1 00:00:00,950 --> 00:00:04,982 So now, we're going to address the same question we addressed in the previous 2 00:00:04,982 --> 00:00:08,620 video. namely the question about the existence 3 00:00:08,620 --> 00:00:12,715 of a bound state in an arbitrarily weak shallow potential wells but now will do 4 00:00:12,715 --> 00:00:18,186 so in more than one dimension. I should mention that there are actually 5 00:00:18,186 --> 00:00:22,260 very few problems in quantum mechanics in 2D and 3D. 6 00:00:22,260 --> 00:00:26,19 We should allow relatively simple analytical solution. 7 00:00:26,19 --> 00:00:30,177 Most often in these cases we have to resolve to either numerical analysis or 8 00:00:30,177 --> 00:00:35,660 rather complicated, technically complicated mathematical calculations. 9 00:00:35,660 --> 00:00:39,224 Now, the solution I am going to present is not exactly trivial but will require a 10 00:00:39,224 --> 00:00:43,37 lot of thinking. But otherwise, it will follow almost one 11 00:00:43,37 --> 00:00:46,169 to one the steps that we used in the solution of the one dimensional 12 00:00:46,169 --> 00:00:50,858 delta-potential. however very importantly, the result of 13 00:00:50,858 --> 00:00:54,970 this steps is going to be quite different in 2D and 3D. 14 00:00:54,970 --> 00:00:58,610 So namely for those of you who want to skip the technical part and just to get 15 00:00:58,610 --> 00:01:03,900 through the bottom lines, I will just tell you the results right away. 16 00:01:03,900 --> 00:01:07,248 So unlike in 1 dimension, where the energy of a bound state, first of all the 17 00:01:07,248 --> 00:01:10,866 bound state always exists and the energy of the bound state in the weak potential 18 00:01:10,866 --> 00:01:16,325 scales sort of as a power law. The second power law of the strength of 19 00:01:16,325 --> 00:01:21,80 the potential, in 2 dimensions, it turns out that the level also always exists. 20 00:01:21,80 --> 00:01:25,590 But the energy of the level is extremely extremely small. 21 00:01:25,590 --> 00:01:29,738 The level is very shallow and the energy of the level scales exponentially in a 22 00:01:29,738 --> 00:01:34,547 very unusual form with the strength of the potential. 23 00:01:34,547 --> 00:01:38,268 And this actually has very important consequences in many cases and in 24 00:01:38,268 --> 00:01:42,477 particular and it release to certain calculation we're going to discuss in the 25 00:01:42,477 --> 00:01:47,230 theory of super conductivity a little later. 26 00:01:47,230 --> 00:01:50,915 And also in three dimensions it turns out that there is no level at all so if you 27 00:01:50,915 --> 00:01:54,810 have a very weak attraction between particles. 28 00:01:54,810 --> 00:01:58,506 For example, a very weak potential well, there is, so these attractions, these 29 00:01:58,506 --> 00:02:02,314 weak attractions are not going to induce a bond state and this fact is also very 30 00:02:02,314 --> 00:02:07,768 important. Now, let me prove the statements I just 31 00:02:07,768 --> 00:02:11,548 made, and the proof, the solution is going to follow very closely as I already 32 00:02:11,548 --> 00:02:15,840 mentioned, the solution that we had in one dimension. 33 00:02:15,840 --> 00:02:19,340 We started from this Schrodinger equation for the delta potential, so now of course 34 00:02:19,340 --> 00:02:22,690 we have to deal with this Schrodinger equation, in higher dimension, where wave 35 00:02:22,690 --> 00:02:27,920 function now is the wave function, is a function of a vector r. 36 00:02:27,920 --> 00:02:31,660 And the kinetic energy now involves this low placement , and, and counters to this 37 00:02:31,660 --> 00:02:35,565 derivative over x, and finally the delta function is the delta function is sort of 38 00:02:35,565 --> 00:02:40,420 a sharply big function in thee d dimensional space. 