 In this lecture, I will continue with the inelastic scattering of neutrons by phonons. As I told you earlier that we have discussed elastic neutron scattering for all structures at various length scales and now we have come to inelastic neutron scattering in which phonons form a very important part of the experimental studies for understanding collective oscillations in crystalline solids known as phonons. I will discuss phonons and later I will also discuss stochastic motions and towards the end I will come to molecular vibrations and also extremely slow motions but now we will continue our discussion on phonons. Now phonons are collective oscillations in a solid when you consider the solid atoms are coupled by springs. So, here basically if there are two atoms we assume that they have a spring connecting them or rather the force behaves like a spring that means the energy force is proportional to displacement in the first order and the simple harmonic oscillator. So, here the this is a picture I borrowed from internet from this source where it is a lithium fluoride crystal which actually has shown as the springs connecting the ions in this crystal and this picture remains valid for all crystalline materials in which we will discuss the phonons. Now the force equation or the equation of motion for the sth part at ion let us say this is the sth ion will given by the force as second differential of the displacement and the sum over all the nearest not nearest all neighbors where p are the neighbors. So, basically the Newtonian equation of d 2 x by dt 2 equal to force that translates here as m into d 2 u s by dt 2 m is the mass of the ion is force is a spring constant. So, it is basically c of p plus s minus c of p depends on that there is a force constant between the two atoms and then sorry I am sorry here I am sorry this is I made a mistake here c p of p plus s into u of s plus p minus u of s summation over p p are the neighbors neighbors of the sth atom. So, this is a simple harmonic force which depends on the displacement of the s plus pth or rather the pth neighbor that means we have just let me just take a linear chain and go to second order force. So, there is a spring here there is also a spring here between the term there is a spring here there is also a spring between the next nearest neighbors. So, if this is the sth atom this is minus 1 this is minus 2 this is plus 1 this is plus 2 and in ideal case I should sum up over all the neighbors and this is the equation of motion for the sth atom and then we get this equation I have to start with I have started with a monatomic lattice c p is the spring constant and then if the sum is over all the neighbors of the sth atom. So, now for this particular case this is the concept of phonon is that I consider a wave like solution where the displacement of u of s plus p is given in terms of a prefactor p k a sum over p please note this here I am the there is a wave vector associated with the displacement which is k a is a lattice constant and I have taken the the displacement as a traveling wave in space and then this is a very general solution that u s plus p in terms of the number u is given by a this is nothing, but a wave like solution and here. So, this is given by a wave like solution and what are the solutions let us get into that now here. So, now in time also I consider dependence of e to the power minus i omega t this is a temporal frequency and this is a spatial frequency and now the displacement I write in terms of u e to the power i s k a is for the sth atom and it is given by a sum over the neighbors you can see I have just use it use the displacement term which is e to the power i s plus p k a minus e to the power minus s k a and sum over neighbors. So, this sum actually here when I do I divide by e to the power i s k plus a what I get is the equation of motion for the atom in this lattice and p is the summation over the nearest neighbors. Now, we know that if I have if these are the atom positions I am showing it as a linear chain, but it can be easily translated into three dimensional that is also. So, fact is that if this is the central atom then if I consider the neighbor here I also have a neighbor at minus p neighbor at or plus p plus 1 also have a neighbor at minus p plus 1. So, they are symmetric the c p and the force constants are symmetric because the force between these two are same as force between these two. Similarly, the force between this and this one is same as force between this and the next nearest neighbor. So, c p equal to minus c minus p not minus c p c of minus p force is the same and then this summation series I can write in terms of because it is e to the power i p k a minus 1 instead of going only for the positive numbers I can sum it over p is greater than 0 and i p k plus e to the power minus i p k because this is minus direction this is a plus direction and p is a number neighbor number which is greater than 0. So, I can write it as e to the power i p k a plus e to the power minus i p k minus 2 and this is equal to cos of 2 cos of p k a. So, 2 cos of p k minus 2 negative. So, ultimately I get this equation is cleaning. So, I have got this equation m omega square equal to c p 2 minus 2 cos p k a. So, that 2 comes out and this is the equation for omega square and if I consider a special case actually often this works well if I consider only nearest neighbor interaction then of all the p's you know only the nearest neighbor only the nearest neighbors they survive and then I have omega square equal to 2 by m c p