 I have been discussing phonons, the collective oscillation of atoms and molecules in a crystalline solid. So far I have gone in the discussion how to measure the phonon dispersion relations in a material and that deals with single crystal samples. Today I will go ahead and I will talk about phonon density of state measurements which is done with powder samples because powder samples are more easily available. And for many physical phenomena density of states is good enough to give us an answer. Then I will stop dealing with phonons and I will go into quasi-elastic neutron scattering for stochastic motions. So I will discuss two different kinds of motions. Phonons were collective oscillations, periodic and their time correlations are known. So this is a coherent effect. So like when you found out crystalline crystallographic structure using coherent neutron scattering cross section, phonon dispersion relations are also found out using coherent scattering cross section. Whereas quasi-elastic neutron scattering is for stochastic motions. Stochastic means statistical motions. So they are more in the realm of diffusion when a molecule diffuses statistically in a liquid or even in some geometrical confinement. So for single crystals we get information about omega versus q and it is an involved process as I discussed with you so far. This is what are the dispersion relations even longitudinal and transverse acoustic modes, optic modes and they have to be found out inside the first Brillouin zone and how we can do the measurements with a choice of g and q for a single crystal. I have discussed these issues and actually towards the end of the previous talk I had shown you results of measurements on a zircon crystal. A zircon is a constituent of the earth's minerals and I showed you that it needs an involved calculation of the phonon dispersion relations in which we also need to use group theoretical considerations where various group theoretical classes or representations are measured at different g and q values. Now from there we go into polycrystalline samples. Polycrystalline samples are easier to get these are powder samples. So they provide phonon density of states and we can work with a sample typically about 10 cc volume. Now what do we mean by density of states? That is the number of states we have between omega and omega plus d omega. So omega 2 omega plus d omega. I will just quickly remind you about what we did in your master's degree positively. Suppose I have a crystal which is having dimensions L. L is quite large. A very large dimension. So for that I know quantum mechanics demands that in one dimension I have gap between the 2k values twice pi by L because there should be a stationary wave inside the sample. And if I consider 3 directions x, y, z, x, y, z then the volume twice pi by Lq will have 1 allowed k values. So then the density will be Lq by 1 upon this twice pi q. So this goes to V by 8 pi q. This is the density in density in k space. Please remember here I drew it in a real space but this is the density in k space. So in k x, k y, k z direction if I talk about a sample of size L then in a unit volume it will be Lq by twice pi q is a density. Because twice pi by L is a volume in which we have got 1 allowed k value. So if I make it into a volume then twice pi by L whole cube will give me 1 k value in the k space. So the density is 1 upon that which is Lq by twice pi cube V by 8 pi cube and I take the example of I know that in the lower limit when omega and k are linear omega is equal to Vk. So in that case what I need to find out one is that in the k space if I consider a sphere of radius k it has got a volume of 4 pi by 3 k cube. This is the volume. And in this volume how many allowed k values will be there? 4 pi by 3 k cube into V by 8 pi cube. Because this is the density this is the volume. So this is the number of omega values. Now I can substitute from this relationship between omega and k that k is equal to omega by V. This is the velocity of the wave like here it is sound velocity when I talk about the k equal to 0 limit in case of phonons. I am just giving a simple example. Then it becomes then n omega will be equal to 4 pi by 3 k cube will be omega cube by V cube into V by 8 8 pi cube. This V is volume of the sample and this V is the velocity of the wave. So then we know that in omega d omega that is the number will be given by omega cube becomes 2 omega square. So 4 pi by 3 2 omega 3 omega square 3 omega square by V cube into V by 8 pi cube. So it goes as omega square. This is the n omega d omega square in which in a dispersion less linear relationship this is the density of states which goes as omega square and we know this is true for free particles also. So n omega d omega square omega square d omega this is my relationship. Now this is in case of a simple case as I took this is a device solid in which it is goes as omega square the density of states. But in actual crystalline solids this relationship is not so simple. So here as I wrote d omega is equal to V omega square by 2 pi square V cube from that relationship and it gives you my omega square relationship. And the internal energy is given as an integration over the density of states numbers present number of particles phonons present at that density of states at a frequency omega temperature t and d omega that gives me internal energy of the system. But in reality in case of real solid we have density of states which might look different and actually we need to measure the density of states which I will show you shortly in case of phonons to find out the internal energy. For example if I know the internal energy I can find out the specific heat which is important for many of the phase transitions from d u by d t. So now one thing is that I know that I have to measure still at various q values. Now for large momentum transfer when q is greater than twice pi by r where r is the inter particle distance then I can make an approximation that the relative motions are incoherent. This will be clear if I make a diagram. As you remember that earlier also I had showed you that if this is the reciprocal lattice then my q and small q together form equal to g because of the conservation relation equal to g which stated earlier. Now if q is large enough now there the delta q or d q around q plus minus d q d q is around few percent few percent. If q is large enough then the resolution unixoid as I showed you earlier allows contribution from various q values. So I am just trying to draw it for you this is a reciprocal so it can be q this q then it is this g this g value this point but it can be q this q so q 1 q 2 it can be this q bringing in this g also g 1 g 2 g 3. So basically now I am getting host of q values that are contributing to my q. So I am getting an sum over all q's and in the