 Welcome to the lecture 9 of quantum mechanics and molecular spectroscopy course. Before we proceed with this lecture, let us have quick recap of the previous lecture. In the previous lecture, we started with the time dependent perturbation theory of many states. So, we had this time independent Hamiltonian H naught, solutions of which were H psi n is equal to En psi n, where n is equal to for example, 1, 2, 3, etc. It is like a particle in a box or something like that for which we already know the solutions. And we know in such a scenario, the function psi n form a complete set. Now, if they form a complete set, then I told you any arbitrary function phi can be written as a linear combination of these functions. Furthermore, since these are stationary states because the Hamiltonian itself is time independent, we can always write psi of n is equal to psi n e to the power of minus i En t by apart from that, the total wave function can always be written as psi as a sum over n An e to the power of minus i En t by h bar multiplied by psi n. Of course, if you have time dependent coefficients, that means the wave function itself is moving in time, then you can always write psi as sigma over n An of t e to the power of minus i En t by h bar to psi n. And I also told you this psi n can be written also as n, this is an equivalent description of it. Then under this condition, we solve the time dependent Schrodinger equation with acting of a perturbation. So, i h bar t by dt of psi equals to psi and in this case h is equal to h naught plus h prime of t. Now, if you have that then you can expand your L h s will be equal to i h bar t by dt of psi which is equal to i h bar t by dt of sigma over n An of t e to the power of minus i En t by h bar and r h s is equal to h naught plus h prime of t psi. This is equal to acting on psi. So, this is equal to h naught plus h prime of t acting on sigma over n An of t e to the power of minus i En t by h bar. So, after expanding this that we looked at in the last class, we come up with the expression An dot t by h bar e to the power of minus i En t by h bar n equals to An of t sigma over n An of t sigma over n h prime of t acting on e to the power of minus i En t by h bar. Now, what you can do is multiply with psi n star on the left in such scenario what you get is i m h bar sigma over n m An dot t e to the power of minus i En t by h bar n is equal to sigma over n An of t h prime t e to the power of minus i En t by h bar. So, this we can rewrite as i h bar sigma over n An dot t e to the power of minus i En t by h bar m n this is equal to n time sigma over n An dot t e to the power of minus i En t by h bar. Now, we know m n is equal to delta m n m h bar t of n is equal to 0 for m is equal to n and not equal to 0 perhaps for m not equal to n. In this case, this is equal to 0 if m not equal to n and is equal to 1 if m is equal to n. So, on the left hand side only one term will survive where m is equal to n. So, what we will get is i h bar dot t e to the power of minus i En t by h bar this is equal to sigma over n not equal to m An of t e to the power of minus i En t by h bar m h prime t. So, this is the important difference. So, there is only one term on the left hand side and there are all the terms n terms on the right hand side except n is equal to m. Now, I can slightly rewrite it as a m dot t equals to if I take i on the other side will become minus i. So, minus i by h bar and e to the power of minus e m t by h bar will become other side will be e to the power plus i En t by h bar. So, this will become sigma over n not equal to m An of t e to the power of minus i En minus e m of t by h bar h prime of t. Now, you can say En minus e m is equal to delta m n m this is nothing but h bar omega n m. So, this will come to be minus i by h bar sigma n not equal to n An of t e to the power of minus i omega n m t m by m this is a m dot t. So, for each mth coefficient the time dependence will depend on the time dependence on the coefficient rest of the coefficients. So, in this case we will have n couple differential equations these have to be solved and you know it is going to be very difficult to solve n couple differential equations. Therefore, we play one and we use something called first order perturbation theory. Let us suppose a system or a molecule atom or whatever is it is in ground state and that I will call it as an i or initial state. So, let us say a of i at time t is equal to 0 is equal to 1 and a of n not equal to i at 0 is equal to 0 or let us say a of f not equal to i is equal of 0 will be equal to 0. So, that means only population in the ground state and others are not populated at all. Now, in the first order perturbation theory we will assume that at time t at time t at later time t when the perturbation has acted and the perturbation is so small that the coefficient do not deviate much. So, what we say is that under weak perturbation limit coefficients deviate from original values. If such is the case then am dot t equals to i by h bar minus sigma over n not equal to m a n of t e to the power of minus i omega n m t m h bar t. Now, I said that for all other coefficients are going to be zeros except the coefficient of i. So, that means a i of a of t is equal to 1 a n not equal to i is equal to 0 of t. So, in such scenario, so if I start from a value of i and go to f of t will be equal to minus i h bar because only one will survive that is a i a i e to the power of minus i omega i f. So, that is the what it is now if this is d by. So, this is nothing but d by dt of a f of t is equal to minus i h bar a i of t rather e to the power of minus i omega f t i. So, if I integrate this a f of t is equal to minus i h bar integral a i of t e to the power of minus i omega i of t f h acting for some time 0 to t prime. So, this is the coefficient that we need to evaluate. Now, in this equation, so a f of t is equal to minus i h bar integral 0 to t prime a i of e to the power of minus i omega i f t f h prime of t. Okay, something that I kind of understood, but this is equal to 1 this will become minus i h bar integral 0 to t prime e to the power of minus i omega i of t, but this is also an integral okay integral psi f star h prime of t psi i d tau. So, it is there is a two integral there is one integral over space this d tau and other is over dt okay. Now, very importantly for this a f of t to not to vanish or become 0 this integral should not be 0 that is one condition if this integral becomes 0 or f h prime of t i if this should not be equal to 0 this integral if this integral becomes 0 of course, the whole a f of t will become 0. Okay, now if this does not have to become 0 which means if you start with some initial state i is a ground state or initial state okay and f is your final state then your h prime t operator that is the time dependent operator or time dependent perturbation operator must be able to project your initial state onto a final state okay. So, what does it mean now if you look at this integral okay this is nothing but integral psi f star h prime of t psi i d tau. Now when h prime acts on psi f it will result in some other function okay let us call it as integral psi f star if h prime t acts on psi i you get to function phi and d tau. Now this is nothing but psi f phi that is nothing but an overlap integral. So, what does it mean it means when h prime t or the perturbation acts on your ground state it should create a new function which will have overlap with your final function. So, overlap is nothing but the projection so that overlap is nothing but whether when one function is projected onto other they have some commonality okay. So, in this case when h prime t acts on your ground state the resulting function must be able to project onto your final state only then this integral will become non-zero okay. So, that is the most important point in the spectroscopy that the perturbation operator should be able to take your initial function and project onto the final function. Otherwise if let us suppose if this is just a operator 1 operator 1 or identity operator then what you get is that f will project onto i because we know f and i are the solutions of the time independent Schrodinger equation they will be orthogonal and they will go to 0. So, that means an identity operator will never be able to cause transitions from initial state i to final state f instead of that you will need something more than that some time dependent operator that will allow that projection of your initial state onto the final state okay. Now that is only the coefficient but in quantum mechanics probability is equal to square of the coefficient okay. So, p of t f is equal to 1 over h bar square modulus of integral e to the power of minus i 0 to t prime omega i f t into f prime of t. So, that is going to be a probability and this probability must be not equal to 0 and more important is this integral must not be 0. If this integral goes to 0 then p of t will go to 0 that will be no probability of having a transition. So, you can now quickly realize the fact that this integral governs the selection rules because whether it is going to be 0 or non-zero in spectroscopy means the selection rules okay and h prime t will have some form which we have not discussed but we will come to that in the next lecture or the lectures following that whatever is the form of h prime t it should make sure that this integral will not go to 0 and thus will correspond to the selection rules okay. We will stop here for this lecture and continue in the next lecture.