1 00:00:00,770 --> 00:00:04,134 In the previous video we saw that with the shallow quantum well, can be 2 00:00:04,134 --> 00:00:08,984 faithfully represented, as an attractive derived delta potential. 3 00:00:08,984 --> 00:00:13,31 and, in this video we are going to solve the Schrodinger equation, with a delta 4 00:00:13,31 --> 00:00:17,590 potential in an arbitrary dimension 1, 2, or 3. 5 00:00:17,590 --> 00:00:21,154 And we're going to use a very powerful method, mathematical method, that, as a 6 00:00:21,154 --> 00:00:24,826 matter of fact, was used by Leon Cooper to solve the key problem in the theory of 7 00:00:24,826 --> 00:00:28,444 superconductivity, and the problem of electron pairing, which brought him 8 00:00:28,444 --> 00:00:31,738 eventually a Nobel prize and now this phenomenon of pairing is now called 9 00:00:31,738 --> 00:00:38,152 Cooper pairing. But well, we'll talk a little bit about 10 00:00:38,152 --> 00:00:41,263 this later but at this stage we're going to follow on this single particle 11 00:00:41,263 --> 00:00:45,600 Schrodinger equation with this sort of short range attraction. 12 00:00:45,600 --> 00:00:49,77 And, so before going through the technicalities I want to give you the 13 00:00:49,77 --> 00:00:54,540 bottom line, the main result if you sort of get lost in these technical details. 14 00:00:54,540 --> 00:00:57,944 So please I would like you to know. the very important outcome of this 15 00:00:57,944 --> 00:01:01,640 compilation we’re going to see, and so the outcome would be the following. 16 00:01:01,640 --> 00:01:05,652 So we will see that in one and two dimensions, any, it doesn't matter how 17 00:01:05,652 --> 00:01:12,430 weak an attractive potential gives rise to bounce the, of the quantum particle. 18 00:01:12,430 --> 00:01:15,852 So it doesn’t matter how shallow a well we have, it always will bind the 19 00:01:15,852 --> 00:01:19,419 particle. But in three dimensions, it turns out 20 00:01:19,419 --> 00:01:22,653 that if you have a very weak potential, very shallow potential, it's not enough 21 00:01:22,653 --> 00:01:26,100 to give the particle, to localize the particle. 22 00:01:26,100 --> 00:01:29,823 And this difference qualitative difference actually have a very important 23 00:01:29,823 --> 00:01:32,985 consequences in various fields of physics, so you may want to remember 24 00:01:32,985 --> 00:01:37,28 that. But now, let me go into the technical 25 00:01:37,28 --> 00:01:38,378 part. Okay? 26 00:01:38,378 --> 00:01:42,596 So, and I will start with a certain math reminder, which basically will remind you 27 00:01:42,596 --> 00:01:46,442 what a Fourier transform is. So, we can solve the problem, we will 28 00:01:46,442 --> 00:01:48,930 solve the problem using only the Fourier transform. 29 00:01:48,930 --> 00:01:51,970 There will be nothing else involved essentially in this solution. 30 00:01:51,970 --> 00:01:54,210 Fourier transform, and inverse Fourier transform. 31 00:01:54,210 --> 00:01:57,888 So what is a Fourier transform? So if you have an arbitrary function, 32 00:01:57,888 --> 00:02:02,512 well reasonable function g of x, in our case it will be way function and various 33 00:02:02,512 --> 00:02:08,117 derivatives of it. So the Fourier transform of this function 34 00:02:08,117 --> 00:02:12,424 g of x, which we'll call f of g of x, so g of ke is an integral of g of x with 35 00:02:12,424 --> 00:02:18,451 this exponential. So physically this k in our case, will 36 00:02:18,451 --> 00:02:26,700 represent the wave vector, and h times k is going to be the particles momentum. 37 00:02:26,700 --> 00:02:31,460 But in principle you can view this wave transform as a mathematical procedure. 