1 00:00:00,650 --> 00:00:03,608 Welcome back, everyone. I'm Charles Clark and this is Exploring 2 00:00:03,608 --> 00:00:06,725 Quantum Physics. In today's lecture, we're going to look 3 00:00:06,725 --> 00:00:10,105 at solving the Schrodinger Equation by a well-known proven technique. 4 00:00:10,105 --> 00:00:12,985 Actually, first used by Schrodinger himself, which is the use of the special 5 00:00:12,985 --> 00:00:19,285 functions of mathematics. I'm using a convenient online reference, 6 00:00:19,285 --> 00:00:22,660 which is freely available around the world for this. 7 00:00:22,660 --> 00:00:25,860 It's a standard repository digital library of information on the so-called 8 00:00:25,860 --> 00:00:29,260 special functions of mathematics, which are the ones that are in most widespread 9 00:00:29,260 --> 00:00:35,537 use in areas of science and engineering. it contains the definitions, properties, 10 00:00:35,537 --> 00:00:38,528 interactive graphics. Something might be of interest to some of 11 00:00:38,528 --> 00:00:41,125 you. TeX encoding of all of all appropriate 12 00:00:41,125 --> 00:00:46,100 material of all the equations and so on. Also, MathML encoding. 13 00:00:46,100 --> 00:00:52,220 And the ability to do Semantic search which recognizes fairly free format 14 00:00:52,220 --> 00:00:59,552 mathematical, mathematical information. Also, I've put some material on the 15 00:00:59,552 --> 00:01:04,191 course page in the additional materials section. 16 00:01:04,191 --> 00:01:08,903 And one of them is a summary of the solutions of frequently encountered 17 00:01:08,903 --> 00:01:14,030 Schrodinger equations in terms of special functions. 18 00:01:14,030 --> 00:01:18,230 and then, something I encourage you to try a demonstration which uses a, a 19 00:01:18,230 --> 00:01:22,070 numerical solution of the Schrodinger equation. 20 00:01:22,070 --> 00:01:25,010 It's coded up very simply on a spreadsheet and basically two lines of 21 00:01:25,010 --> 00:01:29,852 code are all that are required. And this will give you the opportunity to 22 00:01:29,852 --> 00:01:33,680 actually gain some experience with practice in numerical solution which I 23 00:01:33,680 --> 00:01:37,624 think, in my case, has been a really good guide to understanding how things happen 24 00:01:37,624 --> 00:01:43,120 and why and what to expect. I mean, there really is, the only way 25 00:01:43,120 --> 00:01:47,320 that you're going to become comfortable and familiar with solving the Schrodinger 26 00:01:47,320 --> 00:01:52,387 Equations by working on examples. So I do strongly encourage you to take 27 00:01:52,387 --> 00:01:55,735 that opportunity and, you know, perhaps to apply to some of the, the simple 28 00:01:55,735 --> 00:01:59,980 examples we're going to start with now in the lecture. 29 00:01:59,980 --> 00:02:02,740 Well, who was the first to solve the Schrodinger Equation using special 30 00:02:02,740 --> 00:02:07,170 functions? No kidding, it was Schrodinger himself. 31 00:02:07,170 --> 00:02:10,470 What's good enough for Erwin Schrodinger is good enough for us. 32 00:02:10,470 --> 00:02:14,574 in his very first paper on wave mechanics he used special functions to solve the 33 00:02:14,574 --> 00:02:19,114 equations for the hydrogen atom. And I think in the some second or third 34 00:02:19,114 --> 00:02:22,220 paper, he also computed the harmonic oscillator. 35 00:02:22,220 --> 00:02:27,044 And what he did and I think this is one reason why the Schrodinger equation was 36 00:02:27,044 --> 00:02:31,940 so widely adopted quickly is that he found that he could solve the Schrodinger 37 00:02:31,940 --> 00:02:39,420 Equation with standard well-studied equations of mathematical physics. 