 Hello, welcome to the lecture number 4 of the course, Quantum Mechanics and Molecular Spectroscopy. Before we get on with the lecture number 4, we will have a quick recap of lecture number 3. In lecture number 3, we looked at two important points. First is that an average value of A operator is given by psi A psi, this is nothing but integral psi star A operator. This is also equal to psi psi, this is equal to integral psi star A psi t tau where psi equals to small psi into e to the power of minus i E t by h power. And of course, it is only valid if you know psi is a stationary state. Secondly, we said that the time dependence of the average value of operator A is equal to ih bar average value of the commutator h comma A. It simply means that the average value of A or the expectation value of an operator A will have time dependence or not will depend on whether it will commute with the Hamiltonian or not. If it commutes with Hamiltonian, then there will be no time dependence and if it does not commute with Hamilton, it will show some time dependence. And I also told you that this is very useful in spectroscopy because the operator A corresponding to the interaction of light with the matter or white with molecule or atom does not commute with the Hamiltonian of the molecule or the atom. Therefore, one can make a transition from a one expectation value say 1s orbital or 1s energy level and hydrogen atom to some other energy level say 2s or 2p. There is another interesting thing that is if you have two operators f and g, the uncertainty relation of the expectation value of operators f and g is equal to is always going to be greater than half of f and g commutators. For example, we know that Heisenberg's under uncertainty principle says delta x delta px should be greater than equal to half h bar which means the operator x and operator px do not commute. In fact, this is not equal to 0. Now, so now we know the dependence of the, sorry now we know the time dependence of an average value of A will just depend on the that operator will commute with Hamiltonian or not. We will now go to the second part of our course that is the time dependent perturbation theory. To begin with I will show only time dependent perturbation theory of two states because it has some interesting consequences. So, we will start with time perturbation theory two states. Now, what does this mean? It means the following. For example, if you had a Hamiltonian H naught that means some Hamiltonian for which you know the solution and that acts on a wave function psi 1 that will give you E 1 psi 1 and this H naught acts on psi 2 will give you E 2 psi 2 and H naught has only two solutions that is so therefore psi 1 and psi 2 form a complete set. And what did I say about complete set? Any arbitrary function phi can now be written as a linear combination of A 1 psi 1 plus A 2. This is like a plane, okay. If you have a plane you only need two coordinates x 1 sorry x and y. So, any point in the plane is a linear combination of x and y, okay. So, we have a kind of a restricted environment where the Hamiltonian will be restricted to a plane that is consisting of psi 1 and psi 2, okay. Now, I am going to slightly rewrite this, okay. So, you have H naught in terms of bracket notation 1 is equal to E 1 1 and H naught is equal to E 2 where psi 1 is identically equal to 1 and psi 2 is identically equal to. So, if psi 1 and psi 2 form a complete set then there is another complete set that is nothing but I also told you that there is also another complex set that is made up of complex conjugates of 1 and 2. So, that is nothing but set. So, these two are complex conjugates of each other. So, now we have this problem in which H naught of 1 is equal to E 1 1 and H naught of equal to E 2. Now, let there be a some kind of perturbation. So, what will this perturbation do? It will move the states. So, let me write the Hamiltonian corresponding to the, when you have a perturbation the total Hamiltonian H will be nothing but H naught plus some perturbation H prime of T because we are looking at time dependent perturbation, okay. Now, let us suppose, okay, if there is a point or a ball kept on the surface of the surface of a floor and then one can think of playing a football. So, constantly someone is kicking it, okay because then the ball will keep moving all over the floor or if you go to a football match, okay, then there is this floor or the ground in which there is a football is kept and the ball and the players are kicking the ball in different directions and that is your time dependent perturbation. So, what will it happen? It will keep moving its position but it will but let us assume that it will always restricted to the ground, okay. So, but its position is going to change. So, every time you have a new position, it can be written as a linear combination of some new x and y. So, it has to be a linear combination of new 1 and 2 and this position is going to be time dependent. Therefore, the wave function psi can be now written as a 1 of t to 1 e to the power of minus i e 1 t by h bar or less a 2 of t to e to the power of minus i e 2 t by h bar. So, that is my wave function. That wave function works with this entire Hamiltonian. Of course, if h prime t goes to 0, then a 1 and a 2 will collapse to some value. In fact, either a 1 will go to 0 or a 2 will go to 0, okay. So, now we will go back to our Schrodinger equation which is nothing but i h bar d by dt of psi equals to h psi. So, finally what we want to do is we want to solve this Schrodinger equation with this wave function. So, let us do it. So, this will be i h bar t by dt of 