1 00:00:00,050 --> 00:00:01,670 The following content is provided 2 00:00:01,670 --> 00:00:03,820 under a Creative Commons license. 3 00:00:03,820 --> 00:00:06,540 Your support will help MIT OpenCourseWare continue 4 00:00:06,540 --> 00:00:10,120 to offer high quality educational resources for free. 5 00:00:10,120 --> 00:00:12,700 To make a donation or to view additional materials 6 00:00:12,700 --> 00:00:16,603 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,603 --> 00:00:17,565 at ocw.mit.edu. 8 00:00:21,440 --> 00:00:24,130 PROFESSOR: Going to get started right away. 9 00:00:24,130 --> 00:00:29,570 I want to make a comment about energy time uncertainty 10 00:00:29,570 --> 00:00:31,460 relations. 11 00:00:31,460 --> 00:00:35,530 And we talked last time about the fact 12 00:00:35,530 --> 00:00:39,260 that the energy time uncertainty relation really tells you 13 00:00:39,260 --> 00:00:43,760 something about how fast a state can change. 14 00:00:43,760 --> 00:00:48,580 So an interesting way to try to evaluate 15 00:00:48,580 --> 00:00:51,950 that is to consider a system that 16 00:00:51,950 --> 00:00:55,950 has to state that time equals 0 and compute 17 00:00:55,950 --> 00:01:00,550 the overlap with a state at time equals t. 18 00:01:00,550 --> 00:01:06,880 Now this overlap is a very schematic thing. 19 00:01:06,880 --> 00:01:08,210 It's a bracket. 20 00:01:08,210 --> 00:01:09,710 It's a number. 21 00:01:09,710 --> 00:01:11,960 And you know what that means. 22 00:01:11,960 --> 00:01:14,240 You can keep in mind what that is. 23 00:01:14,240 --> 00:01:21,780 It's an integral over space possibly of psi star at t 24 00:01:21,780 --> 00:01:29,600 equals 0 x psi t x. 25 00:01:29,600 --> 00:01:33,820 It's a complete integral, its inner product. 26 00:01:33,820 --> 00:01:39,650 So you may want to understand this, because, in a sense, 27 00:01:39,650 --> 00:01:44,310 this is telling you how quickly a state can change. 28 00:01:44,310 --> 00:01:48,820 At time equals 0, this overlap is just 1. 29 00:01:48,820 --> 00:01:52,020 A little later, this overlap is going to change 30 00:01:52,020 --> 00:01:56,880 and perhaps after some time the overlap is going to be 0. 31 00:01:56,880 --> 00:02:02,120 And we're going to say that we actually have changed a lot. 32 00:02:02,120 --> 00:02:06,300 So this number is very interesting to compute. 33 00:02:06,300 --> 00:02:10,530 And in fact, we might as well square it, 34 00:02:10,530 --> 00:02:12,740 because it's a complex number. 35 00:02:12,740 --> 00:02:16,760 So to understand better what it is we'll square it. 36 00:02:16,760 --> 00:02:21,340 And we'll have to evaluate this. 37 00:02:21,340 --> 00:02:24,310 Now how could you evaluate this? 38 00:02:24,310 --> 00:02:29,670 Well, we'll assume that this system that governs this time 39 00:02:29,670 --> 00:02:32,425 evolution has a time independent Hamiltonian. 40 00:02:39,740 --> 00:02:44,630 Once this evolution is done by a time independent Hamiltonian, 41 00:02:44,630 --> 00:02:46,120 you can wonder what it is. 42 00:02:49,480 --> 00:02:51,640 Now it's quite interesting, and you 43 00:02:51,640 --> 00:02:53,660 will have to discuss that in the homework 44 00:02:53,660 --> 00:02:55,190 because it will actually help you 45 00:02:55,190 --> 00:02:59,810 prove that version of the time energy uncertainty 46 00:02:59,810 --> 00:03:04,650 relationship that says that the quickest time state can turn 47 00:03:04,650 --> 00:03:09,490 orthogonal to itself is bounded by some amount. 48 00:03:09,490 --> 00:03:11,880 Cannot do it infinitely fast. 49 00:03:11,880 --> 00:03:16,250 So you want to know how fast this can change. 