 Now, I will introduce dynamics in our formalism. I request you to recall that we had been working to calculate the scattering amplitude where we evaluated the effect of potential on the incoming and outgoing wave vector of neutron. Now, when you talk about inelastic scattering, there is energy transfer between the neutron and the whole system, the scatterer and neutron and the scatterer forms a total system. So, over here then the magnitude of the scattered vector k prime is no more equal to k, which was true for the diffraction case. So, I need to put a factor of k prime by k because I have to compare the number of neutrons going out for neutrons coming on per second, coming in per second and that is given by the magnitude, this is k prime by k magnitude. Also, if there is energy exchange between the neutron and the scatterer, neutron plus scatter is a total system, then there should be an energy conservation during scattering. That means the energy of the system E lambda and the kinetic energy of the neutron in non-relativistic limit which is h square k square by 2 m for the incoming neutron must be equal to h square k prime square by 2 m plus energy of the scatterer after scattering and this is the energy constants. So, I will write down a long formula, but it is in the same spirit what I did earlier, I will go slowly on this and I will derive it for you. So, now I write down a scattering cross section, earlier I stopped at d sigma dE, if you remember I wrote d sigma dE, d sigma d omega per unit solid angle, now I am writing a cross section which is d to sigma d omega dE that means number of neutrons per unit solid angle per unit energy interval. This first there is a flux normalization, so neutrons coming in passes, neutron going out per second. I had this factor earlier also, if you remember this word. Now I have to sum over initial states, sum over final states, then I have the potential v working on neutron of wave vector k system energy lambda and sigma and square of that and here I have to since I am taking all possible outgoing states I have p lambda p sigma. Now in this expression I need to introduce a delta function, a delta function which takes care of the energy conservation during scattering and it looks like delta this is the energy difference of the neutron before and after and that should be equal to the energy difference of the system because neutron can give or take energy and there should be a balance what neutron gives is taken by the system what system gives should be taken by the neutron. So I write in this I just introduce by hand a delta function to say that in this scattering cross section I have constancy of conservation of energy between neutron and the system. Now I do a small trick my delta function there is a way of writing delta function there are various representations of delta function I use one of them which is the integral formalism which is equal to this is the integral representation of the delta function in time domain you can see e to the power minus i h cross omega e lambda minus e lambda prime t by h. So this also I can write as this is now this integral I will insert inside my statistical averaging expression. So this is what the expression for the delta function I will just mention this I miss this 2 pi h constant term and I will insert this delta function inside my ensemble average. So this is what I have written the delta function and now I have written this part you can bring in the time as I said I will also introduce a Fourier transform over the potential right now I am putting the Fourier transform we know that the Fourier transform will be BJ e to the power i q dot rj for a delta function potential but if I don't have a delta function potential there is reason for this and I will come to it later but if I it's not a delta function then I can write down in general the Fourier transform of the potential as V j q that is the Fourier transform of the potential at site r minus rj and it's V j q is in q space. So now I will write down the expression this expression in which I have k prime k and I know that k prime k gives me e to the power i k minus k prime which is q and V j dagger is a e to the power i q dot r V j r this will come from here so e to the power i k minus k prime in that factor dot r V j r integral gives me i q dot r V j q I write it because I write it as an operator it gives me the Fourier transform of the potential right now I am not assuming it is not a delta function I can put delta function at any time and get back my simple expression for the timing I am not using that so I can write this and now the delta function I have a integration over dt e to the power minus i omega t I have lambda k sigma lambda prime k prime sigma prime now this on on k k prime that integral gives me V j dagger q so I will just write it down please excuse me this long expression but it is important for you this is the first part now I have it is r g algebra so and also we have summation over lambda lambda prime so this is a very long expression I will go one by one so I have used there are these are square so there are two brackets so there are two brackets one complement of each other it starts with lambda lambda prime lambda lambda prime squeezed in between V j prime q and V j q because there are two summations j j prime you can see there are two lambda lambda prime and there are two summations over j and j prime so there are two summations over j and j prime on these two and there are two brackets we can play play around I can take them this way that way it is allowed because these are just integrals which have values physical values and we can use to move them around so with that I have got this expression and also I have got e to the power i e lambda minus e lambda prime by t by h and I have integration over time and e to the power minus i omega t now I will use this thing inside this that means what I am going to do is that your summation over j is there lambda prime is there now this e to the power minus e lambda i e lambda minus e lambda prime t I put it inside this bracket so I write it like this inside this averaging bracket this remains now I have e to the power minus i q dot r j prime I bracket it with e to the power i e lambda prime t by h cross and it is your minus e lambda t by h cross and then rest of the things remain same I have got a summation over lambda summation over lambda prime that means over all possible initial states over all possible final states I have bracketed my r g prime thing between these two and the integral over time remains so this is a very interesting thing what I have done actually if you see I have e to the power minus i e t sorry e lambda e lambda t by h cross then e to the power minus i q dot r j prime i lambda e lambda prime t by h cross and we have a lambda here and a lambda prime there whatever lambda prime and lambda here so now first we will use the fact that this I explained to earlier because this is like a projection operator and for any wave function the total number of total projection is equal to 1 because it is like sum of direction cos cosines equal to 1 but now you concentrate on this part that means these terms I can also turn it around I can bring this one to this side this one to that side so what happens you know this is the same as putting the because h h operating on a state lambda gives the energy of the state into wave vector this is standard rule for eigenfunctions of h in quantum mechanics and similarly h lambda will be equal to e lambda so this inside the bracket I can also write I will say I can write it like this so this is basically this gives me the time evolution of the operator there is a time evolution this is