Download SIMS-2012_Symmetry

MOLECULAR SYMMETRY
AND SPECTROSCOPY
[email protected]
Download ppt file from
http://www.few.vu.nl/~rick
At bottom of page
We began by summarizing
Chapters 1 and 2. Spectroscopy and Quantum Mechanics
Absorption can only occur at resonance hνif = Ef – Ei = ΔEif
νif
M
Integrated absorption coefficient (i.e. intensity) for a line is:
8π3 Na
______
~
~
I(f ← i) = ∫ ε(ν)dν =
ν~if F(Ei )S(f ← i) Rstim(f→i)
line
(4πε0)3hc
Use Q. Mech. to calculate:
ODME of H
ODME of μA
μfi = ∫ (Ψf )* μA Ψi dτ
f
i
P. R. Bunker and Per Jensen:
Fundamentals of Molecular Symmetry,
Taylor and Francis, 2004.
The first 47 pages:
Chapter 1 (Spectroscopy)
Chapter 2 (Quantum Mechanics) and
Section 3.1 (The breakdown of the BO Approx.)
P. R. Bunker and Per Jensen:
To buy it go to:
Molecular Symmetry and Spectroscopy,
http://www.crcpress.com
2nd Edition,
3rd Printing,
NRC Research Press, Ottawa, 2012.
Download pdf file from
www.chem.uni-wuppertal.de/prb
P. R. Bunker and Per Jensen:
Fundamentals of Molecular Symmetry,
Taylor and Francis, 2004.
The first 47 pages:
Chapter 1 (Spectroscopy)
Chapter 2 (Quantum Mechanics) and
Section 3.1 (The breakdown of the BO Approx.)
P. R. Bunker and Per Jensen:
To buy it go to:
Molecular Symmetry and Spectroscopy,
http://www.crcpress.com
2nd Edition,
3rd Printing,
NRC Research Press, Ottawa, 2012.
Download pdf file from
www.chem.uni-wuppertal.de/prb
We then proceeded to discuss Group
Theory and Point Groups
Definitions for groups and point groups:
“Group”
A set of operations that is closed wrt “multiplication”
“Point Group”
All rotation, reflection and rotation-reflection operations
that leave the molecule (in its equilibrium configuration)
“looking” the same.
“Matrix group”
A set of matrices that forms a group.
“Representation”
A matrix group having the same shaped multiplication
table as the group it represents.
“Irreducible representation”
A representation that cannot be written as the sum
of smaller dimensioned representations.
“Character table”
A tabulation of the characters of the irreducible representations.
Character table for the point group C3v
E C3 σ1
C32 (12)
σ2
E (123)
Two 1D
irreducible
representations
of the C3v group
1
2
3σ3
A1
1
1
1
A2
1
1
1
E
2
1
0
The 2D representation M = {M1, M2, M3, ....., M6}
of C3v is the irreducible representation E. In this
table we give the characters of the matrices.
Elements in the same class have the same characters
3 classes and 3 irreducible representations
Character table for the point group C2v
x
(+y)
E (12)
E
C2 E*
σyz (12)*
σxy
z
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1
4 classes and 4 irreducible representations
Spectroscopy
M
MMMM
f
hνif = Ef – Ei = ΔEif
i S(f ← i) = ∑A | ∫ Φf* μA Φi dτ |2
Quantum Mechanics
ODME of H and μA
μfi = ∫ Φf* μA Φi dτ
Group Theory and Point Groups
(Character Tables and Irreducible Representations)
C3v
3
1
E
C3
C 32
σ1
σ2
σ3
2
PH3
8
Point Group symmetry is based on
the geometrical symmetry of the
equilibrium structure.
Point group symmetry not appropriate
when there is rotation or tunneling