38 00:02:39,420 --> 00:02:43,641 So, these were functions that were defined in previous contexts and applied 39 00:02:43,641 --> 00:02:48,220 to things like electromagnetism and acoustics. 40 00:02:48,220 --> 00:02:52,600 And thereby, there was, there was a rich technology that had already been 41 00:02:52,600 --> 00:02:56,140 developed and the properties of the solutions of those functions were 42 00:02:56,140 --> 00:03:02,726 mathematically well-defined in them. So that's a pretty good approach to be 43 00:03:02,726 --> 00:03:08,390 following. So, we're going to look at three 44 00:03:08,390 --> 00:03:13,565 one-dimensional examples in this context. the constant potential, the linear 45 00:03:13,565 --> 00:03:17,645 potential, which describes a particle in electric field, and then the quadratic 46 00:03:17,645 --> 00:03:21,365 potential of which you've already seen a good treatment of by Victor in the 47 00:03:21,365 --> 00:03:26,660 harmonic oscillator section. this is going to look at things in 48 00:03:26,660 --> 00:03:30,270 somewhat more general context. So, the solution is here. 49 00:03:31,400 --> 00:03:36,360 A map on to the exponential, the Airy and the Parabolic Cylinder functions 50 00:03:36,360 --> 00:03:40,866 respectively. and I hope to show you from a few simple 51 00:03:40,866 --> 00:03:45,075 examples some of the generic features that you'd expect when you might be 52 00:03:45,075 --> 00:03:51,990 dealing with a more a system that has parts of it that look like this. 53 00:03:51,990 --> 00:03:57,020 But then, they, they might vary in a piecewise manner or something like that. 54 00:03:57,020 --> 00:04:02,636 For example, if you're designing a a multi-layer semiconductor quantum dot or 55 00:04:02,636 --> 00:04:08,486 a quantum wave guide you will, you will encounter segments of the problem where 56 00:04:08,486 --> 00:04:15,590 the potentials might look piecewise, like they do here. 57 00:04:15,590 --> 00:04:20,806 Well, let's proceed. Okay, now, we look at the constant 58 00:04:20,806 --> 00:04:24,629 potential. Something not terribly interesting in its 59 00:04:24,629 --> 00:04:26,891 own right. But we're going to do it in a more, in 60 00:04:26,891 --> 00:04:30,132 the most general way. So, the, the point is we have a, a given 61 00:04:30,132 --> 00:04:33,203 potential. And now, we have the Schrodinger 62 00:04:33,203 --> 00:04:37,940 equation, which we're defining without any boundary conditions. 63 00:04:37,940 --> 00:04:41,450 And we want to find the solutions for all possible energies. 64 00:04:41,450 --> 00:04:44,852 So we want, in other words, we want to be able to solve this equation in the most 65 00:04:44,852 --> 00:04:50,470 flexible way. And the, the standard approach that 66 00:04:50,470 --> 00:04:56,128 applies to all these examples, and will apply to most other practical problems 67 00:04:56,128 --> 00:05:01,130 you work on, is to transfer, transform the Schrodinger Equation into a, a 68 00:05:01,130 --> 00:05:09,620 standard dimensionless form that defines one of the special functions. 69 00:05:09,620 --> 00:05:14,426 So this one is pretty easy. we can see we, you just, you just move 70 00:05:14,426 --> 00:05:20,483 these constants around, and and scale the variable. 71 00:05:20,483 --> 00:05:26,891 And as you [INAUDIBLE], the, there, there is in fact a dividing line between two 72 00:05:26,891 --> 00:05:33,870 classes a solutions associated with sine of this coefficient. 73 00:05:33,870 --> 00:05:41,124 So, here are, here is the transformation I described choose a scaled variable, and 74 00:05:41,124 --> 00:05:47,310 we get these two the two different solutions. 75 00:05:47,310 --> 00:05:56,150 So the point to note here is this is a second order different actual equation. 76 00:05:56,150 --> 00:06:02,640 And so, it always has two independent solutions. 