39 00:02:40,420 --> 00:02:45,938 So let's say in 3D, so, this delta so far is simply the product of the delta 40 00:02:45,938 --> 00:02:53,470 function facts, delta function of y, and delta function of z. 41 00:02:53,470 --> 00:02:58,420 So the first step in the solution, just like in 1 z, involves retransforming our 42 00:02:58,420 --> 00:03:02,216 equation. So essentially we go from the way 43 00:03:02,216 --> 00:03:06,504 function and the real space to the image of the way function momental space or 44 00:03:06,504 --> 00:03:11,208 more precisely here is a function of the way key. 45 00:03:11,208 --> 00:03:14,600 So the only difference in entire dimension is that the Fourier transform 46 00:03:14,600 --> 00:03:19,15 is slightly more complicated so it involves Fourier transforms along. 47 00:03:19,15 --> 00:03:22,248 Each of the directions, x, y, z, etcetera, so if you're in higher 48 00:03:22,248 --> 00:03:26,155 dimensions. And so if we perform this slightly more 49 00:03:26,155 --> 00:03:30,847 complicated Fourier transform so the kinetic energy which is basically this 50 00:03:30,847 --> 00:03:35,980 derivative of[INAUDIBLE] becomes minus k squared. 51 00:03:35,980 --> 00:03:40,336 The delta function, the Fourier image of the delta function, just like the one I 52 00:03:40,336 --> 00:03:45,15 mentioned is simply 1. And so we get to the following equation, 53 00:03:45,15 --> 00:03:50,360 which is just the linear equation before the image of the wave function. 54 00:03:50,360 --> 00:03:54,56 And as you see there's essentially no difference between one dimensional keys 55 00:03:54,56 --> 00:03:57,696 and D dimensional keys apart from from the fact that this guy now is the reactor 56 00:03:57,696 --> 00:04:03,370 and while this guy was just one dimensional component of the wave effort. 57 00:04:03,370 --> 00:04:09,230 And so again we can of course solve this equation for psi tilde of k. 58 00:04:09,230 --> 00:04:15,240 So we're going to have the same psi of 0 divided by h squared. 59 00:04:15,240 --> 00:04:21,202 K squared over 2m and I will write it as plus absolute value of the energy. 60 00:04:21,202 --> 00:04:25,37 Again, we are looking for a bound state, and we know that the bound state energy 61 00:04:25,37 --> 00:04:30,436 must be negative, if it exists at all. And using the exactly same trick as 62 00:04:30,436 --> 00:04:34,206 before, we're going to integrate this equation, both sides of this equation, 63 00:04:34,206 --> 00:04:40,710 now, over the entire D dimensional space of this wave action. 64 00:04:40,710 --> 00:04:45,966 So if we do so, we, come up with the self consistency equation which again in one d 65 00:04:45,966 --> 00:04:50,930 and in higher dimensions look almost the same. 66 00:04:50,930 --> 00:04:54,950 So the only difference is that the integral here is going to over a higher 67 00:04:54,950 --> 00:05:00,870 dimensional space. So it turns out that this difference the 68 00:05:00,870 --> 00:05:04,390 fact that we now integrating over high dimension space lead a very important 69 00:05:04,390 --> 00:05:08,764 consequences for the actual solution for this equation. 70 00:05:08,764 --> 00:05:11,988 Here, we knew that doesn't matter how small off way is as long as it's non 71 00:05:11,988 --> 00:05:16,430 zero, we can always find an energy that will satisfy this equation. 72 00:05:16,430 --> 00:05:20,34 So now, it' turns out that higher dimension is not going to happen, 73 00:05:20,34 --> 00:05:24,430 necessary for d greater than 2. There will be no solution. 74 00:05:24,430 --> 00:05:28,785 So to see this, we actually have to calculate this integral that appears in 75 00:05:28,785 --> 00:05:33,350 the cell consistency equation that we just derived. 