 We have been discussing the quantum mechanical description of a magnetic resonance condition. Let us recapitulate what we have done so far. We have a Zeeman interaction which gives the main interaction given by this. This is the Zeeman magnetic field, this is the S Z operator and the perturbation which is the small B 1 magnetic field applied in the X Y plane that is given I, this is the way it is. So the total Hamiltonian is H 0 plus H prime. The wave function corresponding to the unproduced Hamiltonian H 0 is written as psi of alpha similarly psi of beta. Here e alpha and e beta are the energy corresponding to the state of alpha and beta. And of course alpha corresponds to the this spin angle on the quantum number of plus half and beta corresponds to the minus half component. So in the presence of this perturbation we say the total wave function can be written as linear combination of this plus this. And we found that we can get an equation of this kind. Here our aim is to find out this coefficients here that tells the equation connecting the evolution of the system from one state to the other. So if you multiply on this side by let us say this one and complex conjugate of that so that I get psi alpha this gives. Now here because of the orthogonality of this wave function psi alpha and psi beta this gives 0 and this term also will give 0. If we evaluate this integral using this form of the perturbation this goes 0. So what I have here is this one is equal to 1. Now keeping in mind that this psi alpha and psi beta have this sort of form that the space part is here and the time part involves the energy this can be written as into e to the power that is the way it looks like. Similarly for the other coefficient we can write now that these equations are very important for that I have shown here in this slide minus h by 2 pi i dc alpha by dt is equal to the functional form given here. Now this can be simplified by taking this h by 2 pi i as h cross and bring it in the denominator. So we have got these two equations which are given here which shows the change of the coefficient alpha as a function of time and given by this. So here it shows that if this term which is essentially same as that if this is 0 these are not going to change. In other words for any change to take place or any transition from let us say alpha to beta to take place or from beta to alpha to take place this must be non-zero in a sense this decides the selection rule for transition. Now we have got this two coupled differential equation this is got a mistake here and this is got a mistake here. Now this e alpha minus e beta is nothing but the energy difference between the alpha state and beta state. So this gives rise to the higher energy and this is lower energy. So this is actually equal to this was missing h cross was missing and other equation is dc. So we have got this coupled equation the c alpha is connected to c beta c beta is connected to alpha. So it is not easy to solve these equations simultaneously. So what we can do is to use a perturbative technique to get an approximate solution. For that we start with this given initial condition that let us say condition is that the beta state is lower alpha state is higher. So the spin system initially is here and then when the perturbation was applied can it make transition from here to there that is what we are going to find out. So at t equal to 0 initial condition was that c alpha was 0 and c of beta equal to 1 and then what happens when t is greater than 0. So with this initial condition now let us try to find out the value of let us say c alpha then we can get d alpha by dt. Now here we have to find out this integral now for that the integral is involves this perturbation expansion. So this integral equal to cosine omega t plus sin omega t. To evaluate this integral we can make use of the lowering operator or raising operator s plus s minus let us see how that is done s plus 1 is sx s minus i xy. So this gives sx is equal to s plus s minus y2 and similarly sy is s minus minus y. So alpha sx beta will be alpha s plus s minus y2 beta. So this gives alpha half of here this is raising operator is lowering operator. So when it raises it can raise beta to alpha and lowering operator s minus operating on beta cannot lower it further where beta is the lowest state. So this gives 0 then you can use the property of the operator. So this give to get the value this belongs exactly half sy component this is alpha sy beta alpha sy beta. So this can written as alpha s plus minus s minus y twice i beta. This will give minus 1 by 2i or is equal to i by 2. So from this I can get the expression of dc alpha by dt in terms of this values exponential minus i it is the power minus i omega t. And similarly dc by dt gives c alpha by i h cross g e beta i b1 by 2 exponential this is this is alpha minus beta by t that is the way it comes up to be ok. Now we have actually started with this equation and then found this solution. If we integrate this now with respect to time then at let us say t equal to 0 c alpha equal to 0 and t greater than 0 this is c alpha. So this gives c alpha function of time and if we integrate this that is where it comes up to be now with the limit put there the value becomes. So this gives the expression of c alpha as a function of time and how it changes and it changes in this fashion. In this system we assume that our spin system to start with was in beta and then this gives the probability that it will be appearing here. But probability is not given by this number alone it will be a square of this number. But since this is a complex number we take a modulus of that which is in this fashion c star alpha into c alpha. So this will be written in terms of the multiply this and it is complex conjugate and simplify this see if I have this here ok. This gives again d c beta by dt in this fashion similarly d c alpha by dt in this fashion and then we multiply this which is complex conjugate and the simplified the equation looks like this. Now in this equation see that sin square and this is the argument inside and the denominator also similar argument here. Now this function becomes very very large when omega is very nearly equal to this g e beta e b 0 by h cross. So what you find here is that this sort of term comes there and the function we have got something like sin square x type of x square something like that which is the function here roughly speak not quite the same. But the t is there but we ignore that for the time being its function is of this kind. So this has a behavior x if you plot it has this sort of behavior that is at x equal to 0 it has very very large amplitude that is probability that the system will go from beta to alpha is very high when this argument becomes 0. Now this is interesting because this also shows that is actually equal to h cross omega and this is nothing but the Larmor precision frequency in disguise why the disguise if you write h by 2 pi and omega is 2 pi nu then this gives h nu equal to b 0 and this is nothing but the resonance condition that we have been talking all the time this also if you write here that g e beta e 0 by h cross which is actually Larmor frequency this is the this is nothing but gamma electron gamma electron b 0 equal to omega this is the Larmor frequency there. So if you have recovered the conclusion that I have derived from the classical description of magnetic resonance exactly same conclusion arrived at but the important difference is that here we have not really brought in the concept of appreciation. Larmor precision involves knowing the direction of the angular momentum or the magnetic moment in some absolute fashion that is I must know the both x y and z component of the three components of the magnetic moment then only the precision can be sort of visualized but quantum magnetic says that that is not possible we can only get one component of the angular momentum or the magnetic moment and the total magnitude of the angular momentum that is the description that you have used here but ultimate conclusion is just the same. Now this can be generalized to spin which is more than half here the description was based on only s equal to half say about two level system but if you have let us say s is equal to 1 then m s will be minus 1 0 plus 1 three level system one can actually do exactly the similar treatment and come to the same conclusion that for delta m s will be plus minus one will be the allowed transition and similar resonance condition will be applicable there also either I can write h equal to so here m s changes by one unit in plus minus half also the spin changes one unit so everything is consistent there so with this analysis we come to the end of this quantum integral description of the resonance in magnetic transition here we have understood that classical picture and quantum picture both give the same answer we end here now.