77 00:06:02,640 --> 00:06:05,310 Now, you can choose those solutions in different ways. 78 00:06:05,310 --> 00:06:08,480 Here is a standard choice for the oscillatory solutions. 79 00:06:08,480 --> 00:06:12,650 The sine and the cosine out of phase by half a wavelength. 80 00:06:12,650 --> 00:06:16,082 and any other solution that you get will, is, can be described as some linear 81 00:06:16,082 --> 00:06:21,460 combination of those two there. So, that's that seems fine. 82 00:06:21,460 --> 00:06:27,164 Now, in the in what we call the classically forbidden region, where the 83 00:06:27,164 --> 00:06:33,420 energy is less than the potential energy, then we get the minus sign in the natural 84 00:06:33,420 --> 00:06:42,019 way of decomposing this equation. And now, here's an important point. 85 00:06:42,019 --> 00:06:47,486 Once again, there are two solutions and actually there are, there are any numbers 86 00:06:47,486 --> 00:06:53,342 of ways you can pick the two solutions. So here, I have the exponent of rho and 87 00:06:53,342 --> 00:06:57,489 the exponent of minus rho. Well, sinh rho and cosh rho would be 88 00:06:57,489 --> 00:07:01,415 equally acceptable. They are independent. 89 00:07:01,415 --> 00:07:10,800 but there is uniquely one solution only up to a multiplicative constant. 90 00:07:10,800 --> 00:07:16,080 One solution only, that, that converges in the, in this, this is what we call the 91 00:07:16,080 --> 00:07:22,358 classically forbidden region. I mean the energy is the potential energy 92 00:07:22,358 --> 00:07:26,405 plus the kinetic energy if the energy is less than the, the potential energy, that 93 00:07:26,405 --> 00:07:30,655 corresponds to the kinetic energy being negative. 94 00:07:30,655 --> 00:07:34,351 Which can't be the case because it's well, you can, but, but you do get, you'd 95 00:07:34,351 --> 00:07:38,285 get, you get you'd get suppressed behavior. 96 00:07:38,285 --> 00:07:42,738 So, if you think about it, this, this area is tricky from the standpoint of 97 00:07:42,738 --> 00:07:47,848 implementation, because if you're, if you are, let's say integrating a normal wave 98 00:07:47,848 --> 00:07:53,253 function. That in, into the forbidden region in the 99 00:07:53,253 --> 00:07:57,633 forbidden region, your solution is going to be a linear combination of one 100 00:07:57,633 --> 00:08:02,790 that's divergent and one that's convergent. 101 00:08:02,790 --> 00:08:06,572 And the chances of your hitting the exactly convergent function are very 102 00:08:06,572 --> 00:08:10,820 small. Okay. 103 00:08:10,820 --> 00:08:14,005 I admit, we're not going to win any Nobel prizes for solving the one-dimensional 104 00:08:14,005 --> 00:08:16,960 Schrodinger equation with a constant potential. 105 00:08:16,960 --> 00:08:20,452 But let me say two things about it. First of all, you can well imagine 106 00:08:20,452 --> 00:08:24,124 simulating, let's say, let's say you have a interesting potential that you want to 107 00:08:24,124 --> 00:08:29,530 solve for some problem a waveguide design, or something like that. 108 00:08:29,530 --> 00:08:32,824 Well, you can think about approximating it by by a step, a stepwise 109 00:08:32,824 --> 00:08:36,230 approximation. So here, I've got these. 110 00:08:36,230 --> 00:08:43,160 I, you sort of like approximate an integral by a, a finite sum and you could 111 00:08:43,160 --> 00:08:49,991 then use use this simple mapping onto functions within each region to advance a 112 00:08:49,991 --> 00:08:58,910 wave function efficiently across an arbitrary potential. 113 00:09:00,530 --> 00:09:05,786 The second is sometimes one thinks of these, these step or square well problems 114 00:09:05,786 --> 00:09:11,384 as being highly contrived. But virtually, every day, you encounter 115 00:09:11,384 --> 00:09:15,999 practical example of such a system and that's when you look, you look through a 116 00:09:15,999 --> 00:09:22,103 window. so let's say, here, you look at a window. 