2 Quantum (Mechanical) Harmonic Oscillator A simple form of Schrödinger equationEigenvalueEigenfunctionHamiltonian OperatorKinetic OperatorPotential OperatorFor One Dimensional Schrödinger Equation
3 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
4 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 equationConsider a given function:1D Schrödinger equation
5 For the left side of Eq.:1st derivative :2nd derivative :
6 Thus from the product:The last two terms must be cancelled in order to satisfy the Schrödinger eigenfunctionTherefore, the energy eigenvalue:
7 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 equationFrom the classical treatment :Substitute β, we get the energy :the energy
8 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 statesEigenfunctionsNormalization constantHerrmite Polynomials
9 Hermite polynomials The recurrence relation for Hermite polynomials For n > 2The first few Hermite polynomialsn = 0n = 1n = 2n = 3n = 4
10 The first four Hermite polynomial wave functions, (n= 0,1, 2, 3)
11 Solving the1D Schrödinger equation IfThe solution for the eigenstate is:IfThe solution for the n eigenstate is:or
12 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. stateground vibrational stateZero point energy
14 The solution of quantum harmonic oscillation gives the energy: Where the vibrational quantum number (n) = 0, 1, 2, 3 …nvibrational stateEvib(harmonic)En-En-1Ground state1/2 h-11st excited st.3/2 hh22nd5/2 h33rd7/2 hNoted 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.
15 Ex. for the Hermite polynomial wave functions at n =0 for HCl What is the normalization constant at this state?
16 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 isThe general form of Hermite polynomialsFor n = 0
17 For n = 1The recurrence relation for Hermite polynomials
18 Hermite polynomials The recurrence relation for Hermite polynomials For n > 2The first few Hermite polynomialsn = 0n = 1n = 2n = 3n = 4
19 Ex. Plot the ground state wave function (n =0) of HCl where observed frequency 2991 cm-1.
20 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 4nEn1/2 h13/2 h25/2 h37/2 h49/2 hE4321hat n=0, E0 0The 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
21 Vibrational transitions 4n0 niTransitions01Fundamental02First overtone03Second overtone04Third overtone321h- The fundamental transitions, n=1, are the most commonly occurring.- The overtone transitions n=2, 3, … are much weaker (low intensities).
22 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=0The 1st excited state vibrational energy, n=1The 2st excited state vibrational energy, n=2
23 Zero point energy = The ground state vibrational energy b) Determine the energy and wavenumber for the fundamental, first-overtone transitionsfundamental transition : n=0 n=11st overtone transition : n=0 n=2
24 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)
25 De=7.41 X J = 8.97 X 1013 s-1h = X J(h)2/4De= X JThe vibrational energies of the ground state and the first and second excited statesFundamental transition : n=0 n=1First overtone transition : n=0 n=2
26 Selection RulesThe 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 directionPermanentDynamics
27 Assuming transition from the ground state (n = 0) to state m n = 0 n = mTruncate at the first derivative:
28 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 functionIf m is even Even functionX Odd functionm must be odd : 1, 3, 5, 7, …
29 Integration give an area under the peaks n=0n=3 Possible transitionsm must be odd : 1, 3, 5, 7, …n=0n=1n=0n=3n=0n=5…n=0n=5Integration give an area under the peaksn=0n=3n=0n=1TransitionsIntegrationn=0n= 0n=0n= = 0n=0n= = 0
30 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 absortionn = 1n = +1 absorptionn = -1 emission
31 Vibration of polyatomic molecules How many vibrational modes are observed for a molecule?A molecule consisting of N atomsPositionRepresentation1 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 – 6Number of vibrational modes for linear molecules = 3 N-5there are only 2 rotational modes
33 H2O- non-linear molecule- N = 3- number of vibrational normal modes = 3 X 3 – 6 = 33685 cm-13506 cm-11885 cm-1H2O has three distinct vibrational frequencies.
34 CO2- linear molecule- N = 3- number of vibrational normal modes = 3 X 3 – 5 = 4Because of xz and xy are degenerate, CO2 has three (not four) distinct vibrational frequencies.
35 Benzene C6H6- non-linear molecule- N = 12- number of vibrational normal modes = 3 X 12 – 6 = 30There are 30 vibrational modes but only 20 distinct vibrational frequencies.
36 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:
37 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)
38 What is the energy difference between (112) and (100)?
42 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 formThe solutions to these 3N-6 equations are the familiar Harmonic Oscillator Wavefunctions and Energies.The total vibrational energy is:
43 Coupled system has two vibrational frequencies: the symmetrical and antisymmetric modes. For symmetrical and asymmetrical, the vibrational frequency is
44 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 .CalculateMorse function of vibrational potential:De = Dissociation energy, Do = bond energyAt the minimum energy (x=xe), the potential energy:
45 The energy levels for the anharmonic correction: The highest value of n consistent with the potential that isDe=7.41 X J = 8.97 X 1013 s-1h = X J(h)2/4De= X J
54 Example for the Hermite polynomial wave functions at n =1 for HCl 1) What is the normalization constant at this state?
55 Example for the Hermite polynomial wave functions at n =1 for HCl 1) What is the normalization constant at this state?
56 EXAMPLEAnswer the following questions.What are the units of A? What role does have in this equationGraph the kinetic and potential energies as a function of timeShow that the sum of the kinetic and potential energies is independent of time.
57 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 .
58 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.
60 SolutionFor 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 .