1 00:00:00,830 --> 00:00:05,660 In this video, I will solve the problem of classical crystal, or an infinite 2 00:00:05,660 --> 00:00:11,490 chain of classical oscillators. And I will determine the normal modes in 3 00:00:11,490 --> 00:00:15,375 this system. and as I mentioned, I'll do it actually 4 00:00:15,375 --> 00:00:19,877 in the chain, which means it's in 1D. And I should mention that actually I 5 00:00:19,877 --> 00:00:23,450 could have done in two-dimensions or three-dimensions. 6 00:00:23,450 --> 00:00:26,254 It would have been a little bit more complicated. 7 00:00:26,254 --> 00:00:29,959 But otherwise, actually the difference, the key features of the solution are 8 00:00:29,959 --> 00:00:34,888 pretty much the same in one-dimension. And higher dimensions, apart from some 9 00:00:34,888 --> 00:00:39,026 polarization effects. and for this reason, I will just 10 00:00:39,026 --> 00:00:43,275 concentrate on the simpler case of 1D. But, you could have done it in any 11 00:00:43,275 --> 00:00:46,149 dimension. So, the specific model I'm going to 12 00:00:46,149 --> 00:00:50,784 consider is presented here. So, we will consider one dimensional 13 00:00:50,784 --> 00:00:54,030 chain. So, we should imagine that this infinite 14 00:00:54,030 --> 00:00:59,070 chain, actually more specifically, I will consider the so-called Periodic Boundary 15 00:00:59,070 --> 00:01:04,275 Conditions for simplicity. But in an infinite system, it doesn't 16 00:01:04,275 --> 00:01:07,781 really matter. And so, in this chain, as you can see, 17 00:01:07,781 --> 00:01:12,553 there are two types of atoms. Well, these guides the brown, the red 18 00:01:12,553 --> 00:01:18,159 circles, they correspond, well in my model to an atom of such 1D crystal. 19 00:01:18,159 --> 00:01:22,767 And the positions of atoms will be labelled by x sub n for lighter atoms and 20 00:01:22,767 --> 00:01:28,086 Y sub n for heavier atoms. And the index n here corresponds to, in 21 00:01:28,086 --> 00:01:33,970 some sense, to an elementary cell. So here, at this elementary cell would be 22 00:01:33,970 --> 00:01:38,617 you know, one lighter atom and one heavier atom. 23 00:01:38,617 --> 00:01:43,639 And, let's say the size of this elementary cell will be two times the 24 00:01:43,639 --> 00:01:49,391 distance between the atoms. And so n can be, you know, 1, 2, 3, 4, so 25 00:01:49,391 --> 00:01:54,890 have an infinite series of these elementary cells. 26 00:01:54,890 --> 00:01:59,050 And also for the sake of simplicity I will be measuring the distances in the 27 00:01:59,050 --> 00:02:04,960 beginning of my solution in the units of this elementary cell size. 28 00:02:04,960 --> 00:02:08,720 So, I will essentially set 2a equals to 1. 29 00:02:08,720 --> 00:02:15,410 So, a is basically the equilibrium length of a spring. 30 00:02:15,410 --> 00:02:20,306 So, 2a is essentially represents symmetry of my, this great symmetry in my 31 00:02:20,306 --> 00:02:24,212 problems. So, if I translate everything by 2a, I 32 00:02:24,212 --> 00:02:28,926 get my crystal. What I'm saying here is I'm going to be 33 00:02:28,926 --> 00:02:33,423 using the units work a set to 1. But basically, this is my problem. 34 00:02:33,423 --> 00:02:37,230 And so, let me write down first the energy of this. 35 00:02:37,230 --> 00:02:40,210 And so, the energy is actually very simple to write down. 36 00:02:40,210 --> 00:02:43,576 So the energy, as usual, will contain the kinetic energy part and the potential 37 00:02:43,576 --> 00:02:47,976 energy part. And so, the kinetic energy part includes 38 00:02:47,976 --> 00:02:54,379 the kinetic energy of all light atoms. So this letter case p sub n over squared, 39 00:02:54,379 --> 00:02:58,268 or root 2m. And the kinetic energy of the heavy 40 00:02:58,268 --> 00:03:02,300 atoms, so which actually have a color code. 41 00:03:02,300 --> 00:03:06,134 The sort of brownish forms corresponds to terms that are associated with lighter 42 00:03:06,134 --> 00:03:09,644 atoms, and the red forms naturally corresponds to terms which are associated 43 00:03:09,644 --> 00:03:14,022 with heavy atoms. And then, there will be potential energy, 44 00:03:14,022 --> 00:03:17,500 which is just the usual Hook's Law if you want. 45 00:03:17,500 --> 00:03:23,735 But I now have Hook's Law for every single spring in this oscillator chain. 