MJ, Feb 7

Vibrationsof

polyatomic molecules

MJ, Feb 7

Outline

* Normal modes

* Selection rules

* Group theory (Tjohooo!)

* Anharmonicity

MJ, Feb 7

Describing the vibrations

Molecule with N atoms has 3N-6 vibrational modes, 3N-5 if linear.

Find expression for potential energy.Taylor expansion around equilibrium positions.

...2

1)0(

,

0

2

0

ji

ji

ji

ii

i

xxxx

Vx

x

VVV

Nji 3...1,

Total energy

ji

jiijxxkV

,2

1

0

2

ji

ij xx

Vk; where

Introduce mass weighted coordinates: iiixmq

ji

jiijqqKV

,2

1

ji

ij

ij mm

kK ; where

i

iq 2

2

1 2

2

1ixmT

ii

Kinetic energy

MJ, Feb 7

Total energy

ji

jiiji

iqqKqE

,

2

2

1

2

1

We can now write the total vibrational energy as:

Nasty cross-terms when ji

What we want is to find set of coordinates where the cross-terms disappear.

Is this at all possible?MJ, Feb 7

A look at CO2

MJ, Feb 7

Vibrations of the individual atoms

Can be broken down as linear combinations of:

These modes do not change the centre of mass,and they are independent.

Normal coordinates

MJ, Feb 7

22

2

1

2

1ii

ii

iQQE

So, we can write the energy as:

where Q are the so called normal coordinates.They can be a bit tricky to find, but at least we know they are there.

Before we see how can use this, lets have a look at the normal modes for our CO2.

Normal modes of CO2

Symmetric stretch

Anti-symmetric stretch

Orthogonal bending

3 x 3 - 5 = 4 vibrational modes

MJ, Feb 7

QM

MJ, Feb 7

Since the total energy is just a sum of terms, so is the Hamiltonian of the vibrations. We write it as:

i

iHH ˆˆ 2

2

2

2

2

1

2

1ˆii

i

iQ

QH

Also the vibrational wavefunction separates into a product of single mode wavefunctions:

)(...)()(63632211

NN

QQQ

Schrödinger equation

MJ, Feb 7

The Scrödinger equation then becomes:

)()(2

1

2

1 2

2

2

2

iiiiii

i

QEQQQ

… and this we recognise, right?

Harmonic oscillator with unit mass and force constant

MJ, Feb 7

Harmonic oscillatorEnergy levels:

iiiiinnnE

i;;½)(

i

i

i

y

innnQyeyHN i

iii ;)( 2/2

Wavefunctions:

Total vibrational energy:

i

iinE ½)(

Harm. Osc. …

We know the ground state:6321

0...00N

i

iE ½Ground state energy:

Ground state wavefunction:

i

y

i

y

i

i yyNeeN 222/2/

0;

22

All normal modes appear symmetrically, and as squares

The ground state is symmetric with respect to all symmetry operations of the molecule.

MJ, Feb 7

MJ, Feb 7

Selection rulesMolecular dipole moment depends on displacements of the atoms in the molecule: Taylor expand...

...0

0

i

ii

QQ

Dipole transition matrix element of a particular mode:

iii

i

iiiinQn

Qnnnn

0

0......000......000

0 if , due to orthogonalityiinn 10

iinnif

MJ, Feb 7

Selection rules

Selection rules for IR absortion: 1in 0

0

iQ

;

Similarly, by observing that

ii

i

innnQ

Qn

ii

0

0

we get selection rules for Raman activity:

1in 0

0

iQ

It can be hard to see which vibrations are IR/Raman active, but, as we have seen before, Group Theory can come to rescue.

MJ, Feb 7

Group theory and vibrations

First find a basis for the molecule. Let’s take the cartesian coordinates for each atom. x1

x3

x2

y1

y2

y3

z1

z2

z3

Water belongs to the C2v group which contains the operations E, C2, v(xz) and v’(yz).

The representation becomes E C2 v(xz) v’(yz)

red9 -1 1 3

The details of a normal mode depend on the strength of the chemical bonds and the mass of the atoms. However the symmetries are just a function of geometry.

Example: H2O (the following stolen from Hedén)

Character table for C2v.

yzRyB

xzRxB

xyRA

zyxzA

yzxzCEC

x

y

vvv

,1111

,1111

1111

,,1111

)(')(

2

1

22

2221

22

3119

)(')(22

redvvv yzxzCEC

Now reduce red to a sum of irreducible representations. Use inspection or the formula.

Continued water example

MJ, Feb 7

yzRyB

xzRxB

xyRA

zyxzA

yzxzCEC

x

y

z

vvv

,1111

,1111

1111

,,1111

)(')(

2

1

2

2221

22

The representation reduces to red=3A1+A2+2B1+3B2

trans= A1+B1+B2

rot=A2+B1+B2

vib=2A1+B2

Continued water example

MJ, Feb 7

Modes left for vibrations

3119

)(')(22

redvvv yzxzCEC

MJ, Feb 7

What to use this for?We know that that the ground state is totally symmetric: (A1)

First excited state of a normal mode belongs to the same irred. repr. as that mode because

iiiQyyH )(

1

001 ii

So for , and must span the same irreducible representation for their product to be in A1. transform as translations, so:

i1

For a transition to be IR active, the normal mode must be parallel to the polarisation of the radiation.

MJ, Feb 7

What more to use this for?

By the same argument one can come the the conclution that

For a transition to be Raman active, the normal mode must belong to the same symmetry species as the components of the polarisability

These scale as the quadratic forms x2, y2, xy etc.

This also leads to the exclusion rule:

In a molecule with a centre of inversion, a mode cannot be both IR and Raman active.

yzRyB

xzRxB

xyRA

zyxzA

yzxzCEC

x

y

z

vvv

,1111

,1111

1111

,,1111

)(')(

2

1

2

2221

22

vib=2A1+B2

MJ, Feb 7

Water again...

A1

A1

B2

All three modes are both IR and Raman active, no centre of inversion.

(a) and (b) are excited by z-polarised light, and (c) by y-polarised.

MJ, Feb 7

Anharmonicity

At this point overtones like can be allowed, since the matrix element containing the quadratic term Qi2 not necessarily vanishes. This can not be determined from group theory, but must be calculated for every molecule.

ii02

...0

0

i

ii

QQ

Electric anharmonicity occurs when our expansion of the dipole moment to first order is not valid.

jiji

ji

ii

i

QQQQ

QQ

0

,

2

0

0½

MJ, Feb 7

Anharmonicity

jiji

ji

ii

i

QQQQ

QQ

0

,

2

0

0½

We also see from the presence of QiQj cross-terms can cause a mixing of normal modes.

In a perfectly harmonic molecule, energy put into one normal mode stays there. Anharmonicity causes the molecule to thermalise.

MJ, Feb 7

Anharmonicity

...!3

1

2

1,,,

k

kjijiijk

jijiij

xxxkxxkV

Also mechanical anharmonicity can lead to mixing of levels if one needs to add cubic and further terms in the expression for the potential.

0a 0b

1a

2a 1b

bbbaa

b

baanbaQQ

QQ

VV

a

a

1002!3

11002 2

2

3

Inversion doublingConsider ammonia: pyramidal molecule with two sets of vibrational levels:Coupling between the levels lead to mixing of up and down wavefunctions which lifts the degeneracy of the levels

Summary

• Harmonic approximation of energy gives transition rules for IR and Raman activity.

• Group theory can help us figure out which transition are active.

• However, anharmonic terms can come in play and mess everything up.