1 00:00:00,025 --> 00:00:03,990 So in this video, we're going to complete the derivation of the spectrum or 2 00:00:03,990 --> 00:00:07,578 dispersion relation of the quantum phonons. 3 00:00:07,578 --> 00:00:11,778 And well, this part as well as the previous part is a bit technical, so I 4 00:00:11,778 --> 00:00:15,558 will actually skip some of the calculations and we'll just show you sort 5 00:00:15,558 --> 00:00:21,548 of the method, illustrate the method. We just called, by the way, in the 6 00:00:21,548 --> 00:00:26,296 Bogoliubov transformation that is used to solve such a problem. 7 00:00:26,296 --> 00:00:30,544 And as a side remark, let me mention that well this material that I'm going to be 8 00:00:30,544 --> 00:00:35,150 covering is actually a pretty advanced, gradient material. 9 00:00:35,150 --> 00:00:37,510 So if you've got to this point, you should be proud of yourself. 10 00:00:37,510 --> 00:00:41,305 This is very complicated, very advanced material of quantum antibody physics 11 00:00:41,305 --> 00:00:45,561 you're actually studying. So in this slide, I would like to review 12 00:00:45,561 --> 00:00:49,095 the calculations we have done so far, sort of tell you where we have been and 13 00:00:49,095 --> 00:00:55,427 also where we're going with this. And as we know, we started with this 14 00:00:55,427 --> 00:01:00,099 classic oscillator chain which is characterized by the coordinates and 15 00:01:00,099 --> 00:01:05,838 momentum for the atoms in this one-dimensional crystal. 16 00:01:05,838 --> 00:01:10,722 And the first step in putting together a quantum problem was to quantize this, 17 00:01:10,722 --> 00:01:15,804 which means basically to replace the variables x and t with the operators, and 18 00:01:15,804 --> 00:01:21,860 so, we just put hats on top of x, p, and h. 19 00:01:21,860 --> 00:01:26,004 So the following step involved expressing this momenta and coordinates with the 20 00:01:26,004 --> 00:01:29,980 corresponding creation and emulation operators, which led to this complicated 21 00:01:29,980 --> 00:01:34,885 looking Hamiltonian. And here, this an and am dagger 22 00:01:34,885 --> 00:01:41,047 correspond to creating an oscillation in a particular site with a particular atom 23 00:01:41,047 --> 00:01:47,750 labeled by m. So the next step that we took in the 24 00:01:47,750 --> 00:01:52,625 previous segment was essentially to Fourier transform this creation and 25 00:01:52,625 --> 00:01:57,500 anti-annihilation operators and to express them in the q space in the wave 26 00:01:57,500 --> 00:02:04,643 vector space if you want. And the corresponding Hamiltonian is 27 00:02:04,643 --> 00:02:09,725 shown here, so it has, as you can see, two types of terms, a dagger and a and 28 00:02:09,725 --> 00:02:15,406 this aq a minus q. And so, now, we have to work with these 29 00:02:15,406 --> 00:02:19,477 operators, these new creation, and annihilation operators, and this 30 00:02:19,477 --> 00:02:23,805 Hamiltonian. So and now I'm coming to where are we 31 00:02:23,805 --> 00:02:27,704 going with this? So if we did not have, let's imagine for 32 00:02:27,704 --> 00:02:32,820 a second that we do not have this term. Let's say we have just this term. 33 00:02:32,820 --> 00:02:37,250 So if this were the case, it would have meant that we already solved the problem. 34 00:02:37,250 --> 00:02:41,498 The reason for this being that this linear combination, the inevitable altura 35 00:02:41,498 --> 00:02:45,052 is a linear combination of this A dagger a's. 36 00:02:45,052 --> 00:02:50,010 independent a dagger a's, they sort of correspond to a collection of independent 37 00:02:50,010 --> 00:02:55,600 harmonic oscillators, the spectrum of which we know perfectly well. 38 00:02:55,600 --> 00:02:58,736 So for each q, we have a, basically a simple harmonic oscillator such as in the 39 00:02:58,736 --> 00:03:02,862 previous lecture. And we know its spectrum and now we just 40 00:03:02,862 --> 00:03:07,218 collect these oscillators and write them as an integral, but all in all, we know 41 00:03:07,218 --> 00:03:11,334 the results. So this is essentially already a 42 00:03:11,334 --> 00:03:13,971 solution. And therefore, this A of q or more 43 00:03:13,971 --> 00:03:17,939 precisely actually do to the omission conjugate part here twice A of q, would 44 00:03:17,939 --> 00:03:24,290 have been the spectrum of our problem, the spectrum of this quantum force. 