 Welcome to the lecture number 6 of the course quantum mechanics and molecular spectroscopy. We will have a quick recap of contents of lecture number 5. In the lecture number 5, we looked at the perturbation theory of 2 states in which if we had a Hamiltonian H naught such that it acted on state 1 and produced E11 and H naught acting on state 2 gives you E22 and the states 1 and 2 form complete set and they are orthonormal. Orthonormal simply means integral of 1 over 2 is equal to 0 and integral of 1 over 1 should be equal to 2 over 2 is equal to 1, okay. Now we said that there is a time dependent perturbation that acts on states 1 and 2 or this system which is nothing but H prime of t and this when added to H naught will give the total Hamiltonian H and it will follow the time dependent Schrodinger equation. In this case the psi is a superposition state one can think of psi is equal to a1 of t e to the power of minus i e1 t by h bar 1 plus a2 of t e to the power of minus i e to t by h bar, okay. So this is the wave function psi. Now by using by plugging the psi in the Schrodinger equation we ended up with following equation after doing some amount of algebra that is nothing but i h bar a1 dot t e to the power of minus i e1 t by h bar should be equal to a2 of t e to the power of minus i e to the power of minus i minus e to t by h bar integral over 1 h prime of t. Similarly, we had i h bar a2 dot t e to the power of minus i e to t by h bar is equal to a1 of t e to the power of minus e1 t by h bar 2 h prime of t, okay. So these are the equations that we ended up with the last lecture. Now let us start looking at this equation little bit more carefully i h bar a1 dot t e to the power of minus i e1 t by h bar equals to a2 t e to the power of minus i a2 t by h bar 1 h prime of t, okay. Now of course this is a definite integral. This definite integral will give you some number, okay. It could be 0 but definitely it could be a number. If this is some number and a2 t is a coefficient and e to the power of minus i this is the phase factor. So all this tells you that the time dependence of a by the way a1 dot t is nothing but t by dt of a1 of t. So the time dependence of a1 coefficient will depend on the a2 coefficient, okay. One can rewrite this little bit in a different way. So a1 dot t is equal to so one can take i h bar on the other side i h bar into a2 of t e to the power of now you can take this exponential function on the other side. So what we will get is minus i e to minus e1 into t by h bar and the integral 1 h prime of t. Of course if you can write e to minus e1 is equal to delta e to 1. So this is equal to h bar omega 2. So if I write that way then I get a1 dot t is equal to 1 over i h bar a2 of t e to the power of minus i omega 2 1 t because this h bar and that h bar will get cancelled, okay into 1 h prime t. Now you can also rewrite this same equation starting from the other equation for a2 dot. So that will come out to be a2 dot of t will be equal to 1 over i h bar a1 of t e to the power of i omega 2 1 t sorry 2. Now we will quickly realize if h prime of t is Hermitian then the integrals 1 h prime t 2 should be equal to 2 h prime t 1. So this is only true when h prime t is Hermitian then you will see that this one and this two will be equal or if they are not even if this h is not Hermitian then this will be complex conjugates of each other. Now we will see that this is if you look at the constants this is i h bar is a constant okay. Now these are some constants because they are definite integrals okay will have either complex conjugate of each other or they are equal to each other. Now all you can see that a1 dot dependence on a2 is with this function e to the power of minus i omega 2 1 while as a2 dot dependence will be on e to the power of i omega 2 1. So these two functions are out of phase with respect to each other is simply something like this okay so that will be our a2. So whenever a2 will go up a1 will come down and whenever a2 will come down a1 will go up okay. So that is the time dependence of the two or a1 and a2 which are out of phase a1 of t and a2 of t are out of phase with respect to each other okay. Now if such is the canary so we need to solve this couple differential equations which are nothing but d a1 of t by dt is equal to a1 a2 of t e to the power of minus i omega 2 1 t a1 h prime t 2 and t a2 t by dt equals to a1 of t e to the power of i omega 2 1 t okay. So these two are couple differential equations okay and such that a1 of t a2 of t are respect to each other okay. Now of course it is kind of difficult to solve but we will make an attempt now let us assume there are two states 1 and 2 with energies e1 and e2 okay. Now a time t is equal to 0 I am going to switch on a constant a perturbation okay and it is constant. So what happens is that a time net if I think of perturbation so I will switch on a time t is equal to 0 this is time is equal to 0 I will switch on a constant perturbation for some time okay until this goes to say t prime okay and I will have some value let us recall it as okay. So this is now in a such a way that this 1 h prime of t 2 becomes h bar v and 2 h prime of t 1 becomes h bar v star okay where v and v star are complex conjugates with respect to each other. In fact if you look at these integrals these integrals are complex conjugates of each other. Now if such is the scenario then your a1 dot t will be equal to 1 over ih bar a2 of t e to the power of minus i omega 21 t and you had this integral 1 h prime 2 h bar t 2 okay and other integral was a2 dot t is equal to 1 by ih bar a1 of t e to the power of i omega 21 t 2 h bar h prime t. Now this integral of course is equal to h bar v and this integral equals to h bar v star okay then what happens I can rewrite these equations as the following a1 dot t is equal to this h bar and