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
a)
X
d)
!
g)
X
X
b)
X
e)
c)
X
Y
X
h)
X
f)
X
X
X
X
X
k)
X
X
X
X
l)
X
X
X
X
X
X
X
Lets say we have a reducible representation !red in C3v which is given below:
C3v
E
2C3
3"v
!1
1
1
1
!2
1
1
-1
!3
2
-1
0
!red
6
0
2
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
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.
!1
!3
E
Intermission: cyclobutane: Cy-butane-1.cmmf
2C3
!
2
!red= 2!1+ 2 !3
Exercise: decompose the following reducible representations
Exercise: Prove, from theorems based on the GOT, that all
irreducible representations of Abelian groups are onedimensional.
Exercise: In the point group D3h, find the following direct
products and identify the irreducible representations that
they contain:
A1ʼ ⊗ A2ʼ
A1ʼ ⊗ A2ʼ
Eʼ⊗ Eʼʼ
Eʼʼ⊗ Eʼʼ
Direct product of matrix representations
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, Kronecker or outer product
"b b %
B = $ 11 12 '
# b21 b22 &
" a11 a12 %
A = $
'
# a21 a22 &
Consider a group G with irreducible
representations
!
# a11b11
%
# a11B a12 B &
a11b21
A " B = C = %
( = %
%
a
B
a
B
a21b11
$ 21
22 '
%
$
!
a11b12
a11b22
a12b12 &
(
(
(
(
...etc.
'
a12b11
a12b21
!
Theorem: The direct (outer) product of two irreducible representations A and B of a group G,
or
# a11b11
%
# a11B a12 B &
a11b21
A " B = C = %
( = %
% a21b11
$ a21B a22 B'
%
$
a11b12
a11b22
a12b11
a12b21
cij,kl= aikbjl
is also a (generally reducible) representation of the group.
!
As always, there is a simplification for the characters:
Theorem: The character of the direct product representation matrix is equal to the
product of the characters of the separate irreducible representation matrices
a12b12 &
(
(
(
(
'
C4v
E
C2
2C4
2"v
2"d
A1
1
1
1
1
1
A2
1
1
1
-1
-1
B1
1
1
-1
1
-1
B2
1
1
-1
-1
1
E
2
-2
0
0
0
A1*A2
1
1
1
-1
-1
B1*E
2
-2
0
0
0
A1*E*B2
2
-2
0
0
0
E*E
4
4
0
0
0
The general decomposition of
a product representation can
be obtained through the
decomposition formula
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
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.
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.
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
Rotations
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"d
Simple
functions
A2
1
1
1
-1
-1
Rz
Higher
functions
same reasoning, applied to dx,y orbital
same reasoning, applied to dz2 and dx2- y2 orbital
Higher functions
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"d
Simple
functions
Higher
functions
B1
1
1
-1
1
-1
z
x2 - y2
2-dimensional representations
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.
C4
$
2"
& cos 4
&
2"
= & #sin
4
&
&& 0
%
2"
4
2"
cos
4
0
sin
'
0)
)
0)
)
1)
)
(
So a function f(x,y,z) = z transforms into itself under the
“matrices” of the irreducible representation A1.
!
C4v
E
C2
2C4
2"v
2"d
A1
1
1
1
1
1
A2
1
1
1
-1
-1
B1
1
1
-1
1
-1
B2
1
1
-1
-1
1
E
2
-2
0
0
0
Simple
functions
Higher
functions
z
(x,y)
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:
E
C4
1 0
C2
0 1
cos 2π/4
sin 2π/4
-sin 2π/4
cos 2π/4
σv(xz)
σd(xz)
1 0
0 -1
0 1
1 0
=
cos π
sin π
-sin π
cos π
0 1
=
10
01
so C4 takes x into y and y into -x
-1 0
does not affect x but
changes y into -y
σv(yz)
takes x into y
and y into x
σdʼ(xz)
-1 0
0 1
0 -1
-1 0
does not affect y but
changes x into -x
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
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”.
C4v
E
C2
2C4
2"v
2"d
A1
1
1
1
1
1
A2
1
1
1
-1
-1
B1
1
1
-1
1
-1
B2
1
1
-1
-1
1
E
2
-2
0
0
0
Simple
functions
Higher
functions
z
(x,y)
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.
Formal derivation of basis functions for an irreducible representation
So far, symmetry operations represented by real orthogonal transformation
matrices R of coordinates
Since the matrix R is real and
also holds.
The operations R form a group. Now we introduce a new group: transformation
operators which affect functions, not coordinates ! These operators are called PR.
PR f (x) = f (R!1x)
and thus
PR f (Rx) = f (x)
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.
2.5 Relation between representation theory and quantum mechanics
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.
Degenerate eigenfunctions
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.
etc.
p orbitals in a 3-dimensional vector space, d orbitals in 5-dim space,
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 !
