Exercise: What is the point group for each of the following substituted cyclobutanes? Assume that (idealized) C4H8 itself has D4h symmetry and that replacing an H by X or Y changes no other structure parameters.

!

X X

XX

X

X

X

XX X

X

Y

X XX

XX

XX

XX

X

X

X

X

a)

b)

c)

d)

e)

f)

g)

k)

h) l)

Lets say we have a reducible representation !red in C3v which is given below:

Then !1= (1·1 ·6 + 2 ·1 ·0 + 3 · 1 ·(2))/6 = 2 times present in !red

!2= (1·1 ·6 + 2 ·1 ·0 + 3 · (-1) ·(2))/6 = 0 times present in !red

!3= (1·2 ·6 + 2 ·(-1) ·0 + 3 · 0 ·(2))/6 = 2 times present in !red

i.e. !red= 2!1+ 2 !3. Such a reduction is always unique!

3"v

E

2C3 !3

!1

!2

!red= 2!1+ 2 !3

If we think about representations as vectors in the vector space made up of symmetry elements, a decomposition into irr. reps means to describe a reducible representation in terms of the basis vectors of that vector space.

C3v E 2C3 3"v

!1 1 1 1

!2 1 1 -1

!3 2 -1 0

!red 6 0 2

Intermission: cyclobutane: Cy-butane-1.cmmf

Exercise: decompose the following reducible representations

Exercise: Prove, from theorems based on the GOT, that all irreducible representations of Abelian groups are one-dimensional.

Exercise: In the point group D3h, find the following direct products and identify the irreducible representations that they contain:A1’ ⊗ A2’A1’ ⊗ A2’E’⊗ E’’E’’⊗ E’’

The most important aspect of our group theory application is the connection between irreducible representations and quantum mechanical basis functions. In this context, often “products” such as

are used, where O is some operator. If we relate basis functions to irreducible representations, we need a method to evaluate their “products”. This is not the usual matrix or “inner” product, but the Kronecker or “outer” product.

Direct product of matrix representations

Consider a group G with irreducible representations

!

A = a11 a12

a21 a22

"

# $

%

& '

!

B = b11 b12

b21 b22

"

# $

%

& '

Direct, Kronecker or outer product

!

A " B = C = a11B a12Ba21B a22B#

$ %

&

' ( =

a11b11 a11b12 a12b11 a12b12

a11b21 a11b22 a12b21

a21b11

#

$

% % % %

&

'

( ( ( ( ...etc.

Theorem: The direct (outer) product of two irreducible representations A and B of a group G,

or

is also a (generally reducible) representation of the group.

As always, there is a simplification for the characters:

cij,kl= aikbjl

!

A " B = C = a11B a12Ba21B a22B#

$ %

&

' ( =

a11b11 a11b12 a12b11 a12b12

a11b21 a11b22 a12b21

a21b11

#

$

% % % %

&

'

( ( ( (

Theorem: The character of the direct product representation matrix is equal to the product of the characters of the separate irreducible representation matrices

The general decomposition of a product representation can be obtained through the decomposition formula

C4v E C2 2C4 2"v 2"d

A1 1 1 1 1 1A2 1 1 1 -1 -1B1 1 1 -1 1 -1B2 1 1 -1 -1 1E 2 -2 0 0 0

A1*A2 1 1 1 -1 -1B1*E 2 -2 0 0 0

A1*E*B2 2 -2 0 0 0E*E 4 4 0 0 0

Why is this important? We will soon make the connection between (wave-) functions, operators, and irreducible representations, and will want to know whether a matrix element forms a basis for an irreducible representation

so we are interested in knowing the irreducible representation of products of functions and hence of products of representations

Functions may have certain symmetry properties under a specific point group. For example, the functions on the right are either gerade or ungerade under inversion.

2.4 Basis functions

Many aspects of the applications of group theory in the present context rely on the connection between irreducible representations and quantum mechanical basis functions. First, the intuitive approach.

