sb, sc) e.g., NH3

C3v E C3 C32 sa sb sc

E E C3 C32 sa sb sc

C3 C3 C32 E sc sa sb

C32 C3

2 E C3 sb sc sa

sa sa sb sc E C3 C32

sb sb sc sa C32 E C3

sc sc sa sb C3 C32 E

First

Seco

nd

In dealing with symmetry operations, product AB is simply written AB. Product RC appear in the row labelled R and column labelled C.

Is C3v Abelian?

C3

a

c b

Looking from above

Confirm that the C3v group satisfies the definition of a group.

Classes in C3v

Check for conjugacy in the C3v multiplication table group

C3v E C3 C32 sa sb sc

E E C3 C32 sa sb sc

C3 C3 C32 E sc sa sb

C32 C3

2 E C3 sb sc sa

sa sa sb sc E C3 C32

sb sb sc sa C32 E C3

sc sc sa sb C3 C32 E

First

Seco

nd

n.b., C3-1 = C3

2 and si

-1 = si

(reflections are their own inverse).

saC3sa -1= C3

2 (same for sb, sc) i.e., C3

= C32 are conjugate.

Similarly, C3

2 saC3 = sc ; C3

2 sbC3 = sa, etc

As E is always in its own class: C3v contains three classes {E}, {C3, C32 }, {sa, sb, sb}

[Physically it was obvious that we had three “types of operation” but it is not always clear that they are in the same class, e.g., The two reflections in C2v (e.g., H2O), are not conjugate as no operator in the group transforms one into the other]

Part IV: Representations

You are familiar with describing a vector quantity, r, in terms of vectors i, j, k as:

by which we really mean

x

x y z y

z

r i j k i j k

x

y

z

r

This is a representation of the vector r in the basis (i, j, k).

Basis vectors are usually chosen to be orthogonal and normalised (but this is not essential). Here i, j, k are unit vectors along the x, y, z directions, respectively.

In the same way, we can represent symmetry operations by considering the effects on a general point or coordinate (x, y):

Matrix representations of symmetry operations

Consider the C2v point group (with it’s elements {E, C2, sv, sv’}) and the effect each operation has on the general point (x, y):

(x, y) C2

(x’, y’)

E (x, y)

x

y

(x’, y’)

y

x x' x x

y y y

1 0

0 1E

1 0

0 1is a representation of operator E in the basis functions (x, y)

x x' -x x

y -y y2

'

1 0

0 1yC

1 0

0 1is a representation of C2 in the basis functions (x, y)

Matrix representations of symmetry operations Similarly,

(x, y)

(x’, y’)

sv

(x, y)

(x’, y’)

sv’

s y

x x' x x

y -y y

1 0

0 1v

1 0

0 1is a representation of sv in the basis functions (x, y)

s

x x' -x x

y y y'

1 0

0 1vy

1 0

0 1is a representation of sv’ in the basis functions (x, y)

So,

1 0( )

0 1E

2

1 0( )

0 1C

s

1 0( )

0 1v

s

1 0( ) ,

0 1v

are the matrix representations of the symmetry operations of the C2v point group in the basis (x, y)

Similarly for C3v:

Consider the effects of group operations on (x,y)

x x'

y3

'yC x’ = r cos (a + 2p/3)

= r cosacos(2p/3) - r sinasin(2p/3) = x cos(2p/3) - y sin(2p/3) = - x/2 – (3/2) y

(x, y)

C3

(x’, y’) 2p/3

a

x = r cosa, y = r sina

Likewise y’ = (3/2) x - y/2

x x' x

y y

312 2

33 1

2 2'y

C

and the matrix representations of the operations of C3v are:

1 0( )

0 1E

s

1 0( )

0 1a

312 2

33 1

2 2

( )C

312 22

33 1

2 2

( )C

s

312 2

3 12 2

( )b

s

312 2

3 12 2

( )c

r

These matrices obey the group multiplication table

C3v E C3 C32 sa sb sc

E E C3 C32 sa sb sc

C3 C3 C32 E sc sa sb

C32 C3

2 E C3 sb sc sa

sa sa sb sc E C3 C32

sb sb sc sa C32 E C3

sc sc sa sb C3 C32 E

i) Clearly matrix multiplication by E changes nothing.

ii) We saw previously that C3sb=sa s

s

3 31 12 2 2 2

33 31 1

2 2 2 2

1 0

0 1

b

a

C

iii) Likewise, scsa = C3:

s s

3 31 12 2 2 2

33 31 1

2 2 2 2

1 0

0 1c a C

Key point - the representation matrices combine as the operations themselves.

Alternative bases: All physical operations on coordinates can be represented by matrices.

We can equally use wavefunctions as basis functions (provided the operation on the set of functions produces a linear combination of these functions)

e.g., consider px and py functions on the N atom in NH3: C3

2p/3

x

y

p p

p p

312 2

33 1

2 2

x x

y y

C

Which is not the same as we found using coordinates but rather its transpose. The same is true for all symmetry operations.

p p

3

312 2

3 13 2 2

' cos(2 / 3) sin(2 / 3)

''

x x y

x y

y x y

C p p p p

p p

C p p p p

The transpose matrices do not obey the group multiplication table because, whilst for operators A=BC, for the transpose matrices AT = CTBT [note reversed order]

Alternative bases:

However, examine

So, by writing the basis as a row vector, post multiplication yields the same transformation matrix as for coordinates and so obeys the group multiplication:

3 31 13 32 2 2 2 and x x y y x yC p p p C p p p

We see

312 2

3 33 1

2 2

( ) ( )x x

x y x yy y

c cC C p p p p

c c

( )( ) ( ) ( )

x x

x y x yy y

c cRS p p p p D RS

c c

3( ) ( )

x

x yy

cp p D C

c

( )( ) ( ) ( ) ( ) ( ) ( )

x x x

x y x y x yy y y

c c cRS p p R p p D S p p D R D S

c c c

i.e., D(RS) = D(R)D(S) – the matrices multiply in the same way as the operations

Matrix representation jargon

A matrix representation of a group is a set of matrices corresponding to the group elements such that D(RS)=D(R)D(S) for all pairs of elements R, S in the group

D(RS)=D(R)D(S) is the homomorphism condition

If all representation matrices are unitary*, then we have a unitary representation.