46 00:03:23,735 --> 00:03:28,775 So for example, in this nth elementary cell, I'm going to have a term here 47 00:03:28,775 --> 00:03:36,680 associated with the spring between this lighter atom and this heavier atom, M. 48 00:03:36,680 --> 00:03:41,040 And so this term essentially as is corresponds to this this spring. 49 00:03:41,040 --> 00:03:44,379 But also, there will be an interaction let's say, of this heavy atom with the 50 00:03:44,379 --> 00:03:47,506 lighter atom from the next elementary cell, the nth pluth, plus 1, ele, 51 00:03:47,506 --> 00:03:52,937 elementary cell. And so there is a term also here in this 52 00:03:52,937 --> 00:03:57,244 Hamiltonian, this classical Hamiltonian, associated with this, in nearest neighbor 53 00:03:57,244 --> 00:04:01,770 interaction. Now, and of course, I have to sum over 54 00:04:01,770 --> 00:04:07,210 all these n's, and this will be will give me the total energy of my classical 55 00:04:07,210 --> 00:04:12,262 crystal. Now, the next step is structurally to 56 00:04:12,262 --> 00:04:16,672 solve this problem is to write down the equations of motion for all the particles 57 00:04:16,672 --> 00:04:20,248 involved. And there's an infinite number of such 58 00:04:20,248 --> 00:04:23,264 particles. And in our case, these equations have 59 00:04:23,264 --> 00:04:25,995 motion, and nothing but Newton's equations. 60 00:04:25,995 --> 00:04:29,780 well, associated with each of these particles. 61 00:04:29,780 --> 00:04:34,132 And, as usual, while the Newton equation is m, m times a, my, mass times 62 00:04:34,132 --> 00:04:39,510 acceleration is equal to the sum of all the forces. 63 00:04:39,510 --> 00:04:43,540 And the force can be found by, as a derivative of radiant, in this case is 64 00:04:43,540 --> 00:04:49,377 just one-dimensional radiant of the corresponding potential energy. 65 00:04:49,377 --> 00:04:55,270 With the respect to the coordinate of a particle where we're in, investigating. 66 00:04:55,270 --> 00:04:59,033 And so, for instance for the light rate, I'm for the nth light rate of this, will 67 00:04:59,033 --> 00:05:02,806 be the Newton law. And for the nth heavy atom, this will be 68 00:05:02,806 --> 00:05:07,512 the Newton Law. And so, if we just formally differentiate 69 00:05:07,512 --> 00:05:12,248 our potential energy, we get these equations of motion for the lighter atoms 70 00:05:12,248 --> 00:05:18,401 and the low, heavy atoms. And it's very nature of that, there are 71 00:05:18,401 --> 00:05:24,360 two terms in each of these equations. Because this essentially corresponds to 72 00:05:24,360 --> 00:05:27,920 having two springs attached to each of the atoms. 73 00:05:27,920 --> 00:05:33,249 So, for example, here in this equation, we can understand the first term as being 74 00:05:33,249 --> 00:05:39,554 associated with force exerted on this heavy atom by this spring. 75 00:05:39,554 --> 00:05:43,702 And the second term is going to be the force exerted on the same atom by this 76 00:05:43,702 --> 00:05:50,352 other spring to the right. So and the goal now is to solve these 77 00:05:50,352 --> 00:05:53,170 equations. Of course, it's an infinite chain of 78 00:05:53,170 --> 00:05:56,460 equations because n goes from minus infinity to plus infinity. 79 00:05:56,460 --> 00:06:02,540 So, this is an integer index which labels our pair of heavy and the light atoms. 80 00:06:02,540 --> 00:06:06,017 So here, I'm showing the same Newton's Law that we derived in the previous 81 00:06:06,017 --> 00:06:09,290 slide, but now just slightly reconstructed. 