1. Vibrational Spectroscopy
Outlines
Vibration of diatomic molecules
Permanent vs Dynamics dipole moment
Force and Harmonic potential energy
Classical harmonic oscillator
Exercise: calculate the force constant
Anharmonicity
Quantum harmonic oscillator

2. Comparison of the bond strength of the following molecules.
1) HF HCl HBr HI
2) H2 HD D2
3) F2 Cl2 Br2 I2 N2 O2

3. Permanent vs Dynamics dipole moment HCl
Diatomic molecules
Permanent vs Dynamics dipole moment
HCl
flexible
Permanent
Dynamics
absorbing energy in the microwave region
Absorb energy in the infrared region
the spring allows the two masses to oscillate about their equilibrium distance.
If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field.
rigid
What’s about homonuclear diatomic molecules?

4. Working model & harmonic potential energy
Two atoms are connected by a spring.
(The strength of the bond is related to the stiffness of the spring.)
stretch
(Equilibrium) bond length
compress
the spring allows the two masses to oscillate about their equilibrium distance.
If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field.
In spectroscopy, the vibrational frequency is related to the (potential) energy.

5. Force and Harmonic Potential Energy
Force acts on the system (Hook’s law):
k : the force constant
- The negative sign shows that the force and the displacement are in opposite directions.
The potential energy (V) of the vibration of the diatomic molecule:
Simple harmonic potential

6. Electric dipole moment oscillates with vibration
Electric field (external force to stretch or compress the molecule)
Electric field interacts with a dipole moment within a molecule.
oscillation of the dipole moment
the spring allows the two masses to oscillate about their equilibrium distance.
If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field.

7. Classical (Physics) Harmonic Oscillator
Consider two masses m1 and m2 that are connected by a coiled spring.
the deviation of the spacing between the masses from its rest position is denoted by ,
the motion of a system composed of 2 bodies was dissociated in the motion for the center of mass and the relative motion of one body with respect to the other one.
Positive and negative values of x correspond to stretching and compression of the spring, respectively.

8. It is more convenient to view the vibration motion in the center of mass coordinates (because we consider only the relative motion).
Introducing “The Center of Mass”, xcm
The two mass is reduced into a single mass
And the reduced mass, 

9. Ex. Find the center of mass of 1H35Cl if the H-Cl bond length is 0
Ex. Find the center of mass of 1H35Cl if the H-Cl bond length is nm. (atomic weight of H and Cl are amu and amu). H Cl z y x H Cl
xcm=0.125

10. Solving the Newton’s second law of motion (to find x)
Try the general solution
Because
therefore
The amplitude of oscillation

11. Particle oscillates between +a and - a with frequency if if
The general solution to the differential equation is
c1 and c2 are arbitrary coefficients

12. Consider the Euler formula:
The equation can be further simplified to

13. b1 and b2 are real because the amplitude of oscillation is real.
If we set the boundary condition x(0) = 0 and v(0) = v0 then

14. Therefore, if x(0) = 0 and v(0) = v0
The specific solution takes the form:
Thus, x(t) is a time-dependence function of sine wave .
Note that if the boundary condition x(0)  0 and v(0) = v0
Then b1 and b2 are nonzero  (start at different phase because of the mixed cosine and sine functions).

15. Last lecture : Important messages
Intro. spectroscopy:
EM : wave vs particle
Spectroscopic techniques, range, type of transitions
Boltzmann population
Vibration of diatomic molecules
Permanent vs dynamics dipole moments
Harmonic oscillator:
Classical physics (Newton’s law) treatment: k
if x(0) = 0 and v(0) = v0

16. Frequency () vs Angular frequency ()

17. Where  is the phase shift.
Therefore, the general solution for harmonic oscillation motion is or
Or equivalent to
Where  is the phase shift.

18. IMPORTANCE
Force constants can be calculated from observed vibrational frequency.
The strength of the bond is related to force constants.

