Symmetry and Spectroscopy. P. R. Bunker and Per Jensen: Molecular Symmetry and Spectroscopy, 2 nd Edition, 2 nd Printing, NRC Research Press, Ottawa,

Symmetry and

Spectroscopy

P. R. Bunker and Per Jensen:

Molecular Symmetry and Spectroscopy, 2nd Edition,

2nd Printing, NRC Research Press, Ottawa, 2006

(ISBN 0-660-19628-X). $49.95 for 747 pages.

paperback. BJ1

P. R. Bunker and Per Jensen:

Fundamentals of Molecular Symmetry,

IOP Publishing, Bristol, 2004

(ISBN 0-7503-0941-5). $57.95

paperback. BJ2

Examples of point group symmetry

H2O

CH3F

C60

C3H4

C2v

C3v

D2d

Ih

Examples of point group symmetry

H2O

CH3F

C60

C3H4

C2v

C3v

D2d

Ih

Point group symmetry of H2O

z

y

(-x)

The point group C2v consists ofthe four operations E, C2y, yz, and xy

The word ´group´ is loaded. To see howwe do two operations in succession

Point groups: Number of rotation axes and reflection planes.

z

y

(-x)

1 2

σyz

C2

z

y

(-x)

1 2

z

y

(-x)

z

y

(-x)

σxy = C2 σyz

1 2

σyz

σxy

C2

z

y

(-x)

1 2

z

y

(-x)

Multiplication Table for H2O Point group

z

y

(-x)C2v = {E, C2, yz, xy }

E C2 yz xy E E C2 yz xy C2 C2 E xy yz yz yz xy E C2 xy xy yz C2 E

Multiplication table (=Rrow Rcolumn, in succession)

Use multiplication table to prove that it is a “group.”

σxy = C2 σyz

{E, C2, yz, xy } forms a “group“ if it obeysthe following GROUP AXIOMS :

•All possible products RS = T belong to the group

•Group contains identity E (which does nothing)

•The inverse of each operation R1 (R1R =RR1 =E ) is in the group

•Associative law (AB )C = A(BC ) holds


C2v

Fermi:

{E, C2, yz, xy } forms a “group“ if it obeysthe following GROUP AXIOMS :






C2v ‘‘Group theory is just a bunch of definitions‘‘






Not a GROUP






Is a GROUP(subgroup of C2v)

Rotationalsubgroup

PH3 at equilibrium

Symmetry operations:C3v = {E, C3, C3

2, 1, 2, 3 }

Symmetry elements:

C3, 1, 2, 3

C3 Rotation axis

k Reflection plane

C32 = 2 1

Multiplying C3v symmetry operations

Reflection

Reflection

Rotation

Multiplication table for C3v

C32 = σ2σ1


C32 = σ2σ1

Note that C3 = σ1σ2


Rotationalsubgroup


3 classes

A matrix group

1 0

0 1

21

23

23

21

21

23

23

21

1 0

0 1

21

23

23

21

21

23

23

21

´M1 = M4 =

´ M5 =

´ M6 =

´M2 =

´M3 =

M1

M2

M3

M4

M5

M6

M1

M1

M2

M3

M4

M5

M6

M2

M2

M3

M1

M6

M4

M5

M3

M3

M1

M2

M5

M6

M4

M4

M4

M5

M6

M1

M2

M3

M5

M5

M6

M4

M3

M1

M2

M6

M6

M4

M5

M2

M3

M1

Multiplication table for the matrix group

Products are Mrow Mcolumn

E C3 C32 σ1 σ2 σ3

This matrix group forms a “representation” of the C3v groupThese two groups are isomorphic.

Multiplication tableshave the ‘same shape’

Irreducible Representations

The matrix group we have just introducedis an irreducible representation of the C3v

point group.

The sum of the diagonal elements (character)of each matrix in an irreducible representationis tabulated in the character table of thepoint group.

The characters of this irreducible rep.

1 0

0 1

21

23

23

21

21

23

23

21

1 0

0 1

21

23

23

21

21

23

23

21

´M1 = E M4 = 1

´ M5 = 2

´ M6 = 3

´M2 = C3

´M3 = C32

C3v C3v

2

-1

-1

0

0

0

The characters of this irreducible rep.

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

3 classes

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

The matrices of the E irreducible rep.

1 0

0 1

21

23

23

21

21

23

23

21

1 0

0 1

21

23

23

21

21

23

23

21

´M1 = E M4 = 1

´ M5 = 2

´ M6 = 3

´M2 = C3

´M3 = C32

C3v C3v

The matrices of the A1 + E reducible rep.

