1 00:00:00,180 --> 00:00:04,328 In this video we're going to consider the same problem of an oscillator chain and a 2 00:00:04,328 --> 00:00:08,974 galactic moves in the chain. But now we're going to start the quantum 3 00:00:08,974 --> 00:00:13,370 collective moves or quantum phonons in a quantum oscillator chain. 4 00:00:14,460 --> 00:00:17,892 So this problem sounds very complicated because we are going to be dealing with 5 00:00:17,892 --> 00:00:21,390 an infinite number of quantum degrees of freedom now. 6 00:00:22,530 --> 00:00:26,622 But as we shall see actually at the end of the day the results are going to be 7 00:00:26,622 --> 00:00:31,242 very similar to those in the classical cases, so namely, the spectrum of quantum 8 00:00:31,242 --> 00:00:35,532 phonons apart from the zero point energy is going to be identical To the spectrum 9 00:00:35,532 --> 00:00:41,926 of classical phonons. Now, to solve the problem, first we need 10 00:00:41,926 --> 00:00:45,623 to define what we want. So let me consider slightly simpler 11 00:00:45,623 --> 00:00:49,222 version of what we looked at in the previous video, and now I'm going to 12 00:00:49,222 --> 00:00:53,309 study an oscillator chain, consisting of the same type of oscillators, and same 13 00:00:53,309 --> 00:00:58,235 type of atoms. So I have here identical atoms with a 14 00:00:58,235 --> 00:01:02,450 mass m and indentical osciallators, identical sort of springs. 15 00:01:02,450 --> 00:01:07,570 Now its going to be quantum springs of sorts with a stiffness key. 16 00:01:07,570 --> 00:01:12,172 So this the x corrdinate and the corresponding classical Hamiltonian 17 00:01:12,172 --> 00:01:18,180 assocaited with this osciallator chain is Presented here. 18 00:01:18,180 --> 00:01:22,410 So we have a sum going over all these atoms. 19 00:01:22,410 --> 00:01:26,430 So n here is sort of lets say this is going to be n equals one, n equals two, n 20 00:01:26,430 --> 00:01:32,112 equals three. So you just label the atoms and x sub n 21 00:01:32,112 --> 00:01:37,760 is the position of the corresponding atom. 22 00:01:37,760 --> 00:01:41,230 Now this is of course the kinetic energy and this the potential energy. 23 00:01:41,230 --> 00:01:45,585 So its essentially the same Hamiltonian we had before but now just for a single 24 00:01:45,585 --> 00:01:50,432 species of atoms if you want. Now to go to quantum problem we need to 25 00:01:50,432 --> 00:01:54,440 quantize this Hamiltonian. We need to make it quantum. 26 00:01:54,440 --> 00:01:58,854 So how do we do this? Well, it turns out that, that it's 27 00:01:58,854 --> 00:02:02,508 actually very easy to do so, so it's as easy as it was in the case of a single 28 00:02:02,508 --> 00:02:07,792 harmonic oscillator. So all we have to do is just to put hats 29 00:02:07,792 --> 00:02:13,949 on top of H. P and X, so which means that we made this 30 00:02:13,949 --> 00:02:21,400 variables in the classical case into operators in the quantum problem. 31 00:02:21,400 --> 00:02:27,300 So, for example if we are working in a position space, so let's say P sub n. 32 00:02:28,680 --> 00:02:34,870 Is simply[UNKNOWN] minus i h bar g over dx m. 33 00:02:34,870 --> 00:02:38,970 So it adds only on the x coordinate of the oscialltor. 34 00:02:38,970 --> 00:02:44,770 Now this[UNKNOWN] satisfied the uh[UNKNOWN] goal of commutation 35 00:02:44,770 --> 00:02:51,200 relations. So if they correspond to the same atom so 36 00:02:51,200 --> 00:02:59,666 x n and p n they commute to i h bar. If the correspond to a different atom, 37 00:02:59,666 --> 00:03:03,257 lets say if n is not equal to, well lets say if we have x n and p n So then is 38 00:03:03,257 --> 00:03:09,520 going to be 0 of n, is not equal to n. So and, we can sort of unify these 39 00:03:09,520 --> 00:03:14,080 computation relations, by writing that delta symbol here. 40 00:03:14,080 --> 00:03:20,800 So which means that it's equal to i h bar 1, if n equals to m, and 0 otherwise. 41 00:03:22,310 --> 00:03:28,850 So this, this is what it means, to, to quantize this on a Hamiltonian. 