1 00:00:00,135 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,690 continue to offer high quality educational resources for free. 5 00:00:10,690 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,250 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,250 --> 00:00:18,220 at ocw.mit.edu. 8 00:00:22,960 --> 00:00:23,780 ROBERT FIELD: OK. 9 00:00:23,780 --> 00:00:30,540 So, today is the first of a pair of lectures taking us 10 00:00:30,540 --> 00:00:33,690 from the Schrodinger picture to the Heisenberg picture. 11 00:00:33,690 --> 00:00:37,580 Or the wave function picture to the matrix picture. 12 00:00:37,580 --> 00:00:41,280 Now almost everyone who does quantum mechanics 13 00:00:41,280 --> 00:00:45,240 is rooted in the Heisenberg picture. 14 00:00:45,240 --> 00:00:49,980 We use the Heisenberg picture because the structure 15 00:00:49,980 --> 00:00:54,930 of the problem is immediately evident when you write down 16 00:00:54,930 --> 00:00:57,140 what you know. 17 00:00:57,140 --> 00:01:03,680 And it's also mostly the way you program your computers 18 00:01:03,680 --> 00:01:06,630 to solve any problem. 19 00:01:06,630 --> 00:01:14,240 So I'm going to get you to the Heisenberg picture by dealing 20 00:01:14,240 --> 00:01:17,150 with the two-level problem, which is-- 21 00:01:17,150 --> 00:01:21,240 it should be called one of the exactly solved problems, 22 00:01:21,240 --> 00:01:25,900 except it's abstract as opposed to harmonic oscillator, 23 00:01:25,900 --> 00:01:29,270 particle in an infinite box, rigid rotor. 24 00:01:29,270 --> 00:01:33,590 It's a exactly-solved problem, and it leads us to-- 25 00:01:33,590 --> 00:01:35,660 or guides us to-- 26 00:01:35,660 --> 00:01:36,500 the new approach. 27 00:01:39,480 --> 00:01:42,630 I'm going to approach this problem 28 00:01:42,630 --> 00:01:45,440 the algebraic or Schrodinger way. 29 00:01:45,440 --> 00:01:50,340 And then I'm going to describe it in a matrix way 30 00:01:50,340 --> 00:01:52,995 and introduce the new language and the new notation. 31 00:01:57,750 --> 00:02:00,750 So I can say, at least at the end of this lecture, 32 00:02:00,750 --> 00:02:03,770 I'll be able to say the word matrix element 33 00:02:03,770 --> 00:02:05,780 and everybody will know what it is, 34 00:02:05,780 --> 00:02:09,389 and I'll stop talking about integrals. 35 00:02:09,389 --> 00:02:13,470 And one of the scary things is, when 36 00:02:13,470 --> 00:02:17,630 we go to the matrix picture, we stop 37 00:02:17,630 --> 00:02:19,520 looking at the wave function. 38 00:02:19,520 --> 00:02:22,250 We never think about the wave functions. 39 00:02:22,250 --> 00:02:26,380 And so there are all sorts of things like phases 40 00:02:26,380 --> 00:02:31,630 that we have to make sure we're not screwing up. 41 00:02:31,630 --> 00:02:36,250 Because we're playing fast and loose with symbols, 42 00:02:36,250 --> 00:02:39,340 and sometimes they can bite you if you don't know what 43 00:02:39,340 --> 00:02:40,670 you're dealing with. 44 00:02:40,670 --> 00:02:44,740 But mostly, we're going to get rid of wave functions, 45 00:02:44,740 --> 00:02:50,470 because we know all the solutions to standard problems, 46 00:02:50,470 --> 00:02:54,010 and we know a lot of integrals involving those solutions 47 00:02:54,010 --> 00:02:56,020 to standard problems. 48 00:02:56,020 --> 00:02:58,510 And so basically, all we need is an index 49 00:02:58,510 --> 00:03:02,760 saying this wave function or this state, this integral, 50 00:03:02,760 --> 00:03:03,980 and you have-- 51 00:03:03,980 --> 00:03:07,200 and then you can write down everything 52 00:03:07,200 --> 00:03:11,460 you need in matrix notation, and then you 53 00:03:11,460 --> 00:03:14,910 can tell your friendly computer, OK, solve the problem for me. 54 00:03:18,420 --> 00:03:18,920 OK. 55 00:03:18,920 --> 00:03:24,770 So once I give you the matrix picture, 56 00:03:24,770 --> 00:03:29,930 then I will talk about how you find the eigenvalues 57 00:03:29,930 --> 00:03:33,890 and eigenvectors of the matrix picture using 58 00:03:33,890 --> 00:03:36,020 a unitary transformation. 59 00:03:36,020 --> 00:03:38,930 And then I'll generalize from the two-level problem, 60 00:03:38,930 --> 00:03:41,420 which is exactly soluble. 61 00:03:41,420 --> 00:03:44,202 You don't need a computer for it, but it's convenient-- 62 00:03:44,202 --> 00:03:46,160 because you don't have to write anything down-- 63 00:03:46,160 --> 00:03:51,420 to N levels where N can be infinite. 64 00:03:51,420 --> 00:03:58,980 And the N-level problem is in principle difficult 65 00:03:58,980 --> 00:04:04,140 because even a computer can't diagonalize an infinite matrix. 66 00:04:04,140 --> 00:04:07,920 And all of your basis sets, all of your standard problems 67 00:04:07,920 --> 00:04:10,950 involve an infinite number of functions. 68 00:04:10,950 --> 00:04:15,120 And so I've introduced a notation and a concept 69 00:04:15,120 --> 00:04:21,180 of solving a matrix equation, but you can't do it unless you, 70 00:04:21,180 --> 00:04:23,190 say, let's make an approximation, 71 00:04:23,190 --> 00:04:26,160 and it's called nondegenerate perturbation theory. 72 00:04:26,160 --> 00:04:28,290 And this is the tool for learning 73 00:04:28,290 --> 00:04:30,270 almost everything you want to know 74 00:04:30,270 --> 00:04:33,080 about complicated problems. 75 00:04:33,080 --> 00:04:37,120 And it's not complicated to apply 76 00:04:37,120 --> 00:04:38,870 nondegenerate perturbation theory. 77 00:04:38,870 --> 00:04:41,870 It's just ugly. 78 00:04:41,870 --> 00:04:43,700 But it's really valuable because you 79 00:04:43,700 --> 00:04:46,040 don't have to remember how to solve 80 00:04:46,040 --> 00:04:48,710 a particular complicated differential equation, 81 00:04:48,710 --> 00:04:50,840 you just write down what you know 82 00:04:50,840 --> 00:04:53,660 and you do some simple stuff, and bang, you've 83 00:04:53,660 --> 00:04:55,460 got a solution to the problem. 84 00:04:55,460 --> 00:04:57,560 And this is really what I want you to come away 85 00:04:57,560 --> 00:04:58,850 from this course with-- 86 00:04:58,850 --> 00:05:03,830 the concept that anything that requires quantum mechanics 87 00:05:03,830 --> 00:05:09,390 you can solve using some form of perturbation theory. 