 So, here I come to some generalized descriptions of formalism of lattice dynamics, the reason being that regarding phonon experiments which I will get on to after a little while that it is important for such experiments to carry out calculations or simulations before. So, we take up the experiment reason being one that the inelastic neutrons scattering or the phonon intensities are much smaller than the Bragg peak intensity. So, it is a low intensity experiments, possibly two orders of magnitude lower the phonon intensities and that is why to have a successful experiment it is necessary that you carry out the lattice dynamical calculations for the sample beforehand. Especially if it is a single crystal sample which is easy to tackle theoretically, but the fact remains that it is not so easy to get single crystals, but for the single crystal samples it is easy to understand and we need to carry out detail lattice dynamical calculations regarding the phonon dispersion curves and then and then only we take up the experiment that is a general trend unlike let us say an extra diffraction experiment where one will make a sample by chemical route or by some other route and first thing they do will put it on extra diffraction instrument. Here before you take up lattice dynamics experiments on an instrument it is important that we understand the phonons and how they behave in this single crystal sample. So, now I with this I start first let me just write down there is something called a dynamical matrix. I will get into the interesting part called dynamical matrix first positions of atoms here the formalism is the protocol is that I am writing X when I say the kth atom in the lth unit cell that means there are unit cells after unit cells in the solid there are unit cells and I am talking about the kth atom in the lth unit cell if there are n atoms atoms per unit cell. So, if there n can be 2 can be free. So, then the kth atom in the lth unit cell is given by its equilibrium position rlk that means kth the lth unit cell kth atom in the unit cell. Plus a displacement vector ulk the ulk is the displacement vector of the kth atom in the lth unit cell and primitive cell there are number of primitive not primitive cell I say unit cells unit cells l equal to 1 to n and this displacement is about its equilibrium position at the lth unit cell for the kth atom. Now earlier also you saw that I took solutions which are wave like here also only the writing is slightly different let me just if you see a travelling a solution is given by e to the power i omega t in time e to the power minus i k dot r in space. And this k is the wavelength of the special variation omega is the temporal variation and this is the travelling wave solution now with this much understanding you see here. So, the alpha is x y z x y z components that is given by one is that e to the power i q dot r just now I wrote i k dot r e to the power i q dot r r is a equilibrium position of kth atom in the lth cell. u alpha kq is that when the this is the x component of the kth atom for the phonon of wave vector q this is for total displacement u x or u y or u z for the kth atom in the lth cell this is coming the from the phonon of wave like q. And then I have earlier also I wrote e to the power i k dot i wrote e to the power i k a if you see this let me just show you this this solution let me just go back to the solution here. You see this is the displacement here I am writing this displacement this was the displacement there is a travelling wave and this is the travelling wave solution with a prefactor here I have written the same in a slightly more generalized platform. I have chosen phonons so the this is the phonon wave vector for the phonon wave vector q this is the kth atom in the unit cell its displacement in one direction alpha can be x y z then this is given by a travelling wave solution which was e to the power i k s plus a. p plus a into a here it is q dot r which is the equilibrium wave vector and the temporal variation so this is the displacement vector. Now in this formalism also I will go for the solution for the equation of motion for example here I have written it that this is what I wrote just now. So now the equation of motion again e to the power i omega t gives m k because m k is the mass m k mass of kth atom omega square q here earlier I just wrote omega square now it is omega as a function of q because depending on the phonon wave vector. Phonon wave vector omega will be different so omega square as a function of q into the displacement vector. So the displacement vector earlier it was u now I am writing u as u x u y u z alpha going from alpha going from alpha is x y z components of the displacement. This is summed over matrix d alpha beta that means it can be d x y d d x x d x y d x z d y y d d so d alpha beta will be d x x d x y d x z sorry that I am sometimes talking too fast. d y z d y y d x y d y y d y z and then similarly d x z d y z d z z so this is so this is the dynamical matrix I will tell you what the dynamical matrix shortly. For the interaction between kth atom and the k prime atoms for a phonon wave vector of q and then this is the u beta component so that means the alpha component or the x component if I say x this depends on x y z displacement and the corresponding dynamical matrix here d alpha beta. That means if it is y then it will be d x y which depends on the phonon wave vector and the interaction between k and k k prime atoms. So dynamical matrix which dictates the force because if you remember in the earlier examples which I used the simpler examples I used a force constant which was c. Now the same I have replaced with the more