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. f. Absorption can only occur at resonance. h ν if = E f – E i = Δ E if.
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AND SPECTROSCOPY
Download ppt file from
http://www.few.vu.nl/~rick
At bottom of page
Chapters 1 and 2. Spectroscopy and Quantum Mechanics
f
Absorption can only occur at resonance
hνif = Ef – Ei = ΔEif
νif
i
M
Integrated absorption coefficient (i.e. intensity) for a line is:
______
8π3 Na
~
~
~
ε(ν)dν
=
I(f ← i) = ∫
F(Ei )
νif
Rstim(f→i)
S(f ← i)
(4πε0)3hc
line
Use Q. Mech. to calculate:
ODME of H
μfi = ∫ (Ψf )* μAΨi dτ
ODME of μA
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:
Molecular Symmetry and Spectroscopy,
2nd Edition, 3rd Printing,
NRC Research Press, Ottawa, 2012.
Download pdf file from
To buy it go to:
http://www.crcpress.com
www.chem.uniwuppertal.de/prb
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:
Molecular Symmetry and Spectroscopy,
2nd Edition, 3rd Printing,
NRC Research Press, Ottawa, 2012.
Download pdf file from
To buy it go to:
http://www.crcpress.com
www.chem.uniwuppertal.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 rotationreflection 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 C3σ1
C32σ2
σ3
Two 1D
irreducible
representations
of the C3v group
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
EC2σyzσxy
(+y)
z
4 classes and 4 irreducible representations
E C3σ1
C32σ2
σ3
C3v
3
2
1
Spectroscopy
M
f
hνif = Ef – Ei = ΔEif
MMMM
i
S(f ← i) = ∑A  ∫ Φf* μAΦi dτ 2
Quantum Mechanics
ODME of H and μA
μfi =
∫ Φf* μAΦi dτ
Group Theory and Point Groups
(Character Tables and Irreducible Representations)
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 spinfree (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 Complete Nuclear Permutation Inversion (CNPI) group
for the water molecule is {E, (12)} x {E,E*} = {E, (12), E*, (12)*}
+
+
e
e
H2
H1
O
O
E*
(12)

e
O
H2
H1
H2
H1
(12)*
Nuclear permutations permute nuclei (coordinates and spins).
Do not change electron coordinates
E* Inverts coordinates of nuclei and electrons.
Does not change spins.
Same CNPI group for CO2, H2, H2CO, HOOD, HDCCl2,…
H
N1N2N3
1
C1
F
3
C2
2
I
C3
O
F
D
2
O
O
1
3
1
3
H H
12C 13C
D H
2
1
2
H
1
+
H
H
3
2
3
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
C1
H4
C2
I
C3
D
The CNPI Group of C3H2ID
GCNPI = {E, (12), (13), (23), (123), (132)}
x{E, (45)}x {E, E*}
= {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
C1
H4
C2
I
C3
D
Number of elements
= 3! x 2! x 2 = 6 x 2 x 2 = 24
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.
H5
The
Allene molecule
C1
H4
C3H4
C2
H7
C3
H6
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
H5
The
Allene molecule
C1
H4
C3H4
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)
H5
00H
The
Allene molecule
C1
H4
C3H4
C2
H7
C3
H6
00H
Number of elements = 3! x 4! x 2 = 6 x 24 x 2 = 288
How many elements?
C3H4O4
H5
00H
The
Allene molecule
C1
H4
C3H4
C2
H7
C3
H6
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
Molecule PG h(PG) h(CNPIG) h(CNPIG)/h(PG)
H2O C2v 4 2!x2=4 1
PH3 C3v 6 3!x2=12 2
Allene D2d 8 4!x3!x2=288 36
C3H4
Benzene D6h 24 6!x6!x2=1036800 43200
C6H6
This number means something!
End of Review of Lecture One
22
ANY QUESTIONS OR COMMENTS?
