L4 - Reducible representations.pdf symmetry And group theory

Reducible representation is on symmetry and group theory

1. Symmetry and Group Theory
Reducible representations.

2. •h: order of the group (number of elements in the group)
•l
i : the dimension of i
th
irrep, (the order of each of the matrices which constitute it)
•R : various operations in the group
•
i(R)
mn : the element of the m
th
row and n
th
column of the matrix corresponding to an operation R in the i
th
irrep
•* : denotes a complex conjugate of 
i(R)
mn, if imaginary or complex numbers are involved.
The great orthogonality theorem (GOT)
Kronecker delta

cd is afunctionof
twovariables, usually just
non-negativeintegers. The
function is 1 if the
variables are equal, and 0
otherwise

ij
i  j (different irreps)

ij = 0 → RHS = 0,
→ vectors
orthogonal
i = j (same irrep)

ij = 1

mm’ 
nn’
m = m’ & n = n’
(same vector)

mm’ = 
nn’ = 1
Note: l > 0, so we can discard the absolute
value sign
Also: mn describes one irrep, therefore 
mm’
& 
nn’ will always return the same value(0,0
or 1,1 but never 1,0 or 0,1)
m  m’ & n  n’

mm’ = 
nn’ = 0 → RHS = 0,
→ vectors orthogonal

3. 4. The sum of the squares of the
characters in any irrep equals h
5. The vectors whose components are the characters of the two different
irreducible representations are orthogonal.
1. In a given representation (reducible or irreducible) the characters of all
matrices belonging to operations in the same class are identical.
2. The number of irreps of a group is equal to the number of classes in the group.
3. The sum of squares of the dimensions (l) of the irreducible
representations of a group is equal to the order of the group (h)
GOT
5 rules
The great orthogonality theorem (GOT)

4. C
3V E C
3
1
C
3
2

v (1)
v (2)
v (3)
A
1 1 1 1 1 1 1 T
z
A
2 1 1 1 -1 -1 -1 R
z
E 2 -1 -1 0 0 0 (Tx, Ty)
(Rx,Ry)
C
3V E 2C
3 3
v
A
1 1 1 1 T
z
A
2 1 1 -1 R
z
E 2 -1 0 (Tx, Ty)
(Rx,Ry)
Property 1: The characters of each element belonging to the same class in
any irreducible representation are identical to each other
Properties of irreducible representationsc
3g=
Example: C
3v Group

5. C
3V E 2C
3 3
v
A
1 1 1 1 T
z
A
2 1 1 -1 R
z
E 2 -1 0 (Tx, Ty)
(Rx,Ry)
Property 2: . The number of irreducible representations of a group equals
the number of classes
Properties of irreducible representations
Example: C
3v Group

6. C
3V E 2C
3 3
v
A
1 1 1 1 T
z
A
2 1 1 -1 R
z
E 2 -1 0 (Tx, Ty)
(Rx,Ry)
Property 3: The sum of the squares of the dimensions of the irreducible
representations of a group equals its order:
Properties of irreducible representations
Example: C
3v Group()
2
2
ii
ii
l Eh= = () () ()
2 2 2
1 2 3
6AE E E  + + =
 2 2 2
1126++=
The order of the group is the
total number of operations
that are in the groupc
c
6hg==

7. C
3V E 2C
3 3
v
A
1 1 1 1 T
z
A
2 1 1 -1 R
z
E 2 -1 0 (Tx, Ty)
(Rx,Ry)
Property 4: The sum of the squares of the characters of any irreducible
representation of a group equals its order:
Properties of irreducible representations
Example: C
3v Group()
2
c
c
i
gch=
 () () ()
2 2 2
3
2 3 6
v
EC  + + =  
   