45 00:03:24,290 --> 00:03:28,760 But, of course, we do have this true. We do have this true. 46 00:03:28,760 --> 00:03:30,700 And we don't know what to do with this, right? 47 00:03:30,700 --> 00:03:33,625 So this, so the spectrum of this Hamiltonian is not at all obvious. 48 00:03:33,625 --> 00:03:40,217 So a way to solve the problem now is to actually to construct a new set of 49 00:03:40,217 --> 00:03:47,680 creation and annihilation process. So here, I have an operator B which is an 50 00:03:47,680 --> 00:03:53,426 annihilation operator, can you type? And there's also a B dagger of q, and so 51 00:03:53,426 --> 00:03:58,532 lets assume that we can construct this B and B dagger as a linear combination of a 52 00:03:58,532 --> 00:04:06,380 and a dagger in such a way that the new Hamiltonian won't have the bad terms. 53 00:04:06,380 --> 00:04:09,914 By bad terms, I mean the terms that we don't know how to deal with and instead 54 00:04:09,914 --> 00:04:13,790 we'll have only this standard combination of b dagger and b for which the spectrum 55 00:04:13,790 --> 00:04:19,371 is obvious. And so this essentially would mean to 56 00:04:19,371 --> 00:04:23,673 solve the problem. The ability to write the Hamiltonian in 57 00:04:23,673 --> 00:04:27,333 this form, in the form of a liner combination of some creation and 58 00:04:27,333 --> 00:04:31,359 annihilation operators of this sort, so would mean to solve the problem in this 59 00:04:31,359 --> 00:04:38,330 why's this a tilde would give us the dispersion relation of these bs. 60 00:04:38,330 --> 00:04:42,908 Which actually are the quantum phonons. So as we will see, so this b dagger hat 61 00:04:42,908 --> 00:04:46,940 and b ha, hat, then correspond to creation, and anagolation operators of 62 00:04:46,940 --> 00:04:51,480 quantum phonons. Essentially, the normal modes, the 63 00:04:51,480 --> 00:04:55,040 quantized normal modes, in this oscillator chain. 64 00:04:55,040 --> 00:04:58,368 But together is well, as you can see, is rather nontrivial for us, we have to do 65 00:04:58,368 --> 00:05:01,748 this quantization and then, well, which was simple, but then we have to do with 66 00:05:01,748 --> 00:05:06,220 this 3 Linear transformations to get to this point. 67 00:05:06,220 --> 00:05:09,570 And this last linear transformation is what is actually known as the Bugoliubov 68 00:05:09,570 --> 00:05:13,952 of the transfrmation. So here's a picture of a young Nikolai 69 00:05:13,952 --> 00:05:21,370 Bogoliubov who was a very influential and talentented mathermatical physicist. 70 00:05:21,370 --> 00:05:25,970 A ration mathermatical physicist. And he was an author of many very 71 00:05:25,970 --> 00:05:30,794 important works and also he came up with a number of very important and practical 72 00:05:30,794 --> 00:05:36,850 mathematical methods that are used up to these days. 73 00:05:36,850 --> 00:05:40,100 And one of these methods is this Bogoliubov transformation that we are 74 00:05:40,100 --> 00:05:44,660 going to be studying. And let me mention that this Bogoliubov 75 00:05:44,660 --> 00:05:49,550 transfomation is still used very actively in research papers. 76 00:05:49,550 --> 00:05:54,900 I myself have used it in a number of of my published works. 