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:
RHn = HRn
The character table of the CNPI
group of the water molecule
(12) E* 1 1
1 1
1 1
1 1
(12)*
1
1
1
1
E 1
1
1
1
A1
A2
B1
B2
It is called C2v(M)
The character table of the CNPI
group of the water molecule
(12) E* 1 1
1 1
1 1
1 1
(12)*
1
1
1
1
E 1
1
1
1
A1
A2
B1
B2
It is called C2v(M)
Now to explain how we label
energy levels using
irreducible representations
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
 Parity
Ψ1+(x)
Ψ2(x)
x
x
Ψ(x) = Ψ(x)
Ψ3+(x)
Eigenfunctions of H
must satisfy
E*Ψ = ±Ψ
x
Ψ+(x) = Ψ+(x)
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
Ψ1+(x)
Ψ2(x)
x
x
Ψ(x) = Ψ(x)
Ψ3+(x)
Eigenfunctions of H
must satisfy
E*Ψ = ±Ψ
x
Ψ+(x) = Ψ+(x)
and (12)ψ = ±ψ
(12) E* 1 1
1 1
1 1
1 1
E 1
1
1
1
(12)*
1
1
1
1
R = ±
A1
A2
B1
B2
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
(12) E* 1 1
1 1
1 1
1 1
E 1
1
1
1
(12)*
1
1
1
1
A1
A2
B1
B2
We are labelling the states using
the irreps of the CNPI group
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
(12) E* 1 1
1 1
1 1
1 1
E 1
1
1
1
(12)*
1
1
1
1
A1
A2
B1
B2
Thus, for example, a wavefunction of “A2 symmetry”
will “generate” the A2 representation:
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
Eψ=+1ψ(12)ψ=+1ψE*ψ=1ψ(12)*ψ=1ψ
(12) E* 1 1
1 1
1 1
1 1
E 1
1
1
1
(12)*
1
1
1
1
A1
A2
B1
B2
Thus, for example, a wavefunction of “A2 symmetry”
will “generate” the A2 representation:
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
Eψ=+1ψ(12)ψ=+1ψE*ψ=1ψ(12)*ψ=1ψ
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 vibrational wavefunction for the v3 = 1
state of the water molecule can be written
approximately as ψ = N(Δr1 – Δr2).
Eψ=+1ψ(12)ψ=1ψE*ψ=+1ψ(12)*ψ=1ψ
This would be labelled as B2.
c = ε, ε2 (=ε*), or ε3 (=1)
If n = 3,
Suppose Rn = E where n > 2.
We still have RΨ = cΨ for nondegenerate Ψ, but now RnΨ = Ψ.
Thus cn = 1 and c = n√1, i.e. c = [ei2π/n]a where a = 1,2,…,n.
where = ei2/3
C3 C32
eiπ = 1
ei2π = 1
For the water molecule ( nondegenerate) :
H = E
RH = RE
HR = ER
Thus R = c since E is nondegenerate.
What about degenerate states?
ℓfold
degenerate energy level with energy En
RΨnk=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 ℓℓ 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 ℓdimensional representation that is generated by the ℓ functions Ψnk
The ℓ functions Ψnktransform according to this representation
We can label energy levels using
the irreps of the CNPI group for any molecule
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
Pages 143149
Pages 99101
34
reducing a representation
Example of using the symmetry operation (12):
(12)
r1´
r2´
´
H1
H2
We have (12) (r1, r2, ) = (r1´, r2´, ´)
We see that (r1´, r2´, ´) = (r2, r1, )
r1´
r2´
r2
r2
r2
r2
r1
r1
r1
r1
r2´
r1´
´
´
3
3
3
3
3
1
2
2
2
2
2
1
1
1
1
3
1
2
2
1
3
2
1
3
E
(12)
´
E*
r1´
r2´
(12)*
´
r2´
r1´
aA1 = ( 13 + 11 + 13 + 11) = 2
aA2 = ( 13 + 11 13 11) = 0
aB1 = ( 13 11 13 + 11) = 0
aB2 = ( 13 11 + 13 11) = 1
Γ = Σ aiΓi
i
A reducible representation
= 2A1 B2
aA1 = ( 13 + 11 + 13 + 11) = 2
aA2 = ( 13 + 11 13 11) = 0
aB1 = ( 13 11 13 + 11) = 0
aB2 = ( 13 11 + 13 11) = 1
Γ = Σ aiΓi
i
i
A reducible representation
= 2A1 B2
aA1 = ( 13 + 11 + 13 + 11) = 2
aA2 = ( 13 + 11 13 11) = 0
aB1 = ( 13 11 13 + 11) = 0
aB2 = ( 13 11 + 13 11) = 1
Γ = Σ aiΓi
i
i
