1 00:00:00,000 --> 00:00:10,240 PROFESSOR: So let's do an example where 2 00:00:10,240 --> 00:00:12,340 we can calculate from the beginning 3 00:00:12,340 --> 00:00:15,490 to the end everything. 4 00:00:15,490 --> 00:00:19,750 Now, you have to get accustomed to the idea of even 5 00:00:19,750 --> 00:00:22,270 though you can calculate everything, 6 00:00:22,270 --> 00:00:25,810 your formulas that you get sometimes are a little big. 7 00:00:25,810 --> 00:00:29,470 And you look at them and they may not 8 00:00:29,470 --> 00:00:34,370 tell you too much unless you plot them with a computer. 9 00:00:34,370 --> 00:00:37,570 So we push the calculations to some degree, 10 00:00:37,570 --> 00:00:40,990 and then at some point, we decide 11 00:00:40,990 --> 00:00:46,840 to plot things using a computer and get some insight 12 00:00:46,840 --> 00:00:47,750 on what's happening. 13 00:00:47,750 --> 00:00:48,666 So here's the example. 14 00:00:54,690 --> 00:01:01,850 We have a potential up to distance, a, to 0. 15 00:01:01,850 --> 00:01:06,750 The wall is always there, and this number is minus v naught. 16 00:01:06,750 --> 00:01:14,250 So it's a well, a potential well. 17 00:01:14,250 --> 00:01:17,140 And we are producing energy. 18 00:01:17,140 --> 00:01:19,050 Eigenstates are coming here. 19 00:01:23,550 --> 00:01:29,040 And the question now is to really calculate the solution 20 00:01:29,040 --> 00:01:33,020 so that we can really calculate the phase shift. 21 00:01:33,020 --> 00:01:35,720 We know how the solutions should read, 22 00:01:35,720 --> 00:01:38,840 but unless you do a real calculation, 23 00:01:38,840 --> 00:01:42,110 you cannot get the phase shift. 24 00:01:42,110 --> 00:01:44,720 So that's what we want to do. 25 00:01:44,720 --> 00:01:49,220 So for that, we have to solve the Schrodinger equation. 26 00:01:49,220 --> 00:01:52,710 Psi of x is equal to what? 27 00:01:55,500 --> 00:01:57,870 Well, there's a discontinuity. 28 00:01:57,870 --> 00:02:00,990 So we probably have to write an answer in which we'll 29 00:02:00,990 --> 00:02:05,420 have a solution in one piece and a solution in the other piece. 30 00:02:05,420 --> 00:02:08,479 But then we say, oh, we wrote the solution 31 00:02:08,479 --> 00:02:13,070 in the outside piece already. 32 00:02:13,070 --> 00:02:14,090 It is known. 33 00:02:14,090 --> 00:02:15,590 It's always the same. 34 00:02:15,590 --> 00:02:16,670 It's universal. 35 00:02:16,670 --> 00:02:19,340 I don't have to think. 36 00:02:19,340 --> 00:02:21,890 I just write this. 37 00:02:21,890 --> 00:02:23,990 E to the i delta. 38 00:02:23,990 --> 00:02:25,980 I don't know what delta is, but that's 39 00:02:25,980 --> 00:02:33,080 the answer, E to the i delta sine kx plus delta should 40 00:02:33,080 --> 00:02:35,870 be the solution for x greater than a. 41 00:02:44,430 --> 00:02:47,640 You know if you were not using that answer, 42 00:02:47,640 --> 00:02:50,610 it has all the relevant information 43 00:02:50,610 --> 00:02:53,850 for the problem, time delays, everything, 44 00:02:53,850 --> 00:02:57,000 you would simply write some superposition of E 45 00:02:57,000 --> 00:03:02,790 to the i kx and E to the minus i kx with two coefficients. 46 00:03:02,790 --> 00:03:10,680 On the other hand, here, we will have, again, a wave. 47 00:03:10,680 --> 00:03:14,460 Now, it could be maybe an E to the i kx or E 48 00:03:14,460 --> 00:03:16,620 to the minus i kx. 49 00:03:16,620 --> 00:03:23,640 Neither one is very good because the wave function 50 00:03:23,640 --> 00:03:27,080 must vanish at x equals 0. 51 00:03:27,080 --> 00:03:31,070 And in fact, the k that represents the kinetic energy 52 00:03:31,070 --> 00:03:37,040 here, k is always related to E by the standard quantity, 53 00:03:37,040 --> 00:03:44,300 k squared equal to mE over h squared or E equal 54 00:03:44,300 --> 00:03:45,560 the famous formula. 55 00:03:45,560 --> 00:03:49,340 On the other hand, there is a different k here 56 00:03:49,340 --> 00:03:52,880 because you have different kinetic energy. 