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.
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).
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.
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
2.6 Examples: symmetry of physical properties, tensor symmetries
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!).
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
r
r
P = "E
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).
!
" P1 %
" (11 (12 (13 %" E1 %
$ '
$
'$ '
P
=
(
(
(
2
21
22
23
$ '
$
'$ E 2 '
$ '
$
'$ '
# P31 &
#( 31 ( 32 ( 33 &# E 3 &
!
Alternative shorthand way of writing
Example: pyroelectricity
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!
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.
Example: polarizability tensor (2nd rank tensor):
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.
" Px %
"( xx ( xy ( xz %" E x %
$ '
$
'$ '
$ Py ' = $( yx ( yy ( yz '$ E y '
$(
'$ '
$P '
# zx ( zy ( zz &# E z &
# &
!
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 z
Hence 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 0 (
%0
0 " zz ('
$
3. Group theory, crystal field splitting and molecular orbitals
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.
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 ! 7dim. 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.
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.
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)
%e$il#
'
' 0
i
" (# ) = ' 0
'
' 0
'
& 0
!
0
e
$i(l$1)#
0
0
0
e$i( l$2)#
0
0
0
0
0 (
*
0 *
0 0 *
*
... 0 *
*
0 e il# )
0
0
And the character is just the sum over m
Theorem: The irreducible representations with characters
are the only odd-dimensional irreducible representations
of the full rotation group for integer l
O
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:
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,
!(E) = 2l + 1
!(C3) = !("(2#/3)) = 1 for l = 0,3
!(C4) = !("(#/2)) = 1 for l = 0,1,4,5
0 forl = 1,4 -1 for l = 2,5
-1 for l = 2,3,6,7
Since the spherical harmonics of order l form bases for the group
of all rotations, they also form them for groups with finite rotations!
irreducible representations of group O
O
E
8C3 3C2 6C2 6C4
1
1 ʼ 1
1
reducible representations of atomic
levels under octahedral symmetry
O
E 8C3 3C2 6C2ʼ 6C4
s
D0
1
1
1
1
1
-1
p
D1
3
0
-1
-1
1
0
0
d
D2
5
-1
1
1
-1
-1
-1
1
f
D3
7
1
-1
-1
-1
-1
1
-1
g
D4
9
0
1
1
1
A1
1
A2
1
1
1
-1
E
2
-1
2
T1
3
0
T2
3
0
atomic physics labels
Using the reduction formula, we can then decompose the higher (and
clearly reducible) representations into the irr. reps of the group O.
D0 A1
D1 T1
An s state cannot split; s states are
always totally symmetric
A p state does not split in a field of cubic
symmetry
D2 E + T2
A d state must split since there are no 5dim irr.representations in the O group
D3 A2 + T1 + T2
An f state splits into one nondegenerate
state and two 3-fold degenerate states
So we can read the
degree of degeneracy
of an atomic level in a
field of known
symmetry from the
character tables!
Crystal field splitting contʼd...
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.
O
E
8C3
6C2
A1
1
1
1
A2
1
1
-1
E
2
-1
0
T1
3
0
-1
T2
3
0
1
D3
E
A1
1
1
1
A2
1
1
-1
E
2
-1
0
Hence we arrive at the following correlation:
Atom
l= 0
s
A1
l=1
p
T1
A2
E
l=2
d
T2
E
A1
E
E
l=3
f
T2
A1
E
A2
E
A2
2C3 3C2
T1
A2
An example
cubic field with
trigonal distortion
cubic field
A1
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.
cubic symmetry (O)
energy
T1
trigonal distortion
D3h
E
A1
Open shells have several
ways of arranging the
electrons in states that
have identical energy,
hence they can be
degenerate. Then the JahnTeller effect may apply energy is lowered when the
symmetry is reduced!
(we ignore spin here maximum of three electrons in
a p level)
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.
3.2 Molecular Orbitals - basics
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:
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
$u= (|a> - |b>) /( Sab)
$g= (|a> + |b>) /( Sab)
$g
$u
Hb
Ha
Hb
Ha
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 C33 = 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:
C5
C55=E
C5
C52
C53
C54
!1
&5=1
&
&2
&3
&4
!2
&10=1
&2
&4
&6
&8
!3
&15=1
&3
&6
&9
&12
!4
&20=1
&4
&8
&12
&16
!5
&25=1
&5=1
&10=1
&15=1
&20=1
With & = e2!im/5
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
C5
A
E1
E2
E=C55 C51
C 52
C 53
C 54
1
1
1
1
1
!5
1
&
&2
&2!
&!
!1
1
&!
&2!
&2
&
!4
1
&
&(
&2
&
!2
1
&2(
&
&(
&2
!3
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.
43
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.
