 Hello, welcome to the lecture number 31 of the course quantum mechanics and molecular spectroscopy. In the previous lecture, we were talking about the harmonic oscillator and its selection rules. So we will continue with the same topic. So if you have a molecule or a AB molecule has a potential something like that, so this I will call it as 0 and this will be some V naught and there is an R equilibrium distance R e that is the if you consider bond AB, the equilibrium distance is R e and this is the potential energy function as a function of R. This is nothing but the potential energy, so the energy. So at the bottom of this well, one can write the potential as a Taylor series expansion. So your V is equal to at equilibrium is nothing but V naught minus dV by dr at R e into R plus 1 over 2 factorial is nothing but 2 d square V by dr square at R e equated R e plus R square plus minus 1 over 3 factorial d cube V by dr cube evaluated R e into R cube plus 1 over 4 factorial d to the power of 4 V by dr to the power of 4 evaluated R e R to the power of 4 plus etc. Now one can always measure energies with respect to some value and if that value is V 0, so measure energies relative to V 0. That means if I measure energies with so I can equate this to be 0. Now the second thing is dV by dr, we are at the bottom of the potential. So when you are at bottom of the potential, the first derivative will go to 0. So this will go to 0 because bottom of the potential are same as a minimum energy in the potential. Then you are left with and if I ignore higher order terms or that means I will ignore third order, higher order means third order which is this order 3, fourth order which is this. So if I ignore that, ignore means basically going to 0. So what I will do is I will also take this to 0 and this to 0. So what I am left with V is equal to half d square V by dr square evaluated R e R square where R is the intermolecular distance or interatomic distance in this case. So if you have a potential where R is the interatomic distance then what you have is the, then what you have is the V equals to half d square V by or approximately equal to dr square evaluated R e R square. This I will write it as half k R square where k is nothing but d square V by dr square evaluated R e. Now if you write the total energy Hamiltonian h is equal to minus h bar square by 2 mu del square R ok now where if you have a and b connected and mu is nothing but m a m b divided by m a plus m b ok and R is the coordinate. So this R is nothing but internal coordinate plus half k R square. So this is my Hamiltonian for harmonic oscillator. Now harmonic oscillator is just a model that represents a e b bond ok. Now one of the problems with harmonic oscillators is that. So when I draw harmonic oscillator, harmonic oscillator is with respect to some R e and it should something like that. So harmonic oscillator never breaks ok it can the potential can raise up to infinity but all chemical bonds between molecule atom a and b they can break they will break eventually if you keep stretching they will break. So harmonic oscillator in some sense is not a good representation for a chemical bond. However at the bottom of the potential that means around the equilibrium geometry harmonic oscillator is a reasonably good model and we use harmonic oscillator to represent bonds when they are at the equilibrium position ok. Now if you have this harmonic oscillator then of course your Hamiltonian as I just said is nothing but minus h bar square by 2 m del square R minus a plus half k R square and where k is nothing but d v d square v by d R square evaluated R e ok this is nothing but your force constant. Now the problem here is the following the thing is that this force constant is only the second order term ok and we are in such case we are ignoring the higher order terms the third order and the fourth order terms which a normal molecule or a normal diatomic molecule a b would have. However harmonic oscillator is an approximation ok. Now if your harmonic oscillator is an approximation and we know R is just a dummy variable. So I am going to rewrite this whole thing in some other variable x. So instead of R I can always use another variable x and that would be minus h bar square by 2 mu del square x plus half k x square this is going to be my Hamiltonian h vibrational. Now you have this Hamiltonian one can solve for it there are two methods to solve this ok one is called ladder operator method and other is the series method. So essentially you are going to solve a second order differential equation ok. So this can be written as minus h bar square by 2 mu t square by dx square plus half k x square this into some function chi of x ok h vibrational into chi of x is equal to e chi of x that is going to be a solution for it. But when you solve either by either of the method ok what have we one can solve by whichever is convenient method and there are textbooks like molecular quantum mechanics by Atkins or quantum chemist by Ira and Levine where you can find the solutions ok you can use either of the method and of course it is only one differential equation. So depending on which method you use or the solution that come out should not depend on the method that you use ok. Now I am just going to write the solutions. So when you solve this using one of these two methods what you get is the following your chi of x ok will depend on a quantum number v ok that is a vibrational quantum number is equal to e v chi of ok. So that is going to be your quantum number where you can show that e v is equal to v plus half h where nu e is nothing but 2 pi omega where nu is nu is frequency and omega is angular frequency. But there is one thing that you must understand that there is this value k and this value k and this nu e should be related in some way ok sorry this is not that should be the right one ok. So how is it related? So the k that is the force constant is related to mu omega e square or this is nothing but 4 pi square mu nu e square. So the force constant k is proportional to nu e square. So that means the vibrational frequency is proportional to square root of the force constant that means if you increase the force