1 00:00:00,790 --> 00:00:03,398 Welcome back everyone. I'm Charles Clark, and this is Exploring 2 00:00:03,398 --> 00:00:05,910 Quantum Physics. In this part we're going to look at 3 00:00:05,910 --> 00:00:09,160 solving the Shroedinger Equation for Systems that have well defined Angular 4 00:00:09,160 --> 00:00:13,576 Momentum. This means typically systems that have 5 00:00:13,576 --> 00:00:19,284 some sort of spherical symmetry. those include, for example, the Hydrogen 6 00:00:19,284 --> 00:00:22,230 Atom and the Isotropic Harmonic oscillator. 7 00:00:22,230 --> 00:00:26,446 We'll touch on both of those during this part, but we'll focus mostly on the 8 00:00:26,446 --> 00:00:30,792 hydrogen atom. And try to bring matters to closure by 9 00:00:30,792 --> 00:00:35,212 generating the Eigenfunctions of the hydrogen atom, really in a very similar 10 00:00:35,212 --> 00:00:42,100 to the way that Schrodinger first used. Just a refresher, here's what we looked 11 00:00:42,100 --> 00:00:47,200 at previously concerning the solution of a Schrodinger equation for a radial 12 00:00:47,200 --> 00:00:52,248 function And we, we happened to choose a function 13 00:00:52,248 --> 00:00:57,240 that corresponds to the ground state of the hydrogen atom. 14 00:00:57,240 --> 00:01:01,193 Now we're going to apply the general result of the previous part, which is 15 00:01:01,193 --> 00:01:04,973 that We have a universal set of functions that 16 00:01:04,973 --> 00:01:09,461 characterize the angular momentum of a system, of a, a particle moving in three 17 00:01:09,461 --> 00:01:15,162 dimension potential. And so we can look at eigenfunctions that 18 00:01:15,162 --> 00:01:19,660 separate the radial from the angular coordinates. 19 00:01:19,660 --> 00:01:23,692 And it doesn't matter, because it going to, quite a remarkable thing is, in 20 00:01:23,692 --> 00:01:28,876 quantum mechanics these, these functions are, are independent of the particular 21 00:01:28,876 --> 00:01:34,425 form of the potential. So for example, this decomposition holds 22 00:01:34,425 --> 00:01:38,512 equally well to the isotropic harmonic oscillator in the column potential and a 23 00:01:38,512 --> 00:01:44,140 number of others. So let's proceed. 24 00:01:44,140 --> 00:01:49,253 Here's the Schrodinger equation. Time independent Schrodinger equation, 25 00:01:49,253 --> 00:01:53,590 usual form. And now we we have this, this identity 26 00:01:53,590 --> 00:01:59,690 that you proved a while ago. That relates the Laplacian, the kinetic 27 00:01:59,690 --> 00:02:04,450 energy operator, to some radial derivatives in the kinetic energy 28 00:02:04,450 --> 00:02:10,832 operator. And so when you when you plug that in, 29 00:02:10,832 --> 00:02:18,077 and just take account of the fact that the L squared operates only on the Yl,m 30 00:02:18,077 --> 00:02:27,654 and returns an eigenvalue. So l squared ylm equal h bar squared l 31 00:02:27,654 --> 00:02:33,180 times l plus 1. that's, that is the eigenvalue, and so 32 00:02:33,180 --> 00:02:38,484 that gives what looks like a sort of an, it has the appearance of a repulsive 33 00:02:38,484 --> 00:02:45,438 inverse square potential there. So here is the here's, here's where we 34 00:02:45,438 --> 00:02:50,046 get when we just substitute the the separated, it's the product into the, 35 00:02:50,046 --> 00:02:58,850 into the Schrodinger equation. And now, my advice is always to go to 36 00:02:58,850 --> 00:03:04,976 dimensionless variables at your first occasion. 37 00:03:04,976 --> 00:03:09,545 It's unwise to just start setting all the coefficients equal to 1. 38 00:03:09,545 --> 00:03:15,000 Because, you're bound to make a mistake, it then becomes untraceable. 39 00:03:15,000 --> 00:03:17,080 It's like sort of a one way hash function. 40 00:03:17,080 --> 00:03:21,112 If you set everything to 1, or do it different times in the calculation, it's 41 00:03:21,112 --> 00:03:26,830 very difficult to back out where the error is when you make one. 42 00:03:26,830 --> 00:03:30,720 and, you know, in these, in these sorts of problems, errors are frequent. 