* i.e., M-1 = M† where M† is the adjoint of the matrix (the complex conjugate of the transpose)

The matrix D(R) representing a symmetry operator R in a basis (1, 2, 3, .... n,

is defined by

( )i j jij

R D R

The set of matrices {D(R)} themselves form a mathematical group.

The various matrices of a representation need not all be different but if they are the representation is said to be faithful.

The dimension of a representation is the dimension of each of its matrices.

Clearly D(R) = D(ER) = D(E)D(R), [homomorphism] D(E) = D(RR-1) = D(R)D(R-1) i.e., D(E) is always the unit matrix and D(R-1) =(D(R))-1

For each matrix representation, {D(R)}, there is a trivial (symmetric) representation for which D(R) = 1 (the 11 unit matrix).

Depending on the symmetry group this may be A1, A1g, Sg+, etc.

Matrix representation jargon (cont.)

Matrix representations, classes and character

Look again the matrix representation of C3v in the basis (x,y)

The trace of a representation (the sum of the elements on the main diagonal) is the same for all operators of the same Class:

tr(E) = 2

tr(C3)= tr(C32)= -1

tr(sa) = tr(sb) = tr(sc) = 0

So, the trace of the representation matrix D(R) is called the character, c(R), of the representation and contains all the information normally needed.

All matrix representations of group elements in the same class have the same trace

1 0( )

0 1E

s

1 0( )

0 1a

312 2

33 1

2 2

( )C

312 22

33 1

2 2

( )C

s

312 2

3 12 2

( )b

s

312 2

3 12 2

( )c

c iii

R tr D R D R

New representations

The number of representations of a particular group is unlimited because:

i) We can choose any number of different basis sets

ii) There is no limit to the dimensions the matrices can be

iii) We can always use a similarity transformation to get a new rep. of the same dimension (corresponds to taking linear combinations of basis functions):

1 1 1 1' ' 'R S QRQ QSQ QRSQ QTQ T

1 1' , ' ,.....S QSQ R QRQ

Suppose T= RS

i.e., the new matrices, S’, R’, T’,... also obey the multiplication table and thus are a new representation of the group

' 1 1( ') ( ) ( ) ( )ii ij jk ki ki ij jk kk

i ijk ijk k

tr R R Q R Q Q Q R R tr R

Importantly, the trace of a matrix is invariant under similarity transformations:

Suppose we have a set of matrices E, R, S,... which form a representation of a group. We can make the same similarity transformation on each matrix:

Change of basis:

However, although there are infinitely many representations there are only a few of any fundamental significance.

Consider the effect of C2v group operations {E, C2, sv, sv’} on a basis of the H atom 1s orbitals in H2O. A set of matrix representations is:

Now take linear combinations of orbitals: The matrix representations in this new basis are:

1 11 2 1 22 2

( ) and ( )s s s s s s

1 0( )

0 1E

2

1 0( )

0 1C

s

1 0( )

0 1v

s

1 0( )

0 1v

Which again are not unique but are all diagonal (note that characters are unchanged by the change of basis; c(E)=2, c(C2)=0, etc. )

2

0 1( )

1 0C

s

1 0( )

0 1v

s

0 1( )

1 0v

1 0( )

0 1E

These are different to the matrices we found earlier as we’ve used a different basis. These matrices are not unique.

n.b., these basis changes are equivalent to similarity transformations:

Start with basis representations

such that

1 2 3( , , ,.... ) n

( )R D R where D(R) is a representation matrix

Now, define new functions as linear combinations of our originals

or i i jij

U U

( ) ( ) ( )R D R R U D R UThen

but

1 1

1

( )

. ., ( ) ( )

U R U D R U

i e D R U D R U

In words: The representation matrix in the new basis is just a similarity transformation of the representation matrix in the original basis.

Similarly for C3v

Consider the H 1s orbitals of NH3 as our basis. The representations in C3v are then:

1 0 0

0 1 0

0 0 1

E

3

0 0 1

1 0 0

0 1 0

C

23

0 1 0

0 0 1

1 0 0

C

s

1 0 0

0 0 1

0 1 0a

s

0 0 1

0 1 0

1 0 0b

s

0 1 0

1 0 0

0 0 1c

Now change basis to an orthogonal set of linear combinations of the 1s orbitals:

1 1 11 1 2 3 2 2 3 3 1 2 32 2 6

( ); ( ); (2 )s s s s s s s s

Now our representations become:

1 0 0

0 1 0

0 0 1

E

313 2 2

3 12 2

1 0 0

0

0

C

s

1 0 0

0 1 0

0 0 1a

s

312 2

3 12 2

1 0 0

0

0

b

2 313 2 2

3 12 2

1 0 0

0

0

C

s

312 2

3 12 2

1 0 0

0

0

c

n.b., • 1 always transforms into itself • 2 and 3 are mixed

So 33 matrices can be simplified into direct sums of 11 matrices (scalars) for 1 and 22 matrices for 2 and 3

Aside: Explanation of Direct sums

In matrix algebra we can define an operation, the direct sum (denoted ) as:

0

0

AA B

B

Similarly we can write the basis as the direct sum 1 2 3 1 2 3, , ,

312 231

2 2 3 12 23 1

2 2

1 0 0

0 1

0

So our matrix rep for C3

(3) (1) (2)D D D

A representation is said to be reducible if all the representation matrices can be simplified into the direct sum of two or more smaller matrices,

Equivalently:

A matrix representation is reducible if any similarity transformation reduces it to the direct sum of smaller (i.e., lower dimension) matrices.

It follows that:

An irreducible representation (or IR) is one which cannot be reduced by a similarity transformation.

Reducible and Irreducible representations

It is the irreducible representations of a group that are of fundamental importance.

The IRs form the building blocks of all representations.

Reduction of representations

C3v E C3 C3

2 sa sb sc

A1 1 1 1 1 1 1 1D rep

spanned by 1

We can go no further in “reducing this representation” – these are called the “irreducible representations” or “irreps” (IR) of the point group.