77 00:05:54,900 --> 00:05:58,987 And this is basically in some sense an everyday tool of theories working In 78 00:05:58,987 --> 00:06:03,135 quantum physics and then condensed matter physics as well as perhaps high energy 79 00:06:03,135 --> 00:06:08,712 physics. So this is an important and important 80 00:06:08,712 --> 00:06:13,068 method but well in the example that we're going to be considering to be sort of 81 00:06:13,068 --> 00:06:17,490 method is rather straight forward and I want to get you advertised, sort of a 82 00:06:17,490 --> 00:06:22,110 motivation for using this method so lets lets look again at this Hamiltonian that 83 00:06:22,110 --> 00:06:27,060 we have for the Original the creation and annihilation operators a sub q dagger and 84 00:06:27,060 --> 00:06:35,160 a sub q. So and again, as I said, so our sort of 85 00:06:35,160 --> 00:06:39,420 strategy would be to correct in some sense, the definition of the creation and 86 00:06:39,420 --> 00:06:46,220 anogolation operators, so that we can get rid of this unwanted true. 87 00:06:46,220 --> 00:06:49,652 And, if you look at, at this Hamiltonian, you will see that this Hamiltonian 88 00:06:49,652 --> 00:06:53,084 involves essentially, so if we didn't have this term again, so we would have 89 00:06:53,084 --> 00:06:58,460 had just a linear combination of independent q operators. 90 00:06:58,460 --> 00:07:03,320 But this term, the appearance of this term couples the q mode with minus q 91 00:07:03,320 --> 00:07:07,306 mode. So whatever transformation we come up 92 00:07:07,306 --> 00:07:12,124 with, it should probably involve a linear combination of at least a q and a minus 93 00:07:12,124 --> 00:07:16,743 q. So let me just preset a form, sort of 94 00:07:16,743 --> 00:07:21,890 knowing the result of course is easy but it will save us some time. 95 00:07:21,890 --> 00:07:25,796 So let me just present a form Well if this generic Bogoliubov transform that 96 00:07:25,796 --> 00:07:31,234 accomplishes what I want. Namely that it introduces new operators 97 00:07:31,234 --> 00:07:36,996 in this case these are these operators b sub q and b daggers of q which are linear 98 00:07:36,996 --> 00:07:41,726 combination of a and a dagger or vice versa a and a dagger are linear 99 00:07:41,726 --> 00:07:51,482 combinations of b and b-dagger is here. And so here I will assume that this 100 00:07:51,482 --> 00:07:56,130 assumption will be verified, sort of a posteriori. 101 00:07:56,130 --> 00:07:59,480 And that, let me assume that this, these coefficients can be chosen to be real 102 00:07:59,480 --> 00:08:05,790 numbers well, functions better to say. And that there even functions of q. 103 00:08:05,790 --> 00:08:10,680 So, u of q is equal to u of minus q. So in this case actually this second 104 00:08:10,680 --> 00:08:14,767 definition here is redundant because the second definition of the creation 105 00:08:14,767 --> 00:08:18,671 operator can be obtained from the Definition of the annihilation operator, 106 00:08:18,671 --> 00:08:24,500 simply by applying the permission conjugation. 107 00:08:24,500 --> 00:08:28,420 But in any case, so, let us assume that we can construct such operator. 108 00:08:28,420 --> 00:08:32,202 So the first thing we need to do, and this is, I'm basically describing the 109 00:08:32,202 --> 00:08:36,480 strategy of constructing Bogoliubov transforms Is to make sure that this b 110 00:08:36,480 --> 00:08:41,680 and b hat ID creation and anti-annihilation operators. 111 00:08:41,680 --> 00:08:44,914 So, I can call them such, but actually, this should have a precise mathematical 112 00:08:44,914 --> 00:08:47,840 meaning. And the precise mathematical meaning 113 00:08:47,840 --> 00:08:52,434 would be that this guy. So, let's see if I have b of sum q and b 114 00:08:52,434 --> 00:08:58,380 of b, let's say dagger. So these guys should be the commutator of 115 00:08:58,380 --> 00:09:02,100 this guy should be zero if q is not equal to b and it should be a delta function 116 00:09:02,100 --> 00:09:06,200 other wise. So infinite otherwise in some sense. 117 00:09:06,200 --> 00:09:10,900 So this what I'm looking for. So my a and a dagger, so the reason I 118 00:09:10,900 --> 00:09:15,625 call the creation and annihilation operators was that they satisfied this 119 00:09:15,625 --> 00:09:19,780 relation. Now I want the new operators which are 120 00:09:19,780 --> 00:09:23,980 actually a linear combination of those to satisfy the same commutation relation so 121 00:09:23,980 --> 00:09:27,700 and this would involve a calculation which would result in a constraint on 122 00:09:27,700 --> 00:09:34,386 these coefficients u and v. So if I calculate this commutation 123 00:09:34,386 --> 00:09:39,322 relation you already know. So this is a the commutator of a sub q 124 00:09:39,322 --> 00:09:45,119 and a dagger sub b. So i simply plug in these definition to 125 00:09:45,119 --> 00:09:51,819 here and this guy to here with replacing q with p and I just calculate the 126 00:09:51,819 --> 00:09:58,754 commutators term by term. So what I can write, I can just write so 127 00:09:58,754 --> 00:10:01,790 we'll have one two three four different terms. 