Kinds of interaction:
" type : symmetric with respect to
internuclear axis
' type antisymmetric with respect to
reflection in a plane containing the
internuclear axis
2px
3px
Bonding and antibonding combination of atomic orbitals
Overlap integral S
antibonding
S<0
bonding
S>0
nonbonding S = 0
Example: CH4
new way of determining
matrix
representations:
B
A
C
D
apply symmetry
operations and write
down which atoms
transforms into which
ones
the matrix indicates which atoms (or s wave functions ) are transformed
E
C3
C2
S4
σd
&1
$
$0
$0
$
$0
%
0 0 0 #& A # & A #
!$ ! $ !
1 0 0 !$ B ! $ B !
=
0 1 0 !$ C ! $ C !
!$ ! $ !
0 0 1 !"$% D !" $% D !"
%(E)= 4
&1
$
$0
$0
$
$0
%
0 0 0 #& A # & A #
!$ ! $ !
0 0 1 !$ B ! $ D !
=
1 0 0 !$ C ! $ B !
!$ ! $ !
0 1 0 !"$% D !" $% C !"
%(C3)= 1
&0
$
$1
$0
$
$0
%
1 0 0 #& A # & B #
!$ ! $ !
0 0 0 !$ B ! $ A !
=
0 0 1 !$ C ! $ D !
!$ ! $ !
0 1 0 !"$% D !" $% C !"
%(C2)= 0
&0
$
$0
$0
$
$1
%
0 1 0 #& A # & C #
!$ ! $ !
0 0 1 !$ B ! $ D !
=
1 0 0 !$ C ! $ B !
!$ ! $ !
0 0 0 !"$% D !" $% A !"
%(S4)= 0
&1
$
$0
$0
$
$0
%
0 0 0 #& A # & A #
!$ ! $ !
1 0 0 !$ B ! $ B !
=
0 0 1 !$ C ! $ D !
!$ ! $ !
0 1 0 !"$% D !" $% C !"
%("d)= 2
number of
atoms that
stay in place
4
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
1
0
0
2
Compare to characters for Td point group
Td
E
8C3
3C2
6S4
6sd
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
!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
Out-of-plane '
orbitals
+
+
+
In-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) transformation 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, dx2-y2)
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.
3.4 Projection and transfer operators
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,
over R:
We multiply by
and sum
Then, using the GOT, we obtain
We call the operator
a transfer operator. If we apply
Qijn to a basis function, the following occurs:
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 Qnij! I.e. one transfers one basis function into the
other, hence the name.
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
QiinF = #in
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
By the way, Qiin is idempotent: Qiin• Qiin= Qiin
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
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.
x
x
y
z
xz + yz + z2
E
x
y
z
xz + yz + z2
C 31
0.5(-x+
y)
0.5(-y+
x)
z
0.5[(-1+
)xz - (
+1)yz]+z2
C 32
0.5(-x-
y)
0.5(-y+
x)
z
0.5[(-1-
)xz + (
-1)yz]+z2
"v’
-x
z
xz - yz + z2
"v’’
0.5(-x+
y)
0.5(-y+
x)
z
0.5[-(1+
)xz + (1 -
)yz]+z2
"v’’
0.5(-x+
y)
0.5(-y+
x)
z
0.5[(-1+
)xz + (1 +
)yz]+z2
Example: benzene, C6H6 under C6 symmetry.
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.
c
d
!
a
e
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)
" a%
"0 1 0
$ '
$
$ b'
$0 0 1
$c'
$0 0 0
C6 $ ' = $
$ d'
$0 0 0
$e'
$0 0 0
$ '
$
#f&
#1 0 0
b
0
0
1
0
0
0
0
0
0
1
0
0
+
-
f
"b%
0%" a %
'$ '
$ '
0'$ b '
$c'
$d'
0'$ c '
'$ ' = $ '
0'$ d '
$e'
$f'
1'$ e '
'$ '
$ '
0&# f &
# a&
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:
D6
E
2C6
2C3
3C2ʼ
C2
3C2ʼʼ
!s
6
0
0
0
2
0
" a%
"0 1 0
$ '
$
$ b'
$0 0 1
$c'
$0 0 0
C6 $ ' = $
$ d'
$0 0 0
$e'
$0 0 0
$ '
$
#f&
#1 0 0
0
0
1
0
0
0
0
0
0
1
0
0
"b%
0%" a %
'$ '
$ '
0'$ b '
$c'
$d'
0'$ c '
'$ ' = $ '
0'$ d '
$e'
$f'
1'$ e '
'$ '
$ '
0&# f &
# a&
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
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
1 2 3 4 5 6
B
-2β
-β
B
E2
E2
0
β
2β
E1
E1
A
A
energy levels and eigenfunctions of benzene
The energy scale comes from simple Hückel molecular orbital theory,
assuming only nearest neighbour interaction.
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/