43 00:03:32,050 --> 00:03:39,740 So generally speaking this is, this is in some sense a generally valid approach. 44 00:03:39,740 --> 00:03:44,150 And that is you identify a length scale, so r has the dimensions of length, you 45 00:03:44,150 --> 00:03:49,010 know, for a real problem it's in nanometers or something. 46 00:03:49,010 --> 00:03:52,597 And you choose a variable row which is dimensionless. 47 00:03:57,250 --> 00:04:03,452 And then a has the units of length. then, then this equation this equation 48 00:04:03,452 --> 00:04:07,130 takes this form. And I've done a few things. 49 00:04:07,130 --> 00:04:13,125 I've preserved this one half here you might say for sentimental reasons. 50 00:04:13,125 --> 00:04:17,367 It helps me rem, it helps remind me of relationships that exist. 51 00:04:17,367 --> 00:04:21,291 you may prefer it a different way. This is really just an arbitrary 52 00:04:21,291 --> 00:04:28,008 convention but you should find that this equation's identical to the original one. 53 00:04:28,008 --> 00:04:32,904 With the redefinition of the energy and the potential, and also something to keep 54 00:04:32,904 --> 00:04:38,780 carefully in mind Is that the wave function has unitive dimension. 55 00:04:38,780 --> 00:04:46,519 their length of minus three halves for a wave function three dimensions. 56 00:04:46,519 --> 00:04:55,990 And so, although you need not necessarily make this this conversion at the start. 57 00:04:55,990 --> 00:04:59,050 You're going to have to do it at the end. So why not, why not do it right off the 58 00:04:59,050 --> 00:05:02,269 bat? So let's just have a quick set check of, 59 00:05:02,269 --> 00:05:08,201 your graph for the concepts here. Okay, I hope that little question was 60 00:05:08,201 --> 00:05:13,622 very easy for you to solve. It's, it's, it just shows you a way, it 61 00:05:13,622 --> 00:05:17,486 gives you a reading of how quickly you can do these sorts of conversions between 62 00:05:17,486 --> 00:05:20,090 [UNKNOWN]. 63 00:05:20,090 --> 00:05:23,402 Again the point of being a physicist is actually to get beyond that, but its very 64 00:05:23,402 --> 00:05:27,280 helpful to start, to always have a grasp on the units in mind. 65 00:05:31,010 --> 00:05:38,218 Now, this equation we've gotten, so far, in dimensionless form, has singular 66 00:05:38,218 --> 00:05:41,960 points. Some of them are apparent. 67 00:05:41,960 --> 00:05:46,621 Near row equals zero there's a singularity structure and then there's 68 00:05:46,621 --> 00:05:53,300 also one at large values of row. Now its nature depends on the potential. 69 00:05:53,300 --> 00:05:56,295 We, we're going to address that momentarily. 70 00:05:56,295 --> 00:06:01,825 But there's really a universal problem of the singularity in this equation near row 71 00:06:01,825 --> 00:06:05,945 equal to zero. And that has to do, you might think about 72 00:06:05,945 --> 00:06:09,585 it. If you have, here's, let's say, circle 73 00:06:09,585 --> 00:06:16,033 surrounding the origin. So this is the point, r equal to zero. 74 00:06:16,033 --> 00:06:21,358 And it, for example, that, that the, the ground state of the Coulomb wave function 75 00:06:21,358 --> 00:06:28,800 we looked at actually has its second derivatives discontinuous at the origin. 76 00:06:28,800 --> 00:06:33,490 It's still allowed solution, because that discontinuity of the second derivative, 77 00:06:33,490 --> 00:06:38,360 doesn't involve a discontinuity in the energy or anything. 78 00:06:38,360 --> 00:06:42,824 This occurs as an isolated point. Nevertheless it's it's, it's really 79 00:06:42,824 --> 00:06:47,916 critical to make sure that you're looking at the wave function in a way that takes 80 00:06:47,916 --> 00:06:52,899 account of any possible singular structure. 81 00:06:54,410 --> 00:07:03,450 And I guess the only way to really do this is to look at some examples. 