A reducible representation
= 2A1 B2
aA1 = ( 13 + 11 + 13 + 11) = 2
aA2 = ( 13 + 11 13 11) = 0
aB1 = ( 13 11 13 + 11) = 0
aB2 = ( 13 11 + 13 11) = 1
Γ = Σ aiΓi
i
i
A reducible representation
= 2A1 B2
We know now that r1, r2, andgenerate the
representation 2A1 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
General for lidimensional irrep i
Diagonal element of representation matrix
Symmetry operation
Simpler for 1dimensional irrep i
Character
1
General for lidimensional irrep i
Simpler for 1dimensional irrep i
Character
Diagonal element of representation matrix
1
Symmetry operation
General for lidimensional irrep i
Simpler for 1dimensional irrep i
Character
Diagonal element of representation matrix
1
Symmetry operation
E (12) E* (12)*
A1 1 1 1 1
PA1 = (1/4) [ E + (12) + E* + (12)* ]
S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1
S3 = P11B2r1= [ E (12) + E* (12)*] r1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S2 = P11A1= [ E + (12) + E*+ (12)* ]
P11B2= [ E (12) + E* (12)* ]
= [ + + + ] =
= [ + ] = 0
Projection operator for A1 acting on r1
PA1
PA1
PB2
PB2
Is „annihilated“ by P11B2
PB2
S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1
S3 = P11B2r1= [ E (12) + E* (12)*] r1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S2 = P11A1= [ E + (12) + E*+ (12)* ]
P11B2= [ E (12) + E* (12)* ]
= [ + + + ] =
= [ + ] = 0
Projection operators for A1 and B2
PA1
PA1
PB2
PB2
Is „annihilated“ by P11B2
PB2
S1 = P11A1r1= [ E + (12) + E*+ (12)* ]r1
S3 = P11B2r1= [ E (12) + E* (12)*] r1
= [ r1 r2 + r1 r2 ] = [ r1 r2 ]
= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]
S2 = P11A1= [ E + (12) + E*+ (12)* ]
= [ + + + ] =
Projection operators for A1 and B2
PA1
PA1
PB2
Aside: S1, S2 and S3 have the symmetry and form of the
normal coordinates.
Labeling is not just bureaucracy. It is useful.
The vanishing integral theorem
Pages 136139
Pages 114117
But first we look at
The symmetry of a product
Pages 109114
1 1
1 1
1 1
THE SYMMETRY OF A PRODUCT
(12)*
1
1
1
1
E 1
1
1
1
A1
A2
B1
B2
A2
Eψ=+1ψ(12)ψ=+1ψE*ψ=1ψ(12)*ψ=1ψ
∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.
B1
Eφ=+1φ(12)φ=1φE*φ=1φ(12)*φ=+1φ
∫ΨaμΨbdτ = 0 if symmetry of product is not A1
The symmetry of the product φψis B1 x A2 = B2.
A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1
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
Ψ+(x)
Ψ(x)
The vanishing integral theorem
x
x
Ψ(x) = Ψ(x)
Ψ+(x)
∫f(τ)dτ = 0 if symmetry of f(τ) is not A1
∫Ψ+ΨΨ+dx = 0
x
 parity
Ψ+(x) = Ψ+(x)
1 1
1 1
1 1
USING THE VANISHING
INTEGRAL THEOREM
E 1
1
1
1
(12)*
1
1
1
1
Symmetry of H
A1
A2
B1
B2
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
vanishes if symmetries not the same
This means that we can
“blockdiagonalize” the
Hamiltonian matrix
A1 A2 B1
ψ1ψ2ψ3ψ4ψ5ψ6ψ7ψ8
0
0
0
0
0
0
0
0
. . .
. . .
. . .
0
Ψ1
Ψ2
Ψ3
Ψ4
Ψ5
Ψ6
Ψ7
ψ8
0
0
A1
A2
B1
0
0
. . .
. . .
. . .
0
0
0
0
0
0
. .
. .
0
0
0
Symmetry is preserved on diagonalization
ODME of μA
μfi = ∫ (Ψf )* μAΨi dτ
Can use symmetry to determine if this ODME = 0
This ODME will vanish if the symmetry
of (Ψf)* μAΨi is not A1
0
0
1 1
1 1
1 1
What is the symmetry ofμA ?
(12)*
1
1
1
1
E 1
1
1
1
A1
A2
B1
B2
μA = ΣCre Ar
r
Charge on
particle r
A coordinate
of particle r
A = spacefixed X, Y or Z
EμA= ?μA (12)μA = ?μA E*μA= ?μA (12)*μA= ?μA
1 1
1 1
1 1
What is the symmetry ofμZ ?
