WEBVTT 00:00:00.135 --> 00:00:02.490 The following content is provided under a Creative 00:00:02.490 --> 00:00:04.030 Commons license. 00:00:04.030 --> 00:00:06.330 Your support will help MIT OpenCourseWare 00:00:06.330 --> 00:00:10.690 continue to offer high quality educational resources for free. 00:00:10.690 --> 00:00:13.320 To make a donation or view additional materials 00:00:13.320 --> 00:00:17.250 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.250 --> 00:00:18.220 at ocw.mit.edu. 00:00:22.960 --> 00:00:23.780 ROBERT FIELD: OK. 00:00:23.780 --> 00:00:30.540 So, today is the first of a pair of lectures taking us 00:00:30.540 --> 00:00:33.690 from the Schrodinger picture to the Heisenberg picture. 00:00:33.690 --> 00:00:37.580 Or the wave function picture to the matrix picture. 00:00:37.580 --> 00:00:41.280 Now almost everyone who does quantum mechanics 00:00:41.280 --> 00:00:45.240 is rooted in the Heisenberg picture. 00:00:45.240 --> 00:00:49.980 We use the Heisenberg picture because the structure 00:00:49.980 --> 00:00:54.930 of the problem is immediately evident when you write down 00:00:54.930 --> 00:00:57.140 what you know. 00:00:57.140 --> 00:01:03.680 And it's also mostly the way you program your computers 00:01:03.680 --> 00:01:06.630 to solve any problem. 00:01:06.630 --> 00:01:14.240 So I'm going to get you to the Heisenberg picture by dealing 00:01:14.240 --> 00:01:17.150 with the two-level problem, which is-- 00:01:17.150 --> 00:01:21.240 it should be called one of the exactly solved problems, 00:01:21.240 --> 00:01:25.900 except it's abstract as opposed to harmonic oscillator, 00:01:25.900 --> 00:01:29.270 particle in an infinite box, rigid rotor. 00:01:29.270 --> 00:01:33.590 It's a exactly-solved problem, and it leads us to-- 00:01:33.590 --> 00:01:35.660 or guides us to-- 00:01:35.660 --> 00:01:36.500 the new approach. 00:01:39.480 --> 00:01:42.630 I'm going to approach this problem 00:01:42.630 --> 00:01:45.440 the algebraic or Schrodinger way. 00:01:45.440 --> 00:01:50.340 And then I'm going to describe it in a matrix way 00:01:50.340 --> 00:01:52.995 and introduce the new language and the new notation. 00:01:57.750 --> 00:02:00.750 So I can say, at least at the end of this lecture, 00:02:00.750 --> 00:02:03.770 I'll be able to say the word matrix element 00:02:03.770 --> 00:02:05.780 and everybody will know what it is, 00:02:05.780 --> 00:02:09.389 and I'll stop talking about integrals. 00:02:09.389 --> 00:02:13.470 And one of the scary things is, when 00:02:13.470 --> 00:02:17.630 we go to the matrix picture, we stop 00:02:17.630 --> 00:02:19.520 looking at the wave function. 00:02:19.520 --> 00:02:22.250 We never think about the wave functions. 00:02:22.250 --> 00:02:26.380 And so there are all sorts of things like phases 00:02:26.380 --> 00:02:31.630 that we have to make sure we're not screwing up. 00:02:31.630 --> 00:02:36.250 Because we're playing fast and loose with symbols, 00:02:36.250 --> 00:02:39.340 and sometimes they can bite you if you don't know what 00:02:39.340 --> 00:02:40.670 you're dealing with. 00:02:40.670 --> 00:02:44.740 But mostly, we're going to get rid of wave functions, 00:02:44.740 --> 00:02:50.470 because we know all the solutions to standard problems, 00:02:50.470 --> 00:02:54.010 and we know a lot of integrals involving those solutions 00:02:54.010 --> 00:02:56.020 to standard problems. 00:02:56.020 --> 00:02:58.510 And so basically, all we need is an index 00:02:58.510 --> 00:03:02.760 saying this wave function or this state, this integral, 00:03:02.760 --> 00:03:03.980 and you have-- 00:03:03.980 --> 00:03:07.200 and then you can write down everything 00:03:07.200 --> 00:03:11.460 you need in matrix notation, and then you 00:03:11.460 --> 00:03:14.910 can tell your friendly computer, OK, solve the problem for me. 00:03:18.420 --> 00:03:18.920 OK. 00:03:18.920 --> 00:03:24.770 So once I give you the matrix picture, 00:03:24.770 --> 00:03:29.930 then I will talk about how you find the eigenvalues 00:03:29.930 --> 00:03:33.890 and eigenvectors of the matrix picture using 00:03:33.890 --> 00:03:36.020 a unitary transformation. 00:03:36.020 --> 00:03:38.930 And then I'll generalize from the two-level problem, 00:03:38.930 --> 00:03:41.420 which is exactly soluble. 00:03:41.420 --> 00:03:44.202 You don't need a computer for it, but it's convenient-- 00:03:44.202 --> 00:03:46.160 because you don't have to write anything down-- 00:03:46.160 --> 00:03:51.420 to N levels where N can be infinite. 00:03:51.420 --> 00:03:58.980 And the N-level problem is in principle difficult 00:03:58.980 --> 00:04:04.140 because even a computer can't diagonalize an infinite matrix. 00:04:04.140 --> 00:04:07.920 And all of your basis sets, all of your standard problems 00:04:07.920 --> 00:04:10.950 involve an infinite number of functions. 00:04:10.950 --> 00:04:15.120 And so I've introduced a notation and a concept 00:04:15.120 --> 00:04:21.180 of solving a matrix equation, but you can't do it unless you, 00:04:21.180 --> 00:04:23.190 say, let's make an approximation, 00:04:23.190 --> 00:04:26.160 and it's called nondegenerate perturbation theory. 00:04:26.160 --> 00:04:28.290 And this is the tool for learning 00:04:28.290 --> 00:04:30.270 almost everything you want to know 00:04:30.270 --> 00:04:33.080 about complicated problems. 00:04:33.080 --> 00:04:37.120 And it's not complicated to apply 00:04:37.120 --> 00:04:38.870 nondegenerate perturbation theory. 00:04:38.870 --> 00:04:41.870 It's just ugly. 00:04:41.870 --> 00:04:43.700 But it's really valuable because you 00:04:43.700 --> 00:04:46.040 don't have to remember how to solve 00:04:46.040 --> 00:04:48.710 a particular complicated differential equation, 00:04:48.710 --> 00:04:50.840 you just write down what you know 00:04:50.840 --> 00:04:53.660 and you do some simple stuff, and bang, you've 00:04:53.660 --> 00:04:55.460 got a solution to the problem. 