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**Vibrational Spectroscopy**

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

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**Quantum (Mechanical) Harmonic Oscillator**

A simple form of Schrödinger equation Eigenvalue Eigenfunction Hamiltonian Operator Kinetic Operator Potential Operator For One Dimensional Schrödinger Equation

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**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

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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

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

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

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**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

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**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

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**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

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

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**Solving the1D Schrödinger equation**

If The solution for the eigenstate is: If The solution for the n eigenstate is: or

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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

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**The Probability (ψ2(x))**

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**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.

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**Ex. for the Hermite polynomial wave functions at n =0 for HCl**

What is the normalization constant at this state?

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**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

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For n = 1 The recurrence relation for Hermite polynomials

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**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

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

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**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

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**Vibrational transitions**

4 n0 ni Transitions 01 Fundamental 02 First overtone 03 Second overtone 04 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).

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**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

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**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

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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)

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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

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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

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**Assuming transition from the ground state (n = 0) to state m**

n = 0 n = m Truncate at the first derivative:

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**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, …

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**Integration give an area under the peaks n=0n=3**

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

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**Selection Rules The only transition from n=0n=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

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**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

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**Vibration Stretching vibration Bending vibration Symmetric Stretching**

Asymmetric Stretching In plane Bending Out of plane Bending

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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.

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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.

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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.

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**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:

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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)

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

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**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:

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**Coupled system has two vibrational frequencies: the symmetrical and antisymmetric modes.**

For symmetrical and asymmetrical, the vibrational frequency is

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**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:

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**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

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Anharmonicity

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**Ex. for the Hermite polynomial wave functions at n =0 for HCl**

What is the normalization constant at this state?

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**Example for the Hermite polynomial wave functions at n =1 for HCl**

What is the normalization constant at this state?

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**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

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**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

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For n = 0

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For n = 1

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For n = 2

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**Example for the Hermite polynomial wave functions at n =1 for HCl**

1) What is the normalization constant at this state?

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**Example for the Hermite polynomial wave functions at n =1 for HCl**

1) What is the normalization constant at this state?

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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.

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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 .

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**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.

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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 .

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**Consider the relative motion via the center of mass coordinates.**

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

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