88 00:05:09,390 --> 00:05:12,125 So hold onto your seats. 89 00:05:25,740 --> 00:05:29,420 So the Schrodinger picture is differential equations. 90 00:05:38,360 --> 00:05:41,270 And they're often coupled differential equations. 91 00:05:41,270 --> 00:05:44,000 And you don't want to go there, usually. 92 00:05:44,000 --> 00:05:46,960 And we're going to replace that with linear algebra. 93 00:05:52,490 --> 00:05:55,880 Now many of you have not had a course in linear algebra. 94 00:05:55,880 --> 00:06:00,834 And so that should make you say, can I do this? 95 00:06:00,834 --> 00:06:02,500 And the answer is, yeah, you can do this 96 00:06:02,500 --> 00:06:04,990 because the linear algebra you are going 97 00:06:04,990 --> 00:06:09,010 to need in this course is based on one concept, 98 00:06:09,010 --> 00:06:14,200 and that is that the solution of coupled linear homogeneous 99 00:06:14,200 --> 00:06:19,495 equations involves diagonalizing or solving a determinant. 100 00:06:22,090 --> 00:06:26,210 And when you solve this determinant of the equation, 101 00:06:26,210 --> 00:06:30,090 it's equivalent to diagonalizing a matrix. 102 00:06:30,090 --> 00:06:33,890 And that's the language we're going to be using all the time. 103 00:06:33,890 --> 00:06:35,660 And so the only thing I'm not going to do 104 00:06:35,660 --> 00:06:38,520 is prove this fundamental theorem of linear algebra, 105 00:06:38,520 --> 00:06:44,480 but the notation is easy to use and the tricks are very simple. 106 00:06:47,800 --> 00:06:52,290 So, we have with exactly-solved problems 107 00:06:52,290 --> 00:06:58,030 complete sets of the energy levels of the wave functions. 108 00:06:58,030 --> 00:07:00,420 And we know a lot of integrals. 109 00:07:00,420 --> 00:07:06,630 Psi i operator psi j d tau. 110 00:07:06,630 --> 00:07:10,150 We know these not by evaluating them, 111 00:07:10,150 --> 00:07:14,130 but because the functions are so simple that we can write these 112 00:07:14,130 --> 00:07:17,470 as simply a function of the initial and final quantum 113 00:07:17,470 --> 00:07:17,970 numbers. 114 00:07:21,170 --> 00:07:24,660 And so all of a sudden, we forget 115 00:07:24,660 --> 00:07:26,580 about the wave functions, because all the work 116 00:07:26,580 --> 00:07:29,390 has been done for us. 117 00:07:29,390 --> 00:07:30,850 We could do it too, but why? 118 00:07:33,600 --> 00:07:36,860 So we start with the two-level problem. 119 00:07:36,860 --> 00:07:39,140 And the two-level problem says we have 120 00:07:39,140 --> 00:07:44,270 two states, psi 1 and psi 2. 121 00:07:44,270 --> 00:07:47,020 Still in the Schrodinger picture. 122 00:07:47,020 --> 00:07:54,270 And this has an energy which we could we could call E1, 123 00:07:54,270 --> 00:08:03,550 and that would be H11, the diagonal integral 124 00:08:03,550 --> 00:08:08,750 of the Hamiltonian between the 1 function and the 1 function, 125 00:08:08,750 --> 00:08:09,520 and this is E2. 126 00:08:13,660 --> 00:08:18,430 Now these two states have an interaction. 127 00:08:18,430 --> 00:08:21,650 They're connected by an interaction term 128 00:08:21,650 --> 00:08:25,570 which causes them to repel each other equal and opposite 129 00:08:25,570 --> 00:08:26,830 amounts. 130 00:08:26,830 --> 00:08:31,880 And so this is H12. 131 00:08:31,880 --> 00:08:36,289 Again, an integral, which you usually don't have to evaluate, 132 00:08:36,289 --> 00:08:39,890 because it's basically done for you. 133 00:08:39,890 --> 00:08:44,760 And then we get E plus and E minus, 134 00:08:44,760 --> 00:08:47,640 and the corresponding eigenfunction. 135 00:08:50,680 --> 00:08:53,200 So this is the two-level problem, 136 00:08:53,200 --> 00:08:56,200 and it's expressed in terms of three parameters-- 137 00:08:56,200 --> 00:09:02,670 H11, H22, and H12. 138 00:09:02,670 --> 00:09:08,540 And we get the two energy levels and the two eigenfunctions. 139 00:09:08,540 --> 00:09:11,290 And we know how to solve this problem in the Schrodinger 140 00:09:11,290 --> 00:09:15,700 picture, and I'll do that first. 141 00:09:15,700 --> 00:09:24,380 And so, H11 is just this integral, psi 1 star H. 142 00:09:24,380 --> 00:09:27,410 And we put a hat on H for the time being-- 143 00:09:27,410 --> 00:09:30,380 1 d tau. 144 00:09:30,380 --> 00:09:41,950 And H22 is a different integral of psi 2 star H psi 2 d tau. 145 00:09:41,950 --> 00:09:50,345 And H12 is psi 1 star H psi 2 tau. 146 00:09:52,990 --> 00:10:00,670 And we know that the Hamiltonian operator is Hermitian, 147 00:10:00,670 --> 00:10:05,440 and so we can write also that this is 148 00:10:05,440 --> 00:10:14,500 psi 2 star H star psi I d tau. 149 00:10:14,500 --> 00:10:19,480 Anyway, we call this thing V. And there's some subtleties. 150 00:10:19,480 --> 00:10:25,550 If the integral between these two things is imaginary, 151 00:10:25,550 --> 00:10:34,190 then the 1-2 and 2-1 integrals have opposite signs. 152 00:10:34,190 --> 00:10:36,710 Because the Hamiltonian is Hermitian. 153 00:10:36,710 --> 00:10:39,380 But let's just think of this as just one number. 154 00:10:39,380 --> 00:10:44,960 So we have E1, E2, and delta. 155 00:10:49,620 --> 00:10:54,330 So the two-level problem is going to let us find 156 00:10:54,330 --> 00:10:56,250 the eigenfunctions-- 157 00:10:56,250 --> 00:10:58,290 the plus and minus eigenfunctions. 158 00:10:58,290 --> 00:10:59,290 C plus 1. 159 00:11:06,330 --> 00:11:11,040 So we have eigenfunctions corresponding to-- which 160 00:11:11,040 --> 00:11:15,960 are eigenfunctions belonging to the eigenvalues, E plus and E 161 00:11:15,960 --> 00:11:17,160 minus. 162 00:11:17,160 --> 00:11:20,690 And there are a linear combination of the two states. 163 00:11:20,690 --> 00:11:23,040 Now, it's a two-level problem. 164 00:11:23,040 --> 00:11:27,730 This is a state space that contains only two states. 165 00:11:27,730 --> 00:11:29,160 It's not an approximation. 166 00:11:31,770 --> 00:11:35,100 And so completeness says we can write any function 167 00:11:35,100 --> 00:11:38,520 we want as a linear combination of the functions and the basis 168 00:11:38,520 --> 00:11:40,440 set. 169 00:11:40,440 --> 00:11:45,180 And so our job is going to be to find these coefficients 170 00:11:45,180 --> 00:11:48,441 and also to find the energy eigenvalues. 