process because this each q talks about a displacement in case of phonons each q talks about a displacement. So this is a longitudinal acoustic phonon this is a optic phonon. Every q values whenever I talk about a phonon every q value corresponds when I touch a dispersion curve a certain displacement for the atoms or one atom. It is for the atoms here a pattern of displacement for the atoms which is coherent but now if so many q's are contributing to my q measurements then it is a sort of averaging over all the q values. And now I can consider that each atom this is a reciprocal lattice but each atom is moving incoherently that means there are motions I can just simply take the motion of a particular atom without bothering about the correlation with other atoms. This is known as incoherent approximation to coherent phonon scattering. So this is I am imposing and correctly so because I have got a q which is large enough so that the delta dqy encircles so many q values. And all the q values added together each q corresponds to a displacement for one particular atom will give a displacement which I consider the total displacement of that particular atom coming from all the phonon branches all the phonon branches and they are adding up incoherently. Actually earlier also we had done it if you remember in case of Debye-Waller factor we talked about a displacement u around r. I talked about r u plus rt in case of Debye-Waller factor and ultimately we talked when I did it that u and rt was g dot r was g r cos theta. And then we found out that square of it came and r square came out to be this came out to be not r square cos theta cos square theta came while calculating the Debye-Waller factor cos square theta comes. And we showed that if it is spherically isotropic then cos square theta comes out to be cos square theta average value comes out to be one third. That is very easy to show from the integration cos square theta d theta 0 to pi when you find out the solid angle average of cos square theta. So, it came one third g square or q square q square in general if I am talking about any value in q space. So, this we got in Debye-Waller factor in that case also we assume that each atom is moving independently. Here starting with phonons and then playing with the q length you find that is sufficiently large q values when sufficiently large means when the resolution in q involves adding up of many many branches then that allows me to consider the movement of each atom. So, each and every atom in the lattice as independent though they are actually hidden after the coherence between the atoms in the lattice when you calculate the phonons. But this phonons themselves at large q for each atom they add up and give me a displacement which is quasi independent. So, now since I am talking about independent atom motions. So, earlier if you remember in phonons my b was b coherent only. I always use b coherent I will come to it the expression that I did b coherent. But now since our motions are incoherent. So, incoherent scattering cross section also will come in if you remember b coherent is basically average of b in a lattice and b incoherent was b square b sorry as I showed you that b square was b square average. So, b square average minus b average square this was the b incoherent square and now instead of using only b coherent because I am assuming that the motions are incoherent and independent of each other my b square will comprise b coherent square plus b incoherent. So, I get a boost for this and now I can put this in my expression and I can get the density of states. So, this I express sometimes back that for many macroscopic thermal information we do not need the details of phonon dispersion, but an average information like density of states will be sufficient. So, now how do I get the density of states for phonons? I took a simple example where phonons at k going to 0 as limit omega is equal to vk dispersion less phonon density of dispersion less phonon curve that gave me as omega square g omega proportional to omega square. So, let me go back to what I showed you earlier this is a phonon single phonon coherent scattering law. Scattering law means it will be scattering amplitude and square of that. So, these are a structure factor part then as I explained to you earlier there is a population factor which tells you that whether it is a neutral or not neutral. So, this is a phonon energy gain or energy loss in case of energy loss it is n omega plus half plus half n omega plus 1 in case of energy gain it is half minus half so it is n omega h cross omega then we had a momentum transfer and energy transfer. Here the structure factor was it was coherent bk coherent this is the displacement part that is the q is the momentum vector transfer for the experiment not phonon wave vector momentum vector transfer this is the phonon wave vector. So, this displacement is for the kth atom contributed by jth branch at momentum vector q let me be clear about it. So, when I talk about xi suppose this is suppose this is a I am just talking about the acoustic branches it is longitudinal this is pi by a. This is longitudinal this is transverse. Now, if I consider this one then I know a displacement pattern for all the atom will be given. So, if I consider a single cell then I know what is the kth atom and what is the displacement. Now, for this transverse phonon I know also that there will be a amplitude in this direction and for the same kth atom coming from this branch I will have one more term. So, I have one motion in this direction one motion in the normal direction considering a longitudinal and the transverse phonon which is contributing to the displacement of the kth atom. That is what I am calling xi and the dot product of that with q the momentum transfer weighted by the mass of that particular atom. So, if there is a atom and a b atom if kth atom is the b atom then the mass of that b atom is there after that you have very simple terms like there is a divivalor factor and this is a structure factor. For this incoherent approximation I am considering each atom is acting independently. So, I need to square this. So, then this will go away this will be minus 2 wq. Now, q dot xi I showed you just now it is one third q square xi square by the same logic by which I did it for calculations in divivalor factor. So, it will be so qi square comes squaring bk square comes here I have weighted it against the average b square value this bk square will include will include bk coherent square plus bk incoherent square. So, now under incoherent approximation because I am considering the motion of the particles independent of each other. So, I have got the total cross section that is what I have written here this is not coherent. Here up to this point it was coherent, but once I have taken the incoherent