4 Exercise - Introduction to Group Theory
4.1 (vibrational selection rules)
To nd vibrational selection rules, we haveto determine whether integrals of the types
R ψ 0 ν f ψ 1 ν
arenon-zero,withfunction
f
beingx, y, z, x 2 , y 2 , z 2 , xy, yz, zx
oranycombinationthereof.Letψ 0 ν
betotallysymmetric and
ψ ν 1
may belong to any irreducible representation. We are now meant to identify the irreducible representationsto whichψ ν 1
may belong in order to give non-zerointegralsformolecules ofsymmetry
C 4 v
andD 3 d
.WestartoutrstdenotingthecharactertablesofC 4 v
:C 4 v E 2C 4 (z) C 2 2σ v 2σ d
linearfcts,rotations quadraticfcts cubicfunctionsA 1 1 1 1 1 1 z x 2 + y 2 , z 2 z 3 , z x 2 + y 2
A 2 1 1 1 − 1 − 1 R z − −
B 1 1 − 1 1 1 − 1 − x 2 − y 2 z x 2 − y 2
B 2 1 − 1 1 − 1 1 − xy xyz
E 2 0 − 2 0 0 (x, y) (R x , R y ) (xz, yz) xz 2 , yz 2
xy 2 , x 2 y
x 3 , y 3
Werststartcalculatingallpossibledirectproducts:
A 1 ⊗ A 1 = A 1
A 1 ⊗ A 2 = A 2
A 1 ⊗ B 1 = B 1 A 1 ⊗ B 2 = B 2
A 1 ⊗ E = E
further:
E ⊗ A 1 = E E ⊗ A 2 = E E ⊗ B 1 = E E ⊗ B 2 = E
E ⊗ E = A 1 + A 2 + B 1 + B 2
goingonwith:
B 1 ⊗ A 1 = B 1
B 1 ⊗ A 2 = B 2
B 1 ⊗ B 1 = A 1 B 1 ⊗ B 2 = A 2
B 1 ⊗ E = E
and
B 2 ⊗ A 1 = B 2
B 2 ⊗ A 2 = B 1
B 2 ⊗ B 1 = A 2 B 2 ⊗ B 2 = A 1
B 2 ⊗ E = E
atlast
A 2 ⊗ A 1 = A 2
A 2 ⊗ A 2 = A 1
A 2 ⊗ B 1 = B 2
A 2 ⊗ B 2 = B 1
A 2 ⊗ E = E
Therefore
A 1
andE
symmetryvibrationswillbeIRactiveandA 1
,B 1
,B 2
andE
symmetryvibrationswill beRamanactive.Weusedthat thedirect product should containtheirreduciblerepresentation of
theirr.rep.wehavelookedat forthetransition.
Nowwedenotethecharatertable of
D 3 d
:D 3d E 2C 3 3C 2 0 i 2S 6 3σ d
linears,rots quadratics cubicfunctionsA 1 g 1 1 1 1 1 1 − x 2 + y 2 , z 2 −
A 2 g 1 1 − 1 1 1 − 1 R z − −
E g 2 − 1 0 2 − 1 0 (R x , R y ) x 2 − y 2 , xy
(xz, yz) −
A 1 u 1 1 1 − 1 − 1 − 1 − − x x 2 − 3y 2
A 2 u 1 1 − 1 − 1 − 1 1 z − y x 2 − 3y 2
, z 3 , z x 2 + y 2 E u 2 − 1 0 − 2 1 0 (x, y) − xz 2 , yz 2 xyz, z x 2 − y 2
x x 2 + y 2
, y x 2 + y 2
Wecanagaincalculateallpossibledirectproducts:
A 1 g ⊗ A 1 g = A 1 g
A 1 g ⊗ A 2 g = A 2 g
A 1 g ⊗ E g = E g
A 1 g ⊗ A 1 u = A 1 u
A 1g ⊗ A 2u = A 2u
A 1 g ⊗ E u = E u
next
A 2 g ⊗ A 1 g = A 2 g
A 2 g ⊗ A 2 g = A 1 g
A 2g ⊗ E g = E g
A 2 g ⊗ A 1 u = A 2 u
A 2 g ⊗ A 2 u = A 1 u
A 2 g ⊗ E u = E u
last
g
E g ⊗ A 1 g = E g
E g ⊗ A 2 g = E g
E g ⊗ E g = A 1 g + A 2 g + E g
E g ⊗ A 1u = E u
E g ⊗ A 2 u = E u
E g ⊗ E u = A 1 u + A 2 u + E u
nowthe
u
-terms:A 1 u ⊗ A 1 g = A 1 u
A 1 u ⊗ A 2 g = A 2 u
A 1 u ⊗ E g = E u
A 1 u ⊗ A 1 u = A 1 g
A 1 u ⊗ A 2 u = A 2 g
A 1 u ⊗ E u = E g
next
A 2u ⊗ A 1g = A 2u
A 2 u ⊗ A 2 g = A 1 u
A 2u ⊗ E g = E u
A 2 u ⊗ A 1 u = A 2 g
A 2 u ⊗ A 2 u = A 1 g
A 2u ⊗ E u = E g
last
u
-term:E u ⊗ A 1 g = E u
E u ⊗ A 2g = E u
E u ⊗ E g = A 1 u + A 2 u + E u
E u ⊗ A 1 u = E g
E u ⊗ A 2 u = E g
E u ⊗ E u = A 1 g + A 2 g + E g
A 2u
andE u
symmetryvibrationswillbeIRactive.A 1g
andE g
symmetryvibrationswillbeRamanactive.
4.2 (magnetic dipole allowed transitions for
T d-symmetry)
Foramoleculeof
T d
symmetrywecandeterminewhat pairsofstatescouldbeconnectedbyamagneticdipole allowedtransition,whileforexampleMethanegot
T d
symmetry. ThecharactertableofT d
:T d E 8C 3 3C 2 6S 4 6σ d
linears,rots quadraticfcts cubicfunctionsA 1 1 1 1 1 1 − x 2 + y 2 + z 2 xyz
A 2 1 1 1 − 1 − 1 − − −
E 2 − 1 2 0 0 − 2z 2 − x 2 − y 2 , x 2 − y 2
− T 1 3 0 − 1 1 − 1 (R x , R y , R z ) −
x z 2 − y 2
, y z 2 − x 2
, z x 2 − y 2 T 2 3 0 − 1 − 1 1 (x, y, z) (xy, xz, yz) x 3 , y 3 , z 3
x z 2 + y 2
, y z 2 + x 2
, z x 2 + y 2
Thedirect productsare:
T 1 ⊗ A 1 = T 1 T 1 ⊗ A 2 = T 2
T 1 ⊗ E = T 1 + T 2
T 1 ⊗ T 1 = A 1 + E + T 1 + T 2
T 1 ⊗ T 2 = A 1 + E + T 1 + T 2
thereforethefollowingtransitionsaremagneticdipole allowed:
T 1 → A 1
T 1 → E T 1 → T 1 T 1 → T 2
this followsfrom thecondition, that we needto have at least one