19. Calculate the force constant, k, for this molecule.
EXAMPLE A strong absorption of infrared radiation is observed for HCl at 2991 cm-1.
Calculate the force constant, k, for this molecule.
1 N = 1 m kg s-2

20. The total energy of the oscillator:
Assuming the total energy is conserved (no heat lost). KE and PE are transformed into each other. Therefore,
At maximum stretching (x=A) : KE = 0
At maximum velocity (x=0) : PE = 0
The total energy is conserved:

21. Ex. When the displacement of the reduced mass of a diatomic molecule is A/2, what fraction of the mechanical energy is kinetic energy and what fraction of it is potential energy?
For x=A/2,
the fraction of its potential energy is
the fraction of its kinetic energy is

22. Ex. At what displacement, as a fraction of A, is the energy half kinetic and half potential?
Ex. Find the velocity v at a point where x = A/2
Use KE(max) for Etotal

23. For the classical treatment, we can assign any values of x (and v) to calculate the potential (and kinetic) energy.
This is why the classical harmonic oscillator has a continuous energy spectrum.

24. EX. Calculate the force constant for HF and HBr if the absorption is observed at 3962 cm-1 for HF and at 2558 cm-1 for HBr.
For HF
For HBr

25. Ex. Based on the force constant of HF, HCl and HBr obtained previously, rearrange an order of these molecules according to the bond strength.
Observed frequencies (cm-1)
Force constant k (N/m) HF
HCl
HBr
kHF > kHCl > kHBr
Ordering the bond strength
HF > HCl > HBr

26. EX. For an IR absorption of 1H35Cl at 2991 cm-1, if hydrogen is replaced by deuterium, by what factor this frequency would be shifted? Will the vibrational frequency of DCl be higher or lower? Assuming the force constant is unaffected by this substitution.
The vibrational frequency for DCl is lower by.

27. Ex. the graphs below illustrate the vibrational potential energy of three diatomic gases. Identify which graphs belongs to HF, HCl and HBr using the force constant of HF, HCl and HBr obtained previously.

28. De = Dissociation energy, Do = bond energy
Anharmonicity
Morse potential
De = Dissociation energy, Do = bond energy
An simple analytical expression, called the Morse potential, represents the main features of the real potential energy for a molecule:
- close to the minimum potential of depth De, the potential is close to be harmonic.
- for large displacement, the potential represents the bond dissociation.

29. Useful mathematics
When x  xe; x-xe very small
Use the expansion series:

30. Energy levels for anharmonic potential (quantum treatment)
Negative sign: suggesting the lower energy than Eharmonic

31. The bond energy
Do = De – E0

32. Ex. The vibration potential of HCl can be described by a Morse potential with De=7.41 X J and  = X 1013 s-1 .
De=7.41 X J
 = 8.97 X 1013 s-1
h = X J
(h)2/4De= X J

33. 1) Calculate the force constants of diatomic gases
Home Work (1 week)
1) Calculate the force constants of diatomic gases
Observed frequencies (cm-1),
Useful formula
F2 892
Cl2 546
Br2 319
I2 213 N O H
HD 3632 D
2) Based on the force constant obtained in 1), rearrange an order of the molecules below according to the bond strength .
a) H2 HD D2
b) F2 Cl2 Br2 I2 N2 O2

34. Calculate the reduced mass in amu and kg for the following molecules using given atomic mass in the table
Atom
Mass (amu) H
1.008 O
15.99 F
18.998 Br
79.904 N
14.007
gases
 (amu)
 (kg)
HBr HF OH N2 O2

35. Calculate the vibrational state populations of diatomic gases in the first excited state (N1) relative to those in the
ground state (N0) at 500K
gases
(cm-1)
N0/N1 at 500K H2
4400 HF
4138
HBr
2649 N2
2358
C-O
2170

44. Calculate the force constants of diatomic gases from the spectrum
Observed frequencies (cm-1) N O

47. Quantum (Mechanical) Harmonic Oscillator
A simple form of Schrödinger equation
Operator
Eigenvalue
Eigenfunction
This is an “Eigenvalue Equation”
One Dimensional Schrödinger Equation