50 00:03:16,250 --> 00:03:22,010 Now it's very surprising what it depends on, this thing. 51 00:03:22,010 --> 00:03:28,740 Because suppose you had an energy eigenstate, 52 00:03:28,740 --> 00:03:33,070 suppose psi at time equals 0 is an energy eigenstate. 53 00:03:33,070 --> 00:03:35,740 What would happen later on? 54 00:03:35,740 --> 00:03:38,160 Well, you know that energy eigenstates 55 00:03:38,160 --> 00:03:41,560 evolve with a phase, an exponential of e 56 00:03:41,560 --> 00:03:45,820 to the minus iht over h bar. 57 00:03:45,820 --> 00:03:51,320 So actually if you had an energy eigenstate, 58 00:03:51,320 --> 00:03:55,235 this thing would remain equal to 1 for all times. 59 00:03:57,740 --> 00:04:00,770 So if this is going to be non-zero, 60 00:04:00,770 --> 00:04:04,375 it's because it's going to have-- 61 00:04:04,375 --> 00:04:07,360 you have to have a state is not an energy eigenstate. 62 00:04:07,360 --> 00:04:10,720 That you're going to have an uncertainty in the energy 63 00:04:10,720 --> 00:04:12,340 and energy uncertainty. 64 00:04:12,340 --> 00:04:17,745 So the curious thing is that you can evaluate this, and expand 65 00:04:17,745 --> 00:04:23,735 it as a power in t, and go, say, to quadratic ordering 66 00:04:23,735 --> 00:04:26,456 in t evaluating what this is. 67 00:04:26,456 --> 00:04:30,670 And this only depends on the uncertainty of h, 68 00:04:30,670 --> 00:04:32,840 and t, and things like that. 69 00:04:32,840 --> 00:04:38,450 So only the uncertainty of h matters at this moment. 70 00:04:38,450 --> 00:04:40,700 So this would be quite interested, I think, 71 00:04:40,700 --> 00:04:45,690 for you to figure out and to explore in detail. 72 00:04:45,690 --> 00:04:50,650 That kind of analysis has been the center 73 00:04:50,650 --> 00:04:56,150 of attention recently, having to do with quantum computation. 74 00:04:56,150 --> 00:04:59,700 Because in a sense, in a quantum computer, 75 00:04:59,700 --> 00:05:03,640 you want to change states quickly and do operations. 76 00:05:03,640 --> 00:05:07,270 So how quickly you can change a state is crucial. 77 00:05:07,270 --> 00:05:10,970 So in fact, the people that proved these inequalities 78 00:05:10,970 --> 00:05:14,760 that you're going to find say that this eventually 79 00:05:14,760 --> 00:05:19,110 will limit the speed of a quantum computer, 80 00:05:19,110 --> 00:05:25,440 and more slow that computers become twice as fast every year 81 00:05:25,440 --> 00:05:29,210 or so, and double the speed. 82 00:05:29,210 --> 00:05:32,440 So this limit apparently, it's claimed, 83 00:05:32,440 --> 00:05:37,870 will allow 80 duplications of the speed 84 00:05:37,870 --> 00:05:43,430 until you hit this limit in a quantum computer 85 00:05:43,430 --> 00:05:44,760 due to quantum mechanics. 86 00:05:44,760 --> 00:05:52,940 So you will be investigating this in next week's homework. 87 00:05:52,940 --> 00:05:56,460 Today we want to begin quickly with an application 88 00:05:56,460 --> 00:06:00,980 of the uncertainty principal to find exact bounds of energies 89 00:06:00,980 --> 00:06:04,830 for quantum systems, for ground states of quantum systems. 90 00:06:04,830 --> 00:06:09,340 So this will be a very precise application of the uncertainty 91 00:06:09,340 --> 00:06:10,970 principle. 92 00:06:10,970 --> 00:06:14,150 Then we'll turn to a completion of these things 93 00:06:14,150 --> 00:06:16,310 that we've been talking about having 94 00:06:16,310 --> 00:06:18,830 to do with linear algebra. 95 00:06:18,830 --> 00:06:24,350 We'll get to the main theorems of the subject. 96 00:06:24,350 --> 00:06:27,070 In a sense, the most important theorems of the subject. 