76 00:05:33,350 --> 00:05:37,185 So to do so we should first know is that the function that we're actually 77 00:05:37,185 --> 00:05:41,345 integrating, does not depend on the directions of the wave action It only 78 00:05:41,345 --> 00:05:47,860 depends on this absolute value. So let's see in two dimensions we can 79 00:05:47,860 --> 00:05:53,332 write d 2 k as a d k times k which is essentially corresponds to an integrating 80 00:05:53,332 --> 00:06:00,478 over the absolute value of k. Times d 5 sub k, which is the angle that 81 00:06:00,478 --> 00:06:07,843 the vector k forms with the x direction. So let's see, this is our two dimensional 82 00:06:07,843 --> 00:06:13,200 k space, this would by k sub y, this would be k sub x. 83 00:06:13,200 --> 00:06:19,710 And so this is our vector k, so this would be the angle 5 sub k. 84 00:06:19,710 --> 00:06:24,130 So, and what I'm saying here is that the equation, this equation that we are 85 00:06:24,130 --> 00:06:31,154 looking at, so can be written as alpha d2k, in this case, over two pi squared. 86 00:06:31,154 --> 00:06:37,535 so this function, and we can write it as simply, well, alpha over 2 pi squared. 87 00:06:37,535 --> 00:06:41,410 Oops. Squared is here. 88 00:06:41,410 --> 00:06:45,915 So dk times k over this function are from zero to infinity. 89 00:06:45,915 --> 00:06:51,159 So we integrate all the possible reactors and then the integral over the angles but 90 00:06:51,159 --> 00:06:57,526 nothing dependent then , so this integral will simply give us the y. 91 00:06:57,526 --> 00:07:01,606 And so in 3D, we can do the same thing so in the new dimension, we can do this 92 00:07:01,606 --> 00:07:05,841 procedure. We can sort of integrate the end of those 93 00:07:05,841 --> 00:07:10,900 and just separate the integral over, the, magnitude of k. 94 00:07:10,900 --> 00:07:16,100 So let me rather for my, self consistency equation in this form. 95 00:07:16,100 --> 00:07:21,255 I will write it as a new dimension. It's alpha s sub d minus one, which I 96 00:07:21,255 --> 00:07:24,394 will, Explain what I mean by that. 97 00:07:24,394 --> 00:07:30,952 two pi, to the power of d. So in here I'm going to write it as d k 98 00:07:30,952 --> 00:07:37,540 to the power of g minus 1 h square k square over 2m plus absolute value of the 99 00:07:37,540 --> 00:07:45,492 energy. So this s sub d minus 1 is the integral 100 00:07:45,492 --> 00:07:52,703 that comes out of the angles. So in two dimensions, it will simply be 2 101 00:07:52,703 --> 00:07:54,126 pi. Okay? 102 00:07:54,126 --> 00:07:58,816 So, and I wrote, in this forum because it actually corresponds sort of to a D minus 103 00:07:58,816 --> 00:08:03,90 1 dimensional theories. In this case, 2 dimensions would be 104 00:08:03,90 --> 00:08:06,490 basically integrating over this circle, in 3 dimensions, it will be integrating 105 00:08:06,490 --> 00:08:10,120 over this solid angles that cover the entire sphere. 106 00:08:10,120 --> 00:08:15,628 So let's say as as 1 is equal to 2pi, which just found out as, s2 is going to 107 00:08:15,628 --> 00:08:20,550 be equal to 4pi. Well we can formally introduce as 0, 108 00:08:20,550 --> 00:08:25,17 which is just 2 et cetera. So many dimension in principle can just 109 00:08:25,17 --> 00:08:28,250 calculated. So for practical purposes, we don't 110 00:08:28,250 --> 00:08:32,510 really need to know anything about these three numbers because in real life we 111 00:08:32,510 --> 00:08:38,514 don't deal with anything about 3 dimensions, 2 dimensions or 1 dimension. 112 00:08:38,514 --> 00:08:42,534 Now, the problem we have that appears here in, already in 2 dimensions is that 113 00:08:42,534 --> 00:08:48,914 well when we integrate let's see. From zero to infinity, so the integral is 114 00:08:48,914 --> 00:08:54,294 actually diversion at large k. So let's see if these equal to 2 and 115 00:08:54,294 --> 00:08:58,386 especially if these equal to 3, so the higher up we go, the more diversion it 116 00:08:58,386 --> 00:09:03,456 becomes as we go to larger k. So this integral is actually equal to 117 00:09:03,456 --> 00:09:07,556 infinity. So, if we look at it sort of without 118 00:09:07,556 --> 00:09:11,350 thinking. And well, when there is infinity so we 119 00:09:11,350 --> 00:09:15,919 have to think about why it appears. So remember that actually this constant 120 00:09:15,919 --> 00:09:19,225 alpha came out, well we are solving essentially The problem for delta 121 00:09:19,225 --> 00:09:23,184 potential. So our v of r is alpha times delta of r 122 00:09:23,184 --> 00:09:30,151 and delta function there are the delta function is a very artificial function. 