 number 5 of the course quantum mechanics and molecular spectroscopy. In the previous lecture we were looking at the time dependent perturbation theory of two states. If we had the time independent Hamiltonian H naught and had two solutions E 2 2 such that 1 and 2 form a complete set and so which also means form a orthonormal complete set. That means integral 1 overlap integral of 1 over 2 will be equal to 0 and overlap integral 1 over 1 equals to 2 over 2 equals to 1, okay. Now in the last class we had ended up with this equation 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 equals to h prime of t e to the power of minus i e 1 t by h bar a 1 1 a 1 of t 1 plus h prime of t e to the power of minus i e 2 t by h bar a 2 of t. So, this is where we stopped in the last lecture, okay and this we came from LHS and this came from the RHS. Now we have to equate this and try to get going to what happens. Now I am going to do one simple trick that is I am going to multiply with psi 1 star on left and integrate. So, we simply means that I will carry out the operation 1. So, when I look at the LHS then it will become 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 to t by h bar 2 that will be your left hand side and the right hand side will be 1 h prime of t e to the power of minus i e 1 t by h bar a 1 of t 1 plus 1 h prime of t e to the power of minus i e to t by h bar a 2 of t. So, these are the four terms that we have. Now if you look at the left hand side then what we have is i h bar fun a 1 dot t e to the power of minus i e 1 t by h bar 1. I will come to second term little bit first let us just you know look at this term. What is a 1 dot t? a 1 dot t is nothing but d by dt of a 1. What is a 1 of t? It is a time dependent a 1 of t is nevertheless even if it is time dependent it is a constant it is going to change but it is still a constant. But you know 1 this wave function 1 is a solution of the time independent Schrodinger equation. So, what was 1? 1 is nothing but h naught 1 is equal to e 1 1. So, the operator in here is time dependent and the wave function is time independent. Therefore, one can write you can bring the operator outside because it is not going to affect the wave function. So, what we will have left hand side we will have i h bar a 1 dot t e to the power of minus i e 1 t by h bar 1 1. And the second term which I have not written here but you know by analogy one can write it as a 2 dot t e to the power of minus i e to t by h bar 1 2. So, that will be your left hand side. Now, let us look at the right hand side. So, let l h s is equal to I will separate them and then equate it later. So, r h s is equal to what you had in the r h s? R h s you had h prime 1 h prime of t e to the power of minus i e 1 t by h bar a 1 of t 1. So, that is what you had one of the terms. Now, you can see that e to the power of minus i e to the power of minus i e 1 t h bar and a 1 of t these are just you know time dependent phase factor and coefficient which you can bring it out. But I cannot bring out h prime of t because it is a time depend perturbation a perturbation always moves the states. So, it will affect your wave functions a perturbation really affects a wave function. And if you have time dependent perturbation its effect will be different in different times. Therefore, one can write this or the other term we had was 1 h prime of t e to the power of minus i e to t by h bar a 2 of t. So, that was your r h s. Now, this I can slightly rewrite because of the arguments that I use will be equal to a 1 of t e to the power of minus i e 1 t by h bar 1 h prime of t 1 plus a 2 of t e to the power of minus i e to t by h bar 1 h prime of t. So, that is your r h s. Now, let us equate r h s and let us see and what we get. So, 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 1 2 should be equal to a 1 of t e to the power of minus i e 1 t by h bar 1 h prime t 2 plus a 2 t e to the power of minus i e to t by h bar 1 h prime t that is what we will get. Now, let us look at this. Now, we know the equations or the wave sorry now we know the wave functions 1 and 2 are orthonormal which means 1 1 integral overlap integral will go to 1 and 1 2 overlap integral will go to 0. That means, on the left hand side only one term will survive. So, what you get is i h bar a 1 dot t e to the power of minus i e 1 t by h bar that 1 1 is just 1 should be equal to a 1 of t e to the power of minus i e 1 t by h bar 1 h prime t 1 plus a 2 of t e to the power of minus i e to t by h bar 1 h prime t. Now, let us suppose you have two solutions this is your 1 with e 1 as the energy and this is 2 with e 2 as energy. Now, I apply so let us suppose there is a molecule which is sitting here. I apply the perturbation and what happens is that the molecule after applied perturbation still is in 1. So, let us say after perturbation I will get other molecule which is this, but it still is in the e 1 state. That means, whether you apply perturbation or not apply perturbation you would not be able to differentiate. So, if I start with state 1 and end up in state 1 you do not even know whether you have started and ended up not. Therefore, any perturbation which leads on to the same wave function e can be neglected ok. It is like transition to the same states you