97 00:06:27,070 --> 00:06:30,240 These are the spectral theorem that tells you 98 00:06:30,240 --> 00:06:35,140 about what operators can be diagonalized. 99 00:06:35,140 --> 00:06:38,520 And then a theorem that leads to the concept 100 00:06:38,520 --> 00:06:41,610 of a complete set of commuting observables. 101 00:06:41,610 --> 00:06:46,030 So really pretty key mathematical ideas. 102 00:06:46,030 --> 00:06:51,500 And the way we're good to do it, I think you will see, 103 00:06:51,500 --> 00:06:56,080 that we have gained a lot by learning the linear algebra 104 00:06:56,080 --> 00:06:59,270 concepts in a slightly abstract way. 105 00:06:59,270 --> 00:07:03,390 I do remember doing this proof that we're going today 106 00:07:03,390 --> 00:07:06,060 in previous years, and it was always 107 00:07:06,060 --> 00:07:10,310 considered the most complicated lecture of the course. 108 00:07:10,310 --> 00:07:12,830 Just taking the [? indices ?] went crazy, 109 00:07:12,830 --> 00:07:16,540 and lots of formulas, and the notation was funny. 110 00:07:16,540 --> 00:07:19,620 And now we will do the proof, and we'll 111 00:07:19,620 --> 00:07:22,170 write a few little things. 112 00:07:22,170 --> 00:07:24,810 And we'll just try to imagine what's going on. 113 00:07:24,810 --> 00:07:29,150 And will be, I think, easier. 114 00:07:29,150 --> 00:07:30,460 I hope you will agree. 115 00:07:30,460 --> 00:07:35,036 So let's begin with an example of a use of the uncertainty 116 00:07:35,036 --> 00:07:35,535 principle. 117 00:07:35,535 --> 00:07:38,800 So example. 118 00:07:38,800 --> 00:07:42,980 So this will be maybe for 20 minutes. 119 00:07:42,980 --> 00:07:48,220 Consider this Hamiltonian for a one dimensional particle 120 00:07:48,220 --> 00:07:50,240 that, in fact, you've considered before, 121 00:07:50,240 --> 00:07:54,510 alpha x to the fourth, for which you did some approximations. 122 00:07:58,600 --> 00:08:02,290 You know that the expectation value 123 00:08:02,290 --> 00:08:06,460 of the energy in the ground state. 124 00:08:06,460 --> 00:08:08,220 You've done it numerically. 125 00:08:08,220 --> 00:08:10,950 You've done it variationally. 126 00:08:10,950 --> 00:08:15,710 And variationally you knew that the energy at every stage 127 00:08:15,710 --> 00:08:18,940 was smaller at a given bound. 128 00:08:18,940 --> 00:08:22,570 The uncertainty principle is not going 129 00:08:22,570 --> 00:08:25,780 to give us an upper bound. 130 00:08:25,780 --> 00:08:27,390 It's going to give us a lower bound. 131 00:08:27,390 --> 00:08:29,540 So it's a really nice thing, because 132 00:08:29,540 --> 00:08:32,610 between the variational principal and the uncertainty 133 00:08:32,610 --> 00:08:37,039 principle, we can narrow the energy of this ground state 134 00:08:37,039 --> 00:08:39,210 to a window. 135 00:08:39,210 --> 00:08:42,140 In one of the problems that you're doing for this week-- 136 00:08:42,140 --> 00:08:44,460 and I'm sorry only today you really 137 00:08:44,460 --> 00:08:48,590 have all the tools after you hear this discussion-- you 138 00:08:48,590 --> 00:08:50,505 do the same for the harmonic oscillator. 139 00:08:50,505 --> 00:08:54,740 You do a variational estimate for the ground state energy. 140 00:08:54,740 --> 00:08:59,420 You do the uncertainty principle bound for ground state energy. 141 00:08:59,420 --> 00:09:02,820 And you will see these two bounds meet. 142 00:09:02,820 --> 00:09:05,320 And therefore after you've done this to bounds, 143 00:09:05,320 --> 00:09:07,480 you found the ground state energy 144 00:09:07,480 --> 00:09:12,100 of the harmonic oscillator, so it's kind of a neat thing. 145 00:09:12,100 --> 00:09:15,570 So we want to estimate the ground state energy. 