22
2
E:221306+−+=

8. C
3V E 2C
3 3
v
A
1 1 1 1 T
z
A
2 1 1 -1 R
z
E 2 -1 0 (Tx, Ty)
(Rx,Ry)
Property 5: The characters of two different irreducible representations are
orthogonal to each other:
Properties of irreducible representations
Example: C
3v Group()()
c
c
0 when
ij
gcc ij =  ()() ()() ()()
1 1 2 1 3 3 3 1 3
: 2 3 0
vv
AEEE CC  + + =  
   ()() ()() ()()
1
:122113100AE + −+ = 
 

9. G.O.T consequences
Example: The C
3v Character Table
C
3V E 2C
3 3
v
Γ
1 1 1 1
Γ
2
Γ
3
1. The characters of each element belonging to the same class in any irreducible
representation are identical to each other
2. The number of irreducible representations of a group equals the number of classes
E, C
3, C
3
2
, σ
v1,σ
v2,
σ
v3

10. G.O.T consequences
C
3V E 2C
3 3
v
Γ
1 1 1 1
Γ
2 1
Γ
3 2
3. The sum of the squares of the dimensions of the irreducible representations of a group
equals its order:()
2
2
ii
ii
l Eh= = () () ()
2 2 2
1 2 3
6E E E   + + =
 2 2 2
1126++=
Example: The C
3v Character Table

11. G.O.T consequences
C
3V E 2C
3 3
v
Γ
1 1 1 1
Γ
2 1
Γ
3 2 ±1 0
4. The sum of the squares of the characters of any irreducible representation of a group
equals its order: ()
2
i
R
Rh=
 () () ()
2 2 2
3
2 3 6
v
EC  + + =  
   
22
2
3
:221306++=
Example: The C
3v Character Table

12. G.O.T consequences
C
3V E 2C
3 3
v
Γ
1 1 1 1
Γ
2 1 1 -1
Γ
3 2 -1 0
5. The characters of two different irreducible representations are orthogonal to each
other:()()0 when
ij
R
RR ij =  ()() ()() ()()
13 1 2 1 3 3 3 1 3
: 2 3 0
vv
EE CC   + + =   
   ()() ()() ()()
13
:122113100 + −+ = 
 
Example: The C
3v Character Table

13. C
3vE 2C
3(z)3
v

1
+1 +1 +1

2
+1 +1 -1

3
+2 -1 0
l
1
2
+ l
2
2
+ l
3
2
= h = 6
# of irreps = equal # of classes
in the group
l
1

= l
2

= 1, l
3

= 2

1 Fully symmetric irrep
Entries under E operation: 1,1,2

2: only 1/-1(half must be
-ve to be orthogonal with 
2 (1
combination possible)

3: we only know the entry under
E(2), so solve equations
Individual irreps (mutually orthogonal)
Check if the vectors are mutually
orthogonal and of equal length
1
2
+ 2(1)
2
+ 3(1)
2
= h = 6
2
2
+ 2(-1)
2
+ 3(0)
2
= h = 6

14. Mulliken Symbols
C
3V E 2C
3 3
v
Γ
1 1 1 1
Γ
2 1 1 -1
Γ
3 2 -1 0
C
3V E 2C
3 3
v
A
1 1 1 1
A
2 1 1 -1
E 2 -1 0
Example: The C
3v Character Table

15. C
3V E 2C
3 3
v
A
1 1 1 1
A
2 1 1 -1
E 2 -1 0
E, C
3, C
3
2
, σ
v1,σ
v2,
σ
v3
Linear vectors
z
x
y
z
x
y
z
x
y0
0
z




 0
0
x



 0
0
y





Z – vector x – vector y – vector
Example: The C
3v Character Table

16. The C
3v Character Table
C
3V E 2C
3 3
v
A
1 1 1 1
A
2 1 1 -1
E 2 -1 0
E, C
3, C
3
2
, σ
v1,σ
v2,
σ
v3
Linear vectorsxx
Ryy
zz


=


 100
010
001
E


=


 1
100
010
001
v



=−


 3
12320
32120
0 01
C
−−

=−


Form representations for the x,y and z vectors
-Generate Γ
xyz
-Block diagonalize to form individual representations

17. When we take x, y, and
z vectors as a basis for
operations, we can
immediately see if the
resulting 3 by 3
matrices are in the
block diagonal form or
not.
Block diagonal matrix : is a
squarediagonal matrixin which
thediagonalelements are
squarematricesof any size and the
off-diagonalelements are 0.