312 2

3 12 2

1 0

0 1

312 2

3 12 2

312 2

3 12 2

312 2

3 12 2

1 0

0 1

2D rep spanned by 2, 3

E

A2 1 1 1 -1 -1 -1

n.b. There is one additional IR in C3v, A2, which does not arise from the basis which we have been using.

Look again at our representations of C3v

One minor irritation with this approach is the non uniqueness of our IRs (compare those above with the ones you derived with Dr Kukura in a different basis). We chose one particular rep of many equivalent (related by a similarity transformation).

Character tables We need some property of the matrices which is invariant to the similarity transformation: the Trace. Instead of writing each IR out fully, we can replace each matrix by its character (i.e., the trace of its representation matrix)

C3v E C3 C32 sa sb sc

A1 1 1 1 1 1 1

A2 1 1 1 -1 -1 -1

E 2 -1 -1 0 0 0

But this contains redundant information because we know (and indeed can see from the table above) all symmetry operations of the same class have the same character. So the character table is usually presented as:

C3v E 2C3 3s h=6

A1 1 1 1 z; (x2 + y2); z2

A2 1 1 -1 Rz

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

Information in Character Tables

1) The top row shows the symmetry operations of the group

2) h is the order of the group (the number of group elements)

3) The first column lists the various IRs of the point group (symbols explained below)

4) The bulk of the table lists the characters of the IRs under each operation

5) The final column lists functions - such as the Cartesian axes, their products or rotations about the Cartesian axes - which transform as (i.e., have the same symmetry as) the various IRs.

C3v E 2C3 3s h=6

A1 1 1 1 z; (x2 + y2); z2

A2 1 1 -1 Rz

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

IR Notation: Mulliken’s convention

Degeneracy: A, or B (S in linear molecules) denote 1-dimensional IRs

E (P in linear molecules) denotes a 2-dimensional IR

T denotes a three dimensional IR

A, or B denote symmetric or antisymmetric, respectively with respect to rotation about the principal axis

primes and denote symmetric or antisymmetric, respectively with respect to reflection in a plane

subscript g or u denote symmetric or antisymmetric, respectively with respect to inversion through a point of symmetry, i

subscript 1 or 2 counting indices denoting symmetric or antisymmetric, respectively with respect to rotation about a perpendicular C2 axis

Part V: The Great Orthogonality Theorem (GOT)

This is the fundamental theorem from which most useful results of group theory (including some we’ve already seen in passing) are derived:

Consider a group of order h, and let D(l)(R) be a matrix representation of operation R in a dl-dimensional IR, (l) of the group. Then:

if

1 if ab

, a b

, a b

0

where * indicates the complex conjugate (allowing for the rep having complex elements) and ab is the Kronecker delta:

( l ) ( l ')ij i ' j ' ll ' ii ' jj '

Rl l '

hD R *D R

d d

Understanding the GOT

A few points follow trivially:

1) if( l ) ( l ')ij ij

R

D R D R l l ' 0

“vectors chosen from different IRs are orthogonal” or “the sum of the products of elements from two different IRs is zero”

2) if and/or ( l ) ( l )ij i ' j '

R

D R D R i i ' j j ' 0

“vectors from different matrix locations in the same IR, are orthogonal”

3) ( l ) ( l )ij ij

R l

hD D

d

“the sum of the squares of the elements of a representation matrix, is the group order divided by the dimension of the IR”

( l ) ( l ')ij i ' j ' ll ' ii ' jj '

Rl l '

hD R *D R

d d

Results following from the GOT

A. The sum of the squares of the dimensions of the IRs of a group equals the group order, h

2 2 2 21 2 3l

l

d d d d ... h

( l )

l

E h c 2

Since c(l)(E), the character of E in the lth IR, is equal to the dimension of the IR, we can equally write this as

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

e.g.,

A A( l ) E

l

E E E E

h

c c c c

1 2

2 22 2

2 2 21 1 2 6

Results following from the GOT

B. The sum of the squares of the characters of any IR equals the group order, h

LHS: Summing over i and i : ( l )( l ) ( l ) ( l )

ii i ' i 'i ' i R R

D (R)D (R) R R c c

RHS: Summing over i and i : ii ' li ' i l l

h hd h

d d

( l )

R

R h c 2

i.e.,

From the GOT, for diagonal elements of an IR : ( l ) ( l )ii i ' i ' ii '

R l

hD (R )D (R )

d

Proof:

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

e.g.,

h

h

h

2 2 2

22 2

2 22

1 2 1 3 1 6

1 2 1 3 1 6

2 2 1 3 0 6

( l )

R

R c 2

Results following from the GOT

For diagonal elements, i = j if ( l ) ( l ')ii ii

R i i

D R D R l l ' 0

We’ve already seen (trivial pt 1) if( l ) ( l ')ij ij

R

D R D R l l ' 0

if ( l ) ( l ')

R

R R l l 'c c 0

C. Vectors whose components are the characters of different IRs are orthogonal, (or the sums of the products of the characters of two different IRs is zero) i.e.,

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

( A ) ( E )

R

R Rc c 1 1 2 2 1 1 3 1 0 0

e.g.,

Results following from the GOT

D. In any given representation (reducible or irreducible), the characters of all matrices belonging to operations of the same class are identical

We saw this in discussion of conjugacy and classes earlier. By definition, all elements in the same class are conjugate as are their matrices within any representation and conjugate matrices have identical characters.

