1 00:00:00,000 --> 00:00:02,350 PROFESSOR: Find our final solution, 2 00:00:02,350 --> 00:00:05,530 we just have to match the equations. 3 00:00:05,530 --> 00:00:12,175 So Psi continues at x equals a. 4 00:00:15,000 --> 00:00:16,270 And what do we have? 5 00:00:16,270 --> 00:00:22,540 Well from two, you have cosine of ka. 6 00:00:22,540 --> 00:00:31,290 And from four you would have a equals e to the minus ka. 7 00:00:31,290 --> 00:00:34,080 This is the value of this-- 8 00:00:34,080 --> 00:00:36,660 so the interior solution at x equals 9 00:00:36,660 --> 00:00:40,760 a must match the value of the exterior solution of k 10 00:00:40,760 --> 00:00:43,260 equals a. 11 00:00:43,260 --> 00:00:52,040 Psi prime must be continuous at x equals a as well. 12 00:00:52,040 --> 00:00:55,750 Well what is the derivative of this function? 13 00:00:55,750 --> 00:01:01,610 It's minus the sine of this. 14 00:01:01,610 --> 00:01:08,600 So it's minus k sine of kx, that becomes ka, 15 00:01:08,600 --> 00:01:11,580 is equal to the derivative of that one, which 16 00:01:11,580 --> 00:01:20,430 is minus Kappa A e to that minus Kappa little a. 17 00:01:23,050 --> 00:01:25,730 Two equations, and how many unknowns? 18 00:01:25,730 --> 00:01:31,140 Well there's A and some information about Kappa and k. 19 00:01:31,140 --> 00:01:35,150 And the easiest way to eliminate that is to divide them. 20 00:01:35,150 --> 00:01:42,210 So you divide the bottom equation by this equation. 21 00:01:42,210 --> 00:01:44,150 So what do we get? 22 00:01:44,150 --> 00:01:47,510 Divide the bottom by the top. 23 00:01:47,510 --> 00:01:51,040 Minus k and and the minuses cancel, 24 00:01:51,040 --> 00:01:53,875 we can cancel those minus signs and you 25 00:01:53,875 --> 00:02:03,330 get k tan ka is equal to Kappa. 26 00:02:06,400 --> 00:02:12,530 But you already are convinced, I hope, on the idea 27 00:02:12,530 --> 00:02:17,110 that we should not use equations that have units. 28 00:02:17,110 --> 00:02:20,860 So I will multiply by little a and a little a 29 00:02:20,860 --> 00:02:26,470 to get a [INAUDIBLE] size, and therefore, the right hand side 30 00:02:26,470 --> 00:02:35,690 becomes Xi equals, and the left side become Eta tan Eta. 31 00:02:48,740 --> 00:02:52,400 OK, I want to make a little comment about these quantities 32 00:02:52,400 --> 00:02:52,900 already. 33 00:02:52,900 --> 00:02:58,720 So all the problem has turned out into the following. 34 00:02:58,720 --> 00:03:03,250 You were given a potential and that determines a number z0. 35 00:03:03,250 --> 00:03:06,970 If you know the width and everything, you know z0. 36 00:03:06,970 --> 00:03:10,470 Now you have to calculate Eta and Xi. 37 00:03:10,470 --> 00:03:16,120 If you know either Eta or Xi, you know Kappa or k. 38 00:03:16,120 --> 00:03:21,250 And if you know either k or Kappa, since you know v0, 39 00:03:21,250 --> 00:03:23,230 you will know the energy. 40 00:03:23,230 --> 00:03:28,690 So it's kind of neat to express this more clearly, 41 00:03:28,690 --> 00:03:36,260 and I think it's maybe easier if one uses Xi. 42 00:03:36,260 --> 00:03:47,310 And look at Xi squared is Kappa squared times a squared. 43 00:03:47,310 --> 00:03:50,350 And what is Kappa squared, it's over there. 44 00:03:50,350 --> 00:03:57,700 2m absolute value of e, a squared over h bar squared. 