18. C
3V E 2C
3 3
v
Linear functions
and rotations
A
1 1 1 1 z
A
2 1 1 -1
E 2 -1 0 (x,y)210E
xy
=−= 1
111A
z
= =

19. 210E
xy
=−= 2
111A
z
= −= C
3VE2C
33
v
Linear functions and rotations
A
111 1 z
A
211-1 R
z
E2-10 (x,y) (R
x,R
y)

20. Reducible representations
•Point group: C
2V
C
2v example

21. •Point group: C
2V
Reducible representations
C
2v example

22. If an operation moves an atom to a symmetry-equivalent position, then all basis vectors to
do with that atom give rise to only off-diagonal elements in the transformation matrix and so
contribute zero to the character for the operation.
Reducible representations
C
2v example

23. C
2v example
’
v (yz)
v (xz)
C
2OO
OO
OO
= 1
= -1
= 1
Total = 1
xx
yy
zz




→
→−
→
C
2:

v (xz)OO
OO
OO
O
= -1
= 1
= 1
Total = 1
xx
yy
zz




→−
→
→ H1 H1
H1 H1
H1 H1
H1
= -1
= 1
= 1
Total = 1
xx
yy
zz




→−
→
→ H2 H2
H2 H2
H2 H2
H2
= -1
= 1
= 1
Total = 1
xx
yy
zz




→−
→
→
’
v (yz)()3
v
= ()
2
1 =−C ()9E=
Reducible representations911 3
xyz
=−

24. C
2V E C
2 
v (xz) ’
v (yz)
No. of stationary atoms 3 1 1 3

xyz
3 -1 1 1

3N
9 -1 1 3
C
2v example
’
v (yz)
v (xz)
C
2100
010
001
−− 
 
− =−
 
 
 
xx
yy
zz
C
2:()3=E ()
2
1 =−C ()1=
v ()1=
v
Reducible representations

25. C
2V E C
2 
v (xz) ’
v (yz)
No. of stationary atoms 3 1 1 3

xyz
3 -1 1 1

3N
9 -1 1 3
’
v (yz)
v (xz)
C
2
Use the following equation to calculate the contribution to the character of the
representation:

xyz = 2cosθ ± 1
θ angle of rotation
+ For proper rotation
-For improper rotation
n
R = No. of unshifted atoms
∴ 
3N = n
R(2cosθ ± 1)
C
2v example

27. The reduction formula

28. C
2VEC
2
v (xz)’
v (yz)

3N
9-1 1 3
C
2VEC
2
v (xz)’
v (yz)
A
111 1 1 zx
2
,y
2
,z
2
A
211 -1 -1 R
z xy
B
11-1 1 -1 x, R
y xz
B
21-1 -1 1 y, R
x yz( ) ( ) n
1
order
# of operations in class(character of RR)character of X
X=   ()()()()()()()()()()()()
1
A
3
1
n1 9 1 1 111 1111
4
3= + + + =

− ()()()()()()()()()()()()
1
B
2
1
n1 9 1 1 111 111 1 3
4
= + −+ + −=

− ()()()()()()()()()()()()
2
B
311
1
n1 9 1 11 11 1 1 31
4
= + + + −

− =

− ()()()()()()()()()()()()
2
A
1
1
n 1111 19 1 1 111
4
3= + + + −

− −=

∴ 
3N = 3 A
1 + 1A
2 + 2B
1 + 3B
2
C
2v example