generalized and mathematically more compact expression so this is the force constant. So force constant means phi alpha beta basically the direction x y x y z so alpha beta will dictate that the force for between kth atom in the lth cell and the k prime atom in the l prime cell. So this is the two atoms which are interacting these are the direction dependent interactions and then I have used the e to the power i q dot r minus r so this is the equilibrium position of the kth atom in the lth cell at site i and this is the and this is the variable r and k prime atom at the l prime cell. Distance if you remember I can write it as r minus r prime I am showing this indicators because this is the distance between the k prime atom in the l prime cell and the kth atom in the lth cell. So when I again write it down as the determinant of coefficient then I have this determinant. So this is the dynamical matrix which basically defines the force between two atoms at two different unit cells with a travelling wave like solution because the displacements have a travelling wave like solution. So the displacement are connected through the force constants and then I get a determinant which is the determinant of the coefficient this part is understandable and this part which was the force part is d alpha beta q means the phonon wave vector q and k and k prime atom so the for the phonon wave vector q the interaction between k and k prime atom is given by d alpha beta in terms of the force constant and this is the m omega square q and delta k k prime delta alpha beta alpha beta because the kth and k prime the diagonal terms will have these terms equal to 1 non diagonal terms will be 0. So now the solutions so that we are trying to solve this equation and actually these force constants we know or we have to know if we have to do a phonon calculation then the force constants the force constants needs to be known should be known because then only we can do a phonon calculation. So we can use various models for the force constants we can use a lennard-jones potential which is known to all of you or if there are ionic solids then there will be ionic potential which you have to use. So lennard-jones ionics we need to know the force constant between the atoms when it had to solve the dynamical matrix and find out the eigenvalues and the eigenvectors. The eigenvalues are the omega, omega-j they set of eigenvalues and the eigenvectors I may call it zai-j now they are vary when I say omega-j that means the eigenvectors it is basically the energy associated with a particular phonon. And for that the phonon zai-j will give the displacement pattern. Dynamical matrix is a Hermitian matrix we know because the forces are real. Dynamical matrix is a Hermitian so its solutions in terms of omega square should be real. So omega-j square is a real but real does not mean positive if omega-j square is less than zero or negative then you have e to the power minus i omega t and then this omega will be imaginary and you will have e to the power minus omega t. So such a phonon you can see with time it decays. So for imaginary omega-j the oscillations cant be sustained. So for solution which are possible for phonons the phonons that are quantized oscillation of the atoms in a crystalline solid they should be greater than zero and the imaginary omega-j they will decay and they will not be able to sustain in the system. Now for n atoms in the unit cell we know that there are 3n degrees of freedom and so there will be 3n eigenvalues and 3n eigenvectors. So after diagonalizing the matrix we get the eigenvalues and the eigenvectors. So what I am trying to do this is too much of mathematical expression basically since there are 3n atoms in the unit cell I am trying to find out 3n number of independent harmonic oscillators. We are familiar with harmonic oscillators which has got solutions like this. So here my attempt is these are coupled oscillators. I will request you to look into coupled harmonic oscillators or coupled pendulums. On internet several videos are available you can see how the coupled pendulums oscillate in space and time. So here I have got 3n normal modes. So my attempt is just like classical mechanics that we try to divide the oscillations in terms of independent oscillations and that is why we have got 3n normal modes and we also have 3n displacement patterns in terms of 3n eigenvectors. I will give an example. For example suppose I have got first the first example that I showed you just one atom per unit cell then I have got 3 modes. So now that means I have got x, y, z components of movement. Now if it is a longitudinal phonon or longitudinal acoustic phonon then this eigenvector will be 1, 0, 0. Because if I consider it is propagating in the x direction, propagation is in the x direction because other displacement 0 and this x displacement is 1. There can be some constant term before or after but the eigenvector is 1, 0, 0. Similarly for the transverse phonon the eigenvector will be 0, 1, 0. If I consider the plane is a y, z plane. So for one transverse mode the displacement is in the y direction and for the other transverse mode the displacement in the z direction. So for acoustic phonons with one atom per unit cell I have got these 3 displacement eigenvectors. So this is the longitudinal one and these 2 are the transverse. So for n number of atoms then I have got a wrong column which is x for first atom, y for first atom, z for first atom. Then x for the second atom, y for the second atom, z for the second atom. Finally if there are n then x for the nth atom, y for the nth atom, z for the nth atom. So I have a long column matrix. So as an example