In a system with point group symmetry Ci, which contains only the identity and the inversion operations, and hence has only two irreducible representations, functions may then transform as the totally symmetric irreducible representation Ag (e.g. the dx2- y2 orbital) or the irreducible representation Au (such as the px or fx(x2-3y2) orbital. So, intuitively we can connect irreducible representations with functions.

basis functions for the irreducible representations of C2v

this example taken from Charles L. B. Macdonald, University of Windsor, Ontario, Canada

this example taken from Charles L. B. Macdonald, University of Windsor, Ontario, Canada

Apart from simple and higher (mostly atomic orbital-) functions, the character tables also contain rotational functions. Consider rotations about the x,y, and z axes Rx, Ry, and Rz. Under C4 symmetry, the Rz rotation transforms as shown in the figure.

Hence we can say that Rztransforms as the irreducible representation A2, or “forms the basis for the irreducible representation A2”.

C4v E C2 2C4 2"v 2"dSimple functions

Higher functions

A2 1 1 1 -1 -1 Rz

Rotations

same reasoning, applied to dx,y orbital

same reasoning, applied to dz2 and dx2- y2 orbital

Consider the function f(x,y,z) = x2- y2, which can be used to represent (together with the radial part) one of the d orbitals of an atom. From the figure, it is clear that the function transforms as the irreducible representation B1.

C4v E C2 2C4 2"v 2"dSimple functions

Higher functions

B1 1 1 -1 1 -1 z x2 - y2

Higher functions

We saw earlier that the operation C4 can be represented by a rotation matrix. Then coordinate x is taken into y, and y into -x, but z is taken into z. The same is true for the C2 and reflection operations.

C4v E C2 2C4 2"v 2"dSimple functions

Higher functions

A1 1 1 1 1 1 z

A2 1 1 1 -1 -1B1 1 1 -1 1 -1

B2 1 1 -1 -1 1

E 2 -2 0 0 0 (x,y)

!

C4 =

cos 2"4

sin 2"4

0

#sin 2"4

cos 2"4

0

0 0 1

$

%

& & & & & &

'

(

) ) ) ) ) )

So a function f(x,y,z) = z transforms into itself under the “matrices” of the irreducible representation A1.

2-dimensional representations

Example: group C4v

In group C3v, and likewise in C4v the coordinate z does not mix with coordinates x and y, hence the representation matrices can be written in block diagonal form. We can write the matrices for operations on coordinates x and y as (2 x 2) matrices:

1 00 1

E =C2cos π sin π

-sin π cos π

1 00 1

=C4cos 2π/4 sin 2π/4

-sin 2π/4 cos 2π/4

0 1-1 0

so C4 takes x into y and y into -x

1 0

0 -1σv(xz)does not affect x but changes y into -y

-1 0

0 1σv(yz) does not affect y but

changes x into -x

0 1

1 0σd(xz)

takes x into y and y into x

0 -1

-1 0σd’(xz) takes x into -y and y

into -x

Now we collect the characters for each operation on the x and y coordinates

E 2C4 C2 2σv 2σd

E Γx,y 2 0 -2 0 0 (x,y)

A1 Γz 1 1 1 1 1 z

Likewise, since the characters of the matrices for the z coordinate are all “1”, we say that the function z (or any function which depends only on the z coordinate) forms the basis for the irreducible representation A1, which is totally symmetric.

C4v E C2 2C4 2"v 2"dSimple functions

Higher functions

A1 1 1 1 1 1 z

A2 1 1 1 -1 -1B1 1 1 -1 1 -1

B2 1 1 -1 -1 1

E 2 -2 0 0 0 (x,y)

Character table: Γx,y has the same characters as the E representation, which is 2-dimensional. Hence we say that “the functions x and y together form the basis (transform as) the irreducible representation E”.

So far, symmetry operations represented by real orthogonal transformation matrices R of coordinates

Since the matrix R is real and

also holds.

and thus PR f (Rx) = f (x) PR f (x) = f (R!1x)

PR changes the shape of a function such that the change of coordinates induced by R is compensated.

The group of the PR is isomorphic to the group of R.