approximation my fjq will have bk total or bk square total when I square them. This is here it will be squared when I do incoherent scattering law and then I have the density of states here. What is the density of states if you see here q square I have absorbed here this gk will have xi again kth atom and its contribution to its motion from the jth branch at the phonon wave vector q. Again I will repeat kth atom contribution to its motion by the jth branch at wave vector q is xi qjk and square and add that means if there are say I have two transverse phonons and one longitudinal phonons say the displacements will be x square plus y square plus z square. This I have to do for all the branches for all the branches when I add them up keeping the energy conservation fixed that means that displacements only when we touch one of the phonon branches and I have to add up over all the momentum vectors. When I do that then what I get is the kth atom for the kth atom the density of states and now if I add up over all the atoms then what I get is actually neutron weighted density of states which is weighted by the scattering cross section of that atom 4 pi bk square by mk and the weighted by the mass. And then it is gk omega which I get from the displacement of all these so I have to do an experiment where I do experiments at large number of qs large number of large number of q values wave vector transfer and then I will cover all the branches and I will get g of omega from the experimental measurement. So this is what so now it is an easier experiment I have given the same thing in terms of an ensemble average of sq omega its average is over omega here and similar over q so they are similar same but I like to show you this because here it is very evident that you get gk omega from your I mean I don't know what I will get is the sum of gk omega in my experiment rather I will get gn omega. In my experiment so phonon density of states I mentioned you in the polycrystalline sample I will do I will use this and this is the but what we measure experimentally you can see dispersion relation is like this but when you talk about phonon density of states there is no q dependence because as I told you I have done measurements at many q values there is an averaging over all the phonon wave vectors because of the dq values. And then when I add up I get this phonon density of states as a function of omega I hope it is clear to you because there is a band gap here between the not band gap a gap in omega spectrum between the acoustic and optic modes. So what to get if I plot g omega versus omega it looks somewhat like this so this is that gap this gap it is just shown in a perpendicular reaction but basically g omega gives me density of states that means d omega by dk for the entire sample. Now there is a summation over all the phonon branches so it is only a function of omega as I showed you earlier and this is what we can measure it is an easier measurement because summation over q and q in a polycrystalline sample the q is not defined because all possible q's are there unlike a single crystal and there is nothing like orienting the crystal in a certain direction. So I have got an average over all the q's and I do an experiment over large number of momentum vector transfers and I get neutron weight dispersion regression which I mentioned here. So how does it look like? Well this is what the experiment is and I am showed you this is how the data will look like let me just show some data. Earlier I had shown you this measurement this is on Zircon you can see this involved group theoretical calculations to find out which class will be measured in an experiment and the corresponding g values earlier I listed out for all these measurements though it is the phonon in q00 direction but various branches were measured in different experiments. Now for the same when I do the density of states it is a much simpler data you can see and this is what we found find that the fit between the experimental data the red dots and the calculated value. This fits unlike your diffraction data this is considered a good fit I mean this is an ethical question regarding how good is a fit but in case of phonon density of states this is considered as a good fit if you do such experiment this will be considered good. And from here I can get information about various macroscopic thermodynamic quantities like specific heat and others. So this is one of the measurements done at Dhruva in on density of states but here you don't have q dependent data q averaged out experiments done only as a function of energy transfer so it is easier to do carry out and we get better results. Now this is another one it is nickel silicon and nickel germanium and this is what again I am showing you this is the calculated value and these are the experimental data this peak doesn't represent very well but this peak reasonably represented I would say. Here the issue is that one big problem is this when I am doing let us say I am writing g in omega as here you can see I have got e to the power minus 2 wkq that means the kth atom and its value of the debauler factor at momentum transfer q. Here in this expression you can see I am putting a wq when it is an average on q square but wq remains same whereas you can see that and this is neutron weighted density of states for all the atoms over here the thing is that in a lattice a heavier atom will have smaller amplitude of motion compared to a lighter atom. This is average wk I am talking about average average and this is an average which I am using here but actually it is not correct because different atoms will have different amplitudes of movement and I am when I am doing density of states I am considering an average w of q. So, this is one source as I am telling you source of error in our density of states calculation also the q as I told you earlier the q has to be q is large q is large because then dq adds up over many q small q values as I showed you in the drawing here many q values. But fact is that if I go to very large q's then again my debauler factor will kill the intensity because debauler factor shows that the intensity decreases with q. So, again I have to take an optimal q it is large enough to work with incoherent approximation but not so large that debauler factor kills my intensity. So, these are the things which has to be taken care of and this is the kind of data that we get experimentally. So, I think here I will call it as top for discussion on phonons. We have discussed the phonon dispersion relations and their measurements and phonon density of states that we can measure using polycrystalline samples and easier experiments. So, the phonons the collective motions has stopped here and the next part I will go on to quasi elastic neutron scattering for stochastic motions.