1

Vibrational Spectroscopy

Outlines

- Quantum harmonic oscillator

- Vibrational transitions

- Selection Rules

- Vibration of polyatomic molecules

2

Quantum (Mechanical) Harmonic Oscillator

A simple form of Schrödinger equation

Hamiltonian Operator

Eigenvalue Eigenfunction

For One Dimensional Schrödinger Equation

VTHH total ˆor

EH

Kinetic Operator Potential Operator

ExVdx

d

m

2 2

22

ExV

dx

d

m

2

22

2

3

1D Schrödinger equation of a diatomic molecule for any states

xExxVdx

xdnnn

n

)(

2 2

22

For solving the Schrödinger Equation, we need to define the Eigen functions associated with the state.

It is commonly known that the exponential function is a good one to work with.

Let see a simple example

4

Consider a given function:

Ex. Show that the function satisfies the Schrödinger equation for the 1D quantum harmonic oscillator. What conditions does this place on ? What is E?

2 xe

2 xe

Schrödinger equation

1D Schrödinger equation

EH

ˆ 22 xxtotal EeeH

2

22 2

22xx EeexV

dx

d

5

For the left side of Eq.:

2

2

2

2

2 2

2

22

2

1 2

2

2

xx

xx

ekxdx

xed

exVdx

ed

ax

ax

axedx

de

2

udvvduuvd )( 22 2d ,

,1d ,xx xevev

uxu

22

22

22

2

2

2

2

2

) 2(

xx

xx

xxx

exe

eex

exexdx

xed

1st derivative :

2nd derivative :

2

2

1kxxV

2 2

22

2xexV

dx

d

6

222

222

2

2

2 222

2

2 2 2

2 2

2

14

2)2(

2

2

1) 2(2

2

2

1 2

2

xxx

xxx

xx

ekxexe

ekxexe

ekxdx

xed

The last two terms must be

cancelled in order to satisfy the

Schrödinger eigenfunction

22

2

2

2 xxx

exedx

xed

ˆ 22 xxtotal EeeH

22

)2(2

E

Thus from the product:

Therefore, the energy eigenvalue:

) 2(2

ˆ 22 2

xxtotal eeH

7

To find β, we use the cancellation of the last two terms

02

14

2

22 2 222

xx ekxex

2 ,

4 22

kk

2

1

22

22

kkE

Substitute β, we get the energy :

Therefore, the condition for β that makes the function satisfies the Schrödinger equation

k

From the classical treatment :

hvvh

E2

1)2(

22

1

2

1

hvE2

1

the energy

8

Unlike classical treatment, quantum harmonic oscillation of diatomic

molecule exhibits several discrete energy states.

They can be better described using Hermite polynomial wave

functions :

,...2,1,0 and ,2/21 2

nexHAx x

nnn

Eigenfunctions

Vibrational energy states

4/1

n ! 2

1

nAn

2/ k

Normalization constant

xH n

21

Herrmite Polynomials 2/2xe

9

HnHH nnn )()1(2)(2 21

Hermite polynomials

.

.

.

124816

128

24

2

1

244

33

22

1

0

H

H

H

H

H

The first few Hermite polynomials

The recurrence relation for Hermite polynomials

For n > 2

n = 0

n = 1

n = 2

n = 3

n = 4

xH n

21

x21

;

10

2

2

2

2

2/13

4/13

3

2/124/1

2

2/1

4/13

1

2/14/1

0

329

124

4

x

x

x

x

exxx

exx

xex

ex

The first four Hermite polynomial wave functions, (n= 0,1, 2, 3)

11

2

2

1

21 x

nnn exHAx

22

2

1

21

2

1

21

total x

nnn

x

nn exHAEexHAH

hvE2

1

2 xe ˆ 22 xxtotal EeeH

hvnE

hvnE

n

n

2

1)12(

)2

1(

The solution for the eigenstate is:

If

Solving the1D Schrödinger equation

If

The solution for the n eigenstate is:

or

12

hvnEn 2

1)12(

Plot of the first few eigenfunctions of the quantum harmonic oscillator (red) together with the potential energy

Zero point energy

ground vibrational state

1st exited vib. state

 The energy of a system cannot have zero energy. 