The operations R form a group. Now we introduce a new group: transformation operators which affect functions, not coordinates ! These operators are called PR.

Formal derivation of basis functions for an irreducible representation

Consider the group of operators which leave the Hamilton operator invariant, i.e. which commute with H, e.g. in an atom where the potential energy V only depends on distance

Let an operator PR act on the function, e.g. a rotation by 90 degrees transforms x into -y, y into x, and z into z:

But V is invariant because it depends only on the squares of the coordinates A similar argument applies to the kinetic energy.

For such an operator PRH = H PR.

Definition: The group of all symmetry operations PR which commute with the Hamilton operator PR H φn= PR E Φn or H PR φn = En PR φn is called the group of the Schrödinger equation An eigenfunction PRφn is also an eigenfunction of H to the same eigenvalue En. All degenerate eigenfunctions can be obtained by applying PR.

2.5 Relation between representation theory and quantum mechanics

Let φ1 and φ2 be eigenfunctions of an operator H which belong to the same eigenvalue En. such as the px and py orbital functions. They are linearly independent and there are no more linearly independent eigenfunctions. Linear combinations are

φ' = c1φ1 + c2φ2

A vector space spanned by the functions φ1, φ2,... φn consists of all functions φʼ= c1φ1 + c2φ2 + … + cnφn. The φ1, φ2,... φn are the basis vectors of the vector space with dimension n.

e. g. p orbitals in a 3-dimensional vector space, d orbitals in 5-dim space, etc.

Degenerate eigenfunctions

Let the eigenvalue be degenerate of order l. Applying PR to one of the φn

we can produce another function of the same energy by a linear combination of basis vectors. This can be expressed by a matrix

PR#1= #1!11+#2!12+#3!13

PR#2= #1!21+#2!22+#3!23

PR#3= #1!31+#2!32+#3!33

This matrix Γ(R ) is a representation of of the group of PRʼs. It is irreducible since we need an l-dimensional matrix to transform a function into a linear combination of basis vectors. The functions φi are the basis functions of that representation !

Degenerate eigenfunctions

Definition: A set of l linearly independent functions φn which are transformed into one another by the matrices of an irreducible representation Γn(R ) are called basis functions of that irreducible representation.

T H φ (qi) = TE φ (qi) and thus

H T φ (qi) = ET φ (qi) hence H φ'(qi)=E φ'(qi). -> φ'(qi) belongs to the same energy level.

-> energy levels are identified according to the irreducible representation for which their eigenfunctions form a basis.

The set of ln degenerate eigenfunctions ψ(n) of energy En form basis functions for an ln –dim. irr. rep. Γ(n) of the group of the Schrödinger equation.

For a group with elements (symmetry operations) R and T, we have

H φ (qi) = Eφ (qi) with H = R H

(the group forms the group of the Schrödinger equation).

Theorem: If a Hamilton operator is invariant with respect to a group of symmetry operations, then eigenfunctions which together belong to an irreducible representations have the same energy.

However, there are many different eigenfunctions, all of which may form a basis for one irr. rep., e.g. Au under Cs symmetry. They may well have different energies. Only those that belong to the same subspace, i.e together form the basis for an irreducible representation, have the same energy.

Theorem: Basis functions which belong to different irreducible representation of a point group are orthogonal.

-> you can read directly from the symmetry group of a problem how many orthogonal classes of basis functions there are.

e.g. in Ci: only gerade and ungerade basis functions

Theorem (Neumannʼs principle): Every physical property of an object must transform as the totally symmetric irreducible representation.

Rephrase this: Every physical property of an object must have at least the symmetry of that object (since this is what symmetry is all about!).

2.6 Examples: symmetry of physical properties, tensor symmetries

Macroscopic properties of objects are described by tensors (array of numbers defined by its transformation properties under coordinate transformation). Example: the polarizability tensor !,which connects the electric field and the polarization that it induces

In an isotropic sample, the spatial variable r is a scalar. In an anisotropic sample, P may not be parallel to E (for example, a polarization in a direction different from the direction of E may be easier).

!

r P = "

r E

Alternative shorthand way of writing

!