E. The number of IRs of a group is equal to the number of classes in the group.

These are extremely powerful results which follow directly from the Great Orthogonality Theorem. However, the GOT itself is excessive for most purposes and a weaker form is much more convenient to use:

The “Little Orthogonality Theorem” (LOT) (or the “first orthogonality theorem for characters”)

Combining results B and C leads to a weaker but more palatable relation:

( l ) ( l ')ll '

R

R R hc c (again, complex conjugate implied)

We can simplify this further by recalling that the all operations of the same class have the same character. If the order of class c is hc, (e.g., in C3v, hC3 = 2) then

if 1

1 if

( l ) ( l ')c

c

, l l 'h c c

h , l l '

c c

0

(The characters of different IRs behave as orthogonal vectors)

and ( l )c

c

h c h c 2

(The sum of the squares of the characters of any IR equals the order of the group)

Example for C3v C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

1 A E

cc

h c ch

c c 11

1 1 2 2 1 1 3 1 0 06

1 E E

cc

h c ch

c c 1

1 2 2 2 1 1 3 0 0 16

The LOT demonstrates that the rows of the character table are orthogonal and thus there can’t be more rows (IRs) than columns (classes)

It can equally be shown (the “second orthogonality theorem for characters”) that the columns too are orthogonal and thus there can’t be more columns than rows (i.e., the number of IRs = number of classes which was our result E)

Examples of using the GOT / LOT

I. Constructing a character table: a) C2v

Start with C2v: We have four elements {E, C2, sv, sv’}, each in a separate class. Hence there are four IRs of this group, 1, 2 , 3 and 4 (result E) From result A, the sum of the squares of the dimensions of these IRs = h Hence, we need four positive integers, dl, such that l l l ld d d d h

1 2 3 4

2 2 2 2 4

C2v E C2 sv sv’

1 1 1 1 1

Work out their characters: 1 (corresponding to A1) is trivial,

which clearly satisfies result B c 2 2 2 2 21 1 1 1 4( l )

R

R h

For which the only solution is

i.e., four one-dimensional IRs

l l l ld d d d 1 2 3 4

1

All other representations must also satisfy

I. Constructing a character table: a) C2v (cont.)

c 2

4( l )

R

R h

They must also be orthogonal to 1 (result C) so each must have two +1 and two -1s. Hence we can identify the four irreducible representations of the C2v point group:

c = 1 ( l ) Ri.e., each IR character

C2v E C2 sv sv’

1 (A1) 1 1 1 1

2 (A2) 1 1 -1 -1

3 (B1) 1 -1 1 -1

4 (B2) 1 -1 -1 1

Repeat for C3v whose elements we’ve seen are {E, 2C3, 3sv}:

I. Constructing a character table: b) C3v

As in any group there must be a one-dimensional IR with characters all 1:

C3v E 2C3 3s h=6

1 A1 1 1 1

The second one-dimensional IR is orthogonal to 1 and has components 1. We must have three +1s and three -1s. Since c(E) must be +1, this can only be

Check against result B (or the LOT): Sc2 = 12 + 2(12) + 3(12) = 6 = h

C3v E 2C3 3s h=6

2 A2 1 1 -1

(n.b. this is the IR we didn’t find earlier using our H 1s basis set in NH3)

We could use the two IRs we derived earlier on symmetry grounds and simply find the third, but, more generally:

Their dimensions, dl must satisfy l l ld d d h 1 2 3

2 2 2 6

i.e., we must have dimensions 1, 1, 2.

I. Constructing a character table: b) C3v (cont.)

The final IR must have dimension 2, i.e., c3(E) = 2.

To find c3(C3) and c3 (sv) we make use of the orthogonality relations (result C):

( ) ( ) ( ) ( ) ( )v

R

( ) ( )v

R R E C

C

c c c c c s

c c s

1 3 3 3 33

3 33

0 1 1 2 1 3 1

2 2 3

( ) ( ) ( ) ( ) ( )v

R

( ) ( )v

R R E C

C

c c c c c s

c c s

2 3 3 3 33

3 33

0 1 1 2 1 3 1

2 2 3

Simultaneous solution yields c3(C3) = -1; c3 (sv) = 0 and thus our table is:

C3v E 2C3 3s h=6

1 A1 1 1 1 z; (x2 + y2); z2

2 A2 1 1 -1 Rz;

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

II. Reducing representations In practice, the most useful application of the GOT to molecular symmetry problems lies in determining which IRs a set of basis functions spans.

Examples of using the GOT / LOT

We know we can write our reducible representation as a direct sum of IRs:

1 2

ll

l

D R D R D R ....

or a

(recall we do this by finding a similarity transformation make the representation matrix block diagonal)

To reduce a representation we must determine the coefficients al - i.e., work out how many times each IR appears in the direct sum. The character is invariant to the similarity transformation so it is convenient to use :

c cl

ll

R a R

II. Reducing representations (cont.)

1

2

3

4

0 0 0

0 0 0

0 0 0

0 0 0

D

D

D

D

D

c c c c c1 2 3 4

To determine the coefficients multiply each side by c(l’)(R) and sum over all elements:

c cl

ll

R a R

c c c c l' l' l

l l ll' l'R R l l

R R a R R h a ha

So the coefficients are given explicitly by the familiar reduction formula:

c c1 l

lR

a R Rh

which, like most of this stuff, is much easier to see in practice:

(from LOT)

or, since all c in the same class are the same c c

1 ll c

c

a h c ch

Reducing representations

Let’s reduce the following representation in C3v:

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0 C3v E 2C3 3s

a 7 1 -3

(often this can be done by inspection but we’ll do it formally)

and so: c c1 l

l cc

a h c ch

1

11 7 2 1 1 3 1 3 0

6a A

2

1 181 7 2 1 1 3 1 3 3

6 6a A

1 122 7 2 1 1 3 0 3 2

6 6a E

Reducing representations

Let’s reduce the following representation in C3v:

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0 C3v E 2C3 3s

a 7 1 -3

(often this can be done by inspection but we’ll do it formally)

and so: c c1 l

l cc

a h c ch

1

11 7 2 1 1 3 1 3 0

6a A

2

1 181 7 2 1 1 3 1 3 3

6 6a A

1 122 7 2 1 1 3 0 3 2

6 6a E

23 2a A Ei.e.,

Reducing representations

Let’s reduce the following representation in C3v:

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0 C3v E 2C3 3s

a 7 1 -3

(often this can be done by inspection but we’ll do it formally)

and so: c c1 l

l cc

a h c ch

1

11 7 2 1 1 3 1 3 0

6a A

2

1 181 7 2 1 1 3 1 3 3

6 6a A

1 122 7 2 1 1 3 0 3 2

6 6a E

23 2a A Ei.e.,

Reducing representations: Example 2

One more example, this time in Td

What symmetry species do the four H 1s orbitals in methane span?