21

23

23

21

21

23

23

21

1 0

0 1

21

23

23

21

21

23

23

21

´M1 = E M4 = 1

´ M5 = 2

´ M6 = 3

´M2 = C3

´M3 = C32

C3v C3v

1 0 00 1 00 0 1

1 0 00

0

1 0 00

0

1 0 00

0

1 0 000

1 0 00

0

‘

‘

‘

‘

‘

‘

‘

The matrices of the A2 + E reducible rep.

21

23

23

21

21

23

23

21

1 0

0 1

21

23

23

21

21

23

23

21

´M1 = E M4 = 1

´ M5 = 2

´ M6 = 3

´M2 = C3

´M3 = C32

C3v C3v

1 0 00 1 00 0 1

-1 0 00

0

1 0 00

0

1 0 00

0

-1 0 000

-1 0 00

0

‘

‘

‘

‘

‘

‘

‘

‘

‘

‘

‘

‘

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

Irreducible representations are “symmetry labels”

Some of Fermi’s definitions• Group

• Subgroup

• Multiplication table of group operations

• Classes

• Representations

• Irreducible and reducible representations

• Character table

See, for example, pp 14-15 and Chapter 5 of BJ1

Some of Fermi’s definitions• Group

• Subgroup

• Multiplication table of group operations

• Classes

• Representations


• Character table

See, for example, pp 14-15 and Chapter 5 of BJ1

Irreducible representationsThe elements of irrep matrices satisfy the„Grand Orthogonality Theorem“ (GOT).

We do not discuss the GOT here, but we list threeconsequences of it: • Number of irreps = Number of classes in the group.

• Dimensions of the irreps, l1, l2, l3 … satisfy

l12 + l2

2 + l32 + … = h,

where h is the number of elements in the group.

• Orthogonality relation


These are used as ‘‘symmetry labels‘‘on energy levels.

Which energy levels can ‘‘interact‘‘and which transitions can occur.

Can also determine whether certain terms are in the Hamiltonian.

IN SOME CIRCUMSTANCES THERE ARE PROBLEMS IF WETRY TO USE POINT GROUP SYMMETRY TO DO THESE THINGS

BUT

How do we use point group symmetry if the molecule rotates and distorts?

H3+

D3h C2v

2

3

113

2

Or if tunnels?

NH3

C3vD3h

What are the symmetriesof B(CH3)3, CH3.CC.CH3, (CO)2, (NH3)2,…?

Nonrigid molecules (i.e. moleculesthat tunnel) are a problemif we try to use a point group.

What should we do if we study transitions (or interactions) between electronic states that have different point group symmetries at equilibrium?

Also

Point groups used for classifying:

The electronic states for any moleculeat a fixed nuclear geometry (see BJ2 Chapter 10), and

The vibrational states for molecules,called “rigid” molecules, undergoing infinitesimal vibrations about a unique equilibrium structure(see BJ2 Pages 230-238).

Rotations andreflections

Permutationsand the inversion

J.T.Hougen, JCP 37, 1422 (1962); ibid, 39, 358 (1963)H.C.Longuet-Higgins, Mol. Phys., 6, 445 (1963)P.R.B. and Per Jensen, JMS 228, 640 (2004) [historical introduction]

To understand how we use symmetrylabels and where the point groupgoes wrong we must studywhat we mean by “symmetry”

See also BJ1 and BJ2

Symmetry not from geometry since molecules are dynamic

• Centrifugal distortion

eg. H3+ or CH4 dipole

moment

• Nonrigid molecules: eg. ethane, ammonia, (H2O)2, (CO)2,…

• Breakdown of BOA: eg. HCCH* - H2CC

Also symmetry appliesto atoms, nuclei and fundamental particles. Geometrical point group symmetry is not possible for them.

We need a more general definition of symmetry

Symmetry Based on Energy Invariance

Symmetry operations are operations that leave the energy of the system (a molecule in our case) unchanged.

Using quantum mechanics we define a symmetry operationas follows:

A symmetry operation is an operation that commutes with the Hamiltonian:

(RH – HR)n = [R,H]n = 0

• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q,s) (-p,-q,s) P(E*) • Reversal symmetry-----Time reversal (p,q,s) (-p,q,-s) T • Ch. conj. Symmetry-----Particle antiparticle C

Symmetry Operations (energy invariance)

• Uniform Space ----------Translation

Symmetry Operations (energy invariance)Separate translation…Translational momentum

Ψtot = Ψtrans Ψint

int = rot-vib-elec.orb-elec.spin-nuc.spin

• Uniform Space ----------Translation• Isotropic Space----------Rotation • Identical electrons-------Permute electrons• Identical nuclei-----------Permute identical nuclei • Parity conservation-----Inversion (p,q) (-p,-q) P(E*) • Reversal symmetry-----Time reversal (p,s) (-p,-s) T • Ch. conj. Symmetry-----Particle antiparticle C


