 Welcome to the lecture number 8 of the course quantum mechanics and molecular spectroscopy. As we usually do, we will take a quick recap of lecture number 7. The lecture number 7, I told you that we start with a n state perturbation theory, which simply means that you start with a Hamiltonian H naught and for a n state, n state psi n, you get value E n n. This is something that you know and your total wave function psi can be written as sum over n A n of t e to the power of minus i E n t by h bar n. And then you have a time dependent perturbation, which is nothing but h prime of t. So, that the total Hamiltonian will be equal to this and then you propagate using time dependent Schrodinger equation, which is nothing but i h bar t by dt is equal of psi equals to h of psi. Now, we use this to get to equation after some rearrangement and algebra that sigma over n i h bar c n dot t e to the power of minus i E n t by h bar n equals to sigma over n, sorry a n, we call it a n t naught c n t a n t a n of t e to the power of minus i E n t by h bar h prime of t ok. Now, what we did is after that we used multiplied with multiply with psi n star on the left. So, that is nothing but multiplied with an integrate. When you do this what you get is i h bar sum over n a n dot t e to the power of minus i E n t by h bar m this is equal to sum over n a n of t e to the power of minus i E n t by h bar h prime of t. So, that is what you get. So, what we had is i h bar sigma over n a n dot t e to the power of minus i E n t by h bar m n equals to sigma over n a n of t to the power of minus i E n t by h bar m h prime of t. And then we argued that since form complete set these are solutions of h naught they are orthogonal or orthonormal that means m n is equal to del m n that is nothing but is equal to equal to 1 when m is equal to n and is equal to 0 when m is not equal to n. Similarly, on the right hand side we said m h prime of t n is equal to 0 if m equal to n and not 0 or may not be equal to 0 if m is not equal to n ok. This is 0 if m is equal to n because we said that you cannot look at perturbation of the state onto itself or you cannot look at the transition from a state to itself ok. Therefore, these are known as self transitions ok, transitions of a state to itself and which cannot be observed. So, we take it to be 0. So, based on this this equation will turn out to be i h bar. Now, we can drop summation because the left hand side of the equation will only survive when m is equal to n or m is equal to m. In that case, we will get a m dot of t e to the power of minus i e m t by h bar ok and m dot m n this integral will go to 1 and this equal to in the right hand side all the terms survive other than when m is equal to n ok. So, that means, sum over n not equal to m a n of t e to the power of minus i e n t by h bar m prime t ok. So, we arrived at this equation towards the end of the last class. Now, we will continue this equation. So, what do you have? i h bar a m dot of t e to the power of minus i e m t by h bar equals to a n sum over n not equal to m a n of t e to the power of minus i e n t by h bar integral m h prime of t ok. Now, this is what is given. Now, we have one issue that we need to look at or we can just rearrange for before we look into it let us do some rearrangement ok. So, i h bar when if I want to take the other side this will become a m dot t equals to 1 over i h sigma over n not equal to m a n of t e to the power of minus i e n t by h bar into e to the power of i e m t by h bar m h prime of t. So, this is equal to now if I write this 1 over i h bar sum over n not equal to n a n of t e to the power of minus i e n minus e m t by h bar m. Now, I can write e n minus e m e n minus e m is equal to delta e is equal to h bar omega ok. So, if I write that this is equal to 1 over i h bar sigma n not equal to m a n of t e to the power of minus i omega n m because this h bar here and this h bar will get cancelled and you will get m h prime of t ok. Now, this is going to be the so, this is nothing but d by d. So, the left hand side nothing but d by dt of a m of t because I told you a m dot is nothing but its time derivative is equal to 1 over i h bar sigma over n not equal to m a n of t e to the power of minus i omega n m d prime ok. So, that is the integral that we have to. Now, we will see that the time dependence of a coefficient m will depend on the all the other coefficients a n's where n is not equal to n that means if you have a 3 let us take m is equal to 3 and there are total number of wave functions of 10 then time dependence of a 3 will depend on a 1, a 2, a 4, a 5, a 6, a 7, a 8, a 9, a 10 ok. So, it will depend on the time dependence of all the other rest of the coefficients. What we have is d by dt of a m of t is equal to i h bar 1 over sigma over n a n of t e to the power of minus