42 00:03:28,850 --> 00:03:32,576 So now the next step we're going to perform is we're going to use our 43 00:03:32,576 --> 00:03:37,061 definition of the creation and annihilation operators that was that were 44 00:03:37,061 --> 00:03:42,305 introduced in the solution of the single harmonic oscillator And we're going to 45 00:03:42,305 --> 00:03:47,687 represent each coordinate xn and momentum pn through the corresponding creation and 46 00:03:47,687 --> 00:03:56,116 annihilation operators a and a dagger. So here are the same, you can check, 47 00:03:56,116 --> 00:04:00,187 going back to the first segment of the lecture today, that as a matter of fact 48 00:04:00,187 --> 00:04:04,199 these are exactly the same definitions We used before but now we just have this 49 00:04:04,199 --> 00:04:10,650 additional index end which refers to the atoms in this chain. 50 00:04:10,650 --> 00:04:16,078 And of course this creation and annihilation operators, they satisfy 51 00:04:16,078 --> 00:04:21,600 their own canonical commutation relations as here. 52 00:04:21,600 --> 00:04:26,610 They commute with one another their commuator is equal to 0. 53 00:04:26,610 --> 00:04:31,722 If they've responded to different atoms, and otherwise if they respond to the same 54 00:04:31,722 --> 00:04:36,780 atom, a n commutation with a dagger n is equal to one. 55 00:04:38,390 --> 00:04:42,599 So what we're going to do now, we're going to plug in this expression of x and 56 00:04:42,599 --> 00:04:47,020 p through a and a dagger into this Hamiltonian. 57 00:04:47,020 --> 00:04:51,390 And let me do that. So if we do so the Hamiltonian takes this 58 00:04:51,390 --> 00:04:55,908 form. Which actually appears to be a more 59 00:04:55,908 --> 00:05:02,670 complicated looking form since there are more terms here than we had before. 60 00:05:02,670 --> 00:05:06,780 But we are on our way to solving the problem. 61 00:05:06,780 --> 00:05:12,060 So the strategy for solving the problem would be actually different from the sort 62 00:05:12,060 --> 00:05:16,416 of typical solution in single particle quantum mechanics where we're dealing 63 00:05:16,416 --> 00:05:21,310 with the Schrodinger equation wave function. 64 00:05:21,310 --> 00:05:26,350 So here of course we still would like to solve the Schrodinger equation. 65 00:05:26,350 --> 00:05:30,250 We have in mind this I shrink your equation, but we don't worry too much at 66 00:05:30,250 --> 00:05:34,230 this stage about the form of the wave functions. 67 00:05:34,230 --> 00:05:39,130 We just want to know the energies that are possible in this in this system. 68 00:05:39,130 --> 00:05:42,485 Just like in the previous video, we just worried about the spectrum of the 69 00:05:42,485 --> 00:05:46,795 dispersion relation of classical wave. So here we just want to know the 70 00:05:46,795 --> 00:05:51,880 dispersion relations. For quantum waves. 71 00:05:51,880 --> 00:05:54,515 So how do we do this? a scenario to solve the problem here 72 00:05:54,515 --> 00:05:57,215 instead of working with the wave functions, we're going to be working with 73 00:05:57,215 --> 00:06:01,970 the operators themselves. And we're going to try to transform the 74 00:06:01,970 --> 00:06:06,254 Hamiltonian into a new, simpler form so that basically we're going to be looking 75 00:06:06,254 --> 00:06:10,905 for a new set of operators. We're going to go from the operators, 76 00:06:10,905 --> 00:06:15,045 let's say a n. And a n dagger, with all possible n's, to 77 00:06:15,045 --> 00:06:18,659 a new operator. So let me ahead, let's say b q, and b q 78 00:06:18,659 --> 00:06:22,764 dagger. Such that the form of the[UNKNOWN], in 79 00:06:22,764 --> 00:06:26,484 terms of this new creation, and annihilation operators, is going to be 80 00:06:26,484 --> 00:06:31,040 much simpler that the form, which is written here. 81 00:06:31,040 --> 00:06:35,931 So basically the challenge here is to find a linear transformation of the 82 00:06:35,931 --> 00:06:41,210 original operators into new forms. Such that the Hamiltonian, which looks 83 00:06:41,210 --> 00:06:45,230 simple, and that such that the solution to Hamiltonian this would be null. 84 00:06:46,270 --> 00:06:51,370 So this is a bit of a technical challenge and a technical exercise. 