Methane belongs to the Td point group whose character table is:

Td E 8C3 3C2 6sd 6S4 h = 24

A1 1 1 1 1 1 x2 + y2 +z2

A2 1 1 1 -1 -1

E 2 -1 2 0 0 (3z2-r2, x2-yy2)

T1 3 0 -1 -1 1 (Rx, Ry, Rz)

T2 3 0 -1 1 -1 (x, y, z); (xy, yz, zx)

The character of each operation in our 4-d basis can be determined by noting the number of members left in their original location by each operation.

We only need consider one operation in each class because all have the same character:

c(E) = 4

c(C3) = 1

c(C2) = 0

c(sd) = 2

c(S4) = 0

Using this method:

a b

c

d

sd

C2, S4

C3

11 24 1 4 8 1 1 3 1 0 6 1 2 6 1 0 1a A

12 24 1 4 8 1 1 3 1 0 6 1 2 6 1 0 0a A

124 2 4 8 1 1 3 2 0 6 0 2 6 0 0 0a E

11 24 3 4 8 0 1 3 1 0 6 1 2 6 1 0 0a T

12 24 3 4 8 0 1 3 1 0 6 1 2 6 1 0 1a T

Td E 8C3 3C2 6sd 6S4

Ha, Hb, Hc, Hd 4 1 0 2 0

i.e.,

And reduce:

The four H 1s orbitals in methane span A1 + T2

or Our representation reduces to A1 + T2

or Our representation contains the IRs A1 and T2

Td E 8C3 3C2 6sd 6S4 h = 24

A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 -1 -1

E 2 -1 2 0 0 (3z2-r2, x2-yy2)

T1 3 0 -1 -1 1 (Rx, Ry, Rz)

T2 3 0 -1 1 -1 (x, y, z); (xy, yz, zx)

The four H 1s orbitals in methane span A1 + T2

The totally symmetrical A1 orbital is essentially s in character The T2 set is either (px, py, pz) or (dxy, dyz, dzx) Hence we can see the origin of the traditional sp3 hybrid orbitals which we would usually discuss for bonding in methane. In fact, the hybrid orbitals must be a linear combination of sp3 and sd3:

That sp3 dominates is inferred from “chemical intuition”.

3 3 hybrid sp sd a

Part VI: Projection Operators and SALCs We’ve seen how we can determine the symmetry species spanned by a particular basis (or how we can reduce a representation to its constituent IRs).

Knowing the IRs of a group we can find linear combinations of basis functions (or symmetry adapted linear combinations) each of which span an IR of a given symmetry. Each SALC is necessarily block-diagonal in form and transforms as one of the IRs of the reducible representation.

Look once more at our basis of H atom 1s orbitals in NH3 and consider the effect of each symmetry operation on one particular s orbital:

C3v E C3 C32 sa sb sc

A1 1 1 1 1 1 1

A2 1 1 1 -1 -1 -1

E 2 -1 -1 0 0 0

R s1 s1 s2 s3 s1 s3 s2

C3

a

c b

s1

s2 s3

Projection Operators and SALCs

Recall that previously we saw that the orthogonal linear combinations

1 1 2 3 3 1 2 3( ); (2 )s s s s s s

led to reduction of the corresponding representation matrices with 1 transforming as A1 symmetry and 3 transforming as E symmetry.

These functions look like the scalar products of the IR character and R s1:

c 1

i i

R

R Rs

C3v E C3 C32 sa sb sc

A1 1 1 1 1 1 1

A2 1 1 1 -1 -1 -1

E 2 -1 -1 0 0 0

R s1 s1 s2 s3 s1 s3 s2

1 2 3 32 1 1e.g., E s s s

This is no fluke, it can be shown from the GOT to be a general result and is the basis of the Projection Operator Method.

General form of the projection formula

Consider the following operator in which the coefficient of symmetry operator R is a component of the representation matrix for the IR l

( l ) ( l )l

ij ijR

dP D R * R

h

Apply this operator to a function j’l’ which transforms as component j’ of IR l’

( l ) ( l )ll' l'j'ij ij

Rj'

dP D R * R

h

( l ) l' ( l')lij j' i' j'

R i'

dD R * D R

h

but l' l' ( l')j' j' i' j'

i'

R D R

( l ) ( l')ij i' j' ll' jj' ii'

R

l' l'l lj' j'

i' i' l'

hD R * D R

d

d d

h h

0

and

otherwise

l( l ) l' l i

ij j' i ll' jj'

if l' l i jP

(GOT)

Interpretation

If l ≠ l’ and/or i ≠ j , then when P acts on some function j’l’ the result is zero

i.e., if the function is not a member of the basis set spanning the IR l

the result of the projection is zero

However, if l = l’ and i = j ,

i.e., the function is a member of the l basis set,

then P projects the function from location j to location i.

This is useful because if we know only one member of a basis of a representation we can project all the others out of it.

Let’s see some simple examples:

0

and

otherwise

l( l ) l' i

ij j'

if l' l i jP

The general projection formula in use

C2v E C2 sv sv’

A1 1 1 1 1

A2 1 1 -1 -1

B1 1 -1 1 -1

B2 1 -1 -1 1

O

H H x

y

z

1s1 1s2

1 1 11 1 2 2 1 1 24 21 1 1 1AP s s s s s s s

2 1 11 1 2 2 1 1 24 21 1 1 1BP s s s s s s s

( l ) ( l )l

ij ijR

dP D R * R

h

(i.e., exactly the linear combinations we saw led to diagonal representations )

A

B

A B

s s s s s

P s s s

P s s s

1

2

1 2

1 11 1 2 1 22 2

11 1 22

11 1 22

0

0

This procedure is straightforward for 1-d IRs

The projection formula for characters

So the projection operator projects out of the original function, the functions which form a basis for the representation. Any component which is abolished cannot contribute to the the basis for that representation.

By summing over all components we can derive a more useful expression for when only the character of the representation is known;

c ( l ) ( l ) ( l )l l

ij iiR i R

d dP D R * R R * R

h h

This is the form most commonly used, its only disadvantage being that it yields only a single function for multi-dimensional IRs.