13

The Probability (ψ2(x))

14

hvnhvnEn )2

1(

2

1)12(

The solution of quantum harmonic oscillation gives the energy:

n vibrational state

Evib

(harmonic)En-En-1

0 Ground state 1/2 h -

1 1st excited st. 3/2 h h

2 2nd 5/2 h h

3 3rd 7/2 h h

Where the vibrational quantum number (n) = 0, 1, 2, 3 …

Noted that the constant Evib= h is good for small n. For larger n, the

energy gap becomes very small. It is better to use anharmonic

approximation.

15

Ex. for the Hermite polynomial wave functions at n =0 for

HCl

4/1

00! 02

1

A4/1

n ! 2

1

nAn

212 1076.8/ k

)cm 2991 kg, 1064.1( -127

2)2( ck

229822

1076.84/121

0

A

What is the normalization constant at this state?

16

Ex. Using Hermite polynomials to show that the ground and first exited states wave function (n =0, n=1) of a diatomic molecule is

2/21 2x

nn exHAx

4/14/1

00! 02

1

A 121

0

xH

2

2

2/1

4/13

1

2/14/1

0

4 x

x

xex

ex

For n = 0

The general form of Hermite polynomials

22 2/14/1

2/121

0

4/1

00! 02

1 xx eexHx

17

4/14/1

112

1

! 12

1

A

22 2/1

4/132/12

14/1

1

4)2(

2

1 xx xeexx

x21

2/21 2x

nn exHAx

For n = 1

xH 21

1 22

The recurrence relation for Hermite polynomials

18

HnHH nnn )()1(2)(2 21

Hermite polynomials

.

.

.

124816

128

24

2

1

244

33

22

1

0

H

H

H

H

H

The first few Hermite polynomials

The recurrence relation for Hermite polynomials

For n > 2

n = 0

n = 1

n = 2

n = 3

n = 4

19

Ex. Plot the ground state wave function (n =0) of HCl where observed frequency 2991 cm-1.

amu 9801.0

212 1076.8/ k

N/m 8.520k

2/4/1

0

2xex

-1cm 2991

20 2

1

2

1kxhvE

hvnEn

2

1

k

Ex 02

20

Ex. Draw the vibrational energy diagram of the first five levels of a diatomic molecule using the solution of quantum harmonic oscillation :

hvnhvnEn

2

1

2

1)12(

The first five levels: n = 0, 1, 2, 3, and 4

The energy of a system cannot have zero energy, even at n=0. It is

called “Zero point energy”.

n En

0 1/2 h

1 3/2 h

2 5/2 h

3 7/2 h

4 9/2 h0 1

2

3 4

h

E

at n=0, E0 0

Zero point energy : the energy of the ground vibrational state

21

Vibrational transitions

h0

1

2

34 n0 ni

Transitions

01 Fundamental

02 First overtone

03 Second overtone

04 Third overtone

- The fundamental transitions, n=1, are the most commonly occurring. - The overtone transitions n=2, 3, … are much weaker (low intensities).

22

Ex. The harmonic vibrational frequency of HCl in wavenumbers is 2991 cm–1.

a) Calculate the energies of the first three vibrational levels in Joules and calculate zero point energy.

