Winter 2013
Chem 356: Introductory Quantum Mechanics
Chapter 6: Rotational and Rovibrational Spectra ....................................................................................... 75
Different Approximations ....................................................................................................................... 80
Spectrum for Harmonic Oscillator + Rigid Rotator ................................................................................. 81
Polyatomic Molecules ............................................................................................................................. 84
Harmonic Oscillator + Rigid Rotor Model to Obtain Quantities from Statistical Mechanics .................. 84
More Conventional Discussion of Rotations for Diatomics .................................................................... 86
Chapter 6: Rotational and Rovibrational Spectra
A) General discussion of two-body problem with central potential
Examples:
a)
Diatomic
b)
 Ze 2
Hydrogen atom with charge Z V (r ) 
4 0 r
V ( R)
Let us consider 2 particles with mass m1 , m2
P2
P2
Hˆ  1  2  V  r1  r2
2m1 2m2
Coordinates:
r1 , p1  m1

dr1
 m1r1
dt
r2 , p2  m1r2
Define center of mass coordinate
Rcm 
m1r1  m2 r2
,
m1  m2
M  m1  m2
Pcm  MRcm  m1r1  m2 r2
Also define the relative coordinate
r  r2  r1
p   r   (r2  r1 )

m1m2
m1  m2
Chapter 6: Rotational and Rovibrational Spectra
75
Winter 2013
Chem 356: Introductory Quantum Mechanics
Then we will show that
Pcm  Pcm
p p
1

 r2
2
2M
2
1
1

(m1 r1  m2 r2 )  (m1 r1  m2 r2 )   (r2  r1 ) 2
2M
2
1

(m12 r12  m2 2 r2 2  2m1 m2 r1 r2  m1 m2 (r12  r2 2  2r1 r2 ))
2M
1
 m1 (m1  m2 )r12  m2 (m1  m2 )r2 2 

2(m1  m2 ) 
T

1
1
m1 r12  m2 r2 2
2
2
P12
P2 2


T
2m1 2m2
Hence the Hamiltonian can be written as
 p2

P
Hˆ  cm  
 V (r ) 
2M  2


2


cm 2   
 2  V (r ) 
2M
 2

2
Now we can apply a separation of variables:
 ( Rcm , r )   ( Rcm ) (r )
 2 2


 cm  ( Rcm )   2    V (r )  (r )

2M

 Etotal
 ( Rcm )
 (r )
2
2
ET , translational energy
E  The energy we are really interested in
If we put the system in a (very large) box, the center of mass problem just yields the particle in the box
solutions
 k xxm 
sin 

 L 
ET 
 l z xm 
sin 

 L 
 m z xm 
sin 

 L 
h 2 2 2 2
 k  l  m2 
2ML2
This describes translational kinetic energy of the center of mass motion [used in stat mech, Chem 350,
Chem356]
Chapter 6: Rotational and Rovibrational Spectra
76
Winter 2013
Chem 356: Introductory Quantum Mechanics
Our interest is the relative motion:

2
2
2 (r )  V (r ) (r )  E (r )
Because V ( r ) only depends on r 
x 2  y 2  z 2 , it is very convenient to use
spherical coordinates
r  x  y  z
2
x  r sin  cos 
2
2

1
2
 y
 

z
  arccos 
 x2  y 2  z 2

y  r sin  sin 
z  r cos 
  arctan  
x




 2
2
2 
To develop this further we need to obtain    2  2  2 
y
z 
 x
2
In spherical coordinates, this is a very tedious exercise using the chain rule.
The basic step in the derivation is like this:
f (r , ,  ) f r f  f 



x
r x  x  x


r     



x
x r x  x 
1
and
r

x
  x2  y 2  z 2  2

1

1 2
x  y2  z2  2  2x

2
x
x r cos  sin 
 
 sin  cos 
r
r
 express everything in spherical coordinates
2
2
2


You might appreciate that deriving the operator
is a lot of work (use MathCad?!)
x 2 y 2 z 2
Let me give you the result for the kinetic energy operator
1   2  
Lˆ2

 
r

2
2 r 2 r  r  2 r 2
2
2
2
Lˆ2  
2
 1  
 
1 2 
sin



 sin   
  sin 2   2 


Chapter 6: Rotational and Rovibrational Spectra
77
Winter 2013
Chem 356: Introductory Quantum Mechanics
Let me also give you
 
