Rigid rotor

Rigid Rotor, useful for MSc and BSc Chemistry students

1. Dr. Pradeep Samantaroy
Department of Chemistry
Rayagada Autonomous College, Rayagada
[email protected]; [email protected]
9444078968

2. r
Prepared by Dr. Pradeep Samantaroy

3. The kinetic energy of the system:

KE = ½ Iω
2

= L
2
/ 2I since: L= Iω

where L = Angular momentum
I = Moment of Inertia
ω= Angular velocity

I = µr
2

where µ is the reduced mass = m
1m
2/(m
1+m
2)
Prepared by Dr. Pradeep Samantaroy

4. Now the Schrodinger equation is
ĤΨ = EΨ
Where Ĥ = T + V

Since the potential energy is assumed to be zero
Ĥ = T = L
2
/ 2I
Now the angular momentum operator L
2
can be written as 2
z
L
2
y
L
2
x
L
2
L 
Prepared by Dr. Pradeep Samantaroy

5. 














y
z
z
yiPzPyL yz
x  












z
x
x
ziPxPzL zx
y
 














x
y
y
xiPyPxL zy
z
 Prepared by Dr. Pradeep Samantaroy

6. TRANSFORMATION OF CO-ORDINATES
x = rsinθcosφ y=rsinθsinφ z= rcosθ
r
2
= x
2
+ y
2
+ z
2

Prepared by Dr. Pradeep Samantaroy

7. 















 coscotsiniLx 















 sincotcosiLy 



iLz 



















2
2
2
22
sin
1
sin
sin
1



L Prepared by Dr. Pradeep Samantaroy

8. Now putting the value of L
2
in the Schrodinger equation E
I
L


2
2 



E
I


























2
2
2
2
sin
1
sin
sin
1
2
1
 0
2
sin
1
sin
sin
1
22
2
2


























E
I
 0
8
sin
1
sin
sin
1
2
2
2
2
2



























E
h
I
Prepared by Dr. Pradeep Samantaroy

9. Now, Ψ is a function of θ and φ and both are independent of each other.
Hence,

Ψ(θ,φ)=Θ(θ). Φ(φ)
Now multiplying Sin
2
θ in the previous equation, 0sin
8
sinsin
2
2
2
2
2



























 E
h
I E
h
I
Assume
2
2
8
,

 2
2
2
sinsinsin
becomesequation theNow




















Prepared by Dr. Pradeep Samantaroy

10. 2
2
2 )(
)(sin)()(
)(
sinsin)(
of value thengSubstituti





















 2
2
2 )(
)(
1
sin
)(
sin
)(
sin

by t throughoudividing Now























 2
2
2
)(
)(
1
mAssume 






 22
sin
)(
sin
)(
sin
mThen 















 As the terms are separated now, they can be equated to one constant, as they
are independent of each other.
Prepared by Dr. Pradeep Samantaroy

11. Now we have two different equation to solve.
Solving individual equations and then combining will provide the
solution for rigid rotor.
θ varies from 0 to π.
φ varies from 0 to 2π.
Prepared by Dr. Pradeep Samantaroy

12. Solution to φ equation 2
2
2
)(
)(
1
have We m






 0)(
)(
2
2
2



m

 constant.ion normalizat theis N
)(
isequation such osolution t general The


im
Ne
Prepared by Dr. Pradeep Samantaroy

13. Normalization of φ equation 







2
1

1d
1d
1)d()(
2
0
2
2
0
2
0
*








N
N
NeNe
imim 2.... 1, 0, m
2
1
)(
becomessolution theHence




im
e
Applying Normalization Condition
Prepared by Dr. Pradeep Samantaroy

14. 0sin
)(
sin
)(
sin

22











mThen 




 Solution to Θ equation
It will become easier to solve, if this equation can be transformed to Legendre equation dx
d
x
dx
d
dx
d
d
dx
d
d
Let
ddxx
xThen
Assume
2
2
22
-1sin.
P(x) )(
sin- sin -1
sin -1
cos x











Prepared by Dr. Pradeep Samantaroy

15. 0)(
1
)(
2
)(
)1(
becomesequation theNow
2
2
2
2
2


















xP
x
m
x
xP
x
x
xP
x  This is similar to Legendre equation, where β = l(l+1)
l is the azimuthal quantum number. ...2,1,0for )(
1
2
1
1
becomesequation Legendre theolution to theNow
2
22



lxPFind
)(x
dx
d
l!
(x)P
(x)P
dx
d
)x((x)P
s
l
l
l
l
ll
l|m|
|m|
|m|/|m|
l
Prepared by Dr. Pradeep Samantaroy

16. Now |m| ≤ l
It can’t be greater than l, as the Legendre equation becomes zero.

Using the value of x, we can find solution to Θ equation as
|m|)!(l
)(l-|m|)!l(
N
N
(x)PN
l,m
ml
|m|
lml




2
12
constant.ion normalizat theis
)(
,
,
 (x) .P
|m|)!(l
)(l-|m|)!l(
|m|
l



2
12
)(
Prepared by Dr. Pradeep Samantaroy

17. lm -l & ,,,where l
e(x).P
|m|)!(l
(l-|m|)!)l(
e(x).P
|m|)!(l
)(l-|m|)!l(
im|m|
l
im|m|
l








3210
.
4
12

2
1
.
2
12

)().(),(
isrotator rigidfor tion wavefunccomplete theNow




 Prepared by Dr. Pradeep Samantaroy

18. number. quantum rotational asknown is
....3210J here w
8
)1(
py SpectroscoIn
....3210 where
8
)1(
8
)1(
2
2
2
2
2
2
J
,,,
I
hJJ
E
,,,l
I
hll
E
h
IE
ll










 Prepared by Dr. Pradeep Samantaroy