5.61 Fall 2004 Lecture #17-19 page 1

THE RIGID ROTOR

Diatomic molecule

+
r1 r2 r0 m1 = r mrmr 1 (center of mass) r2 COM m2

1 2 2 1 2 2 1 2 2 2 K.E. K = m r + m r = (m r + m r ) 2 1 1 2 2 2 2 1 1 2 2

I moment of inertia 1
K = I 2 2

2 2 2 m m I = m r + m r = µr µ = 1 2 (reduced mass) 1 1 2 2 0 + m1 m2

1 1 K = µr 2 2 = µv 2 2 0 2 0 z

Prob l em reduced to a one-body probl em µ with mass µ. r0

y

x 5.61 Fall 2004 Lecture #17-19 page 2

Angular momentum L=r
p

= L = + = + = 2 + 2 = L r1 p1 r2 p2 m1 r 1 v 1 m 2 r 2 v 2 m 1 r 1 m 2 r 2 I

1 L2
K = I 2 = 2 2I

Connection between angular momentum and kinetic energy for rotation Analogous to linear momentum and kinetic energy for translation

Now need to describe this quantum mechanically.

2 ˆ = + ( ) =
+2 ( ) H K V x, y, z xyz V x, y, z 2µ 2 2 2 2 = + + xyz x 2y 2z 2

Change to spherical polar coordinates

Cartesian spherical polar (x, y, z )  (r,,
) x = r sin cos
y = r sin sin
z = r cos

      2 2 = 1 2 + 1  + 1 
 r sin r r 2 r  r r 2 sin    r 2 sin2  
2

so in spherical polar coordinates Hˆ (r,,
) (r,,
) = E (r,,
) is 5.61 Fall 2004 Lecture #17-19 page 3

2 2 ⎪⎧  ⎡ 1 ∂ ⎛ 2 ∂ ⎞ 1 ∂ ⎛ ∂ ⎞ 1 ∂ ⎤ ⎪⎫ − r + sinθ + +V r,θ,φ ψ r,θ,φ ⎨ ⎢ 2 ⎜ ⎟ 2 ⎜ ⎟ 2 2 2 ⎥ ( )⎬ ( ) ⎩⎪ 2µ ⎣r ∂r ⎝ ∂r ⎠ r sinθ ∂θ ⎝ ∂θ ⎠ r sin θ ∂φ ⎦ ⎭⎪ = Eψ r,θ,φ ( )

In our rigid rotor example

V r, , 0 r r ( θ φ) = ( = 0 ) i.e. r is held constant at r0. V r,θ,φ = ∞ r ≠ r ( ) ( 0 )

ψ r,θ,φ → ψ r ,θ,φ ( ) ( 0 ) ∂ r , , 0 ψ ( 0 θ φ) = ∂r

Rewrite Schrodinger eq. without variable r:

2 ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ sin r , , E r , , − 2 ⎢ θ + 2 2 ⎥ψ ( 0 θ φ) = ψ ( 0 θ φ) 2µr sinθ ∂θ ⎝⎜ ∂θ ⎠⎟ sin θ ∂φ 0 ⎣ ⎦ I

Set ψ r ,θ,φ = BY θ,φ ( 0 ) ( ) Need to solve

2 ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ sin Y , EY , − ⎢ θ + 2 2 ⎥ (θ φ) = (θ φ) 2I sinθ ∂θ ⎝⎜ ∂θ ⎠⎟ sin θ ∂φ ⎣ ⎦ 5.61 Fall 2004 Lecture #17-19 page 4

SOLUTIONS TO THE RIGID ROTOR

Solving for the rigid rotor problem, we are now left with solving the diff. eq.

2 ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ sin Y , EY , − ⎢ θ + 2 2 ⎥ (θ φ) = (θ φ) 2I sinθ ∂θ ⎝⎜ ∂θ ⎠⎟ sin θ ∂φ ⎣ ⎦

This is HˆY θ,φ = EY θ,φ ( ) ( )

with Hˆ = Kˆ since the potential Vˆ = 0

Rearranging the diff. eq.

