 # Vibrational Spectroscopy

## Presentation on theme: "Vibrational Spectroscopy"— Presentation transcript:

Vibrational Spectroscopy
Outlines Quantum harmonic oscillator Vibrational transitions Selection Rules Vibration of polyatomic molecules

Quantum (Mechanical) Harmonic Oscillator
A simple form of Schrödinger equation Eigenvalue Eigenfunction Hamiltonian Operator Kinetic Operator Potential Operator For One Dimensional Schrödinger Equation

1D Schrödinger equation of a diatomic molecule for any states
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

Ex. Show that the function satisfies the Schrödinger equation for the 1D quantum harmonic oscillator. What conditions does this place on ? What is E? Schrödinger equation Consider a given function: 1D Schrödinger equation

For the left side of Eq.: 1st derivative : 2nd derivative :

Thus from the product: The last two terms must be cancelled in order to satisfy the Schrödinger eigenfunction Therefore, the energy eigenvalue:

the energy 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 : the energy

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

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

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

Solving the1D Schrödinger equation
If The solution for the eigenstate is: If The solution for the n eigenstate is: or

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

The Probability (ψ2(x))

The solution of quantum harmonic oscillation gives the energy:
Where the vibrational quantum number (n) = 0, 1, 2, 3 … n vibrational state Evib (harmonic) En-En-1 Ground state 1/2 h - 1 1st excited st. 3/2 h h 2 2nd 5/2 h 3 3rd 7/2 h 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.

Ex. for the Hermite polynomial wave functions at n =0 for HCl
What is the normalization constant at this state?

The general form of Hermite polynomials
Ex. Using Hermite polynomials to show that the ground and first exited states wave function (n =0, n=1) of a diatomic molecule is The general form of Hermite polynomials For n = 0

For n = 1 The recurrence relation for Hermite polynomials

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

Ex. Plot the ground state wave function (n =0) of HCl where observed frequency 2991 cm-1.

The first five levels: n = 0, 1, 2, 3, and 4
Ex. Draw the vibrational energy diagram of the first five levels of a diatomic molecule using the solution of quantum harmonic oscillation : The first five levels: n = 0, 1, 2, 3, and 4 n En 1/2 h 1 3/2 h 2 5/2 h 3 7/2 h 4 9/2 h E 4 3 2 1 h at n=0, E0  0 The energy of a system cannot have zero energy, even at n=0. It is called “Zero point energy”. Zero point energy : the energy of the ground vibrational state

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

Ex. The harmonic vibrational frequency of HCl in wavenumbers is 2991 cm–1.
a) Calculate the energies of the first three vibrational levels in Joules and calculate zero point energy. The ground state vibrational energy, n=0 The 1st excited state vibrational energy, n=1 The 2st excited state vibrational energy, n=2

Zero point energy = The ground state vibrational energy
b) Determine the energy and wavenumber for the fundamental, first-overtone transitions fundamental transition : n=0  n=1 1st overtone transition : n=0  n=2

Ex. The vibration potential of HCl can be described by a Morse potential with De=7.41 X J, k = N m-1 and  = X 1013 s-1 . Determine the energy and wavenumber for the fundamental, first-overtone transitions using the vibrational energy function for the anharmonic correction below: First of all, determine the vibrational energies of the ground state (n=0) and the first and second excited states (n= 1 and 2)

De=7.41 X J  = 8.97 X 1013 s-1 h = X J (h)2/4De= X J The vibrational energies of the ground state and the first and second excited states Fundamental transition : n=0  n=1 First overtone transition : n=0  n=2

Selection Rules The transition probability from state n to state m is only nonzero if the transition dipole moment satisfies the following condition: where x = spatial variable μx = dipole moment along the electric field direction Permanent Dynamics

Assuming transition from the ground state (n = 0) to state m
n = 0  n = m Truncate at the first derivative:

Substitute  with Hermite polynomial functions
(so permanent dipole is not relevant for IR absorption) n = 0; n = m; = 0 (Orthogonal) If m is odd Odd function If m is even Even function X Odd function m must be odd : 1, 3, 5, 7, …

Integration give an area under the peaks n=0n=3
Possible transitions m must be odd : 1, 3, 5, 7, … n=0n=1 n=0n=3 n=0n=5 n=0n=5 Integration give an area under the peaks n=0n=3 n=0n=1 Transitions Integration n=0n=  0 n=0n= = 0 n=0n= = 0

Selection Rules The only transition from n=0n=1 is allowed.
For the other an integration is zero. The transition is forbidden. Selection Rules for IR absortion n = 1 n = +1 absorption n = -1 emission

Vibration of polyatomic molecules
How many vibrational modes are observed for a molecule? A molecule consisting of N atoms Position Representation 1 atom : 3 coordinates (x1, y1, z1) N atoms : 3 N coordinates ((x1, y1, z1), (x2,y2,z2), (x3,y3,z3),…..) Degree of freedoms (Molecular motions) * There are only 3 normal modes (coordinates) describing the translational motion. * There are 3 normal modes describing the Rotational motion of the molecules (along 3 axes) Number of vibrational modes = 3N – 6 Number of vibrational modes for linear molecules = 3 N-5 there are only 2 rotational modes

Vibration Stretching vibration Bending vibration Symmetric Stretching
Asymmetric Stretching In plane Bending Out of plane Bending

H2O - non-linear molecule - N = 3 - number of vibrational normal modes = 3 X 3 – 6 = 3 3685 cm-1 3506 cm-1 1885 cm-1 H2O has three distinct vibrational frequencies.

CO2 - linear molecule - N = 3 - number of vibrational normal modes = 3 X 3 – 5 = 4 Because of xz and xy are degenerate, CO2 has three (not four) distinct vibrational frequencies.

Benzene C6H6 - non-linear molecule - N = 12 - number of vibrational normal modes = 3 X 12 – 6 = 30 There are 30 vibrational modes but only 20 distinct vibrational frequencies.

In the normal coordinate system, the vibrational Hamiltonian of polyatomic system can be written as:
Therefore, the wave function is of the form : The total vibrational energy is:

Ex. The water molecule has three normal modes, with fundamental frequencies: 1-bar = 3685 cm-1, 2-bar = 1885 cm-1, 3-bar = 3506 cm-1. a) What is the energy of the (112) state? (i.e. n1=1, n2=1, n3=2)

What is the energy difference between (112) and (100)?

The Schrödinger equation is simplified by separation of variables
The Schrödinger equation is simplified by separation of variables. Therefore, one gets 3N-6 equations of the form The solutions to these 3N-6 equations are the familiar Harmonic Oscillator Wavefunctions and Energies. The total vibrational energy is:

Coupled system has two vibrational frequencies: the symmetrical and antisymmetric modes.
For symmetrical and asymmetrical, the vibrational frequency is

Morse function of vibrational potential:
Ex. The vibration potential of HCl can be described by a Morse potential with De=7.41 X J, k = N m-1 and  = 8.97 X 1013 s-1 . Calculate Morse function of vibrational potential: De = Dissociation energy, Do = bond energy At the minimum energy (x=xe), the potential energy:

The energy levels for the anharmonic correction:
The highest value of n consistent with the potential that is De=7.41 X J  = 8.97 X 1013 s-1 h = X J (h)2/4De= X J

Anharmonicity

Ex. for the Hermite polynomial wave functions at n =0 for HCl
What is the normalization constant at this state?

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

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

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

For n = 0

For n = 1

For n = 2

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

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

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.

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 .

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.

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 .

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