123 00:09:30,151 --> 00:09:35,109 It was basically introduced to model a shallow potential, a physical potential 124 00:09:35,109 --> 00:09:42,350 that has so has a certain with, with equality a and it has a certain depth. 125 00:09:42,350 --> 00:09:45,700 So which means which will be called, you know, u naught. 126 00:09:45,700 --> 00:09:50,398 So, and alpha is of the order of u naught e to the power of d, where d is the 127 00:09:50,398 --> 00:09:56,462 dimentionality of our space. But in any case, so this delta function 128 00:09:56,462 --> 00:10:01,82 potential is sort of introduce, represent real physical potential, that exists in 129 00:10:01,82 --> 00:10:06,590 reality, which has, actually, a finite width and finite depth. 130 00:10:06,590 --> 00:10:10,946 So when we go to a large key, large wave vectors, actually the large wave vectors 131 00:10:10,946 --> 00:10:15,716 respond to small distances. And if we, consider and if we, if we 132 00:10:15,716 --> 00:10:20,10 encounter a divergence at large K which we do here. 133 00:10:20,10 --> 00:10:25,340 So it means that something bad happens in small smaller, small distances. 134 00:10:25,340 --> 00:10:30,30 And so this something bad means well I mean there is no really, there are no 135 00:10:30,30 --> 00:10:36,218 real delta potentials in nature. Okay so all potentials that have a finite 136 00:10:36,218 --> 00:10:39,58 radiance. And so we can copy all of this in some 137 00:10:39,58 --> 00:10:43,145 sense and physical diversions by introducing a parameter 1 over e, where e 138 00:10:43,145 --> 00:10:46,927 is the radius of our real physical potential that we model with the delta 139 00:10:46,927 --> 00:10:51,775 potential. So this is actually a sort of a solution 140 00:10:51,775 --> 00:10:54,783 to this otherwise puzzle because otherwise, it would get infinity and it 141 00:10:54,783 --> 00:10:58,584 shouldn't get infinity. So basically what I'm saying here is that 142 00:10:58,584 --> 00:11:01,436 if you were to put a real potential in the very beginning, and do everything 143 00:11:01,436 --> 00:11:04,518 properly then this cut would appear actually naturally in our calculations 144 00:11:04,518 --> 00:11:09,80 but here we would just introduce it by hand. 145 00:11:09,80 --> 00:11:13,568 So now we're entering a, a final part of our derivation where we will rely on the 146 00:11:13,568 --> 00:11:18,756 integral we just derived. So here, there are no more angle angular 147 00:11:18,756 --> 00:11:23,320 components of the wave vector. There's just one integral involved. 148 00:11:23,320 --> 00:11:27,528 And we cut it off at the 1 over a. So and also, we modified it a little bit, 149 00:11:27,528 --> 00:11:29,860 so there's 1 over alpha. It's not a. 150 00:11:29,860 --> 00:11:32,389 It's alpha is the strength of the potential in the left hand side. 151 00:11:33,480 --> 00:11:37,344 So again, the main question we've been actually asking is whether there is a 152 00:11:37,344 --> 00:11:42,330 bound state in a weak potential. That is, in a potential where, where 153 00:11:42,330 --> 00:11:47,118 alpha goes to zero, and therefore, if alpha goes to zero, 1 over alpha goes to 154 00:11:47,118 --> 00:11:50,906 infinity. And so the question that we are now 155 00:11:50,906 --> 00:11:54,478 asking is whether we can find a solution to this equation when the left-hand side 156 00:11:54,478 --> 00:11:59,225 is very large. So you may ask me, you know, you just got 157 00:11:59,225 --> 00:12:02,510 rid of infinity , so what are you doing? You know? 158 00:12:02,510 --> 00:12:05,390 This is this is really would be reasonable questions let's say from a 159 00:12:05,390 --> 00:12:08,432 mathematician. But you know, this calculation actually 160 00:12:08,432 --> 00:12:11,792 illustrates in a very nice way the difference between doing mathematically 161 00:12:11,792 --> 00:12:16,896 