117 00:09:22,103 --> 00:09:30,220 So here, we have incoming light e to the ikx is representing the incoming wave. 118 00:09:30,220 --> 00:09:34,800 And then, we have the reflected wave of coefficient R, Re to the minus ikx. 119 00:09:34,800 --> 00:09:37,450 And then, there's a transmitted wave into the glass. 120 00:09:37,450 --> 00:09:41,860 And the relationship between the transmitted wave of normal incidence and 121 00:09:41,860 --> 00:09:46,760 the incident one is that the, the wave vector in the glass is multiplied by the 122 00:09:46,760 --> 00:09:54,489 refractive index, which is about 1.5. So my challenge to you is to determine 123 00:09:54,489 --> 00:10:00,726 what that transmission code, what fraction of the light passes from the air 124 00:10:00,726 --> 00:10:07,420 into the glass and what fraction is reflected? 125 00:10:07,420 --> 00:10:12,590 So, I hope you saw that for glass, about 4% of light is transmitted. 126 00:10:12,590 --> 00:10:18,140 So, that, that's a pretty low loss. but it's, it's enough that you can see 127 00:10:18,140 --> 00:10:24,040 your reflection under certain conditions. Now, we look at the linear potential. 128 00:10:24,040 --> 00:10:28,000 So you might say we started with V naught. 129 00:10:28,000 --> 00:10:30,060 Now, we're adding a term, which is constant. 130 00:10:30,060 --> 00:10:32,610 Now, we're adding a term that's linear and x. 131 00:10:32,610 --> 00:10:37,030 So in some respects, this is sort of the next it, in order of, of mathematical 132 00:10:37,030 --> 00:10:40,974 complexity, this is like the next function in the series away from the 133 00:10:40,974 --> 00:10:46,841 constant, the linear. And so there's Schrodinger's equation, 134 00:10:46,841 --> 00:10:52,870 it's very straightforward. And what we do is here is x in terms of 135 00:10:52,870 --> 00:11:02,110 the, the dimensionless variable row. we, we, we take away the offset and 136 00:11:02,110 --> 00:11:08,737 divide by the slope. here is the here is the equation for the 137 00:11:08,737 --> 00:11:16,085 coefficient alpha. Now, are, the standard solution, which is 138 00:11:16,085 --> 00:11:20,812 the defining solution of the area function is here. 139 00:11:20,812 --> 00:11:25,980 It's just the second really, there's a portion of linear multiple, of the wave 140 00:11:25,980 --> 00:11:32,875 function. And again, we have a pair of independent 141 00:11:32,875 --> 00:11:38,516 solutions. So, in the region rho less than zero, the 142 00:11:38,516 --> 00:11:44,281 wave function is always oscillatory. And we see that in that region, we have 143 00:11:44,281 --> 00:11:47,494 these two independent solutions called Ai and Bi, they're both called Airy 144 00:11:47,494 --> 00:11:51,755 functions. So, so any choice, any choice of wave 145 00:11:51,755 --> 00:11:56,810 function in this region is going to be oscillatory. 146 00:11:56,810 --> 00:12:00,888 You might, you might ask yourself, what does that correspond to physically? 147 00:12:00,888 --> 00:12:06,690 Now, once again, in, there's another region. 148 00:12:06,690 --> 00:12:10,722 And I guess you can see, because if rho greater than zero, this is sort, sort of 149 00:12:10,722 --> 00:12:14,187 corresponds to the negative sine equivalent for the the constant 150 00:12:14,187 --> 00:12:20,000 potential. So now, there is one unique solution. 151 00:12:20,000 --> 00:12:22,220 Again, up to an overhaul multiplicative constant. 152 00:12:22,220 --> 00:12:29,564 But one unique solution that converges as, as rho goes to infinity and another 153 00:12:29,564 --> 00:12:35,259 that diverges. So these two have applications. 