⎡ ∂ ⎛ ∂ ⎞ 2IE ⎤ ∂2 sin sin sin2 Y , Y , ⎢ θ θ + 2 θ ⎥ (θ φ) = − 2 (θ φ) ∂θ ⎝⎜ ∂θ ⎠⎟  ∂φ ⎣ ⎦

only θ only φ

We’ve separated the variables, just as in the 3D particle in a box problem.

∴ Try Y θ,φ = Θ θ Φ φ as a solution ( ) ( ) ( )

2IE Define note E β ≡ 2 ( β ∝ ) 

2 ⎡ ∂ ⎛ ∂ ⎞ 2 ⎤ ∂ ⎢sinθ sinθ + β sin θ ⎥Θ θ Φ φ = − Θ θ Φ φ ∂θ ⎜ ∂θ ⎟ ( ) ( ) 2 ( ) ( ) ⎣ ⎝ ⎠ ⎦ ∂φ

Dividing by Θ(θ)Φ(φ) and simplifying

sinθ ∂ ⎛ ∂ ⎞ 1 ∂2 sin sin2 θ Θ(θ) + β θ = − 2 Φ(φ) Θ θ ∂θ ⎝⎜ ∂θ ⎠⎟ Φ φ ∂φ ( ) ( )

only θ only φ 5.61 Fall 2004 Lecture #17-19 page 5

Since and are independent variables, each side of the equation must be equal to a constant m2.

2 1 2 () =
m I ()  2

    sin 2 2 and  sin
() +  sin  = m II
()   

First solve for
() using I

2 () =
m 2 ()  2

() = im
im Solutions are Am e and A
m e

Boundary conditions quantization

( + 2 ) =
()

(+  )
(+  ) im 2 = im im 2 =
im Am e Am e and A
m e A
m e

im(2 )
im(2 )
e = 1 and e = 1

This is only true if m = 0, ±1, ± 2, ± 3,....

m is the “magnetic” quantum number

() = im = ± ± ± Am e m 0, 1, 2, 3,....

 2
Normalization:   ()  () d = 1 0 5.61 Fall 2004 Lecture #17-19 page 6

1 
() = eim m = 0, ±1, ± 2, ± 3,... 2 Now let’s look at
() . Need to solve II sin    sin
() +  sin 2 = m 2
()    dx Change variables: x = cos () = Px() = d
sin

Since 0     
1  x  +1

Also sin2  = 1
cos 2  = 1
x 2

After some rearrangement we obtain the Legendre equation ……

d 2 d  m 2  (
2 ) ()
()+ 
 ()= 1 x 2 Px 2x Px 2 Px 0 dx dx  1
x 

The continuity of
() constraint leads to a quantization for .

= ll( +1) where l = 0, 1, 2,... and m = 0, ±1, ± 2,..., ±l

This directly leads to quantization for the energy!

2 IE
2
=  E =

2 2I

2 E = ll( +1) l = 0, 1, 2,... (for rigid rotor, use J instead of l) 2I

2 = ( + ) = E J JJ 1 J 0, 1, 2,... 2I 5.61 Fall 2004 Lecture #17-19 page 7

m Solutions of Legendre eq. are associated Legendre polynomials Pl m ()= m (
) Pl x Pl cos 1 P 0 (cos) = 1 P 0 (cos) = (3cos2 
1) 0 2 2 0 ( ) =  1 ( ) =   P1 cos cos P2 cos 3cos sin 1 ( ) =  2 ( ) = 2  P1 cos sin P2 cos 3sin etc.

1  (
) 2 m  2l +1 l m ! () = ( ) =   So Alm Pl cos Alm   2 ( + )  l m !

where Alm is the normalization constant
2 2  m  A  P (cos) sind = 1 lm 0  l 

So now putting it all together:

 (  ) = m ( ) =  m ()
() lm r0 , , Yl , l m

1  (
)  2 2l + 1 l m ! m  m ( ) =  ( ) im Yl ,  Pl cos e  4 ( + )  l m !

These functions are called spherical harmonics.