Use energy invariance symmetry
instead. We start by using inversion
symmetry and identical nuclear
permutation symmetry.
The Complete Nuclear Permutation
Inversion (CNPI) Group
Contains all possible permutations
of identical nuclei including E. It also
contains the inversion operation E*
and all possible products of E* with
the identical nuclear permutations.
GCNPI = GCNP x {E,E*}
The spin-free (rovibronic) Hamiltonian
(after separating translation)
Vee + VNN + VNe
THE GLUE
In a world of infinitely powerful computers we could solve the
Sch. equation numerically and that would be that. However,
we usually have to start by making approximations. We then
selectively correct for the approximations made.
The CNPI Group for the Water Molecule
Nuclear permutations permute nuclei (coordinates and spins).
Do not change electron coordinates
The Complete Nuclear Permutation Inversion (CNPI) group
for the water molecule is {E, (12)} x {E,E*} = {E, (12), E*, (12)*}
H1
+
e
H2
O
(12)
H2
+
e
H1
E*
H2
H1
-
e
O
O
(12)*
E* Inverts coordinates of nuclei and electrons.
Does not change spins.
Same CNPI group for CO2, H2, H2CO, HOOD, HDCCl2,…
H
H
1
C1
C2
F
I
N1N2N3
3
2
C3
F
D
O2
O1
O3
1
2
H2
H1
+
3
H1
H2
12C
H3
D
13C
H3
GCNPI = {E, (12), (13), (23), (123), (132)} x {E, E*}
=
GCNP
x {E, E*}
GCNPI = {E, (12), (13), (23), (123), (132)} x {E, E*}
GCNPI={E, (12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
Number of elements = 3! x 2 = 6 x 2 = 12
Number of ways of permuting
three identical nuclei
H5
The CNPI Group of C3H2ID
GCNPI = {E, (12), (13), (23), (123), (132)}
x{E, (45)} x {E, E*}
H4
C1
C2
I
C3
D
= {E, (12), (13), (23), (123), (132),
(45), (12)(45), (13)(45), (23)(45), (123)(45), (132)(45),
E*, (12)*, (13)*, (23)*, (123)*, (132)*,
(45)*, (12)(45)*, (13)(45)*, (23)(45)*, (123)(45)*,
(132)(45)*}
Number of elements = 3! x 2! x 2 = 6 x 2 x 2 = 24
H5
H4
C1
C2
I
C3
Number of elements
= 3! x 2! x 2 = 6 x 2 x 2 = 24
D
If there are n1 nuclei of type 1, n2 of
type 2, n3 of type 3, etc then the total
number of elements in the CNPI group
is n1! x n2! x n3!... x 2.
The CNPI group of allene
The
Allene molecule
C3H4
H5
H4
C1
C2
H7
C3
H6
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
The CNPI group of allene
The
Allene molecule
C3H4
H5
H4
C1
C2
H7
C3
H6
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
Sample elements: (456), (12)(567), (4567), (45)(67)(123)
The CNPI group of allene
The
Allene molecule
C3H4
H5
H4
C1
C2
H7
C3
H6
00H
00H
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
C3H4O4 How many elements?
The CNPI group of allene
The
Allene molecule
C3H4
H5
H4
C1
C2
H7
C3
H6
00H
00H
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
C3H4O4 3! x 4! x 4! x 2 = 6912
The size of the CNPI group depends
only on the chemical formula
Number of elements in the CNPI groups of various
molecules
(C6H6)2
12! x 12! x 2 ≈ 4.6 x 1017
Just need the chemical formula to
determine the CNPI group. Can be BIG
An important number
Molecule
PG
h(PG)
h(CNPIG)
h(CNPIG)/h(PG)
H 2O
C2v
4
2!x2=4
1
PH3
C3v
6
3!x2=12
2
Allene
C 3H 4
D2d
8
4!x3!x2=288