 

Lˆz  xPy  yPx  i  x  y   i
x 

 y
Lˆ2  Lˆ 2  Lˆ 2  Lˆ 2
x
y
z
So L̂2 is precisely the operator corresponding to the total angular momentum
This kinetic energy operator can also be written as
2 
2 
Lˆ2



2  r r r 2  2 r 2
2
Let us note that L̂2 only depends on the angular coordinates  ,  ad so does Lˆ z
In a later lecture I will derive the eigenfunctions and eigenvalues of the L̂2 and Lˆ z operators. We will
use the commutation relations to do this (compare harmonic oscillator)
For now I will just list the solutions
Lˆ2Yl m  ,    l (l  1) 2Yl m  ,  
Lˆz Yl m ( ,  )  m Yl m ( ,  )
m  l.....  l
These functions are the angular parts of the familiar hydrogen orbitals
s-orbitals
l  0, m  0
p-orbitals (3)
l  1 , m  1, 0,1
d-orbitals (5)
l  2 , m  2, 1, 0,1, 2
f-orbitals (7)
l  3 , m  3, 2, 1, 0,1, 2,3
Always (2l  1) l -type orbitals
These functions are normalized as


0
2
sin  d  dYl1 m * ( ,  )Yl m ( ,  )
1
0
  ll1  mm1
 1 for l  l 1 , m  m1
0 otherwise
The extra factor sin  is related to the surface element
Chapter 6: Rotational and Rovibrational Spectra
78
Winter 2013
Chem 356: Introductory Quantum Mechanics
More general for spherical coordinates
r 2 sin  drd d
dxdydz 
Volume element

 


f ( x, y, z)dxdydz 
r
0

2
0
0
r 2 dr  sin  d 
f (r ,  ,  ) d 
To solve H  E in spherical coordinates we try the solution
 (r , ,  )  f (r )Yl m ( ,  )

2 
2 

f (r )Yl m ( ,  )  V (r ) f (r )Yl m ( ,  )

2 
2  r r r 
1
 2 f (r ) Lˆ2Yl m ( ,  )  Ef (r )Yl m ( ,  )
r
2


This yields a radial differential equation:

 2  2


2  r r r 2
2

 f (r ) 

2
l (l  1)
f (r )  V (r ) f (r )  El f (r )
r2
Together with normalization condition


0
f (r ) * r 2 dr  1
This radial equation, depending on l , is a completely general result, valid for any 2-body problem, with a
central potential V  r2  r1

When discussing chapter 7 (Hydrogen atom), I will say more about the spherical solutions Yl m ( ,  ) , and
Ze2
also discuss the radial problem for V (r )  
4 0 r
At this point, I want to return to diatomics. The solutions to the S.E. for a vibrating and rotating molecule
are
 (r, , )  f (r )Yl m ( ,  )  V  r  f  r  
2 
 l (l  1) 
2 

 2  f (r )  
f (r )  V  r  f  r   El f (r )

2 
2  r r r 
 r 
2


0
r 2  f (r )  dr  1
2
The angular functions are the exact solutions related to rotations. The radial equation is “the ‘harmonic
oscillator’ in disguise.”
Chapter 6: Rotational and Rovibrational Spectra
79
Winter 2013
Chem 356: Introductory Quantum Mechanics
Let us bring the equation to a more familiar form:
rf (r )  g (r )
Define
Then
 1 
1  2
f (r ) 
 2 rf (r )  
 f (r )  r 

r  r
r 
 r r 
1  f
 2 f   2 f  2 f 
 2  r 2   


r  r
r   r r r 2 
2

1
2 
1  (l (l  1)
1
rf (r )  
 V (r )  rf (r )   rf (r )  E
Hence

2 
2
r  2 r 
r  r
r

Or using rf (r )  g (r ) :


 2 g  l (l  1)

 V (r )  g (r )  Eg (r )
2
2
2 r
 r

2
This starts to look like a harmonic oscillator equation
x  (r  Re )

 r 



x x r r
r  x  Re
2
 2  2

l (l  1)


 V ( x)  l ( x)  El l ( x)

2
2
 ( x  Re )
 2 x

 normalization:

0  rf (r ) 

If V ( x) 
2
dr  

0
 g (r ) 
2
dr
1 
2
 ( x) dx

2 
1 2
l (l  1)
kx we have harmonic oscillator except for the term
2
 ( x  Re ) 2
Different Approximations
a) Harmonic oscillator + rigid rotor:
l (l  1)
l (l  1)

2
 ( x  Re )
 ( Re ) 2
2
Set
Replace l  J : common variable in rotational spectroscopy
Ev , J
2
1

  v   
J ( J  1)
2   Re 2

 v, J ,m   v ( x)Yl m ( , )
Simplest solution; often used
Chapter 6: Rotational and Rovibrational Spectra
80
Winter 2013
Chem 356: Introductory Quantum Mechanics
b) Solve for anharmonic wavefunction
2


 2 ( x)

 V ( x)  v ( x)  Ev v ( x)

2
 2 x

Then evaluate
2

Bv    v * ( x)

 ( x  Re )2
 v ( x)dx
Ev , J  Ev  Bv J ( J  1)
Rotational constant depends on vibrational level (smaller B , larger R , as V increases)
We only need to calculate few vibrational wavefunctions
c) Solve equation exactly for each v, J ; degeneracy is always 2J  1 , m  J ...J
Spectrum for Harmonic Oscillator + Rigid Rotator
2
1

Ev, J    v    J ( J  1)
2  2I


I   R2 ,  
k



 2I 
Define B  
2
Selection rules (diatomic, H.O./R.R.)
v  1
J  1
  1000 cm 1
B  10 cm 1
Chapter 6: Rotational and Rovibrational Spectra
81
Winter 2013
Chem 356: Introductory Quantum Mechanics
a) Pure rotational transitions
EJ 1  EJ 

hvobs 

2
2I
2
2I
(( J  1)( J  2)  J ( J  1))
 2( J  1) 
2
I
( J  1) 
h2
4 2 I
( J  1)
h2
( J  1)
4 2 I
h
vobs 
( J  1)  2 B( J  1)
4 2 I
v
h
obs  obs  2 ( J  1)  2B( J  1)
c
4 Ic
h
rotational constant B 
(Hertz)
2 2 I
h
B
(cm-1) (use c in cm s-1)
2 2 Ic
Rotational transitions J  1 (selection rule; discussed later)
Set of Equidistant lines
At room temperature both ground and excited rotational levels are occupied, and we get J  J  1
absorption for all J . This leads to a set of equidistant lines in the spectrum.
Let us now consider transitions to different vibrational state
V  1 , J  1
Chapter 6: Rotational and Rovibrational Spectra
82
Winter 2013
Chem 356: Introductory Quantum Mechanics
If we go beyond H.O., then
Ev v 1, J  J 1  Ev  B1 J '( J ' 1)  B0 J ( J  1)
J  J ' ; J '  J  1 and B1  B0 “centrifugal distortions”
Working it out:
E1, J 1  E0, J  VR    2 B1  (3B1  B0 ) J  ( B1  B0 ) J 2
VL  E1, J 1  E0, J    ( B1  B0 ) J  ( B1  B0 ) 2 J 2
( B1  B0 )  0
Also the lines in pure rotational spectrum are not exactly equally spread.
 J  J 1  2 B( J  1)  4 D( J  1)3
EJ  BJ ( J  1)  DJ 2 ( J  1) 2
Why:
solve equation exactly!
Chapter 6: Rotational and Rovibrational Spectra
83
Winter 2013
Chem 356: Introductory Quantum Mechanics
Polyatomic Molecules (briefly)
Assume we know equilibrium geometry, a quadratic force constant matrix, then we can define a center
of mass motion.
Rcm   m j rj /  m j
j
j
And a wavefunction (particle in the box) associated with Rcm
Another 3 coordinates are associated with overall rotation and we can associate classical rigid body
moment of inertia:
I    m j ( R jx 2  R jy 2  R jz 2 )   m j R j R j
j
j
 ,   x, y , z
This symmetric matrix can be diagonalized yielding eigenvalues I a , I b , I c and a corresponding set of
axis. These correspond to displacement of all the nuclei using a rigid rotation.
The remaining (3 N  6) or (3 N  5) coordinates define the normal modes
 1 d2 1 2 
Hˆ  Tˆcm  TˆR   i  
 qi 
2
2 
i
 2 dqi
Jˆa 2 Jˆb 2 Jˆc 2
ˆ
TR 


2I a 2Ib 2Ic
 R j ,k ,m ( , x, )  Pjkm ( )eikx eim
E j ,k ,m
m   j... j
k   j... j
The rotational problem for any I a , I b , I c can be solved using matrix diagonalization: numerically straight
forward
 total   cm  R  1 (q1 )2 (q2 ).....3 N 6 (q3 N 6 )
E  Ecm  ER  Evibr
This separation of variables is valid in the H.O./R.R. approximation. These approximations are
quite good for spectroscopy; Very good for stat mech.
Harmonic Oscillator + Rigid Rotor Model to Obtain Quantities from Statistical Mechanics
I want to recall the basic formulas from Statistical Mechanics here, and show the immense usefulness of
H.O./R.R. for thermochemistry.
Chapter 6: Rotational and Rovibrational Spectra
84
Winter 2013
Chem 356: Introductory Quantum Mechanics
a) Exact formulation of Stat Mech
Define system partition function (e.g. gas of molecules):
  E ( N ,V ) 
Q( N ,V , T )   exp  
 energy levels

 kBT

Connection to thermodynamics:
A( N ,V , T )  kBT ln Q  N ,V , T 
From the absolute value of A the Helmholtz free energy we can get any thermodynamics
function.
How?
dA   sdT  pdV   dN
 A 

  S
 T V , N
 A 
 A 

  P 
 
 V T , N
 N T ,V
U  A  TS , G  A  PV , H  A  TS  PV
 U 
Cv  
etc…

 T V
Now proceed to independent molecules in the gas phase:
Q
1
( qm ) N
N!
qm  q t qV q R q N q e translation, vibration, rotation, nuclear, electronic
qt : from particle in the box quantum solutions
1

 2 Mk B  2
 

2


3
V
qt   T 2
N
Vibrational:

3 N 6
1
i / kT
e 2
qv  

i 1 1  e
i / kt
From Harmonic oscillator Q.M.  sum each level
Rotational + Nuclear spin (complication leads to symmetry factor  ):
1
1
1
1
2
  2  T 2  T 2  T 2
Tx 
qR         
2I x
    TA   TB   TC 
I x : Moments of inertia  from rigid Rotor quantum mechanics
qe  e De / kT  eEi / kT
i
Chapter 6: Rotational and Rovibrational Spectra
85
Winter 2013
Chem 356: Introductory Quantum Mechanics
De :
separated atom limit
Ei : binding energies, bottom of well
One can calculate Q for each molecular species in the gas phase
Using a slight extension one can also calculate QtransitionState and from this one can obtain
reaction rates
 Simple Quantum models
-
Particle in the box
Harmonic oscillator
Rigid rotor
Electronic energies ( no good simple models)
 Accurate thermochemistry for gas phase reactions
More Conventional Discussion of Rotations for Diatomics
Two bodies at fixed distance R rotating around center of mass:
Moment of Inertia:  R 2
Angular momentum: L  LZ   vR (see Bohr atom)
1 2 ( vR)2
L2
v 
v
2
2 R 2
2I
General for rigid body rotation of linear molecules:
T
L2
(more general for polyatomic)
2I
 Relative motion in quantum mechanics
Chapter 6: Rotational and Rovibrational Spectra
86
Winter 2013
Chem 356: Introductory Quantum Mechanics
qx 

k


, 
k

 1 d2 1 2 
Lˆ2
ˆ
H  
 q 
2
2  2 R 2
 2 dq
 v (q)Yl m ( ,  )
Solutions:
E  Ev  ER
2
 1
 v    
J ( J  1)
2
2 R 2

Harmonic Oscillator + Rotational rigid body motion.
 ~ 1000 cm 1
B
2
2I
 1 - 10 cm1
Selection rules for H.O/R.R.
V  1
J  1
Molecules needs permanent dipole to observe rotational transitions
Note: It is hard to see where the approximation comes in or how to improve on it.
Chapter 6: Rotational and Rovibrational Spectra
87