 Welcome to the lecture number 7 of the course quantum mechanics and molecular spectroscopy. Before we proceed with lecture number 7, in the previous 2 lectures, we looked at the perturbation theory of 2 states. So, to begin with we had H naught acting on 1 will give E 1 1 and H naught acting on 2 will give you E 2 2 ok. Then we had perturbation time dependent which was H prime of T and this when added to H naught will give you the total Hamiltonian H and then we went about by solving the Schrodinger equation I H bar d by dt of psi of x comma t is equal to H to psi of x comma t, where psi of x comma t is given as A 1 of t e to the power of minus i E 1 t by H bar 1 plus A 2 of t e to the power of minus i E 2 t H bar. So, after doing all the required mathematics, we arrived at A 1 dot of t is equal to 1 over i H bar A 2 of t e to the power of minus i omega 2 1 t 1 H prime of t A 2 dot of t is equal to 1 over i H prime A 1 of t e to the power of i omega 2 1 t 2 H prime of t 1 ok. So, these two are the couple differential equations between A 1 and A 2 ok. Now, I told you since they are couple differential equation solution of A 1 will depend on A 2 and solution of A 2 will depend on A 1. Now, we used a very simple approximation in which we said there is a perturbation a constant perturbation that is switched on at time t is equal to 0 ok. So, that perturbation is something like this t when it is equal to 0 it. So, some constant value in such scenario then we can written as 1 H prime of t 2 is equal to H bar v and 2 H prime of t 1 is equal to H bar v star theta. Using these conditions and doing little more math we came about the we converted the equations into second order derivatives and finally, came up with a solution that is A 1 of t equals to A e to the power of i omega t minus v e to the power of minus i omega t into e to the power of omega 2 1 t by 2 where omega is given as half of omega 2 1 square less 4 times modulus of v square to the power of half and A 2 of t is equal to A e to the power of i omega t less v e to the power of minus i omega t e to the power of omega 2 1 t by 2. Now, we said A 1 2 and A 2 t A 1 t and A 2 t are phase shifted with respect to each other by pi by 2 because this is like a sin theta function and this is like a cos theta function. So, which means whenever A 1 t increases A 2 t decreases they are out of phase with respect to each other. We know function sin theta and cos theta are out of phase with respect to pi out of phase by pi by 2. So, whenever sin theta goes up cos theta comes down and other way around. So, based on this we came up of what is known as Rabi oscillations. So, Rabi oscillations are between two states even and e to where the population of states 1 and 2 will go up and down out of phase with respect to each other and this is the perturbation theory of the two states. Now, let us start with the perturbation theory of n states generalize the perturbation theory generalize time dependent perturbation theory is what we are going to start. In such scenario we will again start with an assumption that we know an Hamiltonian H naught for which there are n solutions such that this q u e n psi n and n will go from for example, 1, 2, 3, etc. So, where this represents the quantum number. Now, we know that if psi n forms a complete set. So, any arbitrary function chi in the same variable can be written as sum over i c i psi i. This is something we already know. So, a wave function psi can always be written as sum over n c n of t psi n e to the power of minus i e n t by h bar. So, a time dependent wave function is a linear combination of the all the wave functions psi h and their phase factors and their time dependent coefficients. Now, if we have the time dependent perturbation which I will call it as h prime of t, this is my then my whole Hamiltonian H will be nothing but h naught plus h prime. So, when I know the total wave functions psi and h, I can solve the or attempt to solve the time dependent Schrodinger equation which is nothing but i h bar d by dt of psi is equal to h psi. So, we will do the same treatment that we did for the two states, but in a slightly different manner. Now, let us get to it. So, my time dependent perturbation theory is i h bar t by dt of psi is equal to h psi. Now, what is psi? This is nothing but i h bar t by dt of my psi is a n sigma over a n of t e to the power of minus i e n t by h bar n that is the wave function psi n. You can go and look it back. So, my n is this psi is this. So, I am just I have just used this form equals to equals to h is nothing but h naught plus h prime of t this acting on sigma over a n t e to the power of minus i e n t by h bar n. Now, once again here we have two sides LHS this is equal to LHS this is equal to RHS. So, what we will do is we will evaluate LHS and RHS separately and then equate them at a later stage. So, let us start LHS. LHS equals to i h bar d by dt of psi this is nothing but sigma over n t e to the power of minus i e n t by h bar n. So, this is nothing but i h bar. Now, each term will have two things that is a n of t and e to the power of. So, this is function f of t and this is function g of t and of course, the wave function n is itself time independent. So, it will not be able to take a time derivative of it it is just a constant. So, each term. So, this is summation over n. So, you will have as many terms in this as the number of wave functions in your complete set. So, if there are 20 wave functions in the complete set. So, this expansion will be for 20 terms and if there are 100 wave functions in a complete set, then this expansion will be of 100 terms. If there are infinite