38 00:02:31,460 --> 00:02:34,788 An important fact here is that if you, if you know the free transformer of a 39 00:02:34,788 --> 00:02:37,908 function, you can restore the function itself by doing the inverse free 40 00:02:37,908 --> 00:02:43,216 transform. so g of x is equal to an integral of 41 00:02:43,216 --> 00:02:50,30 there should be a g tilde of k, either the power minus ikx dk over 2 pi. 42 00:02:50,30 --> 00:02:53,614 So I should remind you or emphasize that where to put this 2 pi is actually a 43 00:02:53,614 --> 00:02:57,254 matter of convention or convenience so we could have put the 2 pi in the Fourier 44 00:02:57,254 --> 00:03:03,830 transform or we could have put, we could have separated them sort of like this. 45 00:03:03,830 --> 00:03:06,992 So we could have put a 1 over square root of 2 pi here, 1 over square root of 2 pi 46 00:03:06,992 --> 00:03:10,154 here or any other combination of powers so that the total power of 2 pi in the 47 00:03:10,154 --> 00:03:16,109 denominator is equal to 1. And this is a matter of convention in 48 00:03:16,109 --> 00:03:21,730 quantum mechanics the convention is that you know two pi appears in this inverse 49 00:03:21,730 --> 00:03:27,43 reaction which, which features the wave transfer or the momentum so I'll erase 50 00:03:27,43 --> 00:03:32,722 this off here. So anyway, so this what we know now, the 51 00:03:32,722 --> 00:03:35,482 simplest perhaps example of Fourier transform will be exactly Fourier 52 00:03:35,482 --> 00:03:39,852 transform of the delta function. So if we want the Fourier transform, the 53 00:03:39,852 --> 00:03:43,680 delta function, we know that the delta function when integrated in infinity is 54 00:03:43,680 --> 00:03:49,488 just peaks the value of the function is being integrated with at x equals 0. 55 00:03:49,488 --> 00:03:53,568 So, and if we integrate delta of x with the e to the power of ikx from minus 56 00:03:53,568 --> 00:03:57,528 infinity to plus infinity, we simply will get 1, because e exponent of 0 is equal 57 00:03:57,528 --> 00:04:01,860 to 1. So therefore, the Fourier image of the 58 00:04:01,860 --> 00:04:06,290 delta function is 1. Just see the identity in this case base. 59 00:04:06,290 --> 00:04:10,490 So and we can also do the inverse Fourier transform, which basically gives us the 60 00:04:10,490 --> 00:04:15,840 representation of the delta function. It's an integral who minus infinity to 61 00:04:15,840 --> 00:04:19,40 plus infinity is exponential dk over 2 pi. 62 00:04:19,40 --> 00:04:21,330 We have seen it already, actually, before. 63 00:04:21,330 --> 00:04:26,138 So in this identities, actually will be enough again for us to solve the problem. 64 00:04:26,138 --> 00:04:29,792 So, now the problem that we actually want to solve mathematically is the 65 00:04:29,792 --> 00:04:33,360 Schrodinger equation which is presented here. 66 00:04:33,360 --> 00:04:37,229 So the left hand side of this equation here represents the ah,[INAUDIBLE] so the 67 00:04:37,229 --> 00:04:40,833 [INAUDIBLE] in this part is the kinetic energy, the usual kinetic energy written 68 00:04:40,833 --> 00:04:44,13 explicitly in the quardenential presentation, so that's just p square 69 00:04:44,13 --> 00:04:49,70 over 2m. And this guy is our potential, delta 70 00:04:49,70 --> 00:04:51,860 potential. So later on we're going to generalize it 71 00:04:51,860 --> 00:04:54,960 to arbitrary dimension, but at this stage I'm just doing one dimension for 72 00:04:54,960 --> 00:04:58,314 simplicity. So of course we're dealing now with time 73 00:04:58,314 --> 00:05:01,949 and temperature in your equation. And our, the goal of this calculation 74 00:05:01,949 --> 00:05:05,379 basically or the goal of our evaluation would be to determine the value of the 75 00:05:05,379 --> 00:05:08,711 energy, and in particular we will, we will, we would want to figure out whether 76 00:05:08,711 --> 00:05:14,50 or not there are bound states, with negative energy. 77 00:05:14,50 --> 00:05:18,586 So essentially if we want to sort of draw symbolically the potential that we're 78 00:05:18,586 --> 00:05:23,185 dealing with so it's zero everywhere but in the vicinity of x equals zero there is 79 00:05:23,185 --> 00:05:27,217 a sharp sort of well and the question is whether there is a bound state or bound 80 00:05:27,217 --> 00:05:35,737 states in this well with some negative energy So this is basically the question. 