1

x

z

y

2Molecular Orbitals of H O

2

2

2

2

E C xz yz

C E yz xz

xz yz E C

yz xz C E

2 : , , , ,

: , , , ,

: , , , ,

C x y z x y z

xz x y z x y z

yz x y z x y z

H Eigenstates of H odd or even

any basis (1 electron or many-body) for H

22

1two subspaces

2

CC

2

2

11, 4 subspaces

2 2

xzCC xz

22

1 11, , 8 subspaces

2 2 2

xz yzCC xz yz

molecule on xz plane, y axis points towards us

Z axis= molecular axis.

Group multiplication table

Basis states odd or even = symmetry adapted states

2

2 2

1

2

1

2

4

1 1 1 1

1 1 1 1 ,

1 1 1 1 ,

1 1 1 1 ,

v xz yz

z

y

x

C I C g

A z

A xy R

B x R

B y R

Such are the possible symmetry types of the molecular orbitals and vibrations.

Conventional names shown in the first column.

Here is a representation of the Group. If we replace every operator by

1, the multiplication Table is trivially verified; -1 choices are also allowed, as follows:

2

2

2

2

E C xz yz

C E yz xz

xz yz E C

yz xz C E

x

z

y

Group multiplication table

Can a function get a -1 for all operations except I? NO

The multiplication table forbids its existence.

3

a

b c a

b c

a

b c a b

c

C3

3v 3Group (NH )C

2

3 3, , , ,a b cC C Operations

: , , , ,a a b c a c b

The sense of rotation is arbitrary and we may choose C3 as the operation (a, b, c) → (c, a, b);

3v

3v

isomorphous to (3)

(permutations of 3 objects)

means actually (3)

C S

C S

4

a

b c

a

b c

a

a

b c

Convention for multiplying operations

a b

c

aC3 a

c

a b 3

3

: , , , ,a

a b

C a b c c b a

C

the reflection plane through a remains fixed

5

Everything in the Group multiplication table

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

C E C

E C C

C E C

C C E

The rearrangement theorem holds:

Each line and each column contain all R G.

This follows from the definition of Group. In any line or column there are g

elements, and all are distinct since for instance C3R = C3S implies R = S. Every

operation does a permutation of the vertices. In C3v the converse is also true,

so C3v is isomorphous to S(3). Already in C4v the 8 operations are fewer

than the 24 permutations of 4 objects.

3 and are generators

(all elements can be generated by multiplication)

aC

7

If H ⊂ G is a Group itself it is a

subgroup of the Group G.

Example: Orthogonal group O(n) has a special subgroup

( ) matrices | 1TO n A nxn A A 21 det( ) det( )det( ) det( ) det 1T TA A A A A A

( ) ( ), ( ) | det 1 is a subgroupSO n O n SO n A nxn A

Example: the rotation Group has the Group of symmetry rotations of a cube as a subgroup.

Definitions:

Isomorphous Groups: two Groups are isomorphous if they are the same

Group, with different names.

8

If H ⊂ G is a Group itself it is a

subgroup of the Group G.

Let H be a subgroup of G; for a ∈ G

consider the set

H’(a) = {aha−1, h ∈ H}.

Since ah1a−1ah2a

−1 = ah1h2a−1, this is a subgroup of G,

the conjugate

subgroup with respect to a.

Occasionally, it may coincide with H itself (as a set, not

element by element), in which case we write aH = Ha.

If a ∈ H , aH=Ha=H.

If ∀a ∈ G, aH = Ha, then H is called invariant subgroup

of G, or normal divisor of G.

a

H

a

H

aHa-1

Center of the group G Z(G) (from Zentrum=center in German)

( ) |Z G z G zg gz g G

Proof that Z(G) is a subgroup

1

1 1 1

1 1 1 1

Moreover, , ( ) ( ).

indeed,

( ) ( ).

Finally, and (since xg=gx)

x y Z G xy Z G

xyg xgy gxy

x Z G x Z G

gx x x gx

x g xx gx x g

In particular, Z(G) is an invariant subgroup since its elements communte with all elements:

1( ) ( ) .h Z G xhx h Z G x G

9

Example of invariant subgroup:

That is, Z is the set of group elements that commute with all the group.

It is obvious that ( )e Z G

Cayley’s Theorem . Any group of order NG is isomorphous to a subgroup of S(NG)

= symmetric Group of NG objects = Group of permutations.

