7 Exercise - Introduction to Group Theory
7.1 (dipole-allowed transitions for benzene located on an unstructured sur-
face)
Westartwith
C 6symmetry,withthecharactertable:
C 6 E C 6 C 3 C 2 (C 3 ) 2 (C 6 ) 5 linfct quadfct cubfct
A 1 1 1 1 1 1 z x 2 + y 2 , z 2 z 3 , z x 2 + y 2
B 1 − 1 1 − 1 1 − 1 − − y 3x 2 − y 2
, x x 2 − 3y 2 E 1
1 1
ε ε ∗
− ε ∗
− ε
− 1
− 1
− ε
− ε ∗
ε ∗ ε
x + iy
x − iy xz, yz xz 2 , yz 2 x x 2 + y 2
, y x 2 + y 2 E 2 1
1
−ε ∗
−ε
−ε
−ε ∗ 1 1
−ε ∗
−ε
−ε
−ε ∗ − x 2 − y 2 , xy xyz, z x 2 − y 2
with
ε = exp (2πi/6)
weget:ε + ε ∗ = 2 cos π 3
= 1
− ε − ε ∗ = − 2 cos π 3
= − 1
therefore:
C 6 E C 6 C 3 C 2 (C 3 ) 2 (C 6 ) 5
A 1 1 1 1 1 1
B 1 −1 1 −1 1 −1
E 1 2 1 − 1 − 2 − 1 1 E 2 2 − 1 − 1 2 − 1 − 1
Wecannowhavealookatthedipole allowedtransitions(atrstforthe
z
-component):A A A A
B B A A
E 1 E 1 A A
E 2 E 2 A A
B A A B
A B A B
E 2 E 1 A B
E 1 E 2 A B
E 1 A A E 1
E 2 B A E 1
2A + E 2 E 1 A E 1
2B + E 1 E 2 A E 1
E 2 A A E 2
E 1 B A E 2
2B + E 1 E 1 A E 2
2A + E 2 E 2 A E 2
nowfor
x, y
:totalmatrixelement nalstate operator initialstate
E 1 A E 1 A
E 2 B E 1 A
2A + E 2 E 1 E 1 A
2B + E 1 E 2 E 1 A
E 2 A E 1 B
E 1 B E 1 B
2B + E 1 E 1 E 1 B
2A + E 2 E 2 E 1 B
2A + E 2 A E 1 E 1
2B + E 1 B E 1 E 1
2B + 3E 1 E 1 E 1 E 1
2A + 3E 2 E 2 E 1 E 1
2B + E 1 A E 1 E 2
2A + E 2 B E 1 E 2
2A + 3E 2 E 1 E 1 E 2
2B + 3E 1 E 2 E 1 E 2
Thismeansthedipole allowedtransitionsare:
A A A A
A B A B
2A + E 2 E 1 A E 1
2A + E 2 E 2 A E 2
2A + E 2 E 1 E 1 A
2A + E 2 E 2 E 1 B
2A + E 2 A E 1 E 1
2A + 3E 2 E 2 E 1 E 1
2A + E 2 B E 1 E 2
2A + 3E 2 E 1 E 1 E 2
whilethe rstfour are noreal transitions, while theypoint tothe sameorbital.Thereforethe only
realtransitions are:
E 1 ↔ A E 2 ↔ B E 1 ↔ E 2
allfor
E 1,meaningthex, y
dipoleoperator.Thisleadstothefollowingdipole allowedtransitions:
x, y, xz, yz, xz 2 , yz 2 , x x 2 + y 2
, y x 2 + y 2
↔ z, x 2 + y 2 , z 2 , z 3 , z x 2 + y 2 x 2 − y 2 , xy, xyz, z x 2 − y 2
↔ y 3x 2 − y 2
, x x 2 − 3y 2 x, y, xz, yz, xz 2 , yz 2 , x x 2 + y 2
, y x 2 + y 2
↔ x 2 − y 2 , xy, xyz, z x 2 − y 2
7.2 (benzene on Ni(100))
We nowplace the benzene with
C 6v ona Nickel surfacewith C 4v. Therefore the symmetryproperties
willchange.Werstwilltrytodeterminethenewpointgroup,whichhopefullyis
C 2v,wellhopefullywe
didn'tmiss asymmetryrelation.Thecharactertableisgivenwith:
C 2v E C 2 (z) σ v (xz) σ v (yz)
linfct quadfct cubfctA 1 1 1 1 1 z x 2 , y 2 , z 2 z 3 , x 2 z, y 2 z
A 2 1 1 − 1 − 1 − xy xyz
B 1 1 − 1 1 − 1 x xz xz 2 , x 3 , xy 2 B 2 1 − 1 − 1 1 y yz yz 2 , y 3 , x 2 y
z
A 1 A 1 A 1 A 1
A 2 A 2 A 1 A 1
B 1 B 1 A 1 A 1
B 2 B 2 A 1 A 1
A 2 A 1 A 1 A 2
A 1 A 2 A 1 A 2
B 2 B 1 A 1 A 2
B 1 B 2 A 1 A 2
B 1 A 1 A 1 B 1
B 2 A 2 A 1 B 1
A 1 B 1 A 1 B 1
A 2 B 2 A 1 B 1
B 2 A 1 A 1 B 2
B 1 A 2 A 1 B 2
A 2 B 1 A 1 B 2
A 1 B 2 A 1 B 2
nextfor(
y
):totalmatrixelement nalstate operator initialstate
B 2 A 1 B 2 A 1
B 1 A 2 B 2 A 1
A 2 B 1 B 2 A 1
A 1 B 2 B 2 A 1
B 1 A 1 B 2 A 2
B 2 A 2 B 2 A 2
A 1 B 1 B 2 A 2
A 2 B 2 B 2 A 2
A 2 A 1 B 2 B 1
A 1 A 2 B 2 B 1
B 2 B 1 B 2 B 1
B 1 B 2 B 2 B 1
A 1 A 1 B 2 B 2
A 2 A 2 B 2 B 2
B 1 B 1 B 2 B 2
B 2 B 2 B 2 B 2
andatlastfor(
x
):B 1 A 1 B 1 A 1
B 2 A 2 B 1 A 1
A 1 B 1 B 1 A 1
A 2 B 2 B 1 A 1
B 2 A 1 B 1 A 2
B 1 A 2 B 1 A 2
A 2 B 1 B 1 A 2
A 1 B 2 B 1 A 2
A 1 A 1 B 1 B 1
A 2 A 2 B 1 B 1
B 1 B 1 B 1 B 1
B 2 B 2 B 1 B 1
A 2 A 1 B 1 B 2
A 1 A 2 B 1 B 2
B 2 B 1 B 1 B 2
B 1 B 2 B 1 B 2
thereforewethis timegotthetransitions:
totalmatrixelement nalstate operator initialstate
A 1 A 1 A 1 A 1
A 1 A 2 A 1 A 2
A 1 B 1 A 1 B 1
A 1 B 2 A 1 B 2
A 1 B 2 B 2 A 1
A 1 B 1 B 2 A 2
A 1 A 2 B 2 B 1
A 1 A 1 B 2 B 2
A 1 B 1 B 1 A 1
A 1 B 2 B 1 A 2
A 1 A 1 B 1 B 1
A 1 A 2 B 1 B 2
Thismeanswegotthefollowingtransitions: