1
Vibrational Spectroscopy
Harmonic Oscillator
Potential Energy
Vx 
1 2
kx
2
x  R  Re
where Re – equilibrium bond length
Selection Rules
The dipole moment of a molecule can be expanded as a function of x = R – Re.
1   2 
1   3 
  
  x     0    x   2  x 2   3  x3 
2  x 0
6  x 0
 x 0
  0   0 – dipole momentum at equilibrium, permanent dipole moment.
  
  – change in dipole moment along vibrational coordinate
 x 0
– effect of stretching, compression, bending, etc… on dipole moment.
Recall that transition is possible only when transition dipole matrix element is nonzero, that
is f ˆ i  0
For the dipole operator in the transition dipole matrix element, substitute the series
approximation above.
1   2 
1   3 
  
  
f ˆ i  f 0    x   2  x 2   3  x 3  i  f 0    x i
2  x 0
6  x 0
 x 0
 x 0
  
  
  
f 0    x i  f 0 i  f   x i   0 f i    f x i
 x 0
 x 0
 x 0
A transition implies that the final state must be different from the initial state f i  0
Thus f x i  0 for a transition to occur.
  
Also,    0 for a transition to occur. This requirement means that the vibration that is
 x 0
being excited must cause the dipole moment of the molecule to change. This requirement is
true for transitions in infrared vibrational spectroscopy.
2
Recall that   N  H     e

2
2
1
4
 k 
where    2  x
 



  
  
f  i    f x i     Nf Hf    e 2
 x 0
 x 0 
2
Careful:  is reduced mass.
 2 14 
2

 

 N H  e 2
  k   i i


1
 2 4

 d
 k 
1
2 2

   
   
  
f i
 f  i  
 x 0   k 
 x 0 k
1
1
Recall that H   H 1  H 1        1    1
2
2
1
   
    

f  i  
f  i  
f i i 1  f
i 1 

2
 x 0 k
 x 0 k 

1
    

 
i f i 1  f i 1 

2
 x 0 k 

Or perhaps stated a little more clearly.
1
    

f  i  
i f i  1  f i  1 

2
 x 0 k 

The first integral is nonzero only when f  i1    1
The second integral is nonzero only when f  i1    1
Therefore the selection rule for a harmonic oscillator in vibrational spectroscopy is   1
Anharmonic Oscillator
General Potential Energy
The actual potential energy (not idealized) can be written as a Taylor series about the
point R = Re (i.e., x = 0).
1   2 V  2 1   3V  3 1   4 V  4
 V 
V  x   V  0  
x


 x   3 x   4 x 

2  x 2 0
6  x 0
24  x 0
 x 0
The origin of the potential energy can be set arbitrarily (and sensibly) to zero.
That is, V(0) = 0 by definition.
 V 
Also, since the slope of the potential energy curve at Re is zero 
 0
 x 0
3
1 V


2  x 2 0
2
Thus the first nonzero coefficient of the potential energy series above is
This coefficient can be related to the force constant.
 2V 
1  2V  1

k

k



 2
2  x 2 0 2
 x 0
The next nonzero coefficient of the potential energy series causes the oscillator to be
anharmonic. The nonzero coefficients can be manipulated to yield to yield the cubic
anharmonicity constant, ee, the quartic anharmonicity constant, yee, etc, …
Energy levels of the anharmonic oscillator.
2
3
1
1
1



E   h e       e  e      y e e     
2
2
2



Note: The energy levels for the harmonic oscillator are evenly spaced. ( E  h e )
However the energy levels for the anharmonic oscillator are not evenly spaced.
( E  h e ). The energy levels for the anharmonic oscillator get closer together as the
quantum number  increases.
Morse Potential
1 2
kx
2
A better approximation of an actual potential energy curve is the Morse potential.
Electronic potential energy curve is not purely quadratic, i.e., V  x  
1
V  x   De 1  e

 ax 2
 2  2
where a  
 and De is the potential well depth.
 2De 
harmonic (quadratic)
V(x)
x
D0
De
anharmonic (Morse)
D0 – dissociation energy
1
De  D0 
 – zero point energy
2
4
Using the Morse potential, the spectroscopic term becomes
1
1


G     e      e e    
2
2


2
where e 
h e
4De
This expression is an exact solution for the Morse potential.
Worth emphasizing is that the Morse potential is only an approximation (albeit, a good
one).
If the above spectroscopic term, cannot reproduce the experimental spectra, then a more
accurate potential energy function is needed. (Usually based on Taylor series expansion.)
When the polynomial found from the Taylor expansion is used, a more general spectroscopic
term can be written.
2
3
4
1
1
1
1




G      e      e  e      ye e      z e e     
2
2
2
2




Selection Rules of the Anharmonic Oscillator
Nonzero transition dipole matrix elements of the anharmonic oscillator have two origins
1.) Anharmonic wavefunctions are combinations of HO wavefunctions.
2.) Dipole can have quadratic (cubic, etc…) dependence
  2 
i.e.,  2   0 thus f  i and f x 2 i , f x 3 i , f x 4 i , etc…
 x 0
The resultant selection rules are complicated, but transitions where  is a single digit
are all allowed. That is,   1, 2, 3, 4, 5,
Consequences of Anharmonicity
1.) Transitions where   1 are allowed
- Such transitions are called overtones.
- 1  0 transition is called the fundamental.
2.) More than one vibration can be excited with a single transition.
- Such transitions are combination bands.
- Consider water with its three modes of vibration.
H
H
H
H
H
H
H
H
O
O
O
O
symmetric stretch - s
H
antisymmetric stretch - a
H
H
O
bending - b
H
O
5
- The total vibrational wavefunction is product of individual vibrational wavefunctions.
s  a  b   s  a  b
000 - vibrational ground state
10 0  000 - fundamental for symmetric stretching mode
010  000 - fundamental for antisymmetric stretching mode
0 01  000 - fundamental for bending mode
- Examples of combinations bands would be
2 01  000
312  2 21
110  000
0 01  211
Miscellaneous
A hot band is an absorptive transition from an excited state.
Vibration – Rotation Spectra
Analysis for the HCl/DCl experiment is part of this section. The discussion won’t be
repeated here. Refer to the lab handout.
Combination Differences
The technique of combination differences is used to find the rotational constant for an
excited state.
The analysis is similar (but not identical) to the analysis for the band spectrum of nitrogen.
Consider the simplest term for rotovibrational spectroscopy
- harmonic oscillator and rigid rotor only.
1

S  , J    e      B J  J  1
2

However, the rotational constant changes for each vibrational state. We can find the
rotational constant for the excited vibrational state by taking the difference of specific
spectral lines to remove the influence of the rotation of the ground vibrational state.
6
Consider a portion of a molecular energy level diagram.
= 1
P(J+1)
R(J-1)
P(J)
J+1
J
J-1
3
1

  e  B1  J  1 J    e  B0 J  J  1 
2
2

  e   B1  B0  J   B1  B0  J 2
R(J)
J+1
= 0
 P  J   S    1, J  1  S    0, J 
J
J-1
 R  J   S    1, J  1  S    0, J 
3
1

  e  B1  J  1 J  2     e  B0 J  J  1 
2
2

  e   B1  B0  J  1   B1  B0  J  1
2
B1 – rotational constant of the  = 1 state
B0 – rotational constant of the  = 0 state
B – general rotational constant (has dependence on )
Remember P branch  J  1
Remember R branch  J  1
Since the frequencies of the spectral lines depend on B1 and B0, one can take combination
differences of the appropriate lines to eliminate either B1 or B0, thus solving for the other
rotational constant.
1

 R  J  1   P  J  1  4B0  J  
2

1

 R  J    P  J   4B1  J  
2

Thus, both rotational constants can be found.
7
Polyatomic Vibrational Spectroscopy
Normal Vibrational Modes
Vibrations in a polyatomic molecule involve movement of all the atoms in a molecule.
- In contrast, to only two atoms vibrating at a time
Modes of vibration that are independent of each other (i.e., don’t affect each other) are called
normal mode vibrations.
For a molecule with N atoms, 3N total modes of motion exist.
3 translational modes
3 rotational modes
2 modes for linear molecule
3N-6 vibrational modes
3N-5 modes for linear molecule
Symmetry of Polyatomic Vibrations
To find symmetry of a vibration, attach vectors to atoms to indicate atomic movements.
The symmetry label of the vibration is found by noting how these vectors transform when
they are operated upon by symmetry operators.
Example: Consider “puckering” mode of BF3.
F
F
B
above plane
below plane
F
BF3 belongs to the D3h symmetry group
D3h
A1
A2
A1
A2
E
E
E
1
1
1
1
2
2
h
1
1
-1
-1
2
-2
2 C3
1
1
1
1
-1
-1
2 S3
1
1
-1
-1
-1
1
3 C2 3 v
1
1
-1
-1
1
-1
-1
1
0
0
0
0
Note the effect of symmetry operations on vectors.
E  vectors same
=1
h  vectors reversed   = -1
C3  vectors same
=1
vibration has A2 symmetry
S3  vectors reversed   = -1
C2  vectors reversed   = -1
v  vectors same
=1
8
Raman Spectroscopy
Raman Scattering
Raman spectra come from fundamentally different process than “normal”
absorption/emission spectra.
For Raman spectra, light is scattered by molecule, not absorbed or emitted.
Such scattering of light can be considered as a collision between the photon and the
molecule.
1.) If the collision is elastic (elastic scattering), the frequency of the scattered photon is
the same as the incident photon
- Process is known as Rayleigh scattering.
2.) If the collision is inelastic (inelastic scattering), the frequency of the scattered
photon is different than the incident photon. That is, the photon has gained or lost
energy.
- Process is known as Raman scattering.
Vibrational Raman Spectroscopy
anti-Stokes
Stokes
Rayleigh
An incident photon has an electric field that can interact with the charge distribution of a
target molecule.
The applied electric field deforms the electron cloud and the effect of the deformation is
proportional to the strength of the applied electric field.
- In other words, the dipole moment of the molecule changes in response to the electric
field.
 - polarizability
  0  E
The deformed electron cloud can return to
its original state to yield Rayleigh
scattering.
excited electronic
state
virtual electronic
state
= 4
= 3
ground electronic
state
= 2
= 1
= 0
Or the deformed electron cloud can return
to another state.
1.) If the final state has higher energy
than the initial state, the scattering
event (and subsequent spectral
line) is known as a Stokes shift (or
Stokes scattering).
2.) If the final state has lower energy
than the initial state, the scattering
event (and subsequent spectral
line) is known as an anti-Stokes
shift (or anti-Stokes scattering).
9
Selection Rules

1   2  2
  
ˆ
f  i  f  0   E i  f  0   0  
 x  2  x 
2  x 0
 x 0


E i


   
  
 f  0   0  
 x  E i  f 0 i  f 0E i  f 
 xE i
 x 0 
 x 0

  
  
 00
 E f x i 
 E f xi
 x 0
 x 0
Thus for vibrational Raman spectroscopy, the selection rule is   1 for the harmonic
oscillator. For the anharmonic oscillator, the selection rule are complicated as for vibrational
absorption spectroscopy.   1, 2, 3, 4, 5,
Also important to note is that not all vibrations can participate in Raman scattering. Only
those vibrations where the polarizability of the molecule changes during the course of a
  
vibration will have a Raman spectrum (i.e., 
  0)
 x 0
Rotational Raman Spectroscopy
Remember that an applied electric field deforms the electron cloud of a molecule.
As the molecule rotates, the molecular dipole changes because of the applied electric field.
-
+
-
+
-
+
-
+
-
The induced dipole changes at double the rate of rotation.
- The induced dipole changes after half of a rotation.
The consequence of the changing dipole is that the selection rule for rigid-rotor Raman
spectroscopy is J  2 .
Q branch (Rayleigh)
S branch
O branch

+