 Welcome to the lecture number 14 of the course, Quantum Mechanics and Molecular Spectroscopy. As usual, we will have a quick recap of contents of lecture number 13 before we proceed with the lecture number 14. We started with the time depend perturbation and we showed that this was equal to i h bar q by m a del, where del is nothing but d by dr, where r is general coordinate, ok. It could be x, y, z or whatever, ok r, ok. Now, this we could transform to q e naught by omega m cos omega t e dot p, ok, because you should realize i h bar del minus i h bar del is equal to operator p, ok. So, now what you are looking at is the dot product between the electric field vector and the momentum vector. This is nothing but dot product and momentum vector, ok. That means, for this to h prime t to be non-zero, the electric field and momentum should be in the same direction or should have enough projection on to each other, ok. Now, we know that r h naught, h naught commutator, ok, which I derived in the last class is equal to i h bar by m. So, one can replace this momentum operator by r h naught, ok. Having all done, we come across this equation f h prime of t i. So, all I am going to do is, you know, to get the, to do the relevant mathematics, ok, that I have done in the last lecture, you can look it up. Minus i e naught cos omega t divided by h sigma over n q n epsilon n r n i minus e i f sigma over n q n epsilon r n i, ok. These are the integrals. Now, you will see that nuth sigma over n q n r n, this is nothing but dipole moment of the molecule mu or the atom, ok. So, this is just tells you the, how the charge is distributed, ok. So, when we do that and do little bit of rearrangement, what you get is f h prime t i equals to e naught cos omega t omega i f pi omega integral f, where h bar omega i f is equal to delta e is equal to e i minus e f, ok. There is a e f minus e i, but there is a negative sign. So, I kind of cancel that, ok. And there is an i here. So, there is a ratio of omega i f to omega and there is this integral, ok. And this slightly can be written as i cos omega t omega i f pi omega. Now, e naught into e vector is going to give you electric field e. So, f. Now, your integral f h prime of t i will be equal to i cos omega t omega f i omega i f by omega integral f e dot mu. Mu also is a vector by the way the dipole moment is a vector, ok. So, essentially you know if you leave out this, ok, it depends on this integral. So, this integral has been transformed with some prefactor into this integral, ok. So, you can think of it like this f h prime of t i is proportional to f e dot mu, ok. And this integral is called transition moment integral. So, f e dot mu i, ok is one of the most important integrals in spectroscopy. It is a transition moment integral and that will determine whether the transition will take place or not. If transition moment integral goes to 0, ok, the selection rule is decided by this. Now, let us think of a, let me give a simple example, ok. Let us suppose that, ok, there is an electric, there is a electromagnetic radiation which is going like that, ok. So, this is the direction of propagation r, ok. And your electric field vector e is along this, ok. Of course, when it goes down it becomes minus e. So, the electric field vector is going up and down, ok, in this direction e and me. So, it changes sign. Now, if you take a diatomic molecule, let us consider a single diatomic molecule AB. And since the diatomic molecule it is linear, the dipole moment is in this direction. So, if you consider a as a positive and b as negative, the negative direction is it. So, it is, ok, positive, negative, ok. So, a. So, now, you can see that if the propagation direction is r and the dipole moment is along the r. So, now, think of it like this. So, if your light is propagating like that and your molecule AB is like that, ok. Now, what is the e? e is perpendicular. So, your epsilon dot. So, this is your mu and this is the director. So, and your epsilon vector is like this. Epsilon dot mu is equal to 0 because the dipole moment of the molecule AB and your vector epsilon are perpendicular to each other. That means, under the circumstances, the molecule will not absorb light or there will be no transition. On the other hand, if AB is like this, if you have AB like this and your propagation direction is like this and your electric field vector is like this. Now, epsilon and mu are parallel to each other. So, the dot product will be maximum. So, this will be epsilon dot mu will be not equal to 0 or any direction that is has a projection. That means, so this will have some projection on to epsilon. Even that case epsilon dot mu will not be equal to 0, ok. That means, the transition moment integral that is F, transition moment integral is equal to E naught dot mu. This will become nonzero. So, which means, if the dipole moment and epsilon, the electric filter should have projection on to each other, only then the transition moment integral will not to 0. So, that tells you the direction in which the electric field vector and the dipole moment have to be aligned with respect to each other, ok. Now, so what we have is your is basically your F H prime t is equal to minus i plus i E naught cos omega t omega i F by omega integral F. So, as I told you this is nothing but your transition moment integral, ok. And I also told you that the dipole moment and the electric field should have projection on to each other. They must be aligned. If they