K(spatial) group,J, mJ or F,mF labels



Symmetric group Sn

For the BeH molecule (5 electrons) Ψorb-spin transforms as D(0) of S5

PEP

Slater determinant ensures antisymmetry so do not need S5





CNPI group = Complete Nuclear Permutation Inversion Group



CNPI group = Complete Nuclear Permutation Inversion Group

EXAMPLE:The CNPI group for H2O is C2v(M) = {E, (12), E*, (12)*}

The CNPI Group for the Water Molecule

The Complete Nuclear Permutation Inversion (CNPI) group

for the water molecule is C2v(M) = {E, (12), E*, (12)*}

H1 H2

O e+

H2 H1

O e+ H1 H2

Oe-(12) E*

(12)*

We compare C2v and this CNPI groupMultiplication table (Rrow Rcolumn)


E (12) E* (12)* E E (12) E* (12)*

(12) (12) E (12)* E* E* E* (12)* E (12)

(12)* (12)* E* (12) E

C2v and CNPI are isomorphic!

C2v

CNPI

We compare C2v and this CNPI group


E (12) E* (12)* E E (12) E* (12)*

(12) (12) E (12)* E* E* E* (12)* E (12)

(12)* (12)* E* (12) E

Rotationalsubgroup

Permutationsubgroup

CNPI group of water: Character table

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

This group is called C2v(M)

Why RH = HR used to Define Symmetry?

RH = RE

HR = ER

H = E

Thus R = c since E is nondegenerate.However R2 = E, so R(RΨ) = Ψ, but R(RΨ) = c2Ψ. Thus c2 = 1,c = ±1 and R = ±

For the water molecule ( nondegenerate and R2 = E for all R) :

Symmetry of H restricts symmetry of eigenfunctions Ψ

+ 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*

”Why RH = HR used to Define Symmetry?”continued……

RH = RE

HR = ER

H = E

Thus R = c. However R2 = E, so c = ±1 and R = ±

Allows us to SYMMETRY LABEL the energy levels using the irreps of the symmetry group

For the water molecule (with nondegenerate states):

Symmetry of H restricts symmetry of eigenfunctions Ψ

RH=HRimplies

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

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

∫ΨaμΨbdτ = 0 if symmetry of product is not A1

Possible labels would be (1,1), (1,-1), (-1,-1), and (-1,1). However. More generally systematic are the irreducible representation labels (or symmetry labels) from the symmetry group.

The Symmetry Labels of the C2v(M) Group of H2O

(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

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

∫ΨaμΨbdτ = 0 if symmetry of product is not A1


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

R = ±

because of:The vanishing integral theoremBJ1 pp114-117, BJ2 pp 136-139

Labeling is not just bureaucracy…it is useful


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

R = ±

because of:The vanishing integral theoremBJ1 pp114-117, BJ2 pp 136-139

Labeling is not just bureaucracy…it is useful

But first let’s look at three things we overlooked:

Rn=E with n>2, degenerate states, symmetry of a product

Suppose Rn = E where n > 2.

C3(M) E (123) (132)

1 1 1

A 1 1 1

Ea 1 *

Eb 1 *

= ei2/3

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

Thus cn

= 1 and c = n√1

If n = 3 we introduce and c = 1,ε or ε2 (=ε*)

eiπ = -1 C3 C3

2

ei2π = 1

Suppose Rn = E where n > 2.

C3(M) E (123) (132)

1 1 1

A 1 1 1

Ea 1 *

Eb 1 *

= ei2/3

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

Thus cn

= 1 and c = n√1

If n = 3 we introduce and c = 1,ε or ε2 (=ε*)

C3 C32

A pair of separablydegenerateirreps. Degeneratebecause of T

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

Symmetry of a product:C2v(M) example

E (12) E* (12)*

A1 1 1 1 1

A2 1 1 1 1

B1 1 1 1 1

B2 1 1 1 1

B1 x B2, A1 x A2, B1 x A2, B2 x A2, B1 x B1,… A2 A2 B2 B1 A1

The symmetry of the product of two nondegenerate states is easy:

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.See pp 109-114 in BJ1

Reducible representation

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.See pp 109-114 in BJ1

Reducible representation

We say that E x E A1

Back to the vanishing

integral theorem

+ 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) does not contain A1

The vanishing integral theorem

Diagonalizing the molecular Hamiltonian

Schrödinger equation

To apply the vanishing integral rule we look at symmetry of

Eigenvalues and eigenfunctions are found by diagonalization of a matrix with elements

Diagonalizing the molecularHamiltonian

Hmn vanishes if Γ( ) and Γ( ) are different

The Hamiltonian matrix factorizes, for example for H2O

if Γ(integrand) does not contain Γ(s)

= 0

F(Ei ) = [ gie-Ei/kT ] / gje-Ej/kT Boltzmann factor∑j

S(f ← i) = | ∫ Φf* μA Φi dτ |2 Line strength∑A=X,Y,Z

Rstim(f→i) = 1 – exp (-hνif /kT ) Stimulated emission

I(f ← i) = ∫

8π3 Na______(4πε0)3hc2

F(Ei ) S(f ← i) Rstim(f→i)

Integrated absorption intensity for a line is:

= νif

~line

Frequency factor

ε(ν)dν~

Selection rules for transitions

The intensity of a transition is proportional to the square of

For the integral to be non-vanishing, the integrand musthave a totally symmetric component.

