 Welcome to the lecture number 15 of the course quantum mechanics and molecular spectroscopy. As usual, we will have a quick recap of the previous lecture and continue with the present lecture. In the last lecture, we looked at the transition probability to a state f and this was given by E naught square by h bar square omega f i by omega square modulus of integral 0 to t prime dt e to the power of i omega minus i omega i f t cos omega t f yeah. So this is the integral and this we told you is I told you is the transition moment integral. So one can think of p of t to be proportional to the square of the transition moment integral. In the transition moment integral the operator says E dot mu says that the dipole moment of the molecule or the atom okay. By the way dipole moment is not same as the permanent dipole moment okay mu naught okay dipole moment of the molecule and epsilon okay this is epsilon is nothing but your electric field. So electric field must be parallel to the dipole moment or at least should have some projection it cannot be perpendicular okay. Now I will take the other so apart from that there is this integral in this that integral is 0 to t prime dt e to the power of minus i omega i f t cos omega t. I am going to manipulate this integral a little bit. Now we know that cos omega t is equal to half of e to the power of i omega t plus e to the power of minus i omega t. So if I substitute that in here then what I will get is 0 to t prime dt I will write 2 so I will take the half outside this is nothing but e to the power of now omega i f is equal to minus omega f i. This is nothing but e i minus e f and this is nothing but e f minus e i so that is just a reverse of sin omega i f and omega f i. So this can be written as into e to the power of i omega t plus e to the power of minus i omega t. So this will be nothing but half of 0 to t prime dt and then you will have 2 e to the power of i omega f i plus omega into t plus e to the power of i omega f i minus omega into t. So that is the integral. So expanding or continuing what you will have t of f of t is equal to e naught by h bar square e naught square by omega f i by omega whole square to modulus of integral dt 0 to t prime e to the power of i omega f i plus omega t plus 0 to t prime dt e to the power of i omega f i minus omega t integral f e dot mu i whole square. Now the question is what does it say? It says that the omega or the electromagnetic field adds from time 0 to time t prime. So think of it like this. So there is some perturbation time dependent perturbation that starts at 0 and ends at t prime. So 0 to t prime. So this is your limit of integration. So this will be nothing but your limit of integration. But if you consider entire time of minus infinity to plus infinity then what happens is from minus infinite to 0 there is no light. So there is no perturbation ok. And similarly after t prime there is no perturbation. So the effect of perturbation before t is equal to 0 and after t is equal to t prime is going to be non-existent. If that is non-existent then without losing any physical concept this integral can now be written as minus infinity to plus infinity dt e to the power of i omega f i plus omega t. And this integral can be written as integral dt minus infinity to plus infinity e to the power of i omega f i minus omega t. Now this is a case of adding zeros. For example you want to know how much money you have in your bank within say some period of time ok. Say first of a month to fifteenth of a month. But before that you neither you open the account on say first of the month and you close the account on fifteenth of the month and you want to know what is. But if I want to look at the entire time period before the first of the month at the previous month and after time after fifteenth of that month ok you have the account is opened and closed. So before the opening previous month and after fifteenth of month the money that account will have will have zeros will add to zeros. So essentially the entire transaction will try only between the first of the month to fifteenth of the month and transactions before that and transactions after that will not lead to any usefulness. So they are all zeros. So this is a similar scenario where you know the limit of integration is only between zero and t prime. But any extending the integration limits to minus infinity to plus infinity is like adding zeros and has no physical consequence except the fact that it turns out to be a standard integral ok. Now one can write a standard integral of between two variables let us say x and y such that x and y are conjugate. Now what are conjugate variables? Variables that are have inverse unit. So for example if x is length then y will be 1 over length such that product of x and y will give to dimensionless quantity ok. Such variables are called conjugate variables. So in quantum mechanics x and momentum or position and momentum are conjugate variable energy and time are conjugate variable and they are also related by the Schrodinger equation. Now one can write a Fourier integral you can write a Fourier integral of two conjugate variables let us say x, y in such a way that x minus a equals to 1 over 2 pi integral minus infinity to plus infinity dy e to the power of 2x minus a ok. Now what is d? This is nothing but