5 Representasi Redusibel Dan Iredusibel

8/12/2019 5 Representasi Redusibel Dan Iredusibel

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Page 1

S

O O

z2

y2

x2

Generating Reducible Representations

x1

xs y1

y

s

zs

z1

sxz

For the symmetry operationsxz (asv )

x1x2 x2x1 xsxs

y1-y2 y2-y1 ys-ys

z1z2 z2z1 zszs


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s

s

s

s

s

s

s

y

s

xz

z

yx

z

y

x

z

y

x

z

yx

z

y

x

z

y

x

z

yx

z

y

x

z

y

x

1

1

1

2

2

2

2

2

2

1

1

1

2

2

2

1

1

1

)( .

100000000

010000000001000000

000000100

000000010

000000001

000100000

000010000

000001000

s

In matrix form


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Page 3

s

s

s

s

s

s

s

y

s

xz

z

yx

z

y

xz

y

x

z

yx

z

y

xz

y

x

z

yx

z

y

xz

y

x

1

1

1

2

2

2

2

2

2

1

1

1

2

2

2

1

1

1

)( .

100000000

010000000001000000

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000000001000100000

000010000

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s

Only require the characters: The sum of diagonal elements

For s(xz) c= + 1


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s

s

s

s

s

s

s

y

s

yz

z

y

x

z

y

x

zy

x

z

y

x

z

y

x

zy

x

z

y

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z

y

x

zy

x

2

2

2

1

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1

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1

2

2

2

1

1

1

)( .

100000000

010000000

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000100000

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000000100000000010

000000001

s

For s(yz) c= + 3


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Page 5

s

s

s

s

s

s

s

y

s

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

z

y

x

E

2

2

2

1

1

1

2

2

2

1

1

1

2

2

2

1

1

1

.

100000000

010000000

001000000

000100000

000010000

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000000001

For E c= + 9


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s

s

s

s

s

s

s

y

s

zy

x

z

yx

z

y

x

zy

x

z

yx

z

y

x

zy

x

z

yx

z

y

x

C

1

1

1

2

2

2

2

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2

1

1

1

2

2

2

1

1

1

2 .

100000000010000000

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.

For C2 c= -1


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Page 7


C2v

G3n

E C2 s(xz) s(yz)

+9 -1 +1 3

Summarising we get that G3n for this molecule is:

C2v E C2 s(xz) s(yz)

A1 +1 +1 +1 +1 Tz x2, y2, z2

A2 +1 +1 -1 -1 Rz xy

B1 +1 -1 +1 -1 Tx , Rx xz

B2 +1 -1 -1 +1 Ty,Ry yz

To reduce this we need the character table for the point groups


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Page 8

Reducing Reducible Representations

We need to use the reduction formula: RRnga pRRp cc ).(.

1

Where apis the number of times the irreducible representation, p,occurs in any reducible representation.

gis the number of symmetry operations in the group

(R)is character of the reduciblerepresentation

p(R)is character of the irreduciblerepresentation

nRis the number of operations in the class


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Page 9

C2v 1E 1C2 1s(xz) 1s(yz)

A1 +1 +1 +1 +1 Tz x2, y2, z2

A2 +1 +1 -1 -1 Rz xy

B1 +1 -1 +1 -1 Tx , Rx xz

B2 +1 -1 -1 +1 Ty,Ry yz

C2vG3n

E C2s(xz) s(yz)

+9 -1 +1 3

For C2v; g = 4 and nR= 1 for all operations


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Page 10

aA1= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3

RRng

a pR

Rp cc ).(.1

C2v

G3n

E C2s(xz) s(yz)

+9 -1 +1 3

aA2= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x-1) + (1x3x-1)] = (4/4) =1

aB1= (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x1) + (1x3x-1)] = (8/4) =2

aB2= (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x-1) + (1x3x1)] = (12/4) =3

G3n= 3A1+ A2+ 2B1+ 3B2


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Page 11

aA1= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3

Unshifted atoms for Constructing Reducible Representations

The terms in blue represent contributions from the un-shiftedatoms

Onlythese actually contribute to the trace.

If we concentrate only on these un-shifted atoms we can

simplify the problem greatly.

For SO2 (9= 3x 3) ( -1= 1x1) (1= 1x 1) and ( 3= 3x 1)

Number of un-shifted atoms Contr ibution f rom these atoms

E C2 sxz syzRR : 9 -1 1 3


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Page 12

Identity E

E

z

y

x

z

y

x

.

100

010

001

1

1

1

For each un-shifted atom

c(E) = +3

z

y

x

z1

y1

x1


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Page 13

Inversion i

z

y

xz1

y1

x1

i


c(i) = -3

z

y

x

z

y

x

.

100

010

001

1

1

1


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Page 14


z1

y1

x1x

z

y

s(xz)

Reflection s(xz) (Others are same except location of 1 changes)

c(s(xz)) = +1

z

y

x

z

y

x

.

100

010

001

1

1

1


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Page 15

q

q

360/n

x1y1

z1z

y

x

Cn

Rotation Cn

z

y

x

nn

nn

z

y

x

.

100

0360

cos360

sin

0360

sin360

cos

1

1

1

c(Cn) = 1 + 2.cos(360/n)


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Page 16

z

y

x

nn

nn

z

y

x

.

100

0360

cos360

sin

0360

sin360

cos

1

1

1

c(Sn) = -1 + 2.cos(360/n)

Improper rotation axis, Sn

Cn

s(xy)z

y

x

z

yxy1

x1

z1


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Page 17

Summary of contributions from un-shifted atoms to G3n

R (R)

E +3

i -3

s +1

1+ 2.cos(360/n) C2 -1

1+ 2.cos(360/n) C3 ,C32 0

1+ 2.cos(360/n) C4, C43 +1

-1 + 2.cos(360/n) S31,S3

2 -2

-1 + 2.cos(360/n) S41,S4

2 -1

-1 + 2.cos(360/n) S61,S6

5 0


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Page 18

P

O

ClCl

Cl

Worked example: POCl3(C3vpoint group)

R c(R)

E

sv

2C3

+3

+1

0

C3v E 3sv

A1

A2

E

1 1 1

1 1 -1

2 -1 0

C3

Un-shifted

atoms

Contribution

3n

5 2 3

3 0 1

15 0 3

Number of classes,(1 + 2 + 3 = 6)

Order of the group,

g = 6


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Page 19

Reducing the irreducible representation for POCl3

RRng

a pR

Rp cc ).(.1

2C3C3vE 3sv

G3n 15 0 3

a(A1) = 1/6[(1x 15x1) + (2 x 0 x 1) + (3 x 3x 1)] = 1/6 [15 + 0+ 9] = 4

a(A2) = 1/6[(1 x 15 x 1) + ( 2 x 0 x 1) + (3 x 3x1)] = 1/6 [15 + 0 -9] = 1

a(E) = 1/6[ (1 x 15 x 2) + (2 x 0 x 1) + (3 x 3 x 0)] = 1/6[30 + 0 + 0 ] =5

3n= 4A1+ A2+ 5E

For POCl3n= 5 therefore the number of degrees of freedom is 3n =15.

E is doubly degenerate so 3nhas 15 degrees of freedom.

5 Representasi Redusibel Dan Iredusibel

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