128 00:10:01,790 --> 00:10:06,068 So the first term lets say will be the commutator of these two guys so such is 129 00:10:06,068 --> 00:10:10,829 here and using this constraint I will require that the commutator of these two 130 00:10:10,829 --> 00:10:15,521 guys will be two pi delta function of q minus p Now these two commutators which 131 00:10:15,521 --> 00:10:19,799 result in this term should give me zero because these both are action and 132 00:10:19,799 --> 00:10:28,785 emulation operators. So likewise these two guys resulting in 133 00:10:28,785 --> 00:10:33,360 this term should give me zero as well because these are commutators of the 134 00:10:33,360 --> 00:10:38,962 creation operator. And so finally I have a term which 135 00:10:38,962 --> 00:10:43,252 couples b dagger of minus q and b of minus b and well since the commutator of 136 00:10:43,252 --> 00:10:47,476 b b dagger is equal to well this delta function the commutator of b dagger and b 137 00:10:47,476 --> 00:10:51,502 is basically minus the commuator of b and b dagger so it should give me this 138 00:10:51,502 --> 00:10:58,240 function. And let me remind you that, the delta 139 00:10:58,240 --> 00:11:01,750 function volt here, is actually an even function, so it doesn't really matter 140 00:11:01,750 --> 00:11:07,614 that I have here minus signs, etcetera. So, and if I put everything together, so 141 00:11:07,614 --> 00:11:11,683 these guys basically go away, right. So these guys go away, and I have these 142 00:11:11,683 --> 00:11:15,967 two terms, which essentially all involve the same Delta functions and also since 143 00:11:15,967 --> 00:11:19,684 there are delta functions here, Q is actually pint to piece so the delta 144 00:11:19,684 --> 00:11:23,527 function is zero if Q is not equal to P and so there are only terms, here I can 145 00:11:23,527 --> 00:11:27,055 simply write it as let's say UQ U of Q squared and this guy will V of two 146 00:11:27,055 --> 00:11:35,626 squared Now with a minus sign. So if I put everything together from this 147 00:11:35,626 --> 00:11:40,982 expression so this is what I get, okay. And now I have to required that the well 148 00:11:40,982 --> 00:11:46,610 the right hand side reproduce what is supposed to be in the left hand side. 149 00:11:46,610 --> 00:11:49,880 That is this two pi delta function of q minus p. 150 00:11:49,880 --> 00:11:52,690 So which means that this coefficient here. 151 00:11:52,690 --> 00:11:56,350 So this u squared minus v squared must be identically equal to one and this 152 00:11:56,350 --> 00:11:59,890 introduces a constraint, an important constraint on this Bogoliubov 153 00:11:59,890 --> 00:12:06,268 transformation. So basically what we are doing here again 154 00:12:06,268 --> 00:12:09,864 so let me sort of go back to where we are going with this, so we want to put 155 00:12:09,864 --> 00:12:15,829 together this new operators. And so we want to find explicitly the 156 00:12:15,829 --> 00:12:20,169 expressions for these u sub q and v sub q, that first of all wouldn't re-, will, 157 00:12:20,169 --> 00:12:24,633 would ensure that these b and b dagger nd that annihilation and creation operators 158 00:12:24,633 --> 00:12:30,481 of something. And second another constraint on this 159 00:12:30,481 --> 00:12:34,251 this guise would be that the resulting Hamiltonian now expressed in terms of b 160 00:12:34,251 --> 00:12:39,400 and b dagger would be simple enough so that we know its solution. 161 00:12:39,400 --> 00:12:42,690 And this basically the essence of the Bogoliubov transformation. 