146 00:09:15,570 --> 00:09:17,930 So we first write some words. 147 00:09:17,930 --> 00:09:21,180 We just say H in the ground state 148 00:09:21,180 --> 00:09:25,060 will be given by the expectation value of p squared 149 00:09:25,060 --> 00:09:30,290 in the ground state plus alpha times the expectation 150 00:09:30,290 --> 00:09:35,210 value of x to the fourth in the ground state. 151 00:09:35,210 --> 00:09:38,850 Haven't made much progress, but you 152 00:09:38,850 --> 00:09:43,310 have, because you're starting to talk about the right variables. 153 00:09:43,310 --> 00:09:46,920 Now this thing to that you have to know 154 00:09:46,920 --> 00:09:48,950 is that you have a potential that 155 00:09:48,950 --> 00:09:53,530 is like this, sort of a little flatter 156 00:09:53,530 --> 00:09:57,900 than x squared potential. 157 00:09:57,900 --> 00:10:02,280 And what can we say about the expectation value 158 00:10:02,280 --> 00:10:04,420 of the momentum on the ground state 159 00:10:04,420 --> 00:10:09,380 and the expectation value of x in the ground state? 160 00:10:09,380 --> 00:10:14,530 Well, the expectation value of x should be no big problem. 161 00:10:14,530 --> 00:10:17,940 This is a symmetric potential, therefore 162 00:10:17,940 --> 00:10:21,990 wave functions in a one dimensional quantum mechanics 163 00:10:21,990 --> 00:10:25,180 problems are either symmetric or anti symmetric. 164 00:10:25,180 --> 00:10:27,430 It could not be anti symmetric because it's 165 00:10:27,430 --> 00:10:29,790 a ground state and kind of have a 0. 166 00:10:29,790 --> 00:10:31,620 So it's asymmetric. 167 00:10:31,620 --> 00:10:33,180 Has no nodes. 168 00:10:33,180 --> 00:10:35,800 So there's the wave function of the ground 169 00:10:35,800 --> 00:10:38,850 state, the symmetric, and the expectation value 170 00:10:38,850 --> 00:10:40,580 of x in the ground state is 0. 171 00:10:43,310 --> 00:10:47,930 Similarly, the expectation value of the momentum in the ground 172 00:10:47,930 --> 00:10:49,275 state, what is it? 173 00:10:49,275 --> 00:10:49,775 Is? 174 00:10:52,550 --> 00:10:54,120 0 too. 175 00:10:54,120 --> 00:10:58,740 And you can imagine just computing it. 176 00:10:58,740 --> 00:11:02,030 It would be the integral of psi. 177 00:11:02,030 --> 00:11:08,190 Psi is going to be a real d dx h bar over i psi. 178 00:11:08,190 --> 00:11:09,830 This is a total derivative. 179 00:11:09,830 --> 00:11:13,810 If it's a bound state, it's 0 at the ends. 180 00:11:13,810 --> 00:11:14,645 This is 0. 181 00:11:19,400 --> 00:11:23,280 So actually, we have a little advantage here. 182 00:11:23,280 --> 00:11:29,530 We have some control over what p squared is, 183 00:11:29,530 --> 00:11:39,630 because the uncertainty in p in the ground state-- well, 184 00:11:39,630 --> 00:11:45,670 the uncertainty in p squared is the expectation value 185 00:11:45,670 --> 00:11:51,490 of p squared minus the expectation value of p squared. 186 00:11:51,490 --> 00:11:57,120 So in the ground state, this is 0. 187 00:11:57,120 --> 00:12:01,990 So delta p squared in the ground state 188 00:12:01,990 --> 00:12:05,010 is just p squared on the ground state. 189 00:12:07,810 --> 00:12:10,570 Similarly, because the expectation value 190 00:12:10,570 --> 00:12:16,730 of x is equal to 0, delta x squared in the ground state 191 00:12:16,730 --> 00:12:23,130 is equal to expectation value of x squared in the ground state. 192 00:12:23,130 --> 00:12:27,950 So actually, this expectation of p squared is delta p. 193 00:12:27,950 --> 00:12:30,950 And we want to use the uncertainty principle, 194 00:12:30,950 --> 00:12:32,670 so that's progress. 