171 00:11:55,990 --> 00:11:56,500 OK. 172 00:11:56,500 --> 00:11:59,620 So let's start to do some work. 173 00:11:59,620 --> 00:12:01,750 We know this thing. 174 00:12:01,750 --> 00:12:03,550 And let's start some-- 175 00:12:03,550 --> 00:12:08,740 and we know that H psi plus-minus 176 00:12:08,740 --> 00:12:15,010 has to be equal to the eigenenergies psi plus-minus. 177 00:12:15,010 --> 00:12:19,840 Now, in order to get something useful out of this, 178 00:12:19,840 --> 00:12:22,830 we left multiply by psi 1. 179 00:12:27,040 --> 00:12:27,650 OK. 180 00:12:27,650 --> 00:12:30,290 And when you left multiply by psi 1-- well, 181 00:12:30,290 --> 00:12:31,497 let's just write it out. 182 00:12:35,240 --> 00:12:42,590 We're going to get psi 1 HC 1 plus. 183 00:12:42,590 --> 00:12:45,070 So let's just write this out. 184 00:12:45,070 --> 00:12:53,610 We have H11 C1 plus-minus. 185 00:12:53,610 --> 00:13:00,120 Because this is C plus-minus, and we 186 00:13:00,120 --> 00:13:06,360 have the integral of psi 1 with itself or the Hamiltonian 187 00:13:06,360 --> 00:13:07,690 is H11. 188 00:13:07,690 --> 00:13:14,200 And then we get another term. 189 00:13:14,200 --> 00:13:16,080 I'm doing this differently from my notes, 190 00:13:16,080 --> 00:13:28,360 so the other term is C plus-minus psi 2 V. OK. 191 00:13:28,360 --> 00:13:36,420 So plugging in what I have over here and knowing that H11-- 192 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. 193 00:13:45,060 --> 00:13:50,530 We do the same thing, and we have the same left-hand side, 194 00:13:50,530 --> 00:13:59,790 but we can also now express this as that H-- 195 00:14:07,220 --> 00:14:10,172 that psi plus-minus is an eigenfunction. 196 00:14:13,130 --> 00:14:15,710 And so we have the same left-hand side, 197 00:14:15,710 --> 00:14:17,390 but on the right-hand side, we now 198 00:14:17,390 --> 00:14:28,060 get integral psi 1 star H psi plus-minus d tau. 199 00:14:28,060 --> 00:14:41,540 But that's equal to psi star E plus-minus C1 plus-minus psi 1 200 00:14:41,540 --> 00:14:48,640 plus C2 plus-minus psi 2 d tau. 201 00:14:48,640 --> 00:14:49,190 OK. 202 00:14:49,190 --> 00:14:52,700 And so now we're talking about orthonormality of the wave 203 00:14:52,700 --> 00:14:55,880 function, and because we don't have any operators in here, 204 00:14:55,880 --> 00:15:04,980 then this thing becomes simply E plus-minus times C1 plus-minus 205 00:15:04,980 --> 00:15:10,550 plus 0 C2 plus-minus. 206 00:15:14,210 --> 00:15:16,330 There should be a 1 here. 207 00:15:16,330 --> 00:15:16,830 OK. 208 00:15:16,830 --> 00:15:21,040 Because psi 1, psi 2, that's orthogonal. 209 00:15:21,040 --> 00:15:23,320 So now we have an equation that's quite useful. 210 00:15:29,470 --> 00:15:31,980 So we're going to combine these two equations, 211 00:15:31,980 --> 00:15:41,122 and we get C1 plus H11 plus C2 plus-- 212 00:15:41,122 --> 00:15:44,220 I'm sorry, this is plus-minus, plus-minus. 213 00:15:44,220 --> 00:15:51,080 V is equal to E plus-minus C1 plus-minus. 214 00:15:54,870 --> 00:15:57,010 Now we need another equation. 215 00:15:57,010 --> 00:16:00,980 And so we do the same thing and we left multiply by psi 2 star 216 00:16:00,980 --> 00:16:03,170 and integrate. 217 00:16:03,170 --> 00:16:05,870 And we get another equation, and that 218 00:16:05,870 --> 00:16:19,770 is C1 plus-minus the plus C2 plus-minus H22 minus E 219 00:16:19,770 --> 00:16:25,630 plus-minus is equal to 0. 220 00:16:25,630 --> 00:16:33,020 So we combine the two equations that we have derived, solving-- 221 00:16:33,020 --> 00:16:39,920 because both equations can be solved for C1 plus-minus or C2 222 00:16:39,920 --> 00:16:41,210 plus-minus. 223 00:16:41,210 --> 00:16:44,810 So we equate the C1 plus-minus over C2 plus-minus 224 00:16:44,810 --> 00:16:46,430 from the two equations. 225 00:16:46,430 --> 00:16:51,710 And we get this wonderful result, V over H11 minus E 226 00:16:51,710 --> 00:17:04,450 plus-minus is equal to H22 minus E plus-minus over V. 227 00:17:04,450 --> 00:17:07,050 Now you're not going to ever do this. 228 00:17:07,050 --> 00:17:13,800 So yes, you can attempt to reconstruct this 229 00:17:13,800 --> 00:17:16,890 from what I read on the board or what's in your notes, 230 00:17:16,890 --> 00:17:21,359 but the important point is that we're just using what we know. 231 00:17:21,359 --> 00:17:24,450 We say we have eigenfunctions, and we 232 00:17:24,450 --> 00:17:29,790 want to find something about the coefficients of the basis 233 00:17:29,790 --> 00:17:31,900 functions in each of the eigenfunctions. 234 00:17:31,900 --> 00:17:33,850 And we want to find the eigenvalues, 235 00:17:33,850 --> 00:17:36,890 and we get this equation. 236 00:17:36,890 --> 00:17:41,560 So, this is easy. 237 00:17:41,560 --> 00:17:44,230 We relate this to-- 238 00:17:44,230 --> 00:17:47,860 we just multiply through and we get V2, 239 00:17:47,860 --> 00:17:57,989 V squared is equal to H11 minus E plus-minus times H22 240 00:17:57,989 --> 00:17:58,780 minus E plus-minus. 241 00:17:58,780 --> 00:18:07,100 Or we solve, and we know that E plus-minus is equal-- this 242 00:18:07,100 --> 00:18:08,900 is a quadratic equation. 243 00:18:08,900 --> 00:18:14,620 A quadratic and E plus-minus, and so we have this result, 244 00:18:14,620 --> 00:18:26,380 H11 plus H22 plus or minus this H11 squared-- 245 00:18:26,380 --> 00:18:48,680 H11 plus H22 squared minus 4 H11 H22 minus V squared over 2. 246 00:18:48,680 --> 00:18:51,660 This is just a quadratic equation. 247 00:18:51,660 --> 00:18:56,450 So we have the eigenenergies expressed in terms 248 00:18:56,450 --> 00:18:57,680 of the quantities we know. 249 00:19:08,230 --> 00:19:10,210 We simplify the notation. 250 00:19:10,210 --> 00:19:19,410 E bar is H11 plus H22 over 2. 251 00:19:19,410 --> 00:19:26,630 And delta is H11 minus H22 over 2. 252 00:19:29,690 --> 00:19:33,600 So when we do that, we simplify the algebra 253 00:19:33,600 --> 00:19:37,590 and we end up discovering that the eigenvalue equation is 254 00:19:37,590 --> 00:19:46,350 E plus-minus is equal to E bar plus or minus delta squared 255 00:19:46,350 --> 00:19:50,130 plus V squared square root. 256 00:19:50,130 --> 00:19:54,580 That's a simple result. This is something you should remember. 