1 00:00:01,100 --> 00:00:12,970 PROFESSOR: So here is a problem example based 2 00:00:12,970 --> 00:00:19,620 on Griffith's problem 9.3. 3 00:00:19,620 --> 00:00:32,479 So it's a two-state system, and we'll call the states a and b, 4 00:00:32,479 --> 00:00:37,820 with energy's Ea and Eb. 5 00:00:37,820 --> 00:00:45,410 And we'll call omega 0 the difference Ea minus Eb 6 00:00:45,410 --> 00:00:46,535 over h bar. 7 00:00:52,940 --> 00:00:55,970 So those are energy eigenstates, which 8 00:00:55,970 --> 00:01:02,790 means that H0 is really Ea Eb. 9 00:01:07,230 --> 00:01:12,945 We represent the first column vector by the state a. 10 00:01:12,945 --> 00:01:16,170 The second state is the second column vector 11 00:01:16,170 --> 00:01:18,360 with entry 0 and 1-- 12 00:01:18,360 --> 00:01:19,500 is the state b. 13 00:01:19,500 --> 00:01:23,620 That's why the matrix looks like that. 14 00:01:23,620 --> 00:01:26,290 And then you put a perturbation. 15 00:01:26,290 --> 00:01:32,400 And the perturbation, delta H, is 16 00:01:32,400 --> 00:01:38,880 going to take the form U delta t, where U is a matrix that 17 00:01:38,880 --> 00:01:45,540 has Uaa, Uab, Uba, Ubb-- 18 00:01:45,540 --> 00:01:47,240 the matrix elements. 19 00:01:47,240 --> 00:01:48,820 And they're simple enough. 20 00:01:48,820 --> 00:01:55,800 It's a 0 here and an alpha here, and an alpha dagger here. 21 00:01:55,800 --> 00:02:01,450 It should be a Hermitian matrix, and that's the way it is. 22 00:02:01,450 --> 00:02:03,800 The letter U, I guess, is not great, 23 00:02:03,800 --> 00:02:05,520 because it suggests unitary. 24 00:02:05,520 --> 00:02:13,070 But maybe I should put a v. I'll leave it, to not create 25 00:02:13,070 --> 00:02:15,600 problems with my notation. 26 00:02:15,600 --> 00:02:20,150 So here is the question. 27 00:02:20,150 --> 00:02:25,610 This is your perturbation Hamiltonian. 28 00:02:25,610 --> 00:02:28,040 Actually, I wrote it wrong here. 29 00:02:28,040 --> 00:02:30,800 It's a delta function of t. 30 00:02:30,800 --> 00:02:36,740 That's delta H. So in terms of H0 applies everywhere, 31 00:02:36,740 --> 00:02:39,530 H0 applies everywhere, and in the middle 32 00:02:39,530 --> 00:02:48,320 there is a delta H for one little instant. 33 00:02:48,320 --> 00:02:49,810 So what happens here? 34 00:02:52,790 --> 00:02:55,310 We're going to try to figure out what's going on. 35 00:02:55,310 --> 00:02:57,860 But the question will be posed as follows. 36 00:02:57,860 --> 00:03:10,630 Assume the system is in the state a for t less than 0. 37 00:03:13,600 --> 00:03:20,800 Let's ask the probability to be in the state 38 00:03:20,800 --> 00:03:24,610 b for t greater than 0. 39 00:03:30,400 --> 00:03:36,250 Find the probability to be, and be greater than 0. 40 00:03:36,250 --> 00:03:40,600 So what's going on? 41 00:03:40,600 --> 00:03:43,330 Here are the states a and b. 42 00:03:43,330 --> 00:03:46,150 They were eigenstates of H0. 43 00:03:46,150 --> 00:03:50,050 If you had the system in the state a-- that's an energy 44 00:03:50,050 --> 00:03:51,040 eigenstate-- 45 00:03:51,040 --> 00:03:53,800 it would stay in the state a forever. 46 00:03:53,800 --> 00:03:56,110 It wouldn't do anything. 