1-13

1-1-

10

s 10967.8

cm 2991s

cm10998.2

c

JhvE 200 1097.2

2

1

The ground state vibrational energy, n=0

JhvE 201 1091.8

2

3

The 1st excited state vibrational energy, n=1

JhvE 192 1049.1

2

5

The 2st excited state vibrational energy, n=2

hvnEn

2

1

23

JhvE 200 1097.2

2

1

Zero point energy = The ground state vibrational energy

Zero point energy

b) Determine the energy and wavenumber for the fundamental, first-

overtone transitions

1-

1-1301

2001

cm 2991

s 10967.8)(1

10942.5

c

vv

EEh

v

JEEE

fundamental transition : n=0 n=1

1st overtone transition : n=0 n=21-

1902

cm 5982

1019.1

v

JEEE

24

Determine the energy and wavenumber for the fundamental, first-

overtone transitions using the vibrational energy function for the

anharmonic correction below:

22

2

1

4

)(

2

1

n

D

hvhvnE

en

Ex. The vibration potential of HCl can be described by a Morse

potential with De=7.41 X 10-19 J, k = 516.3 N m-1 and = 8.967 X 1013

s-1 .

First of all, determine the vibrational energies of the ground

state (n=0) and the first and second excited states (n= 1 and 2)

25

De=7.41 X 10-19 J = 8.97 X 1013 s-1

h = 5.944 X 10-20 J(h)2/4De= 1.192 X 10-21 JThe vibrational energies of the

ground state and the first and second excited states

22

2

1

4

)(

2

1

n

D

hvhvnE

en

J 1041.1

J 1064.8

J 1094.2

192

201

200

E

E

E

1-

2001

cm 2871

1070.5

v

JEEEFundamental transition : n=0 n=1

1-

1902

cm 5622

1012.1

v

JEEEFirst overtone transition : n=0 n=2

26

Selection Rules

• The transition probability from state n to state m is only nonzero if the transition dipole moment satisfies the following condition:

where x = spatial variable μx = dipole moment along the electric field direction

mnx

0)()()(* xxx nxmmnx

...)())((0

0

x

xxx dx

dtxtx

Permanent Dynamics

0mnx

27

Assuming transition from the ground state (n = 0) to state m

0mx

n = 0 n = m

0)(...)()( 00

0*

x

dx

dtxx

x

xxm

...)())((0

0

x

xxx dx

dtxtx

0)()()( 0*0 xxx xm

mx

dxxdx

dtxxxx

x

xmxm )()()()()( 0

0

*00

*

Truncate at the first derivative:

dxxtxx

dx

ddxxx m

x

xmx )()()()()( 0

*

00

*0

28

2/21 2x

nnn exHAx

dxexHtxexHdx

dAA

dxexHexHAA

xxm

x

xm

xxmxm

mx

22

22

)()()(

)()(

2/10

2/1

00

2/10

2/100

0

Substitute with Hermite polynomial functions

= 0 (Orthogonal)

2/21

000

2xexHAx

2/21 2x

mmm exHAx

n = 0;

n = m;

X Odd functionIf m is odd Odd functionIf m is even Even function

0mnx m must be odd : 1, 3, 5, 7, …

(so permanent dipole is not relevant for IR absorption)

29

Possible transitions )0( mnx

m must be odd : 1, 3, 5, 7, … n=0n=1n=0n=3 n=0n=5 …

n=0n=1

n=0n=3

n=0n=5

Integration give an area under the peaks

Integration

n=0n=1 0

n=0n=3 = 0

n=0n=5 = 0

Transitions

30

Selection Rules

Selection Rules for IR absortion

n = 1

The only transition from n=0n=1 is allowed.

For the other an integration is zero. The transition is forbidden.

n = +1 absorption

n = -1emission

31

How many vibrational modes are observed for a molecule?

A molecule consisting of N atoms

1 atom : 3 coordinates (x1, y1, z1)

N atoms : 3 N coordinates ((x1, y1, z1), (x2,y2,z2), (x3,y3,z3),…..)