81 00:05:35,737 --> 00:05:40,312 And so now, let me do the actual calculation using this reminder in the 82 00:05:40,312 --> 00:05:44,967 Schrodinger equation. So what we're going to do, we're going to 83 00:05:44,967 --> 00:05:48,212 represent this f of x as a Fourier transform of the wave function in 84 00:05:48,212 --> 00:05:53,115 momentum representation. So t's going to be an integral of gk over 85 00:05:53,115 --> 00:05:57,852 2 pi of sound side tilde sub k. I'm going to write it, it's from minus 86 00:05:57,852 --> 00:06:02,272 ikx from minus infinity to plus infinity, and just for the sake of gravity, I'm 87 00:06:02,272 --> 00:06:06,488 going to call this integral I saw so that I don't have to write the standard 88 00:06:06,488 --> 00:06:13,242 factors and limits again. So it's identical equal in my notations 89 00:06:13,242 --> 00:06:16,413 to this. So there are three types of terms that 90 00:06:16,413 --> 00:06:20,573 appear in this equation, so the way function itself appears in the right hand 91 00:06:20,573 --> 00:06:24,928 side, multiply by the energy, in the left and side, what we have for instance this 92 00:06:24,928 --> 00:06:31,810 guy which is the second derivative of the delta Psi double front. 93 00:06:31,810 --> 00:06:35,468 So if we differentiate this free transform twice over the coordinate, 94 00:06:35,468 --> 00:06:39,64 we're going to pull out the, so the beginning of the exponential is the 95 00:06:39,64 --> 00:06:45,660 exponential itself, but they're going to pull out this minus i t squared. 96 00:06:45,660 --> 00:06:51,646 So therefore the we can represent this as a derivative as psi tilde sub k times 97 00:06:51,646 --> 00:06:56,870 minus ik squared, e to the power minus ikx. 98 00:06:56,870 --> 00:07:02,874 So this guy is the Fria transform of psi double prime. 99 00:07:02,874 --> 00:07:09,740 And finally we also have this term, so delta of x, of psi of x. 100 00:07:09,740 --> 00:07:13,880 But since delta of x is a sharply peaked function of x equals 0, so it essentially 101 00:07:13,880 --> 00:07:18,930 pulls out only the value of the wave function of x equals 0. 102 00:07:18,930 --> 00:07:21,790 So all other values of this wave function is In this product it basically don't 103 00:07:21,790 --> 00:07:25,150 matter. So in this theorem is going to be 104 00:07:25,150 --> 00:07:29,888 identical and equal to psi of 0, delta of x. 105 00:07:29,888 --> 00:07:31,965 Okay? And delta of x, we know that the delta of 106 00:07:31,965 --> 00:07:36,12 x, the Fourier image of delta of x as we discussed in the previous the Slight this 107 00:07:36,12 --> 00:07:41,858 is simply equal to 1. So we can represent this as of size 0 and 108 00:07:41,858 --> 00:07:47,732 integral over k again is implied that is for minus infinity to plus infinity e to 109 00:07:47,732 --> 00:07:54,375 the minus ikx that's it. So this is, this is the basically the way 110 00:07:54,375 --> 00:07:58,470 we can write this In this situation. 111 00:07:58,470 --> 00:08:04,385 And so, if you put everything together, so we're going to have, in the left hand 112 00:08:04,385 --> 00:08:09,845 side, we're going to have minus h squared over 2 m, psi tilde sub k minus i k 113 00:08:09,845 --> 00:08:19,126 squared, e to the bar minus i k x. So this term, the second term is going to 114 00:08:19,126 --> 00:08:25,220 be minus alpha Size of 0 which here is simply a constant. 115 00:08:25,220 --> 00:08:28,916 So we have pulled out the value of the wave function x equal 0 is just a 116 00:08:28,916 --> 00:08:34,650 constant and here we have this integral e to the power minus ikx. 117 00:08:34,650 --> 00:08:38,70 In the right hand side we're going to have simply a energy times the wave 118 00:08:38,70 --> 00:08:43,588 function. Coming from here, k psi tilde is to the 119 00:08:43,588 --> 00:08:47,730 power minus ks. So we see that in all three terms we have 120 00:08:47,730 --> 00:08:51,140 this integrals and we have this exponential so that we can simply focus 121 00:08:51,140 --> 00:08:54,385 on the overall coefficients that multiplies all this exponential and 122 00:08:54,385 --> 00:09:02,0 essentially get what we can call it. Schrodinger equation in a momentum. 