195 00:12:32,670 --> 00:12:36,266 We've related something we want to estimate to an uncertainty. 196 00:12:39,580 --> 00:12:42,600 Small complication is that we have an expectation 197 00:12:42,600 --> 00:12:46,850 value of x to the fourth. 198 00:12:46,850 --> 00:12:53,540 Now we learned-- maybe I can continue here. 199 00:12:53,540 --> 00:12:57,510 We learned that the expectations value for an operator squared 200 00:12:57,510 --> 00:13:01,060 is bigger than or equal to the expectation 201 00:13:01,060 --> 00:13:04,300 value of the operator squared. 202 00:13:04,300 --> 00:13:10,300 So the expectation value of x to the fourth 203 00:13:10,300 --> 00:13:16,930 is definitely bigger than the expectation value of x squared 204 00:13:16,930 --> 00:13:19,590 squared. 205 00:13:19,590 --> 00:13:22,270 And this is true on any state. 206 00:13:24,800 --> 00:13:27,900 This was derived when we did uncertainty. 207 00:13:27,900 --> 00:13:30,530 We proved that the uncertainty squared 208 00:13:30,530 --> 00:13:33,150 is positive, because the norm of a vector, 209 00:13:33,150 --> 00:13:35,820 and that gave you this thing. 210 00:13:35,820 --> 00:13:39,340 So here you think of the operator as x squared. 211 00:13:39,340 --> 00:13:42,050 So the operator squared is x to the fourth. 212 00:13:42,050 --> 00:13:45,630 And here's the operator expectation value squared. 213 00:13:45,630 --> 00:13:50,120 So this is true for the ground state. 214 00:13:50,120 --> 00:13:55,070 It's also here true for any state, so is the ground state. 215 00:13:55,070 --> 00:13:58,580 And this x squared now is delta x. 216 00:13:58,580 --> 00:14:05,120 So this is delta x on the ground state to the fourth. 217 00:14:05,120 --> 00:14:07,425 So look what we have. 218 00:14:07,425 --> 00:14:12,460 We have that the expectation value of H on the ground state 219 00:14:12,460 --> 00:14:17,890 is strictly equal to delta p on the ground state squared 220 00:14:17,890 --> 00:14:23,120 over 2m plus alpha. 221 00:14:23,120 --> 00:14:28,660 And we cannot do a Priorean equality here, so we have this. 222 00:14:28,660 --> 00:14:30,700 This is so far an equality. 223 00:14:30,700 --> 00:14:37,320 But because of this, this thing is bigger than that. 224 00:14:37,320 --> 00:14:39,850 Well, alpha is supposed to be positive. 225 00:14:39,850 --> 00:14:46,990 So this is bigger than delta p ground state squared over 226 00:14:46,990 --> 00:14:53,555 2m plus alpha delta x on the ground state to the fourth. 227 00:14:58,400 --> 00:15:00,040 OK, so far so good. 228 00:15:00,040 --> 00:15:02,850 We have a strict thing, this. 229 00:15:02,850 --> 00:15:06,270 And the order of the inequality is already showing up. 230 00:15:06,270 --> 00:15:09,830 We're going to get, if anything, a lower bound. 231 00:15:09,830 --> 00:15:12,575 You're going to be bigger than or equal to something. 232 00:15:18,080 --> 00:15:19,610 So what is next? 233 00:15:19,610 --> 00:15:22,000 Next is the uncertainty principle. 234 00:15:22,000 --> 00:15:26,780 We know that delta p delta x is greater than or equal 235 00:15:26,780 --> 00:15:30,540 to h bar over 2 in any state. 236 00:15:30,540 --> 00:15:36,090 So the delta p ground state and delta x on the ground state 237 00:15:36,090 --> 00:15:38,260 still should be equal to that. 238 00:15:38,260 --> 00:15:43,660 Therefore delta p ground state is bigger than 239 00:15:43,660 --> 00:15:51,170 or equal than h over 2 delta x in the ground state like that. 240 00:15:55,020 --> 00:15:58,480 So this inequality still the right direction. 241 00:15:58,480 --> 00:16:05,140 So we can replace this by something 242 00:16:05,140 --> 00:16:08,050 that is bigger than this quantity day 243 00:16:08,050 --> 00:16:10,620 without disturbing the logic. 