257 00:19:57,810 --> 00:19:59,780 So if you have a two-level problem, 258 00:19:59,780 --> 00:20:03,547 the energy eigenvalues are the average energy plus or minus 259 00:20:03,547 --> 00:20:04,130 this quantity. 260 00:20:08,390 --> 00:20:11,130 This is an exact solution for a two-level problem. 261 00:20:11,130 --> 00:20:12,944 For all two-level problems. 262 00:20:16,430 --> 00:20:16,970 OK. 263 00:20:16,970 --> 00:20:19,280 And we even simplify the notation more. 264 00:20:19,280 --> 00:20:25,080 We call X delta squared plus V squared. 265 00:20:25,080 --> 00:20:35,560 And so this becomes E bar plus-minus X squared. 266 00:20:35,560 --> 00:20:37,600 Pretty compact. 267 00:20:37,600 --> 00:20:40,600 Now, the reason for simplifying the notation 268 00:20:40,600 --> 00:20:43,090 is that we're going to derive the values 269 00:20:43,090 --> 00:20:47,820 for the eigenfunctions, and they involve a lot of symbols. 270 00:20:47,820 --> 00:20:51,020 And we want to make this as compact as possible, 271 00:20:51,020 --> 00:20:54,630 and so we'll do that. 272 00:20:54,630 --> 00:20:55,130 OK. 273 00:20:55,130 --> 00:20:57,020 So when we started out, it looked 274 00:20:57,020 --> 00:21:02,060 like we were going to solve for these quantities, 275 00:21:02,060 --> 00:21:06,830 but we took a detour and solved for the eigenenergies first. 276 00:21:06,830 --> 00:21:10,100 And this is one of the things that happens in linear algebra. 277 00:21:10,100 --> 00:21:12,830 You get something you can get easily-- 278 00:21:12,830 --> 00:21:15,430 quickly before you get the other stuff that you want. 279 00:21:18,510 --> 00:21:23,410 OK, but the second part of the job 280 00:21:23,410 --> 00:21:26,660 is to find these mixing coefficients. 281 00:21:26,660 --> 00:21:31,240 And so one thing you do is you say, 282 00:21:31,240 --> 00:21:45,330 well, we insist that the wave functions be orthog-- 283 00:21:45,330 --> 00:21:47,980 normalized. 284 00:21:47,980 --> 00:21:51,820 After that a lot of algebra ensues. 285 00:21:51,820 --> 00:21:58,780 And I'm not going to even attempt to work through it. 286 00:21:58,780 --> 00:22:02,020 You may want to work through it, but what I recommend doing 287 00:22:02,020 --> 00:22:06,400 is looking at the solution and then 288 00:22:06,400 --> 00:22:08,530 checking to see whether it does things 289 00:22:08,530 --> 00:22:11,270 that you expect it has to do. 290 00:22:11,270 --> 00:22:15,250 Because one of the things that kind of inspection 291 00:22:15,250 --> 00:22:22,520 leads you to is factors of two errors and sign errors. 292 00:22:22,520 --> 00:22:23,030 OK. 293 00:22:23,030 --> 00:22:26,840 But, C plus-minus-- C1 plus-minus 294 00:22:26,840 --> 00:22:33,960 is equal to 1/2 times 1 plus or minus 295 00:22:33,960 --> 00:22:42,880 delta over the square root of X. Two square roots. 296 00:22:42,880 --> 00:22:43,380 Whoops. 297 00:22:49,970 --> 00:22:51,970 OK, no square root in here because we 298 00:22:51,970 --> 00:22:53,850 get square root there. 299 00:22:53,850 --> 00:23:02,440 OK and C2 plus-minus is 1/2 1 minus or plus 300 00:23:02,440 --> 00:23:07,720 delta square root, x square root. 301 00:23:07,720 --> 00:23:09,640 OK? 302 00:23:09,640 --> 00:23:13,810 So these two things are kind of simple-looking, 303 00:23:13,810 --> 00:23:16,570 but there's an awful lot of compression 304 00:23:16,570 --> 00:23:19,900 and select and clever manipulation 305 00:23:19,900 --> 00:23:21,250 in order to get this. 306 00:23:21,250 --> 00:23:23,320 But the important thing to notice 307 00:23:23,320 --> 00:23:27,850 is that we have the energy difference divided 308 00:23:27,850 --> 00:23:33,640 by this thing, this delta squared plus V 309 00:23:33,640 --> 00:23:39,130 squared that expresses the importance 310 00:23:39,130 --> 00:23:40,810 of the two-level interaction. 311 00:23:40,810 --> 00:23:43,855 And these two things enter with opposite signs. 312 00:23:50,130 --> 00:23:50,770 OK. 313 00:23:50,770 --> 00:23:53,350 And so one of the checks that's really easy to do 314 00:23:53,350 --> 00:24:00,770 is let's let V go to 0, and let's let V go to infinity. 315 00:24:00,770 --> 00:24:03,940 OK, remember, our Hamiltonian-- 316 00:24:03,940 --> 00:24:04,900 the two-level problem-- 317 00:24:08,290 --> 00:24:10,130 I can't write it as a matrix yet, 318 00:24:10,130 --> 00:24:18,670 so remember that we had E1, E2, and V. 319 00:24:18,670 --> 00:24:22,210 And this was the higher energy, this was the lower energy, 320 00:24:22,210 --> 00:24:28,760 and so delta is E1 minus E2-- 321 00:24:28,760 --> 00:24:30,190 I better not write that. 322 00:24:34,040 --> 00:24:35,980 OK. 323 00:24:35,980 --> 00:24:46,540 So we normally write E1 over E2, and we have this V interaction 324 00:24:46,540 --> 00:24:48,220 between them. 325 00:24:48,220 --> 00:24:49,120 OK. 326 00:24:49,120 --> 00:24:58,120 And so if the interaction integral between the two 327 00:24:58,120 --> 00:25:06,100 functions is 0, then E1 is the higher energy, 328 00:25:06,100 --> 00:25:08,500 and it corresponds to-- 329 00:25:08,500 --> 00:25:12,550 it should correspond to psi 1 alone. 330 00:25:12,550 --> 00:25:20,560 Well, so the higher energy we're taking the plus combinations 331 00:25:20,560 --> 00:25:28,640 here, and if V is equal to 0, then this is delta over delta. 332 00:25:28,640 --> 00:25:32,110 And this is either 2 or 0. 333 00:25:32,110 --> 00:25:33,580 And when it's the higher-- 334 00:25:33,580 --> 00:25:37,870 the upper sine, it's 2 divided by 2 or 1, 335 00:25:37,870 --> 00:25:40,030 exactly what you expect. 336 00:25:40,030 --> 00:25:44,200 And this one is 1 minus 1. 337 00:25:44,200 --> 00:25:46,930 So these two are behaving, right? 338 00:25:46,930 --> 00:25:50,380 And the V to 0 rate. 339 00:25:50,380 --> 00:25:57,720 Now, in the V going to infinity, then this is infinite. 340 00:25:57,720 --> 00:26:16,120 So that just means that this term goes away, right? 341 00:26:16,120 --> 00:26:24,230 And so we have C1 plus and minus is equal to square root of 2-- 342 00:26:24,230 --> 00:26:27,260 1 over the square root of 2. 