123 00:09:02,0 --> 00:09:06,636 So in some sense, we can just get rid of these integrals and let me have some 124 00:09:06,636 --> 00:09:11,940 space here. Let me erase most of this stuff. 125 00:09:11,940 --> 00:09:17,88 So as a result, if we basically look at this equation without the integrals for a 126 00:09:17,88 --> 00:09:21,752 particular key. So this term, the first term on the 127 00:09:21,752 --> 00:09:26,383 left-hand side, is going to give us now plus h squared over 2m. 128 00:09:26,383 --> 00:09:31,396 k squared psi tilde of k. the second term is going to give us minus 129 00:09:31,396 --> 00:09:35,754 alpha times side of 0. So this psi of 0 is 130 00:09:35,754 --> 00:09:39,490 [UNKNOWN]. 131 00:09:39,490 --> 00:09:43,679 And the right hand side is going to be a the energy that we're looking for so this 132 00:09:43,679 --> 00:09:48,710 is our unknown, this is what we're looking for, psi tilde of k . 133 00:09:48,710 --> 00:09:52,285 So we see that instead of solving the complicated differential equation as it 134 00:09:52,285 --> 00:09:57,465 looked in the beginning, we've now reduced the problem to solve essentially. 135 00:09:57,465 --> 00:10:02,140 A linear equation that, we know how to do from high school math. 136 00:10:02,140 --> 00:10:06,892 So we simply can put this psi tilde in the left hand side, this guy in the right 137 00:10:06,892 --> 00:10:11,644 hand side and so we're going to have h squared k squared over 2m minus energy 138 00:10:11,644 --> 00:10:16,930 times psi tilde. So these two guys. 139 00:10:16,930 --> 00:10:20,826 And in the right hand side we're going to have plus alpha psi of 0. 140 00:10:20,826 --> 00:10:26,658 So, the way function, the other unknown, the way function therefore is simply, 141 00:10:26,658 --> 00:10:31,680 equal to alpha side of zero way function in real space divided by h squared t 142 00:10:31,680 --> 00:10:39,738 square over two m minus energy. And, well in some sense this is already, 143 00:10:39,738 --> 00:10:46,892 partially, solution to our problem. It's a solution in the sense that it 144 00:10:46,892 --> 00:10:50,836 determines the wave function in momentum space and if we want for instance to find 145 00:10:50,836 --> 00:10:54,316 the wave function in real space, all we have to do is just inverse Fourier 146 00:10:54,316 --> 00:11:00,380 transform it back to x. But our main interest again is to answer 147 00:11:00,380 --> 00:11:05,302 the question of whether or not there are bounce states in this potential. 148 00:11:05,302 --> 00:11:13,300 And, so this, well I just copied here, the same equation we derived previously. 149 00:11:13,300 --> 00:11:18,57 So emphasize that this is really what we're after here. 150 00:11:18,57 --> 00:11:22,152 So, the question is, are there any available energy levels, leveled in this 151 00:11:22,152 --> 00:11:26,572 shallow quantum well, which responds to particle beam localized in the vicinity 152 00:11:26,572 --> 00:11:30,732 of the So it turns out that this question can be answered from this equation in 153 00:11:30,732 --> 00:11:36,686 just two simple steps. And those of you like mathematical 154 00:11:36,686 --> 00:11:39,866 thought puzzles, sort of, sort of aficionados of, of you know, these 155 00:11:39,866 --> 00:11:45,21 puzzles, may just want to stop the presentation and stare at this equation. 156 00:11:45,21 --> 00:11:51,294 And try to come up with a trick that will allow you to figure out this enumeration. 157 00:11:51,294 --> 00:11:55,850 But if you just want to know the trick, let me show you how it works. 158 00:11:55,850 --> 00:11:59,136 And it's actually very neat, it doesn't require any super complicated 159 00:11:59,136 --> 00:12:03,180 calculations. So this stage, let me recall that psi of 160 00:12:03,180 --> 00:12:08,652 x basically the can be represented as a the inverse Fourier transform of psi 161 00:12:08,652 --> 00:12:15,335 tilde of k, e to the power minus ikx. So therefore psi of 0, which appears in 162 00:12:15,335 --> 00:12:20,292 the right hand side of this equation. Psi of 0 corresponds simply to setting x 163 00:12:20,292 --> 00:12:24,507 to 0 in this equation. So it's going to be an integral of all, 164 00:12:24,507 --> 00:12:28,78 over all k over 2 pi of this psi tilde of k. 165 00:12:28,78 --> 00:12:32,397 So let me apply this integral in both sides of this equation. 