244 00:16:10,620 --> 00:16:14,800 So we have H ground state now is greater 245 00:16:14,800 --> 00:16:20,490 than or equal to replace the delta p by this thing 246 00:16:20,490 --> 00:16:27,860 here, h squared over 8, because this is squared 247 00:16:27,860 --> 00:16:37,230 and there's another 2m delta x ground state squared plus alpha 248 00:16:37,230 --> 00:16:42,060 delta x ground state to the fourth. 249 00:16:42,060 --> 00:16:43,010 And that's it. 250 00:16:43,010 --> 00:16:46,970 We've obtained this inequality. 251 00:16:52,340 --> 00:17:00,830 So here you say, well, this is good but how can I use it? 252 00:17:00,830 --> 00:17:04,270 I don't know what delta x is in the ground state, 253 00:17:04,270 --> 00:17:07,099 so what have I gained? 254 00:17:07,099 --> 00:17:11,540 Well, let me do a way of thinking 255 00:17:11,540 --> 00:17:14,290 about this that can help you. 256 00:17:14,290 --> 00:17:21,599 Plot the right hand side as a function of delta 257 00:17:21,599 --> 00:17:23,380 x on the ground. 258 00:17:23,380 --> 00:17:25,270 So you don't know how much it is, 259 00:17:25,270 --> 00:17:29,840 delta x on the ground state, so just plot it. 260 00:17:29,840 --> 00:17:32,400 So if you plot this function, there 261 00:17:32,400 --> 00:17:37,910 will be a divergence as this delta x goes to 0, 262 00:17:37,910 --> 00:17:39,250 then it will be a minimum. 263 00:17:39,250 --> 00:17:43,420 It will be a positive minimum, because this is all positive. 264 00:17:43,420 --> 00:17:45,125 And then it will go up again. 265 00:17:47,880 --> 00:17:52,100 So the right hand side as a function of delta x is this. 266 00:17:52,100 --> 00:17:53,950 So here it comes. 267 00:17:53,950 --> 00:17:57,490 You see I don't know what delta x is. 268 00:17:57,490 --> 00:18:00,640 Suppose delta x happens to be this. 269 00:18:00,640 --> 00:18:02,760 Well, then I know that the ground state 270 00:18:02,760 --> 00:18:06,460 energy is bigger than that value. 271 00:18:06,460 --> 00:18:09,070 But maybe that's not delta x. 272 00:18:09,070 --> 00:18:14,280 Delta x may be is this on the ground state. 273 00:18:14,280 --> 00:18:18,280 And then if it's that, well, the ground state energy 274 00:18:18,280 --> 00:18:21,830 is bigger than this value over here. 275 00:18:21,830 --> 00:18:25,570 Well, since I just don't know what it is, 276 00:18:25,570 --> 00:18:30,570 the worst situation is if delta x is here, 277 00:18:30,570 --> 00:18:38,150 and therefore definitely H must be bigger than the lowest value 278 00:18:38,150 --> 00:18:41,130 that this can take. 279 00:18:41,130 --> 00:18:47,000 So the claim is that H of gs, therefore 280 00:18:47,000 --> 00:18:53,070 is greater than or equal than the minimum of this function 281 00:18:53,070 --> 00:18:57,620 h squared over 8m, and I'll just write here 282 00:18:57,620 --> 00:19:06,900 delta x squared plus alpha delta x to the fourth over delta x. 283 00:19:06,900 --> 00:19:12,995 The minimum of this function over that space is the bound. 284 00:19:15,560 --> 00:19:22,770 So I just have to do a calculus problem here. 285 00:19:22,770 --> 00:19:25,760 This is the minimum. 286 00:19:25,760 --> 00:19:29,260 I should take the derivative with respect to delta x. 287 00:19:29,260 --> 00:19:32,060 Find delta x and substitute. 288 00:19:32,060 --> 00:19:34,140 Of course, I'm not going to do that here, 289 00:19:34,140 --> 00:19:38,170 but I'll tell you a formula that will do that for you. 290 00:19:38,170 --> 00:19:43,800 A over x squared plus Bx to the fourth 291 00:19:43,800 --> 00:19:50,660 is minimized for x squared is equal to 1 over 2-- 292 00:19:50,660 --> 00:19:52,250 it's pretty awful numbers. 