number of wave function then this expansion will be of infinite number of terms. Now, in such a scenario, each term of course, will be of two functions. So, when you take a derivative use the product rule and so what you get is i h bar. Now, I will take the derivative with respect to n. So, sum over n a n dot of t e to the power of minus i e n t by h bar n sigma over n i e n by h bar a n of t e to the power of minus i e n t by h bar acting on n something like that. So, I am just going to rewrite it in a way is equal to i h bar sigma over n a n dot t e to the power of minus i e n t by h bar n and here this i and this i will become minus 1 and there is already minus n it will become plus. So, plus and this h bar and this h bar will cancel. So, what we will get is a n of t e to the power of sigma over n n minus i e n t by h bar e n. So, you will get two terms. But of course, each of these terms is summation of many, many terms depending on the value of n. So, you remember this my LHS is this. Now, let us look at the RHS. What is the RHS? RHS equals to h naught plus h prime of t acting on sigma over n a n of t e to the power of minus i e n t by h bar n. So, this is nothing but when h naught acts h naught is time independent. This is just not kinetic energy and potential energy operators of the original Hamiltonian and we know in chemistry we said that the original Hamiltonian that is is independent of time. So, h naught when acts on this. So, this will not act on a n or e to the power of minus i n t. So, it just acts on n and gives you sigma over n a n of t e to the power of minus i e n t by h bar when h naught acts on n we will get e n that is a first and the second term will be plus a n of t e to the power of minus i e n t by h bar and h prime of t will act on. So, that is the two values of course, there is sum over n here. Now, let us equate the LHS and RHS. Now, my LHS was. So, we have to now make LHS equals to RHS. So, LHS was sigma over n a n dot t i h bar e to the power of minus i e n t by h bar n that is my LHS plus sigma over n a n of t e to the power of minus i e n t by h bar e n that should be equal to my RHS which is nothing but sigma over n a n of t e to the power of minus i e n t by h bar e n n plus sigma over n a n of t e to the power of minus i e n t by h bar h prime of t acting on n. Now, we will see that the second term of the LHS is equal to the first term of the RHS. So, we can cancel these two. So, what we end up in this case is i h bar sigma over n a n dot t e to the power of minus i e n t by h bar n equals to sigma over n a n of t e to the power of minus i e n t by h bar h prime of t. So, this is what you get. You can go back and quickly look at the two state perturbation theory. This equation looks very similar when n is equal to just two state when n is equal to 1 comma 2 then this equation is exactly equal to what we got for the two state perturbation theory. However, in this case since we are starting for n states or many states a generalized form can be written like this. Now, what I am going to do is that I will start with the equation i h bar sigma over n a n dot of t e to the power of minus i e n t by h bar to n is equal to sigma over n a n of t e to the power of minus i e n t by h bar h prime t. Now, let us quickly do the multiply with psi m star on the left. So, if I do that I will get i h bar sigma over n a n dot t e to the power of minus i e n t by h bar m of course, psi m star equals to this is equal to not to n a n of t e to the power of i e n t by h bar h prime of t. So, this is my this now we know that the wave functions psi n is nothing, but they form complete set they are the solutions of the Hamiltonian h dot. Therefore, the integral m n is equal to delta m n that means, when m is equal to n this integral the overlap integral will go to 1 otherwise it will go to 0. Now, if you plug in that so, which means for anything other than value of n is equal to m. So, left hand side if you take left hand side for anything for n is equal to m this term will survive LHS for n is equal to m LHS will survive else it will go to 0. So, which means that H s will now be equal to sigma over n i h bar a n dot t e to the power of minus i e n t by h bar m. So, only when m is equal to n or n is equal to m this will survive that means, you can drop the summation this will become i h bar a m dot t e to the power of minus i e m t by h bar because for n is equal to m this will go to m n will go to 1 and m is not equal to n this integral m n will go to 0. That means, only one survive term will survive all the rest of the terms will go to 0. Now, if you look at the RHS this will be equal to sigma over n a n of t e to the power of minus i e n t by h bar m h prime t n. Now, if you take the integral m h prime t n this integral will be 0 if m is equal to m. Now, you can see that this is going to be because what you are looking at is you are looking at the transients onto itself. You are looking at the transients of itself when m is equal to n you know that we cannot look at transients of from state to from a given state to itself. That means, only the terms m h prime t n will survive if m is not equal to n. Now, which means I can rewrite this equation as i h bar a m dot t e to the power of minus i e m t by h bar should be equal to sum over n not equal to m. So, now you can take all these terms except that when m is not equal to n. So, when m is equal to n of course, I told you this will not survive m not equal to n a n of t e to the power of minus i e n t by h bar. So, this is the equation that we need to deal with. We will stop here for this lecture and continue in the next lecture. Thank you.