your Kronecker delta that is one that means when x is equal to a this will go to 1 else will go to 0 ok. Now this is the now let us look at the integrals that we had what are the two integrals that we had in the previous case that those were minus infinity to plus infinity dt e to the power of i omega fi plus omega t plus minus infinity to plus infinity dt e to the power of i omega fi minus omega. Now you can look at these two integrals ok you can compare this integral with these two integrals they look very similar except the fact that the 2 pi is missing but 2 pi can always be multiplied and divided. So, if I do that then P f of t will become e not square ok. So, what I will do is I will multiply by pi square by h bar square omega fi by omega square modulus of integral dt 1 over 2 pi sorry I will multiply by 4 pi square ok 1 over 2 pi dt minus infinity to plus infinity e to the power of i omega fi plus omega plus 1 over 2 pi integral minus infinity to plus infinity dt e to the power of i omega fi minus omega t square f mu dot epsilon square ok. Now if you have this now if you will now look at this to the this can be written as this function can be written as del omega fi plus omega and this function can be written as del omega fi minus omega. So, your P of t will now become 4 e not square pi square by h bar square omega fi by omega square modulus of del omega fi plus omega plus del omega fi minus omega whole square f epsilon dot mu i whole square ok. So, this is your probability of transition from state i to state f. Now this involves 2 factors that are the Kronecker deltas omega fi plus omega and omega fi minus omega. Now let us consider a very simple scenario. The scenario is this ok this is your E i that will be nothing but with h omega i and this is E f this is what h bar omega f and this energy difference is delta E this is equal to h bar into omega f minus omega i this is equal to h bar omega fi that is the energy separation. So, omega fi this is nothing but the frequency of this energy separation ok. Now what is this? So, which means in this case if you look at these 2 if you have del omega fi minus omega plus omega and you have del omega fi minus omega del omega fi plus omega. Now remember we showed when you have del x minus a ok. So, this implies if x is equal to a then this function will go to 1 and x is not equal to a this function will go to 0. If I apply the same principle omega fi plus omega this means when omega fi is equal to minus omega this function will go to 1 when omega fi is not equal to minus omega this function will go to 0. Similarly, if you have del omega fi minus omega then omega fi is equal to omega this function will go to 1 and omega fi is not equal to omega this function will go to 0. Now there is one issue is that what is this condition? Omega is angular frequency how can angular frequency be negative? So, here I am looking at a condition omega fi is equal to minus omega ok which means this has to be negative and negative angular frequency does not exist ok. One way to look at the negative angular frequency is emission of light when positive. So, which means when you go from top to bottom that is E i to E f E f one can think of the frequency to be positive and when you go down the frequency. So, this will correspond to omega fi and this will correspond to minus omega fi ok. So, when the frequency negative simply means it is a case of stimulated emission. So, this corresponds to stimulated and this will correspond to absorption. Now let us go back to our probability that is nothing but f of t is equal to 4 pi square E naught square by h bar square omega fi pi omega square modulo of. Now there are two terms one corresponds to. So, there are two terms let me write down omega fi plus omega plus delta omega fi minus omega square integral f epsilon dot mu i whole square ok. Now you look at there are two terms this one and this. So, I told you this is nothing but emission and this is nothing but absorption. Now it turns out that it is not possible for omega fi to be positive of omega and omega fi to be negative of omega simultaneously ok. So, omega fi can equal to minus omega and omega fi is equal to omega is not simultaneous that means stimulated emission and absorption cannot happen simultaneously they have to happen one after the other ok. So, in that case if since they cannot happen simultaneously you cannot have the both the conditions you can have only one condition and if you are dealing with absorption then this condition will happen and when you are dealing with stimulated emission this condition will be valid. So, for absorption initial state i to a final state the probability of this transition is equal to 4 pi square epsilon naught square by h bar square omega fi by omega whole square del omega fi minus omega whole square integral f epsilon dot mu ok. So, this is the probability of transition for the absorption from a initial state i to a final state f ok. Now, there are several factors here one is the transient moment integral T mi and then there is this chronicle and then there is a ratio of omega fi to omega ok and many other factors are dependent, but this is the probability of transition and then if you look at this particular equation in the next lecture more carefully I am going to stop it here and thank you very much.