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MOLECULAR SYMMETRYAND SPECTROSCOPY

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I(f ← i) = ∫8π3 Na______

(4πε0)3hcF(Ei ) S(f ← i) Rstim(f→i) = νif

~line

ε(ν)dν~ ~

Integrated absorption coefficient (i.e. intensity) for a line is:

hνif = Ef – Ei = ΔEif Absorption can only occur at resonance f

iM

νif

ODME of H

μfi = ∫ (Ψf )* μA Ψi dτODME of μA

Use Q. Mech. to calculate:

Chapters 1 and 2. Spectroscopy and Quantum Mechanics

We began by summarizing

P. R. Bunker and Per Jensen:

Fundamentals of Molecular Symmetry,

Taylor and Francis, 2004.


Molecular Symmetry and Spectroscopy,

2nd Edition, 3rd Printing,

NRC Research Press, Ottawa, 2012.

To buy it go to:

http://www.crcpress.comDownload pdf file from

www.chem.uni-wuppertal.de/prb

Chapter 1 (Spectroscopy)Chapter 2 (Quantum Mechanics) andSection 3.1 (The breakdown of the BO Approx.)

The first 47 pages:

http://www.crcpress.com/


http://www.chem.uni-wuppertal.de/prb


Fundamentals of Molecular Symmetry,

Taylor and Francis, 2004.


Molecular Symmetry and Spectroscopy,

2nd Edition, 3rd Printing,

NRC Research Press, Ottawa, 2012.

To buy it go to:

http://www.crcpress.comDownload pdf file from

www.chem.uni-wuppertal.de/prb

Chapter 1 (Spectroscopy)Chapter 2 (Quantum Mechanics) andSection 3.1 (The breakdown of the BO Approx.)

The first 47 pages:

We then proceeded to discuss GroupTheory and Point Groups



http://www.chem.uni-wuppertal.de/prb

“Group”

A set of operations that is closed wrt “multiplication”

“Point Group”

All rotation, reflection and rotation-reflection operationsthat leave the molecule (in its equilibrium configuration)“looking” the same.

“Matrix group”

A set of matrices that forms a group.

“Irreducible representation”

A representation that cannot be written as the sumof smaller dimensioned representations.

Definitions for groups and point groups:

“Representation”

A matrix group having the same shaped multiplicationtable as the group it represents.

“Character table”A tabulation of the characters of the irreducible representations.

Character table for the point group C3v

E (123) (12)

1 2 3

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 thistable we give the characters of the matrices.

E C3 σ1

C32 σ2

σ3

Elements in the same class have the same characters

Two 1Dirreduciblerepresentationsof the C3v group

3 classes and 3 irreducible representations

Character table for the point group C2v

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

E C2 σyz σxy

4 classes and 4 irreducible representations

z

x

(+y)

E C3 σ1

C32 σ2

σ3

C3v

1

3

2

hνif = Ef – Ei = ΔEif

S(f ← i) = ∑A | ∫ Φf* μA Φi dτ |2

ODME of H and μA μfi =

Spectroscopy

Quantum Mechanics

Group Theory and Point Groups

PH3

f

iMMMM

M

∫ Φf* μA Φi dτ

(Character Tables and Irreducible Representations)

8

Point Group symmetry is based onthe geometrical symmetry of theequilibrium structure.

Point group symmetry not appropriate when there is rotation or tunneling

Use energy invariance symmetryinstead. We start by using inversionsymmetry and identical nuclear permutation symmetry.

The Complete Nuclear Permutation Inversion (CNPI) Group

Contains all possible permutationsof identical nuclei including E. It also contains the inversion operation E*and all possible products of E* withthe identical nuclear permutations.

GCNPI = GCNP x {E,E*}

The spin-free (rovibronic) Hamiltonian

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 thenselectively correct for the approximations made.