23 3 3 3vZ = {E,C , C } C is an invariant Abelian subgroup.

2

2

3 3

2

3 3

2

3

3 3

2

3 3

2

3 3

3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

E C C

C E C

C

C

C

C

E

E

HaaHGa ,

Invariant subgroup H

A Group G is called simple if it does not have invariant subgroups. So, Z3 is simple,

but C3v is not.

A Group G is called semi-simple if it does not have abelian

invariant subgroups. So, Z3 is semi-simple, but C3v is not.

10

11

In the C3v example, it is natural to think that

rotations and reflections are two essentially

different types of operations. Moreover, a closer

inspection shows that the product of two rotations is

a rotation, the product of two reflections is a

rotation, while the product of a rotation and a

reflection is a reflection. Thus, there is structure

within groups, and by dividing operations into types

one can gain understanding about the Group.

Equivalence classes in set theory and quotient set

In everyday life, we find it useful to classify individuals according

to some criterion, e.g. their nationality, and then discuss about

relations among the cathegories. In the case of nationality, this

would mean international relations.

In set theory, one would say that there is an equivalence

relation ~ in the set X.

Example:

Consider the equivalence relation on the set of integers Z: x~y if and only if x-y is

even. This relation gives rise to exactly two equivalence classes: [0] consisting of all

even numbers, and [1] consisting of all odd numbers. Under this relation [7] [9] and

[1] all represent the same element of the quotient set {[0] , [1] }=Z / ~.

Equivalence relation: it is denoted by ~

Defining properties are:

A ~A

A ~B B ~A A ~B , B ~C A ~C

That is, you ~ me if we both belong to the same nation; nationality

is an equivalence class (my equivalence class is the set of all

individuals such that they are equivalent to me, that is, the Italians).

12

Quotient set

The set of Nations is the quotient set X / ~ i.e. Nations=

individuals/nationality

In italiano: le nazioni sono gli individui suddivisi per nazionalita’. Il fatto che in

inglese subdivide=divide ha portato probabilmente a questa notazione.

Quotient set= insieme degli insiemi ottenuti con delle suddivisioni.

subgroup, gH gG G H

, right coset (laterale destro)

, left coset (laterale sinistro).

Hg hg h H

gH gh h H

Cosets (laterali in Italian)

13

g

H

gH

Let us consider right cosets (similar statements hold for the left ones)

Given G and H, different elements g and g’ may end in the same coset.

which means g’h’ = gh for some h, h’ in H. This is equivalent to g’= g h’’

for some h’’=h h’^(-1).

Otherwise g and g’ end in different cosets.

So this is a way to divide the Group G, establishing equivalence classes.

The set of cosets is the quotient set G/H.

One can think of the right coset as an equivalence class:

, G group, H G subgroup.

.

is indeed an equivalence class since :

x y

x y x Hy

1 1

since

since , with .

, since , ' ' and ' .

x x e H x Hx

x y y x x hy y h x h H

x y y z x z x hy y h z x hh z hh H

subgroup, gH gG G H

, right coset (laterale destro)

, left coset (laterale sinistro).

Hg hg h H

gH gh h H

14

g

H

gH

Let us consider right cosets (similar statements hold for the left ones)

15

2

3 3 3subgroup , ,Z E C C

2

3 3 3coset , , , ,a a a b cright Z E C C

3v 3 3 aC Z Z

3vCExample:

How many elements in a coset?

choos

su

e

bgro

a fi

up of ord

xed but

er NH

g G g

H

H

G

, right coset

, left coset

Hg hg h H

gH gh h H

1 2 1 2

For each both cosets have elements :

indeed, , so each h yields an element.

Hg G N

h g h g h

H

h

g

2

2

3 3

2

3 3

2

3

3 3

2

3 3

2

3 3

3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

E C C

C E C

C

C

C

C

E

E

15

We have split the Group: !

16

h H G H of order NH HhgHg

1 1g H hg H because

1

1 1' 'hg h H g h h H

Coset Hg1 of order NH is totally disjoint from H

G

g1

H

Hg1

NH elements

1 G H 3vIf G = H Hg , N = 2N . (e.g.C )

17

H

G

g1

Hg1 g2

Hg2

2 2 2 1

-1

1 1 1

jj

e form the coset Hg with g H, and g Hg ;

Hg2 is disjoint from H and from Hg1 (since hg2 = h'g g2 = h h'g Hg ).