P1

P2

P31

"

#

$ $ $

%

&

' ' ' =

(11 (12 (13

(21 (22 (23

(31 (32 (33

"

#

$ $ $

%

&

' ' '

E1

E2

E3

"

#

$ $ $

%

&

' ' '

A crystal can have a permanent dipole moment only if one of the Cartesian coordinates forms a basis for the totally symmetric irreducible representation, since otherwise, all directions are mixed and μ vanishes.

C4v: The coordinate z forms a basis for the totally symmetric irreducible representation A1. Crystals or molecules with point group C4v can have a permanent dipole moment along their z axis.

If the primitive unit cell of a crystal has a dipole moment

the crystal is said to be pyroelectric; the sum is over all charges in the unit cell. Dipole moment μ is a vector!

Example: pyroelectricity

Under C2h symmetry, which product of two coordinates transforms as the totally symmetric irreducible representation?

To read from the character tables, αxx transforms as x2; αyy transforms as y2; αzz transforms as z2. αxz transforms as xz etc.

The operation C2 transforms x into - x, y into -y, and z into zHence xy into xy; xz into -xz, yz into -yz.

Hence for a crystal with C2h point symmetry the polarizability tensor looks like this:

axy = ayx

-> only four independent components

!

" xx " xy 0" yx " yy 00 0 " zz

#

$

% % %

&

'

( ( (

It can be shown that αij=αji, i.e. tensor is symmetric - only six independent components. αxx connects Ex with Px, i.e. α transforms as the product of two spatial coordinates. If a symmetry operation results in x1 x2= -x1x2, then the corresponding tensor element must vanish.

Example: polarizability tensor (2nd rank tensor):

!

PxPyP

"

#

$ $ $

%

&

' ' '

= ( xx ( xy (xz

( yx ( yy (yz

( zx ( zy ( zz

"

#

$ $ $

%

&

' ' '

Ex

Ey

Ez

"

#

$ $ $

%

&

' ' '

3.1 The Full Rotation Group and its irreducible representations (elementary level)

The symmetry group of an atom in free space is that of the full rotation group. Here - elementary level. We restrict ourselves to odd-dimensional representations, i.e. consider only integer spin quantum numbers.

In the full rotation group there are infinitely many classes - every rotation by θ is in a class by itself. All rotations by θ are in the same class though, since one can take a third rotation to make them identical. There are then infinitely many classes and infinitely many irreducible representations.

For an atom in free space, the group of the Schrödinger equation is the full rotation group. Its representations are s ! 1-dim, p ! 3-dim, d ! 5-dim, f ! 7-dim. etc. For these irreducible representations the spherical harmonics

are basis functions. Plm are the associated Legendre polynomials, Nlm are normalization constants. For each l there are 2l+1 solutions with m = (-l,-l+1,..,0,..,+l-1,l). Hence there are 2l+1 dimensional irr. reps.

3. Group theory, crystal field splitting and molecular orbitals

Theorem: A rotation around an axis results in a transformation of the spherical harmonics such that they can be expressed as a linear combination

Let us choose a simple way to evaluate the representations of odd dimensions (1,3,5,7...). Consider a rotation around an axis by an angle !:

without loss of generality since all rotations by θ are in the same class.

The representation matrix is a diagonal matrix (for each m)

And the character is just the sum over m !

" i(#) =

e$ il# 0 0 0 00 e$i(l$1)# 0 0 00 0 e$ i( l$2)# 0 00 0 0 ... 00 0 0 0 eil#

%

&

' ' ' ' ' '

(

)

* * * * * *

Theorem: The irreducible representations with characters

are the only odd-dimensional irreducible representations of the full rotation group for integer l

Example: atom in a field of lower symmetry, for example from a surrounding octahedral crystal lattice.

For the octahedral point group O, the character table looks like this:

O E 8C3 3C2 6C2ʼ 6C4

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

We now calculate the characters for state in an atom with angular momentum l for specific rotations. From the theorem,

Since the spherical harmonics of order l form bases for the group of all rotations, they also form them for groups with finite rotations!