82 00:06:09,290 --> 00:06:13,510 So, I just grouped together this 2x sub n and 2Y sub n. 83 00:06:13,510 --> 00:06:18,020 So, of course it's a very complicated looking system of equations. 84 00:06:18,020 --> 00:06:22,640 We actually have an infinite system of coupled differential equations and well 85 00:06:22,640 --> 00:06:27,790 naively, we would you could just say that it's hopeless. 86 00:06:27,790 --> 00:06:30,820 You know, there's no way we could solve such a complicated model. 87 00:06:30,820 --> 00:06:36,064 But it turns out that using combination of a powerful mathematical method which 88 00:06:36,064 --> 00:06:42,760 we'll actually use the rate of transform and the reasonable physical guess. 89 00:06:42,760 --> 00:06:48,200 We can solve the problem in a few relatively straight-forward steps. 90 00:06:48,200 --> 00:06:53,761 So, the method involved essentially a free transform where we represent our 91 00:06:53,761 --> 00:07:01,751 time dependent coordinates of the light and heavy atoms as linear combination. 92 00:07:01,751 --> 00:07:07,130 Or an integral over a wave vector of a wave like excitation. 93 00:07:07,130 --> 00:07:10,350 So here, I have essentially a plain wave in one dimension. 94 00:07:10,350 --> 00:07:15,634 And so, remember that I introduced this unions of 2a equals to 1. 95 00:07:15,634 --> 00:07:21,255 Therefore, this wave vector multiplies just an integer label n, which labels my 96 00:07:21,255 --> 00:07:27,475 importance of my light atoms. And, I represent the heavy atoms as a 97 00:07:27,475 --> 00:07:34,380 linear combination of their own wave is some new amplitudes Y sub q. 98 00:07:34,380 --> 00:07:40,165 And here I use q times n plus 1 half because this guy is shifted by 1 half in 99 00:07:40,165 --> 00:07:46,597 my units of length relative to the light atom. 100 00:07:46,597 --> 00:07:50,952 And importantly, here I have the time dependence again which is consistent with 101 00:07:50,952 --> 00:07:55,650 the wave like behavior. So, I assume essentially that whatever 102 00:07:55,650 --> 00:07:59,990 solutions there are, this solution represent waves running through this 103 00:07:59,990 --> 00:08:04,065 crystal. And now, the next step in the solution is 104 00:08:04,065 --> 00:08:09,082 to plug in this trial functions into the Newton equations. 105 00:08:09,082 --> 00:08:13,632 And to see if we can make, if we can find a self consistent solution, which is 106 00:08:13,632 --> 00:08:20,283 consistent both with the Newton equations, and with this representation. 107 00:08:20,283 --> 00:08:23,731 I'm going a bit ahead myself. I would say that, of course, the reason 108 00:08:23,731 --> 00:08:27,147 I'm presenting the solution is because we will indeed find that it's self 109 00:08:27,147 --> 00:08:32,165 consistent. But the self consistency will include in 110 00:08:32,165 --> 00:08:37,640 a very essential way a relation between Omega and q, between the frequency of the 111 00:08:37,640 --> 00:08:43,300 wave. And the wave vector. 112 00:08:43,300 --> 00:08:47,347 So and this is the,this is going to be the dispersion of the waves, the possible 113 00:08:47,347 --> 00:08:52,900 dispersion of the waves. And basically, to solve the problem in 114 00:08:52,900 --> 00:08:57,170 this context means to find the this recursion basically to find Omega as a 115 00:08:57,170 --> 00:09:01,590 function. So now, let me consider every single 116 00:09:01,590 --> 00:09:05,616 term, let's say in the first Newton equation, corresponding to the light 117 00:09:05,616 --> 00:09:10,112 atoms. So, the first term on the left-hand side 118 00:09:10,112 --> 00:09:14,789 the second derivative of the coordinate with respect to the time. 119 00:09:14,789 --> 00:09:20,128 The acceleration here is going to involve the integral over all wave vectors. 