45 00:04:01,370 --> 00:04:09,320 Now you want to find e, you're going to get in some units. 46 00:04:09,320 --> 00:04:13,310 Even e is nice to have it without units. 47 00:04:13,310 --> 00:04:19,050 So I will multiply and divide by v0. 48 00:04:19,050 --> 00:04:28,040 2m v0 a squared over h squared, absolute value of e over v0. 49 00:04:28,040 --> 00:04:31,510 After all, you probably prefer to know e 50 00:04:31,510 --> 00:04:36,260 over v0, which tells you how proportional the energy is 51 00:04:36,260 --> 00:04:38,240 to the depth of the potential. 52 00:04:38,240 --> 00:04:41,260 And this is your famous constant z0. 53 00:04:41,260 --> 00:04:52,500 So e over v0 is actually equal to Xi over z0 squared. 54 00:04:52,500 --> 00:04:54,650 And this is something just to keep in mind. 55 00:04:54,650 --> 00:04:58,970 If you know Xi, you certainly must know z0, 56 00:04:58,970 --> 00:05:01,140 because that's not in your potential, 57 00:05:01,140 --> 00:05:04,820 and then you know how much is the energy. 58 00:05:04,820 --> 00:05:08,030 All very convenient things. 59 00:05:08,030 --> 00:05:11,750 So punchline for solutions. 60 00:05:27,910 --> 00:05:29,080 So what do we have? 61 00:05:29,080 --> 00:05:31,170 We have two equations. 62 00:05:31,170 --> 00:05:33,525 This equation maybe should be given a number. 63 00:05:39,110 --> 00:05:45,351 Xi equals 8 at tan Eta and Eta and Xi squared giving you z 64 00:05:45,351 --> 00:05:45,850 squared. 65 00:05:45,850 --> 00:05:47,690 So how do we solve it? 66 00:05:47,690 --> 00:05:49,950 We solve it graphically. 67 00:05:49,950 --> 00:05:55,120 We have Psi, Eta, and then we say, 68 00:05:55,120 --> 00:05:59,890 oh, let's try to plot the two equations. 69 00:05:59,890 --> 00:06:01,820 Well this is a circle. 70 00:06:01,820 --> 00:06:03,440 Eta squared plus Psi squared. 71 00:06:03,440 --> 00:06:06,630 Now Xi and Eta must be positive, so we 72 00:06:06,630 --> 00:06:09,960 look at solutions just in this quadrant. 73 00:06:09,960 --> 00:06:13,860 Let's put here pi over 2. 74 00:06:13,860 --> 00:06:25,850 Pi 3 pi over 2, 2 pi, and here is Eta and there is Xi. 75 00:06:25,850 --> 00:06:29,800 Well this is a circle, as we said, but let's look at this. 76 00:06:29,800 --> 00:06:34,335 Xi equals Eta tan Eta. 77 00:06:34,335 --> 00:06:37,885 That vanished as Eta goes to 0 and will diverge 78 00:06:37,885 --> 00:06:40,990 at Eta equals pi over 2. 79 00:06:40,990 --> 00:06:44,370 So this part, at least, looks like this. 80 00:06:49,430 --> 00:06:52,640 And then it will go negative, which don't care, 81 00:06:52,640 --> 00:06:59,670 from this region, and then reach here at pi. 82 00:07:04,380 --> 00:07:08,370 And after pi, it will go positive again 83 00:07:08,370 --> 00:07:14,050 and it will reach another infinity here. 84 00:07:14,050 --> 00:07:21,960 And then at 3 pi, at 2 pi, it will go again and reach 85 00:07:21,960 --> 00:07:25,090 not another infinity like that. 86 00:07:25,090 --> 00:07:27,770 So these are these curves. 87 00:07:27,770 --> 00:07:33,970 And the other curve, the circle, is just a circle here. 88 00:07:33,970 --> 00:07:40,190 So, for example, I could have a circle like this. 89 00:07:40,190 --> 00:07:45,990 So the radius of this circle is radius z0. 90 00:07:48,750 --> 00:07:51,600 And there you go, you've solved the problem. 91 00:07:51,600 --> 00:07:55,260 At least intuitively you know the answer. 