start from ground state and you go to back to the ground state. So, you do not even know whether the transition state has taken place or not taken place. Therefore, one can equate this term to be 0 because perturbation here acts on state 1 and overlaps with the state 1. That means, you have started with state 1 and ended with state 1 you do not even know whether the perturbation has taken place or not ok. So, effectively the first term will go to 0 ok. So, this is nothing but transitions or perturbations to the same state do not count ok. Now, in such scenario then your first term is gone. So, what you get is i h bar a 1 dot t e to the power of minus i e 1 t by h bar equals to a 2 of t e to the power of minus i e 2 t by h bar 1 h prime t ok. So, that is one equation that we will get ok. Now, let us this do the 6 let us go back to the first equation that I wrote in the beginning of the lecture that is nothing but 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 2 t by h bar 2 should be equal to a 1 t h prime of t e to the power of minus e 1 t by h bar 1 plus a 2 t h prime of t e to the power of minus e to t by h bar 2. So, that was the equation that we started with multiply with psi 2 star and integrate ok this is equivalent of you know ok. Now, remember last time we multiplied with psi 1 star and integrated now we are multiplying with psi 2 star and integrated. So, quickly I will write skip couple of steps because you already know how we did with psi 1 star. So, what we get is this i h bar a 1 dot t e to the power of minus i e 1 t by h bar 2 1 plus a 2 dot t e to the power of minus i e to t by h bar 2 2 overlap integral should be equal to a 1 of t e to the power of minus i e 1 t by h bar 2 h prime of t 1 plus a 2 dot a 2 not a 2 dot t a 2 t e to the power of minus i e to t by h bar 2 h prime t ok. Now, we use the same analogy as last time. So, 2 1 overlap integral will go to 0 2 2 1 will be 1. So, this integral we do not know we have to evaluate, but once again transitions from 2 to 2 will be equal to 0. So, this will go to 0 ok. So, after this we can rewrite this equation as i h bar a 2 dot t e to the power of minus i e to t by h bar should be equal to a 1 t e to the power of minus i e 1 t by h bar 2 h prime of t ok, this is the second equation ok. Now, what I will do is I will correct both the equations together. So, first equation was i h bar a 1 dot t e to the power of minus i e 1 t by h bar should be equal to a 2 of t e to the power of minus i e to t by h bar 1 h prime of t 2 and i h bar a 2 dot t e to the power of minus i e to the power of minus i e 1 t h bar 2 h prime t ok. Now, there is something else that I can do is I will take the i h bar slightly rearrange these 2 equations ok. So, a 1 dot t equals to the i h bar I can take on the other side. So, i becomes 1 over i h bar e to the power of e 1 t i e 1 t h bar I will by h bar I will take the other side. So, that will become e to the power of minus i e 2 minus e 1 by h bar 1 h prime of t and the other equation will become a 2 dot t is equal to 1 over i h bar e to the power of ok. Now, this is e to minus. So, that will become i e 2 minus e 1 by h bar 2 h prime. Now, let us say e 1 e 2 minus e 1 is equal to delta e this is equal to h bar omega 2 1 ok omega 2 1 will be the angular frequency thus corresponds the energy difference between the e 2 and e 1 states. So, therefore, now you can see when I replace e 2 minus e 1 as h h bar omega 2 1 this h bar in the numerator and this h bar in the numerator will get cancelled. So, what you finally end up with the following equations a 1 dot t equals to 1 over i h bar a 2 t e to the power of minus i omega 2 1 t 1 h prime of t and a 2 dot t is equal to 1 over i h bar e to the power of a 1 of t a 2 power of i omega 2 1 t 2 h prime of t ok. So, these are so which means the time dependence of a 1 will depend on a 2 and time dependence of a 2 will depend on a 1 that means these two equations are couple differential equations ok. Now more importantly one thing that you can look at is the following a 1 changes ok with respect to a 2 and a 2 changes with respect to a 1. Now there is one thing that is very interesting is this here this is e to the power of minus i omega t and this e to the power of plus i omega t. So, which means these two are phased out what it means it means when a 1 goes up a 2 comes down and a 2 goes up a 1 comes down. So, these are phased out or out of phase with respect to each other. So, coefficients a 1 t and a 2 t are out of phase with respect to each other ok. So, what is a 1 and t psi of x comma t is equal to 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 2. So, this was our total wave function ok and this is the coefficients time different coefficients of the a 1 and a 2 and we know the square of the coefficients gives you the probability therefore, ok. So, when you see that a 1 and a 2 are out of phase with respect to each other that means the probability of finding a 1 state if it goes up the probability of finding a 2 state will go down. Similarly, if the probability of a 2 state will go up and the probability of a 1 state will go on. So, they are going to be with respect to each other phased out or you know out of phase will stop here and continue in the next lecture. Thank you. .