120 00:09:20,128 --> 00:09:24,344 and that will just label this integral as as integral sub q, not to write the 121 00:09:24,344 --> 00:09:30,136 differentials. and here I'm going to have x sub q. 122 00:09:30,136 --> 00:09:35,076 And the derivative of the exponential with respect to time is going to produce 123 00:09:35,076 --> 00:09:41,220 minus i Omega squared exponential of iqn minus i Omega t. 124 00:09:41,220 --> 00:09:46,658 So, this is just minus Omega squared. Now, I also have here this term of Y sub 125 00:09:46,658 --> 00:09:53,584 n plus Y n plus 1, which essentially corresponds here in this picture. 126 00:09:53,584 --> 00:09:59,354 to having this red atom to the left and the red atom to the right. 127 00:09:59,354 --> 00:10:04,026 And so, if I now calculate this term, Y sub n plus T n minus 1, I simply can 128 00:10:04,026 --> 00:10:10,980 write this expression. So I have an integral sub q, Y sub q. 129 00:10:10,980 --> 00:10:17,217 And here I have a sum of two terms, e to the power iq n plus 1 half plus e to the 130 00:10:17,217 --> 00:10:25,135 power iqn minus 1 half. Because now this corresponds to n minus 131 00:10:25,135 --> 00:10:31,261 1, e to the bar minus i Omega t. Or I can factor out this e to the bar 132 00:10:31,261 --> 00:10:35,805 iqn, and here I will have e to the bar iq over 2, plus e to the bar minus iq over 133 00:10:35,805 --> 00:10:39,516 2. And this is equal to the cosine, 134 00:10:39,516 --> 00:10:44,616 proportional to the cosine. So, I can write it identically as an 135 00:10:44,616 --> 00:10:52,284 integral over q, Y sub q twice, cosine of q over 2 multiplying the same exponential 136 00:10:52,284 --> 00:11:01,090 as in this term here. And if I put everything together. 137 00:11:01,090 --> 00:11:05,446 So what I'm going to get is and basically if I focus only on the coefficients, 138 00:11:05,446 --> 00:11:10,066 which multiply the same plane waves in each of these integrals, I can ride the 139 00:11:10,066 --> 00:11:18,597 essentially the same Newton's Law. But now, in the [UNKNOWN] space as minus 140 00:11:18,597 --> 00:11:31,450 m Omega squared x of q from this guy is equal to k twice Y sub q cosine q over 2. 141 00:11:31,450 --> 00:11:37,190 This term, which we just derived minus 2x sub q. 142 00:11:37,190 --> 00:11:43,190 And this is essentially the Newton Law corresponding to the light] atoms written 143 00:11:43,190 --> 00:11:48,490 in the Fourier space. Of course, we can repeat the same kind of 144 00:11:48,490 --> 00:11:52,850 procedure for the second Newton Law for the heavy atoms. 145 00:11:52,850 --> 00:11:57,475 And if we do so we are going to get the following system of equations. 146 00:11:57,475 --> 00:12:00,895 So, the first one we just derived, and the second one I didn't explicitly 147 00:12:00,895 --> 00:12:05,850 derive, but it's derived in a very similar fashion. 148 00:12:05,850 --> 00:12:10,778 So and well the good thing about these equations first of all that instead of an 149 00:12:10,778 --> 00:12:16,676 infinite chain of coupled equations. Actually just have two coupled equations, 150 00:12:16,676 --> 00:12:19,039 okay? You see that of course, I have all 151 00:12:19,039 --> 00:12:22,478 different q's involved. So, I can change my q, but there is no 152 00:12:22,478 --> 00:12:26,360 coupling between different q's. Unlike in real space where we need to 153 00:12:26,360 --> 00:12:30,304 have coupling between neighboring size n, and another thing which is good is that 154 00:12:30,304 --> 00:12:36,390 its no longer a differential equation. Instead of the derivative, I now have I 155 00:12:36,390 --> 00:12:42,780 mean I got squared here. So, it's much easier now to deal with 156 00:12:42,780 --> 00:12:48,070 this essentially linear system of equations. 157 00:12:49,090 --> 00:12:52,930 And to solve the remaining part, let me just rewrite the same equation in a 158 00:12:52,930 --> 00:12:57,140 slightly different form. I will write it as symmetrics. 