36
Benzene
C 6H 6
D6h
24
6!x6!x2=1036800
43200
This number means something!
End of Review of Lecture One
ANY QUESTIONS OR COMMENTS?
22
CNPI group symmetry is based on
energy invariance
Symmetry operations are operations
that leave the energy of the system
(a molecule in our case) unchanged.
Using quantum mechanics:
A symmetry operation is an operation that
commutes with the Hamiltonian:
RHn = HRn
The character table of the CNPI
group of the water molecule
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
It is called C2v(M)
The character table of the CNPI
group of the water molecule
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
It is called C2v(M)
Now to explain how we label
energy levels using
irreducible representations
Labelling energy levels
For the water molecule (no degeneracies, and R2 = identity for all R) :
H = E
RH = RE
Since RH = HR and E is a number, this leads to HR = ER.
H(R) = E(R)
E is nondegenerate. Thus RΨ = cΨ.
But R2 = identity. Thus c2 = 1, so c = ±1 and R = ±
R = (12), E* or (12)*
The eigenfunctions have symmetry
+ Parity
R = E*
Ψ1+(x)
- Parity
Ψ2-(x)
x
x
Ψ-(-x) = -Ψ-(x)
Ψ3+(x)
x
Ψ+(-x) = Ψ+(x)
Eigenfunctions of H
must satisfy
E*Ψ = ±Ψ
E*ψ(xi) = ψE*(xi), a new function.
ψE*(xi) = ψ(E*xi) = ψ(-xi) = ±ψ(xi)
Since E*ψ(xi) can only be ±ψ(xi)
This is different from Wigner’s approach
See PRB and Howard (1983)
+ Parity
- Parity
Ψ1+(x)
Ψ2-(x)
x
x
Ψ-(-x) = -Ψ-(x)
Ψ3+(x)
x
Ψ+(-x) = Ψ+(x)
Eigenfunctions of H
must satisfy
E*Ψ = ±Ψ
and (12)ψ = ±ψ
There are four symmetry types of H2O wavefunction
R = ±
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
The Symmetry Labels of the CNPI Group of H2O
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
We are labelling the states using
∫ΨaHΨ
0 if symmetries of Ψa and Ψb are different.
bdτ =
the
irreps
of the CNPI group
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
The Symmetry Labels of the CNPI Group of H2O
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
Thus, for example, a wavefunction of “A2 symmetry”
will “generate” the A2 representation:
=+1ψ of of
E*ψ=-1ψ
HΨbbdτ
dτ == 00 if(12)ψ
if symmetry
symmetries
Ψa andis Ψ
different.
∫ΨEψ=+1ψ
product
not
A(12)*ψ=-1ψ
aμΨ
b are
a
1
The Symmetry Labels of the CNPI Group of H2O
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
Thus, for example, a wavefunction of “A2 symmetry”
will “generate” the A2 representation:
=+1ψ of of
E*ψ=-1ψ
HΨbbdτ
dτ == 00 if(12)ψ
if symmetry
symmetries
Ψa andis Ψ
different.
∫ΨEψ=+1ψ
product
not
A(12)*ψ=-1ψ
aμΨ
b are
a
1
For the water molecule we can, therefore label
the energy levels as being A1, A2, B1 or B2
using the irreps of the CNPI group.
The labelling business
The vibrational wavefunction for the v3 = 1
state of the water molecule can be written
approximately as ψ = N(Δr1 – Δr2).
Eψ=+1ψ
(12)ψ =-1ψ
E*ψ=+1ψ
(12)*ψ=-1ψ
This would be labelled as B2.
n
Suppose R = E where n > 2.
n
We still have RΨ = cΨ for nondegenerate Ψ, but now R Ψ = Ψ.
n
Thus c = 1 and c =
If n = 3, c =
n
√1, i.e. c = [ei2π/n]a where a = 1,2,…,n.
ε, ε2 (=ε*),
or
ε3
(=1) where

