1

Chapter 7. A Quantum Mechanical Model for the

Vibration and Rotation of Molecules

Harmonic oscillator:

Hooke’s law:

kxF , x is displacement

Harmonic potential:

2

2

1)( kxFdxxV

k is force constant:

02

2

xdx

Vdk (curvature of V at equilibrium)

Newton’s equation:

kxFdt

xdm

2

2

(diff. eqn)

Solutions:

x = Asint, (position)

21)( /k/m (vibrational frequency)

2

Verify:

rhskxtAm

km

tmAtdt

dmA

dt

tAdmlhs

sin

sincossin 2

2

2

Momentum:

tAmdt

dxmp cos

Energy:

22

2

)sin(2

1

2

)cos(

2

tAkm

tAm

Vm

pE

22

)(sin)(cos2

222

2222

kAAm

ttAm

When t = 0, x = 0 and p = pmax

= mA

When t = , p = 0 and x = xmax

= A

Energy conservation is maintained by oscillation between

kinetic and potential energies.

3

Schrödinger equation for harmonic oscillator:

Ekx

dx

d

m

2

2

22

2

1

2

Energy is quantized:

2

1E ... 1, 0,

where the vibrational frequency

2/1

m

k

Restriction of motion leads to uncertainty in x and p, and

quantization of energy.

Wavefunctions:

2/2

)()( yeyHNx

where

4/1

2,

mkxy

Normalization factor

4

2/1

!2

N

)(yH is the Hermite polynomial

1)(0 yH

yyH 2)(1

24)( 22 yyH

...

11 22 HyHH (recursion relation)

Let’s verify for the ground state

2/

00

22

)( xeNx

2/22

242

0

2/222/222/2

0

2/222/2

0

2/2

2

22

0

22

222222

2222

22

222

2

1

2

2

1

2

2

1

2

x

xxx

xx

x

em

xm

kN

ekxexem

N

ekxxedx

d

mN

ekxdx

d

mNlhs

Note that

5

4/1

2

mk

It is not difficult to prove that the first term is zero, and the

second term as 2/ .

2/0

22

2

1 xeNrhs

So, lhsrhs , it is a solution.

One can also show normalization:

14

222

0

20

2

0

22

dxeNdx x

where we have taken advantage of the even symmetry of the

Gaussian

22axxa ee

and used the integral formula

a

dxe ax

40

2

For higher states, the wavefunctions are essentially the product

of a Gaussian and a polynomial:

6

The Gaussian is always an even function, but the polynomial

can either be even )()( xfxf or odd )()( xfxf . As a

result, the wavefunctions are either even or odd, depending on .

Properties of the oscillator

i. Zero point energy ( 0 )

2

10 E (quantum effect)

Even at 0 K, there is still vibration.

ii. Orthonormality:

dx*

iii. Average displacement

0)()(*

dxxxx

7

No matter if )(x is even or odd, 2

)(x is even, and thus the

integrant is odd. Integration of an odd function in , is

always zero.

iv. Average momentum

0)()(*

dxxdx

d

ix

Because the derivative of an even function is an odd function,

and vice versa.

v. Nodes

The state has 1 nodes (due to Hermite polynomial)

vi. Energy separation

)2/1(2/111 EEE

(equal separation).

vii. Tunneling

Wavefunction penetrates into classically forbidden regions.

A quantum phenomenon.

Example: A harmonic oscillator has a force constant of 475 N/m

and a mass of 1.61×10-27 kg. Calculate the ZPE and the photon

frequency needed to bring the oscillator from ground state to the

first excited state.

8

114

271043.5

1061.1

/475

s

kg

mN

m

k

( 2// skgmN )

J

sJsE

20

114340

1086.2

1043.510054.15.02

1

JEE 200 1072.52

nm

sJs

J

h

E

3476

1063.810626.6

1072.5 113

34

20

( c )

Example: The IR spectrum of H35Cl is dominated by a peak at

2886 cm-1. Calculate the force constant of the molecule.

Reduced mass

kgamukgamu 2727 1061.1/1066.197.0351

351

114110 1043.52886/100.314.32

~22

scmscm

c

mNskg

skgk

/475/475

1043.51061.1

2

2114272

9

What about DCl?

HCl

kgamukgamu

2

1014.3/1066.189.1352

352 2727

HClskg

skgk

2

11089.3

1014.3

/475 114

27

2

HClcmscm

s

c

~

2

12063

/100.314.32

1089.3

2~ 1

10

114

Isotope effect is important to assign spectral lines.