162 00:12:42,690 --> 00:12:45,777 In this case this Bogoliubov transformation is relatively simple, and 163 00:12:45,777 --> 00:12:49,158 actually most cases you will see in the literature, even the research literature, 164 00:12:49,158 --> 00:12:53,780 it is relatively simple, but in general, it may be more complicated. 165 00:12:53,780 --> 00:12:57,401 So it doesn't necessarily have to include just two, two, two, two terms, it may 166 00:12:57,401 --> 00:13:01,555 include more terms. But in all these cases, so the strategy 167 00:13:01,555 --> 00:13:05,145 will be the same. So basically to ensure that the operators 168 00:13:05,145 --> 00:13:09,114 you introduce are indeed creation annihilation operators and second then 169 00:13:09,114 --> 00:13:14,524 they simply our Hamiltonian. So now we can sort keep this constraint 170 00:13:14,524 --> 00:13:19,210 in mind or we can resolve it once and for all by just using new barometrization if 171 00:13:19,210 --> 00:13:25,718 you want of this u and v. So and this can be achieved by using the 172 00:13:25,718 --> 00:13:30,666 following mathematical identities. So if we have, if we use, the if we 173 00:13:30,666 --> 00:13:34,690 introduce u of q being a hyperbolic cosine. 174 00:13:34,690 --> 00:13:39,204 So let me remind you that the hyperbolic cosine x is equal to e to the power x 175 00:13:39,204 --> 00:13:43,866 plus e to the power minus x divided by two and the hyperbolic sign of x is equal 176 00:13:43,866 --> 00:13:50,320 to e super x, minus e super minus x, divided by 2. 177 00:13:50,320 --> 00:13:55,645 So, this is a real and sometimes real version of the sign and cosine in, in 178 00:13:55,645 --> 00:14:00,260 which case I would have had And the metric[UNKNOWN] but the main case of this 179 00:14:00,260 --> 00:14:04,662 guy's so satisfy the following identity which is sort of an analogue of the 180 00:14:04,662 --> 00:14:09,348 identity that you most of you know that sin squared plus cosine squared is equal 181 00:14:09,348 --> 00:14:16,522 to one. So here for the hyperbolic functions I 182 00:14:16,522 --> 00:14:20,274 have the hyperbolic cosine squared minus hyperbolic sin squared is equal to one 183 00:14:20,274 --> 00:14:24,601 and this is exactly the identity that I want For my ??? 184 00:14:24,601 --> 00:14:29,728 if I use this ??? we should resolve this identity once and 185 00:14:29,728 --> 00:14:33,427 for all. Which basically reduces the problem with 186 00:14:33,427 --> 00:14:36,960 2 parameter, to a problem with a single parameter. 187 00:14:36,960 --> 00:14:41,436 In this case this is [UNKNOWN]. So my sort of expression for the creation 188 00:14:41,436 --> 00:14:46,160 and angulation operators. A and a dagger now becomes this ok? 189 00:14:46,160 --> 00:14:49,923 So I have a of q is equal to cosine hyperbolic of Lambda b q plus sin 190 00:14:49,923 --> 00:14:56,425 hyperbolic of Lambda b of minus q dagger and the corresponding a dagger. 191 00:14:56,425 --> 00:15:01,737 Alright so now what needs to be done is that we ahve to just do a brute force 192 00:15:01,737 --> 00:15:06,394 analysis. We have just go ahead and plug in this 193 00:15:06,394 --> 00:15:11,086 expression this expression sent to the Hamiltonian, and I will leave it for you 194 00:15:11,086 --> 00:15:16,740 for those of you who are interested. For those of you for instance who are 195 00:15:16,740 --> 00:15:20,030 interested in becoming theorists, or who are theorists, or who are mathematicians 196 00:15:20,030 --> 00:15:22,850 or are just curious about the technical part so you can go ahead and do the 197 00:15:22,850 --> 00:15:25,717 simple, relatively straight, at least I'm not going to call it simple, its a 198 00:15:25,717 --> 00:15:31,450 straightforward exercise of just plugging in lets say this and this guy. 199 00:15:31,450 --> 00:15:35,350 Here we're going to put this linear combination, instead of this guy we're 200 00:15:35,350 --> 00:15:39,445 going to put this linear combonation, etc and so if you do so well you will see 201 00:15:39,445 --> 00:15:46,340 some unpleasant algebra but this is an exercise I'm sure you can do it. 202 00:15:46,340 --> 00:15:49,626 So here I'm showing the result of the calculation, if you did everything 203 00:15:49,626 --> 00:15:53,124 correctly this is what you should get and the result here is a Hamiltonian which 204 00:15:53,124 --> 00:15:57,300 now is expressed in terms of these new operators. 