C3(M) E (123)
C3 (132)
C 32
1
1
1
A
1
1
1
Ea
1

*
Eb
1
*

iπ
e
= -1
ei2π = 1
=
ei2/3
For nondegenerate states we had
this as the effect of a symmetry
operation on an eigenfunction:
For the water molecule (  nondegenerate) :
H = E
RH = RE
HR = ER
Thus R = c since E is nondegenerate.
What about degenerate states?
ℓ-fold degenerate energy level with energy En
R Ψnk = D[R ]k1Ψn1 + D[R ]k2Ψn2 + D[R ]k3Ψn3 +…+ D[R ]kℓΨnℓ
For each relevant symmetry operation R, the constants
D[R ]kp form the elements of an ℓℓ matrix D[R ].
ForT = RS it is straightforward to show that
D[T ] = D[R ] D[S ]
The matrices D[T ], D[R ], D[S ] ….. form an ℓ-dimensional
representation that is generated by the ℓ functions Ψnk
The ℓ functions Ψnk transform according to this
representation
Labelling energy levels using the CNPI Group
We can label energy levels using
the irreps of the CNPI group for any molecule
∫ΨaaμΨ
HΨbbdτ
dτ == 00 ifif symmetry
symmetries
Ψa andis Ψ
of of
product
not
A1 different.
b are
Pages 143-149
Pages 99-101
34
Determining symmetry and
reducing a representation
Example of using the symmetry operation (12):
(12)

r 1´
r2´
´
H2
We have (12) (r1, r2, ) = (r1´, r2´, ´)
We see that (r1´, r2´, ´) = (r2, r1, )
H1
3
E
r1

3
r2
r 1´
2
1
´
1
3
(12)
r1

3
r2
r 2´
2
1
r 2´
´
2
r1 ´
r2 ´
´
1
r1 ´
r2 ´
´
r 1´
2
=
r1
r2

=
r2
r1

=
r1
r2

=
r2
r1

3
E*
r1

r2
2
1
´
2
r2´
1
r 1´
r1 ´
r2 ´
´
3
3
(12)*
r1
1

1
r2
2
2
´ r ´
2
r 1´
3
r1 ´
r2 ´
´
R
a =
E
r1
r2

(12)
r1
r2

E*
r1
r2

(12)*
r1
r2

a´
=
r1 ´
r2 ´
´
=
r1 ´
r2 ´
´
=
r1 ´
r2 ´
´
=
r1 ´
r2 ´
´
= D[R] a
=
r1
r2

=
r2
r1

=
r1
r2

=
r2
r1

=
1
0
0
0
1
0
0
0
1
r1
r2

=3
=
0
1
0
1
0
0
0
0
1
r1
r2

=1
=
1
0
0
0
1
0
0
0
1
r1
r2

=
0
1
0
1
0
0
0
0
1
r1
r2

=3
=1
E (12) E* (12)*
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1

3
1
3
1
aA1 =
1
4
Γ = Σ aiΓi
i
A reducible representation
( 13 + 11 + 13 + 11) = 2
( 13 + 11  13  11) = 0
aB1 =
1
4
1
4
aB2 =
1
4
( 13  11 + 13  11) = 1
aA2 =
( 13  11  13 + 11) = 0
 = 2 A1  B2
E (12) E* (12)*
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1

3
1
3
1
aA1 =
1
4
Γ = Σ aiΓi
i
i
A reducible representation
( 13 + 11 + 13 + 11) = 2
( 13 + 11  13  11) = 0
aB1 =
1
4
1
4
aB2 =
1
4
( 13  11 + 13  11) = 1
aA2 =
( 13  11  13 + 11) = 0
 = 2 A1  B2
E (12) E* (12)*
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1

3
1
3
1
aA1 =
1
4
Γ = Σ aiΓi
i
i
A reducible representation
( 13 + 11 + 13 + 11) = 2
( 13 + 11  13  11) = 0
aB1 =
1
4
1
4
aB2 =
1
4
( 13  11 + 13  11) = 1
aA2 =
( 13  11  13 + 11) = 0
 = 2 A1  B2
E (12) E* (12)*
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1