(after separating translation)

The CNPI Group for the Water Molecule

The Complete Nuclear Permutation Inversion (CNPI) group

for the water molecule is {E, (12)} x {E,E*} = {E, (12), E*, (12)*}

H1 H2

O e+

H2 H1

O e+ H1 H2

Oe-(12) E*

(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,…

I

F

H

D

C1

C2

C3

O OO

1 3

2

H H12C 13C

D H

1 2

3

1

32

1

32

F

H N1N2N3

H1

H2 H3

+

GCNPI = {E, (12), (13), (23), (123), (132)} x {E, E*} = GCNP x {E, E*}

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

GCNPI = {E, (12), (13), (23), (123), (132)} x {E, E*}

Number of elements = 3! x 2 = 6 x 2 = 12

Number of ways of permutingthree identical nuclei

GCNPI = {E, (12), (13), (23), (123), (132)} x{E, (45)} x {E, E*}

(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

The CNPI Group of C3H2ID

I

H5

D

C1

C2

C3

H4

= {E, (12), (13), (23), (123), (132),

I

H5

D

C1

C2

C3

H4 Number of elements = 3! x 2! x 2 = 6 x 2 x 2 = 24

If there are n1 nuclei of type 1, n2 oftype 2, n3 of type 3, etc then the totalnumber of elements in the CNPI groupis n1! x n2! x n3!... x 2.

H5TheAllene molecule C1

C2

C3

H4


H7

H6

C3H4

The CNPI group of allene


C2

C3

H4


H7

H6

C3H4


Sample elements: (456), (12)(567), (4567), (45)(67)(123)


C2

C3

H4


H7

H6

C3H4


00H

00H

C3H4O4How many elements?


C2

C3

H4


H7

H6

C3H4


00H

00H

3! x 4! x 4! x 2 = 6912 C3H4O4

Number of elements in the CNPI groups of variousmolecules

The size of the CNPI group depends only on the chemical formula

(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 numberMolecule 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 43200C6H6

22

This number means something!

End of Review of Lecture One

ANY QUESTIONS OR COMMENTS?

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:

RHn = HRn

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

The character table of the CNPIgroup of the water molecule

It is called C2v(M)

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

The character table of the CNPIgroup of the water molecule

It is called C2v(M)

Now to explain how we labelenergy levels using

irreducible representations

Labelling energy levels

RH = RE

Since RH = HR and E is a number, this leads to HR = ER.

H = E

E is nondegenerate. Thus RΨ = cΨ.

But R2 = identity. Thus c2 = 1, so c = ±1 and R = ±

For the water molecule (no degeneracies, and R2 = identity for all R) :

H(R) = E(R)

The eigenfunctions have symmetry

R = (12), E* or (12)*

+ Parity - Parity

x

Ψ1+(x)

x

Ψ3+(x)

x

Ψ2-(x)

Ψ+(-x) = Ψ+(x)

Ψ-(-x) = -Ψ-(x)

Eigenfunctions of Hmust satisfyE*Ψ = ±Ψ

R = E*

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 approachSee PRB and Howard (1983)

+ Parity - Parity

x

Ψ1+(x)

x

Ψ3+(x)

x

Ψ2-(x)

Ψ+(-x) = Ψ+(x)

Ψ-(-x) = -Ψ-(x)

Eigenfunctions of Hmust satisfyE*Ψ = ±Ψ

and (12)ψ = ±ψ

There are four symmetry types of H2O wavefunction

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

R = ±

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

The Symmetry Labels of the CNPI Group of H2O

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

We are labelling the states using the irreps of the CNPI group


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different. ∫ΨaμΨbdτ = 0 if symmetry of product is not A1Eψ=+1ψ (12)ψ =+1ψ E*ψ=-1ψ (12)*ψ=-1ψ

Thus, for example, a wavefunction of “A2 symmetry” will “generate” the A2 representation:


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different. ∫ΨaμΨbdτ = 0 if symmetry of product is not A1Eψ=+1ψ (12)ψ =+1ψ E*ψ=-1ψ (12)*ψ=-1ψ

Thus, for example, a wavefunction of “A2 symmetry” will “generate” the A2 representation:

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).

This would be labelled as B2.

Eψ=+1ψ (12)ψ =-1ψ E*ψ=+1ψ (12)*ψ=-1ψ

Suppose Rn = E where n > 2.

C3(M) E (123) (132)

1 1 1

A 1 1 1

Ea 1 *

Eb 1 *

where =

ei2/3

We still have RΨ = cΨ for nondegenerate Ψ, but now Rn Ψ = Ψ.

Thus cn

= 1 and c = n√1, i.e. c = [ei2π/n]a where a = 1,2,…,n.

If n = 3, c = ε, ε2 (=ε*), or ε3 (=1)

eiπ = -1 C3 C3

2

ei2π = 1

For nondegenerate states we hadthis as the effect of a symmetry operation on an eigenfunction:

RH = RE

HR = ER

H = E

Thus R = c since E is nondegenerate.

For the water molecule ( nondegenerate) :

What about degenerate states?