if there are 2 matrices, 2, I am sorry 2 atoms. If I have 2 atoms then for the longitudinal it will look like this 1, 0, 0. First atom 1, 0, 0 for the second atom. So for 2 atoms this is 1, 0, 0, 1, 0, 0 is the longitudinal one. Similarly for the 2 atoms for the acoustic mode it will be 0, 0, sorry 0, 1, 0 then it will be 0, 1, 0 for 2 atoms. For the longitudinal mode. So now that is what I wrote just now that in general it is like this. But for as I told that for 1, 0 direction in the Brillouin zone this is the longitudinal acoustic mode. This is the transverse acoustic mode. Now let us consider at q equal to 0 a longitudinal optical phonon. In case of optical phonon the atoms do not move in the same direction. I explained to you that sorry I explained to you that they move opposite to each other. So here you can see the eigenvector will be 1, sorry eigenvector should be 1, 0, 0. Then the second atom if this atom is moving in the x direction the other atom is longitudinal phonon. The other atom is moving in the opposite direction minus 1, 0, 0. So this is the displacement vector. So this gives you a sense that for a given number of atoms in unit cell how to write down the eigenvectors and the eigenvalues are omega j square q. That means the jth branch for the q Brillouin zone value, q value of the momentum vector in the Brillouin zone. So now instead of deriving the entire scattering law let me just give it to you and then explain the terms in it. So this is the scattering law. If you remember earlier we had talked about the scattering law in which we talked about e to the power minus i q dot rj 0 e to the power i q dot rj prime. t ensemble average apart from other terms. So this is the space time correlation function. That gives me here this is a phonon structure factor. This is a number of phonons and there apart from the constant there are two delta functions. The first delta function if you remember when we did diffraction because there is no energy transfer we said that q should be equal to g a reciprocal weight vector. Now we not only have a q which is a momentum transfer q is equal to 4 pi by lambda sin theta. But we also associated the phonon weight vector and all of them have associated momentum with them and that tells me that in this case q plus q. So the momentum transfer in the experiment at the depending on the angle at which you are measuring and the phonon weight vector. Phonon weight vector means when I plot my omega versus q this is the q acoustic mode, optic mode. This is the q this is the q these two together should be equal to some reciprocal weight vector. And this is the conservation law for momentum and this is the conservation law for energy. The neutron in the process of scattering either the nuclear can gain energy or lose energy. So neutrons neutron can gain energy energy or lose energy lose energy energy. So that's why delta omega plus minus omega j which is the phonon energy. Now interestingly here the neutron when it loses energy then it excites one phonon. And when neutron gains energy then one phonon deexcites and gives the energy to the neutron. So that means there are two ways but here because of the Boltzmann factor which is e to the power minus e by kt at any temperature the lower levels are lower energy levels are more populated than the higher energy levels. So that means the population in lower levels is much higher compared to an upper energy level. And that means the neutron which is interacting with it has a chance of exciting from lower to higher level and lose energy. That means neutron gives the energy and that energy is utilized in exciting a phonon. That means the loss of neutron energy is more probable. So this gives a momentum conservation law and the energy conservation law together with a structure factor, one phonon structure factor, one phonon structure factor weighed by the number of phonons. So if we know that if there is a gain in the number of phonons then it is a plus so n plus 1 you know n plus 1 h omega by 2 is the number of h cross n plus 1 h cross omega is the number of phonons of normal mode with temporal frequency omega j. And when it loses it is n so it is minus of it goes to 0 n h cross omega. So in this summation this takes care of the loss and gain of the energy together with a structure factor. Now let me explain to you the structure factor in case of phonon measurement. I have written it down here. Please note that apart from this part in rate this is exactly same as structure factor that we derive for a finite temperature diffraction. Let me just recall the expression I wrote structure factor Bj. This was a structure factor we evaluated for 0 degree Kelvin structure that means there is no temperature for a system of crystals in a lattice. And for a finite temperature we argue that this atom position they start getting because of the vibration of around the mean position they become larger. And then we get a Debye Waller factor which is dependent to actually the Debye Waller factor between the minus q square then average value of u square by 3 I evaluated. Now let us look at the expression here. Apart from this phonon displacement term q dot xi xi is the eigenvector for the qj for the jth phonon with wave vector q and this is the kth atom its mass is root over mk. So displacement vector for the kth atom for the phonon vector qj it must be a little confusing let me say. So that means what I mean is this there are phonons there are phonons. So this is the q value possibly for the jth phonon for this phonon the displacement vector is q I wrote it as qj and this phonon has a displacement pattern for all the atoms. So this is the qj for the kth atom in