 So, we are back to this but before I go further I just want to quickly recap what we have done so far in the previous four lectures. One is that I started with Fermi Golden Rule which is the top line where I narrated that psi k is a starting wave function with wave vector k goes to psi k prime under the influence of potential V, Vr it is a scattering system not just one but in general is a scattering system and rho k prime E because when you talk about transition from one state to another you need to involve the density of state so rho k prime E is the density of states and the final wave vector rho k prime that is what I wrote and from there I went ahead and derived the scattering amplitude f k k prime which is minus m by 2 pi h square square k prime Vr k it is in the bracket notation the same the way I have written above in integral form this is in a bracket t-rex bracket notation where k is equal to e to the power ik dot r apart from some constant and ket which is k prime here you can see which is e to the power minus ik prime dot r and it is an integration over r for both of them. I assume a delta function potential known as Fermi pseudo potential for the scattering system because here so far I have assumed just neutron and nucleus interaction is acting and because nucleus is very very tiny a femtometer size compared to the wavelength of the neutron which is angstrom I can assume it as a delta function potential and where b sigma is a spin isotope dependent scattering length for the site r where the delta function is existing and with these two I found that the elastic scattering elastic means here elastic means no energy transfer between the neutron and the then the scattering system d sigma by d omega there is some mistake it will be this is the expression this is also equal to if I write f of k k prime is also the square of this same as this so the one bracket was wrong here it will be i q dot rl minus rl prime q dot rl prime here so instead I or the bracket will shift q dot rl minus rl prime so this is the elastic scattering if obtained and there I showed you that when I am calculating the elastic scattering there is an averaging over b l prime b l b l prime complex conjugate b l over the all the isotopes spin in the lattice and that gives me two parts one is one b average square and that is the mean value b average is the mean potential seen by the neutron when average over spin an isotope of the scattering system and the fluctuation around the mean which is given by delta l l prime b square average minus b average square and when I write down the scattering cross section in terms of these two I get two parts one is as I showed here b average square e to the power i q dot rl minus rl prime which is the coherent part this causes diffraction because here we have q dot rl minus rl prime so we need to know that the position of the l prime with respect to position of the l and in a crystallographic lattice these are fixed and whenever diffraction occurs then we know that q will be equal to reciprocal lattice vector g and rl minus rl prime will be given by the miller indices or the of the positions or the planes and another part is n into b square average minus b average square this is an incoherent part and this incoherent part has no angle dependence the coherent part we can see to the power i q dot rl minus rl prime brings in the angle variation with respect to the q values so so I told you earlier that e to the power i we have outside this so this we could write that in terms of an average scattering and in terms of b square average minus b average square so this is an incoherent part this is a coherent part and the coherent part we have this has got angle dependence and the incoherent part has got which is a fluctuation around the main which has got no angle dependence and I also showed you how to find out b average for a distribution of isotopes and spins if I have excuse me I a k isotope having nuclear spin ik then the b average is given by the relative weight of the two states when the neutron spin is parallel anti-parallel to the nuclear state so you have nutrient this if this is the nuclear spin then the neutron spin can be plus minus half with respect to this and the number of states are 2 into this plus 1 that's the twice s plus 1 and this is equal to for plus half it is twice i plus 2 and for minus half this is equal to twice i so adding these two the total weight becomes 4 i plus 2 so relative weight for i plus half states is twice i plus 2 divided by 4 i plus 2 equal to i plus 1 upon twice i plus 1 and this if the number of isotopes then i becomes ik for the k isotope and then this will have another summation over the concentration of the kth one kth isotope and it goes like this so now I can write the average scattering length as ck concentration of the kth isotope and then the spin weightage ik plus 1 upon twice i k plus 1 into bk plus which is scattering amplitude for the neutron spin in direction of the nuclear spin and bk minus when the neutron spin is in opposite direction to the neutron spin so i plus half and i minus half now the question is how do I find out bk plus and bk minus of course experimentally but I am not going to tell you right now how the experiments are done similarly b square average a very simple extension instead of bk plus I write bk plus square and bk minus become bk minus square and the weightages become same so we can find out b average we can find out b square average and we can find out b square average minus b average square and we can find out the coherent scattering cross sections. We also told you that the scattering amplitude shkl for x-rays it is fj e to the power minus twice pi i xj h yj k zj l for an hkl plane where the position of the atoms are given by a xj plus b yj plus c za and for neutron this fj gets replaced