47 00:03:56,110 --> 00:04:01,000 Indeed if whoever prepared that state, 48 00:04:01,000 --> 00:04:04,960 ever since he or she prepared it, it has stayed in the state 49 00:04:04,960 --> 00:04:07,210 a until time equals 0. 50 00:04:07,210 --> 00:04:10,450 But then the system is kicked and it's 51 00:04:10,450 --> 00:04:14,260 kicked with some delta function strength, 52 00:04:14,260 --> 00:04:16,990 and in the off-diagonal terms. 53 00:04:16,990 --> 00:04:18,670 That's pretty important. 54 00:04:18,670 --> 00:04:22,250 It's not kicked on the diagonal terms, it's kicked here. 55 00:04:22,250 --> 00:04:26,110 So these terms suggest a transition from state a 56 00:04:26,110 --> 00:04:28,390 to b here. 57 00:04:28,390 --> 00:04:35,890 And presumably, due to this, the system will transition. 58 00:04:35,890 --> 00:04:43,100 And when it transitions, it must transition instantaneously 59 00:04:43,100 --> 00:04:43,940 in a sense. 60 00:04:43,940 --> 00:04:48,080 Because if it takes time to transition, 61 00:04:48,080 --> 00:04:50,420 the delta function asks for zero time. 62 00:04:50,420 --> 00:04:53,450 If it takes time to transition, then it would not transition. 63 00:04:53,450 --> 00:04:58,010 Because after the delta function has come and gone, 64 00:04:58,010 --> 00:05:00,050 the system doesn't change anymore. 65 00:05:00,050 --> 00:05:03,870 So it really has to transition instantaneously. 66 00:05:03,870 --> 00:05:08,450 So at time equals 0, you must generate 67 00:05:08,450 --> 00:05:13,115 a probability or an amplitude to be in the state b already. 68 00:05:15,770 --> 00:05:19,640 So there must be an immediate transition 69 00:05:19,640 --> 00:05:24,740 in which the state, originally, at time less than 0 70 00:05:24,740 --> 00:05:30,620 was the state a, but at time a little bit after zero 71 00:05:30,620 --> 00:05:37,340 must be a little bit of a plus a little bit of b. 72 00:05:37,340 --> 00:05:45,080 So this is t0 minus, 0 minus, and 0 plus. 73 00:05:45,080 --> 00:05:48,080 At 0 minus, it's a, and at 0 plus 74 00:05:48,080 --> 00:05:51,920 must be something like that. 75 00:05:51,920 --> 00:05:55,910 Because if at 0 plus it's still a, 76 00:05:55,910 --> 00:05:57,800 it's going to stay a forever. 77 00:05:57,800 --> 00:06:02,120 So it must have some amplitude to be at b at 0 plus. 78 00:06:02,120 --> 00:06:07,650 So you're having your first transition amplitude problem-- 79 00:06:07,650 --> 00:06:11,270 how does an interaction make you jump from one state to another? 80 00:06:14,930 --> 00:06:20,750 Now my letters here, alpha and beta, are pretty bad-- 81 00:06:20,750 --> 00:06:22,790 U and V maybe-- 82 00:06:22,790 --> 00:06:27,810 because I have an alpha there in the interaction. 83 00:06:27,810 --> 00:06:32,090 And maybe, if you think perturbatively, 84 00:06:32,090 --> 00:06:35,540 you would say, look, if alpha is very small, 85 00:06:35,540 --> 00:06:40,130 there must be a small transition probability. 86 00:06:40,130 --> 00:06:42,830 If the perturbation is of order alpha, 87 00:06:42,830 --> 00:06:47,030 maybe I expect the perturbed state to be of order alpha. 88 00:06:47,030 --> 00:06:51,470 So maybe, after all, this is like alpha times 89 00:06:51,470 --> 00:06:56,960 b, or proportional to alpha times b, a number here. 90 00:06:56,960 --> 00:07:02,790 And this one-- well, if alpha is very small, 91 00:07:02,790 --> 00:07:05,660 that's probably roughly about 1 still-- 92 00:07:05,660 --> 00:07:08,690 a little bit less, alpha squared less, 93 00:07:08,690 --> 00:07:10,580 for probability conservation. 