293 00:19:52,250 --> 00:19:58,550 2 to the 1/3 A over B to the 1/3. 294 00:19:58,550 --> 00:20:12,390 And its value at that point is 2 to the 1/3 times 3/2 times 295 00:20:12,390 --> 00:20:16,760 A to the 2/3 times B to the 1/3. 296 00:20:19,470 --> 00:20:22,335 A little bit of arithmetic. 297 00:20:30,420 --> 00:20:33,210 So for this function, it turns out 298 00:20:33,210 --> 00:20:37,390 that A is whatever coefficient is here. 299 00:20:37,390 --> 00:20:39,280 B is whatever coefficient is there, 300 00:20:39,280 --> 00:20:42,130 so this is supposed to be the answer. 301 00:20:42,130 --> 00:20:46,130 And you get H on the ground state 302 00:20:46,130 --> 00:20:52,830 is greater than or equal to 2 to the 1/3 3/8 h squared 303 00:20:52,830 --> 00:21:02,430 square root of alpha over m to the 2/3, which is about 0.4724 304 00:21:02,430 --> 00:21:09,030 times h squared square root of alpha over m to the 2/3. 305 00:21:09,030 --> 00:21:11,410 And that's our bound. 306 00:21:11,410 --> 00:21:13,740 How good or how bad is the bound? 307 00:21:13,740 --> 00:21:14,540 It's OK. 308 00:21:14,540 --> 00:21:17,190 It's not fabulous. 309 00:21:17,190 --> 00:21:30,710 The real answer is done numerically is 0.668. 310 00:21:30,710 --> 00:21:33,580 I think I remember variational principal gave you 311 00:21:33,580 --> 00:21:38,170 something like 0.68 or 0.69. 312 00:21:38,170 --> 00:21:43,790 And this one says it's bigger than 0.47. 313 00:21:43,790 --> 00:21:45,190 It gives you something. 314 00:21:45,190 --> 00:21:50,700 So the important thing is that it's completely rigorous. 315 00:21:50,700 --> 00:21:54,310 Many times people use the uncertainty principle 316 00:21:54,310 --> 00:21:57,190 to estimate ground state energies. 317 00:21:57,190 --> 00:21:59,560 Those estimates are very hand wavy. 318 00:21:59,560 --> 00:22:02,800 You might as well just do dimensional analysis. 319 00:22:02,800 --> 00:22:04,570 You don't gain anything. 320 00:22:04,570 --> 00:22:06,500 You don't know the factors. 321 00:22:06,500 --> 00:22:08,185 But this is completely rigorous. 322 00:22:08,185 --> 00:22:12,240 I never made an approximation or anything here. 323 00:22:12,240 --> 00:22:14,160 Every step was logical. 324 00:22:14,160 --> 00:22:16,580 Every inequality was exact. 325 00:22:16,580 --> 00:22:20,250 And therefore, this is a solid result. 326 00:22:20,250 --> 00:22:22,980 This is definitely true. 327 00:22:22,980 --> 00:22:25,970 It doesn't tell you an estimate of the answer. 328 00:22:25,970 --> 00:22:31,400 If you dimensional analysis, you say the answer is this times 1, 329 00:22:31,400 --> 00:22:36,350 and that's as good as you can do with dimensional analysis. 330 00:22:36,350 --> 00:22:37,530 It's not that bad. 331 00:22:37,530 --> 00:22:41,480 The answer turns out to be 0.7. 332 00:22:41,480 --> 00:22:44,110 But the uncertainty principle really, 333 00:22:44,110 --> 00:22:47,660 if you're careful, sometimes, not for every problem, 334 00:22:47,660 --> 00:22:52,190 you can do a rigorous thing and find the rigorous answer. 335 00:22:52,190 --> 00:22:54,400 OK, so are there any questions? 336 00:23:00,280 --> 00:23:01,980 Your problem in the homework will 337 00:23:01,980 --> 00:23:04,900 be to do this for the harmonic oscillator 338 00:23:04,900 --> 00:23:06,911 and find the two bounds. 339 00:23:06,911 --> 00:23:07,410 Yes? 340 00:23:07,410 --> 00:23:08,910 AUDIENCE: How does the answer change 341 00:23:08,910 --> 00:23:10,492 if we don't look at the ground state? 342 00:23:10,492 --> 00:23:11,900 PROFESSOR: How do they what? 343 00:23:11,900 --> 00:23:12,890 PROFESSOR: How does the answer change 344 00:23:12,890 --> 00:23:15,320 if we look at a state different from the ground state? 