i omega n m d m h prime of t. Now, I just want to find out a m of t. So, a m of t will be equal to 1 over h bar integral of sum over n a n of t e to the power of minus i omega n m d integral m ok. So, that is the integrated form of this coefficient and this is the most important. Now, to get to this equation I have not made any approximations because all the math that we did was completely rigorous. No approximation was made to get to this equation ok. Unfortunately, this equation involves couple differential equations of you know n n couple differential equations ok. So, if there are n total n coefficients each coefficient m coefficient mth coefficient will depend on rest of all the coefficients. So, we will get a series of couple different total if you have 10 such coefficients or 10 such wave function then you will get 10 couple differential equations ok. If you have 20 of them then you will get 20 different couple differential equations which are going to be very difficult to solve ok. So, then at this point of time let us make a small approximation and that approximation I will call till it as first order approximation. What do I do in the first order approximation? Ok, to begin with when there was no perturbation what did you have when there was no perturbation then we had H naught of n is equal to En. Now, let us suppose there is a initial state i ok. So, which means H naught of i equals to E i i ok. What is this initial state i? Let us state this initial state as a ground state or we can say this is H naught of ground state G is equal to E of ground state into ground state. Now, if you had a ground state all the population would be in the ground state and all the excited states will not be having any population. That means at t equals to 0 without perturbation only the initial state was populated. That means C i of 0 equals to 1 and this will corresponds state H naught i is equal to E i i and C of any other state let us say f of 0 is equal to 0 and this will corresponds to H naught of f is equal to E f of f where f is not equal to 1 ok is 0 that is the initial. So, only the ground state is populated or only the initial state is populated or could we could say initial as ground. So, C i of 0 is equal to 1 and C f of 0 is equal to 0 ok. Of course, here means all the other states other than the colonial states. So, if let us say counter number 1 is i initial state is counter number 1 then H 1 or C 1 is 1 and C 2, C 3, C 4, C 5, etc are 0s ok. If as in the case of you know harmonic oscillator sorry how is the case of harmonic oscillator 0 state is populated as in the case of Patkin's box only n is equal to 1 state is populated ok. Now and anything other than 1 that means states 2, 3, 4, etc they are not populated ok. So, that is the scenario when you switch on the which we do when you do not have the perturbation or without the perturbation. Now, what you want to do is that let us consider a weak perturbation limit. Now what does weak perturbation limit says if you have a weak perturbation limit we will assume that under such limit thus even when you switch on the perturbation the coefficients do not change too much ok. So, weak perturbations what happens is that the coefficients ok this is the approximation that I am using. So, we have using a weak perturbation limit and then we are saying that the coefficients do not change that means C i of t will still be equal to 1 and C f of t will be equal to 0 for f not equal to i ok. Now let us go back to our initial equation what was that I said C m of t equals to 1 over i h bar integral sum over n sorry a n not C a n a n of t to the power of minus i omega n m of now what I am going to say. So, a i is equal to a i of t is equal to 1 and a f of t is equal to 0 as long as or a n of t is equal to 0 if n is not equal to 1 in such an error or not then I can write a a f of t that is what you want to relate is equal to 1 over i h bar integral of all of them will go to 0 because the coefficients will 0 only one coefficient will survive that will be a i of t e to the power of minus i omega f i prime of t a of t this is equal to 1. So, MUL has no meaning in multiplying. So, we will get a f of t is equal to 1 over i h bar integral of course, we have to integrate over some time t t prime e to the power of minus i omega f i t and this will become f h prime this will become f and this will become i of t. So, that is my coefficient of f state and if I want to consider the probability of f state p of f of t then I will take modulus of a f of t whole square. So, this will become 1 over h bar square integral 0 to t e to the power of minus i omega f i t f h prime of t i. So, this is the probability of finding the fth wave function. So, that allows the transition from i state to the f state and the probability of such transition. So, we will stop it here.