94 00:07:10,580 --> 00:07:13,040 But I expect an alpha b here. 95 00:07:16,600 --> 00:07:25,030 So OK, when you're given a problem like that, 96 00:07:25,030 --> 00:07:28,000 under normal conditions it's a good idea 97 00:07:28,000 --> 00:07:32,510 to think about it before trying to solve it. 98 00:07:32,510 --> 00:07:34,630 And I think we've done a little bit of thinking. 99 00:07:34,630 --> 00:07:38,740 We've realized the state stays in a after a while, 100 00:07:38,740 --> 00:07:42,775 and immediately must transition into some other state. 101 00:07:45,970 --> 00:07:48,390 And once it's in that other state, its going 102 00:07:48,390 --> 00:07:51,960 to remain in that state forever because those are energy 103 00:07:51,960 --> 00:07:55,425 eigenstates, so you're going to get exponentials of E 104 00:07:55,425 --> 00:07:56,860 to the minus iEt. 105 00:07:56,860 --> 00:07:59,580 But this is going to be that. 106 00:07:59,580 --> 00:08:02,620 OK, so let's solve this. 107 00:08:02,620 --> 00:08:05,350 Here is our equation. 108 00:08:05,350 --> 00:08:07,580 Let's try to write it for this case. 109 00:08:07,580 --> 00:08:15,660 So I would have i h bar C a dot of t. 110 00:08:15,660 --> 00:08:18,240 We have two states-- 111 00:08:18,240 --> 00:08:20,850 1 and 2-- but here is a and b. 112 00:08:20,850 --> 00:08:29,160 So you would have i omega a then something, t and delta H 113 00:08:29,160 --> 00:08:31,320 a with something. 114 00:08:31,320 --> 00:08:36,000 Well, the delta H is clear that it can only 115 00:08:36,000 --> 00:08:41,100 be a delta H from a to b. 116 00:08:41,100 --> 00:08:43,140 Because there's Uab. 117 00:08:43,140 --> 00:08:45,750 There's only that transition amplitude. 118 00:08:45,750 --> 00:08:51,420 And then it would be an i omega ab, 119 00:08:51,420 --> 00:09:06,900 which was defined as omega 0, delta Hab of time, cb of time. 120 00:09:06,900 --> 00:09:11,970 And that's the only term that exists here from the sum. 121 00:09:11,970 --> 00:09:16,420 You could sum over a and b, but we saw that over a, you get 0. 122 00:09:16,420 --> 00:09:18,570 So there's just this term. 123 00:09:18,570 --> 00:09:25,235 And then we have i H bar Cb dot of [? t ?] 124 00:09:25,235 --> 00:09:28,560 is equal to E to the i. 125 00:09:28,560 --> 00:09:31,410 And now you're going to have ba-- 126 00:09:31,410 --> 00:09:38,910 so delta Hba of t Ca of t. 127 00:09:38,910 --> 00:09:42,990 And here I would have omega ba, which 128 00:09:42,990 --> 00:09:48,180 is the opposite sign as omega ab, which we anyway 129 00:09:48,180 --> 00:09:49,840 call omega 0. 130 00:09:49,840 --> 00:09:53,850 So here's minus i omega 0 t. 131 00:09:56,360 --> 00:09:57,260 Yes. 132 00:09:57,260 --> 00:10:01,250 AUDIENCE: [INAUDIBLE] perturbation disappears? 133 00:10:01,250 --> 00:10:02,134 PROFESSOR: Sorry. 134 00:10:02,134 --> 00:10:03,912 AUDIENCE: Does the perturbation vanish? 135 00:10:03,912 --> 00:10:04,620 PROFESSOR: Right. 136 00:10:04,620 --> 00:10:14,480 Because delta Haa is equal to 0, and therefore there's 137 00:10:14,480 --> 00:10:16,840 no coupling to Ca. 138 00:10:16,840 --> 00:10:24,200 And delta Hbb is 0 from that matrix element. 