The projection formula for characters

C3v sN s1 s2 s3

E sN s1 s2 s3

C3 sN s2 s3 s1

C32 sN s3 s1 s2

sa sN s1 s3 s2

sb sN s3 s2 s1

sc sN s2 s1 s3

We’ve seen before that this spans 2A1 + E, so: 1. Draw up a table showing the effect on each

operator on each basis function

2. Multiply each member by the character of the relevant operation

3. Add all column entries

4. Finally multiply by dl/h

c ( l ) R R

R

c ( l )

R

R R

c

( l )l

R

dR R

h

C3

a

c b

s1

s2 s3

Construct the SALCs for C3v in the full s-orbital basis {sN, s1, s2, s3}:

c ( l ) ( l )l

R

dP R R

h

sN

So, for A1 symmetry SALC

1 16

AN N N N N N N NP s s s s s s s s

1 1 11 1 2 3 1 3 2 1 2 36 3

AP s s s s s s s s s s

For IR A1, d = 1 and all c(R) = 1, hence column 1 (sN) gives:

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

Likewise, column 2 (s1) gives:

[as do columns 3, 4]

C3v sN s1 s2 s3

E sN s1 s2 s3

C3 sN s2 s3 s1

C32 sN s3 s1 s2

sa sN s1 s3 s2

sb sN s3 s2 s1

sc sN s2 s1 s3

Steps 2-4:

c ( l ) ( l )l

R

dP R R

h

And for the E symmetry SALC

26 2 0 0 0 0E

N N N NP s s s s

For the IR E, d = 2 and column 1 gives:

Column 2 gives:

2 11 1 2 3 1 2 36 32 0 0 0 2EP s s s s s s s

Columns 3, 4 give 12 2 3 13 2

EP s s s s 13 3 1 23 2

EP s s s sand

Not linearly Independent (sum = 0)

C3v E 2C3 3s h=6

A1 1 1 1

A2 1 1 -1

E 2 -1 0

C3v sN s1 s2 s3

E sN s1 s2 s3

C3 sN s2 s3 s1

C32 sN s3 s1 s2

sa sN s1 s3 s2

sb sN s3 s2 s1

sc sN s2 s1 s3

c ( l ) ( l )l

R

dP R R

h

So, in summary:

1

2

1 2

0 0

0 0

0 0 0

0 0 0

N N

AN N

AN

EN

A A E

s s

P s s

P s

P s

i.e., The only projection of sN is onto A1

1

2

1 2

1 11 1 2 3 1 2 33 3

11 1 2 33

1

11 1 2 33

0 2

0 0

0 0 0

0 0 2

A

A

E

A A E

s s s s s s s

P s s s s

P s

P s s s s

2EP s 0 1

2 3 13 2s s s+ +

13 1 23 2s s s3

EP s 0 + +

and similar for s2, s3

0

0

and similar for s2, s3

and similar for s2, s3

Schmidt Orthogonalisation

Using the projection operator for E in C3v we obtained three different functions from the functions s1, s2, s3 (once normalised):

11 1 1 2 36

2E Es P s s s s

12 2 2 3 16

2E ES P s s s s

But these are not orthogonal: e.g.,

+

- -

1Es

-

+ -

2Es

1 11 2 6 22 2 1 0E E

s S

It is desirable to generate a second function orthogonal to the first:

13 2 3 1 26

2E ES P s s s s

+ -

3Es

-

Schmidt Orthogonalisation

Let 2 1E Es s'i.e., we want such that 1 0E

s'

1 1

2 1 6 22 2 1E Es s d

12 12

1 1 12 1 3 1 2 326 6

3 312 32 26

2 2

So E Es s'

s s s s s s

s s

or, upon normalisation, 12 32

' s s

+ -

Determine : We require

1

2 1 1 1 2 1

0

0

Es

E E E E E Es s s s s s

' d

d d d1

So, our SALCs for C3v in an s-orbital basis are:

1 1

12 1 1 2 33

13 1 2 36

14 2 32

2

NA s

A s s s

E s s s

E s s 1

1A 1

2A

- -

+

3E

- +

4E

Consistent with the fact that this basis spans 2A1 + E

Part VII: Direct Product Representations

Having seen previously how single functions (x, y, z, etc.) transform, we might reasonably ask how quadratic functions (e.g., xy) transform.

Consider a set of functions i1 transforming as IR 1 and j

2 transforming as IR 2.

The direct product representation is written:

= 1 2

and has basis functions i1j

2 running over all i, j.

n.b.,

If 1 is n dimensional and 2 m dimensional, is nm dimensional (often reducible)

Direct Product Representations: Example 1

1 1 1 2 2 2 c c and A A A A A AR R R R

1 2 1 2 1 2 c c A A A A A AR R R

i.e., in this case, the character of the product function is 1 2 2c c c =A A AR R R

For example, consider the following sets of functions:

transforming as A1 symmetry and

transforming as A2 symmetry in C3v

1 A

2A

Both functions are non-degenerate (represented by 1-D representations) so

Direct Product Representations

If, however, the functions are degenerate:

1 2 1 2 1 1 2 2 = i j i j k ki l lj

k l

R R R D R D R

1 2

1 2

c c c

ii jji j

D R D R

R R R

The RHS is a complicated matrix but we are only interested in diagonal elements (both k = i and l = j). Hence, the trace (the sum over all i and j of these diagonal elements) is

1 2 1 2 1 2 i j k l ki lj

k l

R D R D R

The characters of the operations of a direct product basis are the products of the corresponding characters for the original bases.

This allows reduction of the direct product representation

Rearranging:

Reducing direct product representations: Example I

Determine the symmetry of the IRs spanned by the quadratic forms x2, y2, z2 in C3v.

C3v E 2C3 3s h=6

A1 1 1 1 z; (x2 + y2); z2

A2 1 1 -1 Rz

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

The (x, y, z) basis spans a reducible representation with characters

3

1 2 3

1 1 0

1 0 1

c

c

c s

E

C

The x2, y2, z2 basis thus spans a representation with characters

3

9

0

1

c

c

c s

E

C

Which reduces to 2A1 + A2 + 3E (go on, convince yourself)

1 2 c c cR R R

Reducing direct product representations: Example II

Confirm the symmetry species of the IRs spanned by the basis (xz, yz) in C3v.