00:04:55.460 --> 00:04:57.560 And this is really what I want you to come away 00:04:57.560 --> 00:04:58.850 from this course with-- 00:04:58.850 --> 00:05:03.830 the concept that anything that requires quantum mechanics 00:05:03.830 --> 00:05:09.390 you can solve using some form of perturbation theory. 00:05:09.390 --> 00:05:12.125 So hold onto your seats. 00:05:25.740 --> 00:05:29.420 So the Schrodinger picture is differential equations. 00:05:38.360 --> 00:05:41.270 And they're often coupled differential equations. 00:05:41.270 --> 00:05:44.000 And you don't want to go there, usually. 00:05:44.000 --> 00:05:46.960 And we're going to replace that with linear algebra. 00:05:52.490 --> 00:05:55.880 Now many of you have not had a course in linear algebra. 00:05:55.880 --> 00:06:00.834 And so that should make you say, can I do this? 00:06:00.834 --> 00:06:02.500 And the answer is, yeah, you can do this 00:06:02.500 --> 00:06:04.990 because the linear algebra you are going 00:06:04.990 --> 00:06:09.010 to need in this course is based on one concept, 00:06:09.010 --> 00:06:14.200 and that is that the solution of coupled linear homogeneous 00:06:14.200 --> 00:06:19.495 equations involves diagonalizing or solving a determinant. 00:06:22.090 --> 00:06:26.210 And when you solve this determinant of the equation, 00:06:26.210 --> 00:06:30.090 it's equivalent to diagonalizing a matrix. 00:06:30.090 --> 00:06:33.890 And that's the language we're going to be using all the time. 00:06:33.890 --> 00:06:35.660 And so the only thing I'm not going to do 00:06:35.660 --> 00:06:38.520 is prove this fundamental theorem of linear algebra, 00:06:38.520 --> 00:06:44.480 but the notation is easy to use and the tricks are very simple. 00:06:47.800 --> 00:06:52.290 So, we have with exactly-solved problems 00:06:52.290 --> 00:06:58.030 complete sets of the energy levels of the wave functions. 00:06:58.030 --> 00:07:00.420 And we know a lot of integrals. 00:07:00.420 --> 00:07:06.630 Psi i operator psi j d tau. 00:07:06.630 --> 00:07:10.150 We know these not by evaluating them, 00:07:10.150 --> 00:07:14.130 but because the functions are so simple that we can write these 00:07:14.130 --> 00:07:17.470 as simply a function of the initial and final quantum 00:07:17.470 --> 00:07:17.970 numbers. 00:07:21.170 --> 00:07:24.660 And so all of a sudden, we forget 00:07:24.660 --> 00:07:26.580 about the wave functions, because all the work 00:07:26.580 --> 00:07:29.390 has been done for us. 00:07:29.390 --> 00:07:30.850 We could do it too, but why? 00:07:33.600 --> 00:07:36.860 So we start with the two-level problem. 00:07:36.860 --> 00:07:39.140 And the two-level problem says we have 00:07:39.140 --> 00:07:44.270 two states, psi 1 and psi 2. 00:07:44.270 --> 00:07:47.020 Still in the Schrodinger picture. 00:07:47.020 --> 00:07:54.270 And this has an energy which we could we could call E1, 00:07:54.270 --> 00:08:03.550 and that would be H11, the diagonal integral 00:08:03.550 --> 00:08:08.750 of the Hamiltonian between the 1 function and the 1 function, 00:08:08.750 --> 00:08:09.520 and this is E2. 00:08:13.660 --> 00:08:18.430 Now these two states have an interaction. 00:08:18.430 --> 00:08:21.650 They're connected by an interaction term 00:08:21.650 --> 00:08:25.570 which causes them to repel each other equal and opposite 00:08:25.570 --> 00:08:26.830 amounts. 00:08:26.830 --> 00:08:31.880 And so this is H12. 00:08:31.880 --> 00:08:36.289 Again, an integral, which you usually don't have to evaluate, 00:08:36.289 --> 00:08:39.890 because it's basically done for you. 00:08:39.890 --> 00:08:44.760 And then we get E plus and E minus, 00:08:44.760 --> 00:08:47.640 and the corresponding eigenfunction. 00:08:50.680 --> 00:08:53.200 So this is the two-level problem, 00:08:53.200 --> 00:08:56.200 and it's expressed in terms of three parameters-- 00:08:56.200 --> 00:09:02.670 H11, H22, and H12. 00:09:02.670 --> 00:09:08.540 And we get the two energy levels and the two eigenfunctions. 00:09:08.540 --> 00:09:11.290 And we know how to solve this problem in the Schrodinger 00:09:11.290 --> 00:09:15.700 picture, and I'll do that first. 00:09:15.700 --> 00:09:24.380 And so, H11 is just this integral, psi 1 star H. 00:09:24.380 --> 00:09:27.410 And we put a hat on H for the time being-- 00:09:27.410 --> 00:09:30.380 1 d tau. 00:09:30.380 --> 00:09:41.950 And H22 is a different integral of psi 2 star H psi 2 d tau. 00:09:41.950 --> 00:09:50.345 And H12 is psi 1 star H psi 2 tau. 00:09:52.990 --> 00:10:00.670 And we know that the Hamiltonian operator is Hermitian, 00:10:00.670 --> 00:10:05.440 and so we can write also that this is 00:10:05.440 --> 00:10:14.500 psi 2 star H star psi I d tau. 00:10:14.500 --> 00:10:19.480 Anyway, we call this thing V. And there's some subtleties. 00:10:19.480 --> 00:10:25.550 If the integral between these two things is imaginary, 00:10:25.550 --> 00:10:34.190 then the 1-2 and 2-1 integrals have opposite signs. 00:10:34.190 --> 00:10:36.710 Because the Hamiltonian is Hermitian. 00:10:36.710 --> 00:10:39.380 But let's just think of this as just one number. 00:10:39.380 --> 00:10:44.960 So we have E1, E2, and delta. 00:10:49.620 --> 00:10:54.330 So the two-level problem is going to let us find 00:10:54.330 --> 00:10:56.250 the eigenfunctions-- 00:10:56.250 --> 00:10:58.290 the plus and minus eigenfunctions. 00:10:58.290 --> 00:10:59.290 C plus 1. 00:11:06.330 --> 00:11:11.040 So we have eigenfunctions corresponding to-- which 00:11:11.040 --> 00:11:15.960 are eigenfunctions belonging to the eigenvalues, E plus and E 00:11:15.960 --> 00:11:17.160 minus. 00:11:17.160 --> 00:11:20.690 And there are a linear combination of the two states. 00:11:20.690 --> 00:11:23.040 Now, it's a two-level problem. 00:11:23.040 --> 00:11:27.730 This is a state space that contains only two states. 00:11:27.730 --> 00:11:29.160 It's not an approximation. 