139 00:10:24,200 --> 00:10:26,750 And therefore, you cannot couple b to b. 140 00:10:31,340 --> 00:10:40,690 And now I'll just write it i H bar Ca dot of t, 141 00:10:40,690 --> 00:10:49,030 i H bar Cb dot of t is equal E to the i omega 0t. 142 00:10:49,030 --> 00:10:55,150 Delta Hab was alpha delta of time-- 143 00:10:55,150 --> 00:10:57,820 Cb of t. 144 00:10:57,820 --> 00:11:00,850 And this is E to the minus i omega 145 00:11:00,850 --> 00:11:06,400 0t, alpha star delta of time-- 146 00:11:06,400 --> 00:11:09,450 Ca of t. 147 00:11:09,450 --> 00:11:15,411 Hba was Uba, which is alpha star times the delta function 148 00:11:15,411 --> 00:11:15,910 of time. 149 00:11:22,260 --> 00:11:25,020 Here I have a delta function of time. 150 00:11:25,020 --> 00:11:30,390 It says that this only matters when time is equal to 0. 151 00:11:30,390 --> 00:11:34,080 So things are a little delicate here. 152 00:11:34,080 --> 00:11:35,610 But let's see. 153 00:11:35,610 --> 00:11:38,530 Can I write here-- 154 00:11:38,530 --> 00:11:44,340 well, if I have f of x times delta of x, 155 00:11:44,340 --> 00:11:49,170 we know this is f of 0 times delta of x. 156 00:11:49,170 --> 00:11:51,600 Because anyway, only zeros [INAUDIBLE].. 157 00:11:51,600 --> 00:11:54,840 So I can put time equals 0 here, and forget 158 00:11:54,840 --> 00:12:02,850 about this exponential-- have alpha delta of t, cb of t, 159 00:12:02,850 --> 00:12:09,960 and here have alpha star delta of t Ca of t. 160 00:12:15,120 --> 00:12:17,130 Everybody happy so far? 161 00:12:20,710 --> 00:12:28,200 But here, I have Cb of t and Ca of t, and a delta t. 162 00:12:28,200 --> 00:12:34,800 So I should put alpha delta of t Cb at 0, 163 00:12:34,800 --> 00:12:42,401 and alpha star delta of t Ca at 0. 164 00:12:42,401 --> 00:12:42,900 Right? 165 00:12:45,865 --> 00:12:46,365 Yes? 166 00:12:50,110 --> 00:12:53,860 People look less convinced, which I congratulate you 167 00:12:53,860 --> 00:12:56,110 for doubting this. 168 00:12:56,110 --> 00:12:58,030 This would be bad. 169 00:12:58,030 --> 00:13:00,880 It seems like I'm just following the logic, 170 00:13:00,880 --> 00:13:04,720 but I've gone too far now. 171 00:13:04,720 --> 00:13:08,260 Why did I go too far? 172 00:13:08,260 --> 00:13:12,190 First, this is going to be a disaster. 173 00:13:12,190 --> 00:13:22,000 Because in many ways, we don't know what these numbers are. 174 00:13:22,000 --> 00:13:24,190 You see, we are argued that this is going 175 00:13:24,190 --> 00:13:27,850 to change this continuously. 176 00:13:27,850 --> 00:13:30,910 Those numbers are going to change this continuously. 177 00:13:30,910 --> 00:13:32,950 At time equals 0-- 178 00:13:32,950 --> 00:13:38,290 before time equals 0, Ca was 1 and Cb was 0. 179 00:13:38,290 --> 00:13:42,010 But after time equals 0, Cb is going to be a number-- 180 00:13:42,010 --> 00:13:45,070 something that we don't know. 181 00:13:45,070 --> 00:13:47,040 So the delta function, which does it 182 00:13:47,040 --> 00:13:51,250 hiy-- the one before 0, the one after 0, the average? 183 00:13:51,250 --> 00:13:52,240 What does it do? 184 00:13:52,240 --> 00:13:56,440 This is a case where you have a delta function that we 185 00:13:56,440 --> 00:13:58,400 don't know what it's doing. 