343 00:26:27,260 --> 00:26:29,710 And same here. 344 00:26:29,710 --> 00:26:31,920 So what we get is what we call 50/50 mixing. 345 00:26:37,410 --> 00:26:39,740 OK, so this also works. 346 00:26:42,250 --> 00:26:45,640 So I don't say that this can-- 347 00:26:45,640 --> 00:26:49,720 this confirms that I have not made an algebraic mistake, 348 00:26:49,720 --> 00:26:56,120 I haven't, but it's a good test because it's really easy to do. 349 00:26:56,120 --> 00:26:58,220 And if you're not getting what you expect, 350 00:26:58,220 --> 00:27:01,100 you know you've either blown a sign or a factor 351 00:27:01,100 --> 00:27:07,260 or two, which are the two things that you fear most in a wave 352 00:27:07,260 --> 00:27:10,630 function free approach. 353 00:27:10,630 --> 00:27:14,290 Because you've got nothing to do except check your algebra, 354 00:27:14,290 --> 00:27:18,320 and usually, you made the mistake because it was subtle, 355 00:27:18,320 --> 00:27:20,710 and you'll make it again when you're checking it. 356 00:27:20,710 --> 00:27:23,050 So these kinds of checks are really valuable. 357 00:27:28,365 --> 00:27:28,865 OK. 358 00:27:33,310 --> 00:27:37,750 So once you know that this is likely to be correct, 359 00:27:37,750 --> 00:27:39,850 then you do some other things, and you 360 00:27:39,850 --> 00:27:45,760 check the wave functions you've derived with your C plus-minus 361 00:27:45,760 --> 00:27:55,460 1 and C plus-minus two are both normalized and orthogonal. 362 00:28:01,340 --> 00:28:10,160 And that the energy you get is such that psi plus is E plus. 363 00:28:10,160 --> 00:28:12,560 And you know what E plus was. 364 00:28:12,560 --> 00:28:16,550 And so you just go in and you calculate 365 00:28:16,550 --> 00:28:20,930 what the energy should be giving using the values for PSI 366 00:28:20,930 --> 00:28:28,370 plus-minus And so we plug those into the original equations. 367 00:28:28,370 --> 00:28:34,160 And you can also, again, show that psi plus-minus 368 00:28:34,160 --> 00:28:50,590 a star H psi plus-minus is equal to 0 369 00:28:50,590 --> 00:28:56,240 because the eigenfunctions are orthogonal. 370 00:28:56,240 --> 00:28:59,450 We already did that, but we plugged it into an equation 371 00:28:59,450 --> 00:29:01,190 here, and we got it a second time. 372 00:29:07,240 --> 00:29:14,770 So in the lecture notes, there is a lengthy algebraic proof 373 00:29:14,770 --> 00:29:17,380 or demonstration that E plus-minus 374 00:29:17,380 --> 00:29:20,540 is equal to plus or minus square of x, 375 00:29:20,540 --> 00:29:22,570 which are already derived, but then I 376 00:29:22,570 --> 00:29:23,830 just did it the long way. 377 00:29:23,830 --> 00:29:24,730 OK . 378 00:29:24,730 --> 00:29:26,710 So this is it-- 379 00:29:26,710 --> 00:29:31,090 we are about to move from the Schrodinger picture 380 00:29:31,090 --> 00:29:34,700 to the matrix picture. 381 00:29:34,700 --> 00:29:46,970 So the trick now is to go to linear algebra, 382 00:29:46,970 --> 00:29:49,520 go to the matrix picture and learn 383 00:29:49,520 --> 00:29:53,600 how to just write the equations and what the language is 384 00:29:53,600 --> 00:29:54,575 and show that it works. 385 00:29:57,460 --> 00:30:04,480 So suppose you have two square matrices, A and B, 386 00:30:04,480 --> 00:30:09,490 they're n by n squared matrices, you know the rules 387 00:30:09,490 --> 00:30:11,307 for matrix multiplication? 388 00:30:15,290 --> 00:30:23,000 So if you wanted, the mn element of this product or two 389 00:30:23,000 --> 00:30:28,550 matrices, you would go, n equals-- 390 00:30:28,550 --> 00:30:29,640 i equals 1. 391 00:30:32,320 --> 00:30:38,850 Let's use the same notation. j equals 1 to n of Am-- 392 00:30:38,850 --> 00:30:40,544 because I want the mn-- 393 00:30:40,544 --> 00:30:42,270 j Bjn. 394 00:30:47,220 --> 00:30:50,040 And so when you multiply two square matrices, 395 00:30:50,040 --> 00:30:51,600 you get back a square matrix. 396 00:30:51,600 --> 00:30:53,950 And this picture is not a bad one. 397 00:30:53,950 --> 00:30:57,020 So you can say, all right. 398 00:31:03,410 --> 00:31:08,510 So if you need a little q to remind yourself, 399 00:31:08,510 --> 00:31:14,390 you take this row and multiply term by term 400 00:31:14,390 --> 00:31:16,940 and add the results-- 401 00:31:16,940 --> 00:31:20,630 this row in that column, and you get a number here. 402 00:31:20,630 --> 00:31:24,260 And you just repeat that, and it's really easy 403 00:31:24,260 --> 00:31:25,820 to tell your computer to do this, 404 00:31:25,820 --> 00:31:28,130 and it's rather tedious to do it yourself 405 00:31:28,130 --> 00:31:30,080 if it's more than a two-by-two matrix. 406 00:31:33,060 --> 00:31:35,420 OK. 407 00:31:35,420 --> 00:31:40,290 So what about this thing C? 408 00:31:40,290 --> 00:31:51,940 Well, C is a N row column matrix. 409 00:31:51,940 --> 00:31:54,120 So it's N rows, 1 column. 410 00:31:56,800 --> 00:32:00,280 It looks like this-- 411 00:32:00,280 --> 00:32:04,090 C1, C2, Cn. 412 00:32:06,910 --> 00:32:13,290 And so if you want to multiply a square matrix by a vector, 413 00:32:13,290 --> 00:32:15,880 we know the rules too, OK? 414 00:32:15,880 --> 00:32:19,930 And again, this picture is a useful one. 415 00:32:19,930 --> 00:32:22,900 So let's just draw something like this. 416 00:32:22,900 --> 00:32:29,320 We do this and this, and that gives you 417 00:32:29,320 --> 00:32:30,800 one element in the column. 418 00:32:34,970 --> 00:32:36,490 OK. 419 00:32:36,490 --> 00:32:40,420 Now, the last thing that I want to remind you of-- 420 00:32:40,420 --> 00:32:41,860 I better use this board. 421 00:32:46,680 --> 00:32:59,570 If we have a two-state problem, then the vector c1 is 1, 0; 422 00:32:59,570 --> 00:33:02,960 the vector c2-- 423 00:33:02,960 --> 00:33:05,840 this should be a lower case c, because we 424 00:33:05,840 --> 00:33:10,100 tend to use lowercase letters for vectors and uppercase 425 00:33:10,100 --> 00:33:11,135 letters for matrices. 426 00:33:16,040 --> 00:33:32,340 And so if we do c1 dagger c2, or c1 dagger c1. 