00:11:31.770 --> 00:11:35.100 And so completeness says we can write any function 00:11:35.100 --> 00:11:38.520 we want as a linear combination of the functions and the basis 00:11:38.520 --> 00:11:40.440 set. 00:11:40.440 --> 00:11:45.180 And so our job is going to be to find these coefficients 00:11:45.180 --> 00:11:48.441 and also to find the energy eigenvalues. 00:11:55.990 --> 00:11:56.500 OK. 00:11:56.500 --> 00:11:59.620 So let's start to do some work. 00:11:59.620 --> 00:12:01.750 We know this thing. 00:12:01.750 --> 00:12:03.550 And let's start some-- 00:12:03.550 --> 00:12:08.740 and we know that H psi plus-minus 00:12:08.740 --> 00:12:15.010 has to be equal to the eigenenergies psi plus-minus. 00:12:15.010 --> 00:12:19.840 Now, in order to get something useful out of this, 00:12:19.840 --> 00:12:22.830 we left multiply by psi 1. 00:12:27.040 --> 00:12:27.650 OK. 00:12:27.650 --> 00:12:30.290 And when you left multiply by psi 1-- well, 00:12:30.290 --> 00:12:31.497 let's just write it out. 00:12:35.240 --> 00:12:42.590 We're going to get psi 1 HC 1 plus. 00:12:42.590 --> 00:12:45.070 So let's just write this out. 00:12:45.070 --> 00:12:53.610 We have H11 C1 plus-minus. 00:12:53.610 --> 00:13:00.120 Because this is C plus-minus, and we 00:13:00.120 --> 00:13:06.360 have the integral of psi 1 with itself or the Hamiltonian 00:13:06.360 --> 00:13:07.690 is H11. 00:13:07.690 --> 00:13:14.200 And then we get another term. 00:13:14.200 --> 00:13:16.080 I'm doing this differently from my notes, 00:13:16.080 --> 00:13:28.360 so the other term is C plus-minus psi 2 V. OK. 00:13:28.360 --> 00:13:36.420 So plugging in what I have over here and knowing that H11-- 00:13:36.420 --> 00:13:42.390 we know that V is psi 1 H psi 2, and so this is what we get. 00:13:45.060 --> 00:13:50.530 We do the same thing, and we have the same left-hand side, 00:13:50.530 --> 00:13:59.790 but we can also now express this as that H-- 00:14:07.220 --> 00:14:10.172 that psi plus-minus is an eigenfunction. 00:14:13.130 --> 00:14:15.710 And so we have the same left-hand side, 00:14:15.710 --> 00:14:17.390 but on the right-hand side, we now 00:14:17.390 --> 00:14:28.060 get integral psi 1 star H psi plus-minus d tau. 00:14:28.060 --> 00:14:41.540 But that's equal to psi star E plus-minus C1 plus-minus psi 1 00:14:41.540 --> 00:14:48.640 plus C2 plus-minus psi 2 d tau. 00:14:48.640 --> 00:14:49.190 OK. 00:14:49.190 --> 00:14:52.700 And so now we're talking about orthonormality of the wave 00:14:52.700 --> 00:14:55.880 function, and because we don't have any operators in here, 00:14:55.880 --> 00:15:04.980 then this thing becomes simply E plus-minus times C1 plus-minus 00:15:04.980 --> 00:15:10.550 plus 0 C2 plus-minus. 00:15:14.210 --> 00:15:16.330 There should be a 1 here. 00:15:16.330 --> 00:15:16.830 OK. 00:15:16.830 --> 00:15:21.040 Because psi 1, psi 2, that's orthogonal. 00:15:21.040 --> 00:15:23.320 So now we have an equation that's quite useful. 00:15:29.470 --> 00:15:31.980 So we're going to combine these two equations, 00:15:31.980 --> 00:15:41.122 and we get C1 plus H11 plus C2 plus-- 00:15:41.122 --> 00:15:44.220 I'm sorry, this is plus-minus, plus-minus. 00:15:44.220 --> 00:15:51.080 V is equal to E plus-minus C1 plus-minus. 00:15:54.870 --> 00:15:57.010 Now we need another equation. 00:15:57.010 --> 00:16:00.980 And so we do the same thing and we left multiply by psi 2 star 00:16:00.980 --> 00:16:03.170 and integrate. 00:16:03.170 --> 00:16:05.870 And we get another equation, and that 00:16:05.870 --> 00:16:19.770 is C1 plus-minus the plus C2 plus-minus H22 minus E 00:16:19.770 --> 00:16:25.630 plus-minus is equal to 0. 00:16:25.630 --> 00:16:33.020 So we combine the two equations that we have derived, solving-- 00:16:33.020 --> 00:16:39.920 because both equations can be solved for C1 plus-minus or C2 00:16:39.920 --> 00:16:41.210 plus-minus. 00:16:41.210 --> 00:16:44.810 So we equate the C1 plus-minus over C2 plus-minus 00:16:44.810 --> 00:16:46.430 from the two equations. 00:16:46.430 --> 00:16:51.710 And we get this wonderful result, V over H11 minus E 00:16:51.710 --> 00:17:04.450 plus-minus is equal to H22 minus E plus-minus over V. 00:17:04.450 --> 00:17:07.050 Now you're not going to ever do this. 00:17:07.050 --> 00:17:13.800 So yes, you can attempt to reconstruct this 00:17:13.800 --> 00:17:16.890 from what I read on the board or what's in your notes, 00:17:16.890 --> 00:17:21.359 but the important point is that we're just using what we know. 00:17:21.359 --> 00:17:24.450 We say we have eigenfunctions, and we 00:17:24.450 --> 00:17:29.790 want to find something about the coefficients of the basis 00:17:29.790 --> 00:17:31.900 functions in each of the eigenfunctions. 00:17:31.900 --> 00:17:33.850 And we want to find the eigenvalues, 00:17:33.850 --> 00:17:36.890 and we get this equation. 00:17:36.890 --> 00:17:41.560 So, this is easy. 00:17:41.560 --> 00:17:44.230 We relate this to-- 00:17:44.230 --> 00:17:47.860 we just multiply through and we get V2, 00:17:47.860 --> 00:17:57.989 V squared is equal to H11 minus E plus-minus times H22 00:17:57.989 --> 00:17:58.780 minus E plus-minus. 00:17:58.780 --> 00:18:07.100 Or we solve, and we know that E plus-minus is equal-- this 00:18:07.100 --> 00:18:08.900 is a quadratic equation. 00:18:08.900 --> 00:18:14.620 A quadratic and E plus-minus, and so we have this result, 00:18:14.620 --> 00:18:26.380 H11 plus H22 plus or minus this H11 squared-- 00:18:26.380 --> 00:18:48.680 H11 plus H22 squared minus 4 H11 H22 minus V squared over 2. 00:18:48.680 --> 00:18:51.660 This is just a quadratic equation. 00:18:51.660 --> 00:18:56.450 So we have the eigenenergies expressed in terms 00:18:56.450 --> 00:18:57.680 of the quantities we know. 00:19:08.230 --> 00:19:10.210 We simplify the notation. 00:19:10.210 --> 00:19:19.410 E bar is H11 plus H22 over 2. 00:19:19.410 --> 00:19:26.630 And delta is H11 minus H22 over 2. 