3
1
3
1
aA1 =
1
4
Γ = Σ aiΓi
i
i
A reducible representation
( 13 + 11 + 13 + 11) = 2
( 13 + 11  13  11) = 0
aB1 =
1
4
1
4
aB2 =
1
4
( 13  11 + 13  11) = 1
aA2 =
( 13  11  13 + 11) = 0
 = 2 A1  B2
We know now that r1, r2, and  generate the
representation 2 A1  B2
Consequently, we can generate from r1, r2, and
 three „symmetrized“ coordinates:
S1 with A1 symmetry
S2 with A1 symmetry
S3 with B2 symmetry
For this, we need projection operators
Projection operators:
General for li-dimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1-dimensional irrep i
1
Character
Projection operators:
General for li-dimensional irrep i
Simpler for 1-dimensional irrep i
Character
Diagonal element of representation matrix
1
Symmetry operation
Projection operators:
General for li-dimensional irrep i
Simpler for 1-dimensional irrep i
Character
Diagonal element of representation matrix
1
A1
E
1
(12)
1
E*
1
Symmetry operation
(12)*
1
PA1 = (1/4) [ E + (12) + E* + (12)* ]
Projection operator for A1 acting on r1
S1 = PP11A1A1r1 =
1
4
[ E + (12) + E* + (12)* ]r1
=
1
4
[ r1 + r2 + r1 + r2 ] =
S2 = PP11A1A1 =
1
4
[ E + (12) + E* + (12)* ]
=
1
4
[ + + +] =
S3 = P11 r1 =
1
4
[ E  (12) + E*  (12)*] r1
=
1
4
[ r1  r2 + r1  r2 ] =
P11  =
1
4
[ E  (12) + E*  (12)* ]
=
1
4
[   +    ] =
2
PBB2
2
PBB2
1
2
[ r1 + r2 ]

1
2
[ r1  r2 ]
0
 Is „annihilated“ by PP11BB2
2
Projection operators for A1 and B2
S1 = PP11A1A1r1 =
1
4
[ E + (12) + E* + (12)* ]r1
=
1
4
[ r1 + r2 + r1 + r2 ] =
S2 = PP11A1A1 =
1
4
[ E + (12) + E* + (12)* ]
=
1
4
[ + + +] =
S3 = P11 r1 =
1
4
[ E  (12) + E*  (12)*] r1
=
1
4
[ r1  r2 + r1  r2 ] =
P11  =
1
4
[ E  (12) + E*  (12)* ]
=
1
4
[   +    ] =
2
PBB2
2
PBB2
1
2
[ r1 + r2 ]

1
2
[ r1  r2 ]
0
 Is „annihilated“ by PP11BB2
2
Projection operators for A1 and B2
S1 = PP11A1A1r1 =
1
4
[ E + (12) + E* + (12)* ]r1
=
1
4
[ r1 + r2 + r1 + r2 ] =
S2 = PP11A1A1 =
1
4
[ E + (12) + E* + (12)* ]
=
1
4
[ + + +] =
S3 = P11 r1 =
1
4
[ E  (12) + E*  (12)*] r1
=
1
4
[ r1  r2 + r1  r2 ] =
2
PBB2
1
S3 have
4
Aside: S1, S2 and
normal coordinates.
1
4
1
2
[ r1 + r2 ]

1
2
[ r1  r2 ]
the symmetry and form of the
E (12) E* (12)*
A1
1
1
1
1
A2
1
1
1
1
B1
1
1 1
1
B2
1
1
1
1

