classical model of rigid rotor a particle rotating around a fixed point, as shown below, has angular...
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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.