into 1 minus cos p k a where it is p is equal to 1. So, I can call it c. So, this will be equal to 2 by m 1 only. So, it is c into 1 minus cos k a and cos k a equal to cos k by 2 plus k a by sorry let me write it here and cos a plus b is cos square k by 2 sin square k by 2 omega square and finally, it is very easy to show that the solution is omega equal to 4 c by m to the power half it is a free factor sin k by 2. Now, let us look at this solution. Now, this sin k by 2 as k goes to 0 k goes to 0 sin k by 2 goes as k by 2 you know sin x equal to x and when I consider this then this solution gives me k going to 0 omega equal to 4 c by m to the power half k by 2 which comes to c by m k. So, omega is proportional to k as a linear relationship and that means, when I go to center of the zone which is at k equal to 0 then the solution rather omega k relationship it goes down to 0 as 0 linearly. So, this is similar to let us say a sound wave propagating in a continuum media we know that sound wave velocity of 330 meters per second per second and in this range there is no dispersion when I mean dispersion let me just mean it here that omega versus k for sound wave it goes like this goes like this for our case which is phonons it is going like this. So, when k goes to 0 it behaves like a sound wave or a or rather continuum media and the wave propagating and mechanical wave propagating in that and that is why omega is proportional to k this branch is called acoustic branch because as omega k goes to 0 omega also goes to 0 like sound wave and actually speaking when you tend towards k equal to 0 the acoustic wave or the acoustic phonon goes into sound wave. This is basically an elastic wave in an elastic medium where we have the microscopic structure of atoms sitting at various sides, but when you go to k equal to 0 or long wavelength limit the whole medium behaves like a continuum because again as we had discussed earlier when the wavelength is so large it cannot distinguish between the sides and it acts like a continuum the medium acts like a continuum and what we get is sound wave. Now, another interesting thing is that let us consider the relative displacements of the neighbors. Now, from our solution u s plus 1 by u s is equal to e to the power i k a because we wrote the solution as u s plus p equal to c e to the power i s plus p k a and this was c p and it was sum over p. So, that is what we wrote earlier oh sorry here it instead of that we read u this was what we wrote earlier. So, now when I do that s plus 1 and u s the ratio is e to the power i k a and when we use this value consider the boundary of the first Brillouin zone. What is the Brillouin zone? If let me just consider a linear lattice of spacing a or spacing a real space its reciprocal space is again a linear lattice of spacing twice pi by a. So, because this is a here the spacing will be twice pi by a that is how we make the reciprocal lattice. And now the Brillouin zone is nothing but we check connect the nearest neighbors and perpendicular to them at half the position and that is the Wigner side cell we call it the Wigner side cell in reciprocal space. So, it is pi by a here and minus pi by a in the negative direction and this is known as first Brillouin zone this is the first Brillouin zone. So, first Brillouin boundary comes at pi by a for more complicated for example, if I have other kinds of lattices then the Brillouin zone shapes will be different, but Brillouin zone at Brillouin zone boundary we will have special properties of the phonon. So, here let me see when in this chain where k is equal to pi by a what we find now u s plus 1 the neighbors and u s at the zone boundary k equal to pi by a it becomes e to the power i pi equal to minus 1 that means, at the zone boundary the neighbors the nearest neighbors are exactly pi phase apart in their movement. Now, once they are pi phase away any further difference will not give us a new independent displacement set because already they are pi away after that they can only repeat something which has happened less than pi by a. So, this is one reason we plot phonon dispersion curves in the range of 0 to pi by a for phonons also minus pi by a this I mentioned earlier today I mentioned it let me just do it. So, this is the zone center this is pi by a minus pi by a we try to evaluate the phonons up to the zone boundary. Now, because any k beyond zone boundary let us say zone boundary is pi by a if I goes marginally outside then this can be if this is the phonon dispersion curve in the second zone it will look like this I will come to again later it will come like this. So, this is same as a point if I move it by minus twice pi by a and get back to the first drill war zone. So, all phonons by all phonons I mean that all the displacement patterns if I can call it pattern all the displacement vectors all the possible displacements are contained inside the first drill war zone and outside the first drill war zone whatever we find can be brought back to the first drill war zone. That is why I try to get the phonon dispersion curve within the first drill war zone where this is the phonon wave vector please remember again and again this is the phonon wave vector this is the phonon frequency and this is the dispersion curve. So, because at the zone boundary we complete the set where u s plus 1 any two neighboring atoms are pi phase apart and after that we cannot have any new motion new motion pattern that will give us a new phonon. So, this is already contained in the first drill war zone now. So, in the