3
1
3
1
The three
Normal modes
of the water
molecule
A1
A1
B2
Labeling is not just bureaucracy. It is useful.
PAUSE
50
Labeling is not just bureaucracy. It is useful.
The vanishing integral theorem
Pages 114-117
Pages 136-139
But first we look at
The symmetry of a product
Pages 109-114
THE SYMMETRY OF A PRODUCT
A1
A2
B1
B2
Eψ=+1ψ
E
1
1
1
1
(12)ψ =+1ψ
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
E*ψ=-1ψ
(12)*ψ=-1ψ
∫ΨaHΨφbdτ (12)
= 0 if symmetries of Ψa and Ψb are different.
Eφ =+1
φ =-1φ E*φ =-1φ (12)*φ =+1φ
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
A2
B1
The
the product φψ is B1 =
xA
A12 = B2.
A2 xsymmetry
B1 = B2, Bof
1 x B2 = A2, B1 x A2 x B2
B1 x B2, A1 x A2, B1 x A2, B2 x A2, B1 x B1,…
A2
A2
B2
B1
A1
Symmetry of a product. Example: C3v
A1  A1 = A1
A1  A2 = A2
A2  A2 = A1
A1  E = E
A2  E = E
E  E = A1  A2  E
E  E:
4
1
0
Reducible representation
Characters of the product representation are the products
of the characters of the representations being multiplied.
Symmetry of triple product is obvious extension
+ Parity
- Parity
Ψ+(x)
Ψ-(x)
x
x
Ψ-(-x) = -Ψ-(x)
Ψ+(x)
x
Ψ+(-x) = Ψ+(x)
∫Ψ+Ψ-Ψ+dx = 0
- parity
+ Parity
- Parity
Ψ+(x)
Ψ-(x)
The vanishing integral
theorem
x
Ψ
(-x)
=
-Ψ
(x)
+
Ψ (x)
∫f(τ)dτ = 0 if symmetry of f(τ) is not A1
x
Ψ+(-x) = Ψ+(x)
∫Ψ+Ψ-Ψ+dx = 0
- parity
x
USING THE VANISHING
INTEGRAL THEOREM
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)* Symmetry of H
1
-1
1
-1
Using symmetry labels and the vanishing
integral theorem we deduce that:
∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,
Integral vanishes if Ψa and Ψb have different symmetries
ODME of H
vanishes if symmetries not the same
This means that we can
“block-diagonalize” the
Hamiltonian matrix
A1
A1
A2
B1
Ψ1
0
Ψ2
0
Ψ3
0
Ψ4
0
Ψ5
0
Ψ6
Ψ70
ψ80
B1
ψ1ψ2ψ3ψ4ψ5ψ6ψ7ψ8
. . .
. . .
. . .
. . .
. . .
. . .
. .
. .
0
0
A2
0
0
0
0
0
0
0
0 0
0
0
0 0
Symmetry is preserved on diagonalization
ODME of μA
μfi = ∫ (Ψf )* μA Ψi dτ
Can use symmetry to determine if this ODME = 0
This ODME will vanish if the symmetry
0
0
of (Ψf )* μA Ψi is not A1
What is the symmetry of μA ?
A1
A2
B1
B2
E
1
1
1
1
(12) E* (12)*
1
1
1
1
-1 -1
1
-1
-1
-1
1 -1
μA = Σ
Cre Ar
r
Charge on
particle r
A coordinate
of particle r
A = space-fixed X, Y or Z
EμA= ?μA
(12)μA = ?μA
E*μA= ?μA
(12)*μA= ?μA
What is the symmetry of μZ ?
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)*
1
-1
1
-1
μA = Σ
Cre Ar
r
Charge on
particle r
A coordinate
of particle r
A = space-fixed X, Y or Z
EμA= +1μA
(12)μA = +1μA
E*μA= -1μA
μA has symmetry A2
(12)*μA= -1μA
The Symmetry Labels of the CNPI Group of H2O
Γ(H) = A1
R = ±
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)* Symmetry of H
1
-1
Symmetry of μA
1
-1
Γ(μA) = A2
Using symmetry labels and the vanishing
integral theorem we deduce that:
∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,
∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,
The Symmetry Labels of the CNPI Group of H2O
Γ(H) = A1
R = ±
A1
A2
B1
B2
E
1
1
1
1
(12) E*
1
1
1
-1
-1
-1
-1
1
(12)* Symmetry of H