We go on until G has been totally partitioned:

G = Hg .

W

This proves the famous Lagrange Theorem:

Theorem 4. The order of any subgroup of G is a divisor of NG.

G HIf G = H Hg, N = 2N .

Otherwise, :iterate

The set of cosets is a partitioning of the Group elements, the quotient set G/H .

The elements of G can be partitioned in stacks and each stack contains an

equal number of elements; each stack is labelled by an element and contais

those that are equivalent, i.e. belong to the same coset according to H.

18

Giuseppe Lodovico Lagrangia (Lagrange)

19

Picking (ah) in aH, ∀h,h’ , one sees that (ah) h’ = ah’’ is again in aH. :

Let H invariant subgroup of G

That is, ∀a ∈ G, aH = Ha (All its right cosets are equal to the left cosets, )

Define abstract set multiplication of coset by H: aH × H = aH, meaning the above

Z3 is an invariant subgroup of C3v and the left coset σaZ3 ={σa, σb, σc} is the set of

reflections; multiplying by the rotations in Z3 we

get {σa, σb, σc} × Z3 = {σa, σb, σc} .

HaaHGa ,

Invariant subgroup H

2

2

3 3

2

3 3

2

3

3 3

2

3 3

2

3 3

3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

E C C

C E C

C

C

C

C

E

E

Example:

20

We also define abstract coset multiplication:

1 2 1 2Definition: , , removing all duplicatesaH bH ahbh h H h H

1 2 1 2 3 2 3 2, , , ,

,

aH bH ahbh h H h H abh h h H h H

abh h H abH

1 3 1 3H subgroup invarian :t ,b h b h hb bhH Hb

coset(a) X coset(b)=coset(ab)

1 1in particular, aH a H a a H H The set of cosets is a group with this multiplication: it is the quotient

Group G/H

where H=Identity

3v 3 3 3/ , rotations,reflectionsaC Z Z Z

Rotation X rotation=rotation, rotationXreflection=reflection, reflectionX

reflection=rotation

Invariant subgroups and quotient Groups

2

2

3 3

2

3 3

2

3

3 3

2

3 3

2

3 3

3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

E C C

C E C

C

C

C

C

E

E

If the set of operators can be replaced by numbers that agree with

the multiplication table, that is a representation of the Group.

It is interesting to find representations of Z3

21 3 3

2 i -2 i23 3

1 3 3

* 23 3

A representation: E=C =C =1

Besides A one can represent C by =e and C by *=e ;

in a third representation, C by C

by ( complex conjugate representations)

and

21

2

1 3 3

3

A representation: E=C =C =1

Electron wavefunctions such that

C = , = = =

are totalsymmetric.

The symmetry allows that.

a b c

22

: ( , , ) ( 1,0,1)

1 cos sin

1 cos sin

i

i

basis x y z m

m e i x iy

m e i x iy

3

2

33

2

2 *3

:

:

ii i i

ii i i

C e e e e

C e e e e

Take the p orbitals of an atom centered at the origin : (x,y,z). The triangle

Is on the (x,y) plane, and the z orbital, belongs to A1.

23 3 3

1

*

*

1 1 1

1,

1

Z I C C

A z

E x y

2 i

3 =e

Representations:

Under C3:

x + iy is multiplied by

and x−iy is multiplied by ∗ ; thus the (x,y) pair is a basis for the conjugate

representations

23

Paul Adrien Maurice Dirac

Important contributors to the Group Theory

24

Group Algebra

When G is not Abelian, the elementary method allows diagonalizing H simultaneously

with one or a few compatible R ∈ G. In the C3v example, one can choose

C3 (and C23 , but this adds nothing) or one reflection. In this way, one is neglecting

most of the symmetry- related information.

Although this does not

cause mistakes, the use of Group Theory is much more rewarding.

We must find linear combinations of the operators R ∈ G such that they commute

with all G elements (besides H; but this we take for granted, by definition of

a symmetry Group).

We could think about powers or products, but since G

is closed the most general function is a linear combination.

For any abstract Group, the linear combinations of the elements

constitute the Group algebra.

One has multiplication and linear combinations.

Dirac characters and Irreducible Representations

25

Commuting linear combination

, (rearrangement)

is totalsymmetric.

It is compatible with all T G and commutes with H.

