WEBVTT 00:00:01.100 --> 00:00:12.970 PROFESSOR: So here is a problem example based 00:00:12.970 --> 00:00:19.620 on Griffith's problem 9.3. 00:00:19.620 --> 00:00:32.479 So it's a two-state system, and we'll call the states a and b, 00:00:32.479 --> 00:00:37.820 with energy's Ea and Eb. 00:00:37.820 --> 00:00:45.410 And we'll call omega 0 the difference Ea minus Eb 00:00:45.410 --> 00:00:46.535 over h bar. 00:00:52.940 --> 00:00:55.970 So those are energy eigenstates, which 00:00:55.970 --> 00:01:02.790 means that H0 is really Ea Eb. 00:01:07.230 --> 00:01:12.945 We represent the first column vector by the state a. 00:01:12.945 --> 00:01:16.170 The second state is the second column vector 00:01:16.170 --> 00:01:18.360 with entry 0 and 1-- 00:01:18.360 --> 00:01:19.500 is the state b. 00:01:19.500 --> 00:01:23.620 That's why the matrix looks like that. 00:01:23.620 --> 00:01:26.290 And then you put a perturbation. 00:01:26.290 --> 00:01:32.400 And the perturbation, delta H, is 00:01:32.400 --> 00:01:38.880 going to take the form U delta t, where U is a matrix that 00:01:38.880 --> 00:01:45.540 has Uaa, Uab, Uba, Ubb-- 00:01:45.540 --> 00:01:47.240 the matrix elements. 00:01:47.240 --> 00:01:48.820 And they're simple enough. 00:01:48.820 --> 00:01:55.800 It's a 0 here and an alpha here, and an alpha dagger here. 00:01:55.800 --> 00:02:01.450 It should be a Hermitian matrix, and that's the way it is. 00:02:01.450 --> 00:02:03.800 The letter U, I guess, is not great, 00:02:03.800 --> 00:02:05.520 because it suggests unitary. 00:02:05.520 --> 00:02:13.070 But maybe I should put a v. I'll leave it, to not create 00:02:13.070 --> 00:02:15.600 problems with my notation. 00:02:15.600 --> 00:02:20.150 So here is the question. 00:02:20.150 --> 00:02:25.610 This is your perturbation Hamiltonian. 00:02:25.610 --> 00:02:28.040 Actually, I wrote it wrong here. 00:02:28.040 --> 00:02:30.800 It's a delta function of t. 00:02:30.800 --> 00:02:36.740 That's delta H. So in terms of H0 applies everywhere, 00:02:36.740 --> 00:02:39.530 H0 applies everywhere, and in the middle 00:02:39.530 --> 00:02:48.320 there is a delta H for one little instant. 00:02:48.320 --> 00:02:49.810 So what happens here? 00:02:52.790 --> 00:02:55.310 We're going to try to figure out what's going on. 00:02:55.310 --> 00:02:57.860 But the question will be posed as follows. 00:02:57.860 --> 00:03:10.630 Assume the system is in the state a for t less than 0. 00:03:13.600 --> 00:03:20.800 Let's ask the probability to be in the state 00:03:20.800 --> 00:03:24.610 b for t greater than 0. 00:03:30.400 --> 00:03:36.250 Find the probability to be, and be greater than 0. 00:03:36.250 --> 00:03:40.600 So what's going on? 00:03:40.600 --> 00:03:43.330 Here are the states a and b. 00:03:43.330 --> 00:03:46.150 They were eigenstates of H0. 00:03:46.150 --> 00:03:50.050 If you had the system in the state a-- that's an energy 00:03:50.050 --> 00:03:51.040 eigenstate-- 00:03:51.040 --> 00:03:53.800 it would stay in the state a forever. 00:03:53.800 --> 00:03:56.110 It wouldn't do anything. 00:03:56.110 --> 00:04:01.000 Indeed if whoever prepared that state, 00:04:01.000 --> 00:04:04.960 ever since he or she prepared it, it has stayed in the state 00:04:04.960 --> 00:04:07.210 a until time equals 0. 00:04:07.210 --> 00:04:10.450 But then the system is kicked and it's 00:04:10.450 --> 00:04:14.260 kicked with some delta function strength, 00:04:14.260 --> 00:04:16.990 and in the off-diagonal terms. 00:04:16.990 --> 00:04:18.670 That's pretty important. 