Lecture 8/11

2

3

1 1 3

2

The feasibility of an operation depends

on potential energy barriers and

experimental resolution

Should we extend this definition?

H2Te Rigid Rotor Energy Levels [E(JKaKc)-E(JJ0)]

JJ,0

JJ,1

JJ-1,1

JJ-1,2

JJ-2,2

JJ-2,3

JJ-3,3

JJ-3,4

JJ-4,4

JJ,0

JJ,1

JJ-1,1

JJ-1,2

J

Ae=6.26, Be=6.11, Ce=3.09 cm-1

Actual H2Te Energy Levels [E(JKaKc)-E(JJ0)]

J

JJ,0

JJ,1

JJ-1,1

JJ-1,2

JJ,0

JJ,1

JJ-1,1

JJ-1,2

JJ-2,2

JJ-2,3

JJ-3,3

JJ-3,4

JJ-4,4

1R 1L

2L 2R

1

2

1

1 1

2 2

2

JJ,0

JJ,1

JJ-1,1

JJ-1,2

in C2v(M)

-6

-5

-4

-3

-2

-1

0

0 10 20 30 40 50 60 70 80

Six-fold enery clusters in the

vibrational ground state of PH3

A1/A2

A2/A1

E

E

Cluster = A1 A2 2E in C3v(M)

1 PCS

= E 1 PCS

3 PCS

= (132)1 PCS

4 PCS

= (23)*1 PCS

2 PCS

= (123)1 PCS

5 PCS

= (12)*1 PCS

6 PCS

= (13)*1 PCS

Cluster = A1 A2 2E in C3v(M)

Clusters can be explained by

semi-classical theory

=0

Stationary states

J

Rotational energy surface

The rotational energy surface is a radial plot of

EJ EJ(min) as a function of J,J

Classical Hamiltonian function

z

x

y

PH3 rotational energy surface, J = 100

Six equivalent maxima

Each cluster state at highest

energy corresponds to stable

angular momentum trajectory

around a maximum

Cluster states are „localized“

at their respective maxima

„No tunneling between

cluster states“

Extension of feasibility definition:

If energy splittings resulting from potential

energy tunneling cannot be resolved,

permutation-inversion operations accompanied

by this tunneling are considered unfeasible

If energy splittings resulting from rotational

energy surface tunneling cannot be resolved,

permutation-inversion operations accompanied

by this tunneling are considered unfeasible

Old:

New = Old together with:

See also Per Jensen and P. R. Bunker: The Molecular Symmetry Group

for Molecules in High Angular Momentum States, J. Mol. Spectrosc. 164,

315-317 (1994).

Application to H2X:

MS group is C2v(M) = { E, (12), E*, (12)* }

1

2

1

1 1

2 2

2

When tunneling between cluster states

is neglected, (12), E*, (12)* are

unfeasible, only E is feasible.

New MS group is C1 = { E} with one irrep A

A

Reverse correlation C1 C2v(M)

A

Application to PH3:

MS group is C3v(M) = { E, (123),(132) , (12)*,(13)*,(23)* }

When tunneling between cluster states

is neglected, (123), (132), (12)*,

(13)*, (23)* are unfeasible,

only E is feasible.

New MS group is C1 = { E} with one irrep A

A

Reverse correlation C1 C3v(M)

A

etc.

A1 A2 2E

A Question To Think About:

The MS group of C60 is

the icosahedral group

with 120 elements

The MS group of the ammonia dimer (NH3)2 has 144

elements when the effect of inversion of the two

NH3 moieties is observable

Does (NH3)2 then

have higher

symmetry than C60 ?

A Question To Think About:

Symmetry and molecules - summary:

Symmetry operations (e.g., permutation-

inversion operations) for a molecule leave the

molecular Hamiltonian invariant.

The operations can be organized in symmetry

groups.

The symmetry groups have irreducible

representations.

The wavefunctions for the molecular eigenstates

transform according to the irreducible

representations of the symmetry group.

How does symmetry help us?

A.

In practice, we solve the molecular Schrödinger

by diagonalizing matrix representations of

the Hamiltonian. These calculations are

facilitated by symmetry.

B.

Symmetry imposes selection rules on

molecular transitions (for example, absorption

and emission transitions).

Vanishing integral rule

The quantum mechanical integral

must vanish (i.e., be = 0) unless the integrand

contains a totally symmetric component in the

symmetry group(s) of the Hamiltonian

𝐼 = 𝜓′ 𝑂𝜓′′𝑑𝜏 *

𝜓′ 𝑂𝜓′′ *

Parenthetic remark:

symmetry of a product

Given two sets of wavefunctions

s-fold degenerate,

irrep n

r-fold degenerate,

irrep m

Which representation is generated by the set of

products Φ𝑛𝑖 Φ𝑚𝑗, i= 1, 2, 3, ..., s; j = 1, 2, 3, ..., r ?