85 00:06:51,370 --> 00:06:55,290 So and the first stop in this exercise is just to write the Hamiltonian in terms of 86 00:06:55,290 --> 00:06:59,620 this a and a dagger as is. So we have to basically expand these 87 00:06:59,620 --> 00:07:03,396 brackets and calculate the square of these brackets, which leads to a rather 88 00:07:03,396 --> 00:07:09,402 lengthy expression, as here. So and I have omitted here an overall 89 00:07:09,402 --> 00:07:12,832 constant. There could be a constant which 90 00:07:12,832 --> 00:07:17,120 corresponds essentially to an energy shift And also I have written it in a 91 00:07:17,120 --> 00:07:22,422 slightly more[UNKNOWN] form. well its not as long as [UNKNOWN] 92 00:07:22,422 --> 00:07:28,850 covering it because of this term which is corresponds to your mission conjugate. 93 00:07:28,850 --> 00:07:32,205 So basically I have this guy so you can just work it out and we can work it out 94 00:07:32,205 --> 00:07:37,116 as an exercise if you want. So and you will see that we have many 95 00:07:37,116 --> 00:07:41,532 many terms and each term here but be compensate with the permission congregate 96 00:07:41,532 --> 00:07:45,948 terms somewhere else because Hamiltonian in the end of the day must ask permission 97 00:07:45,948 --> 00:07:50,300 of so, and in this case, it simplifies out life a little bit because we don't 98 00:07:50,300 --> 00:07:58,240 have to write two lengthy expressions but still it looks pretty bad. 99 00:07:58,240 --> 00:08:00,070 So there are a lot of terms we don't want to see. 100 00:08:00,070 --> 00:08:04,060 So we don't know what the spectrum, what the solution of this Hamiltonian is. 101 00:08:04,060 --> 00:08:07,718 We see a bunch of this a and a daggers in various combinations and we see also 102 00:08:07,718 --> 00:08:12,700 couplings between different sides, et cetra, et cetra, et cetra. 103 00:08:12,700 --> 00:08:17,116 So in order to solve this Hamiltonian in the way I just described, that is to find 104 00:08:17,116 --> 00:08:21,404 new set of operators that simplifies form so just like in the classical case it is 105 00:08:21,404 --> 00:08:25,116 convenient to perform a for your transform but now its going to before 106 00:08:25,116 --> 00:08:33,840 your transform for eh the creation And, annihilation operators themselves. 107 00:08:33,840 --> 00:08:38,384 So what we have here is a representation of the annihilation here, and the 108 00:08:38,384 --> 00:08:42,573 creation operators, in terms of some other operators a sub q, and a sub q 109 00:08:42,573 --> 00:08:47,324 dagger. So, I use the same symbols, but, this is 110 00:08:47,324 --> 00:08:52,540 really a different operators. And so this a sub q are in a sense 111 00:08:52,540 --> 00:08:57,560 Fourier images of this a sub n. And so the two are related by this 112 00:08:57,560 --> 00:09:02,500 Fourier transform, which now is an integral going from minus pi to pi. 113 00:09:02,500 --> 00:09:06,790 So and it's not from minus infinity to the plus infinity because we have a 114 00:09:06,790 --> 00:09:11,410 periodic. Array of these oscillators. 115 00:09:11,410 --> 00:09:14,730 Now at the continuum of oscillators. So its more like actually very serious 116 00:09:14,730 --> 00:09:19,801 than Fourier transform that we're using. In any case I'll be using this symbol 117 00:09:19,801 --> 00:09:23,655 here. So itegral with the substrate q with two 118 00:09:23,655 --> 00:09:28,613 sort of label all the the terms that I have here, and in this notations, for 119 00:09:28,613 --> 00:09:36,500 example, the Fourier-transform of the creation operator is presented here. 120 00:09:36,500 --> 00:09:42,280 So it can be obtained from this one simply by Conjugating, both sides. 