166 00:12:32,397 --> 00:12:38,20 So if I have an equation, I can take integral of both sides of this equation. 167 00:12:38,20 --> 00:12:42,570 And if I do so in there, left hand side, I'm going to have psi tilde of k which 168 00:12:42,570 --> 00:12:47,77 per this relation, gives me simply psi of 0. 169 00:12:49,350 --> 00:12:56,82 While in the right-hand side I will have this coefficient, which is simply a 170 00:12:56,82 --> 00:13:03,18 constant alpha psi of 0, an integral, over k 1 over h squared, k squared over 171 00:13:03,18 --> 00:13:09,792 2m minus epsilon. So the neat thing that happens here is 172 00:13:09,792 --> 00:13:13,16 that this, we see that there is a psi of zero that we don't really know at this 173 00:13:13,16 --> 00:13:16,180 stage. But this sign of zero appears in both 174 00:13:16,180 --> 00:13:18,950 sides of this equation so we can simply cancel it. 175 00:13:18,950 --> 00:13:23,159 And so we are where we end up with an equation that doesn't have the wave 176 00:13:23,159 --> 00:13:27,298 function anymore. So it has only one unknown and this 177 00:13:27,298 --> 00:13:30,458 unknown is exactly the energy we are after. 178 00:13:30,458 --> 00:13:35,81 So, if we can solve our equation and find the value of the energy, this would imply 179 00:13:35,81 --> 00:13:39,700 that there exists a bound state in our potential. 180 00:13:39,700 --> 00:13:43,924 And remember that the bound state has negative energy so it must be below the 181 00:13:43,924 --> 00:13:47,170 continuum. And this would correspond sort of to 182 00:13:47,170 --> 00:13:49,985 localize particles. So we can simply let us say models here 183 00:13:49,985 --> 00:13:53,919 and the plus sign. So this would correspond to our image 184 00:13:53,919 --> 00:13:58,445 we're looking for and also if we want sort of restore normal form of the 185 00:13:58,445 --> 00:14:05,56 integral, so we can Recall that my notation was simply to 186 00:14:05,56 --> 00:14:10,750 save some space. So this is really what, what we have. 187 00:14:10,750 --> 00:14:14,301 So let me skip the remaining technical step which essentially boils down just to 188 00:14:14,301 --> 00:14:18,100 copulating this integral which is an elementary integral. 189 00:14:18,100 --> 00:14:23,482 And just present the result. And so here I sort of rewritten this 190 00:14:23,482 --> 00:14:28,580 equation in Sort of slightly different form and this 191 00:14:28,580 --> 00:14:33,499 is the result of the calculations. If you calculate the integral, you will 192 00:14:33,499 --> 00:14:38,210 see that there is, exists one and only one solution for the energy. 193 00:14:38,210 --> 00:14:42,990 And the energy is equal to minus m alpha squared over 2h squared. 194 00:14:42,990 --> 00:14:46,443 So alpha here, remember, alpha here is the parameter. 195 00:14:46,443 --> 00:14:51,573 Which multiplies the delta potential. So, in some sense, alpha controls the 196 00:14:51,573 --> 00:14:55,347 strength of the potential. The smaller the alpha, the weaker the 197 00:14:55,347 --> 00:15:00,146 potential that acts on the particle. And the result here is that, in the 198 00:15:00,146 --> 00:15:04,702 shallow quantum well, in one dimension, or in the corresponding delta potential 199 00:15:04,702 --> 00:15:08,789 well, there exists one, and only one, bound state with this energy, which 200 00:15:08,789 --> 00:15:14,540 scales so/g, with the strength of the potential. 201 00:15:14,540 --> 00:15:18,566 So in the second part of this video we're going to discuss how this result is 202 00:15:18,566 --> 00:15:22,470 modified in a very essential way if we go to more physical dimensions to 2 and 3 203 00:15:22,470 --> 00:15:25,283 dimensions.