constant vibrational frequency will increase. So that is what we already know. So if the bond becomes more stiffer the vibrational frequency will increase ok. Stiffness is measured by the force content x. So this is a measure of stiffness of a bond ok. Now the problem so what we are going to solve the Schrodinger equation at hand for the vibration h v i b vibration chi v of x is equal to minus h bar square by 2 mu d square by dx square plus half k x square into chi vibration of x ok is equal to e vibration into chi vibration of x. So that is your chi v. Now means I just said that e v equals to half plus v h nu e is also equal to half plus v bar omega v and here v is the vibrational quantum number and that goes from 0 1 2 3 2. Now the solutions for this would be ok. So you have what you have you have h vibration of chi v of x is equal to e v of chi v of and your chi v of x will be some normalization of n of v exponential minus alpha x square by 2 h v alpha to the power of half x ok. This is the form ok. This is the form of the solution ok. Now where this one is the normalization constant and this is your Gaussian function and this is the Hermite polynomial ok. Now one thing that I have written is your alpha which I have not written is equal to mu omega by alpha is some kind of a some kind of a weighting weighted coordinate see because you are multiplying with x. So it is kind of a weighting of the coordinate. So this is actually is called mass weighting or mass weighing ok. So one can always write or generally in the text to book it is written that alpha to the power of half x equals to y. If you write if I make that transformation then what happens your chi v will now become function of y by the way x and y are not different functions like orthogonal variables like what we are using Cartesian coordinate. X is y is just a transformation of x ok chi of y is equal to n v exponential y square by 2 exponential y square by 2 h v of y ok. So just trying to become more easier to represent or write ok and I told you this is nothing but your normalization constant and this is nothing but your Gaussian function and this is the Hermite polynomial. When you solve it turns out that the Hermite polynomials have a dependences that means there is generating function the recursion relationship is h e plus 1 of y is equal to 2 y h v of y minus 2 mu. So let me h v plus 1 of y is equal to 2 y h v of y plus 2 v h v minus 1 of y and I can get the generating functions. So what I get is h 0 of y is equal to 1 h 1 of y is equal to 2 y h 2 of y is equal to 4 y square minus 2 h 3 of y is equal to 8 y cube minus 12 y h 4 of y will give me 16 y to the power of 4 minus 48 y square plus 12 h 5 of y is equal to 32 y to the power of 5 minus 160 y cube plus 125 ok. Now you can see h 0 of y is only 1 h y of y is 2 y. So it y is equal to 0 it will go to 0. So this will have 1 node, this will have 2 nodes because the quadratic equation, this will have 3 nodes because it is a cubic equation, it will have 4 nodes it is a quarter cube and this is a will have 5 nodes. So when I draw ok but all of them are the Hermite polynomials which are riding over your Gaussian function. So when I draw them so v is equal to 0 I will get a Gaussian function ok. Now for v is equal to 1 I will get a but this is so one can imagine so this is y is equal to 0 or equilibrium ok. So we will get this function and the third case we will get. So if you look at the functions they will look like particle in a box function. So they can always be represented like this. So something like that ok. They look like particle in a box function but cartesian box functions are sinusoidal function. Here they are Gaussian functions that are modulated by the Hermite polynomials ok. So this is v is equal to 0, v equals to 1, v is equal to 2. So you can always see that the functions become odd and even as the quantum number ok. So this v is equal to 0 is an even function, v is equal to 1 will be odd function, v is equal to will be even function and so on ok. So for the even quantum numbers the function is an even function and for odd quantum number it is odd function. So if I look back my h, my not h but chi of y is equal to or v is equal to some constant v that is normalization constant e to the exponential minus y square by 2 ok. I am not sure oh they should have been a negative sign here. I am sorry about that. So there should have been a negative sign here ok there is a negative sign here already there. So I just missed. So minus y square by 2. Now one thing that I want to tell you is very simply is that when you have the Hermite wave function chi of v ok. Now I am writing in terms of y. This is nothing but some normalizations constant n of v it is power of exponential minus y square by 2 h v of y. But we said y is nothing but alpha to the power of half into x where alpha is equal to what was the value of alpha mu omega by h bar yeah that is what it is alpha is this ok. Now mu is the if you have a and b masses mu is a constant omega is also constant because omega is nothing but is related to omega is related to square root of k ok and k is related to the potential. So this also is a constant h bar is a constant. So alpha is a constant. So alpha to the power of half will also be a constant that means when you have y to the power of alpha x ok. The behavior of y or the shape of the function of y will be any variable y will be same as the variable x ok. So h v of x transform to h v of y will have a same shape except that the y will get multiplied by so that means the depending on the value of alpha the scale will either elongate or shrink. If alpha is greater than 1 then the scale will expand and alpha if so this will lead to expansion of the scale and alpha is less than 1 this will be instead of expansion it will the scale will shrink. But it will not change the way the function will look like the shape of the function will remain constant ok. So you have the functions of or Eigen functions of your harmonic oscillators as exponent Gaussian functions multiplied by the Hermite polynomials ok and will look like oscillator is a representation of vibrational wave function which we will continue in the next lecture. Thank you.