are perpendicular to each other, there will be no the transition moment integral will go to 0, ok. Now, there is another thing that you can do is that if you now this is just a transition, but your transition probability P F of t is given by, ok, 1 by H bar square integral of 0 to t prime dt e to the power of minus i omega i F t integral F H prime of t i whole square. But now we have evaluated this integral as this, ok. So, this will be equal to E naught square E naught square by H bar square omega i F by omega whole square 0 to t prime dt e to the power of minus i omega i F t cos omega t integral F epsilon dot mu i whole square. Now, this is a integral that does not involve time and this is a integral. So, I can separate it out as two integrals. So, this is nothing but E naught square by H bar square omega i F by omega square integral 0 to t prime dt e to the power of minus i omega i F t cos omega t square to F epsilon dot mu i whole square, ok. So, we will know how to evaluate this integral if you know E dot mu operator and if you know the states f and i, ok. And these are just the pre factors. So, now it comes to evaluating this integral, now let us look at that integral. So, the integral 0 to t dt e to the power of minus i omega i F t cos omega t is something that I need to evaluate, ok. Now, this integral this is equal to 0 to t prime dt e to the power of minus i omega i F t cos omega t cos omega t is can always written as e to the power of i omega t minus e to sorry plus e to the power of minus i omega t by 2, is not it cos theta definition is that, e to the power of i theta plus e to power of minus i theta by 2. Now, if I slightly rearrange this equation will become 0 to t prime dt e to the power of if I take i common then I get omega minus omega i F t plus e to the power of minus e to the power of minus i if I take minus i common then I will get omega plus omega i F and there is a half outside I will get 2 integrals. Now, I am going to slightly rearrange this. So, this is equal to half integral 0 to t prime. So, omega i F will become. So, this will become e to the power of i omega F i plus omega t 2 plus e to the power of my, ok. Now, this omega F i if it becomes minus omega F i minus of omega F i and I can take minus of here from outside. So, it will become i into omega F i minus of ok. So, I have these 2 equations now. So, which means I have this is equal to you will get 2 integrals, integral 0 to 2 prime e to the power of i omega F i plus omega t dt plus half integral 0 to t prime e to the power of i omega F i minus omega t, ok. So, this is our next step half of integral 0 to t prime e to the power of i omega F i plus omega t dt plus half integral 0 to t prime e to the power of i omega F i minus omega t dt. Now, what I will do is I will multiply pi and divide by pi for both of them. So, this is equal to pi into 1 by 2 pi integral 0 to t prime e to the power of i omega F i plus omega t dt plus over 2 pi 0 to t prime e to the power of i omega F i minus omega t. Now, I will make a small mathematical trick. Let us suppose you have a some atom or a molecule some system, ok. So, let us say some atom or a molecule and a time t is equal to 0 you switch on the light, ok. So, and a time t is equal to t is equal to t prime you switch off the light. So, what you do is that you start switching on the light and then a time t is equal to t prime equal. So, this is 0 and this is t prime. So, you switch on the perturbation at time t is equal to 0 and switch off the perturbation at time t is equal to 0. And you will see now this will be minus infinity and this will be plus infinity, ok. Once the perturbation is switched off or not switched on, ok, time does not the perturbation does not really add that means, ok. Instead of integrating over from 0 to t because then what happens here it is all zeros no perturbation, ok. So, which means if you have no perturbation it does not add to the integral. So, which means instead of 0 to t prime as you are integral definite integral ends, ok. You can write it as minus infinity to plus infinity. Even though it is from minus infinity to plus infinity the actual region in which the light adds or the light acts on the molecule is only 0 to t prime rest of the time you are just adding zeros, ok. So, it does not really matter, ok. So, I want to make this transformation it will not change any physical principle. However, it will deduce these equation to some standard integrals. So, when I have done that then I can write it as pi into 1 over 2 pi minus infinity to plus infinity e to the power of i omega f i minus omega plus omega into t dt plus 1 over 2 pi integral minus infinity to plus infinity e to the power of i omega f i minus omega t, ok. So, what I have done is I have just changed the limits of your integral instead of going from 0 to t prime I have taken from minus infinity to plus infinity, minus infinity to 0 there is no light acting on it, ok. So, that integral essentially the contribution to that integral is essentially 0, but after t prime to infinity again there is no light. So, the contribution to the integral from t prime to infinity will also be 0. So, essentially what you are doing you are adding lot more zeros and the number is not going to change by adding zeros, but this transformation or this simple mathematical trick allows you to get to a standard integral, ok. So, this is our integral that we need to evaluate, ok. So, I will stop here and continue in the next lecture. Thank you.