calculation and physics and math. So well, the infinity I just got in rid 162 00:12:16,896 --> 00:12:21,10 off was an unphysical infinity, it's not real. 163 00:12:21,10 --> 00:12:25,554 It was sort of an artifact of our model. And also this infinity was not dependent 164 00:12:25,554 --> 00:12:28,500 on the energy. So whether or not we put energy to 0 or 165 00:12:28,500 --> 00:12:32,512 to for something finite, you know, these so called ultraviolet diversions would 166 00:12:32,512 --> 00:12:35,894 have existed. And so we cut it off, we sort of 167 00:12:35,894 --> 00:12:39,890 remembered where these things came from and got rid of these errors. 168 00:12:39,890 --> 00:12:43,602 Now, the question is whether we can have, and on arbitrary, large integral by 169 00:12:43,602 --> 00:12:48,813 adjusting the value of the energy, okay. And so here is basically this integral, 170 00:12:48,813 --> 00:12:53,947 so I don't write here at this overall coefficient because it's just a number. 171 00:12:53,947 --> 00:12:57,746 It can't be made large or small, it is just a number assigned. 172 00:12:57,746 --> 00:13:01,46 2D is 1 over 2 pi and 3D is 1 over 2 pi squared, it's just, because it's 173 00:13:01,46 --> 00:13:04,876 unimportant. So, how do we make how do we answer this 174 00:13:04,876 --> 00:13:07,277 question. So, if we look at it we see that the 175 00:13:07,277 --> 00:13:10,699 energy, absolute value of energy appears in the denominator, okay? 176 00:13:10,699 --> 00:13:13,430 So the larger the energy, the smaller this integral. 177 00:13:13,430 --> 00:13:18,830 It's very clear cut tendency. So these integral is the largest when 178 00:13:18,830 --> 00:13:23,730 energy's equal to zero. So therefore, the questions we're really 179 00:13:23,730 --> 00:13:27,78 asking is whether we can, you know, if we set this energy to zero, and the 180 00:13:27,78 --> 00:13:30,642 remaining integral will be k to the power of d minus 1 divided by k squared, or 181 00:13:30,642 --> 00:13:35,717 just simply this. So, the question we're asking whether the 182 00:13:35,717 --> 00:13:38,830 integral below can diverge. So we, we got rid of divergence in the 183 00:13:38,830 --> 00:13:41,406 previous slide. In this slide, we're asking whether we 184 00:13:41,406 --> 00:13:44,236 can get it back. But the important thing here is that we 185 00:13:44,236 --> 00:13:48,410 want to get it back not from a large momentum but from small momentum. 186 00:13:48,410 --> 00:13:52,322 So the latter is called infrared divergence. 187 00:13:52,322 --> 00:13:55,127 Okay. And in physics, for those of you who will 188 00:13:55,127 --> 00:13:58,20 be thinking about becoming theoretical physicists. 189 00:13:58,20 --> 00:14:01,125 You may remember that infrared diversions is if you find something like that well, 190 00:14:01,125 --> 00:14:04,230 either you made the mistake or you should if it's new, you should think about 191 00:14:04,230 --> 00:14:08,360 number of parameters. Okay, so it's very important. 192 00:14:08,360 --> 00:14:12,8 Infrared diversion is real unlike ultraviolet diversions which are sort of 193 00:14:12,8 --> 00:14:16,163 artificial. So and, this boils down basically this 194 00:14:16,163 --> 00:14:19,567 question. Now, clearly if if let's say d is equal 195 00:14:19,567 --> 00:14:24,875 to 3, so the dimension of space is equal to 3, the space in which we live. 196 00:14:24,875 --> 00:14:28,657 S then we simply have a integral from 0 to 1 or a, or where d k is finite, there 197 00:14:28,657 --> 00:14:33,470 is no diversions. And therefore the answer to the question 198 00:14:33,470 --> 00:14:37,430 is the question of existence of the bound state is that in 3D there is no bound 199 00:14:37,430 --> 00:14:41,206 state. Okay, so if alpha is sufficiently small 200 00:14:41,206 --> 00:14:44,740 the right hand side can't possibly compete with the large value in the left 201 00:14:44,740 --> 00:14:48,418 hand side. But if we go to the lower dimension and 202 00:14:48,418 --> 00:14:51,960 among the reasonable dimensions 2 comes to mind. 203 00:14:51,960 --> 00:14:56,514 So unless we consider some fractals or whatever, so that's the only thing we can 204 00:14:56,514 --> 00:15:01,754 have 1 and 2. So in 2 dimensions, this integral becomes 205 00:15:01,754 --> 00:15:05,570 an integral from 0 to 1 over a, dk over k. 206 00:15:05,570 --> 00:15:10,898 And we get a very weak simple logarithmic divergence so at the lower limit we're 207 00:15:10,898 --> 00:15:15,774 going to get 1 over 0. So it's equal to infinity. 