00:04:18.670 --> 00:04:22.250 It's not kicked on the diagonal terms, it's kicked here. 00:04:22.250 --> 00:04:26.110 So these terms suggest a transition from state a 00:04:26.110 --> 00:04:28.390 to b here. 00:04:28.390 --> 00:04:35.890 And presumably, due to this, the system will transition. 00:04:35.890 --> 00:04:43.100 And when it transitions, it must transition instantaneously 00:04:43.100 --> 00:04:43.940 in a sense. 00:04:43.940 --> 00:04:48.080 Because if it takes time to transition, 00:04:48.080 --> 00:04:50.420 the delta function asks for zero time. 00:04:50.420 --> 00:04:53.450 If it takes time to transition, then it would not transition. 00:04:53.450 --> 00:04:58.010 Because after the delta function has come and gone, 00:04:58.010 --> 00:05:00.050 the system doesn't change anymore. 00:05:00.050 --> 00:05:03.870 So it really has to transition instantaneously. 00:05:03.870 --> 00:05:08.450 So at time equals 0, you must generate 00:05:08.450 --> 00:05:13.115 a probability or an amplitude to be in the state b already. 00:05:15.770 --> 00:05:19.640 So there must be an immediate transition 00:05:19.640 --> 00:05:24.740 in which the state, originally, at time less than 0 00:05:24.740 --> 00:05:30.620 was the state a, but at time a little bit after zero 00:05:30.620 --> 00:05:37.340 must be a little bit of a plus a little bit of b. 00:05:37.340 --> 00:05:45.080 So this is t0 minus, 0 minus, and 0 plus. 00:05:45.080 --> 00:05:48.080 At 0 minus, it's a, and at 0 plus 00:05:48.080 --> 00:05:51.920 must be something like that. 00:05:51.920 --> 00:05:55.910 Because if at 0 plus it's still a, 00:05:55.910 --> 00:05:57.800 it's going to stay a forever. 00:05:57.800 --> 00:06:02.120 So it must have some amplitude to be at b at 0 plus. 00:06:02.120 --> 00:06:07.650 So you're having your first transition amplitude problem-- 00:06:07.650 --> 00:06:11.270 how does an interaction make you jump from one state to another? 00:06:14.930 --> 00:06:20.750 Now my letters here, alpha and beta, are pretty bad-- 00:06:20.750 --> 00:06:22.790 U and V maybe-- 00:06:22.790 --> 00:06:27.810 because I have an alpha there in the interaction. 00:06:27.810 --> 00:06:32.090 And maybe, if you think perturbatively, 00:06:32.090 --> 00:06:35.540 you would say, look, if alpha is very small, 00:06:35.540 --> 00:06:40.130 there must be a small transition probability. 00:06:40.130 --> 00:06:42.830 If the perturbation is of order alpha, 00:06:42.830 --> 00:06:47.030 maybe I expect the perturbed state to be of order alpha. 00:06:47.030 --> 00:06:51.470 So maybe, after all, this is like alpha times 00:06:51.470 --> 00:06:56.960 b, or proportional to alpha times b, a number here. 00:06:56.960 --> 00:07:02.790 And this one-- well, if alpha is very small, 00:07:02.790 --> 00:07:05.660 that's probably roughly about 1 still-- 00:07:05.660 --> 00:07:08.690 a little bit less, alpha squared less, 00:07:08.690 --> 00:07:10.580 for probability conservation. 00:07:10.580 --> 00:07:13.040 But I expect an alpha b here. 00:07:16.600 --> 00:07:25.030 So OK, when you're given a problem like that, 00:07:25.030 --> 00:07:28.000 under normal conditions it's a good idea 00:07:28.000 --> 00:07:32.510 to think about it before trying to solve it. 00:07:32.510 --> 00:07:34.630 And I think we've done a little bit of thinking. 00:07:34.630 --> 00:07:38.740 We've realized the state stays in a after a while, 00:07:38.740 --> 00:07:42.775 and immediately must transition into some other state. 