159 00:12:57,140 --> 00:13:00,426 Well, it's the same equation I just had before, but now I wrote in a slightly 160 00:13:00,426 --> 00:13:04,280 different way. I wrote it asymmetric, 2x2 metrics acting 161 00:13:04,280 --> 00:13:08,568 on a vector if you want, which contains the Fourier component of the coordinates 162 00:13:08,568 --> 00:13:12,344 of the light atoms and the Fourier components of the coordinates of the 163 00:13:12,344 --> 00:13:17,820 heavy atoms. And the, the right-hand side is zero. 164 00:13:17,820 --> 00:13:21,084 Because if we go back, we see that all the terms here are proportional to one or 165 00:13:21,084 --> 00:13:25,810 the other coordinate, there is not free terms here which is just constants. 166 00:13:25,810 --> 00:13:29,008 So I can write it like that and it's equal to zero. 167 00:13:29,008 --> 00:13:32,846 And maybe most of your, or some of, you know, from linear algebra. 168 00:13:32,846 --> 00:13:36,674 That in order for me to be able to have a known trivial solution to this equation, 169 00:13:36,674 --> 00:13:41,956 which is not identically equal to zero. Basically, a solution where x and q are 170 00:13:41,956 --> 00:13:47,174 not identically equal to zero. I have to ensure that the determinant of 171 00:13:47,174 --> 00:13:51,840 this matrix, so the determinate of this matrix vanishes. 172 00:13:51,840 --> 00:13:56,600 And well this is determinate of a 2x2 matrix, which I can calculate. 173 00:13:56,600 --> 00:14:01,496 And well, if you do so, you will get the following equation, which is the 174 00:14:01,496 --> 00:14:06,034 algebraic equation. And essentially, it reduces to a 175 00:14:06,034 --> 00:14:08,850 quadratic equation. Well, it's, formally, it's an equation of 176 00:14:08,850 --> 00:14:10,795 the 4th order. But there is only Omega squared involved. 177 00:14:11,980 --> 00:14:16,228 So all, all in all, it's just a quadratic equation that I hope most of you know how 178 00:14:16,228 --> 00:14:23,020 to deal with. And well, what is this equation for? 179 00:14:23,020 --> 00:14:26,260 Okay, this is again, an equation for the frequency. 180 00:14:26,260 --> 00:14:31,151 Or more precisely, for the dependence of the frequency on the wave vector, which 181 00:14:31,151 --> 00:14:36,528 is known as a dispersion relation. Well, we have seen actually many 182 00:14:36,528 --> 00:14:40,752 dispersion relations already in this course in the context of Quantum Physics 183 00:14:40,752 --> 00:14:45,594 and also classical Physics. For instance, when we talk about the 184 00:14:45,594 --> 00:14:50,683 energy of a particle, the free particle being proportional to p squared. 185 00:14:50,683 --> 00:14:54,715 So, this is actually dispersional relation, say of an electron or any other 186 00:14:54,715 --> 00:14:58,900 free particle. When we talk about the dispersional 187 00:14:58,900 --> 00:15:04,131 relation of a photon where Omega scales linearly with the wave vector. 188 00:15:04,131 --> 00:15:08,870 This is well the expression relation of electromagnetic waves. 189 00:15:08,870 --> 00:15:14,230 So now, we have here an equation which determines the dispersion relation of 190 00:15:14,230 --> 00:15:19,596 elastic waves, other, otherwise known as phonons. 191 00:15:19,596 --> 00:15:26,500 Which can exist in this sort of simple model of a one-dimensional crystal. 192 00:15:26,500 --> 00:15:30,910 So, now before solving this equation, let me go back to the sort of physical, 193 00:15:30,910 --> 00:15:37,090 regional physical units where the dimension of length is restored. 194 00:15:37,090 --> 00:15:42,493 So, here we have this q, the wave vector which is measured in 1 over distance. 195 00:15:42,493 --> 00:15:47,317 And remember that my so the convention was that 2a, a being basically the size 196 00:15:47,317 --> 00:15:53,334 of the length of a spring was equal to 1. So, if we want to restore the right 197 00:15:53,334 --> 00:15:57,618 physical dimension, I should multiply this q here by 2e and also of course, Y 198 00:15:57,618 --> 00:16:02,589 minus cosine squared is equal to sine squared. 