154 00:12:35,259 --> 00:12:37,380 We'll talk about this one, the quantum bouncing ball. 155 00:12:37,380 --> 00:12:42,580 and the Bi's is, is arises in field ionization, and scanning tunneling 156 00:12:42,580 --> 00:12:47,733 microscopy. Well, just important, how important is 157 00:12:47,733 --> 00:12:51,023 the quantum mechanical bouncing ball? Well, might you ask. 158 00:12:51,023 --> 00:12:53,732 Well first off, it's, it's a good example, because it, it it's, it's kind 159 00:12:53,732 --> 00:12:56,570 of like the, you might say in some sense the simplest case of a balanced state 160 00:12:56,570 --> 00:13:01,533 problem one has. But in fact, it's been the subject of a 161 00:13:01,533 --> 00:13:07,081 very fascinating recent experiment of trapping neutrons the neutral counterpart 162 00:13:07,081 --> 00:13:12,290 to the proton in the earth's gravitational field. 163 00:13:12,290 --> 00:13:17,911 So, in this experiment very cold neutrons are put into a chamber and they, they 164 00:13:17,911 --> 00:13:22,656 they're reflected by a mirror, and so, they bounce back and forth in the earth, 165 00:13:22,656 --> 00:13:29,330 earth's gravitational field. Now, you remember this map of all the 166 00:13:29,330 --> 00:13:32,722 theories that Victor showed in the introductory lectures, that showed that 167 00:13:32,722 --> 00:13:35,690 gravity was something kind of disconnected from the main themes of 168 00:13:35,690 --> 00:13:42,780 quantum mechanics and, and relativity. While here in this experiment, they're 169 00:13:42,780 --> 00:13:47,330 actually looking at the quantized energy levels of a particle in the earth's, 170 00:13:47,330 --> 00:13:52,780 subject to the earth's gravitational field. 171 00:13:52,780 --> 00:13:57,516 So this is a test of whether a quantum particle performs as expected under the 172 00:13:57,516 --> 00:14:01,854 influence of gravity. In principle, this is a very, in my 173 00:14:01,854 --> 00:14:06,100 opinion, very significant, intellectually interesting experiment. 174 00:14:06,100 --> 00:14:10,445 Now actually, I mentioned a predecessor, predecessor in earlier lecture, which is 175 00:14:10,445 --> 00:14:14,104 neutron interferometery in the gravitational field. 176 00:14:14,104 --> 00:14:18,046 well, no surprises were found here, but you might, you might consult this paper, 177 00:14:18,046 --> 00:14:21,510 it's rather interesting. And we're now going to solve the, solve 178 00:14:21,510 --> 00:14:27,565 the problem associated. Okay, just some numbers for you to keep 179 00:14:27,565 --> 00:14:31,627 in mind. here's the standard value for the earth's 180 00:14:31,627 --> 00:14:36,468 gravity, the mass of a neutron. And now, in the way this problem is set 181 00:14:36,468 --> 00:14:40,300 up, here is the, here is the Schrodinger equation that we're solving. 182 00:14:41,340 --> 00:14:46,790 And so this part includes the the linear potential. 183 00:14:46,790 --> 00:14:50,318 So, x is zero on the surface of the mirror, and as you go up, you're in a 184 00:14:50,318 --> 00:14:56,038 potential which is Mgx. and then, the boundary condition on the 185 00:14:56,038 --> 00:15:02,743 mirror it's a bouncing ball. So we require the wave function to vanish 186 00:15:02,743 --> 00:15:06,414 there. All right, so now I just have a sanity 187 00:15:06,414 --> 00:15:10,560 check for you to answer in terms of an in video quiz. 188 00:15:13,690 --> 00:15:16,025 Okay, I admit, that was another unit check. 189 00:15:16,025 --> 00:15:20,660 But to tell you the truth, I always check units. 