 we will come to the lecture number 16 of the course quantum mechanics and molecular microscopy. We will just look at the final equation that we derived in the previous lecture and proceed with that. Toward the end of the last lecture we showed that the probability of transition to a final state F is given by 4 pi square E naught square by H bar square E naught bar H square omega F phi pi omega del omega F phi plus omega plus del omega F phi minus omega square F epsilon dot mu dot i whole square. I made a small mistake in the last lecture is that this 4 does not exist because you know when you write cos omega t you have 1 over 2. So that when comes out of this square square or absolute square cancel the score. So there is a small mistake but that is not going to change the way we look at the entire problem. So P of t is just given by pi square E naught square by H bar square into omega F phi by omega square del omega F phi plus omega del omega F phi omega square less of half epsilon mu i whole square ok. Now there is one more thing is when you are squaring this. So omega F i square of course is equal to omega i f square ok. So the transition whether it is going from the state F to state i or state i to state f this will remain the same ok. Now the other thing that we said that when you have del omega F i plus omega this will correspond to stimulated emission and del omega F i minus omega this will correspond to absorption ok. And we said this cannot happen simultaneously either stimulated emission will happen or absorption will happen ok. So only one of the process can happen at given point of time and both cannot happen simultaneously. So if we consider for absorption process from initial state i to a final state f then P F of t is given by pi square E naught square by H bar square omega F i by omega square omega F i minus omega square F epsilon dot mu that is the transient probability for absorption process going from an initial state i to a final state f ok. Now one can also write very simply slight rearrangement. So P F of t now we know H bar is nothing but H by 2 pi ok. So when we had pi square by H bar square. So H bar square is equal to H square by 4 pi square or 1 over H square is equal to 4 pi square by H bar square ok. So one can think of this as if I take this 4 so 1 over 4 H bar square is equal to pi square by H bar square. So I can always write this so your P F of t can also be written as E naught square omega F i by omega square del omega F i minus omega square F epsilon dot mu whole square. So this can also be written as E naught square by 4 H square omega F i by omega whole square del omega F i minus omega whole square F epsilon dot mu. Now you can see you can write in two different ways essentially the functional form still remains the same. So it will have the square of the transient moment integral and then we will have this delta function and some constants. Now the problem here is this that how this equation ok will behave ok that is what we want to look at. Let us look at the delta function delta of some function x ok. This can be written as limits n tends to infinity 1 over 2 pi sin n x by 2 divided by x by 2. This is a way one can write a delta function as well ok. Now one can also write delta omega F i minus omega as delta omega if one writes that then your of delta omega can be written as limit n tends to infinity 1 over 2 pi sin of delta omega to n by 2 divided by delta omega. But what is your n here n is just the time ok. So I can still rewrite as delta omega is equal to limit t tends to infinity is equal to 1 over 2 pi sin of delta omega t by 2 by delta omega. Now why I am using limit t tends to infinity because you can I we have all but this is a valid limit because remember when we drew this perturbation curve from 0 to t prime and goes to infinity. So time can go up to infinity without any consequence because after t prime it does not the perturbation does not exist. So one can think of this limit to be taken over. So this same as extending our integral. So which means I can rewrite P of t of F this is equal to pi square E naught square by H bar square omega F i by omega square del of omega F i minus omega square theta F epsilon dot mu i whole square. Now this I will write in terms of the limit ok. So this will come out to be there was a 2 pi 1 or 2 pi so that I can bring it outside. So pi square E naught square by H bar square omega F i by omega whole square into 1 over 4 pi square because and this pi square and this pi square will get cancelled and this one will be sin of delta omega t by 2 divided by delta omega by 2 whole square F epsilon mu pi where delta omega is nothing but omega F i minus omega ok. Now so I am going to slightly rewrite this equation. So P F of t is equal to E naught square by 4 H bar square omega F i pi omega square modulus of sin del omega t by 2 by del omega by 2 square whole square. So this is the probability. So now you can see the probability of a transition or an absorption from initial state to final state will depend on this will depend on this transition moment integral and will get modulated by this function. So this is nothing but T mi transition moment integral and this is nothing but your modulating function. I will come back to what modulation function really means but just you know for a minute let us keep it let us look at it. So I am going to tell now what does it really mean? It means that your probability of function F of t equals to ok. Now as I told you this will depend on E naught square by 4 H bar square omega F i by omega square sin square del omega t by 2 divided by del omega by 2 because the square so I have just bought it out of square and multiplied by modulus of F epsilon dot mu i square. Now this I told you the transition moment integral it is a definite integral I will just write it as t and I take a square of it so it will be t square. So your P F of t is equal to E naught square by 4 H bar square omega F i by omega square sin del omega t by 2 square of that and del omega by 2 square of that to t square. Now you can see for a given omega ok omega F i is fixed because the energy difference between the two states is fixed. So your initial state and the final states are fixed so omega F i is a constant omega will vary because if your electromagnetic radiation varies that will vary because you can tune the radiation you can go from some frequency to some other frequency or some wavelength to some other wavelength. For example if you are recording an absorption spectrum in the visible light with the visible light then your wavelength will change from 400 nanometer to 800 nanometer and correspondingly the frequencies will also change ok. So omega will is a varying function ok. So this will get affected because omega F i is fixed but omega could vary. So this ratio will vary apart from that this del this sin function will vary because you know as delta omega varies sin also will varies. But you see sin function can only go from 0 to 1 ok. So the sin square function also goes from only from 0 to 1 ok in fact sin function goes from minus 1 to plus 1 but sin square function can only go from 0 to 1. So this function is basically modulating between 0 and 1 and now if I plot delta omega this is 0 ok that means omega F i is equal to omega this is equal to delta omega is equal to 0 and then you can think of some units 1 2 minus 1 minus 2 minus 3 and then 1 2 3 etc. And if I plot this function here that is nothing but sin square del omega t by 2 whole square divided by del omega by 2 rather there is no t I will just plot the without the t then this function will look something like this ok or this is kind of exaggerated view actually these will be even lower ok. So this will hit the roof this will go much more and this will so which means your absorption ok will also happen away from the resonance but so this is the resonance delta omega is equal to 0 is also called resonance condition it can also happen away from the resonance ok. However you will see that this will be very very low. So essentially away from the resonance will still not be able to see transitions. Major transition will happen or the maximum probability of transition will only happen at the resonance that is nothing but the omega F i is equal to omega and this is also called Bohr condition. So essentially your P f of t equals to E naught square by 4 h bar square omega F i by omega whole square sin square del omega t by 2 divided by del omega whole square this is gives you probability of transition or absorption from initial state I to a final state F something like that ok. Now one interesting fact is that this has got a width ok it has got a width and this width has some important consequence you know we will go to the line widths and then we will discuss this and the line width comes from this function ok and this function will tell us selection rules ok. So forever and this will tell you effectively tell you intensity. So essentially there are three factors that determine the probability of a transition from a ground state or initial state I to a final state one is the transition moment integral that will tell you whether this action will be allowed or not if allowed what its value second thing is your modulating function which will tell you how the line widths will come about and third one will be intensity that will depend on the how much light you are shining. So we will stop at this point of time and continue in the next lecture. Thank you.