Rigid rotors

Angular momentum vector

l r p I (r fixed)

where the angular velocity and moment of inertia are

d

dt

, 2I r

The rotational energy (classical):

2

2

lE

I

10

Schrödinger eqn. for 2D motion:

E

yxm 2

2

2

22

2

Use polar coordinate system ( sincos r, yrx , it becomes

for 2D rigid rotor:

Ed

d

I 2

22

2

General solution:

2/1

2

2/12

,2

1)(

IEmeim

m

Cyclic boundary condition:

)2()(

So,

)()1()(

)2(

22

2)2(

mimmi

immiim

eeN

eNeNe

2m has to be even integers, i.e.

, ..., , m 210

11

Quantization

I

mEm

2

22 (2D)

Angular momentum:

mlz (quantized angular momentum)

Angular momentum operator

ixy

yx

i

pypxL xyz

ˆˆˆˆˆ

is a common eigenfunction of both operators:

me

iime

iL imim

z

2/12/1

2

1

2

1ˆ

I

me

IH im

2

)(

2

1

2ˆ

22/1

2

22

Uncertainty principle (just like p and x)

2/ zl

12

3D rotors:

Spherical coordinates:

cossinrx

sinsinry

cosz r

ddrdrdxdydzd sin2

Hamiltonian:

22

2ˆ

mH

2

2

2

22

2

2

2

2

2

22

2

ˆ1

2

2

mr

Lr

rrm

zyxm

The differential operator 2 is called Laplacian.

If r is fixed, the 3D rigid rotor Hamiltonian:

I

LH

2

ˆˆ

2

(rigid rotor)

where

13

sin

sin

1

sin

1ˆ2

2

2

22 L

Schrödinger eqn.

EI

L

2

ˆ2

The solution is the spherical harmonics:

)()(cos),(),( mm

llmm

l PNY

where

)!(

)!(

2

)12(

ml

mllNlm

im

m e

2/1

2

1)(

(2D rotor)

and )(cosm

lP is the associated Legendre function.

The quantization of two angular coordinates leads to two

quantum numbers:

l = 0, 1, 2, ...

m = 0, ±1, ±2, ..., ±l (2l+1)

14

Rotational energy (3D rotor)

I

llEl

2

)1( 2

Since E is independent of m, each l level is 2l + 1 degenerate,

which can be removed by an external field (Zeeman effect).

When 0m , )(cosm

lP becomes the Legendre polynomial:

1)(cos0 P

cos)(cos1 P

2/1cos3)(cos 22 P

…

For 0m

)(coscos

sin)(cos

lm

mmm

l Pd

dP

Examples of spherical harmonics:

100 Y

cos01 Y

ieY

2

1sin1

1

1cos32

1 202 Y

15

They are eigenfunctions for both 2L̂ and zL̂ :

),()1(),(ˆ 22 ml

ml YllYL

),(),(ˆ ml

mlz YmYL

Verification for cos01 Y

22

22

2

2

2

22

2cossin2sin

sinsin

10

cossinsin

1

sin

1ˆ

L

II

LH

2

2

2

ˆˆ

22

0cosˆ i

Lz

The spherical harmonics are orthonormal

mmllm

lm

l YYdd

2

0

*

0

),(),(sin

Note the volume element and ranges.

16

Spherical harmonics have nodes due to Legendre functions.

For cos01 Y , the angular node can be determined by

cos 0 , o90

Vector model:

The angular momentum can be considered as a vector L

. In

quantum mechanics, it is quantized in

i. magnitude:

)1( llL

ii. orientation: mLz

Only certain directions of L

are allowed.

Example (l = 1, ml = 0, ±1):

Example: Calc. amplitude of L, Lz and E for a rotor with I =

4.6x10-48 kg m2 and l = 1.

17

JsJsllL 3434 1049.110055.12)1(

JsmL lz3410055.1 ,0 (3-fold degen.)

The angle between L

and the z-axis for 1lm :

1

cos2

zL

L , o45

Jkgm

Js

I

LE 21

248

2342

104.2)106.4(2

)1049.1(

2

Example. A 3D rotor absorbs at 11 cm-1. Calculate its

moment of inertia.

I

EEE2

)02( 2

01

~hchE

so

247

110

34

2

1006.5

11/100.314.32

1005.1

~2~

kgm

cmscm

Js

chcI