208 00:15:15,774 --> 00:15:18,176 Okay? So the Integral divergence, which is 209 00:15:18,176 --> 00:15:20,770 great. It means that we have a bound state. 210 00:15:20,770 --> 00:15:25,852 So this really the summary of what we learned from just simply considering this 211 00:15:25,852 --> 00:15:30,392 integral to which we simplified our problem. 212 00:15:30,392 --> 00:15:34,227 So, if our dimensionality is smaller or equal to 2, so the integral diversions 213 00:15:34,227 --> 00:15:37,839 are physical. It means that there exists always exists 214 00:15:37,839 --> 00:15:41,570 a bond state, doesn't matter how weak the attraction is. 215 00:15:41,570 --> 00:15:45,680 So otherwise the integral is finite and they're no bond states. 216 00:15:45,680 --> 00:15:48,30 So even though there's a well there's nothing in there. 217 00:15:48,30 --> 00:15:51,58 There's nothing in this well. So the last thing I'm going to do is that 218 00:15:51,58 --> 00:15:54,675 I'll estimate the value of the energy of the bond. 219 00:15:54,675 --> 00:15:57,733 State in two dimensions. So we just found there is one, so let's 220 00:15:57,733 --> 00:16:00,612 estimate its energy. And of course, for this I have to solve 221 00:16:00,612 --> 00:16:04,737 this equation, specifically in 2D. So, let me do so, let me just calculate 222 00:16:04,737 --> 00:16:07,920 this integral, actually estimate this integral. 223 00:16:07,920 --> 00:16:11,310 I can, of course, calculate it exactly. But let me estimate it, it's easier. 224 00:16:11,310 --> 00:16:14,550 So in the first thing I'm going to do, I'm going to factor out this coefficient 225 00:16:14,550 --> 00:16:18,896 here in front of the integral. I'm going to have m over pi a squared. 226 00:16:18,896 --> 00:16:25,60 And so I will get an integral of 0 to 1 over a, and let me write it in this form, 227 00:16:25,60 --> 00:16:31,599 d k times k. And so here I'm going to have k squared 228 00:16:31,599 --> 00:16:39,240 plus square root of 2 m e over h squared. So I just wrote in this form, you will 229 00:16:39,240 --> 00:16:42,654 understand why. So if I set energy to 0 I have the 230 00:16:42,654 --> 00:16:47,346 logarithmic diversions that well, as we know allows me allows the existence of 231 00:16:47,346 --> 00:16:54,209 the of the bounce state. Now but, it is finite so these diversions 232 00:16:54,209 --> 00:16:57,29 disappear. So since the smaller the e, the larger 233 00:16:57,29 --> 00:17:00,83 the diversion. So now to estimate the value of the 234 00:17:00,83 --> 00:17:04,241 integral, I can simply move this sort of cutoff of these diversions from the 235 00:17:04,241 --> 00:17:09,410 denominator to the lower limit of integration. 236 00:17:09,410 --> 00:17:13,343 This method is called well this is asymptotic method is called logarithmic 237 00:17:13,343 --> 00:17:16,670 accuracy. So if we have integrals of this type with 238 00:17:16,670 --> 00:17:21,210 logarithmic diversions, if we don't want to perform exact calculations. 239 00:17:21,210 --> 00:17:24,906 Sometimes it's not possible so one can just do this trick and this provides a 240 00:17:24,906 --> 00:17:28,230 reasonable estimate. So if we do that. 241 00:17:28,230 --> 00:17:33,565 So what we're going to get is approximately, let me write it this way, 242 00:17:33,565 --> 00:17:39,676 approximately m over pi h squared in integral, now from this 2 m e, h squared, 243 00:17:39,676 --> 00:17:45,662 1 over e. And so here is I will call simply dk over 244 00:17:45,662 --> 00:17:49,444 k and so this of course is a logarithm so the result is then m over pi h squared 245 00:17:49,444 --> 00:17:55,418 log of 1 over a the upper limit times h over 2me, the lower limit. 