48. 1D Schrödinger equation of a diatomic molecule
For solving the Schrödinger Equation, we need to define the Eigen functions associated with the state.
It is commonly known that the exponential function is a good one to work with.
Let see a simple example

49. the Schrödinger equation
Ex. Show that the function satisfies the Schrödinger equation for the quantum harmonic oscillator. What conditions does this place on ? What is E?
Solution:
and
the Schrödinger equation
Differential of production function

50. Thus from the product:
Consider the solution for the Schrödinger eq.
The last two terms must be cancelled in order to satisfy the Schrödinger eigenfunction
Therefore, the energy eigenvalue:

51. To find β, we use the cancellation of the last two terms
Therefore, the condition for β that makes the function satisfies the Schrödinger equation
From the classical treatment :
Substitute β, we get the energy :

52. They can be better described using Hermite polynomial wave functions :
Unlike classical treatment, quantum harmonic oscillation of diatomic molecule exhibits several discrete energy states.
They can be better described using Hermite polynomial wave functions :
Vibrational energy states
Eigenfunctions
Normalization constant
Herrmite Polynomials

53. The solution of quantum harmonic oscillation gives the energy:
Where the vibrational quantum number (n) = 0, 1, 2, 3 … n
vibrational state
Evib
(harmonic)
Ground state
1/2 h 1
1st excited st.
3/2 h 2
2nd
5/2 h 3
3rd
7/2 h
It should be noted that the constant Evib= h is good for small n. For larger n, the energy gap becomes very small. It is better to use anharmonic approximation.

54. Vibrational transitionsh 1 2 3 4
n0 ni
Transitions
01
Fundamental
02
First overtone
03
Second overtone
04
Third overtone
- The fundamental transitions, n=1, are the most commonly occurring.
- The overtone transitions n=2, 3, … are much weaker.

55. To plot the vibrational wave functions, we separate the functions into three terms, An , Hn and
An and are not difficult to calculate
The most complicate one is Herrmite Polynomials

56. Hermite polynomials The recurrence relation for Hermite polynomials
For n > 2
The first few Hermite polynomials
n = 0
n = 1
n = 2
n = 3
n = 4

57. For n = 0

58. For n = 1

59. For n = 2

60. The first four Hermite polynomial wave functions, (n= 0,1, 2, 3)

61. Plot of the first few eigenfunctions of the quantum harmonic oscillator (red) together with the potential energy
 The energy of a system cannot have zero energy. 
1st exited vib. state
ground vibrational state
Zero point energy

62. The Probability (ψ2(x))

63. Example for the Hermite polynomial wave functions at n =1 for HCl
What is the normalization constant at this state?

64. Example for the Hermite polynomial wave functions at n =1 for HCl
1) What is the normalization constant at this state?

65. Example for the Hermite polynomial wave functions at n =1 for HCl
1) What is the normalization constant at this state?

66. Example for the Hermite polynomial wave functions (n =0) for HCl
The corresponding energy:
Where n=0, 1, 2, 3, …,

69. EXAMPLE
Answer the following questions.
What are the units of A? What role does have in this equation
Graph the kinetic and potential energies as a function of time
Show that the sum of the kinetic and potential energies is independent of time.

70. What are the units of A?
Because x(t) has the units of length and the sine function is dimensionless, A must have the units of length.
The quantity  sets the value of x at t = 0,because .

73. b) Graph the kinetic and potential energies as a function of time
Show that the sum of the kinetic and potential energies is independent of time.

75. Solution
For the J=0 → J=2 transition,
The preceding calculations show that the J=0 → J=1 transition is allowed and that the J=0 → J=2 transition is forbidden. You can also show that is also zero unless MJ=0 .

77. Consider the relative motion via the center of mass coordinates.
Angular Momentum (L) m 
Consider the relative motion via the center of mass coordinates.