00:07:45.970 --> 00:07:48.390 And once it's in that other state, its going 00:07:48.390 --> 00:07:51.960 to remain in that state forever because those are energy 00:07:51.960 --> 00:07:55.425 eigenstates, so you're going to get exponentials of E 00:07:55.425 --> 00:07:56.860 to the minus iEt. 00:07:56.860 --> 00:07:59.580 But this is going to be that. 00:07:59.580 --> 00:08:02.620 OK, so let's solve this. 00:08:02.620 --> 00:08:05.350 Here is our equation. 00:08:05.350 --> 00:08:07.580 Let's try to write it for this case. 00:08:07.580 --> 00:08:15.660 So I would have i h bar C a dot of t. 00:08:15.660 --> 00:08:18.240 We have two states-- 00:08:18.240 --> 00:08:20.850 1 and 2-- but here is a and b. 00:08:20.850 --> 00:08:29.160 So you would have i omega a then something, t and delta H 00:08:29.160 --> 00:08:31.320 a with something. 00:08:31.320 --> 00:08:36.000 Well, the delta H is clear that it can only 00:08:36.000 --> 00:08:41.100 be a delta H from a to b. 00:08:41.100 --> 00:08:43.140 Because there's Uab. 00:08:43.140 --> 00:08:45.750 There's only that transition amplitude. 00:08:45.750 --> 00:08:51.420 And then it would be an i omega ab, 00:08:51.420 --> 00:09:06.900 which was defined as omega 0, delta Hab of time, cb of time. 00:09:06.900 --> 00:09:11.970 And that's the only term that exists here from the sum. 00:09:11.970 --> 00:09:16.420 You could sum over a and b, but we saw that over a, you get 0. 00:09:16.420 --> 00:09:18.570 So there's just this term. 00:09:18.570 --> 00:09:25.235 And then we have i H bar Cb dot of [? t ?] 00:09:25.235 --> 00:09:28.560 is equal to E to the i. 00:09:28.560 --> 00:09:31.410 And now you're going to have ba-- 00:09:31.410 --> 00:09:38.910 so delta Hba of t Ca of t. 00:09:38.910 --> 00:09:42.990 And here I would have omega ba, which 00:09:42.990 --> 00:09:48.180 is the opposite sign as omega ab, which we anyway 00:09:48.180 --> 00:09:49.840 call omega 0. 00:09:49.840 --> 00:09:53.850 So here's minus i omega 0 t. 00:09:56.360 --> 00:09:57.260 Yes. 00:09:57.260 --> 00:10:01.250 AUDIENCE: [INAUDIBLE] perturbation disappears? 00:10:01.250 --> 00:10:02.134 PROFESSOR: Sorry. 00:10:02.134 --> 00:10:03.912 AUDIENCE: Does the perturbation vanish? 00:10:03.912 --> 00:10:04.620 PROFESSOR: Right. 00:10:04.620 --> 00:10:14.480 Because delta Haa is equal to 0, and therefore there's 00:10:14.480 --> 00:10:16.840 no coupling to Ca. 00:10:16.840 --> 00:10:24.200 And delta Hbb is 0 from that matrix element. 00:10:24.200 --> 00:10:26.750 And therefore, you cannot couple b to b. 00:10:31.340 --> 00:10:40.690 And now I'll just write it i H bar Ca dot of t, 00:10:40.690 --> 00:10:49.030 i H bar Cb dot of t is equal E to the i omega 0t. 00:10:49.030 --> 00:10:55.150 Delta Hab was alpha delta of time-- 00:10:55.150 --> 00:10:57.820 Cb of t. 00:10:57.820 --> 00:11:00.850 And this is E to the minus i omega 00:11:00.850 --> 00:11:06.400 0t, alpha star delta of time-- 00:11:06.400 --> 00:11:09.450 Ca of t. 00:11:09.450 --> 00:11:15.411 Hba was Uba, which is alpha star times the delta function 00:11:15.411 --> 00:11:15.910 of time. 00:11:22.260 --> 00:11:25.020 Here I have a delta function of time. 00:11:25.020 --> 00:11:30.390 It says that this only matters when time is equal to 0. 00:11:30.390 --> 00:11:34.080 So things are a little delicate here. 00:11:34.080 --> 00:11:35.610 But let's see. 00:11:35.610 --> 00:11:38.530 Can I write here-- 00:11:38.530 --> 00:11:44.340 well, if I have f of x times delta of x, 00:11:44.340 --> 00:11:49.170 we know this is f of 0 times delta of x. 00:11:49.170 --> 00:11:51.600 Because anyway, only zeros [INAUDIBLE].. 