199 00:16:02,589 --> 00:16:08,029 So, I can just rewrite in the usual physical union, sorry, I can write this 200 00:16:08,029 --> 00:16:15,182 term as sine squared q times a. And now well again, this is just the a 201 00:16:15,182 --> 00:16:22,100 quadratic equation Omega, for Omega. So, if I solve it I will get two branches 202 00:16:22,100 --> 00:16:27,780 with a plus and minus sign. And well, I get they're actually 203 00:16:27,780 --> 00:16:32,680 non-linear and non-trivial relation between m lambda and q. 204 00:16:32,680 --> 00:16:36,145 There are two relations as a matter, two types of expectations, two types of 205 00:16:36,145 --> 00:16:39,350 waves. A normal moves that I get out of this 206 00:16:39,350 --> 00:16:44,306 solution. So, and if I plug this dispersional 207 00:16:44,306 --> 00:16:51,520 relations, so this would be my Omega of q, and q here is defined. 208 00:16:51,520 --> 00:16:59,430 I should have mentioned it before between minus pi over 80 up to pi over 80. 209 00:16:59,430 --> 00:17:04,248 So, we'll get two terms. There will be your range corresponding to 210 00:17:04,248 --> 00:17:09,483 plus sign, the plus range which will look like this. 211 00:17:09,483 --> 00:17:14,385 And there will be a range corresponding to minus sign which will look 212 00:17:14,385 --> 00:17:18,917 approximately as so. So, this is Omega minus and this is Omega 213 00:17:18,917 --> 00:17:23,798 plus. And the Omega minus is called Acoustic 214 00:17:23,798 --> 00:17:30,170 Phonons, and the Omega plus is called Optical Phonons. 215 00:17:30,170 --> 00:17:34,785 So the, as you can see, there is a gap separating this optical phonons from zero 216 00:17:34,785 --> 00:17:38,921 energy. So, if we hit our is, our oscillator 217 00:17:38,921 --> 00:17:44,696 chain with a hammer and it's going to be a non-linear perturbation. 218 00:17:44,696 --> 00:17:49,368 We're going to exact all kinds of moles. But if we just touch it very slightly, 219 00:17:49,368 --> 00:17:52,848 very gently and apply just a weak perturbation, there is no way we can 220 00:17:52,848 --> 00:17:58,820 excite an optical phonon. We're only going to excite these energy 221 00:17:58,820 --> 00:18:02,870 excitations in the vicinity of zero energy. 222 00:18:02,870 --> 00:18:06,710 So, this is basically the energy associated energy that we would need to 223 00:18:06,710 --> 00:18:11,032 spend in order to excite in this case acoustic moles. 224 00:18:11,032 --> 00:18:15,256 And the dispersion of the acoustic moles as q goes to zero, we'll also discuss it 225 00:18:15,256 --> 00:18:22,087 in the next slide is linear dispersion. And this type of linear dispersion is 226 00:18:22,087 --> 00:18:28,235 called sound, sound-like dispersion. And the coefficient of proportionality, 227 00:18:28,235 --> 00:18:32,424 basically Omega as a function of q, in the q to 0 limit, for this acoustic 228 00:18:32,424 --> 00:18:36,805 phonons. And this coefficient of proportionality 229 00:18:36,805 --> 00:18:39,989 is the speed of sound. I actually forgot to mention a couple of 230 00:18:39,989 --> 00:18:42,958 things here. So in this equation, there appears this 231 00:18:42,958 --> 00:18:46,197 variable mu. So, mu here is simply the reduced mass 232 00:18:46,197 --> 00:18:50,240 we, we actually have seen before, quite a few times. 233 00:18:50,240 --> 00:18:54,965 Which is, in this case, the product of the mass of light atoms and heavy atoms 234 00:18:54,965 --> 00:19:01,438 and divided by their sum. And the last thing I'm going to mention 235 00:19:01,438 --> 00:19:06,040 on this slide is something about the optical phonons. 236 00:19:06,040 --> 00:19:10,330 Namely that the existence of this optical branch is related to the fact that we 237 00:19:10,330 --> 00:19:15,840 have two type of two types of atoms present in our crystal. 