427 00:33:35,410 --> 00:33:41,570 Now this dagger means conjugate transpose, except, well, 428 00:33:41,570 --> 00:33:47,450 there isn't anything to do except convert a matrix-- 429 00:33:47,450 --> 00:33:58,250 a vector into-- so that becomes a row and this 430 00:33:58,250 --> 00:33:59,390 becomes a column. 431 00:34:03,110 --> 00:34:05,740 And so a row times the column gives a number. 432 00:34:08,360 --> 00:34:11,770 And that number is going to be-- well, let's do it. 433 00:34:11,770 --> 00:34:16,719 We have 1, 0; 0, 1. 434 00:34:16,719 --> 00:34:21,790 1 times 0 is 0, 0 times 1 is 0, and so this is 0. 435 00:34:21,790 --> 00:34:25,000 Well that's orthogonality. 436 00:34:25,000 --> 00:34:32,730 This, we have 1 times 1 is 1. 437 00:34:32,730 --> 00:34:34,030 0 times 0 is 0. 438 00:34:34,030 --> 00:34:37,760 So it's 1 plus 0. 439 00:34:37,760 --> 00:34:38,770 Right. 440 00:34:38,770 --> 00:34:40,429 OK, all right. 441 00:34:46,960 --> 00:34:47,460 OK. 442 00:34:50,400 --> 00:34:58,530 So the Schrodinger equation becomes in matrix language-- 443 00:34:58,530 --> 00:35:01,900 and now, one notation is-- 444 00:35:01,900 --> 00:35:04,020 I'm going to stop using the double underline. 445 00:35:04,020 --> 00:35:06,360 That means boldface. 446 00:35:06,360 --> 00:35:09,500 And we don't use hats anymore. 447 00:35:09,500 --> 00:35:12,340 Or at least if we were really consistent, 448 00:35:12,340 --> 00:35:15,400 when we go away from the Schrodinger picture, 449 00:35:15,400 --> 00:35:20,320 we don't put hats on operators, we make them boldface letters. 450 00:35:20,320 --> 00:35:21,000 OK. 451 00:35:21,000 --> 00:35:26,580 Now, the thing is we're so comfortable in matrix land, 452 00:35:26,580 --> 00:35:29,740 that we don't use either. 453 00:35:29,740 --> 00:35:30,260 OK. 454 00:35:30,260 --> 00:35:33,780 But remember, we're talking about different things. 455 00:35:33,780 --> 00:35:36,950 So the Schrodinger equation looks like this. 456 00:35:45,600 --> 00:35:53,070 And so we have a matrix, delta v, v delta times 1, 0. 457 00:35:56,810 --> 00:36:06,350 And that's delta, v; or delta times 1, 0 plus v times 0, 1. 458 00:36:06,350 --> 00:36:07,760 Isn't that interesting? 459 00:36:07,760 --> 00:36:12,410 Remember, the Hamiltonian or any operator operating 460 00:36:12,410 --> 00:36:16,640 on a function gives rise to a linear combination 461 00:36:16,640 --> 00:36:19,280 of the functions and the basis set. 462 00:36:19,280 --> 00:36:22,921 And so here's one of the functions, here's the other. 463 00:36:22,921 --> 00:36:23,420 OK. 464 00:36:23,420 --> 00:36:27,990 Nothing very mysterious has happened here. 465 00:36:27,990 --> 00:36:36,990 So this very equation is going to be HC is equal to EC. 466 00:36:43,180 --> 00:36:47,340 OK, so how do we approach this? 467 00:37:01,550 --> 00:37:11,220 Well I have to introduce a new symbol, 468 00:37:11,220 --> 00:37:15,840 and that's going to be this symbol T. It's a matrix. 469 00:37:18,390 --> 00:37:20,670 It's a unitary matrix. 470 00:37:20,670 --> 00:37:24,470 And we want it to have a special properties. 471 00:37:24,470 --> 00:37:27,620 Those special properties will be shown here. 472 00:37:27,620 --> 00:37:31,230 OK, so first of all, we have this matrix-- 473 00:37:31,230 --> 00:37:34,860 T11, T-- 474 00:37:34,860 --> 00:37:37,420 AUDIENCE: Your Hamiltonian matrix is incorrect. 475 00:37:37,420 --> 00:37:41,320 So I think you mean to say E bar plus-minus? 476 00:37:41,320 --> 00:37:43,240 ROBERT FIELD: I'm sorry? 477 00:37:43,240 --> 00:37:44,470 AUDIENCE: I think the diagonal elements of the Hamiltonian 478 00:37:44,470 --> 00:37:45,430 matrix should be-- 479 00:37:45,430 --> 00:37:46,388 ROBERT FIELD: Yeah, OK. 480 00:37:46,388 --> 00:37:54,420 If we wanted the eigenvalues, OK? 481 00:37:54,420 --> 00:37:55,470 This is-- 482 00:37:55,470 --> 00:37:59,470 AUDIENCE: So the infinite matrix in this case should be-- 483 00:37:59,470 --> 00:38:02,560 the diagonal elements should be E1 and E2. 484 00:38:02,560 --> 00:38:05,120 So I think you meant E bar plus delta, 485 00:38:05,120 --> 00:38:09,130 and then E bar minus delta. 486 00:38:09,130 --> 00:38:10,080 ROBERT FIELD: Yeah. 487 00:38:10,080 --> 00:38:11,100 OK. 488 00:38:11,100 --> 00:38:13,530 There is something in the notes and something 489 00:38:13,530 --> 00:38:15,840 in my notes which-- 490 00:38:15,840 --> 00:38:18,240 we can always write the Hamiltonian 491 00:38:18,240 --> 00:38:30,780 as E bar 0, 0, E bar plus delta v v minus delta. 492 00:38:30,780 --> 00:38:35,360 We can always take out this constant term. 493 00:38:35,360 --> 00:38:39,020 And this is the thing we're always working on. 494 00:38:39,020 --> 00:38:42,730 And so rather than-- 495 00:38:42,730 --> 00:38:46,720 and so we could call this H prime. 496 00:38:46,720 --> 00:38:49,570 Or we can simply say, oh, we have these two things-- 497 00:38:49,570 --> 00:38:53,260 this always gets added in, and it's not affected. 498 00:38:53,260 --> 00:38:59,050 If we take this diagonal constant matrix 499 00:38:59,050 --> 00:39:03,830 and apply a transformation to it, a unitary transformation, 500 00:39:03,830 --> 00:39:07,000 you get that matrix again, you get-- it's nothing. 501 00:39:07,000 --> 00:39:23,060 OK, so we wanted to describe some special matrix 502 00:39:23,060 --> 00:39:27,710 where the transpose of that matrix 503 00:39:27,710 --> 00:39:30,790 is equal to the inverse of that matrix. 504 00:39:30,790 --> 00:39:34,060 Now-- yes? 505 00:39:34,060 --> 00:39:37,110 AUDIENCE: So shouldn't there be a minus on there? 506 00:39:37,110 --> 00:39:39,660 On the prior equation? 507 00:39:39,660 --> 00:39:42,230 For the bottom right entry. 508 00:39:42,230 --> 00:39:44,470 The primary delta and the-- 509 00:39:44,470 --> 00:39:46,330 first line. 510 00:39:46,330 --> 00:39:48,036 First line. 511 00:39:48,036 --> 00:39:49,660 ROBERT FIELD: This is delta minus the-- 512 00:39:49,660 --> 00:39:50,243 AUDIENCE: Yes. 513 00:39:50,243 --> 00:39:52,621 But up there, you just had delta delta. 514 00:39:52,621 --> 00:39:53,371 ROBERT FIELD: Yep. 515 00:39:56,710 --> 00:39:57,664 Thank you. 516 00:40:01,020 --> 00:40:01,750 OK. 517 00:40:01,750 --> 00:40:04,950 Now, when you have a matrix, you really 518 00:40:04,950 --> 00:40:11,940 like to have T minus 1T is equal to the unit matrix. 