1
-1
Symmetry of μA
1
-1
Γ(μA) = A2
Using symmetry labels and the vanishing
integral theorem we deduce that:
∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,
that is, if the symmetry of Ψa is not the same as Ψb
∫
Ψa*μAΨbdτ
= 0 if symmetry of ψa*μAψb is not A1,
that is, if the symmetry of the product ΨaΨb is not A2
Pages 113-114
Symmetry of rotational levels of H2O
b
JKaKc
a
c
KaKc
e
o
e
o
e
o
o
e
Γrot
A1
A2
B1
B2
Allowed transitions
So we can use the CNPI group to:
1. Symmetry label energy levels
and
2. Determine which ODME vanish.
Ch. 7
Ch. 6
65
BUT BUT BUT...
There are problems with the CNPI Group
Number of elements in the CNPI groups of various
molecules
Huge groups. Size bears no relation to geometrical symmetry
C6H6, for example, has a 1036800-element CNPI group,
but a 24-element point group at equilibrium, D6h
Often gives SUPERFLUOUS multiple symmetry labels
PH3
3
2
1
There are two VERSIONS
of this molecule
PH3
TWO VERSIONS: Distinguished
by numbering the identical nuclei
Very, very
high
potential
barrier
2
3
1
2
1
3
Bone et al., Mol. Phys., 72, 33 (1991)
~12000 cm-1
The number of versions of the minimum is given by:
(order of CNPI group)/(order of point group)
For H2O this is 4/4 = 1
H
For H3+ this is 12/12= 1
F
For PH3 or CH3F this is 12/6 = 2
C1
C2
I
C3
For O3 this is 12/4 = 3
For HN3 this is 12/2 = 6
D
12/1 = 12
The six versions of HN3
H
N1 N2 N3
1 3 2
2
2
3
3
1
3
1
2
3
1
2
1
H
F
C1
C2
I
C3
D
12 versions
The number of versions of the minimum is given by:
(order of CNPI group)/(order of point group)
For H2O this is 4/4 = 1
H
For H3+ this is 12/12= 1
F
For PH3 or CH3F this is 12/6 = 2
C1
C2
I
C3
For O3 this is 12/4 = 3
For HN3 this is 12/2 = 6
D
12/1 = 12
The number of versions of the minimum is given by:
(order of CNPI group)/(order of point group)
For H2O this is 4/4 = 1
H
For H3+ this is 12/12= 1
F
For PH3 or CH3F this is 12/6 = 2
C1
C2
I
C3
For O3 this is 12/4 = 3
For HN3 this is 12/2 = 6
D
12/1 = 12
For C6H6 this is (6!x6!x2)/24
= 1036800/24 = 43200
Using the CNPI Group to symmetry
label the energy levels of a molecule
that has more than one version.
Character Table of CNPI group of PH3
GCNPI
GCNPI
12 elements
6 classes
6 irred. reps
E (123) (23) E* (123)* (23)*
(132) (31)
(132)* (31)*
(23)
(23)*
PH3 (or CH3F) Using the CNPI Group
A1’
E’ + E’’
A1’’
A1’’ + A2’
A 2’
A2’’
A1’ + A2’’
E’
E’’
PH3 (or CH3F) Using the CNPI Group
A1’
E’ + E’’
A1’’
A1’’ + A2’
A 2’
A2’’
A1’ + A2’’
E’
E’’
Why this
Degeneracy?
This double labeling results from the fact that there
are two versions of the PH3 molecule and the
tunneling splitting between these versions is not
observed.
For understanding the spectrum
this is a “superfluous” degeneracy
(particularly for CH3F)
The number of versions of the minimum is given by:
(order of CNPI group)/(order of point group)
For H2O this is 4/4 = 1
For PH3 or CH3F this is 12/6 = 2
For O3 this is 12/4 = 3
For C3H4 this is 288/8 = 36
For C6H6 this is 1036800/24 = 43200,
and using the CNPI group each energy level would