190 00:15:20,660 --> 00:15:23,440 I never, I hope never to forget to do that. 191 00:15:23,440 --> 00:15:28,524 You catch so many errors that way. And what I hope many of your saw is that 192 00:15:28,524 --> 00:15:37,055 you didn't really need to do deep analysis to figure out what alpha is. 193 00:15:37,055 --> 00:15:42,775 Because you see in this equation, alpha rho is equal to E, and rho is 194 00:15:42,775 --> 00:15:48,060 dimensionless. So, that means that alpha must be equal 195 00:15:48,060 --> 00:15:53,569 to energy or another way of noting that is when you put Mgx. 196 00:15:53,569 --> 00:15:57,921 That's got to equal to that's got to be equal to alpha rho, and rho again has 197 00:15:57,921 --> 00:16:02,145 been chosen manifestly to be dimensionless. 198 00:16:04,990 --> 00:16:10,192 Okay, now, I have another brief quiz. basically, the same sort of idea of 199 00:16:10,192 --> 00:16:17,900 getting familiar with the details of the problem as it's laid out here. 200 00:16:17,900 --> 00:16:23,440 Okay, so this, the answer was rho is this classical turning point. 201 00:16:23,440 --> 00:16:28,110 It's the highest point reached by the neutron's trajectory. 202 00:16:28,110 --> 00:16:34,062 And I guess you can see that, here, in the sense that for rho equal to zero, we 203 00:16:34,062 --> 00:16:42,210 have E is equal to Mgx. So that means all the energy is in the 204 00:16:42,210 --> 00:16:48,912 potential energy. So that, that rho of zero is the turning 205 00:16:48,912 --> 00:16:54,450 point to the classical motion. So now, you can see the influence of the 206 00:16:54,450 --> 00:16:58,840 turning point here, that's where the wave functions starts to die off. 207 00:16:58,840 --> 00:17:02,374 The sort of the, the lore is there's like a quarter of a wavelength of damped 208 00:17:02,374 --> 00:17:06,669 oscillation there at the just beyond the turning point. 209 00:17:07,820 --> 00:17:13,276 And now, do you see how we're going to solve the the, the problem with the 210 00:17:13,276 --> 00:17:18,588 boundary condition? We, we're going to require that the wave 211 00:17:18,588 --> 00:17:22,270 function vanished at the surface of the mirror. 212 00:17:22,270 --> 00:17:25,040 Well, the surface of the mirror is given by this. 213 00:17:25,040 --> 00:17:32,640 And so, and, but the wave function is, is just is the most general solution here is 214 00:17:32,640 --> 00:17:41,036 Ai of rho minus the sum rho naught. And so, when we choose the values of rho 215 00:17:41,036 --> 00:17:46,668 to correspond to the nodes of the area function here, then you might say, this 216 00:17:46,668 --> 00:17:54,466 here, the profile from here on out. That is the amplitude of the ground 217 00:17:54,466 --> 00:18:02,140 state, because it's the lowest energy solution that vanishes on the mirror. 218 00:18:02,140 --> 00:18:08,720 Now, to find the next one, you're basically moving the mirror down. 219 00:18:08,720 --> 00:18:13,060 You might say you could think of a different experiment where you drop, you 220 00:18:13,060 --> 00:18:17,680 drop the neutron from a certain height, and then you, you keep moving the mirror 221 00:18:17,680 --> 00:18:24,498 down until you get, until you get sort of a perfect reflection. 222 00:18:24,498 --> 00:18:28,658 And so, the sequence of excited states can be, can be determined by picking off 223 00:18:28,658 --> 00:18:35,788 the area functions. And so once again, it's sort of, here's 224 00:18:35,788 --> 00:18:40,496 the, here's the energy. It's, it's the scaling constant times 225 00:18:40,496 --> 00:18:44,034 these rhos, the lowest one, you know, it's, it's 2.34, it's of the order of 226 00:18:44,034 --> 00:18:47,952 [UNKNOWN]. So, this is the case where, just 227 00:18:47,952 --> 00:18:52,168 calculating the scaling constant gives you a good prediction in the energy scale 228 00:18:52,168 --> 00:18:59,709 of the system. Now, here is the the quadratic equation. 229 00:18:59,709 --> 00:19:04,061 now, I'm just going to take the, the the case where it has a positive quadratic 230 00:19:04,061 --> 00:19:10,050 coefficient, which does correspond to the, the harmonic oscillator. 