186 00:14:01,580 --> 00:14:03,600 So this is totally wrong. 187 00:14:03,600 --> 00:14:07,940 So I will just stop there. 188 00:14:07,940 --> 00:14:11,600 And I will consider, however these two equations. 189 00:14:14,820 --> 00:14:23,070 So we have the equations now in a nice way. 190 00:14:23,070 --> 00:14:25,390 I'll reconsider them. 191 00:14:25,390 --> 00:14:29,090 And we are facing a difficulty, the difficulty 192 00:14:29,090 --> 00:14:35,120 of the delta function not allowing us to treat nicely 193 00:14:35,120 --> 00:14:46,510 the transition of i h bar Ca dot is equal to alpha delta of Cb, 194 00:14:46,510 --> 00:14:55,250 and i h bar Cb dot equal alpha star delta of t Ca. 195 00:15:01,770 --> 00:15:08,880 Well now, the only way to resolve this difficulty 196 00:15:08,880 --> 00:15:12,270 is to think about this physically, 197 00:15:12,270 --> 00:15:16,740 and introduce a regulator that is physical. 198 00:15:16,740 --> 00:15:22,070 You see, no signal in the world really is a delta function. 199 00:15:22,070 --> 00:15:27,720 Nothing becomes infinite, and nothing lasts for zero time. 200 00:15:27,720 --> 00:15:32,910 So this interaction, this delta function, 201 00:15:32,910 --> 00:15:36,120 represents a function of time-- 202 00:15:36,120 --> 00:15:37,940 delta of time. 203 00:15:37,940 --> 00:15:44,880 It's a spike, but we can think of it as a regulated thing. 204 00:15:44,880 --> 00:15:47,970 You've regulated delta functions in all kinds of ways. 205 00:15:47,970 --> 00:15:51,330 It might be useful to regulate it as follows, 206 00:15:51,330 --> 00:15:57,720 as a function of time that is 0 before time equals 0, 207 00:15:57,720 --> 00:16:01,980 it's 0 after some time, t star, and it 208 00:16:01,980 --> 00:16:04,620 has height 1 over t star. 209 00:16:08,360 --> 00:16:10,705 That's a picture of a delta function. 210 00:16:14,250 --> 00:16:17,640 It has the right area at least. 211 00:16:17,640 --> 00:16:21,060 And we're going to think of this system 212 00:16:21,060 --> 00:16:28,770 as exactly doing that in between, for t, 213 00:16:28,770 --> 00:16:33,910 in the interval 0 to t star. 214 00:16:33,910 --> 00:16:35,850 The delta function is going to be 215 00:16:35,850 --> 00:16:39,090 replaced by this function, this constant function. 216 00:16:39,090 --> 00:16:46,170 So it will have i h bar h Ca dot is 217 00:16:46,170 --> 00:16:53,645 equal to alpha over t star Cb. 218 00:16:57,300 --> 00:17:10,089 And i h bar Cb dot is equal to alpha star over t sub star. 219 00:17:10,089 --> 00:17:13,780 Now the star doesn't mean complex conjugation for t. 220 00:17:13,780 --> 00:17:18,040 It's just a number, a glyph called t0 or something 221 00:17:18,040 --> 00:17:18,970 like that. 222 00:17:18,970 --> 00:17:23,569 And this is true for this time interval. 223 00:17:23,569 --> 00:17:29,740 And then you would say, look, I have two problems with this. 224 00:17:29,740 --> 00:17:32,980 First, you've entered this at t star, 225 00:17:32,980 --> 00:17:36,170 and now your calculation is going to depend on t star. 226 00:17:36,170 --> 00:17:39,090 And then the answer is going to depend on t star. 227 00:17:39,090 --> 00:17:42,780 And what are you going to put for t star? 228 00:17:42,780 --> 00:17:47,720 Well, that better not happen. 