205 00:15:57,300 --> 00:16:01,407 B of q and b dagger of q. And as you can see it has the form which 206 00:16:01,407 --> 00:16:05,226 is pretty much the same as we had before just with some new coefficients which I 207 00:16:05,226 --> 00:16:09,102 call now i a tilde and b tilde and here is the expressions for these coefficients 208 00:16:09,102 --> 00:16:12,921 involving hyperbolic cosines of two Lamba and hyperbolic sins of two Lambda and 209 00:16:12,921 --> 00:16:21,120 these are the regional coefficients. And the Hamiltonian for A's. 210 00:16:21,120 --> 00:16:24,684 And so, you may ask me what, you know, why did we bother to you know, introduce 211 00:16:24,684 --> 00:16:28,960 new operators, if we got exactly the same thing? 212 00:16:28,960 --> 00:16:32,092 So, well, the answer to this, and actually this brings me to the second 213 00:16:32,092 --> 00:16:35,872 step in this Bogoliubov transformation is that well lambda here is a free 214 00:16:35,872 --> 00:16:40,230 parameter, so, better to say a free fun, function. 215 00:16:40,230 --> 00:16:44,120 We can choose this function Anyway we see fit. 216 00:16:44,120 --> 00:16:47,618 In particular we can choose it such that this unwanted term, the bad term that we 217 00:16:47,618 --> 00:16:51,050 don't want to see is identically equal to zero. 218 00:16:51,050 --> 00:16:54,590 So we can esssentially can set by hand this b tilde of q equal to zero which 219 00:16:54,590 --> 00:16:59,575 leads to an equation for lambdance. So this equation as you can see from here 220 00:16:59,575 --> 00:17:03,095 if we just move this guy to the, to the right-hand side and divide by cosine, so 221 00:17:03,095 --> 00:17:08,160 sine hyperbolic over cosine hyperbolic is tangent hyperbolic. 222 00:17:08,160 --> 00:17:15,349 So the tangent hyperbolic of 2 lambda of Q, is equal to minus B of Q versus over A 223 00:17:15,349 --> 00:17:20,602 of Q. And so this essentially gives us a 224 00:17:20,602 --> 00:17:24,880 specific result for this Lambda of q which ensures that our Hamiltonian in 225 00:17:24,880 --> 00:17:30,800 terms of this new creation, annihilation operators looks simple. 226 00:17:32,050 --> 00:17:36,576 So now let me clean it up a little bit and sort of summarize things again. 227 00:17:36,576 --> 00:17:40,175 So this is the result that we got and also let me remind you that this a of q 228 00:17:40,175 --> 00:17:44,650 and b of q, the regional coefficients we already know. 229 00:17:44,650 --> 00:17:47,914 So we derived them in the previous part of this lecture and so let me remind you 230 00:17:47,914 --> 00:17:51,680 of their explicit forms. So here is their explicit form. 231 00:17:51,680 --> 00:17:55,450 So this explicitly b of q and this is explicitly a of q and well I wrote it 232 00:17:55,450 --> 00:17:59,415 sort of explicitly this way but as you can see lot of things here cancel out 233 00:17:59,415 --> 00:18:06,530 including this H omega and there loads of other simplifications involved. 234 00:18:06,530 --> 00:18:10,436 And if you play around with this, well, with this The coefficients if you do a 235 00:18:10,436 --> 00:18:14,404 little bit of algaebra you will see that well first of all Lambda of q actually 236 00:18:14,404 --> 00:18:20,623 determines as you can see from here. So this guy is zero as we wanted it to be 237 00:18:20,623 --> 00:18:25,350 but now this guy is determined by Lambda and we do know Lambda. 238 00:18:25,350 --> 00:18:30,138 So we can simply plug it Y'know in here and if you, once everything is said and 239 00:18:30,138 --> 00:18:37,200 done you actually has an expilict expression for this a tilde of q. 240 00:18:37,200 --> 00:18:40,640 Which becomes in the end of the day surprisingly simple. 