first drill war zone so now that is why one thing is that I will find out the motions at the first drill here the solution for a monatomic lattice I have showed that as k goes to 0 the k omega was proportional to k which is acoustic, but now when k equal to pi by a the frequency is given by here you can see I had written a sine k a by 2 when k equal to pi by a it is sine pi by 2 1. So, omega is equal to 4 c by m to the power half and the relation is sine. So, that means for a monatomic case let me just plot the 0 to pi by a. So, this is the phonon dispersion curve. So, here it is 4 c by m to the power half 4 c by m to the power half and here it is 0 and for this simple case it is a sinusoidal curve, but now this is a movement now there is one atom in the unit cell unit cell. So, this has got three degrees of freedom three degrees of freedom. So, of these three I have shown you the dispersion relation for one now basically when you have particles or planes of particles this wave which I wrote as e to the power i s plus p a. So, this displacement can be longitudinal or transverse. So, first let me point out one by one we have phonons only in the first Brillois zone we need to do a measurement where the q the phonon wave vector goes from 0 to pi by a and maybe negative side. Second if I have a monatomic crystal monatomic crystal that means one atom per unit cell then I have three degrees of freedom and then these three are one can be longitudinal and two can have the two transverse motion transverse motions. So, this makes it three waves or these all three are called acoustic waves why I have come to it shortly. Now, suppose there are two atoms. So, if there are two atoms per unit cell an example let us say CSCL. So, let me just let me just go. So, what is the CSCL structure let me just try to draw it for you as an example sample. So, you have chlorine at the end edges. So, eight edges gives me one chlorine per unit cell and one body centered rescission. Now, that means if I consider let us say a propagation in this direction which is 1 0 0 direction in reciprocal please remember in reciprocal space then or in real space this will have planes like this I will have planes of chlorine followed by planes of cesium then again planes of chlorine. So, this is how if I go in 1 0 0 direction I will get the planes which now I am showing as a chain basically the planes these chains are actually let me see I can make something back on you this is basically a plane plane of atoms moving. So, now there are two movements say here in CSCL case one is CS one is CL in general there are two atoms of different mass, but they are also part of a traveling wave like solution this traveling like wave like solution this concept was brought in I think by Chris Igor Tom, but this is a basic concept of phonons these are the frequency pattern or the displacement pattern is a wave and actually when there are a number of atoms in a unit cell we can find out what are the number of waves that are possible which you can consider as independent of each other that means we can break down all the frequencies and all the displacement in terms of eigenvalues and eigenvectors I will complete shortly. So, now one is that let us consider this atom we are considering nearest number interaction. So, when you have a nearest number interaction that motion of equation of motion is given its mass is m 1 so m 1 into dt us by dt 2 which is the time derivative is equal to the force which is the nearest number is v s minus 1 and v s. So, v s plus v s minus 1 minus 2 us because these are sum of two terms v s minus us plus and v s minus 1 minus us. So, for the two terms and that is how it comes v s plus v s 1 minus 2 us similarly for so, this is what we wrote here and we take solutions for us and v s again earlier I had chosen a travel even solution like this now I have chosen two solutions here us equal to u and then this is the wave vector that dictates the wavelength of the motion and this is for the second atom. So, now substituting this in this equation I have omega square m m 1 u you just this is e to the power i omega t for that time variation and then two differentiation using minus omega square m 1 u and then this is the solution which is 1 plus e to the power minus i k minus 2 c u because this us comes as u e to the power i k s k a minus i omega t. So, it comes like this and similarly for the other one the second atom also it will be minus omega square m 2 v equal to c u into 1 minus i k. So, now, I have got two equations in u and v and I can write down the coefficient form the determinant for a real solution this determinant should be zero. You can see from here I can take u on this side. So, it becomes c minus e to the power i k and 2c minus m 1 omega square here 2c minus m 1 omega square and this is c into e to the power i k. Now, so, u and v here related. So, that is why I wrote a u v this coefficient of this is v and the second one minus omega square m 2 properly the same temporal minus omega square m 2 v equal to c u 1 plus e to the power minus i k a minus 2c v this will be the second equation. So, adding these two equations I get 2c minus m 1 omega square u minus c into e to the power plus 1 plus e to the power minus i k v and here it is minus c 1 plus e to the power i k v and this is sorry u into v. So, that is why I said u v into this determinant should be solved and then for this to have real solutions it should be the determinant of the coefficient should go to zero and then this is the equation which I obtained for the motion of the