229 00:17:47,720 --> 00:17:50,210 We have to go on a limb. 230 00:17:50,210 --> 00:17:52,490 Many times when you do a calculation of something, 231 00:17:52,490 --> 00:17:53,540 you go on a limb. 232 00:17:53,540 --> 00:17:56,540 You say something, and see if it works. 233 00:17:56,540 --> 00:17:58,220 This seems reasonable. 234 00:17:58,220 --> 00:18:01,040 And what we expect is that we will 235 00:18:01,040 --> 00:18:06,710 find an answer that this independent on this t star. 236 00:18:06,710 --> 00:18:11,870 So that we can take this star to 0 and make sense of it. 237 00:18:11,870 --> 00:18:14,240 The other question that could come 238 00:18:14,240 --> 00:18:18,330 is that, oh, that was your delta function. 239 00:18:18,330 --> 00:18:23,420 So one seconds, you used delta function for this. 240 00:18:23,420 --> 00:18:28,460 Maybe we have to put back these terms, these exponentials. 241 00:18:28,460 --> 00:18:31,580 We set them to 1. 242 00:18:31,580 --> 00:18:34,520 But then you could say, look, I don't 243 00:18:34,520 --> 00:18:36,630 think I have to put them back. 244 00:18:36,630 --> 00:18:37,760 Why? 245 00:18:37,760 --> 00:18:43,850 Because I can choose t star as small as I want, 246 00:18:43,850 --> 00:18:48,980 such that omega t star, which is the possible value 247 00:18:48,980 --> 00:18:54,470 that this can get, at most, is like 10 to the minus 60. 248 00:18:54,470 --> 00:18:57,805 And therefore, this phase is going to be 0, essentially, 249 00:18:57,805 --> 00:18:59,300 and that's going to be 1. 250 00:18:59,300 --> 00:19:04,890 So I claim that I don't have to worry about this anymore. 251 00:19:04,890 --> 00:19:06,180 So what do we have? 252 00:19:06,180 --> 00:19:12,320 We have two simple differential equations, like this. 253 00:19:12,320 --> 00:19:19,940 With the conditions that Ca at time equals 0 is 1, 254 00:19:19,940 --> 00:19:25,600 and Cb at time equals 0 is 0. 255 00:19:25,600 --> 00:19:32,120 The state before the delta function hits is given by that. 256 00:19:32,120 --> 00:19:35,240 And now what we want to know is, how much is 257 00:19:35,240 --> 00:19:41,810 Ca after the delta function and Cb after the delta function? 258 00:19:46,330 --> 00:19:53,020 So for that, we have all the tools needed. 259 00:19:53,020 --> 00:19:55,910 I can take this equation. 260 00:19:55,910 --> 00:19:57,745 This is a coupled system of equations. 261 00:19:57,745 --> 00:19:59,290 It's a simple system. 262 00:19:59,290 --> 00:20:03,150 I could differentiate again here. 263 00:20:03,150 --> 00:20:10,250 And I get a Ca double-dot, Cb dot, and use this equation. 264 00:20:10,250 --> 00:20:17,740 So you get the equation-- if you differentiate again, take a dot 265 00:20:17,740 --> 00:20:25,180 to form Ca double-dot, you get that Ca double-dot is 266 00:20:25,180 --> 00:20:35,875 equal to minus alpha over h bar t star squared Ca. 267 00:20:41,360 --> 00:20:47,090 You can see that you put the i h bar here, take the derivative, 268 00:20:47,090 --> 00:20:50,090 you get an alpha times an alpha star, which 269 00:20:50,090 --> 00:20:52,880 is length of alpha squared. 270 00:20:52,880 --> 00:20:57,230 And the h bar appears two times, t star appears two times. 