They are the eigenfunctions of HYˆ = EY for the rigid rotor. 5.61 Fall 2004 Lecture #17-19 page 8

1 2 1 3   Y 0 = Y 1 = sine i 0 12 1   (4 ) 8

1 1 2 2 3 
3 
 Y 0 =  cos Y 1 =  sine i 1  4 1  8 etc. SPHERICAL HARMONICS

m ( ) =  m ()
() Yl , l m 1  + (l
m )! 2 m 2l 1  m im Y (,) =  P (cos)e l  4 ( + )  l l m !  l = 0, 1, 2,... m = 0, ±1, ± 2, ± 3,... ± l

m ˆ
=
Yl ’s are the eigenfunctions to H E for the rigid rotor problem. 1 1 5  2 Y 0 = Y 0 = (3cos2 
1) 0 12 2   (4 ) 16

1 1 2 2 3  ± 15  ±  Y 0 =  cos Y 1 =  sin cose i 1  4 2  8 1 1 2 2 3   ± 15  ±  Y 1 =  sinei Y 2 =  sin2 e 2i 1  8 2  32 1 2
3 
 Y 1 =  sine i 1  8

m 
Y ’s are orthonormal: Y m (,)Y m (,)sindd =   l  l  l ll  mm  =   = 
= 1 if l l
= 1 if m m normalization Krönecker delta     ll 0 if l l mm 0 if m m orthogonality 5.61 Fall 2004 Lecture #17-19 page 9

ˆ m = m Energies: (eigenvalues of HYl Elm Y l ) Switch l
J conventional for molecular rotational quantum # 2 IE Recall
= = ll((+ 1)  JJ+ 1) J = 0, 1, 2,...
2

2
E = JJ( + 1) J 2I

J =2

2
± J =1 E = Y 0 , Y 1 (3x degenerate ) 1 I 1 1 J =0 = 0 ( ) E0 0 Y0 nondegenerate

= ( + ) Degeneracy of each state gJ 2J 1 from m = 0, ± 1, ± 2,..., ± J

Spacing between states as J

2
2
= ( + )( + )
( + ) = ( + ) E J +1 E J  J 1 J 2 JJ 1  J 1 2I I

Transitions between rotational states can be observed through spectroscopy, i.e. through absorption or emission of a photon

+ +  h Absorption

EJ EJ+1 - -

+ +

or Emission h EJ EJ-1 - - 5.61 Fall 2004 Lecture #17-19 page 10

Molecules need a permanent dipole for rotational transitions. Oscillating electric field grabs charges and torques the molecule.

2 µ   d 
 µ  Strength of transition I   ( ) dx JJ dx  J J

electric field dipole moment of radiation of rotor

Leads to selection rule for rotational transitions:
J =±1

Recall angular momentum is quantized (in units of
). Photon carries one quantum of angular momentum. Conservation of angular momentum
J =±1 Angular momentum of molecule changes by 1 quantum upon absorption or emission of a photon.
2 =  =  =
= ( + )
= h ( + ) E photon h photon E rot E J +1 E J J 1 photon 2 J 1 J  J +1 J  J +1 I J  J +1 4 I

Define the rotational constant B h h B  (Hz) or B  (cm-1) 8
2 I 8
2cI

= ( + )
-1 = ( + ) J  J +1 (Hz) 2 BJ 1 J  J +1 (cm ) 2BJ 1 E

J =3 = E3 12 Bhc

= E2  3 6Bhc

J =2 = E2 6Bhc
= E1 2 4Bhc J =1 E = 2 Bhc
= 1 E0 1 2 Bhc 5.61 Fall 2004 Lecture #17-19 page 11

This gives rise to a rigid rotor absorption spectrum with evenly spaced lines.

2 B

01 12 23 34 45 56

-1 Spacing between transitions is 2 B (Hz) or 2B (cm )


 = 2 BJ( + 1) + 1
2BJ( + 1) = 2B J +1 J +2 J  J +1  

Use this to get microscopic structure of diatomic molecules directly from the absorption spectrum!

Get B directly from the separation between lines in the spectrum.

Use its value to determine the bond length r0 !

h m m 2 B = I = µr 2 µ = 1 2
2 0 + 4 cI m1 m 2

1 1  h  2  h  2
= -1 = r0  2  (B in cm ) or r0  2  (B in Hz)  8
cBµ   8
Bµ