00:19:29.690 --> 00:19:33.600 So when we do that, we simplify the algebra 00:19:33.600 --> 00:19:37.590 and we end up discovering that the eigenvalue equation is 00:19:37.590 --> 00:19:46.350 E plus-minus is equal to E bar plus or minus delta squared 00:19:46.350 --> 00:19:50.130 plus V squared square root. 00:19:50.130 --> 00:19:54.580 That's a simple result. This is something you should remember. 00:19:57.810 --> 00:19:59.780 So if you have a two-level problem, 00:19:59.780 --> 00:20:03.547 the energy eigenvalues are the average energy plus or minus 00:20:03.547 --> 00:20:04.130 this quantity. 00:20:08.390 --> 00:20:11.130 This is an exact solution for a two-level problem. 00:20:11.130 --> 00:20:12.944 For all two-level problems. 00:20:16.430 --> 00:20:16.970 OK. 00:20:16.970 --> 00:20:19.280 And we even simplify the notation more. 00:20:19.280 --> 00:20:25.080 We call X delta squared plus V squared. 00:20:25.080 --> 00:20:35.560 And so this becomes E bar plus-minus X squared. 00:20:35.560 --> 00:20:37.600 Pretty compact. 00:20:37.600 --> 00:20:40.600 Now, the reason for simplifying the notation 00:20:40.600 --> 00:20:43.090 is that we're going to derive the values 00:20:43.090 --> 00:20:47.820 for the eigenfunctions, and they involve a lot of symbols. 00:20:47.820 --> 00:20:51.020 And we want to make this as compact as possible, 00:20:51.020 --> 00:20:54.630 and so we'll do that. 00:20:54.630 --> 00:20:55.130 OK. 00:20:55.130 --> 00:20:57.020 So when we started out, it looked 00:20:57.020 --> 00:21:02.060 like we were going to solve for these quantities, 00:21:02.060 --> 00:21:06.830 but we took a detour and solved for the eigenenergies first. 00:21:06.830 --> 00:21:10.100 And this is one of the things that happens in linear algebra. 00:21:10.100 --> 00:21:12.830 You get something you can get easily-- 00:21:12.830 --> 00:21:15.430 quickly before you get the other stuff that you want. 00:21:18.510 --> 00:21:23.410 OK, but the second part of the job 00:21:23.410 --> 00:21:26.660 is to find these mixing coefficients. 00:21:26.660 --> 00:21:31.240 And so one thing you do is you say, 00:21:31.240 --> 00:21:45.330 well, we insist that the wave functions be orthog-- 00:21:45.330 --> 00:21:47.980 normalized. 00:21:47.980 --> 00:21:51.820 After that a lot of algebra ensues. 00:21:51.820 --> 00:21:58.780 And I'm not going to even attempt to work through it. 00:21:58.780 --> 00:22:02.020 You may want to work through it, but what I recommend doing 00:22:02.020 --> 00:22:06.400 is looking at the solution and then 00:22:06.400 --> 00:22:08.530 checking to see whether it does things 00:22:08.530 --> 00:22:11.270 that you expect it has to do. 00:22:11.270 --> 00:22:15.250 Because one of the things that kind of inspection 00:22:15.250 --> 00:22:22.520 leads you to is factors of two errors and sign errors. 00:22:22.520 --> 00:22:23.030 OK. 00:22:23.030 --> 00:22:26.840 But, C plus-minus-- C1 plus-minus 00:22:26.840 --> 00:22:33.960 is equal to 1/2 times 1 plus or minus 00:22:33.960 --> 00:22:42.880 delta over the square root of X. Two square roots. 00:22:42.880 --> 00:22:43.380 Whoops. 00:22:49.970 --> 00:22:51.970 OK, no square root in here because we 00:22:51.970 --> 00:22:53.850 get square root there. 00:22:53.850 --> 00:23:02.440 OK and C2 plus-minus is 1/2 1 minus or plus 00:23:02.440 --> 00:23:07.720 delta square root, x square root. 00:23:07.720 --> 00:23:09.640 OK? 00:23:09.640 --> 00:23:13.810 So these two things are kind of simple-looking, 00:23:13.810 --> 00:23:16.570 but there's an awful lot of compression 00:23:16.570 --> 00:23:19.900 and select and clever manipulation 00:23:19.900 --> 00:23:21.250 in order to get this. 00:23:21.250 --> 00:23:23.320 But the important thing to notice 00:23:23.320 --> 00:23:27.850 is that we have the energy difference divided 00:23:27.850 --> 00:23:33.640 by this thing, this delta squared plus V 00:23:33.640 --> 00:23:39.130 squared that expresses the importance 00:23:39.130 --> 00:23:40.810 of the two-level interaction. 00:23:40.810 --> 00:23:43.855 And these two things enter with opposite signs. 00:23:50.130 --> 00:23:50.770 OK. 00:23:50.770 --> 00:23:53.350 And so one of the checks that's really easy to do 00:23:53.350 --> 00:24:00.770 is let's let V go to 0, and let's let V go to infinity. 00:24:00.770 --> 00:24:03.940 OK, remember, our Hamiltonian-- 00:24:03.940 --> 00:24:04.900 the two-level problem-- 00:24:08.290 --> 00:24:10.130 I can't write it as a matrix yet, 00:24:10.130 --> 00:24:18.670 so remember that we had E1, E2, and V. 00:24:18.670 --> 00:24:22.210 And this was the higher energy, this was the lower energy, 00:24:22.210 --> 00:24:28.760 and so delta is E1 minus E2-- 00:24:28.760 --> 00:24:30.190 I better not write that. 00:24:34.040 --> 00:24:35.980 OK. 00:24:35.980 --> 00:24:46.540 So we normally write E1 over E2, and we have this V interaction 00:24:46.540 --> 00:24:48.220 between them. 00:24:48.220 --> 00:24:49.120 OK. 00:24:49.120 --> 00:24:58.120 And so if the interaction integral between the two 00:24:58.120 --> 00:25:06.100 functions is 0, then E1 is the higher energy, 00:25:06.100 --> 00:25:08.500 and it corresponds to-- 00:25:08.500 --> 00:25:12.550 it should correspond to psi 1 alone. 00:25:12.550 --> 00:25:20.560 Well, so the higher energy we're taking the plus combinations 00:25:20.560 --> 00:25:28.640 here, and if V is equal to 0, then this is delta over delta. 00:25:28.640 --> 00:25:32.110 And this is either 2 or 0. 00:25:32.110 --> 00:25:33.580 And when it's the higher-- 00:25:33.580 --> 00:25:37.870 the upper sine, it's 2 divided by 2 or 1, 00:25:37.870 --> 00:25:40.030 exactly what you expect. 00:25:40.030 --> 00:25:44.200 And this one is 1 minus 1. 00:25:44.200 --> 00:25:46.930 So these two are behaving, right? 00:25:46.930 --> 00:25:50.380 And the V to 0 rate. 00:25:50.380 --> 00:25:57.720 Now, in the V going to infinity, then this is infinite. 