271 00:20:57,230 --> 00:21:01,140 This is an oscillating solution. 272 00:21:01,140 --> 00:21:05,860 So that's simple enough. 273 00:21:09,410 --> 00:21:16,430 So what happens here is that Ca will oscillate in time, 274 00:21:16,430 --> 00:21:19,440 and you will have a solution of the form 275 00:21:19,440 --> 00:21:31,850 Ca is equal to a constant, beta 0 cosine of alpha h bar t star 276 00:21:31,850 --> 00:21:42,390 t, plus beta 1 sine of alpha p, over h bar t star. 277 00:21:45,030 --> 00:21:53,090 And if you wish, Cb from the first equation 278 00:21:53,090 --> 00:21:57,230 is proportional to Ca dot. 279 00:22:01,460 --> 00:22:06,570 And Ca dot is going to be of the form beta 280 00:22:06,570 --> 00:22:14,570 0 sine off this thing, plus beta 1 cosine of this thing. 281 00:22:21,980 --> 00:22:26,120 OK, so we've turned this into a tractable problem because 282 00:22:26,120 --> 00:22:29,270 of flattening the delta function. 283 00:22:29,270 --> 00:22:36,050 And your initial conditions-- again, Cb must vanish at time 284 00:22:36,050 --> 00:22:37,910 equals 0. 285 00:22:37,910 --> 00:22:44,000 So if Cb must vanish at time equals 0, beta 1 must be 0. 286 00:22:48,420 --> 00:22:52,260 If beta 1 is zero, this term is gone, 287 00:22:52,260 --> 00:22:57,480 and Ca was 1 for time equals 0. 288 00:22:57,480 --> 00:22:59,820 So beta 0 is 1. 289 00:23:04,030 --> 00:23:18,570 Therefore, Ca is equal to cosine alpha t, over h star t star 290 00:23:18,570 --> 00:23:19,900 Ca of t. 291 00:23:23,580 --> 00:23:31,150 And Cb of t, you can calculate it from the top equation. 292 00:23:31,150 --> 00:23:34,450 You take the derivative of Ca, and divide 293 00:23:34,450 --> 00:23:38,080 by alpha multiplied by t star. 294 00:23:41,240 --> 00:23:42,730 I'll let you do it. 295 00:23:42,730 --> 00:23:54,460 You get minus i alpha, over alpha, sine, alpha t 296 00:23:54,460 --> 00:23:58,930 over ht star. 297 00:23:58,930 --> 00:24:05,740 OK, we solved for the functions. 298 00:24:05,740 --> 00:24:08,160 We know what's going to happen. 299 00:24:08,160 --> 00:24:12,640 And now, what did we want? 300 00:24:12,640 --> 00:24:17,020 We wanted to know what are those things for later times, 301 00:24:17,020 --> 00:24:20,350 for times t star or more. 302 00:24:20,350 --> 00:24:24,340 Now you cannot use this equation beyond time t star, 303 00:24:24,340 --> 00:24:27,470 because they assumed the delta function is there. 304 00:24:27,470 --> 00:24:29,230 So what you have to figure out is 305 00:24:29,230 --> 00:24:32,690 what are these coefficients at time t star. 306 00:24:32,690 --> 00:24:37,150 And those would be the coefficients at any later time, 307 00:24:37,150 --> 00:24:39,100 because there's no more Hamiltonian. 308 00:24:39,100 --> 00:24:45,800 So Ca at t greater than t star is, in fact, 309 00:24:45,800 --> 00:24:49,840 equal to Ca at t star. 310 00:24:49,840 --> 00:24:54,850 And Ca at t star, happily, is a number 311 00:24:54,850 --> 00:24:57,640 that doesn't depend on t star-- 312 00:24:57,640 --> 00:25:01,570 cosine alpha over h bar. 313 00:25:04,140 --> 00:25:11,900 And Cb at t greater than t star is Cb at t star. 314 00:25:11,900 --> 00:25:15,710 It doesn't change any more after the delta 315 00:25:15,710 --> 00:25:18,500 function has turned off. 316 00:25:18,500 --> 00:25:27,740 And Cb at t star is minus i alpha over alpha-- 317 00:25:27,740 --> 00:25:40,235 alpha over alpha, like this, times sine of alpha over h bar. 318 00:25:45,800 --> 00:25:48,290 And this is really what you wanted. 