So, the characters of the direct product basis, 1c c c

A ER R R

are: E: 1 x 2 = 2 C3: 1 x -1 = -1 and s: 1 x 0 = 0, i.e., the characters of E so the direct product basis (xz, yz) spans E In other words A1 E = E

Basis (xz, yz) is a direct product of z (which spans A1) and (x, y) which spans E,

C3v E 2C3 3s h=6

A1 1 1 1 z; (x2 + y2); z2

A2 1 1 -1 Rz

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

1 2 c c cR R R

Direct product tables Similarly the product E E has characters (4, 1, 0) which reduces to = A1 + [A2] + E (consider the IRs spanned by (x, y)(x, y) =(x2, xy, yx, y2)).

We can tabulate such decompositions of direct products once and for all. They are often as useful as character tables e.g.,:

In cases in which the IRs have additional symmetry labels we also need the following tables;

n.b., direct product tables are symmetric about the main diagonal so only the upper half is shown

Important points to note:

The direct product l l’ only contains the totally symmetric IR (TSIR) if l = l’, i.e., if they have the same symmetry

In many cases the presence or absence of the TSIR in a direct product is all we need to know (e.g., to determine if particular integrals are identically zero- see below)

Symmetry species denoted in square brackets, e.g., [A2] represent antisymmetrised products.

These are useful in determining the symmetry of particular combinations of functions which can help in establishing which potential combinations satisfy rules such as the Pauli Exclusion Principle (instead of using microstate tables as we will do in the Spectroscopy course next term).

Part VIII: Full Rotation Groups

I. In 2D: R2 (symmetry of a circular system)

Consider the operation comprising an infinitesimal rotation about z:

Now, consider the angular momentum operator on (x,y):

z i x yy x

l

z x ,y i x y x ,y i y ,xy x

l

C x ,y r cos ,r sin

r cos r sin ...,r sin r cos ...

x y ...,y x ... x ,y y ,x ... (1)

Substituting for (-y, x) into (1) we see 1 izC x ,y ... x ,yl

and we can identify the “generator of rotation” about z axis from which all rotations can be , well, “generated”.

Similarly...

Now consider consecutive rotations about different axes:

1 1 a ..., and i ix y ...l l

are generators of rotations about the x and y axes, respectively.

2

1 1

1

a a

a a

y x i iy x

i iy x y x

C C ... ...

....

l l

l l l l

But

2

1 1

1

a a

a a

x y i ix y

i iy x x y

C C ... ...

...

l l

l l l l

And the difference

2

a a a

a

y x x y iy x x y

iz

C C C C l l l l

l [recall commutation relations!]

In other words the origin of the commutation of angular momentum operators can be understood from composite rotations

x y

z

a

a

β

β

lz

II. R3 the full rotation group in 3d Fact: Under rotation, the spherical harmonic functions, Ylml for a given l transform into linear combinations of one another (i.e., pp, dd but not pd).

q

immY P e l

ll and

1 2 , ,Y ,Y ,Y ,....Yll l l l,l l l form a basis for a 2l+1 dimensional representation of R3

Hence, the result of a rotation by a around the z-axis is

1

a a a

a q q q1z i i i

, ,C Y ,Y ,....,Y P e ,P e ,...,P el l- l

ll l l l l

and its representation matrix will have diagonal elements aa 1 iie ,e , etc.l-l

It’s character will thus be the sum of these:

0

1 2 2 2 2 1 2

aa aac

a a a

1

ii iC e e .. e .. e

cos cos .... cos cos x

l-l l

l l

which, in the limit a0 is 2l + 1 as expected (energy levels with quantum number l are 2l+1-fold degenerate)

R3 Character Table and Russell-Saunders Coupling

By spotting which terms arise from which ml states our character table becomes

R3 E C(a)

ml = 0 S 1 1

ml = 0, 1 P 3 1 + 2cos(a)

ml = 0, 1, 2 D 5 1 + 2cos(a) + 2cos(2a)

ml = 0, 1, 2, 3 F 7 1 + 2cos(a) + 2cos(2a) + 2cos(3a)

An infinite group

Alternatively (e.g., in the OUP symmetry Tables by Peter Atkins and Mark Child) you may see it in condensed form as

IRs labelled as j S 1; P 2, etc

Russell-Saunders Coupling in Atoms

R3 S P D F

S S P D F

P S + [P] + D P + D + F D + F + G

D S + [P] + D + [F] + G P + D + F + G + H

F S + [P] + D + [F] + G + [H] + I

All atoms exhibit spherical symmetry and thus transform as R3 Consider a p2 electronic configuration and think of the terms which might arise:

2 21 2 1 4 4

1 4 2 1 2

1 1 2 1 2 2 2

a a a

a a

a a a

p p cos cos cos

cos cos

cos cos cos S P D

l1=1, l2=1 couple to give total orbital angular momentum quantum number, L = 0, 1, 2 (as expected by the Clebsch-Gordon Series, L = l1+l2, l1+l2-1,... |l1-l2|).

n.b., The antisymmetric [P] function here alerts us to a symmetry constraint (Pauli) which restricts the S, L combinations permitted.

In this case, p2 1S, 3P, 1D (c.f. Tedious microstate table we will meet in the Spectroscopy course in HT)

Of course, we can see this immediately from the direct product table:

Part IX: Symmetry Properties of Common Integrals A. Volume Integrals d

d R d R d

Let = , the basis of a 1-d representation

c c

c R R

R d R d R d

R d R d

LH : S R R

R d R d h d

1

10 c c c Now, unless = A

R R

R R R A

10 So: unless = d A The integral is zero unless transforms as the TSIR

With GOT we can show that the same result extends to higher dimensional functions.

These are simple numbers and are thus unaffected by an operation R, so

Td E 8C3 3C2 6sd 6S4

A1 1 1 1 1 1

A2 1 1 1 -1 -1

E 2 -1 2 0 0

T1 3 0 -1 -1 1

T2 3 0 -1 1 -1

B. Overlap Integrals

Here we are concerned with the symmetry of the integrand which transforms as the direct product 1 2.