00:25:57.720 --> 00:26:16.120 So that just means that this term goes away, right? 00:26:16.120 --> 00:26:24.230 And so we have C1 plus and minus is equal to square root of 2-- 00:26:24.230 --> 00:26:27.260 1 over the square root of 2. 00:26:27.260 --> 00:26:29.710 And same here. 00:26:29.710 --> 00:26:31.920 So what we get is what we call 50/50 mixing. 00:26:37.410 --> 00:26:39.740 OK, so this also works. 00:26:42.250 --> 00:26:45.640 So I don't say that this can-- 00:26:45.640 --> 00:26:49.720 this confirms that I have not made an algebraic mistake, 00:26:49.720 --> 00:26:56.120 I haven't, but it's a good test because it's really easy to do. 00:26:56.120 --> 00:26:58.220 And if you're not getting what you expect, 00:26:58.220 --> 00:27:01.100 you know you've either blown a sign or a factor 00:27:01.100 --> 00:27:07.260 or two, which are the two things that you fear most in a wave 00:27:07.260 --> 00:27:10.630 function free approach. 00:27:10.630 --> 00:27:14.290 Because you've got nothing to do except check your algebra, 00:27:14.290 --> 00:27:18.320 and usually, you made the mistake because it was subtle, 00:27:18.320 --> 00:27:20.710 and you'll make it again when you're checking it. 00:27:20.710 --> 00:27:23.050 So these kinds of checks are really valuable. 00:27:28.365 --> 00:27:28.865 OK. 00:27:33.310 --> 00:27:37.750 So once you know that this is likely to be correct, 00:27:37.750 --> 00:27:39.850 then you do some other things, and you 00:27:39.850 --> 00:27:45.760 check the wave functions you've derived with your C plus-minus 00:27:45.760 --> 00:27:55.460 1 and C plus-minus two are both normalized and orthogonal. 00:28:01.340 --> 00:28:10.160 And that the energy you get is such that psi plus is E plus. 00:28:10.160 --> 00:28:12.560 And you know what E plus was. 00:28:12.560 --> 00:28:16.550 And so you just go in and you calculate 00:28:16.550 --> 00:28:20.930 what the energy should be giving using the values for PSI 00:28:20.930 --> 00:28:28.370 plus-minus And so we plug those into the original equations. 00:28:28.370 --> 00:28:34.160 And you can also, again, show that psi plus-minus 00:28:34.160 --> 00:28:50.590 a star H psi plus-minus is equal to 0 00:28:50.590 --> 00:28:56.240 because the eigenfunctions are orthogonal. 00:28:56.240 --> 00:28:59.450 We already did that, but we plugged it into an equation 00:28:59.450 --> 00:29:01.190 here, and we got it a second time. 00:29:07.240 --> 00:29:14.770 So in the lecture notes, there is a lengthy algebraic proof 00:29:14.770 --> 00:29:17.380 or demonstration that E plus-minus 00:29:17.380 --> 00:29:20.540 is equal to plus or minus square of x, 00:29:20.540 --> 00:29:22.570 which are already derived, but then I 00:29:22.570 --> 00:29:23.830 just did it the long way. 00:29:23.830 --> 00:29:24.730 OK . 00:29:24.730 --> 00:29:26.710 So this is it-- 00:29:26.710 --> 00:29:31.090 we are about to move from the Schrodinger picture 00:29:31.090 --> 00:29:34.700 to the matrix picture. 00:29:34.700 --> 00:29:46.970 So the trick now is to go to linear algebra, 00:29:46.970 --> 00:29:49.520 go to the matrix picture and learn 00:29:49.520 --> 00:29:53.600 how to just write the equations and what the language is 00:29:53.600 --> 00:29:54.575 and show that it works. 00:29:57.460 --> 00:30:04.480 So suppose you have two square matrices, A and B, 00:30:04.480 --> 00:30:09.490 they're n by n squared matrices, you know the rules 00:30:09.490 --> 00:30:11.307 for matrix multiplication? 00:30:15.290 --> 00:30:23.000 So if you wanted, the mn element of this product or two 00:30:23.000 --> 00:30:28.550 matrices, you would go, n equals-- 00:30:28.550 --> 00:30:29.640 i equals 1. 00:30:32.320 --> 00:30:38.850 Let's use the same notation. j equals 1 to n of Am-- 00:30:38.850 --> 00:30:40.544 because I want the mn-- 00:30:40.544 --> 00:30:42.270 j Bjn. 00:30:47.220 --> 00:30:50.040 And so when you multiply two square matrices, 00:30:50.040 --> 00:30:51.600 you get back a square matrix. 00:30:51.600 --> 00:30:53.950 And this picture is not a bad one. 00:30:53.950 --> 00:30:57.020 So you can say, all right. 00:31:03.410 --> 00:31:08.510 So if you need a little q to remind yourself, 00:31:08.510 --> 00:31:14.390 you take this row and multiply term by term 00:31:14.390 --> 00:31:16.940 and add the results-- 00:31:16.940 --> 00:31:20.630 this row in that column, and you get a number here. 00:31:20.630 --> 00:31:24.260 And you just repeat that, and it's really easy 00:31:24.260 --> 00:31:25.820 to tell your computer to do this, 00:31:25.820 --> 00:31:28.130 and it's rather tedious to do it yourself 00:31:28.130 --> 00:31:30.080 if it's more than a two-by-two matrix. 00:31:33.060 --> 00:31:35.420 OK. 00:31:35.420 --> 00:31:40.290 So what about this thing C? 00:31:40.290 --> 00:31:51.940 Well, C is a N row column matrix. 00:31:51.940 --> 00:31:54.120 So it's N rows, 1 column. 00:31:56.800 --> 00:32:00.280 It looks like this-- 00:32:00.280 --> 00:32:04.090 C1, C2, Cn. 00:32:06.910 --> 00:32:13.290 And so if you want to multiply a square matrix by a vector, 00:32:13.290 --> 00:32:15.880 we know the rules too, OK? 00:32:15.880 --> 00:32:19.930 And again, this picture is a useful one. 00:32:19.930 --> 00:32:22.900 So let's just draw something like this. 00:32:22.900 --> 00:32:29.320 We do this and this, and that gives you 00:32:29.320 --> 00:32:30.800 one element in the column. 00:32:34.970 --> 00:32:36.490 OK. 00:32:36.490 --> 00:32:40.420 Now, the last thing that I want to remind you of-- 00:32:40.420 --> 00:32:41.860 I better use this board. 00:32:46.680 --> 00:32:59.570 If we have a two-state problem, then the vector c1 is 1, 0; 00:32:59.570 --> 00:33:02.960 the vector c2-- 00:33:02.960 --> 00:33:05.840 this should be a lower case c, because we 00:33:05.840 --> 00:33:10.100 tend to use lowercase letters for vectors and uppercase 00:33:10.100 --> 00:33:11.135 letters for matrices. 00:33:16.040 --> 00:33:32.340 And so if we do c1 dagger c2, or c1 dagger c1. 