121 00:09:42,280 --> 00:09:45,670 That's why I have a minus sign here as opposed to a plus sign here. 122 00:09:45,670 --> 00:09:49,766 So there are a couple of identities that I want to emphasize without deriving 123 00:09:49,766 --> 00:09:52,720 them. So the first one is the sum over all 124 00:09:52,720 --> 00:09:57,510 integers, basically over all the oscillators in our problem. 125 00:09:57,510 --> 00:10:02,630 And, so n goes from zero, plus minus 1, plus minus 2, etc. 126 00:10:02,630 --> 00:10:07,980 So and if I sum over all all these n's. These exponentials will give rise to two 127 00:10:07,980 --> 00:10:12,555 pi times the delta function and sort of the inverse identity to it, if we 128 00:10:12,555 --> 00:10:17,205 integrate the exponential from minus pi to pi so then the right hand side is 129 00:10:17,205 --> 00:10:24,570 going to be delta Simple, so this integral is equal to zero. 130 00:10:24,570 --> 00:10:28,346 If n is not equal to zero and it's equal to one, well, in a trivial way, 131 00:10:28,346 --> 00:10:33,405 otherwise. I'm mentioning these identities because 132 00:10:33,405 --> 00:10:40,150 they're very useful in deriving a form of the, Hamiltonian in the q-space. 133 00:10:40,150 --> 00:10:44,630 From the Hamiltonian we derived in the previous slide in real space, which is 134 00:10:44,630 --> 00:10:50,826 which was dependent on the index n. And this derivation, which takes a little 135 00:10:50,826 --> 00:10:54,360 bit of time and effort, I will leave mostly to you, and this will be one of 136 00:10:54,360 --> 00:10:58,293 problems and you'll Upcoming homework for self-assessment, and probably this is 137 00:10:58,293 --> 00:11:03,490 going to be the most complicated problem there. 138 00:11:03,490 --> 00:11:08,818 However, there is one important identity that I want to prove explicitly here, and 139 00:11:08,818 --> 00:11:16,018 this will be an identity involving a commutator of two q space operators. 140 00:11:16,018 --> 00:11:20,976 So again our goal would be to rewrite the annihilation in terms of this a sub q and 141 00:11:20,976 --> 00:11:24,996 a sub q[INAUDIBLE] and I refer to this guys as annihilation and creation 142 00:11:24,996 --> 00:11:31,779 operators but I haven't proven that they are indeed such operation. 143 00:11:32,850 --> 00:11:35,940 So what does it mean for them to be creation annihilation operators? 144 00:11:35,940 --> 00:11:40,158 So they , it would mean that they satisfy certain type of computational relations 145 00:11:40,158 --> 00:11:43,977 similar to the economical computational relations we have discussed in the 146 00:11:43,977 --> 00:11:50,195 previous lecture. So, and these commutation relations for 147 00:11:50,195 --> 00:11:55,370 the real space operators a sub n, let's say a sub m dagger, we know that this is 148 00:11:55,370 --> 00:12:01,750 delta symbol of n and m. So now we want to check what kind of 149 00:12:01,750 --> 00:12:06,700 limitation relations these guys satisfy and in order for me to work it out, work 150 00:12:06,700 --> 00:12:10,825 out these commutation relations I'm going to use an inverse Fourier 151 00:12:10,825 --> 00:12:17,718 transform. The first term in this commutation 152 00:12:17,718 --> 00:12:24,924 relation is sub q q. It can be written as a sum over n1 a sub 153 00:12:24,924 --> 00:12:33,492 n1 e to the power minus iq1n1. So this is an inverse Fourier transform 154 00:12:33,492 --> 00:12:39,696 sometimes opposite to this guy here and the second term here is going to involve 155 00:12:39,696 --> 00:12:47,880 for some let's say over n2 a dagger' n2 E to the bar plus iq2m2. 156 00:12:47,880 --> 00:12:53,603 And this corresponds to the Fourier-transform sort of reversing this 157 00:12:53,603 --> 00:12:58,035 relation. So now the commutator only cares about 158 00:12:58,035 --> 00:13:02,140 the operators. Oh, and this is a linear operation. 