get as symmetry label the sum of 43200 irreps.
Clearly using the CNPI group gives
very unwieldy symmetry labels. 77
The CNPI Group approach
works, in principle, and can be used
to determine which ODME vanish.
BUT IT IS OFTEN HOPELESSLY UNWIELDY
In 1963 Longuet-Higgins figured out
how to set up a sub-group of the CNPI
Group that achieves the same result
without superfluous degeneracies.
This subgroup is called
The Molecular Symmetry (MS) group.
PH3 (or CH3F)
Ab initio calc with neglect of tunneling
Ab Initio CALC IN HERE
Very, very
high
potential
barrier
2
3
1
2
1
3
It would be
superfluous
to calc points
No observed tunneling through barrier in other min
PH3 (or CH3F)
Only NPI OPERATIONS FROM IN HERE
Very, very
high
potential
barrier
2
3
1
2
1
3
No observed tunneling through barrier
PH3 (or CH3F)
Only NPI OPERATIONS FROM IN HERE
Very, very
high
potential
barrier
2
3
1
2
1
3
(12) superfluous
E*
superfluous
(123), (12)* useful
No observed tunneling through barrier
If we cannot see any effects
of the tunneling through the
barrier then we only need
NPI operations for one
version. Omit NPI elements
that connect versions since
they are not useful; they are
superfluous
superfluous.
For CH3F:
superfluous
useful
GCNPI={E, (12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
useful elements are
The six feasible
GMS ={E, (123), (132), (12)*, (13)*,(23)*}
IfIf we
we cannot
cannot see
see any
any effects
effects
of
of the
the tunneling
tunneling through
through the
the
barrier
barrier then
then we
we only
only need
need
NPI
NPI operations
operations or
forone
one
version.
version. Omit
Omit NPI
NPI elements
elements
that
that take
connect
us between
versions since
versions
they are since
not useful;
they are
theynot
are
useful; they unfeasible.
are unfeasible.
superfluous
For PH3
or CH3F:
GCNPI={E, (12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
The six feasible elements are
GMS ={E, (123), (132), (12)*, (13)*,(23)*}
Character Table of CNPI group of PH3 or CH3F
GCNPI={E, (12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
GGCNPI
CNPI
12 elements
6 irred. reps
E (123) (23) E* (123)* (23)*
(132) (31)
(132)* (31)*
(23)
(23)*
H
A1’
μA
A1”
Using the CNPI Group
E’ + E’’
A1’
E’ + E’’
A1’’
A1’’ + A2’
A 2’
A2’’
A1’ + A2’’
E’
E’’
2 versions → sum or two irrep labels on each level
Character table of the MS group of PH3 or CH3F
E (123) (12)*
(123) (13)*
(12)
E (132)
1
2 (23)*
3
A1
1
1
1
A2
1
1
1
E
2
1
0
H A1
μA A2
Using the MS Group
E
E
A1
A2
A2
E
A1
The MS group gives all the information we need
as long as there is no observable tunneling, i.e.,
as long as the barrier is insuperable.
Using CNPIG versus MSG for PH3
E’ + E’’
E
E’ + E’’
E
A1’’ + A2’
A2
A1’ + A2’’
A1
CNPIG
MSG
Can use either to determine if an ODME vanishes.
But clearly it is easier to use the MSG.
superfluous
Unfeasible elements of the CNPI group
interconvert versions that are separated
by an insuperable energy barrier
useful
The subgroup of feasible elements forms a group called
THE MOLECULAR SYMMETRY GROUP
(MS GROUP)
89