238 00:19:15,840 --> 00:19:19,683 So, if we had just one type of atom, let's say these blue atoms were the only 239 00:19:19,683 --> 00:19:23,603 type of atoms. Then, we would not of had the optical 240 00:19:23,603 --> 00:19:28,440 branch and we only would of had the acoustic branch. 241 00:19:28,440 --> 00:19:31,960 Which by the way, the acoustic branch response to what I was discussing in the 242 00:19:31,960 --> 00:19:35,370 end of the previous lecture to the presence of this sort of broken symmetry 243 00:19:35,370 --> 00:19:41,288 in the crystal. So finally, let me calculate the speed of 244 00:19:41,288 --> 00:19:46,293 sound, that is the coefficient of proportionality in this this portion of 245 00:19:46,293 --> 00:19:55,056 the acoustic wave as q goes to zero. So as we can see here so if we just focus 246 00:19:55,056 --> 00:20:04,008 on the minus sign in this equation, okay? So and if we consider the limit of q very 247 00:20:04,008 --> 00:20:11,652 small, then this guy can be expanded in Taylor series just up to first two order. 248 00:20:11,652 --> 00:20:17,300 So, we're just going to have qa squared instead of the sine. 249 00:20:17,300 --> 00:20:21,986 And also well, this whole thing, the square root of 1 minus this 4 mu squared 250 00:20:21,986 --> 00:20:27,048 over m M qa squared in the limit of small q. 251 00:20:27,048 --> 00:20:32,022 It can be written again using the Taylor series if we have 1 minus epsilon. 252 00:20:32,022 --> 00:20:36,048 And the epsilon is small, we can approximately write it as 1 minus epsilon 253 00:20:36,048 --> 00:20:39,440 over 2. So, in this case, epsilon is a useful 254 00:20:39,440 --> 00:20:42,424 thing. And this thing is small because we are 255 00:20:42,424 --> 00:20:47,596 near the zero momentum and zero energy. So, we can write that approximately is 1 256 00:20:47,596 --> 00:20:53,556 minus 2 mu squared mM qa squared. So, if we plug it back in here, so we're 257 00:20:53,556 --> 00:20:59,174 going to see that 1 minus 1 gives us zero, and we have just this guy 258 00:20:59,174 --> 00:21:07,082 appearing. So, the spectrum of the acoustic phonons 259 00:21:07,082 --> 00:21:14,804 in the q to zero limit is approximately equal to k over mu 2 mu squared and mM qa 260 00:21:14,804 --> 00:21:22,505 squared. Or we can cancel these guys and we're 261 00:21:22,505 --> 00:21:30,227 going to have simply recalling the definition of the reduced mass two times 262 00:21:30,227 --> 00:21:40,304 the stiffness divided by the sum of two masses, qa squared. 263 00:21:40,304 --> 00:21:47,444 Or, if we take finally the square root of this expression, we're going to get Omega 264 00:21:47,444 --> 00:21:53,849 minus, in the q to zero limit, is approximately equal to the square root of 265 00:21:53,849 --> 00:22:03,270 2k a squared, m plus M times q. And this is our speed of sound. 266 00:22:05,620 --> 00:22:09,520 So in the next video, I'm going to redo the problem. 267 00:22:09,520 --> 00:22:13,845 Actually, in the quantum limit. And, the, surprisingly, we're going to 268 00:22:13,845 --> 00:22:18,170 see that the solution is not more complicated than the classical limit. 269 00:22:18,170 --> 00:22:22,010 Well, I actually find that the quantum solution actually easier than the 270 00:22:22,010 --> 00:22:26,452 classical solution. But in any case however you classify the 271 00:22:26,452 --> 00:22:31,209 solution, easier or more complicated, the solution is actually going to be quite 272 00:22:31,209 --> 00:22:37,233 similar to what we found here. And the spectrum of the quantum phonons 273 00:22:37,233 --> 00:22:42,907 will turn out to be exactly equal to the spectrum of the classical formulas. 274 00:22:42,907 --> 00:22:47,329 So which is another sort of striking manifestation of quantum to classical 275 00:22:47,329 --> 00:22:52,070 correspondence, and actually a rather complicated system. 276 00:22:52,070 --> 00:22:55,200 In this case, this system models a crystal.