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 to t by h bar. This should be equal to h bar sorry h into a 1 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 to t by h bar, okay. Now this is a long equation. So, what I am going to do is I am going to call this as LHS that is left hand side and I am going to call this as RHS. I will evaluate the left hand side and right hand side independently and then equate them later. So, let us start with LHS. LHS equals to i h bar d by dt of 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 to the power of minus i e to t by h bar. Now let us take the derivative. Now we will see that in each of these terms, okay, term 1 and term 1 and term 2, there are two time dependent parts a 1 t and e to the power of minus i e 1 t h bar. Similarly, a 2 t e to the power of minus e to t by h bar. So, you have to use a product toll of differentiation. What you get is i h bar a 1 t dot, okay, where a 1 t dot is equal to d a 1 t by dt. Similarly, a 2 dot of t is equal to d a 2 t by dt. It is just a notation, okay, just keep in mind. When I say put a dot over something, it means it is a time derivative, okay. e to the power of minus i e 1 t by h bar 1 plus a 1 t. Now a 1 t, now I will take the time derivative of the other one. This will be minus i e 1 by h bar e to the power of minus i e 1 t by h bar to 1. Similarly, you get a 2 dot t e to the power of minus i e to t by h bar 2 plus a 2 t minus i e to t by h bar into e to the power of minus i e to t by h bar to 2. So, this I can rewrite as is equal to i h bar a 1 dot t e to the power of minus i e 1 t by h bar 1 plus a 2 dot t e to the power of minus i e to t by h bar 2. Now, you can see there is an i here and there is an i here, i square that will be minus 1 and there is a minus sign as well. So, that will become plus 1. There is a h bar in the numerator and there is an h bar in the denominator. So, these two get cancelled. So, essentially what to get is that plus e 1 into a 1 t e to the power of minus i e 1 t by h bar 1 plus e 2 to a 2 t e to the power of minus i 2 t by h bar. So, there are four terms. I will call this as L 1 that is LHS 1, L2, LHS 2, L3, LHS 3, L4, LHS 4. Now, let us look at the RHS. This is nothing but your h acting on your wave function that is nothing but a 1 of t e to the power of minus i e 1 t by h bar 1 plus a 2 t e to the power of minus i e to t by h bar 2. I want to quickly go back and see that there are four terms. The first two terms are the time derivatives with respect to a 1 and a 2 and the second two terms have this energy values e 1 and e 2. So, something that you must remember. Now, this is equal to nothing, this is nothing but is equal to your h is nothing but h naught plus h prime t. Now, this must act on a 1 of t e to the power of minus i e 1 t by h bar 1 plus a 2 t e to the power of minus i e 1 t by h bar 2. So, this will be nothing but a 1 of t e to the power of minus i e 1 t by h bar h naught will only act on 1. So, h naught acting on 1 plus a 2 times t e to the power of minus i e 2 t by h bar h naught acting on 2. You see because h naught is your initial Hamiltonian which is nothing but your time independent Hamiltonian. So, it will not act on anything that is time dependent. So, either a 1 t or e to the power of minus i e 1 t by h bar it will only act on the state a 1 and give you e 1 1. Similarly, the second part of the equation plus a 1 t h prime t now acting on e to the power of minus i e 1 t by h bar acting on 1 plus a 2 times t h prime t acting on e to the power of minus i e 2 t by h bar. So, these are the 4 terms that you will get now I will act. Now, when h naught acts on 1 you will give me e 1 1. So, this will become e 1 e 1 of t e to the power of minus i e 1 t by h bar 1 plus second one h naught acting on 2 will give e 2 t. So, e 2 a 2 of t e to the power of minus i e 2 t by h bar 2 plus the 2 terms that is a naught a 1 t h prime of t acting on e to the power of minus i e 1 t by h bar 1 plus a 2 of t h prime of t acting on e to the power of minus i e 2 t by h bar. So, here also we get 4 terms this I will call it as right hand side 1 R 1, right hand side term 2 R 2, right hand side term R 3, right hand side term 4 R 4. So, but you know LHS and RHS must be equal. So, LHS must be equal to RHS. Now, you will quickly see that L 3 term is equal to R 1 term and L 4 term is equal to R 2 term. So, this one and this one here R 1, R 2 will be equal to L 3, L 3 and L 4. So, since L 3 and L 4 are L 3 and L 4 are equal to R 1 plus R 2. So, which means you can cancel out each other. So, what you will be left out with is i h bar e to the power of minus i e 1 t by h bar 1 plus a 2 dot t e to the power of minus i e 2 t by h bar is equal to a 1 t h prime t acting on e to the power of minus i e 1 t by h bar 1 plus a 2 t h prime t acting on e to the power of minus i e 2 t by h bar. So, this is what you get from that equation. Now, let us do some little bit of math. Now, what I am going to do is the following. I am going to multiply psi 1 star on the left. This simply means that I will put a function like that and multiply and integrate. Now, if I do that I will get i h bar 1 a 1 dot t e to the power of minus i e 1 t by h bar 1 plus 1 a 2 dot t e to the power of minus i e 2 t by h bar this is equal to 1 a 1 t h prime t e to the power of minus i e 1 t by h bar 1 plus 1 a 2 t h prime t e to the power of minus i e 2 t by h bar 2. Okay. I will have four integrals that I have to evaluate. If I can evaluate these four integrals, then I will know the time dependence of this perturbation. We will carry on the remaining part in the next lecture. I am going to stop now. Thank you.