!(E) = 2l + 1

!(C3) = !("(2#/3)) = 1 for l = 0,3 0 forl = 1,4 -1 for l = 2,5

!(C4) = !("(#/2)) = 1 for l = 0,1,4,5 -1 for l = 2,3,6,7

Using the reduction formula, we can then decompose the higher (and clearly reducible) representations into the irr. reps of the group O.

So we can read the degree of degeneracy of an atomic level in a field of known symmetry from the character tables!

OO E 8C3 3C2 6C2ʼ 6C4

s D0 1 1 1 1 1p D1 3 0 -1 -1 1d D2 5 -1 1 1 -1f D3 7 1 -1 -1 -1g D4 9 0 1 1 1

D0 A1 An s state cannot split; s states are always totally symmetric

D1 T1 A p state does not split in a field of cubic symmetry

D2 E + T2 A d state must split since there are no 5-dim irr.representations in the O group

D3 A2 + T1 + T2 An f state splits into one nondegenerate state and two 3-fold degenerate states

O E 8C3 3C2 6C2ʼ

6C4

A1 1 1 1 1 1A2 1 1 1 -1 -1E 2 -1 2 0 0T1 3 0 -1 -1 1T2 3 0 -1 1 -1

irreducible representations of group O reducible representations of atomic levels under octahedral symmetry

atomic physics labels

What happens if we lower the symmetry further? Consider a lowering from octahedral symmetry to D3 symmetry by elongating one of the body diagonal axes.

Hence we arrive at the following correlation:

Atom cubic field with trigonal distortion

l= 0 s A1 A1

l=1 p T1 A2 E

l=2 d T2 A1 E E

E

l=3 f T2 A1 E

T1 A2 E

A2 A2

cubic field

O E 8C3 6C2

A1 1 1 1A2 1 1 -1E 2 -1 0T1 3 0 -1T2 3 0 1

D3 E 2C3 3C2

A1 1 1 1

A2 1 1 -1

E 2 -1 0

Crystal field splitting contʼd...

An example

An example of molecular symmetry lowering by the environment

from St. Böttcher, Ph.D. student in AG Horn

Another kind of symmetry-induced level splitting

The Jahn–Teller effect, sometimes also known as Jahn–Teller distortion, or the Jahn–Teller theorem, describes the geometrical distortion of non-linear molecules under certain situations. This electronic effect is named after Hermann Arthur Jahn and Edward Teller, who proved, using group theory, that orbital non-linear spatially degenerate molecules cannot be stable.[1] The theorem essentially states that any non-linear molecule with a spatially degenerate electronic ground state will undergo a geometrical distortion that removes that degeneracy, because the distortion lowers the overall energy of the complex.

Generalized Unsöld Theorem: The sum Σ | φkj |2 over all functions which form a basis for a degenerate irreducible representation is invariant under all operations of the group.

This means, among other things, that closed electronic shells have spherical symmetry, i.e. a closed shell state forms the basis for the totally symmetric irreducible representation.

Open shells have several ways of arranging the electrons in states that have identical energy, hence they can be degenerate. Then the Jahn-Teller effect may apply - energy is lowered when the symmetry is reduced!

energy

T1 E

A1

cubic symmetry (O) trigonal distortion D3h

(we ignore spin here - maximum of three electrons in a p level)

As soon as we have more than one center, we lose the spherical symmetry of the atom. We want to construct orthonormal orbitals from the atomic states which have the symmetry of the molecule -> a linear combination of atomic orbitals “LCAO”.

Consider the classical case of the H2+ - molecule. We have two atomic orbitals | a > and | b > on atoms Haand Hb. There are two combinations, Ψg and Ψu:

3.2 Molecular Orbitals - basics

Ψg= ca|a> + cb|b>, ca and cb constants. Symmetry requires that |ca|= |cb|. Normalisation requires that 1 = <Ψ|Ψ>dτ = (ca< a | + cb< b |)(ca| a > + cb| b >) = 2|ca|2± 2|cb|2< a | b >, since the |a > and | b > are normalised and real.