by the average scattering amplitude bj which is a basically a delta function potential so fj again is a form factor which has got a q dependence and the intensity in case of x-ray peaks they go down at the higher angles because of fj which is not the case for neutron because bj it is does not have any angle dependence it is constant all over q space then I introduced you to the Debye-Weller factor what happens when the atoms are at a finite temperature or the system is at a finite temperature t they oscillate around a mean position and the amplitude of oscillation u square around the mean position tells us the intensity falls as i 0 e to the power minus one-third g square u square g is the reciprocal lattice vector at which we are measuring the intensity the Bragg peak intensity and i 0 is the 0 degree Kelvin intensity so with increase in temperature so long as the lattice is there the mean position around the mean position of the lattice is not destroyed then the vibration around this mean position causes a drop in intensity but not disappearance of the Bragg peak is only when the lattice starts melting the Bragg peak disappears followed by this I brought in the time dependence through a delta function and this is what I wrote again and again so we have an average of the potential at j and j prime q summed over j j prime and we also have actually I have repeated it both of them need not be written this sum this is I assume there so and one bolts one way for every state and it's a statistical average of 82 per minus i q dot r j 0 it is equal to i q dot r j prime t which is a correlation function I narrated and they also showed you that s of q omega is a Fourier transform over this correlation function as written in the expression above and then I showed you that s of q omega is a Fourier transform of intermediate scattering law i of q t and i of q t is just the opposite way is a Fourier transform from s of q omega now you see from these expressions I can write at t if I put t equal to 0 then you see i of q t is d omega e to the power i omega t s q omega here if I put t equal to sorry here if I put t equal to 0 then i of q 0 is given by d omega s of q omega that means t equal to 0 is not a fixed origin that means at any instant of time the intermediate scattering function is seen as an integration of s of q omega over energy space in the other way round if I put omega equal to 0 that means process of scattering in which energy transfer is equal to 0 this is an integration of our i of l I have to put omega equal to 0 then s of q 0 is an integration over d t i q t here that means when I look at a diffraction with perfect 0 energy transfer that gives me an average picture over time so let me these are integration over omega t remains and s of q omega is time integration and then s of q 0 for 0 energy transfer this is i q omega omega equal to 0 d d t and when I talk about this means integration over energy you are looking at the system over infinite time whereas if I consider i q t then at any instant of time which I call it origin of time is given by integration over all the energies now what does it mean we know that we do diffraction how do diffraction let me see I have got an incident beam I have got a sample and then I have got a detector here let me show it as a detector I will also tell you later how the detector works so at an angle I have a detector and I will count the neutrons on the detector now here I don't do any energy analysis usually energy analysis so in this case what I am doing is actually s of q omega d omega I am integrating over whatever energy is coming to the detector and I told you just now s of q omega d omega if I integrate what I get is an instantaneous picture on the other hand if I do an experiment with the same configuration but now let's say after the sample at an angle theta I ensure that I measure only those neutrons which have got omega equal to zero how do I do it I can do it by putting another crystal to select the energy of the neutrons and then I can put the detector here if I do that then what I have is actually integration over all the time now are they different will he get two different diffraction patterns no because the integrated picture as I told you integration it is nothing but instantaneous picture quantum mechanically but then we are doing the experiment over finite time finite time means our time scales of the experiments time scale of the experiment is much much larger than the time scales in the system in sorry count much much larger than the time scales of the system with this that means we keep putting one frame on one frame on one frame on one frame and that becomes a statistical average on the other hand when I do an experiment like this where I do a absolute zero and register for experiment that is equivalent to average over infinite time and these two things are exactly same statistical average at any instant over the entire ensemble of atoms taken one after another after another is the same as the statistical ensemble average over the entire time t so they are both of them are same but the conceptually they are slightly different because in one case we can this kind of experiments are also done usually we do this which is an integration over s of q omega d omega and there's a diffraction pattern I'm sure not many of you have used this concept but this concept can be used in neutron scattering and we do experiments in two various ways or conceptually it is possible to do it in two different ways so once one is an instantaneous picture and one is an integrated picture but both of them should converge to the same structure otherwise we are in trouble so but both of them do convert this one structure because ensemble average and time average for such a large number statistical system should be same so I stop the module here