* There are only 3 normal modes (coordinates) describing the translational motion.* There are 3 normal modes describing the Rotational motion of the molecules (along 3 axes)

Number of vibrational modes = 3N – 6Number of vibrational modes for linear molecules = 3 N-5

there are only 2 rotational modes

Vibration of polyatomic molecules

Position Representation

Degree of freedoms (Molecular motions)

32

Vibration

Stretching vibration Bending vibration

SymmetricStretching

AsymmetricStretching

In plane Bending

Out of plane Bending

33

H2O

- non-linear molecule

- N = 3

- number of vibrational normal modes = 3 X 3 –

6 = 3

3685 cm-1 3506 cm-1 1885 cm-1

H2O has three distinct vibrational frequencies.

34

CO2

- linear molecule

- N = 3

- number of vibrational normal modes = 3 X 3 – 5

= 4

Because of xz and xy are degenerate, CO2 has three (not four) distinct vibrational frequencies.

35

Benzene C6H6

- non-linear molecule

- N = 12

- number of vibrational normal modes = 3 X 12 – 6

= 30

There are 30 vibrational modes but only 20 distinct

vibrational frequencies.

36

In the normal coordinate system, the vibrational Hamiltonian of polyatomic system can be written as:

2

1

2

63

1

22

22

N

iii

i

QkQ

H

... 2

1

2

2

1

2

2

1

22332

3

222222

2

222

1121

22

QkQ

QkQ

QkQ

H

...)()()()(63

1332211

N

iii QHQHQHQHH

Therefore, the wave function is of the form :

...)()()( )( 332211

63

1

QQQQN

iii

The total vibrational energy is:

...2

1

2

1

2

1

2

1...)( 332211

63

1331

hvnhvnhvnhvnnnnEN

iii

37

Ex. The water molecule has three normal modes, with

fundamental frequencies: 1-bar = 3685 cm-1, 2-bar =

1885 cm-1, 3-bar = 3506 cm-1.

a) What is the energy of the (112) state?

2

12

2

11

2

11)112( 321 hvhvhvE

2

5

2

3

2

3321 vhcvhcvhc

)cm 3506(2

5)cm 1885(

2

3)cm 3685(

2

3 1-1-1- hchchc

J 1040.3 19

(i.e. n1=1, n2=1, n3=2)

38

What is the energy difference between (112) and (100)?

J 1040.3)112( 19E

J 1063.1)100( 19E

J 101.77 10)63.14.3(

)100()112(1919

EEE

39

40

41

42

The solutions to these 3N-6 equations are the familiar Harmonic Oscillator Wavefunctions and Energies.

hvnEn 2

1)12(

The total vibrational energy is:

The Schrödinger equation is simplified by separation of variables. Therefore, one gets 3N-6 equations of the form

2

1

22

2

22

iiiiii

i EQkQ

43

• Coupled system has two vibrational frequencies: the

symmetrical and antisymmetric modes.

• For symmetrical and asymmetrical, the vibrational

frequency is

1

2

1 kvsymmetric

21 2

2

1 kkv ricqntisymmet

44

Ex. The vibration potential of HCl can be described by a Morse

potential with De=7.41 X 10-19 J, k = 516.3 N m-1 and = 8.97 X 1013 s-

1 .

Calculate

ee DxxV )(

Morse function of vibrational potential:

2)(1)( exxe eDxV

At the minimum energy (x=xe), the potential energy:

De = Dissociation energy, Do = bond energy

45

en DE

The energy levels for the anharmonic correction:

The highest value of n consistent with the potential that is

ee

DnD

hvhvn

22

2

1

4

)(

2

1

22

2

1

4

)(

2

1

n

D

hvhvnE

en

De=7.41 X 10-19 J = 8.97 X 1013 s-1

h = 5.944 X 10-20 J(h)2/4De= 1.192 X 10-21 J

ee

DnD

hvhvn

22

2

1

4

)(

2

1

46

Anharmonicity

2)(1)( exxe eDxV

eD

k

2

exxdx

Vdk

2

2

47

Ex. for the Hermite polynomial wave functions at n =0 for

HCl

4/1

00! 02

1

A4/1

n ! 2

1

nAn

212 1076.8/ k

)cm 2991 kg, 1064.1( -127

2)2( ck

28928

102.2

2

14/118

1

A

What is the normalization constant at this state?