82 00:07:03,450 --> 00:07:07,319 So here, here are two relevant examples pictures taken from the digital library 83 00:07:07,319 --> 00:07:11,110 of mathematical functions chapter on Bessel functions. 84 00:07:11,110 --> 00:07:15,202 The Bessel function, is a, a motion, a wave function describes a motion of a 85 00:07:15,202 --> 00:07:19,122 particle in two dimensions. That's just, these are pretty good 86 00:07:19,122 --> 00:07:22,480 images. And so there are, there are, there are 87 00:07:22,480 --> 00:07:28,207 two types as we saw in examples I covered back in the first discussion of solving 88 00:07:28,207 --> 00:07:35,223 the Schrodinger equation. there are functions that are oscillatory, 89 00:07:35,223 --> 00:07:41,060 and then they're, in this case they're called j and y, are the two families. 90 00:07:41,060 --> 00:07:45,616 And they I want to say j0 and y0, I guess are the things that should be compared 91 00:07:45,616 --> 00:07:50,932 here. So if they oscillate together out of 92 00:07:50,932 --> 00:07:56,060 phase but they're both always oscillatory. 93 00:07:56,060 --> 00:07:59,450 But when one ap, approaches the singular point. 94 00:07:59,450 --> 00:08:11,860 There's one family, the j is regular, near r equals 0 and y is irregular. 95 00:08:16,460 --> 00:08:20,177 Then on the other hand, there are functions which travel in the so called 96 00:08:20,177 --> 00:08:24,736 forbidden regions. This is, let's say, what happens when a 97 00:08:24,736 --> 00:08:28,860 parquil encounters a barrier and can't pass over the barrier. 98 00:08:28,860 --> 00:08:33,360 Well, physically we know the wave function has to has to die off as one 99 00:08:33,360 --> 00:08:38,801 goes into the barrier. But, very often, the wave function that 100 00:08:38,801 --> 00:08:44,520 dies off there, near the singular point, will, will always diverge. 101 00:08:44,520 --> 00:08:49,500 In fact, this is, this is always the case for a free particle. 102 00:08:49,500 --> 00:08:53,847 And then correspondingly, a wave function that is regular near the singular point 103 00:08:53,847 --> 00:08:59,020 will necessarily always diverge. So it's, it's building the appropriate 104 00:08:59,020 --> 00:09:05,466 behavior into the wave function. It's very important at the initial stage. 105 00:09:05,466 --> 00:09:11,946 Okay, now how do we isolate the singular points near the, near, near the near the 106 00:09:11,946 --> 00:09:17,204 cordon origin. Well, I'm emphasizing this because it 107 00:09:17,204 --> 00:09:22,028 turns out that for most potentials, this analysis is rather, is independent of the 108 00:09:22,028 --> 00:09:27,645 details of the potential. It's really sort of the very high energy 109 00:09:27,645 --> 00:09:32,145 limit of the kinetic energy operator that's involved in this very close 110 00:09:32,145 --> 00:09:37,955 approach to the origin. And unless, unless, a potential is more 111 00:09:37,955 --> 00:09:42,530 singular than the inverse square and we don't really know of any physical 112 00:09:42,530 --> 00:09:47,286 potential Any potential that can be made in the 113 00:09:47,286 --> 00:09:53,990 laboratory that has that type of singular behavior all the way to the origin. 114 00:09:53,990 --> 00:09:59,170 unless it would have that, then it will be dominated at small rho by the effects 115 00:09:59,170 --> 00:10:05,860 of kinetic energy. And so what one does near the singular 116 00:10:05,860 --> 00:10:14,500 point is to assume a a a form of this type where this, this unknown exponent nu 117 00:10:14,500 --> 00:10:25,027 will supply the desire. By an appropriate choice, will supply the 118 00:10:25,027 --> 00:10:30,970 desired behavior. And so, your row equal to 0, you just, 119 00:10:30,970 --> 00:10:36,180 you retain the leading order term in this expansion. 