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 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 ℓ-dimensional representation that is generated by the ℓ functions Ψnk

The ℓ functions Ψnk transform according to this representation

degenerate energy level with energy Enℓ-fold

Labelling energy levels using the CNPI Group

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different. ∫ΨaμΨbdτ = 0 if symmetry of product is not A1

We can label energy levels using the irreps of the CNPI group for any molecule

34

Pages 143-149 Pages 99-101

Example of using the symmetry operation (12):

H1

H2

r1´r2´

´(12)

We have (12) (r1, r2, ) = (r1´, r2´, ´)

We see that (r1´, r2´, ´) = (r2, r1, )

Determining symmetry and reducing a representation

2

3

1

1

3

2

3

12

1

3

23

1 2

1

3

2

1

3

21

3

2

r2´r1´

´

r1r2

r1r2

r1r2

r1r2

r1´r2´

´

r1´r2´´

r2´r1´´

(12)

E

E*

(12)*

r1´ r1 r2´ = r2 ́

r1´ r2 r2´ = r1 ́

r1´ r1 r2´ = r2 ́

r1´ r2 r2´ = r1 ́

(12)

E

E*

(12)*

r1 r1´ r1 1 0 0 r1 r2 = r2´ = r2 = 0 1 0 r2 ́ 0 0 1

r1 r1´ r2 0 1 0 r1 r2 = r2´ = r1 = 1 0 0 r2 ́ 0 0 1

r1 r1´ r1 1 0 0 r1 r2 = r2´ = r2 = 0 1 0 r2 ́ 0 0 1

r1 r1´ r2 0 1 0 r1 r2 = r2´ = r1 = 1 0 0 r2 ́ 0 0 1

R a = a´ = D[R] a

= 3

= 1

= 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 = ( 13 + 11 + 13 + 11) = 24

1

aA2 = ( 13 + 11 13 11) = 04

1

aB1 = ( 13 11 13 + 11) = 04

1

aB2 = ( 13 11 + 13 11) = 14

1

= 2 A1 B2

A reducible representation

Γ = Σ aiΓi i

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 = ( 13 + 11 + 13 + 11) = 24

1

aA2 = ( 13 + 11 13 11) = 04

1

aB1 = ( 13 11 13 + 11) = 04

1

aB2 = ( 13 11 + 13 11) = 14

1

= 2 A1 B2

A reducible representation

Γ = Σ aiΓi

i

i

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 Character

1

Projection operators:

General for li-dimensional irrep i

Diagonal element of representation matrix

Symmetry operation

Simpler for 1-dimensional irrep i Character

1

E (12) E* (12)* A1 1 1 1 1

PA1 = (1/4) [ E + (12) + E* + (12)* ]

Projection operator for A1 acting on r1

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4

1

4

1

2

1

= [ + + + ] =

S2 = P11A1 = [ E + (12) + E* + (12)* ] 4

1

4

1

= [ r1 r2 + r1 r2 ] = [ r1 r2 ]

S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4

1

4

1

2

1

= [ + ] = 0

P11B2 = [ E (12) + E* (12)* ] 4

1

4

1

Is „annihilated“ by P11B2

PA1

PA1

PB2

PB2

PB2

Projection operators for A1 and B2

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4

1

4

1

2

1

= [ + + + ] =

S2 = P11A1 = [ E + (12) + E* + (12)* ] 4

1

4

1

= [ r1 r2 + r1 r2 ] = [ r1 r2 ]

S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4

1

4

1

2

1

= [ + ] = 0

P11B2 = [ E (12) + E* (12)* ] 4

1

4

1

Is „annihilated“ by P11B2

PA1

PA1

PB2

PB2

PB2

Projection operators for A1 and B2

= [ r1 + r2 + r1 + r2 ] = [ r1 + r2 ]

S1 = P11A1r1 = [ E + (12) + E* + (12)* ]r1 4

1

4

1

2

1

= [ + + + ] =

S2 = P11A1 = [ E + (12) + E* + (12)* ] 4

1

4

1

= [ r1 r2 + r1 r2 ] = [ r1 r2 ]

S3 = P11B2r1 = [ E (12) + E* (12)*] r1 4

1

4

1

2

1

4

1

4

1

PA1

PA1

PB2

Aside: S1, S2 and S3 have the symmetry and form of thenormal coordinates.

A1

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

The threeNormal modesof the watermolecule

Labeling is not just bureaucracy. It is useful.

PAUSE50

The vanishing integral theorem

Labeling is not just bureaucracy. It is useful.