00:11:51.600 --> 00:11:54.840 So I can put time equals 0 here, and forget 00:11:54.840 --> 00:12:02.850 about this exponential-- have alpha delta of t, cb of t, 00:12:02.850 --> 00:12:09.960 and here have alpha star delta of t Ca of t. 00:12:15.120 --> 00:12:17.130 Everybody happy so far? 00:12:20.710 --> 00:12:28.200 But here, I have Cb of t and Ca of t, and a delta t. 00:12:28.200 --> 00:12:34.800 So I should put alpha delta of t Cb at 0, 00:12:34.800 --> 00:12:42.401 and alpha star delta of t Ca at 0. 00:12:42.401 --> 00:12:42.900 Right? 00:12:45.865 --> 00:12:46.365 Yes? 00:12:50.110 --> 00:12:53.860 People look less convinced, which I congratulate you 00:12:53.860 --> 00:12:56.110 for doubting this. 00:12:56.110 --> 00:12:58.030 This would be bad. 00:12:58.030 --> 00:13:00.880 It seems like I'm just following the logic, 00:13:00.880 --> 00:13:04.720 but I've gone too far now. 00:13:04.720 --> 00:13:08.260 Why did I go too far? 00:13:08.260 --> 00:13:12.190 First, this is going to be a disaster. 00:13:12.190 --> 00:13:22.000 Because in many ways, we don't know what these numbers are. 00:13:22.000 --> 00:13:24.190 You see, we are argued that this is going 00:13:24.190 --> 00:13:27.850 to change this continuously. 00:13:27.850 --> 00:13:30.910 Those numbers are going to change this continuously. 00:13:30.910 --> 00:13:32.950 At time equals 0-- 00:13:32.950 --> 00:13:38.290 before time equals 0, Ca was 1 and Cb was 0. 00:13:38.290 --> 00:13:42.010 But after time equals 0, Cb is going to be a number-- 00:13:42.010 --> 00:13:45.070 something that we don't know. 00:13:45.070 --> 00:13:47.040 So the delta function, which does it 00:13:47.040 --> 00:13:51.250 hiy-- the one before 0, the one after 0, the average? 00:13:51.250 --> 00:13:52.240 What does it do? 00:13:52.240 --> 00:13:56.440 This is a case where you have a delta function that we 00:13:56.440 --> 00:13:58.400 don't know what it's doing. 00:14:01.580 --> 00:14:03.600 So this is totally wrong. 00:14:03.600 --> 00:14:07.940 So I will just stop there. 00:14:07.940 --> 00:14:11.600 And I will consider, however these two equations. 00:14:14.820 --> 00:14:23.070 So we have the equations now in a nice way. 00:14:23.070 --> 00:14:25.390 I'll reconsider them. 00:14:25.390 --> 00:14:29.090 And we are facing a difficulty, the difficulty 00:14:29.090 --> 00:14:35.120 of the delta function not allowing us to treat nicely 00:14:35.120 --> 00:14:46.510 the transition of i h bar Ca dot is equal to alpha delta of Cb, 00:14:46.510 --> 00:14:55.250 and i h bar Cb dot equal alpha star delta of t Ca. 00:15:01.770 --> 00:15:08.880 Well now, the only way to resolve this difficulty 00:15:08.880 --> 00:15:12.270 is to think about this physically, 00:15:12.270 --> 00:15:16.740 and introduce a regulator that is physical. 00:15:16.740 --> 00:15:22.070 You see, no signal in the world really is a delta function. 00:15:22.070 --> 00:15:27.720 Nothing becomes infinite, and nothing lasts for zero time. 00:15:27.720 --> 00:15:32.910 So this interaction, this delta function, 00:15:32.910 --> 00:15:36.120 represents a function of time-- 00:15:36.120 --> 00:15:37.940 delta of time. 00:15:37.940 --> 00:15:44.880 It's a spike, but we can think of it as a regulated thing. 00:15:44.880 --> 00:15:47.970 You've regulated delta functions in all kinds of ways. 00:15:47.970 --> 00:15:51.330 It might be useful to regulate it as follows, 00:15:51.330 --> 00:15:57.720 as a function of time that is 0 before time equals 0, 00:15:57.720 --> 00:16:01.980 it's 0 after some time, t star, and it 00:16:01.980 --> 00:16:04.620 has height 1 over t star. 00:16:08.360 --> 00:16:10.705 That's a picture of a delta function. 