48

Example for the Hermite polynomial wave functions at n =1

for HCl

4/1

11! 12

1

A4/1

n ! 2

1

nAn

182 102.2/ k

)cm 2991 kg, 1064.1( -127

2)2( ck

28928

102.2

2

14/118

1

A

-110 rads1063.5

What is the normalization constant at this state?

c2 2

49

2/21 2x

nn exHAx

To plot the vibrational wave functions, we separate the functions

into three terms, An , Hn and2/2xe

xH n

21

The most complicate one is Herrmite Polynomials

An and are not difficult to calculate 2/2xe

50

HnHH nnn )()1(2)(2 21

Hermite polynomials

.

.

.

124816

128

24

2

1

244

33

22

1

0

H

H

H

H

H

The first few Hermite polynomials

The recurrence relation for Hermite polynomials

For n > 2

n = 0

n = 1

n = 2

n = 3

n = 4

51

2/21 2x

nn exHAx

4/14/1

00! 02

1

A

22 2/14/1

2/121

0

4/1

00! 02

1 xx eexHx

x21

For n = 0

10 H

52

4/14/1

112

1

! 12

1

A

22 2/1

4/132/12

14/1

1

4)2(

2

1 xx xeexx

x21

2/21 2x

nn exHAx

For n = 1

xH 21

1 22

53

4/14/1

2222

1

! 22

1

A

x21

2/21 2x

nn exHAx

For n = 2

22 2/124/1

2/124/1

2 124

)24(22

1 xx exexx

242)(424 22212

2 xxH

4/14/1

22

4/1

n 22

1

! 22

1;

! 2

1

An

An

54

22/14/1

0xex

Example for the Hermite polynomial wave functions at n =1

for HCl

1) What is the normalization constant at this state?

1 ,2/21 2

nexHAx x

nn

4/1

11! 12

1

A

2/ k

4/1

n ! 2

1

nAn

2/ k

)cm 2991 kg, 1064.1( -127

55

Example for the Hermite polynomial wave functions at n =1

for HCl

1) What is the normalization constant at this state?4/1

11! 12

1

A4/1

n ! 2

1

nAn

2/k

)cm 2991 kg, 1064.1( -127

2-

34

27-110-1

s kg 8.520

10626.6

)kg 1067.19801.0)(s cm10998.22)(cm 2991(

k

222 )2( ck

4/14/1

11 2

1

! 12

1

A

)cm 2991 kg, 1064.1( -127

s rad1063.5

)cm 2991s cm10998.22(

2 2

1-10

1-1-10

c

56

EXAMPLE

k

tAtx ),sin()(

Answer the following questions.

a) What are the units of A? What role does have in this equation

b) Graph the kinetic and potential energies as a function of time

c) Show that the sum of the kinetic and potential energies is independent of time.

57

a) What are the units of A?

Because x(t) has the units of length and the sine function is dimensionless, A must have the units of length.

The quantity sets the value of x at t = 0,because . )sin()0sin()0( AAx

58

b) Graph the kinetic and potential energies as a function of timea) Show that the sum of the kinetic and potential

energies is independent of time.

59

SolutionFor the J=0 → J=2 transition,

The preceding calculations show that the J=0 → J=1 transition is allowed and that the J=0 → J=2 transition is forbidden. You can also show that is also zero unless MJ=0 .

04

1

4

1

8

5

2

cos

4

cos3

8

5sincos1cos3

8

5

0

242

0

22

0

20

ddz

61

62

Consider the relative motion via the center of mass coordinates.

Angular Momentum (L)

m

63

64

65

66

67

68

69

70

71

72