μZ = Σ qi Zi

Z

i

Product of symmetries of Φs must contain that of μZ

Symmetry of Z

Z has symmetry *

* ???

What is

Symmetry of Z

Z has symmetry *

* has character +1 under all permutations P

1 under all permutation-inversions P*

Symmetry of Z for H2O

* = A2


(12) E* 1 1

1 -1 -1 -1 -1 1

E 1111

(12)* 1 -1 1-1

A1

A2

B1

B2

Symmetry of μZ

R = ±

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

As a result the Hamiltonian matrix is block diagonal.

∫Ψa*μΨbdτ = 0 if symmetry of product ΨaΨb is not Γ*

Symmetry of H

Using symmetry labels and the vanishingintegral theorem we deduce that:

Γ(μZ) = Γ*

0 0 00

Γ(H) = Γ(s)

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


Γ = Σ aiΓi

i

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


Γ = Σ aiΓi

i

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


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 operatorsPages 102-109 of BJ1

Projection operators:

General for li-dimensional irrep i

Diagonal element of representation matrix

Symmetry operation

Simpler for 1-dimensional irrep i Character

1




Symmetry operation


1




Symmetry operation


1

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

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




Symmetry operation


1

E (12) E* (12)* A1 1 1 1 1 B2 1 -1 1 -1

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

PB2 = (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. See pp 269-277 in BJ1, and 232-233 in BJ2

[H,R] Symmetry and conservation laws(see chapter 14 of BJ2)

iħ ∂Ψ/∂t = HΨ where Ψ is a function of q and t

Does symmetry change with time?

∂<Ψ|R|Ψ>/∂t = <∂Ψ/∂t|R|Ψ> + <Ψ|∂(RΨ)/∂t>

= <∂Ψ/∂t|R|Ψ> + <Ψ|R|∂Ψ/∂t> = [<HΨ|R|Ψ> - <Ψ|R|HΨ>]

iħ__

= <Ψ|[H,R]|Ψ> (H is Hermitian)

= 0

iħ__

So Far:• Point group (geometrical) symmetry• H2O and PH3 point groups used as examples• Group theory definitions: Irreducible reps and Ch. Tables• Reducible representations and projection operators• Problems using point groups: Rotation, tunneling,…• Use [H,R]=0 to define R as a symmetry operation• Introduce the CNPI group• Explain why [H,R]=0 used to define symmetry• Can label energy levels (the Ψ generate a representation.)• Vanishing integral theorem • Forbidden interactions and forbidden transitions• Conservation of symmetry

Where are we going?

HΨn = EnΨn

H = H0 + H’

where H0Ψn0 = En

0Ψn0

En0 is -fold degenerate: Eigenfunctions are Ψn1

0,Ψn20,…,Ψn

0

We want to symmetry label the energy levels using theirreducible representations of a symmetry group.

We do this because it helps us to do many things:

Which En0 can be mixed by H’: Block diagonalize H-matrix.

Selection rules for transitions: Only if connected by Γ*.Nuclear spin statistics and intensity ratiosTunneling splittings, Stark effect, Zeeman effect,Breakdown of Born-Oppenheimer approximation…

ℓ ℓ

The ℓ-fold degenerate eigenfunctions Ψn10,Ψn2

0,…,Ψnℓ0

GENERATE an ℓ-fold irreducible representation of thesymmetry group (this labels the energy level En

0).

The basis of what we do using symmetry is that:

Ψn10

Ψn20

.

.

Ψnℓ0

Ψn10

Ψn20

.

.

Ψnℓ0

R

= D(R)

The matrices D(R) form an irreducible representation

To obtain the matrices D(R), and hence the irreduciblerep. label, we need to know the Ψni

0 and to know how the symmetry ops transform the coordinates in the Ψni

0.

The above follows from the fact that [H,R] = 0.

But first of all we must decide on the symmetry group that we are going to use. It could be the CNPI group

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 and (as we shall see) multiple symmetry labels

Documents

Symmetry and Spectroscopy. P. R. Bunker and Per Jensen: Molecular Symmetry and Spectroscopy, 2 nd Edition, 2 nd Printing, NRC Research Press, Ottawa,