231 00:19:10,050 --> 00:19:16,826 And what you do is you you, you, you, you transform, so this, this potential always 232 00:19:16,826 --> 00:19:23,313 has a minimum somewhere. You just move to that minimum then 233 00:19:23,313 --> 00:19:28,478 there's an energy offset here. And then as we know from the analytic 234 00:19:28,478 --> 00:19:33,734 solution of the harmonic oscillator, the, the spectrum is given by this, but again, 235 00:19:33,734 --> 00:19:39,720 let's look at it from the general, the general perspective. 236 00:19:39,720 --> 00:19:44,356 So, the standard equation for the quadratic potential is what's called a 237 00:19:44,356 --> 00:19:49,752 parabolic cylinder function. And in particular, there is a class of 238 00:19:49,752 --> 00:19:55,990 well-studied solutions traveling under this name, the D, D sub n's. 239 00:19:55,990 --> 00:20:05,384 And the two solutions to the, this equation are Dn of rho and Dn of minus 240 00:20:05,384 --> 00:20:11,582 rho. And it so happens that these are chosen 241 00:20:11,582 --> 00:20:17,050 so that for positive n, Dn of rho always converges. 242 00:20:17,050 --> 00:20:20,830 So this would be something quite useful to know if you were dealing with a 243 00:20:20,830 --> 00:20:24,880 problem, which was sort of piece wise quadratic. 244 00:20:24,880 --> 00:20:29,232 And you wanted to make sure that you were getting a solution to tail off in the in 245 00:20:29,232 --> 00:20:36,656 the class of forbidden region. on the other hand, there in the region 246 00:20:36,656 --> 00:20:46,730 rho less than zero both of these diverge unless n happens to be an integer. 247 00:20:46,730 --> 00:20:50,650 And that of course, is the here's, here's the n, you see. 248 00:20:50,650 --> 00:20:53,786 And that, that, that of course corresponds to the solution from the 249 00:20:53,786 --> 00:20:57,460 harmonic oscillator. So those are just shown here. 250 00:20:57,460 --> 00:21:00,042 Here is n equals zero. It's the ground state of the harmonic 251 00:21:00,042 --> 00:21:03,306 oscillator n equal to three. But now, he sees, n, for n equal 3 252 00:21:03,306 --> 00:21:07,454 halves, you get something that's kind of harmonic oscillator looking, but do you 253 00:21:07,454 --> 00:21:13,471 see it's not crossing? It's not crossing it's not, it's not 254 00:21:13,471 --> 00:21:19,220 intersecting rho equals zero, so it can't be an odd solution. 255 00:21:19,220 --> 00:21:22,441 And it doesn't have a math a local maximum of rho equals zero. 256 00:21:22,441 --> 00:21:25,321 So it can't it doesn't have a zero derivative, rho equal to zero, so it 257 00:21:25,321 --> 00:21:30,704 can't be infinite an even solution. So, these, these all, these all diverge 258 00:21:30,704 --> 00:21:35,404 at large values of negative rho. Okay, I think, now is a good time to 259 00:21:35,404 --> 00:21:39,612 leave off. And just imagine again that I think 260 00:21:39,612 --> 00:21:44,235 you'll, those of you who like to solve problems should find some benefit from 261 00:21:44,235 --> 00:21:50,400 working out the numerical solutions to some of these equations. 262 00:21:50,400 --> 00:21:54,370 You can modify this spreadsheet to treat potentials of your own choosing. 263 00:21:54,370 --> 00:21:58,150 By the way, it's not, it's not a highly competitive method of solving an ordinary 264 00:21:58,150 --> 00:22:03,050 differential equations. It's just something that happens to be 265 00:22:03,050 --> 00:22:07,060 easy to code up, it's [UNKNOWN] when understood. 266 00:22:07,060 --> 00:22:11,250 And you could, you know, you can see the solution happening in real time. 267 00:22:11,250 --> 00:22:17,154 again, there's also a a summary of the cases that are most readily treated by 268 00:22:17,154 --> 00:22:20,710 use of special functions.