(12)*
1
1
1
1
E 1
1
1
1
A1
A2
B1
B2
μA = ΣCre Ar
r
Charge on
particle r
A coordinate
of particle r
A = spacefixed X, Y or Z
EμA= +1μA (12)μA = +1μA E*μA= 1μA (12)*μA= 1μA
μA has symmetry A2
(12) E* 1 1
1 1
1 1
1 1
Γ(H) = A1
E 1
1
1
1
(12)*
1
1
1
1
Symmetry of H
R = ±
A1
A2
B1
B2
Symmetry of μA
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,
(12) E* 1 1
1 1
1 1
1 1
Γ(H) = A1
E 1
1
1
1
(12)*
1
1
1
1
Symmetry of H
R = ±
A1
A2
B1
B2
Symmetry of μA
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
Pages 113114
∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,
that is, if the symmetry of the product ΨaΨbis not A2
Symmetry of rotational levels of H2O
b
JKaKc
a
c
Γrot
KaKc
e e A1
o o A2
e o B1
o e B2
Allowed transitions
65
Ch. 7
Ch. 6
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 1036800element CNPI group,
but a 24element point group at equilibrium, D6h
Often gives SUPERFLUOUS multiple symmetry labels
PH3
TWO VERSIONS: Distinguished
by numbering the identical nuclei
Very, very
high
potential
barrier
2
~12000 cm1
3
1
1
2
3
Bone et al., Mol. Phys., 72, 33 (1991)
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
C1
F
C2
For PH3 or CH3F this is 12/6 = 2
I
C3
For O3 this is 12/4 = 3
D
12/1 = 12
For HN3 this is 12/2 = 6
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
C1
F
C2
For PH3 or CH3F this is 12/6 = 2
I
C3
For O3 this is 12/4 = 3
D
12/1 = 12
For HN3 this is 12/2 = 6
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
C1
F
C2
For PH3 or CH3F this is 12/6 = 2
I
C3
For O3 this is 12/4 = 3
D
12/1 = 12
For HN3 this is 12/2 = 6
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
E (123) (23) E* (123)* (23)*
(132) (31) (132)* (31)*
(23) (23)*
GCNPI
GCNPI
12 elements
6 classes
6 irred. reps
PH3 (or CH3F)
Using the CNPI Group
A1’
E’ + E’’
A1’’
A1’’ + A2’
A2’
A2’’
A1’ + A2’’
E’
Why this
Degeneracy?
E’’
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.
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works, in principle, and can be used
to determine which ODME vanish.
BUT IT IS OFTEN HOPELESSLY UNWIELDY
In 1963 LonguetHiggins figured out how to set up a subgroup 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
1
2
3
It would be
superfluous
to calc points
in other min
No observed tunneling through barrier
PH3 (or CH3F)
Only NPI OPERATIONS FROM IN HERE
Very, very
high
potential
barrier
2
3
1
1
2
3
No observed tunneling through barrier
PH3 (or CH3F)
Only NPI OPERATIONS FROM IN HERE
Very, very
high
potential
barrier
2
3
1
1
2
3
(12) superfluous
E* superfluous
(123), (12)* useful
No observed tunneling through barrier
GMS ={E, (123), (132), (12)*, (13)*,(23)*}
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
useful
superfluous
GCNPI={E,(12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
For CH3F:
useful
GMS ={E, (123), (132), (12)*, (13)*,(23)*}
If we cannot see any effects
of the tunneling through the
barrier then we only need
NPI operations or one
version. Omit NPI elements
that take us between
versions since they are not
useful; they are unfeasible.
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
unfeasible.
superfluous
For PH3
or CH3F:
GCNPI={E,(12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
Character Table of CNPI group of PH3 or CH3F
GCNPI={E, (12), (13), (23), (123), (132),
E*, (12)*, (13)*,(23)*, (123)*, (132)*}
H
A1’
E (123) (23) E* (123)* (23)*
(132) (31) (132)* (31)*
(23) (23)*
GCNPI
GCNPI
12 elements
μA
6 irred. reps
A1”
E’ + E’’
A1’
E’ + E’’
A1’’
A1’’ + A2’
A2’
A2’’
A1’ + A2’’
E’
E’’
2 versions → sum or two irrep labels on each level
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.
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)
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