246 00:17:55,418 --> 00:18:00,698 And so now I am in the position to find the energy so I can simply solve this 247 00:18:00,698 --> 00:18:06,594 equation for e by exponentiating both sides of this equation which gets results 248 00:18:06,594 --> 00:18:13,640 of the logarithm. And so the result is going to be so the 249 00:18:13,640 --> 00:18:18,860 energy or the absolute value of the energy is going to be, let me write out 250 00:18:18,860 --> 00:18:24,950 the order, a squared over m a squared exponential of, two pi h squared divided 251 00:18:24,950 --> 00:18:34,610 by alpha with a minus. So this is really the final result. 252 00:18:34,610 --> 00:18:40,319 And the result is very interesting. So for a theoretical physicist at least, 253 00:18:40,319 --> 00:18:44,200 because the function that we got is a very strange function. 254 00:18:44,200 --> 00:18:47,973 So remember that this parameter alpha controls the strength of the potential. 255 00:18:47,973 --> 00:18:51,572 And, we, consider the case when it's very small. 256 00:18:51,572 --> 00:18:57,312 So, e to the power minus 1 over alpha is really exponentially small, very shallow 257 00:18:57,312 --> 00:19:04,332 level and already shallow potential. But the function is strange because well 258 00:19:04,332 --> 00:19:08,500 if you write it, either to the power minus Something over alpha. 259 00:19:08,500 --> 00:19:12,270 So this function does not have a Taylor Expansion. 260 00:19:12,270 --> 00:19:16,404 So remember that, you know? If we have a, function, f of alpha. 261 00:19:16,404 --> 00:19:20,634 Normally, at small alpha, we can approximate it by f of 0. 262 00:19:20,634 --> 00:19:28,490 Plus, you know, f prime of zero, times alpha plus f2 prime subzero over 2. 263 00:19:28,490 --> 00:19:31,867 alpha square root, et cetera. So this is a regular Taylor expansion of 264 00:19:31,867 --> 00:19:36,94 a function. So in this case, all the derivatives, all 265 00:19:36,94 --> 00:19:40,820 the terms in this expansion are equal to 0. 266 00:19:40,820 --> 00:19:44,996 And this actually would have pose the problem if we had try to solve the 267 00:19:44,996 --> 00:19:50,585 problem using so called petrubation theory which is[UNKNOWN]. 268 00:19:50,585 --> 00:19:55,625 Because oftentimes, it gives us access to solutions that are otherwise impossible 269 00:19:55,625 --> 00:19:58,558 to get. So we start with a normal solution and 270 00:19:58,558 --> 00:20:01,820 then do a Taylor expansion in the vicinity of the solution. 271 00:20:01,820 --> 00:20:04,130 So here, it wouldn't have been impossible. 272 00:20:04,130 --> 00:20:08,240 Exactly because of these weird properties of the function exists, we can plot it. 273 00:20:08,240 --> 00:20:10,539 We can look at it, but there is no Taylor expansion at all. 274 00:20:11,780 --> 00:20:16,64 So -- and it turns out that in the second lecture this week, we're going to study 275 00:20:16,64 --> 00:20:20,660 some basics about the theory of superconductivity. 276 00:20:20,660 --> 00:20:23,453 And there is a classic formula in this Nobel Prize-winning 277 00:20:23,453 --> 00:20:27,329 Barneen-Cooper-Schrieffer theory of superconductivity, which features this 278 00:20:27,329 --> 00:20:31,208 function. This sort of function and actually the 279 00:20:31,208 --> 00:20:34,571 reason, it took people, as, as I will talk about in the next lecture, the 280 00:20:34,571 --> 00:20:38,333 reason it took people almost 50 years to come up with a solution to the problem of 281 00:20:38,333 --> 00:20:44,990 superconductivity is partially because of this circumstance. 282 00:20:44,990 --> 00:20:49,20 That the result actually is something that can only be obtained by an exact 283 00:20:49,20 --> 00:20:52,985 non-interpretive calculation but it couldn't be obtained using the 284 00:20:52,985 --> 00:20:55,863 Perturbation theory.