Classical Model of Rigid Rotor A particle rotating around a fixed point, as shown below, has angular...

Classical Model of Rigid Rotor

A particle rotating around a fixed point, as shown below, has angular momentum and rotational kinetic energy (“rigid rotor”)

The classical kinetic energy is given by:

If the particle is rotating about a fixed point at radius r with a frequency ʋ (s−1 or Hz), the velocity of the particle is given by:

where ω is the angular frequency (rad s−1 or rad Hz). The rotational kinetic energy can be now expressed as:

Also

where

Consider a classical rigid rotor corresponding to a diatomic molecule. Here we consider only rotation restricted to a 2-D plane where the two masses (i.e., the nuclei) rotate about their center of mass.

The rotational kinetic energy for diatomic molecule in terms of angular momentum

Note that there is no potential energy involved in free rotation.

Momentum Summary

21

2K r p

I

22ˆ

2K

L̂ i r

Linear

Classical QM

Rotational (Angular)

Momentum

Energy

Momentum

Energy 2

2ˆ2

K rI

p̂ i

L r p

drp mv m

dt

2 2

22 2

p m d rK

m dt

Angular Momentum

L r p

x y zx y z p p p L

x y z

x y z

p p p

i j k

L

Angular Momentum

x y zL L L L i j k

x z y

y x z

z y x

L yp zp

L zp xp

L xp yp

Angular Momentum

ˆ ˆ ˆ ˆi L r p r

d d dx y z i

dx dy dz

L

Angular Momentum

x y zL L L L i j k

ˆ

ˆ

ˆ

x

y

z

d dL i y z

dz dy

d dL i z x

dx dz

d dL i x y

dy dx

Two-Dimensional Rotational Motion

cos( )x r

x

y

r sin( )y r

d d

dx dy

i j

Polar Coordinates

2 22

2 2

d d

dx dy

Two-Dimensional Rotational Motion

22

2 2

1 1d d drdr r dr r d

2 2 2

2 2 2 2

1 1d d d d drdr r dr r d dx dy

2 2 22

2 2

1 1ˆ2 2

d d dH r

dr r dr r d

Two-Dimensional Rigid Rotor

22ˆ ( , ) ( , ) ( , )

2H r r E r

Assume r is rigid, ie. it is constant

2 2 22

2 2

1 1ˆ2 2

d d dH r

dr r dr r d

2 2 22

2 2

1ˆ2 2r

dH

r d

Two-Dimensional Rigid Rotor

2 2

2( ) 0

2

dE

I d

2

2 2

2( ) 0

d IE

d

2 2

2ˆ ( ) ( ) ( )

2

dH E

I d

0)(22

2

m

d

d

I

mE

EI

m

2

2

22

22

Solution of equation

Energy and Momentum

mL

I

m

I

L

I

mE

Z

Z

22

2

222

22

As the system is rotating about the z-axis

Two-Dimensional Rigid Rotor

2 2

2m

mE

I

zmL m

E

mzmLmEm

6

5

4

3

2

1

2

I

18.0

12.5

8.0

4.5

2.00.5

6

5

4

3 2

6

5

4

321

Only 1 quantum number is require to determine the state of the system.

Spherical coordinates

Spherical polar coordinate

Hamiltonian in spherical polar coordinate

22 2

2 2 2 2 2

1 1 1sin

sin sin

d d d d dr

r dr dr r d d r d

Transition from the above classical expression to quantum mechanics can be carried out by replacing the total angular momentum by the corresponding operator:

Rigid Rotor in Quantum Mechanics

Wave functions must contain both θ and Φ dependence:

are called spherical harmonics

Schrondinger equation

22 2

sinby gMultiplyinI

Two equations

Solution of second equation

Solution of First equation

mJP

Associated Legendre Polynomial

)1( JJ

Associated Legendre Polynomial

21(cos ) (cos 1)

2 ! (cos )

l

ll l

dP

l d

(cos ) sin (cos )(cos )

m

m ml l

dP P

d

00Y

For l=0, m=0

0

2 00

1(cos ) (cos 1) 1

2 0! (cos )l

dP

d

0

0 00 (cos ) sin 1 1

(cos )

dP

d

410

0 Y

First spherical harmonicsSpherical Harmonic, Y0,0

origin thefrom surface of distance20

0 constY

l= 1, m=0

0

0 01 (cos ) sin cos cos

(cos )

dP

d

01,0

1 3 3( , ) 1 cos( ) cos( )

22 2iY e

l= 1, m=0

origin thefrom surface of distancecos2201 constY

θ cos2θ

0 1

30 3/4

45 1/2

60 1/4

90 0

l=2, m=02

2,0

5( , ) (3cos ( ) 1)

4Y

θ cos2θ 3cos2θ-10 1 2

30 3/4 (9/4-1)=5/4

45 1/2 (3/2-1)=1/2

60 1/4 (3/4-1)=-1/4

90 0 -1

l = 1, m=±1

1, 1

3( , ) sin( )

2 2iY e

If Ф1 and Ф2 are degenerateeigenfunctions, their linear combinations are also an eigenfunction with the same eigenvalue.

Complex Value??

l=1, m=±1

1, 1 1, 1

1 3 3( , ) ( , ) sin( ) sin( )cos( )

2 4 2 2 2i iY Y e e

Along x-axis

1, 1 1, 1

1 3 3( , ) ( , ) sin( ) sin( )sin( )

2 4 2 2 2i iY Y e e

i i

Three-Dimensional Rigid Rotor States

E

l zmLlE,..,lm mY

33,2,1,0, 1, 2, 3Y

22,1,0, 1, 2Y

11,0, 1Y

2

I

6.0

3.0

1.0

0.5

0

3

2

10

Only 2 quantum numbers are required to determine the state of the system.

2

( 1)2lE l lI

( 1)lL l l zL m

12

6

2

Lm

0

1 0 -1 00Y

1 0-1 -2

2

1 0-1 -2

2

-3

3

0

2

0

2

32

0

22

Rotational Spectroscopy2

2( 1)

2Jo

E J Jr

1J JE E E

J : Rotational quantum number

2

2( 1)( 2) ( 1)

2 o

J J J Jr

2

( 1)2JE J JI

IhcB

JhcBJI

E

2

121

2

2

Rotational Constant

Rotational Spectroscopy

hcE h hc

2

( 1)

4

h J

Ic

2 ( 1)B J

2 28 o

hB

r c

Wavenumber (cm-1)

Rotational Constant

1J Jv c c

2 ( 1 1) 2 ( 1) 2c B J B J cB

Frequency (v)

vv

Line spacing

Bond length

• To a good approximation, the microwave spectrum of H35Cl consists of a series of equally spaced lines, separated by 6.26*1011

Hz. Calculate the bond length of H35Cl.