319 00:25:48,290 --> 00:25:54,130 Now you know what the state is doing after the delta 320 00:25:54,130 --> 00:25:55,810 function has turned on. 321 00:25:55,810 --> 00:26:02,240 It has this amplitude to be in the b state. 322 00:26:02,240 --> 00:26:04,950 And it is, as we predicted-- remember, 323 00:26:04,950 --> 00:26:11,100 this is like the face of alpha, because the alpha is alpha 324 00:26:11,100 --> 00:26:12,720 times its face-- 325 00:26:12,720 --> 00:26:15,690 length of alpha times it face, and it will cancel. 326 00:26:15,690 --> 00:26:19,270 And here you have sine of this quantity. 327 00:26:19,270 --> 00:26:22,380 So it is proportional. 328 00:26:22,380 --> 00:26:26,020 Cb is proportional to alpha. 329 00:26:26,020 --> 00:26:28,900 And that's exactly what you would expect. 330 00:26:28,900 --> 00:26:34,785 So just to complete the story, let's write the answer. 331 00:26:50,460 --> 00:26:52,410 And what is the answer? 332 00:26:52,410 --> 00:27:02,490 Phi of t for t greater than 0 is the coefficient Ca 333 00:27:02,490 --> 00:27:09,630 of t, which was cosine, alpha over h bar, 334 00:27:09,630 --> 00:27:17,850 times E to the minus iEat, over h bar a. 335 00:27:17,850 --> 00:27:23,400 Remember, the solution is Ca E to the minus iEta. 336 00:27:23,400 --> 00:27:32,490 And then the other one, which is minus i alpha, over alpha, 337 00:27:32,490 --> 00:27:47,280 sine of alpha over h bar, e to the minus iEbt over h bar, b. 338 00:27:47,280 --> 00:27:54,160 That is the state after the delta function has turned on. 339 00:27:54,160 --> 00:27:57,090 And what is the probability to be 340 00:27:57,090 --> 00:28:01,935 found in the state b, time t? 341 00:28:01,935 --> 00:28:09,870 It would be b times the state squared. 342 00:28:13,140 --> 00:28:20,370 So you get-- b will couple just to that. b with b gives you 1. 343 00:28:20,370 --> 00:28:21,940 This is a phase. 344 00:28:21,940 --> 00:28:23,730 This is a phase. 345 00:28:23,730 --> 00:28:27,240 Sine squared of this thing is sine 346 00:28:27,240 --> 00:28:33,570 squared of alpha over h bar. 347 00:28:33,570 --> 00:28:38,730 What is the probability to be in a? 348 00:28:38,730 --> 00:28:42,420 It's a psi of t. 349 00:28:42,420 --> 00:28:48,460 And in this case, it will be cosine squared 350 00:28:48,460 --> 00:28:52,370 of alpha over h bar. 351 00:28:52,370 --> 00:28:54,910 And it's very handy that cosine squared 352 00:28:54,910 --> 00:28:58,180 plus sine squared is equal to 1, because it 353 00:28:58,180 --> 00:29:02,680 has to be either in a or on b. 354 00:29:02,680 --> 00:29:06,490 And it's kind of interesting, if you 355 00:29:06,490 --> 00:29:08,890 were doing this in perturbation-- we solved it 356 00:29:08,890 --> 00:29:09,610 exactly. 357 00:29:09,610 --> 00:29:12,680 If you were doing this in perturbation theory, 358 00:29:12,680 --> 00:29:19,300 we would have found P of b proportional to alpha squared. 359 00:29:19,300 --> 00:29:22,840 Because the first term in the expansion here, 360 00:29:22,840 --> 00:29:25,020 in terms of the interaction.