00:33:35.410 --> 00:33:41.570 Now this dagger means conjugate transpose, except, well, 00:33:41.570 --> 00:33:47.450 there isn't anything to do except convert a matrix-- 00:33:47.450 --> 00:33:58.250 a vector into-- so that becomes a row and this 00:33:58.250 --> 00:33:59.390 becomes a column. 00:34:03.110 --> 00:34:05.740 And so a row times the column gives a number. 00:34:08.360 --> 00:34:11.770 And that number is going to be-- well, let's do it. 00:34:11.770 --> 00:34:16.719 We have 1, 0; 0, 1. 00:34:16.719 --> 00:34:21.790 1 times 0 is 0, 0 times 1 is 0, and so this is 0. 00:34:21.790 --> 00:34:25.000 Well that's orthogonality. 00:34:25.000 --> 00:34:32.730 This, we have 1 times 1 is 1. 00:34:32.730 --> 00:34:34.030 0 times 0 is 0. 00:34:34.030 --> 00:34:37.760 So it's 1 plus 0. 00:34:37.760 --> 00:34:38.770 Right. 00:34:38.770 --> 00:34:40.429 OK, all right. 00:34:46.960 --> 00:34:47.460 OK. 00:34:50.400 --> 00:34:58.530 So the Schrodinger equation becomes in matrix language-- 00:34:58.530 --> 00:35:01.900 and now, one notation is-- 00:35:01.900 --> 00:35:04.020 I'm going to stop using the double underline. 00:35:04.020 --> 00:35:06.360 That means boldface. 00:35:06.360 --> 00:35:09.500 And we don't use hats anymore. 00:35:09.500 --> 00:35:12.340 Or at least if we were really consistent, 00:35:12.340 --> 00:35:15.400 when we go away from the Schrodinger picture, 00:35:15.400 --> 00:35:20.320 we don't put hats on operators, we make them boldface letters. 00:35:20.320 --> 00:35:21.000 OK. 00:35:21.000 --> 00:35:26.580 Now, the thing is we're so comfortable in matrix land, 00:35:26.580 --> 00:35:29.740 that we don't use either. 00:35:29.740 --> 00:35:30.260 OK. 00:35:30.260 --> 00:35:33.780 But remember, we're talking about different things. 00:35:33.780 --> 00:35:36.950 So the Schrodinger equation looks like this. 00:35:45.600 --> 00:35:53.070 And so we have a matrix, delta v, v delta times 1, 0. 00:35:56.810 --> 00:36:06.350 And that's delta, v; or delta times 1, 0 plus v times 0, 1. 00:36:06.350 --> 00:36:07.760 Isn't that interesting? 00:36:07.760 --> 00:36:12.410 Remember, the Hamiltonian or any operator operating 00:36:12.410 --> 00:36:16.640 on a function gives rise to a linear combination 00:36:16.640 --> 00:36:19.280 of the functions and the basis set. 00:36:19.280 --> 00:36:22.921 And so here's one of the functions, here's the other. 00:36:22.921 --> 00:36:23.420 OK. 00:36:23.420 --> 00:36:27.990 Nothing very mysterious has happened here. 00:36:27.990 --> 00:36:36.990 So this very equation is going to be HC is equal to EC. 00:36:43.180 --> 00:36:47.340 OK, so how do we approach this? 00:37:01.550 --> 00:37:11.220 Well I have to introduce a new symbol, 00:37:11.220 --> 00:37:15.840 and that's going to be this symbol T. It's a matrix. 00:37:18.390 --> 00:37:20.670 It's a unitary matrix. 00:37:20.670 --> 00:37:24.470 And we want it to have a special properties. 00:37:24.470 --> 00:37:27.620 Those special properties will be shown here. 00:37:27.620 --> 00:37:31.230 OK, so first of all, we have this matrix-- 00:37:31.230 --> 00:37:34.860 T11, T-- 00:37:34.860 --> 00:37:37.420 AUDIENCE: Your Hamiltonian matrix is incorrect. 00:37:37.420 --> 00:37:41.320 So I think you mean to say E bar plus-minus? 00:37:41.320 --> 00:37:43.240 ROBERT FIELD: I'm sorry? 00:37:43.240 --> 00:37:44.470 AUDIENCE: I think the diagonal elements of the Hamiltonian 00:37:44.470 --> 00:37:45.430 matrix should be-- 00:37:45.430 --> 00:37:46.388 ROBERT FIELD: Yeah, OK. 00:37:46.388 --> 00:37:54.420 If we wanted the eigenvalues, OK? 00:37:54.420 --> 00:37:55.470 This is-- 00:37:55.470 --> 00:37:59.470 AUDIENCE: So the infinite matrix in this case should be-- 00:37:59.470 --> 00:38:02.560 the diagonal elements should be E1 and E2. 00:38:02.560 --> 00:38:05.120 So I think you meant E bar plus delta, 00:38:05.120 --> 00:38:09.130 and then E bar minus delta. 00:38:09.130 --> 00:38:10.080 ROBERT FIELD: Yeah. 00:38:10.080 --> 00:38:11.100 OK. 00:38:11.100 --> 00:38:13.530 There is something in the notes and something 00:38:13.530 --> 00:38:15.840 in my notes which-- 00:38:15.840 --> 00:38:18.240 we can always write the Hamiltonian 00:38:18.240 --> 00:38:30.780 as E bar 0, 0, E bar plus delta v v minus delta. 00:38:30.780 --> 00:38:35.360 We can always take out this constant term. 00:38:35.360 --> 00:38:39.020 And this is the thing we're always working on. 00:38:39.020 --> 00:38:42.730 And so rather than-- 00:38:42.730 --> 00:38:46.720 and so we could call this H prime. 00:38:46.720 --> 00:38:49.570 Or we can simply say, oh, we have these two things-- 00:38:49.570 --> 00:38:53.260 this always gets added in, and it's not affected. 00:38:53.260 --> 00:38:59.050 If we take this diagonal constant matrix 00:38:59.050 --> 00:39:03.830 and apply a transformation to it, a unitary transformation, 00:39:03.830 --> 00:39:07.000 you get that matrix again, you get-- it's nothing. 00:39:07.000 --> 00:39:23.060 OK, so we wanted to describe some special matrix 00:39:23.060 --> 00:39:27.710 where the transpose of that matrix 00:39:27.710 --> 00:39:30.790 is equal to the inverse of that matrix. 00:39:30.790 --> 00:39:34.060 Now-- yes? 00:39:34.060 --> 00:39:37.110 AUDIENCE: So shouldn't there be a minus on there? 00:39:37.110 --> 00:39:39.660 On the prior equation? 00:39:39.660 --> 00:39:42.230 For the bottom right entry. 00:39:42.230 --> 00:39:44.470 The primary delta and the-- 00:39:44.470 --> 00:39:46.330 first line. 00:39:46.330 --> 00:39:48.036 First line. 00:39:48.036 --> 00:39:49.660 ROBERT FIELD: This is delta minus the-- 00:39:49.660 --> 00:39:50.243 AUDIENCE: Yes. 00:39:50.243 --> 00:39:52.621 But up there, you just had delta delta. 00:39:52.621 --> 00:39:53.371 ROBERT FIELD: Yep. 00:39:56.710 --> 00:39:57.664 Thank you. 00:40:01.020 --> 00:40:01.750 OK. 00:40:01.750 --> 00:40:04.950 Now, when you have a matrix, you really 00:40:04.950 --> 00:40:11.940 like to have T minus 1T is equal to the unit matrix. 