We can find useful new operators by introducing conjugation.

G GN N

T G T G

T S G S TS

S S

1

Conjugated Group elements.

, . : :

This conjugation is an equivalence: ( )

,

A B G Def A B X G A X BX

A A X E

A B B A A B B C C A

Meaning: X is a change of reference that converts B into A

Classes

26

All elements conjugated together are a class. The class concept, that

allows to put together operations that do essentially the same thing: such

operators are called conjugated

Example:

C3v has 3 classes {E} , {C3, C3 2}, {σa, σb, σc} .

In Abelian Groups each element is itself a class.

2

3 3

2

3 3

2

3 3

3

1

1 2

3

2

3 3

2

3 3

2

3

3

2

3

1

3 3

a b c

c a b

b c a

a b c

b c a

c a b

b c

b b

b b c b

b b

E C C

C E C

C C E

C

C

E C C

C C E

C

C

E C

C C

27

1

Define Dirac's characters

commutes with all G: G, .

Cn

CT C

C C C

T

X X X

1 1

C

1 1

1

,

Sum of n terms conjugated with ;

those terms are all diff

I

eren

ndeed,

t si

nce ' '

G

CC n

C

T C

C C C C

n

C

T C

X G X X X TX

T

X TX X T X T T

X X X X X

T

A symmetry-induced quantum number for each class

cDirac characters commute with all X G , so they commute with each other

Proof

28

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

2

3 3

a b c

c a b

b c a

a b c

b c a

c a b

E C C

C C E

C E C

E C C

C E C

C C E

2

3 3R

a b c

C C

c

c c'

Any function f( ) of Dirac characters commutes with all X G,

and any f( )=linear combination of all

2 2 2 2 4

3 3 3 3

2 2

( ) 2 2

( ) 3 3( ) 3 3

R R

a b c a b a c R

C C C C

Using the multiplication table,we can evaluate

22

21

R R R

2 2

2

3 3 9 3

R

R

2 2

1

3 3 0 0

R

R

Same equations must be obeyed by their eigenvalues:

Characters are not

independent, but occur in

combinations

29

One can label the eigenfunctions of the

Hamiltonian by the irreps: they are sets of good

quantum numbers. A and B are used for one-

dimensional irreps, E for two-dimensional, T for

three-dimensional ones

The allowed combinations correspond to the possible symmetry types allowed

by C3v; they are called Irreducible Representations or Irreps for short:

1

2

Irrep

2 3

2 3

1 0

R

A

A

E

irreps are never mixed by symmetry operations since

C C C C C

G, .X X X X

1 2

1 1 1 2 2 2 1 2

1 2

1 2

2

and belong to different eigenvalues of the symmetry operators,

, and , with this implies not only

0

but al

The diagonalization problem is broken.

Proo

so

f

0.

If

T t T t t t

H

TH

2 2 2 2 1, so and have different symmetry quantum numbers;

since is unitary, orthogonality follows.

HT t H H

T

Important

31

A1 is a trivial representation with all operators represented by 1 and Dirac characters equal to nC for each class. This exists for any Group.

A2 is similar but odd for reflections.

In all non-Abelian Groups there are irreps that require

matrices, rather than numbers, to represent operators, like the irrep E . Just consider transforming coordinates (x, y) in the plane (or orbitals (px, py)):

a

c b

1

2

Irrep

2 3

2 3

1 0

R

A

A

E

3

1 0 1 0generators : ( ) ( ) ( )

0 1 0 1a

c sD E D D C

s c

One can work out all operations from the

3

2 1 2 3( ) , cos( ) , sin( )

3 2 3 2

c sD C c s

s c

The matrix which rotates (x,y) is just

32 32

The other D(R) matrices

by multiplication

3 3

2( ) ( ( ) ( )( ) () )c a b a

c sD D

c sD D D C

s cD C

s c

3 3

2 1( ) ( )c s

D C D Cs c

3

1 0 1 0generators: ( ) ( ) ( )

0 1 0 1

2 1 2 3cos( ) , sin( )

3 2 3 2

a

c sD E D D C

s c

c s

a

c b

What did we gain so far? A set of good quantum numbers (Dirac’s characters) and representations of the symmetry operators. We want practical methods to use all that.

x

'x

d'd

Piani di riflessione

4 C multiplication Tablev

Exercise:

Repeat the

procedure

For this case