The integral Sab= < a | b > is the overlap integral between | a > and | b >; it is a measure for “not being orthogonal”. The wave functions Ψgand Ψuare then

$g= (|a> + |b>) /( Sab)

Ha

Hb

$u $g

Ha

$u= (|a> - |b>) /( Sab)

Hb

The $’s will have different energies, $u will be higher - less charge between the protons. We thus have two molecular orbitals of different symmetry.

The discussion of a linear molecule gives us the opportunity to discuss the character tables of cyclic groups. Cyclic groups are those that are formed by the repetitive application of one element A, e.g. a rotation by 120° i.e. C3. This then gives rise to the

group consisting of C3, C32, and C3

3 = E. All cyclic groups are Abelian. In a cyclic group

every element is in a class by itself, since a similarity transformation does not create a new element.

Hence there are as many 1-dimensional irreducible representations as there are group elements. Since Ah= E, !(Ah) = 1 and thus %(!m(A)) = e2!im/h with m = 1,2,...,h. For other operations %(!m(Ap)) = e2!imp/h. Take as an example the group C5:

With & = e2!im/5

&25=1

&20=1

&15=1

&10=1

&5=1

&20=1 &15=1 &10=1 &5=1 !5

&16 &12 &8 &4 !4

&12 &9 &6 &3 !3

&8 &6 &4 &2 !2

&4 &3 &2 & !1

C54 C5

3 C52 C5 C5

C55=E

43

We note, however, that the two & having exponents add up to 5 are complex conjugates to one another. For example, (&4)*= (cos(2'4/5)+i sin(2'4/5) )* = cos(2' 4/5) -i sin(2'4/5) = cos(2'1/5) +i sin(2'1/5) = &. We then rearrange the table, replacing &3 by (&2)( and &4 by &", to obtain

Some representations are associated in pairs, such that the elements of one row are the complex conjugates of the other. This is useful for certain physical applications.

C5 E=C55 C5

1 C52 C5

3 C54

A 1 1 1 1 1 !5

1 & &2 &2! &! !1

1 &! &2! &2 & !4

1 & &( &2 & !2

1 &2( & &( &2 !3

E1

E2

3.3 Symmetry of LCAO orbitals

Using the p orbitals of the carbon atom, we want to construct LCAO molecular orbitals that have the symmetry of a hydrocarbon molecule. We will analyze trigonal and tetrahedral bonds, and find suitable directed orbitals which transform into themselves under the symmetry operations of the molecule, i.e. which form the basis for an irreducible representation of the point group of the molecule.

2px 3px

Kinds of interaction:" type : symmetric with respect to internuclear axis

' type antisymmetric with respect to reflection in a plane containing the internuclear axis

Bonding and antibonding combination of atomic orbitals

Overlap integral S

nonbonding S = 0

bonding S > 0

antibonding S < 0

Example: CH4

A

B

CD

apply symmetry operations and write down which atoms transforms into which ones

new way of determining matrix representations:

%(E)= 4

%(C3)= 1

%(C2)= 0

%(S4)= 0

%("d)= 2

We note that the character is derived from those orbitals that transform into + or - themselves; hence it is easy to determine % without writing down the matrices

!!!!!

"

#

$$$$$

%

&

=

!!!!!

"

#

$$$$$

%

&

!!!!!

"

#

$$$$$

%

&

D

C

B

A

D

C

B

A

1000

0100

0010

0001

!!!!!

"

#

$$$$$

%

&

=

!!!!!

"

#

$$$$$

%

&

!!!!!

"

#

$$$$$

%

&

C

B

D

A

D

C

B

A

0100

0010

1000

0001

!!!!!

"

#

$$$$$

%

&

=

!!!!!

"

#

$$$$$

%

&

!!!!!

"

#

$$$$$

%

&

C

D

A

B

D

C

B

A

0100

1000

0001

0010

!!!!!

"

#

$$$$$

%

&

=

!!!!!

"

#

$$$$$

%

&

!!!!!