120 00:10:37,480 --> 00:10:40,952 And what you find is that nu has got to be either, either equal to l, which is a 121 00:10:40,952 --> 00:10:46,478 positive number, or minus l plus 1. And these correspond to the solutions 122 00:10:46,478 --> 00:10:53,261 that are regular and irregular near the origin, respectively, so we choose the 123 00:10:53,261 --> 00:11:06,343 We choose we choose nu equal to l near rho equal to 0. 124 00:11:06,343 --> 00:11:11,179 And that, that then leads to this, that then leads to this equation for phi, 125 00:11:11,179 --> 00:11:19,090 which is a scaled version of psi. Okay, now we face the issue of large rho 126 00:11:19,090 --> 00:11:22,630 behavior, and this is completely determined by the behavior of the 127 00:11:22,630 --> 00:11:26,820 potential. So, a basic message to keep in mind is 128 00:11:26,820 --> 00:11:31,300 when we're, we're computing a normalizable state. 129 00:11:31,300 --> 00:11:36,316 That is a, a, eigenfunction whose, the interval of whose square of all spaces 130 00:11:36,316 --> 00:11:41,332 equal to one. That means that it must converge at large 131 00:11:41,332 --> 00:11:45,480 distances from the origin. And so, in other words, it's gotta be 132 00:11:45,480 --> 00:11:49,759 confined by something. And the confinement typic, condition, is 133 00:11:49,759 --> 00:11:55,546 typically written this way. it's, it's confined in the, in the region 134 00:11:55,546 --> 00:11:59,706 where the energy Is less than the potential energy. 135 00:11:59,706 --> 00:12:03,366 sorry, the barri, the barrier begins where the energy is less than the 136 00:12:03,366 --> 00:12:06,615 potential energy. It's confined to a region where the 137 00:12:06,615 --> 00:12:09,500 energy's greater than the potent, the potential energy. 138 00:12:09,500 --> 00:12:13,709 And then, that's where we need to make sure that we don't pick up diverging 139 00:12:13,709 --> 00:12:23,656 solutions. We actually saw this behavior before and 140 00:12:23,656 --> 00:12:29,464 it, it was in the context of something that was put as a problem in the week six 141 00:12:29,464 --> 00:12:36,462 homework. The quantum mechanical bouncing ball, and 142 00:12:36,462 --> 00:12:45,320 so in that case this direction here is let's say, sorry, the other way. 143 00:12:50,660 --> 00:12:55,330 That's the direction of gravity pointing towards negative rho. 144 00:12:55,330 --> 00:12:58,680 And really this is just, this just has to do with convention for defining the area 145 00:12:58,680 --> 00:13:01,790 function. You could make it the other way if you 146 00:13:01,790 --> 00:13:05,806 like. so basically as you may recall, if we 147 00:13:05,806 --> 00:13:11,482 have a classical ball and this point defines the classical turning point of 148 00:13:11,482 --> 00:13:17,290 its trajectory. It bounces back and forth. 149 00:13:17,290 --> 00:13:20,810 It goes, it goes, slightly into the forbidden region and as described by this 150 00:13:20,810 --> 00:13:24,695 airy function Ai which decays in the forbidden region. 151 00:13:24,695 --> 00:13:29,760 The, the, the of the complementary function, always diverges. 152 00:13:29,760 --> 00:13:34,840 In fact, the Ai is unique. You could add a small amount of Ai to a 153 00:13:34,840 --> 00:13:38,545 Bi to get another solution, and you'd never know the difference, because it's 154 00:13:38,545 --> 00:13:45,452 so rapidly divergent. So we want to end our discussion of 155 00:13:45,452 --> 00:13:52,086 motion and potential. we, we, you know, kind of easily generate 156 00:13:52,086 --> 00:13:57,040 the motion in the classically allowed region. 157 00:13:57,040 --> 00:14:01,396 But then there's a quantization condition that, that only picks up the sorts of 158 00:14:01,396 --> 00:14:07,298 energies that match onto the converging solutions in the forbidden region. 159 00:14:07,298 --> 00:14:14,431 Okay, so at this stage, we actually have to discuss specific potential. 