the biggest change now comes in why so because we know the time dependence either we can put to the wave function or we can put to the operator here I am putting this time dependence on operator and we know in a static state what is the time variation you know at any time t I can write suppose a wave function is psi at time t then any physical operator at time t will be given by this now for stationary states psi t is given by and psi 0 that means if I consider the wave functions are unchanged then I can write the same thing as so if the set of eigenfunctions are stationary then the time dependence is absorbed in the operator and I can find out the average at any time using this and that's what I have done I have done over here so I can write it as e to the power iq dot rj prime t instead of rj so I have brought in the time dependence a position vector in this so let me go back I introduced a delta function I am going slow I introduced a delta function to conserve energy in the scattering process then I used a specific expression or representation of delta function in terms of time integral because it's in the energy space then I also introduced the Fourier transform the potential now I have input this in the large expression where I have put the e to the power minus i e lambda minus e lambda prime t by h cross from the integral over time inside the bracket for vj prime q and that gives me rj prime t because I showed you that for any operator the time dependence is given by e to the power i h t by h cross into the operator into the wave function which is stationary so either we can take the time dependence on the wave function or on the operator here we take a stationary set of wave functions and the operator is changing and that's how we get the expression which is at zero time now are summation and sorry I should write down the specifically this goes to if I take it out this will remain for lambda for the other things and I get out the average of that outside should be treating so you can see that I have done the same thing over here in that summation I had probability lambda of state lambda if I consider the summation over lambda prime of lambda prime lambda prime over all the projections then I can remove the summation over lambda prime and because summation over lambda prime summation over this summation gives me one so that summation I am removing from there I am removing from there I have brought in e to the power i lambda prime i t e lambda prime h cross by divided by h cross and e to the power minus i t e lambda upon h cross from this integral from the integral you see for delta function inside the summation sign and inside the ensemble average picture when I do that there are two things one is that in the classical picture which will be valid for most of the cases the probability of energy e lambda is given as a Maxwell distribution then which is e to the power minus beta e lambda by z and this part I will emphasize once again for any wave function you can see if phi is an eigen function of the Hamiltonian which is lambda here this part this part this part in red I told that this gives the time dependence in the operator variable rj and now I can write down the whole expression it's a Fourier transform over this whole part where we have a summation over all the sides for the Fourier transforms at specific sides this is a p lambda which is an statistical weight for the state with energy lambda and there is an averaging of e to the power minus i q dot rj 0 e to the power i q dot rj prime t and this is an ensemble average of this value and summation over all the atoms what does it mean physically let me just what I have actually I have a summation aj prime I have got an ensemble average of so one is that we have the energy state lambda the probability of that e to the power minus beta e lambda by z partition function we are aware that this is the Maxwell weight Maxwellian weight e to the minus e by kt upon sum over all the states and this thing this part is about dynamics of the system why because I am seeking the ensemble average of a quantity which has I am saying if the jth atom is at the origin at time t equal to zero what is the position of the j prime that term at time t under the given dynamics it correlates the two and this is called a correlation function this is a correlation function and this correlation function is directly related to the double differential scattering cross section that are right d to sigma d omega d prime so you can see the double differential scattering cross section is given by the Fourier transform over a time Fourier transform over a large expression in which I am looking at the ensemble average probabilistically weighted with the energy e to the power of beta e lambda by z is a statistical weight and I am seeking the average value ensemble average value of one atom molecule whatever it is is the at one side and what is the probability of another scattering unit I call it in general scattering unit at another side at time t if there is no correlation then we can take them out of this average separately we can do the averaging separately in ensemble I can tell you but if they are correlated for example phonons in a solid motion of one is related to the motion of another one and then the ensemble average needs to know the correlation to get the double differential scattering cross section I haven't told you so far how to find out e to the power i q j q dot I haven't given any prescription to find out the ensemble average and summation over all sides I have not given any prescription we need a prescription if you want to know the double differential scattering cross section we do need a prescription how to calculate that in next part and later I will give you at least some prescription I can tell you all the prescription because there are various kinds of dynamics but to know that I have to have a prescription to figure out how to calculate r j 0 minus r j prime t and that is the game that I have to either I find a model because direct Fourier transform is often not possible but here I what I want to highlight to you that the time Fourier transfer over this large quantity on the right hand side you see gives me d to sigma d omega d e prime that means number of neutrons per unit solid angle per unit energy interval d prime because the energy the prime of the outgoing neutron so now I can shorten the expression this whole expression I can shorten it I can write n number of scatters k prime by k this is an average over energy spin v square q because vj dagger v dagger and summation has been taken care and I write something called sq omega what is sq omega the sq omega is as I showed you wrote wrote down over here the sq omega is 1 by twice by h cross n I have absorbed the n here so I have brought in n here it's a Fourier transform of summation over all the sides j j prime e to the power i q dot r j 0 and e to the power i q dot r j prime t I have given the signature of operators over here because in quantum mechanics these are called operators but for all our calculations I can even take them as classical positions and I will not be too wrong when I do the derivation so now this part the right part it contains the dynamics of the system and sq omega you can see this is in time and in q space and over that when you took a Fourier transfer over time I go to sq omega in q and omega space where I do the experiment that means I do my experiment where I find out the wavelength of the outgoing neutron at a certain angle from there I find out the energy transfer and the wave vector transfer and that experimental data is related to the dynamics of the system and how to get them out is in the next part