57 00:03:52,880 --> 00:04:02,200 There must be a k prime here, which is 2m E plus v naught. 58 00:04:02,200 --> 00:04:05,180 That's a total kinetic energy over h squared. 59 00:04:05,180 --> 00:04:09,195 And yes, the solutions could be E to the ik prime 60 00:04:09,195 --> 00:04:14,740 x equal to minus ik prime x minus ik prime x, 61 00:04:14,740 --> 00:04:20,769 but since they must vanish at 0, should be a sine function. 62 00:04:20,769 --> 00:04:22,690 So the only thing we can have here 63 00:04:22,690 --> 00:04:33,650 is a sine of k prime x for x less than a and a coefficient. 64 00:04:33,650 --> 00:04:36,830 We didn't put the additional normalization here. 65 00:04:36,830 --> 00:04:38,810 We don't want to put that, but then we 66 00:04:38,810 --> 00:04:41,210 must put the number here, so I'll put it here. 67 00:04:44,130 --> 00:04:46,770 That's the answer, and that's k and k prime. 68 00:04:52,280 --> 00:04:56,320 Now we have boundary conditions that x equals a. 69 00:04:56,320 --> 00:05:04,780 So psi continues at x equals a. 70 00:05:04,780 --> 00:05:07,060 What does it give you? 71 00:05:07,060 --> 00:05:14,680 It gives you a sine of ka is equal to E to the i delta 72 00:05:14,680 --> 00:05:19,500 sine of ka plus delta. 73 00:05:19,500 --> 00:05:27,020 And psi prime continues at x equals a will give me 74 00:05:27,020 --> 00:05:35,530 ka cosine ka equal-- 75 00:05:35,530 --> 00:05:40,800 I have primes missing; I'm sorry, primes-- 76 00:05:40,800 --> 00:05:50,025 equals k E to the i delta cosine ka plus delta. 77 00:05:55,570 --> 00:05:57,640 What do we care for? 78 00:05:57,640 --> 00:06:00,330 Basically we care for delta. 79 00:06:00,330 --> 00:06:04,420 That's what we want to find out because delta tells us 80 00:06:04,420 --> 00:06:06,475 all about the physics of the scattering. 81 00:06:06,475 --> 00:06:12,690 It tells us about the scattering amplitude, sine squared delta. 82 00:06:12,690 --> 00:06:19,270 It tells us about the time delay, and let's calculate it. 83 00:06:19,270 --> 00:06:22,150 Well, one way to calculate it is to take 84 00:06:22,150 --> 00:06:28,360 a ratio of these two equations so that you get rid 85 00:06:28,360 --> 00:06:29,770 of the a constant. 86 00:06:29,770 --> 00:06:32,710 So from that side of the equation, 87 00:06:32,710 --> 00:06:40,090 you get k cotangent of ka plus delta 88 00:06:40,090 --> 00:06:46,210 is equal to k prime cotangent of k prime a. 89 00:06:51,610 --> 00:06:58,660 Or cotangent of ka plus delta is k prime over k. 90 00:06:58,660 --> 00:07:04,130 We'll erase this. 91 00:07:04,130 --> 00:07:07,700 And now you can do two things. 92 00:07:07,700 --> 00:07:12,820 You can display some trigonometric wizardry, 93 00:07:12,820 --> 00:07:21,415 or you say, OK, delta is arc cotangent of this minus ka. 94 00:07:24,000 --> 00:07:27,670 That is OK, but it's not ideal. 95 00:07:27,670 --> 00:07:32,860 It's better to do a little bit of trigonometric identities. 96 00:07:32,860 --> 00:07:35,140 And the identity that is relevant 97 00:07:35,140 --> 00:07:48,970 is the identity for cotangent of a plus b is cot a cot b minus 1 98 00:07:48,970 --> 00:07:54,760 over cot a plus cot b. 99 00:07:54,760 --> 00:08:01,990 So from here, you have that this expression 100 00:08:01,990 --> 00:08:18,990 is cot ka cot delta minus 1 over cot ka plus cot delta. 101 00:08:18,990 --> 00:08:27,430 And now, equating left-hand side to this right-hand side, 102 00:08:27,430 --> 00:08:31,870 you can solve for cotangent of delta. 103 00:08:31,870 --> 00:08:34,740 So cotangent of delta can be solved for-- 104 00:08:34,740 --> 00:08:37,230 and here is the answer. 105 00:08:37,230 --> 00:08:48,180 Cot delta is equal to tan ka plus k prime over k cot k 106 00:08:48,180 --> 00:09:00,260 prime a over 1 minus k prime over k cot k prime a tan ka. 107 00:09:03,030 --> 00:09:07,460 Now, who would box such a complicated equation? 108 00:09:07,460 --> 00:09:10,361 Well, it can't be simplified any more. 109 00:09:10,361 --> 00:09:10,860 Sorry. 110 00:09:10,860 --> 00:09:13,640 That's the best we can do.