159 00:13:02,140 --> 00:13:05,332 So we can factor out in some sense the sum, and we can factor out these 160 00:13:05,332 --> 00:13:10,180 exponentials. So[UNKNOWN] commutation relation as so. 161 00:13:10,180 --> 00:13:16,257 So its going to be a sum where n one and n two e to the power i q two n two minus 162 00:13:16,257 --> 00:13:23,210 q one n one. In here I'm going to the commutator of a 163 00:13:23,210 --> 00:13:27,890 sub n 1a dagger sub n 2 and so if we now compare this canonical commutation 164 00:13:27,890 --> 00:13:32,960 relation with this guy we're going to see that this is nothing but delta symbol of 165 00:13:32,960 --> 00:13:37,952 n one and n two which means that it sort of extracts just one term out of this sum 166 00:13:37,952 --> 00:13:42,788 where n one is equal to n two and what we can do we can simply set m one equals to 167 00:13:42,788 --> 00:13:55,160 m two and lets say We will set n 1 equals n 2, and equals to some other n. 168 00:13:55,160 --> 00:13:57,320 And so, we're going to write it as follows. 169 00:13:57,320 --> 00:14:03,425 Is going to be equal to the sum, over n now, e to the power i, q 2 minus q 1, 170 00:14:03,425 --> 00:14:09,690 times n. And that's it, so the operators are gone. 171 00:14:09,690 --> 00:14:15,234 And now at this stage we're going to use finally this identity, the first identity 172 00:14:15,234 --> 00:14:21,310 that I just defined and we'll write the final results. 173 00:14:21,310 --> 00:14:25,204 So let me actually write it in red and its going to be equal to two pi delta 174 00:14:25,204 --> 00:14:30,650 function of q one minus q two, of q two minus q one. 175 00:14:30,650 --> 00:14:34,580 So it needs to be compared with this conical commutation relation. 176 00:14:34,580 --> 00:14:39,444 So and as you can see of course here this is in some since a continuum version of 177 00:14:39,444 --> 00:14:44,688 this conical commutation relation and this means actually that these guys a sub 178 00:14:44,688 --> 00:14:49,552 q one well actually a sub q and a sub dagger The two represent annihilation and 179 00:14:49,552 --> 00:14:54,644 creation operators, but what do these creation and annihilation operators 180 00:14:54,644 --> 00:15:03,383 create or destroy? So like a sub n and a sub n dagger, which 181 00:15:03,383 --> 00:15:08,128 sort of create local oscillations of a particular side, of a particular 182 00:15:08,128 --> 00:15:12,946 oscillator somewhere In the well defined position in space so these a's of q and 183 00:15:12,946 --> 00:15:20,285 a's of q dagger they anagolate and create waves if you want. 184 00:15:20,285 --> 00:15:26,316 Propogating through the oscillator chain. So these guys are not local in space They 185 00:15:26,316 --> 00:15:30,120 instead create waves with a certain wave vector. 186 00:15:30,120 --> 00:15:34,970 A cube basically is the wave vector. So here again now I'm using the units 187 00:15:34,970 --> 00:15:41,770 where the lengths are measured in the units of the oscillator lengths. 188 00:15:41,770 --> 00:15:46,469 So the distance between the neighboring. Equilibrium size but in any case so this 189 00:15:46,469 --> 00:15:50,922 is really the interpretation of what we are doing and now if we plug in this 190 00:15:50,922 --> 00:15:56,035 expression. This expression sent to Hamiltonian in 191 00:15:56,035 --> 00:16:00,610 the previous slide and use these identities and other simple algeabraic 192 00:16:00,610 --> 00:16:05,560 manipulations, we're going to arrive At the following expression for the same 193 00:16:05,560 --> 00:16:10,390 Hamiltonian. So this is actually, this is the very 194 00:16:10,390 --> 00:16:13,910 same Hamiltonian we started with but now rewritten in terms of this a sub q and 195 00:16:13,910 --> 00:16:17,265 sub q dagger and well its not at all obvious that this is the result and well 196 00:16:17,265 --> 00:16:20,950 this a and captial a and capital b, these coefficients here have this form and its 197 00:16:20,950 --> 00:16:27,598 not obvious at all. So this actually To derive this guy would 198 00:16:27,598 --> 00:16:33,660 be challenge for you in the bonus problem of your homework here.