00:40:11.940 --> 00:40:16.820 Getting the inverse of a matrix in a general problem 00:40:16.820 --> 00:40:19.700 is really awful. 00:40:19.700 --> 00:40:25.420 But for a unitary matrices, all you do to get the inverse 00:40:25.420 --> 00:40:26.950 is to take-- 00:40:26.950 --> 00:40:28.465 you flip it on the diagonal. 00:40:31.270 --> 00:40:40.980 Now strictly, the conjugate transpose of a unitary matrix 00:40:40.980 --> 00:40:42.660 is the inverse. 00:40:42.660 --> 00:40:45.480 But often we have real matrices. 00:40:45.480 --> 00:40:50.190 But the important thing is always this flipping along-- 00:40:50.190 --> 00:40:52.500 flipping the matrix on the diagonal. 00:40:52.500 --> 00:40:56.180 And that gives you the inverse unless there 00:40:56.180 --> 00:40:58.940 is stuff here that is complex. 00:40:58.940 --> 00:40:59.440 OK. 00:41:02.690 --> 00:41:06.650 So this conjugate transpose, it would look like-- 00:41:18.482 --> 00:41:20.870 OK? 00:41:20.870 --> 00:41:21.370 OK. 00:41:21.370 --> 00:41:25.630 Now, what we want to do is derive the matrix form 00:41:25.630 --> 00:41:28.060 of the Schrodinger equation using 00:41:28.060 --> 00:41:33.920 this unitary transformation. 00:41:33.920 --> 00:41:38.010 So we start out again with HC is equal to EC. 00:41:44.260 --> 00:41:52.190 And now, we insert TT dagger. 00:41:55.010 --> 00:41:59.030 And this is one of the things where you screw up. 00:41:59.030 --> 00:42:02.180 Whether you insert TT dagger or T dagger T, 00:42:02.180 --> 00:42:03.680 because they're both 1. 00:42:03.680 --> 00:42:06.830 And if you use the wrong one, all of your phases, 00:42:06.830 --> 00:42:08.870 everything is wrong. 00:42:08.870 --> 00:42:11.330 But it's still correct, the equations are correct, 00:42:11.330 --> 00:42:15.050 but the things you've memorized are no longer correct. 00:42:15.050 --> 00:42:18.620 OK, so we're going to insert this unit 00:42:18.620 --> 00:42:32.950 matrix between H and C. And of course, we have to-- 00:42:32.950 --> 00:42:37.050 we would have to insert this on the other side. 00:42:37.050 --> 00:42:47.850 T T dagger C and E. But that's one. 00:42:47.850 --> 00:42:50.570 OK, so we don't do anything on the left-hand side. 00:42:53.310 --> 00:42:55.790 And now we left multiply by T dagger. 00:43:05.340 --> 00:43:06.550 OK. 00:43:06.550 --> 00:43:12.740 And we call this H twiddle, and we call this C twiddle. 00:43:15.430 --> 00:43:24.160 So we have H twiddle, C twiddle is equal to EC twiddle. 00:43:24.160 --> 00:43:26.850 Now we're cooking. 00:43:26.850 --> 00:43:32.160 OK, because we construct this unitary matrix 00:43:32.160 --> 00:43:37.350 to cause this Hamiltonian, the transformed Hamiltonian 00:43:37.350 --> 00:43:38.460 to be in diagonal form. 00:43:48.850 --> 00:43:58.450 So we say that H twiddle, which is T dagger T, 00:43:58.450 --> 00:44:06.800 is equal to E plus, E minus, 0, 0 for the 2 by 2 problem. 00:44:11.870 --> 00:44:14.730 OK, that leads to some requirements. 00:44:14.730 --> 00:44:16.450 What are the elements of the T matrix? 00:44:19.620 --> 00:44:30.930 But the important thing is we say that T diagonalizes H. 00:44:30.930 --> 00:44:35.624 So this magic matrix gives you the energy eigenvalues. 00:44:41.060 --> 00:44:49.450 But we also have T dagger C is equal to C twiddle. 00:44:49.450 --> 00:44:54.510 And so this gives you the linear combination 00:44:54.510 --> 00:44:59.752 of the column vectors that correspond to the eigenvector. 00:45:06.840 --> 00:45:08.210 So what is it? 00:45:08.210 --> 00:45:31.520 We have here T dagger, which is TT, T12 T, T21 T, T22 T 00:45:31.520 --> 00:45:46.970 times C. And that's supposed to be equal to C twiddle or C1 00:45:46.970 --> 00:45:50.060 twiddle, C2 twiddle. 00:45:50.060 --> 00:45:56.070 So we put everything together and we discover that OK, 00:45:56.070 --> 00:45:59.430 when we multiply this by that, we get 00:45:59.430 --> 00:46:01.410 the element that goes up here. 00:46:01.410 --> 00:46:20.690 And so the element on top is C1 T11 dagger plus C2 T12-- 00:46:23.720 --> 00:46:27.920 I'm using T's and daggers independently, 00:46:27.920 --> 00:46:29.610 and we do the same thing here. 00:46:29.610 --> 00:46:34.980 So we have a column vector, and it's 00:46:34.980 --> 00:46:39.780 composed of the original state. 00:46:39.780 --> 00:46:45.030 And so anyway, when we do everything, 00:46:45.030 --> 00:46:50.010 we discover that using the solution 00:46:50.010 --> 00:46:54.990 to the two-level problem from the Schrodinger picture, 00:46:54.990 --> 00:46:56.580 we know everything. 00:46:56.580 --> 00:47:08.620 So H twiddle, C twiddle plus is H plus C twiddle plus. 00:47:08.620 --> 00:47:18.330 And that's just E plus 1, 0. 00:47:18.330 --> 00:47:35.180 And H twiddle C twiddle minus is E minus times 0, 1. 00:47:38.355 --> 00:47:38.855 OK. 00:47:42.630 --> 00:47:47.200 This is all very confusing because the notation 00:47:47.200 --> 00:47:49.890 is unfamiliar. 00:47:49.890 --> 00:47:52.770 But the important thing is that everything 00:47:52.770 --> 00:47:55.800 you can do in the Schrodinger picture you 00:47:55.800 --> 00:47:58.350 can do in this matrix picture. 00:47:58.350 --> 00:48:05.700 And you can find these elements of this T matrix, 00:48:05.700 --> 00:48:09.600 and it turns out that what you-- 00:48:12.810 --> 00:48:23.730 the matrix that diagonalizes the Hamiltonian, the columns of T 00:48:23.730 --> 00:48:26.610 dagger are the eigenvectors. 00:48:29.700 --> 00:48:34.520 So you find this matrix, it diagonalizes 00:48:34.520 --> 00:48:39.270 H. The computer tells you the eigenenergies, 00:48:39.270 --> 00:48:42.520 and it also tells you T dagger or T plus-- 00:48:42.520 --> 00:48:43.020 TT. 00:48:46.260 --> 00:48:49.140 And so with that, you know how to write 00:48:49.140 --> 00:48:53.970 in the original basis set what the eigenbasis is 00:48:53.970 --> 00:48:55.280 for each eigenvalue. 