In direct product tables the TSIR is only found on the main diagonal, which means

1 2

i j d

An overlap integral vanishes unless the two functions transform as the same symmetry.

Proof: Reduction formula gives us the no. of times the TSIR appears in our integrand:

1 2 1 1 2

1 2 1 2

1

1 1

c c c

c c c

but

A

R R

a R R Rh h

R R R

Hence, from the LOT (GOT): 1 2

1 21

1

c c R

a R Rh

e.g., the overlap integral for an s and a p orbital (with the same origin), 0 ps

or i j

C. More complex integrals such as “Matrix Elements” 31 2

i j d

In which is some quantum mechanical operator which transforms as 3. For non zero matrix elements we thus require that: 1 2 contains 3 or, equivalently, that 1 2 3 contains the TSIR

3

H , the total energy operator must have the full symmetry of the molecule (as we’ve seen). Hence, it must transform as the totally symmetric IR. Our integrand thus transforms as 1 2 which, by the arguments we used in the discussion of overlap integrals, can only contain the TSIR if 1 = 2.

3

HExample 1: The energy or resonance integral,

An energy integral can only be non

zero if i and j transform as (belong to) the same IR

1 2

i jH d

It follows (perfectly generally) that

i jH

C. More complex integrals such as “Matrix Elements” 31 2

i j d

The intensity of any observed transition is proportional to Rij2 and thus Rij encodes

all spectroscopic selection rules including those imposed by symmetry constraints (see HT Atomic and Molecular Spectroscopy Course).

If we re-cast the dipole moment operator as i i i i i ii i i

q x q y q zˆ

it is clear that the vector Rij has scalar components along the x, y, z directions of

The transition dipole moment will be non zero (resulting in an allowed transition) provided at least one of these matrix elements is non zero.

1 2 1 2 1 2 and i j i j i jx y z, , respectively.

3

Example 2: Selection rules for electric dipole transitions,

is the transition dipole moment, in which is the dipole moment operator.

i ii

q rˆ

i jˆ

ij i jR ˆ

Hence fundamental transitions (v=0 v=1) are allowed provided the vibrational mode involved transforms as one of the Cartesian coordinates.

Example 2: Selection rules for electric dipole transitions, (cont.)

1 2 x and /or 1 2 y and /or 1 2 z

So, for an electric dipole transition to be allowed

must contain the totally symmetric irreducible representation.

By convention, we choose the z-axis to be the principal axis of the molecule. If it is the term which is non zero, we describe the transition as a parallel transition. The other two terms would lead to perpendicular transitions.

1 2 z

In vibrational (IR) spectroscopy this often simplifies even further because most transitions are out of the vibrational ground state and:

i) the ground (zero-point, v = 0) vibrational wavefunction transforms as the TSIR. ii) v = 1 wavefunctions transform as the symmetry of the vibrational mode

MOs and dipole allowed transitions in benzene

D6 E 2C6 2C3 C2 3C2 3C2 h=12

A1 1 1 1 1 1 1

A2 1 1 1 1 –1 –1 z

B1 1 –1 1 –1 1 –1

B2 1 –1 1 –1 –1 1

E1 2 1 –1 –2 0 0 x, y

E2 2 –1 –1 2 0 0

pz orbitals

6 0 0 0 –2 0

Benzene has D6h symmetry and D6h = D6 Ci

Consider the symmetry species spanned by the basis set comprising the pz orbitals on each of the carbon atoms:

The representation reduces to A2 + B2 + E1 + E2 and, upon account of the inversion symmetry (Ci) this gives: A2u + B2g + E1g + E2u.

The form of the MOs can be obtained from the projection operator method:

And the relative energies of these can be determined from the number of nodes of each wavefunction:

So the ground state configuration is (a2u)2(e1g)4: [by virtue of being “closed shell”, this is a 1A1gstate, of which more later]:

Dipole allowed transitions in benzene

Recall that for a transition between two states, to be allowed, the transition dipole moment must be non zero:

From the character table we note that the coordinate axes (and thus the various components of the dipole moment) transform as follows: x, y transform as E1u and z transforms as A2u

But of all the conceivable one-photon electronic transitions, which are symmetry allowed?

1 2 0

i j dˆ or

Since g g = g; u u = g; and g u = u g = u, it is clear that the only allowed transitions are those which entail a change of parity i.e., g u

(n.b., this is generally true for one-photon transitions – the photon can be considered as an “odd” function).

0 i jˆ

Possible one-electron transitions

E1 E2 = B1 + B2 + E1 and the transition is allowed with transition moment along x, y

u g

But B2g A2u ? A2 B2 = B1 (i.e., neitherE1u nor A2u and so this is symmetry forbidden)

And B2g E1g is forbidden on parity grounds (as is E2u A2u)

Also E1g A2u would be allowed with transition moment along x, y and B2g E2u would be allowed with transition moment along x, y

x = y = E1u z = A2u

So; E2u E1g

The HOMO-LUMO transition would be e2u e1g : consult direct product table:

But that’s not the full story:

The ground state configuration is (a2u)2(e1g)4 which is a 1A1g state (singlet as all electrons are paired and totally symmetric spatial function) The HOMO-LUMO transition is a e2u e1g promotion which generates an excited configuration of (a2u)2(e1g)3(e2u)1. A number of states arise from this configuration:

E1g E2u = B1u + B2u + E1u

As the ground state transforms as the TSIR, we can immediately see that

1B2u 1A1g is forbidden (since neither x, y, nor z transform as 1B2u) 1B1u 1A1g is forbidden (since neither x, y, nor z transform as 1B1u)

but 1E1u 1A1g is symmetry allowed with transition moment along x, y

So, given that we can have singlet and triplet spin states of each, generates a total of 6 excited electronic states: 1B1u, 1B2u, 1E1u, and 3B1u, 3B2u ,

3E1u.

The DS= 0 selection rule means only the singlet states are accessible from the 1A1g

ground state.

Which is more than enough for anyone, so…

…thank you and have a very merry Christmas!