the unit cell that gives me the xi the displacement vector for the jth phonon jth phonon and corresponding to that the displacement for the kth atom. So this is what the phonon displacement vector and then this term exactly what we had for the 0 degree Kelvin structure factor and this is the Debye Waller factor as I discussed earlier. So now this is the one phonon structure factor this is also one phonon because there is also possibility those small probability you can have two phonons getting excited by a neutron or three phonons or multi phonons. But we will restrict ourselves to one phonon excitation that means which is the largest probability that the phonon can excite I mean sorry the neutron can excite a phonon. Now let us consider the case of a longitudinal phonon longitudinal phonon means the wave vector transfer q this q is not the phonon vector please remember. This is the momentum transfer in a diffraction experiment or in a phonon scattering experiment q is equal to 4 pi by lambda sin theta. This q is the phonon wave vector which is here these two I will show you how they are related. So now q and xi are parallel when they are parallel then q dot xi it is a longitudinal phonon because the displacement vector and the momentum vector the propagation direction they are parallel then this is a longitudinal phonon. But in that case q dot xi is a constant some constant and then this expression apart from some constants apart is exactly same as the Bragg structure factor so that means wherever I have got a large value of the Bragg peak for the longitudinal phonon I should also have a good coherent dynamical scattering or scattering from the phonon. So now I have written down two conservation laws one is that energies delta omega plus minus omega j and the other one is delta of q plus q their vectors minus reciprocal that is vector and I said that here the momentum conservation demands that g is equal to q plus q. Now let me look at the reciprocal lattice reciprocal this is the reciprocal lattice please note reciprocal lattice this is the reciprocal lattice. Now here I need not restrict myself to first Braille warzone the phonon wave vector needs to be restricted by first Braille warzone but my g can go over several Braille war zones and that is why I showed the g here it is 1 2 3 apart but this g in general should be equal to q plus q. So this is a vectorial representation this is the momentum transfer 4 pi by lambda sin theta and this is the q vector for the phonon within the first Braille warzone so g and q q can be much much larger than the q which is less than pi by a in its length. So now let me just show you two cases one is that longitudinal wave vector then g and q are parallel so q also can be parallel and there is a longitudinal phonon and this is how the vector looks like q plus q is equal to g. If it is a transverse phonon then this phonon wave vector q is perpendicular to the g vector or the reciprocal lattice vector and then I show you the vector diagram again q plus q equal to g but now the q and g they are perpendicular to each other. So that means for our experiment if I am using a single crystal the orientation of the sample and the values of q and g has to be chosen in a manner when I am either I am measuring a longitudinal phonon or an optical phonon we have to choose the g and q accordingly for a single crystal sample. So I will just stop here and get into experiments later but this is the triple axis spectrometer at Dhruva. I will try to give you example from some other sources but typically the triple axis spectrometer is like this. This is the huge monochromator drum at the center of which there is a monochromator. This I have shown you earlier also and you have seen that the monochromator drum contains a monochromator at the center of it. Here is the sample. Here is the sample. So this is the second axis. This is the first axis. This is the second axis and then around the sample you have got an analyzer which rotates along with the detector in a theta to theta mode. So these are the three axis. So this is how the diagram looks like. So now you can see that in the triple axis spectrometer we have first axis a monochromator, second axis a sample because you need to orient the sample and the third axis is the analyzer for energy analysis of the scattered beam. Earlier we didn't have this because we were doing diffraction. We were integrating sq omega over omega and we only had a detector but diffraction works. Now we cannot do this. We have to do the energy analysis. So now we have to have conservation of energy. We have to have conservation of the momentum in this experiment. So scans are performed in a certain path in the q, e or q omega space. It is energy transfer space. So either if this is the phonon dispersion relations either I can do a scan with a constant q or I can do a scan on constant omega in the space. So now mostly e i and e f may be kept fixed and we can keep varying the q by going to different angles or we can manipulate in a way where q is kept fixed and we go to different final energies. Usually incident energy is kept fixed and the final energy can be varied by changing the analyzer angle. So both the scans are possible which we will discuss on the next day but this is the basic arrangement of the triple axis spectrometer. This is the most used spectrometer for phonons but there are other spectrometers which I will come to like filter detector spectrometer and quasi elastic neutron scattering spectrometer where we look for stochastic motion or molecular vibrations. But triple axis spectrometer is generally most favored for phonon measurements with this I stopped today.