519 00:40:11,940 --> 00:40:16,820 Getting the inverse of a matrix in a general problem 520 00:40:16,820 --> 00:40:19,700 is really awful. 521 00:40:19,700 --> 00:40:25,420 But for a unitary matrices, all you do to get the inverse 522 00:40:25,420 --> 00:40:26,950 is to take-- 523 00:40:26,950 --> 00:40:28,465 you flip it on the diagonal. 524 00:40:31,270 --> 00:40:40,980 Now strictly, the conjugate transpose of a unitary matrix 525 00:40:40,980 --> 00:40:42,660 is the inverse. 526 00:40:42,660 --> 00:40:45,480 But often we have real matrices. 527 00:40:45,480 --> 00:40:50,190 But the important thing is always this flipping along-- 528 00:40:50,190 --> 00:40:52,500 flipping the matrix on the diagonal. 529 00:40:52,500 --> 00:40:56,180 And that gives you the inverse unless there 530 00:40:56,180 --> 00:40:58,940 is stuff here that is complex. 531 00:40:58,940 --> 00:40:59,440 OK. 532 00:41:02,690 --> 00:41:06,650 So this conjugate transpose, it would look like-- 533 00:41:18,482 --> 00:41:20,870 OK? 534 00:41:20,870 --> 00:41:21,370 OK. 535 00:41:21,370 --> 00:41:25,630 Now, what we want to do is derive the matrix form 536 00:41:25,630 --> 00:41:28,060 of the Schrodinger equation using 537 00:41:28,060 --> 00:41:33,920 this unitary transformation. 538 00:41:33,920 --> 00:41:38,010 So we start out again with HC is equal to EC. 539 00:41:44,260 --> 00:41:52,190 And now, we insert TT dagger. 540 00:41:55,010 --> 00:41:59,030 And this is one of the things where you screw up. 541 00:41:59,030 --> 00:42:02,180 Whether you insert TT dagger or T dagger T, 542 00:42:02,180 --> 00:42:03,680 because they're both 1. 543 00:42:03,680 --> 00:42:06,830 And if you use the wrong one, all of your phases, 544 00:42:06,830 --> 00:42:08,870 everything is wrong. 545 00:42:08,870 --> 00:42:11,330 But it's still correct, the equations are correct, 546 00:42:11,330 --> 00:42:15,050 but the things you've memorized are no longer correct. 547 00:42:15,050 --> 00:42:18,620 OK, so we're going to insert this unit 548 00:42:18,620 --> 00:42:32,950 matrix between H and C. And of course, we have to-- 549 00:42:32,950 --> 00:42:37,050 we would have to insert this on the other side. 550 00:42:37,050 --> 00:42:47,850 T T dagger C and E. But that's one. 551 00:42:47,850 --> 00:42:50,570 OK, so we don't do anything on the left-hand side. 552 00:42:53,310 --> 00:42:55,790 And now we left multiply by T dagger. 553 00:43:05,340 --> 00:43:06,550 OK. 554 00:43:06,550 --> 00:43:12,740 And we call this H twiddle, and we call this C twiddle. 555 00:43:15,430 --> 00:43:24,160 So we have H twiddle, C twiddle is equal to EC twiddle. 556 00:43:24,160 --> 00:43:26,850 Now we're cooking. 557 00:43:26,850 --> 00:43:32,160 OK, because we construct this unitary matrix 558 00:43:32,160 --> 00:43:37,350 to cause this Hamiltonian, the transformed Hamiltonian 559 00:43:37,350 --> 00:43:38,460 to be in diagonal form. 560 00:43:48,850 --> 00:43:58,450 So we say that H twiddle, which is T dagger T, 561 00:43:58,450 --> 00:44:06,800 is equal to E plus, E minus, 0, 0 for the 2 by 2 problem. 562 00:44:11,870 --> 00:44:14,730 OK, that leads to some requirements. 563 00:44:14,730 --> 00:44:16,450 What are the elements of the T matrix? 564 00:44:19,620 --> 00:44:30,930 But the important thing is we say that T diagonalizes H. 565 00:44:30,930 --> 00:44:35,624 So this magic matrix gives you the energy eigenvalues. 566 00:44:41,060 --> 00:44:49,450 But we also have T dagger C is equal to C twiddle. 567 00:44:49,450 --> 00:44:54,510 And so this gives you the linear combination 568 00:44:54,510 --> 00:44:59,752 of the column vectors that correspond to the eigenvector. 569 00:45:06,840 --> 00:45:08,210 So what is it? 570 00:45:08,210 --> 00:45:31,520 We have here T dagger, which is TT, T12 T, T21 T, T22 T 571 00:45:31,520 --> 00:45:46,970 times C. And that's supposed to be equal to C twiddle or C1 572 00:45:46,970 --> 00:45:50,060 twiddle, C2 twiddle. 573 00:45:50,060 --> 00:45:56,070 So we put everything together and we discover that OK, 574 00:45:56,070 --> 00:45:59,430 when we multiply this by that, we get 575 00:45:59,430 --> 00:46:01,410 the element that goes up here. 576 00:46:01,410 --> 00:46:20,690 And so the element on top is C1 T11 dagger plus C2 T12-- 577 00:46:23,720 --> 00:46:27,920 I'm using T's and daggers independently, 578 00:46:27,920 --> 00:46:29,610 and we do the same thing here. 579 00:46:29,610 --> 00:46:34,980 So we have a column vector, and it's 580 00:46:34,980 --> 00:46:39,780 composed of the original state. 581 00:46:39,780 --> 00:46:45,030 And so anyway, when we do everything, 582 00:46:45,030 --> 00:46:50,010 we discover that using the solution 583 00:46:50,010 --> 00:46:54,990 to the two-level problem from the Schrodinger picture, 584 00:46:54,990 --> 00:46:56,580 we know everything. 585 00:46:56,580 --> 00:47:08,620 So H twiddle, C twiddle plus is H plus C twiddle plus. 586 00:47:08,620 --> 00:47:18,330 And that's just E plus 1, 0. 587 00:47:18,330 --> 00:47:35,180 And H twiddle C twiddle minus is E minus times 0, 1. 588 00:47:38,355 --> 00:47:38,855 OK. 589 00:47:42,630 --> 00:47:47,200 This is all very confusing because the notation 590 00:47:47,200 --> 00:47:49,890 is unfamiliar. 591 00:47:49,890 --> 00:47:52,770 But the important thing is that everything 592 00:47:52,770 --> 00:47:55,800 you can do in the Schrodinger picture you 593 00:47:55,800 --> 00:47:58,350 can do in this matrix picture. 594 00:47:58,350 --> 00:48:05,700 And you can find these elements of this T matrix, 595 00:48:05,700 --> 00:48:09,600 and it turns out that what you-- 596 00:48:12,810 --> 00:48:23,730 the matrix that diagonalizes the Hamiltonian, the columns of T 597 00:48:23,730 --> 00:48:26,610 dagger are the eigenvectors. 598 00:48:29,700 --> 00:48:34,520 So you find this matrix, it diagonalizes 599 00:48:34,520 --> 00:48:39,270 H. The computer tells you the eigenenergies, 600 00:48:39,270 --> 00:48:42,520 and it also tells you T dagger or T plus-- 601 00:48:42,520 --> 00:48:43,020 TT. 602 00:48:46,260 --> 00:48:49,140 And so with that, you know how to write 603 00:48:49,140 --> 00:48:53,970 in the original basis set what the eigenbasis is 604 00:48:53,970 --> 00:48:55,280 for each eigenvalue. 