"

#

$$$$$

%

&

A

B

D

C

D

C

B

A

0001

0010

1000

0100

!!!!!

"

#

$$$$$

%

&

=

!!!!!

"

#

$$$$$

%

&

!!!!!

"

#

$$$$$

%

&

C

D

B

A

D

C

B

A

0100

1000

0010

0001

the matrix indicates which atoms (or s wave functions ) are transformed

EE

C3

C2

S4

!d

number of atoms that stay in place

4

1

0

0

2

Compare to characters for Td point group

Td E 8C3 3C2 6S4 6sd

A1 1 1 1 1 1A2 1 1 1 -1 -1E 2 -1 2 0 0T1 3 0 -1 1 -1T2 3 0 -1 -1 1!red 4 1 0 0 2 !red = A1 + T2

Looking at the basis functions for A1 and T2, we see that basis functions which transform as A1 and T2 are those for which the s and px,y,z atomic orbitals form a basis. This is called an sp3 hybrid orbital.

So, taking one function that transforms as the A1 irr. rep. and three functions that together form the T2 irr. rep., we can form a linear combination function that has a shape directed along the four tetrahedral directions.

Example: cyclopropenyl

Let us consider a planar trigonal molecule with symmetry D3h, e.g. the cyclopropenyl group. There are three sigma type orbitals, i.e. the bonds between the carbon atoms and the hydrogen atoms, and six p-type orbitals, three in plane, three out of plane.

" orbitals

In-plane ' orbitals

+

+

+ -

-

-

Out-of-plane ' orbitals

+

- - -

+ +

The s orbitals transform into themselves under E, i.e. %(E) = 3, and so do they for "h.

For other opreations one can calculate this as follows: the transformation matrix transforms the m.o. into itself or a linear combination of its partners. But the trace (character) only shows whether it transforms into + or - itself. E.g.

Thus we can see by “direct inspection” what the character of the (reducible) trans-formation matrix is for each group of orbitals. For the ' orbitals %(E) = 6, %("h) =

0, because the out-of-plane ones go into minus themselves, the in-plane ones go into themselves (in simple language...). The character table then looks like this:

" orbitals

so for cyclopropenyl the " and the ! type reducible representation have the above characters

If we decompose the reducible representation for the s orbitals we find that "= A1’ + E’. From a look at the basis functions listed in the character table we see that the following combinations are possible:

sp2: s + (px,py)

sd2: s + (dxy, dx2

-y2)

dp2: dz2+ (px,py)

d3 : dz2+ (dxy, dx

2-y

2)

These are the basis functions for the respective irr.reps. from the character tables.

However, for an analysis of more complex molecular orbitals, we need a systematic approach. So as usual, after this intuitive example, we do it properly using a formal approach.

Consider a degenerate representation, i.e. l > 1. We can then assign each basis function to one row of the representation matrix, and need two indices for the basis function, one for the irreducible representation and the other for the row of the matrix as follows: We then have the

Theorem: Functions which are partners of an irreducible representation, i.e. which together form a basis for the irreducible representation, are orthogonal if they belong to different rows of the same unitary representation matrix. Example: px and py as basis functions for E’ in D3h

dxz, dyz as basis functions for E’’ of D3h

We can describe such a function )j by a linear combination of its partners when operating on it, We multiply by and sum over R:

3.4 Projection and transfer operators

Qijn #j = 0 if # j does not belong to the i-th row of the n-th irr.rep. !n

Qijn # j= # i if # j does belong to the i-th row of the n-th irr.rep. !n

In other words,if one has one basis function, one can generate all the others by applying Qn

ij! I.e. one transfers one basis function into the other, hence the name.

Then, using the GOT, we obtain

Qijn to a basis function, the following occurs:

We call the operator a transfer operator. If we apply

There is a version of the transfer operator which only uses characters:

Projection operators

Theorem: Let !1, !2, !3, ... !n be all irreducible representations of a group. Then any function in the vector space of the group can be decomposed into a sum

where the #in are basis functions which

belong to the i-th row of the n-th irreducible representation.