this h bar will cancel i in the numerator will go to sorry in the denominator will go to numerator as minus i. So minus i v a2 of t e to the power of minus i omega 21 t that will be a1 dot and your a2 dot t will be equal to minus i v star a1 of t e to the power of i omega 21 t. So I am going to rewrite them again so what you had is a1 dot t equals to minus i v a2 of t e to the power of minus i omega 21 t okay and a2 dot t is equal to minus i v star a2 of a1 of t e to the power of i omega 21 t. So what I am going to do is I am going to rewrite this equation as slightly different way so what I will do is this I will make a I will write in terms of a1 and a2. So one can rewrite this equation as this will become a2 equals to minus 1 by i v a1 dot t e to the power of i omega 21 t because if this I take on the other side then minus i omega 21 will become less and this will become a1 of t it also is of t will become minus 1 by i v star a2 dot t e to the power of minus i omega 21 t okay. What I will do is I will differentiate once more differentiate equations with respect to time okay. So what I will get is if I differentiate once more and rearrange okay what I will get is the following a1 double dot t equals to minus i v a2 dot t e to the power of minus i omega 21 t minus omega 21 v a2 of t e to the power of minus i omega 21 t okay. Similarly I can get a2 dot t equals to minus i v star a1 dot t e to the power of i omega 21 t plus omega 21 to v star a1 of t e to the power of i omega 21 t okay. Then what I can do is instead of a1 and a2 I can plug this here and finally what I can get is the following after rearrangement what I get is a1 double dot t equals to minus a1 of t v v star minus i omega 21 a1 dot t. Similarly a2 dot t is equal to minus a2 of t v v star plus i omega 21 a2 dot t. So one can do you know you can all use this four equations okay and rearrange okay because wherever there is a a1 dot you can plug it in wherever there is an a1 you can plug it in a2 dot and then you can do amount of rearrangement it is not very easy one can do it as a homework. So this is conversion of these to this okay you can try it as a homework it is not very easy it takes little more than 10 or 15 minutes or this is just simple math so one can be able to do it. So what we have is these two equations okay so the two equations that we have is very simply that a1 double dot t is equal to minus a1 of t v star minus omega 21 minus i omega 21 a1 dot of t a2 double dot of t is equal to minus a2 of t v v star minus i omega 21 of a2 dot t. Now by writing this we have done one thing very simple now the equation 1 will only consist of a1 of t a1 of t its first derivative and its second derivative and this equation 2 will have only a2 of t. So a2 its first derivative and its second derivative with respect to. So what we have done by doing this we have separated out the couple differential equations in a1 and a2 as second order differential equations a1 and a2 respectively. So they are no longer coupled okay but while from going from first order equations to given to second order equation differential equations to uncouple them. Now if you have such a case one can find a general solution a1 of t will be equal to a e to the power of i omega t minus b e to the power of i omega t minus b e to the power of minus i omega t e to the power of omega 21 t by 2 and a2 of t is equal to a e to the power of i omega t plus b e to the power of minus i omega t e to the power of omega 21 t by 2 and where omega will be equal to half of omega 21 square plus 4 times modulus of e square whole to the power of half and a and b are constants which will depend on initial conditions. Now one can solve this with the condition that at t is equal to 0, a1 of 0 is equal to 1 and a2 of 0 is equal to 0. Why are we doing that? Because if we had two states 1 and 2 and their coefficients being a1 of t and a2 of t a time t is equal to 0 we are assuming that the population is only in the state 1 and state 2 has no population and when you do that these equations will run on to a1 of t will be equal to cos omega t plus i omega 21 by 2 omega sin omega t whole to i e to the power of minus i omega 21 t by 2 and a2 of t will be equal to minus i by modulus of v by omega sin e to the power of i omega 21 t by 2. So these are the two equations that end up and one can then one can then equate the probability, get the probability. So p1 of t will be nothing but modulus of a1 of t whole square and p2 of t that is probability of finding the state the system in state 2 is a2 of t modulus square. In such scenario what we will get is the following p1 of t after doing some math what we will get is 1 minus 4 modulus of v square divided by omega 21 square plus 4 times modulus of v square into sin square half of omega 21 square plus 4 times modulus of v whole square to the power of half and p2 of t will be equal to modulus of v by omega square sin omega t which will be nothing but 4 times modulus of v square divided by omega 21 square plus 4 times modulus of v whole square sin square half of omega 21 square plus 4 times modulus of v square to the power of half. Now one can think of it this is p2 of t. Now think of it like this so there is some function that is sin square x and some multiplier. Now let us look at p2 of t. Now this is sin square of some function multiplied by something. So let us call it as a sin square. So p1 of t is nothing but 1 minus a sin square of x. So at time t is equal to 0. Let us say this is 0 and this is 1 that is a population. Now if I take population then my a1 will go down like that. So this is p1 of t and a2 will whenever will go like this. So this is p2 of t such that p1 of t plus p2 of t is equal to 1. These oscillations of states 1 and 2 are called Rabi oscillations. So we will stop here and continue in the next lecture. Thank you.