241 00:18:40,640 --> 00:18:43,630 So again this calculation is not obvious from here. 242 00:18:43,630 --> 00:18:47,650 It requires a little bit of work and due to the format of the lectures And lack of 243 00:18:47,650 --> 00:18:51,370 time I'm not presenting it but those of you who are interested can of course 244 00:18:51,370 --> 00:18:58,390 check that this is indeed the result. So this is by construction but a of q has 245 00:18:58,390 --> 00:19:02,121 this form. So if we plot it so this a of q has this 246 00:19:02,121 --> 00:19:06,606 form which is actually a very familiar form so this is a of q which we have seen 247 00:19:06,606 --> 00:19:10,953 before in the classical problem in the context of acoustic forms and this is 248 00:19:10,953 --> 00:19:18,050 exactly what they are but now they are quantum formula. 249 00:19:18,050 --> 00:19:21,625 So sort of to summarize the result and also to emphasize a very important point 250 00:19:21,625 --> 00:19:26,280 so this is what we actually got. We have found explicitly a series of 251 00:19:26,280 --> 00:19:30,768 linear transformations Of the regional Hamiltonian, of the regional operators 252 00:19:30,768 --> 00:19:36,020 involved with the Hamiltionian. So I should have the Hamiltonian can be 253 00:19:36,020 --> 00:19:41,078 sort of was casted in this form. Which is the so called diagonal form of 254 00:19:41,078 --> 00:19:44,502 the Hamiltonian. Sometimes if you happen to read research 255 00:19:44,502 --> 00:19:49,785 papers, the reasearchers call about diagonalization of a Hamiltonian. 256 00:19:49,785 --> 00:19:52,912 So today in lies the Hamiltonian, and, and in sort of in the problem, in a 257 00:19:52,912 --> 00:19:56,622 problem which involves mainly quantum particles, essentially means, the ability 258 00:19:56,622 --> 00:20:00,067 to write it, as a linear combination of independent oscillators or some other 259 00:20:00,067 --> 00:20:05,970 particles, the sort of behavior of which we know very well. 260 00:20:05,970 --> 00:20:10,154 And so, this is exactly what we accomplished here. 261 00:20:10,154 --> 00:20:16,189 And so here we have also this Hermitian conjugate, so if we couple it, it will be 262 00:20:16,189 --> 00:20:22,139 simply the same term, so we can write it integral over q, well twice A tilde of q 263 00:20:22,139 --> 00:20:29,384 b dagger q b q. So, and well, I should also mention one 264 00:20:29,384 --> 00:20:33,532 thing that I sort of swept under the rug and this thing is that I haven't paid 265 00:20:33,532 --> 00:20:37,924 attention to constants that appear in the Hamiltonian so, there could be constant 266 00:20:37,924 --> 00:20:43,569 here. If I had done it, so the correct result 267 00:20:43,569 --> 00:20:47,349 that I would have found is well, this is correct, but there also is a constant 268 00:20:47,349 --> 00:20:52,060 term here, which corresponds to zero point motion. 269 00:20:52,060 --> 00:20:55,895 And this term happens to be exactly one half, so if you remember the solution to 270 00:20:55,895 --> 00:20:59,907 the simple harmonic oscillator, we have essentially h omega and plus one half and 271 00:20:59,907 --> 00:21:06,089 so similarly it appears here. So in this two a tilde of q is exactly 272 00:21:06,089 --> 00:21:12,432 the energy modes of the oscillator. So we can call it so this guy is equal to 273 00:21:12,432 --> 00:21:17,256 h m n of q and this would be the spectrum of the quantum phonons, and this is as 274 00:21:17,256 --> 00:21:22,264 presented here. So the spectrum of the quantum phonons 275 00:21:22,264 --> 00:21:25,388 here is twice omega. Omega here, is the frequency of a single 276 00:21:25,388 --> 00:21:29,066 oscillation in some sense. In the original classical picture it 277 00:21:29,066 --> 00:21:32,258 would have been the frequency, the natural frequency of this spring 278 00:21:32,258 --> 00:21:36,898 connecting the atoms in some sense. And so, the remarkable thing, that I want 279 00:21:36,898 --> 00:21:39,859 to emphasize in the very end, and this will be the end of the, today's lecture 280 00:21:39,859 --> 00:21:43,102 is that, it turns out that this spectrum of quantum phonons, is actually identical 281 00:21:43,102 --> 00:21:48,533 to the spectrum of classical phonons. There