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Φ𝑛1, Φ𝑛2, Φ𝑛3, Φ𝑛4, … , Φ𝑛𝑟

Φ𝑚1, Φ𝑚2, Φ𝑚3, Φ𝑚4, … , Φ𝑚𝑠

Effect of symmetry operation R on the

functions Φ𝑛𝑖 and Φ𝑚𝑗:

Effect of R on the product Φ𝑛𝑖 Φ𝑚𝑗:

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Effect of R on the product ni mj:

Short-hand notation:

Elements of transformation „supermatrix“:

Diagonal elements:

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Diagonal elements of transformation „supermatrix“:

Characters of „product representation“ nm:

Notation (direct product):

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Example: C3v(M)

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

Special case:

n1, n2 have E symmetry in C3v(M)

Which representation is generated by the three

products n1n1, n2n2, n1n2= n2n1?

Symmetric product representation:

with characters:

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Special case:

Characters:

[E E]: 3 0 1

E2 = E

(123)2 = (132)

((12)*)2 = E

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Antisymmetric product representation:

with characters:

{E E}: 1 1 -1

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nth order symmetric product

representation [E]n

has the characters

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Its dimension is 𝑛 + 1 and it is generated, for example,

by the 𝑛 + 1 products:

n1𝑛, n1

𝑛−1n2, n1𝑛−2 n2

2, n1𝑛−3n2

3, …

…n13n2

𝑛−3, n12n2

𝑛−2, n1 n2𝑛−1, n2

𝑛

Vanishing integral rule

The quantum mechanical integral

must vanish (i.e., be = 0) unless the integrand

contains a totally symmetric component in the

symmetry group(s) of the Hamiltonian

𝐼 = 𝜓′ 𝑂𝜓′′𝑑𝜏 *

𝜓′ 𝑂𝜓′′ *

Totally symmetric component?

All groups have a totally symmetric irreducible

representation (s), e.g.

In the totally symmetric

representation, each group

element is represented by

the 11 matrix 1. The group

{1} is homomorphic to any

group.

The representation of must be

= n(s) (s) .... with n(s) 0

in order that the integral can be non-vanishing

𝜓′ 𝑂𝜓′′ *

Diagonalizing the molecular

Hamiltonian

Schrödinger equation

We apply the vanishing integral rule

Eigenvalues and –functions are found by

diagonalization of a matrix with elements

𝜓𝑗 = 𝑐𝑗𝑛

𝑛

𝜓𝑛 0

𝐻𝜓𝑗 = 𝐸𝑗𝜓𝑗 ^

𝐻𝑚𝑛 = 𝜓𝑚∗𝐻𝜓𝑛𝑑𝜏

0 0 ^

Diagonalizing the molecular

Hamiltonian

The Hamiltonian is invariant under symmetry operations

so the integrand in Hmn generates the product

characters

*

The number of times that the totally symmetric

representation occurs is

m

*

𝐻𝑚𝑛 = 𝜓𝑚∗𝐻𝜓𝑛𝑑𝜏

0 0 ^

Diagonalizing the molecular

Hamiltonian

Hmn can only be non-vanishing if and belong to

the same irreducible representation

The Hamiltonian matrix

factorizes, for example for H2O

Computing time for diagonalization

𝑁3 without factorization

4 (𝑁4)3 =

𝑁3

16 with factorization;

m

*

Total matrix

dimension N

𝜓𝑚 𝜓𝑛 0 0

Intensities

A (A = X, Y, Z) is a space-fixed component of the

molecular dipole moment

Cre is the charge, Ar the A

coordinate of particle r 𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟

Selection rules for transitions

So the intensity of a rotation-vibration transition is

proportional to the square of

Vanishing integral rule: For the integral to be non-

vanishing, the integrand must have a totally symmetric

component.

𝐼TM = Φrve 𝜇𝐴 Φrve d𝜏 ´* ´´

Symmetry of A for H2O

Axis system XYZ with origin in molecular center of

mass; the protons are labeled 1, 2

Both protons have the charge +1e

X

Y

Z

(Xi,Yi,Zi)

(12) (X1,Y1,Z1) = (X2,Y2,Z2)

(12) (X2,Y2,Z2) = (X1,Y1,Z1)

(12) (X3,Y3,Z3) = (X3,Y3,Z3)

(12) A = A

𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟

Symmetry of A for H2O

X

Y

Z

(Xi,Yi,Zi)

(12)* (X1,Y1,Z1) = (X2, Y2,Z2)

(12)* (X2,Y2,Z2) = (X1, Y1,Z1)

(12)* (X3,Y3,Z3) = (X3, Y3,Z3)

E* (X1,Y1,Z1) = (X1, Y1,Z1)

E* (X3,Y3,Z3) = (X3, Y3,Z3)

E* (X2,Y2,Z2) = (X2, Y2,Z2)

E* A = A

(12)* A = A

𝜇𝐴 = 𝐶𝑟𝑒𝐴𝑟 𝑟

Symmetry of A for H2O

E* A = A

(12)* A = A

(12) A = A

* = A2

Symmetry of A generally

A has symmetry *

P A = A; P „pure“ permutation

P* A = A; P* permutation-inversion

* has character +1 under all „pure“ permutations P,

1 under all permutation-inversions P*