160 00:14:14,431 --> 00:14:20,801 Because, for example, for the harmon, isotropic harmonic oscillator, this new 161 00:14:20,801 --> 00:14:27,012 rho goes as rho squared. At large rho for the coolant potential 162 00:14:27,012 --> 00:14:32,970 goes as minus 1 over rho, which, which dies off. 163 00:14:32,970 --> 00:14:37,074 So we'll treat the hydrogen atom and then I'll make some comments in the end that 164 00:14:37,074 --> 00:14:40,779 will enable you to think about how you would apply it to the isotropic harmonic 165 00:14:40,779 --> 00:14:46,977 oscillator. So we're going to do do the the classical 166 00:14:46,977 --> 00:14:53,010 problem assuming that the, the nucleus has infinite mass. 167 00:14:53,010 --> 00:14:56,574 That just means that the, all the orbital motion is tied up in the electron, so we 168 00:14:56,574 --> 00:14:59,760 use for the mass of the par, of the moving particle, we use the mass of the 169 00:14:59,760 --> 00:15:05,018 electron. And this actually, this is, I showed you 170 00:15:05,018 --> 00:15:10,430 how to, how to get away from this restriction. 171 00:15:10,430 --> 00:15:15,330 In the material on the Bohr model, but its very useful for matching answers to 172 00:15:15,330 --> 00:15:22,658 precise experiment data. And so the interaction is in, this choice 173 00:15:22,658 --> 00:15:27,986 Gaussian units, this cooling interaction, and when we take the mass, the, the mass 174 00:15:27,986 --> 00:15:33,600 of the electron and the scaling parameter A. 175 00:15:33,600 --> 00:15:38,800 Which originally was just chosen, it wasn't identified what it was, we take it 176 00:15:38,800 --> 00:15:44,339 to be the bohr radius. Then as it turns out when you go through 177 00:15:44,339 --> 00:15:49,650 these these identities, you might try that. 178 00:15:49,650 --> 00:15:53,310 You get this very simple solution for the potential in the dimensionless 179 00:15:53,310 --> 00:15:58,626 coordinates. So, here is the equation with what, with 180 00:15:58,626 --> 00:16:04,641 which we must deal. I dunno, maybe it, you know, I guess it's 181 00:16:04,641 --> 00:16:07,920 a little bit more specific than what you've seen previously. 182 00:16:07,920 --> 00:16:12,272 Nothing here is very complicated, but it's solution takes a little bit of 183 00:16:12,272 --> 00:16:15,752 thought. So let's get started. 184 00:16:15,752 --> 00:16:23,012 Now, as we discussed before, a bound state in the hydrogen atom is one for 185 00:16:23,012 --> 00:16:30,839 which the energy. Well, it's for which the energy is less 186 00:16:30,839 --> 00:16:39,888 than 0, and that means that this parameter epsilon is less than 0. 187 00:16:39,888 --> 00:16:48,116 So I'm writing epsilon as in this form, minus kappa squared over 2, where kappa 188 00:16:48,116 --> 00:16:54,046 is real. And again, this is completely this is a 189 00:16:54,046 --> 00:16:58,360 conventional designation. You could you, you could use another 190 00:16:58,360 --> 00:17:02,370 terminology if you like. It doesn't really matter, what you call 191 00:17:02,370 --> 00:17:05,297 this. It's just the, the, the, the point is, 192 00:17:05,297 --> 00:17:10,639 you want to, It is important to retain in the, 193 00:17:10,639 --> 00:17:14,548 equation. A meaningful signature that epsilon is 194 00:17:14,548 --> 00:17:16,949 negative. And maybe, maybe this is the way to see 195 00:17:16,949 --> 00:17:21,864 why, why this choice is made. Because, when, for epsilon, less than 196 00:17:21,864 --> 00:17:26,930 zero, if we go to large values of row in this equation. 197 00:17:26,930 --> 00:17:30,800 Then everything else dies off with a power one over row. 198 00:17:30,800 --> 00:17:35,424 And basically asymptotically, you, you know, there are very large values in a 199 00:17:35,424 --> 00:17:39,564 row. You just have to solve the equation that 200 00:17:39,564 --> 00:17:45,284 contains these two terms, and of course that is one, for which there is also an, 201 00:17:45,284 --> 00:17:54,858 an exponential plus kappa row solution. So what we are we are setting up this 202 00:17:54,858 --> 00:18:00,910 condition and finding it a new function, f, which is