00:48:58.340 --> 00:49:06.140 And so the next step, which will happen in the next lecture, 00:49:06.140 --> 00:49:17.450 is the general unitary transformation for two-level. 00:49:17.450 --> 00:49:22.310 Now, you expect that there is going 00:49:22.310 --> 00:49:26.540 to be a general solution for the two-level problem, 00:49:26.540 --> 00:49:28.910 because the two-level problem in the Schrodinger picture 00:49:28.910 --> 00:49:32.360 led to a quadratic equation. 00:49:32.360 --> 00:49:35.300 And that had an analytical solution. 00:49:35.300 --> 00:49:40.910 And so there is going to be a general and exact solution 00:49:40.910 --> 00:49:42.800 the two-level problem. 00:49:42.800 --> 00:49:45.920 And this unitary transformation is 00:49:45.920 --> 00:49:55.730 going to be written in terms of cosine theta sine theta 00:49:55.730 --> 00:50:01.310 minus sine theta cosine theta. 00:50:01.310 --> 00:50:07.910 This is a matrix which is unitary, 00:50:07.910 --> 00:50:12.440 and it, when applied to your basis set, 00:50:12.440 --> 00:50:20.090 conserves normalization and orthogonality. 00:50:20.090 --> 00:50:22.540 And so the trick is to be able to find 00:50:22.540 --> 00:50:27.070 the theta that causes the Hamiltonian in question 00:50:27.070 --> 00:50:28.930 to be diagonalized. 00:50:28.930 --> 00:50:32.790 And the algebra for that will happen in the next lecture. 00:50:32.790 --> 00:50:33.800 OK. 00:50:33.800 --> 00:50:39.270 Remember, you're not going to do this ever. 00:50:39.270 --> 00:50:44.730 You're going to use the idea of this unitary transformation, 00:50:44.730 --> 00:50:48.180 and you're going to use that to get-- 00:50:48.180 --> 00:50:51.130 to diagonalize the matrix. 00:50:51.130 --> 00:50:53.710 And this will lead to some formulas 00:50:53.710 --> 00:50:56.080 which you're going to like, because you 00:50:56.080 --> 00:50:58.630 can forget sines and cosines. 00:50:58.630 --> 00:51:00.130 Everything is going to be expressed 00:51:00.130 --> 00:51:03.160 in terms of things like this-- 00:51:03.160 --> 00:51:07.590 V over delta. 00:51:07.590 --> 00:51:09.840 So we have matrix element-- 00:51:09.840 --> 00:51:16.000 off diagonal matrix element over the energy difference. 00:51:16.000 --> 00:51:20.290 And that's the basic form of nondegenerate perturbation 00:51:20.290 --> 00:51:23.927 theory, which is applied not just in 2-by-2 problems, 00:51:23.927 --> 00:51:24.760 but to all problems. 00:51:29.400 --> 00:51:34.380 And so you want to remember this lecture as, 00:51:34.380 --> 00:51:37.570 this is how we kill the two-level problem. 00:51:37.570 --> 00:51:40.860 Then we can discover what we did and apply it 00:51:40.860 --> 00:51:43.320 to a general problem where it's not 00:51:43.320 --> 00:51:44.760 a two-level problem anymore, it's 00:51:44.760 --> 00:51:47.290 an infinite number of levels. 00:51:47.290 --> 00:51:51.560 And when certain approximations are met, 00:51:51.560 --> 00:51:57.860 it applies, and it gives you the most accurate energy levels 00:51:57.860 --> 00:52:00.530 and wave functions you could want. 00:52:00.530 --> 00:52:03.710 And you know how accurate they are going to be. 00:52:03.710 --> 00:52:08.870 And this is liberating, because now, you 00:52:08.870 --> 00:52:12.510 can take a not exactly-solved problem 00:52:12.510 --> 00:52:15.680 and you can solve it approximately. 00:52:15.680 --> 00:52:19.410 And you can use the solution to determine, OK, 00:52:19.410 --> 00:52:22.980 there's going to be some function of the quantum numbers 00:52:22.980 --> 00:52:24.850 which describes the energy levels. 00:52:24.850 --> 00:52:26.870 Well what is that function? 00:52:26.870 --> 00:52:31.050 And what are the coefficients in that function? 00:52:31.050 --> 00:52:32.820 And how do they relate to the things 00:52:32.820 --> 00:52:36.280 you know from the Hamiltonian? 00:52:36.280 --> 00:52:40.240 So it's incredibly powerful. 00:52:40.240 --> 00:52:44.590 And once you're given the formulas 00:52:44.590 --> 00:52:46.990 for nondegenerate perturbation theory, 00:52:46.990 --> 00:52:50.780 you can solve practically any problem in quantum mechanics. 00:52:50.780 --> 00:52:55.360 Not just numerically, but with insight. 00:52:55.360 --> 00:53:04.440 It tells you, if we make observations of some system, 00:53:04.440 --> 00:53:06.720 we determine a set of energy levels, which is called 00:53:06.720 --> 00:53:09.760 the spectrum of that operator. 00:53:09.760 --> 00:53:15.490 And the spectrum of the operator is an explicit function 00:53:15.490 --> 00:53:19.630 of the physical constants-- the unique interactions 00:53:19.630 --> 00:53:21.830 between states. 00:53:21.830 --> 00:53:30.200 And so you will then know how the experimental data 00:53:30.200 --> 00:53:34.002 determines the mechanism of all the interactions, 00:53:34.002 --> 00:53:39.790 and you can calculate the wave function. 00:53:39.790 --> 00:53:41.650 You can't observe the wave function, 00:53:41.650 --> 00:53:46.900 but you can discover its traces in the energy levels. 00:53:46.900 --> 00:53:49.790 And then you can reproduce the energy levels, 00:53:49.790 --> 00:53:53.050 and if you have the energy levels-- the eigenstates, 00:53:53.050 --> 00:53:58.390 you can also describe any dynamics using 00:53:58.390 --> 00:54:00.530 the same formulas in here. 00:54:00.530 --> 00:54:07.640 So this is an incredible enablement. 00:54:07.640 --> 00:54:11.310 And you don't have to look at my derivations. 00:54:11.310 --> 00:54:12.750 I'm not proud of the derivations, 00:54:12.750 --> 00:54:14.650 I'm proud of the results. 00:54:14.650 --> 00:54:20.070 And if you can handle these results that 00:54:20.070 --> 00:54:22.416 come from perturbation theory as well as you've 00:54:22.416 --> 00:54:23.790 done so far in the course, you're 00:54:23.790 --> 00:54:30.840 going to be at the research level very soon. 00:54:30.840 --> 00:54:33.680 OK, see you on Friday.