two particles. Interestingly now I have got six degrees of freedom six degrees of freedom. Now, I know that out of which three will be acoustic acoustic because that will mean one longitudinal I will show you structurally and two transverse wave longitudinal means the particle displacement and the wave propagation direction they are parallel k and particle displacement they are parallel in case of transverse k and the particle displacements they are perpendicular to each other and that can happen in two planes. Now, for this equation I have got two solutions one is when k goes to zero let us discuss this solutions then omega square is equal to 2c 1 by m 1 plus 1 by m 2 and omega square equal to because there it is a equation in omega to the power 4 I can put that equal to some say some I do not know alpha square where alpha equal to omega square. So, it becomes a binary equation second order equation in omega square and these are the two solutions when k goes to zero and when k goes to pi I have got the solutions 2c by m 1 2c by m 2 m 1 is greater than m 2. So, this will be the solution when I have two atoms per unit cell. So, how does the displacement they look like when I said longitudinal waves please note that the k vector this is the k vector and this is the because in this direction the wave is propagating and I show the displacement of this plane this plane of atoms. So, you can see that the second plane of atom is displaced this much the third plane may be displaced a little more the fourth plane little more then again it keeps reducing and finally it comes to a plane after the wavelength that I get back to the central plane which I have assumed the central plane. So, that is a longitudinal wave. So, here the displacements follow a sine curve whereas in this case also here the second plane is transverse if this is the k direction the displacement is in this direction. So, the second plane you can see this is displaced upward little more little less and ultimately it comes down to the displacement here. So, that means, the sinusoidal wave here I can draw like this. So, here if I consider this as a direction then I have got two particular direction because there are planes of atom in this direction also. So, there are two transverse waves in this case and there is a longitudinal wave and in case of only one atom these two things together longitudinal and transverse acoustic waves they make up for the total degrees of freedom which is free. When we have two atoms then also the number of acoustic waves should be free. So, this is a acoustic waves which will be free and also there will be optical waves optical waves why optical I will complete shortly why call them optical. You can see that that k equal to pi by a because m1 is greater than m2 2c by m1 is smaller than m2 this is the solution and this is the solution for the optical branch. But at k equal to 0 you can see this is the optical branch and this part because omega square is proportional to k square and omega is proportional to k barring a constant factor. So, it goes to 0. So, now we have got three optical branches and three acoustic branches for two atoms. So, I hope now I have able to make you understand the displacement patterns in case of monatomic and diatomic lattice and this can be further generalized for any kind of lattice structures. Now, this I called acoustic because I said earlier that they the omega goes to 0 as k goes to 0 and as k goes to 0 the medium behaves like a continuum. So, what about optical part? Here I show you the displacement model of the acoustic mode and the optical mode borrowed from here. So, here please see this is the acoustic mode. The atoms they move in the same direction and the amplitude of their oscillation varies as per the wave vector k which is the wavelength of this oscillation, but they are all they all are falling on the same wave pattern. So, when this is one atom going down the other one also going down when it is going up all of them go up and this is the displacement pattern in case of optic mode optic mode. You please see the same pattern here I have got the wavelength remains same, but the nearest atoms they are moving in opposite direction. When this one is moving in this direction the next nearest neighbor is moving in this direction this is in this direction and the next to next nearest neighbor again in goes in this direction. So, the atoms are moving moving against each other each other. This part this part of my description I borrowed from Kittel's introduction to solid state physics the phonons. So, now I have talked to you about acoustic modes and when atoms are moving against each other when they are moving against each other if the center of mass is fixed and if there are ions like as I showed you incisium chloride then this when this ions are moving against each other it looks like a dipole oscillation dipole oscillating oscillation and this dipole oscillation is able to couple with the electric field of an electromagnetic wave and can absorb or emit electromagnetic waves and that is why these are called the optical modes when the atoms are moving against each other with the center of mass fixed. So, acoustic mode go in the low K continuum limit goes to sound wave and optical mode because they are capable of absorbing or releasing electromagnetic radiation because in case of ionic solids they will be like oscillating dipoles with this I stop in the next part I will go into more generalized description of phonons and then how to talk about various measurements in the next part.