associated with the decaying 203 00:18:00,910 --> 00:18:06,832 exponential. And we want f to behave so that, 204 00:18:06,832 --> 00:18:14,530 basically, it exhausts the function. In other words, there can always be a, a 205 00:18:14,530 --> 00:18:20,034 small amount of the diverging solution that gets into the description of our 206 00:18:20,034 --> 00:18:26,510 wave function. And that would, that would poison it. 207 00:18:26,510 --> 00:18:33,354 So the way that this is done in As you'll see the approach, it has some 208 00:18:33,354 --> 00:18:36,578 similarities with the numerical integration method that's discussed in 209 00:18:36,578 --> 00:18:43,956 the additional materials. Okay, so there with this substitution, we 210 00:18:43,956 --> 00:18:51,337 now, we obtain this equation, this differential equation for the the 211 00:18:51,337 --> 00:19:03,182 function f, this one here. Now, you must be wondering what's the 212 00:19:03,182 --> 00:19:06,152 point? It just doesn't look, it doesn't look any 213 00:19:06,152 --> 00:19:09,385 simpler than, than, than, than the original equation, but it's got these 214 00:19:09,385 --> 00:19:14,720 different parts. But here's something to notice. 215 00:19:14,720 --> 00:19:20,920 The terms underlined in red here, you can see it's a second derivative. 216 00:19:20,920 --> 00:19:24,516 So this is basically like a 1 over rho squared acting on f, and this is a first 217 00:19:24,516 --> 00:19:30,785 derivative divided by 1 over rho. So there's a scaling down by a a rho 218 00:19:30,785 --> 00:19:37,006 squared. So if you think, for example, of if you 219 00:19:37,006 --> 00:19:44,188 have the power series expansion for f and you had a term, you know, which was f 220 00:19:44,188 --> 00:19:53,919 sub, sub j, x to the j. Then these terms here would, would, would 221 00:19:53,919 --> 00:20:03,172 take you down to x to the j minus 2. And then the other terms here are, the 222 00:20:03,172 --> 00:20:09,428 first order in rho, a derivative and, and a division, so for the same thing x of j 223 00:20:09,428 --> 00:20:15,906 x to the j. This takes you down to x to the j minus 224 00:20:15,906 --> 00:20:22,405 one, so this form gives 1 a 2 terms recurrence relationship, between the, 225 00:20:22,405 --> 00:20:30,590 between the coefficients in this, in this expa, expansion. 226 00:20:30,590 --> 00:20:33,490 And that, that recurrence relationship is solvable. 227 00:20:33,490 --> 00:20:39,806 It's basically, you know? If you have ai plus 1 is equal f, f sub 228 00:20:39,806 --> 00:20:46,080 i, ai. That's like a geometric series. 229 00:20:46,080 --> 00:20:49,561 You can very often find an exact solution, or, at the least, you can, you 230 00:20:49,561 --> 00:20:54,076 can numerically solve it quite readily. If you have a three term recurrence 231 00:20:54,076 --> 00:20:57,450 formula, on the other hand. It becomes, much more, much more 232 00:20:57,450 --> 00:21:04,038 demanding. Okay, I recognize that there is a lot of 233 00:21:04,038 --> 00:21:10,185 mathematical detail in this presentation. And I can only say this is a sort of 234 00:21:10,185 --> 00:21:14,773 thing that if you go through it and learn it, you will get insight into how to 235 00:21:14,773 --> 00:21:20,969 solve a larger class of problems. But basically, you can see that 236 00:21:20,969 --> 00:21:24,332 everything that's in the chain of reasoning here, they're all these 237 00:21:24,332 --> 00:21:29,635 transformations. and what they lead to is the discovery 238 00:21:29,635 --> 00:21:35,267 that if this, if this power, if this power series terminates at a finite value 239 00:21:35,267 --> 00:21:41,140 of n. Then it turns out that that corresponds 240 00:21:41,140 --> 00:21:48,636 to energy eigenvalue, which gives back the, the result obtained from the 241 00:21:48,636 --> 00:21:54,884 Well actually we didn't derive the result obtained from the symmetry arguments, but 242 00:21:54,884 --> 00:22:01,058 we outlined t hose