92 00:07:55,260 --> 00:07:59,490 And there's a lot of things that come out of this calculation. 93 00:07:59,490 --> 00:08:05,680 If the radius z0 is 3 pi over 2, for example, 94 00:08:05,680 --> 00:08:09,500 and the radius z0 represents some potential of some depth 95 00:08:09,500 --> 00:08:13,990 and width, there will be just two solutions. 96 00:08:13,990 --> 00:08:16,620 These are these solutions. 97 00:08:16,620 --> 00:08:21,420 These points represent values of Xi and values of Eta, 98 00:08:21,420 --> 00:08:23,470 from which you could read the energy. 99 00:08:23,470 --> 00:08:27,570 In fact, you can look at that state and say, 100 00:08:27,570 --> 00:08:32,549 that's the state of largest Xi, and therefore 101 00:08:32,549 --> 00:08:36,990 it's the state of the largest absolute value of the energy. 102 00:08:36,990 --> 00:08:39,849 It's the most deeply bound state. 103 00:08:42,520 --> 00:08:44,820 Then this is next deeply bound state. 104 00:08:44,820 --> 00:08:48,440 So there's two bounce states in this case. 105 00:08:48,440 --> 00:08:54,110 Interestingly, however shallow this potential might be, 106 00:08:54,110 --> 00:08:59,720 however small, z0, the circle, will always 107 00:08:59,720 --> 00:09:02,270 have one intersection, so there will always 108 00:09:02,270 --> 00:09:05,870 be at least one solution. 109 00:09:05,870 --> 00:09:07,790 That's the end of that story. 110 00:09:07,790 --> 00:09:13,100 Let me say that for the odd case, odd solutions, 111 00:09:13,100 --> 00:09:14,330 I will not solve it. 112 00:09:14,330 --> 00:09:17,450 It's a good thing to do in recitation or as part 113 00:09:17,450 --> 00:09:20,690 of the home work as well. 114 00:09:20,690 --> 00:09:28,360 The answer for the odd case is that Psi is equal to minus Eta 115 00:09:28,360 --> 00:09:32,150 [? Cot ?] Eta. 116 00:09:32,150 --> 00:09:36,800 And in that case, I'll give you a little preview of how 117 00:09:36,800 --> 00:09:40,040 the this [? Cot ?] looks. 118 00:09:40,040 --> 00:09:41,560 It looks like this. 119 00:09:41,560 --> 00:09:46,310 And then there are more branches of this thing. 120 00:09:46,310 --> 00:09:51,110 So for the odd solution, you have these curves. 121 00:09:51,110 --> 00:09:53,420 And if you have a circle, sometimes you 122 00:09:53,420 --> 00:09:55,070 don't have a solution. 123 00:09:55,070 --> 00:09:57,320 It doesn't intersect this. 124 00:09:57,320 --> 00:10:02,060 So these odd solutions, you will see and try to understand, 125 00:10:02,060 --> 00:10:04,220 they don't always exist. 126 00:10:04,220 --> 00:10:08,040 You meet a potential that is sufficiently deep 127 00:10:08,040 --> 00:10:10,590 to get an odd solution. 128 00:10:10,590 --> 00:10:12,830 And then the odds and even solutions 129 00:10:12,830 --> 00:10:16,070 will interweave each other and there 130 00:10:16,070 --> 00:10:18,290 will be a nice story that you will 131 00:10:18,290 --> 00:10:20,220 explore in a lot of detail. 132 00:10:20,220 --> 00:10:25,700 But the is, you've reviewed the problem to unit free 133 00:10:25,700 --> 00:10:29,080 calculation, in which you can get the intuition of when 134 00:10:29,080 --> 00:10:32,320 solutions exist and and when they don't. 135 00:10:32,320 --> 00:10:34,890 But solving for the particular numbers 136 00:10:34,890 --> 00:10:37,470 are transcendental equations, and you 137 00:10:37,470 --> 00:10:40,130 need a computer to solve.