00:16:14.250 --> 00:16:17.640 It has the right area at least. 00:16:17.640 --> 00:16:21.060 And we're going to think of this system 00:16:21.060 --> 00:16:28.770 as exactly doing that in between, for t, 00:16:28.770 --> 00:16:33.910 in the interval 0 to t star. 00:16:33.910 --> 00:16:35.850 The delta function is going to be 00:16:35.850 --> 00:16:39.090 replaced by this function, this constant function. 00:16:39.090 --> 00:16:46.170 So it will have i h bar h Ca dot is 00:16:46.170 --> 00:16:53.645 equal to alpha over t star Cb. 00:16:57.300 --> 00:17:10.089 And i h bar Cb dot is equal to alpha star over t sub star. 00:17:10.089 --> 00:17:13.780 Now the star doesn't mean complex conjugation for t. 00:17:13.780 --> 00:17:18.040 It's just a number, a glyph called t0 or something 00:17:18.040 --> 00:17:18.970 like that. 00:17:18.970 --> 00:17:23.569 And this is true for this time interval. 00:17:23.569 --> 00:17:29.740 And then you would say, look, I have two problems with this. 00:17:29.740 --> 00:17:32.980 First, you've entered this at t star, 00:17:32.980 --> 00:17:36.170 and now your calculation is going to depend on t star. 00:17:36.170 --> 00:17:39.090 And then the answer is going to depend on t star. 00:17:39.090 --> 00:17:42.780 And what are you going to put for t star? 00:17:42.780 --> 00:17:47.720 Well, that better not happen. 00:17:47.720 --> 00:17:50.210 We have to go on a limb. 00:17:50.210 --> 00:17:52.490 Many times when you do a calculation of something, 00:17:52.490 --> 00:17:53.540 you go on a limb. 00:17:53.540 --> 00:17:56.540 You say something, and see if it works. 00:17:56.540 --> 00:17:58.220 This seems reasonable. 00:17:58.220 --> 00:18:01.040 And what we expect is that we will 00:18:01.040 --> 00:18:06.710 find an answer that this independent on this t star. 00:18:06.710 --> 00:18:11.870 So that we can take this star to 0 and make sense of it. 00:18:11.870 --> 00:18:14.240 The other question that could come 00:18:14.240 --> 00:18:18.330 is that, oh, that was your delta function. 00:18:18.330 --> 00:18:23.420 So one seconds, you used delta function for this. 00:18:23.420 --> 00:18:28.460 Maybe we have to put back these terms, these exponentials. 00:18:28.460 --> 00:18:31.580 We set them to 1. 00:18:31.580 --> 00:18:34.520 But then you could say, look, I don't 00:18:34.520 --> 00:18:36.630 think I have to put them back. 00:18:36.630 --> 00:18:37.760 Why? 00:18:37.760 --> 00:18:43.850 Because I can choose t star as small as I want, 00:18:43.850 --> 00:18:48.980 such that omega t star, which is the possible value 00:18:48.980 --> 00:18:54.470 that this can get, at most, is like 10 to the minus 60. 00:18:54.470 --> 00:18:57.805 And therefore, this phase is going to be 0, essentially, 00:18:57.805 --> 00:18:59.300 and that's going to be 1. 00:18:59.300 --> 00:19:04.890 So I claim that I don't have to worry about this anymore. 00:19:04.890 --> 00:19:06.180 So what do we have? 00:19:06.180 --> 00:19:12.320 We have two simple differential equations, like this. 00:19:12.320 --> 00:19:19.940 With the conditions that Ca at time equals 0 is 1, 00:19:19.940 --> 00:19:25.600 and Cb at time equals 0 is 0. 00:19:25.600 --> 00:19:32.120 The state before the delta function hits is given by that. 00:19:32.120 --> 00:19:35.240 And now what we want to know is, how much is 00:19:35.240 --> 00:19:41.810 Ca after the delta function and Cb after the delta function? 00:19:46.330 --> 00:19:53.020 So for that, we have all the tools needed. 