Pages 114-117 Pages 136-139

The symmetry of a product Pages 109-114

But first we look at

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

A2 x B1 = B2, B1 x B2 = A2, B1 x A2 x B2 = A1

∫ΨaHΨbdτ = 0 if symmetries of Ψa and Ψb are different.

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

Eψ=+1ψ (12)ψ =+1ψ E*ψ=-1ψ (12)*ψ=-1ψ

The symmetry of the product φψ is B1 x A2 = B2.

THE SYMMETRY OF A PRODUCT

A2

Eφ =+1φ (12)φ =-1φ E*φ =-1φ (12)*φ =+1φ B1

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

E E: 4 1 0

A1 A1 = A1

A1 A2 = A2

A2 A2 = A1

A1 E = E

A2 E = E

E E = A1 A2 E

Characters of the product representation are the products of the characters of the representations being multiplied.

Reducible representation

Symmetry of triple product is obvious extension

+ Parity - Parity

x

Ψ+(x)

x

Ψ+(x)

x

Ψ-(x)

Ψ+(-x) = Ψ+(x)

Ψ-(-x) = -Ψ-(x)

∫Ψ+Ψ-Ψ+dx = 0

- parity

+ Parity - Parity

x

Ψ+(x)

x

Ψ+(x)

x

Ψ-(x)

Ψ+(-x) = Ψ+(x)

Ψ-(-x) = -Ψ-(x)

∫Ψ+Ψ-Ψ+dx = 0

- parity

∫f(τ)dτ = 0 if symmetry of f(τ) is not A1

The vanishing integral theorem

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,

Symmetry of H

Using symmetry labels and the vanishingintegral theorem we deduce that:

USING THE VANISHING INTEGRAL THEOREM

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” theHamiltonian matrix

ψ1ψ2ψ3ψ4ψ5ψ6ψ7ψ8

Ψ1

Ψ2

Ψ3

Ψ4

Ψ5

Ψ6

Ψ7

ψ8

A1 A2 B1

A1

A2

B1

0 0

000 0

0 0 0 0 0 0 0 0

0

0

0

0

0

0

0

0

. . .

. . .

. . .. . .. . .. . .

. .

. .

Symmetry is preserved on diagonalization

μfi = ∫ (Ψf )* μA Ψi dτ

ODME of μA

Can use symmetry to determine if this ODME = 0

This ODME will vanish if the symmetryof (Ψf )* μA Ψi is not A1

0 0

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

μA = Σ Cre Arr

Charge onparticle r

A coordinateof particle r

What is the symmetry of μA ?

A = space-fixed X, Y or Z

EμA= ?μA (12)μA = ?μA E*μA= ?μA (12)*μA= ?μA

(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

μA = Σ Cre Arr

Charge onparticle r

A coordinateof particle r

What is the symmetry of μZ ?

EμA= +1μA (12)μA = +1μA E*μA= -1μA (12)*μA= -1μA

μA has symmetry A2

A = space-fixed X, Y or Z


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

Symmetry of μA

R = ±

∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,

Symmetry of H


Γ(μA) = A2

Γ(H) = A1

∫Ψa*HΨbdτ = 0 if symmetry of Ψa*HΨb is not A1,


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

Symmetry of μA

R = ±

∫Ψa*μAΨbdτ = 0 if symmetry of ψa*μAψb is not A1,

that is, if the symmetry of the product ΨaΨb is not A2

Symmetry of H


Γ(μA) = A2

Γ(H) = A1

∫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 113-114

Symmetry of rotational levels of H2O

JKaKc

KaKc

e e A1

o o A2

e o B1

o e B2

Γrot

a

b

c

Allowed transitions

So we can use the CNPI group to:

1.Symmetry label energy levels

and

2. Determine which ODME vanish.

65Ch. 7 Ch. 6

BUT BUT BUT...There are problems with the CNPI Group

Number of elements in the CNPI groups of variousmolecules

C6H6, for example, has a 1036800-element CNPI group,but a 24-element point group at equilibrium, D6h

Huge groups. Size bears no relation to geometrical symmetry

Often gives SUPERFLUOUS multiple symmetry labels

PH3

1

2

3

There are two VERSIONSof this molecule

2

3113

2

PH3

Very, veryhighpotentialbarrier

~12000 cm-1

TWO VERSIONS: Distinguishedby numbering the identical nuclei

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 PH3 or CH3F this is 12/6 = 2