is absolutely no difference 282 00:21:48,533 --> 00:21:51,418 whatsoever. So if you go back to the problem that we 283 00:21:51,418 --> 00:21:55,636 solved earlier today, in today's lecture, so namely the classical problem, so it 284 00:21:55,636 --> 00:21:59,056 was slightly more complicated in the sense that it involved two types of 285 00:21:59,056 --> 00:22:02,647 atoms, but if we have just one type of atom so this is the result which we would 286 00:22:02,647 --> 00:22:08,650 of got. So an important conclusion of this sort 287 00:22:08,650 --> 00:22:12,610 of discovery that quantum and classical phonons turn out to be exactly the same 288 00:22:12,610 --> 00:22:16,570 is the following so we can say that we could have quantized our problem at any 289 00:22:16,570 --> 00:22:21,534 stage. So in the solution I present it we did so 290 00:22:21,534 --> 00:22:25,900 at the very beginning. So we essentially converted the original 291 00:22:25,900 --> 00:22:30,054 coordinates and momento of the atoms into quantum operators, x hat sub n and p hat 292 00:22:30,054 --> 00:22:33,774 sub n and then we went ahead with this quantum calculation with the newer 293 00:22:33,774 --> 00:22:39,000 transform. But instead of doing that sort of there 294 00:22:39,000 --> 00:22:43,944 is a second version of the solution. One could have solved the problem, the 295 00:22:43,944 --> 00:22:48,105 classical problem first and which means finding the trajectories of particles and 296 00:22:48,105 --> 00:22:52,952 the corresponding waves. And so one who could have quantized this 297 00:22:52,952 --> 00:22:57,110 normal mode. So basically so the normal modes can be 298 00:22:57,110 --> 00:23:02,470 converted then into this Essentially operators, h omega q, sum b dagger q, b q 299 00:23:02,470 --> 00:23:09,360 plus one half. So we could have done that. 300 00:23:09,360 --> 00:23:12,556 And the results, basically we have proven that the result are, are exactly the 301 00:23:12,556 --> 00:23:15,760 same. And this is very important. 302 00:23:15,760 --> 00:23:19,192 It shows us something about the quantum to classical correspondence, and this 303 00:23:19,192 --> 00:23:22,676 quantum to classical correspondence sort of persists Also in more complicated 304 00:23:22,676 --> 00:23:25,838 problem. Let's say problems in three dimensions 305 00:23:25,838 --> 00:23:29,290 with more degrees of freedom. With different types of atoms. 306 00:23:29,290 --> 00:23:32,700 And this is true as long as we have non-interacting theory, which in the 307 00:23:32,700 --> 00:23:36,275 language of this creation and annihilation operators essentially means 308 00:23:36,275 --> 00:23:41,600 a theory which is bi-linear in this b's and b daggers. 309 00:23:41,600 --> 00:23:45,382 So as long as we have this bi-linear forms we're fine with this quantum two 310 00:23:45,382 --> 00:23:49,598 plus equal correspondents but once we have y'know more complicated nonlinear 311 00:23:49,598 --> 00:23:54,700 terms, so this situation becomes much more complicated. 312 00:23:54,700 --> 00:23:57,625 So the last thing I'm going to mention is that the appearance of such terms, which 313 00:23:57,625 --> 00:24:00,730 we're not going to discuss at all but you know if we had this quadratic terms, b 314 00:24:00,730 --> 00:24:06,318 dagger b dagger b b for instance. So they would have meant that the photons 315 00:24:06,318 --> 00:24:10,626 interact with one another. So in some sense it would have responded 316 00:24:10,626 --> 00:24:16,550 to scattering of waves off of each other, waves running through the crystal. 317 00:24:16,550 --> 00:24:20,171 And this is a very complicated challenge to understand the behavior of such an 318 00:24:20,171 --> 00:24:24,667 interacting system. And there is still on going theoretical 319 00:24:24,667 --> 00:24:29,232 challenges in describing it. So it's definitely beyond the scope of 320 00:24:29,232 --> 00:24:33,324 this course, but this theory is well understood and I hope some of you got an 321 00:24:33,324 --> 00:24:39,840 idea about phonons and collective modes and classical and quantum physics.