symmetry arguments. those, those are things you really have 243 00:22:01,058 --> 00:22:04,692 to do by yourself to see how they, work. But we got the same, we got the same 244 00:22:04,692 --> 00:22:12,518 result from this solution. The differential equation as is done by 245 00:22:12,518 --> 00:22:20,949 the symmetry argument. The other thing that is inferred from the 246 00:22:20,949 --> 00:22:26,169 analysis is that if the, if the power series doesn't terminate, then the 247 00:22:26,169 --> 00:22:35,200 solution will diverge exponentially. Because you will have in it, you'll have 248 00:22:35,200 --> 00:22:41,660 terms in this, the higher order terms in the power series here as you, as you 249 00:22:41,660 --> 00:22:49,725 increase the the value of the index indefinitely 250 00:22:49,725 --> 00:22:56,560 Will turn out to build up into an exponentially diverging function. 251 00:22:56,560 --> 00:23:02,727 Now I haven't made it an assignment. But I put on the in the additional 252 00:23:02,727 --> 00:23:08,915 materials part of the course page, a spreadsheet that enables you to solve the 253 00:23:08,915 --> 00:23:17,249 Schrodinger Equation For the even parity states, the harmonic 254 00:23:17,249 --> 00:23:23,459 oscillator and the isotropics, the s states, the l equals zero states of the 255 00:23:23,459 --> 00:23:29,902 hydrogen atom. Solve the Schrodinger equation by 256 00:23:29,902 --> 00:23:35,836 numerical integration using a boundary condition that either has the value of 257 00:23:35,836 --> 00:23:43,800 the function or its derivative vanishing. At a point, and then Well, I maybe need 258 00:23:43,800 --> 00:23:50,558 to going into the spreadsheet to see it. Using the energies and iterative 259 00:23:50,558 --> 00:23:55,574 parameter, and seeing what happens as you vary the, the energy around the exact 260 00:23:55,574 --> 00:24:01,230 value, and what you find here, here's an example. 261 00:24:01,230 --> 00:24:05,073 Where 1 is solving the Schrodinger Equation with the hydrogen, but with a 262 00:24:05,073 --> 00:24:09,511 slightly wrong energy. This isn't a representation of whether 263 00:24:09,511 --> 00:24:12,425 there's an additional factor of r out of a wave functional order to make the 264 00:24:12,425 --> 00:24:16,054 boundary condition there. But basically what you do is you start 265 00:24:16,054 --> 00:24:19,156 out with uh,with a wave function that looks like its going to converge, then 266 00:24:19,156 --> 00:24:22,298 you, as you get to larger and larger r you. 267 00:24:22,298 --> 00:24:27,043 It, it diverges as you get closer to the energy, then you follow the converging 268 00:24:27,043 --> 00:24:33,970 trajectory for longer time. But eventually, eventually you'll take 269 00:24:33,970 --> 00:24:37,515 off and diverge. So finding the exact solution, or finding 270 00:24:37,515 --> 00:24:41,114 a good approximation of the, a good numerical approximation of the exact 271 00:24:41,114 --> 00:24:45,820 solution of these cases Is done by iterating the energy so that 272 00:24:45,820 --> 00:24:49,655 you basically postpone this divergence, which is, just because you're using 273 00:24:49,655 --> 00:24:54,421 finite precision arithmetic. You're always going to get divergence of 274 00:24:54,421 --> 00:24:57,310 some sort. You're sort of postponing it, inevitably. 275 00:24:57,310 --> 00:25:03,400 And this the, solving this equation is sort of like, solving the power series 276 00:25:03,400 --> 00:25:09,210 termination method that we just discussed. 277 00:25:09,210 --> 00:25:13,640 which was in fact first used by Schrodinger. 278 00:25:13,640 --> 00:25:18,531 So I, I suggest you use other time you might spend listening to additional 279 00:25:18,531 --> 00:25:24,079 lectures to to, acquainting yourself with the, with what happens when you actually 280 00:25:24,079 --> 00:25:29,990 try to solve a Schrodinger equation numerically. 281 00:25:29,990 --> 00:25:32,150 OK that's it, thanks for attending.