For H2O this is 4/4 = 1

For O3 this is 12/4 = 3

For H3+ this is 12/12= 1

For HN3 this is 12/2 = 6

I

F

H

D

C1

C2

C3

12/1 = 12

The six versions of HN3

H N1 N2 N3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

I

F

H

D

C1

C2

C3

12 versions








I

F

H

D

C1

C2

C3

12/1 = 12





For C6H6 this is (6!x6!x2)/24 = 1036800/24 = 43200




I

F

H

D

C1

C2

C3

12/1 = 12

Character Table of CNPI group of PH3

12 elements 6 classes

6 irred. reps

GCNPI E (123) (23) E* (123)* (23)* (132) (31) (132)* (31)* (23) (23)*

GCNPI

Using the CNPI Group to symmetry label the energy levels of a molecule that has more than one version.

A1’

A1’’

A2’

A2’’

E’

E’’

Using the CNPI Group

E’ + E’’

A1’’ + A2’

A1’ + A2’’

PH3 (or CH3F)

A1’

A1’’

A2’

A2’’

E’

E’’


E’ + E’’

A1’’ + A2’

A1’ + A2’’

PH3 (or CH3F)

Why thisDegeneracy?

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 C6H6 this is 1036800/24 = 43200,


For C3H4 this is 288/8 = 36

and using the CNPI group each energy level wouldget as symmetry label the sum of 43200 irreps.

77

Clearly using the CNPI group givesvery unwieldy symmetry labels.

This subgroup is called The Molecular Symmetry (MS) group.

The CNPI Group approachworks, in principle, and can be used to determine which ODME vanish.

In 1963 Longuet-Higgins figured out how to set up a sub-group of the CNPI Group that achieves the same resultwithout superfluous degeneracies.

BUT IT IS OFTEN HOPELESSLY UNWIELDY

2

3113

2


No observed tunneling through barrier

Ab Initio CALC IN HERE

Ab initio calc with neglect of tunnelingPH3 (or CH3F)

It would besuperfluousto calc pointsin other min

2

3113

2



Only NPI OPERATIONS FROM IN HERE

PH3 (or CH3F)

2

3113

2



Only NPI OPERATIONS FROM IN HERE

PH3 (or CH3F)

(12) superfluousE* superfluous(123), (12)* useful

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}For CH3F:

GMS ={E, (123), (132), (12)*, (13)*,(23)*}

The six feasible elements are

superfluous

superfluous

useful

useful

If we cannot see any effectsof the tunneling through thebarrier then we only needNPI operations for one version. Omit NPI elementsthat connect versions since they are not useful; they are

superfluous.

If we cannot see any effectsof the tunneling through thebarrier then we only needNPI operations or one version. Omit NPI elementsthat take us betweenversions since they are not useful; they are unfeasible.

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

For PH3

or CH3F:

GMS ={E, (123), (132), (12)*, (13)*,(23)*}

The six feasible elements are

superfluous

If we cannot see any effectsof the tunneling through thebarrier then we only needNPI operations for one version. Omit NPI elementsthat connect versions since they are not useful; they are

unfeasible.

GCNPI={E, (12), (13), (23), (123), (132),

E*, (12)*, (13)*,(23)*, (123)*, (132)*}

Character Table of CNPI group of PH3 or CH3F

12 elements

6 irred. reps

GCNPI E (123) (23) E* (123)* (23)* (132) (31) (132)* (31)* (23) (23)*

GCNPI

μA

A1”

HA1’

A1’

A1’’

A2’

A2’’

E’

E’’


E’ + E’’

A1’’ + A2’

A1’ + A2’’

2 versions → sum or two irrep labels on each level

E’ + E’’

Character table of the MS group of PH3 or CH3F

E (123) (12)

1 2 3

A1 1 1 1

A2 1 1 1

E 2 1 0

E (123) (12)* (132) (13)* (23)*

μA A2

H A1

A1

A2

E

Using the MS Group

E

A2

A1

The MS group gives all the information we needas long as there is no observable tunneling, i.e.,as long as the barrier is insuperable.

E

Using CNPIG versus MSG for PH3

E

A2

A1

Can use either to determine if an ODME vanishes.But clearly it is easier to use the MSG.

E’ + E’’

A1’’ + A2’

A1’ + A2’’

CNPIG MSG

E’ + E’’ E

The subgroup of feasible elements forms a group called

THE MOLECULAR SYMMETRY GROUP(MS GROUP)

Unfeasible elements of the CNPI groupinterconvert versions that are separated

by an insuperable energy barrier

89

superfluous

useful