00:19:53.020 --> 00:19:55.910 I can take this equation. 00:19:55.910 --> 00:19:57.745 This is a coupled system of equations. 00:19:57.745 --> 00:19:59.290 It's a simple system. 00:19:59.290 --> 00:20:03.150 I could differentiate again here. 00:20:03.150 --> 00:20:10.250 And I get a Ca double-dot, Cb dot, and use this equation. 00:20:10.250 --> 00:20:17.740 So you get the equation-- if you differentiate again, take a dot 00:20:17.740 --> 00:20:25.180 to form Ca double-dot, you get that Ca double-dot is 00:20:25.180 --> 00:20:35.875 equal to minus alpha over h bar t star squared Ca. 00:20:41.360 --> 00:20:47.090 You can see that you put the i h bar here, take the derivative, 00:20:47.090 --> 00:20:50.090 you get an alpha times an alpha star, which 00:20:50.090 --> 00:20:52.880 is length of alpha squared. 00:20:52.880 --> 00:20:57.230 And the h bar appears two times, t star appears two times. 00:20:57.230 --> 00:21:01.140 This is an oscillating solution. 00:21:01.140 --> 00:21:05.860 So that's simple enough. 00:21:09.410 --> 00:21:16.430 So what happens here is that Ca will oscillate in time, 00:21:16.430 --> 00:21:19.440 and you will have a solution of the form 00:21:19.440 --> 00:21:31.850 Ca is equal to a constant, beta 0 cosine of alpha h bar t star 00:21:31.850 --> 00:21:42.390 t, plus beta 1 sine of alpha p, over h bar t star. 00:21:45.030 --> 00:21:53.090 And if you wish, Cb from the first equation 00:21:53.090 --> 00:21:57.230 is proportional to Ca dot. 00:22:01.460 --> 00:22:06.570 And Ca dot is going to be of the form beta 00:22:06.570 --> 00:22:14.570 0 sine off this thing, plus beta 1 cosine of this thing. 00:22:21.980 --> 00:22:26.120 OK, so we've turned this into a tractable problem because 00:22:26.120 --> 00:22:29.270 of flattening the delta function. 00:22:29.270 --> 00:22:36.050 And your initial conditions-- again, Cb must vanish at time 00:22:36.050 --> 00:22:37.910 equals 0. 00:22:37.910 --> 00:22:44.000 So if Cb must vanish at time equals 0, beta 1 must be 0. 00:22:48.420 --> 00:22:52.260 If beta 1 is zero, this term is gone, 00:22:52.260 --> 00:22:57.480 and Ca was 1 for time equals 0. 00:22:57.480 --> 00:22:59.820 So beta 0 is 1. 00:23:04.030 --> 00:23:18.570 Therefore, Ca is equal to cosine alpha t, over h star t star 00:23:18.570 --> 00:23:19.900 Ca of t. 00:23:23.580 --> 00:23:31.150 And Cb of t, you can calculate it from the top equation. 00:23:31.150 --> 00:23:34.450 You take the derivative of Ca, and divide 00:23:34.450 --> 00:23:38.080 by alpha multiplied by t star. 00:23:41.240 --> 00:23:42.730 I'll let you do it. 00:23:42.730 --> 00:23:54.460 You get minus i alpha, over alpha, sine, alpha t 00:23:54.460 --> 00:23:58.930 over ht star. 00:23:58.930 --> 00:24:05.740 OK, we solved for the functions. 00:24:05.740 --> 00:24:08.160 We know what's going to happen. 00:24:08.160 --> 00:24:12.640 And now, what did we want? 00:24:12.640 --> 00:24:17.020 We wanted to know what are those things for later times, 00:24:17.020 --> 00:24:20.350 for times t star or more. 00:24:20.350 --> 00:24:24.340 Now you cannot use this equation beyond time t star, 00:24:24.340 --> 00:24:27.470 because they assumed the delta function is there. 00:24:27.470 --> 00:24:29.230 So what you have to figure out is 00:24:29.230 --> 00:24:32.690 what are these coefficients at time t star. 00:24:32.690 --> 00:24:37.150 And those would be the coefficients at any later time, 00:24:37.150 --> 00:24:39.100 because there's no more Hamiltonian. 