If we set i=j in the definition of the transfer operator and apply Qiin to F, then

i.e. Qiin is an operator which projects out that part of a

function that belongs to the i-th row of the n-th irred. representation

QiinF = #i

n

By the way, Qiin is idempotent: Qii

n• Qiin= Qii

n

Example for the application of a projection operator

We obtain, as usual, a simplification if we consider just the characters, not the representation matrices themselves: y

x

Let us project out a basis function for an irreducible representation of the group C3v, from the function f(x,y,z) = xz + yz + z2. For this we need the transformation properties of functions under the symmetry operations of the group.

z

z

z

z

z

z

z

0.5[(-1+ )xz + (1 + )yz]+z2 0.5(-y+ x) 0.5(-x+ y) "v’’

0.5[-(1+ )xz + (1 - )yz]+z2 0.5(-y+ x) 0.5(-x+ y) "v’’

xz - yz + z2 -x "v’

0.5[(-1- )xz + ( -1)yz]+z2 0.5(-y+ x) 0.5(-x- y) C32

0.5[(-1+ )xz - ( +1)yz]+z2 0.5(-y+ x) 0.5(-x+ y) C31

xz + yz + z2 y x E

xz + yz + z2 y x

Obtain the characters for the transformation matrices for this set of orbitals by considering which orbital transforms into + or - itself (as before). The E operation has the identity matrix as always, hence the character is 6 (6 atomic orbitals). Applying C6 we obtain hence %(C6) = 0.

A similar reasoning can be applied to all the other operations:The entry in the main diagonal of the matrix is +1 if the orbital is unchanged upon execution of the symmetry operation, 0 if the orbital moves to another site, and -1 if the parity of the orbital is inverted(for p and d orbitals)

Example: benzene, C6H6 under C6 symmetry.

+

-

a

bc

d

ef

!

C6

abcdef

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

=

0 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 11 0 0 0 0 0

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

abcdef

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

=

bcdefa

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

E.g. %(C2’) = 2, since two orbitals remain in the same position A reducible representation for the basis that is formed by the six s-type orbitals as follows:

We reduce this representation to find that !s= A1+ B1+ E1+ E2, i.e. the reducible representation can be decomposed into two nondegenerate and two doubly degenerate representations. The basis functions for these are formed from the atomic orbitals on the atomic sites a…f

D6 E 2C6 2C3 3C2ʼ C2 3C2ʼʼ

!s 6 0 0 0 2 0

!

C6

abcdef

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

=

0 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 11 0 0 0 0 0

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

abcdef

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

=

bcdefa

"

#

$ $ $ $ $ $ $

%

&

' ' ' ' ' ' '

Now we apply the projection operator to find the functions that form a basis for the irreducible representations. We apply it to one of the basis functions, i.e. an orbital on lattice site a. The projection operator will include all functions that form a basis for the respective irr. rep.

For the irr. rep. B1 we apply the projection operator again to orbital a

This M.O.has alternating signs on adjacent atom sites.

For the irr. rep. E1 we need two basis functions, since this representation is twofold degenerate. They can be obtained by using the projection operator on atomic orbitals a and b separately. The resulting molecular orbitals then need to be orthonormalized. The result is

Similarly, for the E2 representation we obtain

Since all wave functions belong to different irreducible representations, or to different rows of the same irreducible representation, their overlap integrals are all zero

The wave functions of the p molecular orbitals made up out of the pz atomic orbitals on each carbon atom site in the benzene ring are thus

The energy scale comes from simple Hückel molecular orbital theory, assuming only nearest neighbour interaction.

A

E1

E2

B-2β

-β

β

2β

0

1 2 3 4 5 6

B

E2

E1

A

energy levels and eigenfunctions of benzene

Correlation tables We have used the reduction formula several times in order to reduce a reducible representation to the irreducible representations which it contains.

In the case of symmetry reduction (e.g. when a symmetry operation is no longer valid because of a new ligand being added), we need to do the same thing: relate an irreducible representation in the higher group, which now is reducible in the lower one, to its new irreducible representation in the lower point group.

http://symmetry.jacobs-university.de/