605 00:48:58,340 --> 00:49:06,140 And so the next step, which will happen in the next lecture, 606 00:49:06,140 --> 00:49:17,450 is the general unitary transformation for two-level. 607 00:49:17,450 --> 00:49:22,310 Now, you expect that there is going 608 00:49:22,310 --> 00:49:26,540 to be a general solution for the two-level problem, 609 00:49:26,540 --> 00:49:28,910 because the two-level problem in the Schrodinger picture 610 00:49:28,910 --> 00:49:32,360 led to a quadratic equation. 611 00:49:32,360 --> 00:49:35,300 And that had an analytical solution. 612 00:49:35,300 --> 00:49:40,910 And so there is going to be a general and exact solution 613 00:49:40,910 --> 00:49:42,800 the two-level problem. 614 00:49:42,800 --> 00:49:45,920 And this unitary transformation is 615 00:49:45,920 --> 00:49:55,730 going to be written in terms of cosine theta sine theta 616 00:49:55,730 --> 00:50:01,310 minus sine theta cosine theta. 617 00:50:01,310 --> 00:50:07,910 This is a matrix which is unitary, 618 00:50:07,910 --> 00:50:12,440 and it, when applied to your basis set, 619 00:50:12,440 --> 00:50:20,090 conserves normalization and orthogonality. 620 00:50:20,090 --> 00:50:22,540 And so the trick is to be able to find 621 00:50:22,540 --> 00:50:27,070 the theta that causes the Hamiltonian in question 622 00:50:27,070 --> 00:50:28,930 to be diagonalized. 623 00:50:28,930 --> 00:50:32,790 And the algebra for that will happen in the next lecture. 624 00:50:32,790 --> 00:50:33,800 OK. 625 00:50:33,800 --> 00:50:39,270 Remember, you're not going to do this ever. 626 00:50:39,270 --> 00:50:44,730 You're going to use the idea of this unitary transformation, 627 00:50:44,730 --> 00:50:48,180 and you're going to use that to get-- 628 00:50:48,180 --> 00:50:51,130 to diagonalize the matrix. 629 00:50:51,130 --> 00:50:53,710 And this will lead to some formulas 630 00:50:53,710 --> 00:50:56,080 which you're going to like, because you 631 00:50:56,080 --> 00:50:58,630 can forget sines and cosines. 632 00:50:58,630 --> 00:51:00,130 Everything is going to be expressed 633 00:51:00,130 --> 00:51:03,160 in terms of things like this-- 634 00:51:03,160 --> 00:51:07,590 V over delta. 635 00:51:07,590 --> 00:51:09,840 So we have matrix element-- 636 00:51:09,840 --> 00:51:16,000 off diagonal matrix element over the energy difference. 637 00:51:16,000 --> 00:51:20,290 And that's the basic form of nondegenerate perturbation 638 00:51:20,290 --> 00:51:23,927 theory, which is applied not just in 2-by-2 problems, 639 00:51:23,927 --> 00:51:24,760 but to all problems. 640 00:51:29,400 --> 00:51:34,380 And so you want to remember this lecture as, 641 00:51:34,380 --> 00:51:37,570 this is how we kill the two-level problem. 642 00:51:37,570 --> 00:51:40,860 Then we can discover what we did and apply it 643 00:51:40,860 --> 00:51:43,320 to a general problem where it's not 644 00:51:43,320 --> 00:51:44,760 a two-level problem anymore, it's 645 00:51:44,760 --> 00:51:47,290 an infinite number of levels. 646 00:51:47,290 --> 00:51:51,560 And when certain approximations are met, 647 00:51:51,560 --> 00:51:57,860 it applies, and it gives you the most accurate energy levels 648 00:51:57,860 --> 00:52:00,530 and wave functions you could want. 649 00:52:00,530 --> 00:52:03,710 And you know how accurate they are going to be. 650 00:52:03,710 --> 00:52:08,870 And this is liberating, because now, you 651 00:52:08,870 --> 00:52:12,510 can take a not exactly-solved problem 652 00:52:12,510 --> 00:52:15,680 and you can solve it approximately. 653 00:52:15,680 --> 00:52:19,410 And you can use the solution to determine, OK, 654 00:52:19,410 --> 00:52:22,980 there's going to be some function of the quantum numbers 655 00:52:22,980 --> 00:52:24,850 which describes the energy levels. 656 00:52:24,850 --> 00:52:26,870 Well what is that function? 657 00:52:26,870 --> 00:52:31,050 And what are the coefficients in that function? 658 00:52:31,050 --> 00:52:32,820 And how do they relate to the things 659 00:52:32,820 --> 00:52:36,280 you know from the Hamiltonian? 660 00:52:36,280 --> 00:52:40,240 So it's incredibly powerful. 661 00:52:40,240 --> 00:52:44,590 And once you're given the formulas 662 00:52:44,590 --> 00:52:46,990 for nondegenerate perturbation theory, 663 00:52:46,990 --> 00:52:50,780 you can solve practically any problem in quantum mechanics. 664 00:52:50,780 --> 00:52:55,360 Not just numerically, but with insight. 665 00:52:55,360 --> 00:53:04,440 It tells you, if we make observations of some system, 666 00:53:04,440 --> 00:53:06,720 we determine a set of energy levels, which is called 667 00:53:06,720 --> 00:53:09,760 the spectrum of that operator. 668 00:53:09,760 --> 00:53:15,490 And the spectrum of the operator is an explicit function 669 00:53:15,490 --> 00:53:19,630 of the physical constants-- the unique interactions 670 00:53:19,630 --> 00:53:21,830 between states. 671 00:53:21,830 --> 00:53:30,200 And so you will then know how the experimental data 672 00:53:30,200 --> 00:53:34,002 determines the mechanism of all the interactions, 673 00:53:34,002 --> 00:53:39,790 and you can calculate the wave function. 674 00:53:39,790 --> 00:53:41,650 You can't observe the wave function, 675 00:53:41,650 --> 00:53:46,900 but you can discover its traces in the energy levels. 676 00:53:46,900 --> 00:53:49,790 And then you can reproduce the energy levels, 677 00:53:49,790 --> 00:53:53,050 and if you have the energy levels-- the eigenstates, 678 00:53:53,050 --> 00:53:58,390 you can also describe any dynamics using 679 00:53:58,390 --> 00:54:00,530 the same formulas in here. 680 00:54:00,530 --> 00:54:07,640 So this is an incredible enablement. 681 00:54:07,640 --> 00:54:11,310 And you don't have to look at my derivations. 682 00:54:11,310 --> 00:54:12,750 I'm not proud of the derivations, 683 00:54:12,750 --> 00:54:14,650 I'm proud of the results. 684 00:54:14,650 --> 00:54:20,070 And if you can handle these results that 685 00:54:20,070 --> 00:54:22,416 come from perturbation theory as well as you've 686 00:54:22,416 --> 00:54:23,790 done so far in the course, you're 687 00:54:23,790 --> 00:54:30,840 going to be at the research level very soon. 688 00:54:30,840 --> 00:54:33,680 OK, see you on Friday.