00:24:39.100 --> 00:24:45.800 So Ca at t greater than t star is, in fact, 00:24:45.800 --> 00:24:49.840 equal to Ca at t star. 00:24:49.840 --> 00:24:54.850 And Ca at t star, happily, is a number 00:24:54.850 --> 00:24:57.640 that doesn't depend on t star-- 00:24:57.640 --> 00:25:01.570 cosine alpha over h bar. 00:25:04.140 --> 00:25:11.900 And Cb at t greater than t star is Cb at t star. 00:25:11.900 --> 00:25:15.710 It doesn't change any more after the delta 00:25:15.710 --> 00:25:18.500 function has turned off. 00:25:18.500 --> 00:25:27.740 And Cb at t star is minus i alpha over alpha-- 00:25:27.740 --> 00:25:40.235 alpha over alpha, like this, times sine of alpha over h bar. 00:25:45.800 --> 00:25:48.290 And this is really what you wanted. 00:25:48.290 --> 00:25:54.130 Now you know what the state is doing after the delta 00:25:54.130 --> 00:25:55.810 function has turned on. 00:25:55.810 --> 00:26:02.240 It has this amplitude to be in the b state. 00:26:02.240 --> 00:26:04.950 And it is, as we predicted-- remember, 00:26:04.950 --> 00:26:11.100 this is like the face of alpha, because the alpha is alpha 00:26:11.100 --> 00:26:12.720 times its face-- 00:26:12.720 --> 00:26:15.690 length of alpha times it face, and it will cancel. 00:26:15.690 --> 00:26:19.270 And here you have sine of this quantity. 00:26:19.270 --> 00:26:22.380 So it is proportional. 00:26:22.380 --> 00:26:26.020 Cb is proportional to alpha. 00:26:26.020 --> 00:26:28.900 And that's exactly what you would expect. 00:26:28.900 --> 00:26:34.785 So just to complete the story, let's write the answer. 00:26:50.460 --> 00:26:52.410 And what is the answer? 00:26:52.410 --> 00:27:02.490 Phi of t for t greater than 0 is the coefficient Ca 00:27:02.490 --> 00:27:09.630 of t, which was cosine, alpha over h bar, 00:27:09.630 --> 00:27:17.850 times E to the minus iEat, over h bar a. 00:27:17.850 --> 00:27:23.400 Remember, the solution is Ca E to the minus iEta. 00:27:23.400 --> 00:27:32.490 And then the other one, which is minus i alpha, over alpha, 00:27:32.490 --> 00:27:47.280 sine of alpha over h bar, e to the minus iEbt over h bar, b. 00:27:47.280 --> 00:27:54.160 That is the state after the delta function has turned on. 00:27:54.160 --> 00:27:57.090 And what is the probability to be 00:27:57.090 --> 00:28:01.935 found in the state b, time t? 00:28:01.935 --> 00:28:09.870 It would be b times the state squared. 00:28:13.140 --> 00:28:20.370 So you get-- b will couple just to that. b with b gives you 1. 00:28:20.370 --> 00:28:21.940 This is a phase. 00:28:21.940 --> 00:28:23.730 This is a phase. 00:28:23.730 --> 00:28:27.240 Sine squared of this thing is sine 00:28:27.240 --> 00:28:33.570 squared of alpha over h bar. 00:28:33.570 --> 00:28:38.730 What is the probability to be in a? 00:28:38.730 --> 00:28:42.420 It's a psi of t. 00:28:42.420 --> 00:28:48.460 And in this case, it will be cosine squared 00:28:48.460 --> 00:28:52.370 of alpha over h bar. 00:28:52.370 --> 00:28:54.910 And it's very handy that cosine squared 00:28:54.910 --> 00:28:58.180 plus sine squared is equal to 1, because it 00:28:58.180 --> 00:29:02.680 has to be either in a or on b. 00:29:02.680 --> 00:29:06.490 And it's kind of interesting, if you 00:29:06.490 --> 00:29:08.890 were doing this in perturbation-- we solved it 00:29:08.890 --> 00:29:09.610 exactly. 00:29:09.610 --> 00:29:12.680 If you were doing this in perturbation theory, 00:29:12.680 --> 00:29:19.300 we would have found P of b proportional to alpha squared. 00:29:19.300 --> 00:29:22.840 Because the first term in the expansion here, 00:29:22.840 --> 00:29:25.020 in terms of the interaction.