345 00:23:15,320 --> 00:23:19,190 PROFESSOR: Different from the ground state? 346 00:23:19,190 --> 00:23:21,970 So the question was how would this change 347 00:23:21,970 --> 00:23:24,650 if I would try to do something different from the ground 348 00:23:24,650 --> 00:23:25,720 state. 349 00:23:25,720 --> 00:23:30,640 I think for any state, you would still 350 00:23:30,640 --> 00:23:35,470 be able to say that the expectation 351 00:23:35,470 --> 00:23:37,700 value of the momentum is 0. 352 00:23:37,700 --> 00:23:42,430 Now the expectation value of x still would be 0. 353 00:23:42,430 --> 00:23:48,460 So you can go through some steps here. 354 00:23:48,460 --> 00:23:52,760 The problem here being that I don't have a way 355 00:23:52,760 --> 00:23:55,260 to treat any other state differently. 356 00:23:55,260 --> 00:23:57,690 So I would go ahead, and I would have 357 00:23:57,690 --> 00:24:03,830 said for any stationary state, or for any energy eigenstate, 358 00:24:03,830 --> 00:24:07,660 all of what I said is true. 359 00:24:07,660 --> 00:24:09,110 So I don't get a new one. 360 00:24:12,210 --> 00:24:16,140 These things people actually keep working and writing papers 361 00:24:16,140 --> 00:24:17,880 on this stuff. 362 00:24:17,880 --> 00:24:21,810 People sometimes find bounds that are a little original. 363 00:24:21,810 --> 00:24:23,124 Yes? 364 00:24:23,124 --> 00:24:25,415 AUDIENCE: How do you know the momentum expectation is 0 365 00:24:25,415 --> 00:24:25,914 again? 366 00:24:25,914 --> 00:24:30,170 PROFESSOR: The momentum expectation for a bound state 367 00:24:30,170 --> 00:24:30,940 goes like this. 368 00:24:30,940 --> 00:24:35,379 So you want to figure out what is psi p psi. 369 00:24:35,379 --> 00:24:36,420 And you do the following. 370 00:24:36,420 --> 00:24:39,350 That's integral. 371 00:24:39,350 --> 00:24:42,230 Now psi in these problems can be chosen 372 00:24:42,230 --> 00:24:44,710 to be real, so I won't bother. 373 00:24:44,710 --> 00:24:52,240 It's psi of x h bar over i d dx of psi. 374 00:24:52,240 --> 00:25:01,525 So this is equal to h bar over 2i the integral dx of d dx 375 00:25:01,525 --> 00:25:02,765 of psi squared. 376 00:25:07,340 --> 00:25:13,960 So at this moment, you say well that's just h bar over 2i, 377 00:25:13,960 --> 00:25:19,800 the value of psi squared at infinity and at minus infinity. 378 00:25:19,800 --> 00:25:23,360 And since it's a bound state, it's 0 here, 0 there, 379 00:25:23,360 --> 00:25:27,020 and it's equal to 0. 380 00:25:27,020 --> 00:25:30,800 A state that would have expectation value of momentum 381 00:25:30,800 --> 00:25:34,520 you would expect it to be moving. 382 00:25:34,520 --> 00:25:37,610 So this state is there is static. 383 00:25:37,610 --> 00:25:41,630 It's stationary It doesn't have expectation value of momentum. 384 00:25:41,630 --> 00:25:42,250 Yes? 385 00:25:42,250 --> 00:25:45,172 AUDIENCE: Is the reason that you can't get a better estimate 386 00:25:45,172 --> 00:25:47,607 for the things that are on the ground state, 387 00:25:47,607 --> 00:25:50,042 because if you consider the harmonic oscillator, 388 00:25:50,042 --> 00:25:53,451 the uncertainty delta x [? delta p ?] from the ground 389 00:25:53,451 --> 00:25:57,370 state [INAUDIBLE] you go up to higher states. 390 00:25:57,370 --> 00:25:59,770 PROFESSOR: Right, I think that's another way. 391 00:25:59,770 --> 00:26:05,580 AUDIENCE: [INAUDIBLE] higher state using the absolute. 392 00:26:05,580 --> 00:26:06,920 PROFESSOR: Yeah. 393 00:26:06,920 --> 00:26:08,150 That's a [INAUDIBLE]. 394 00:26:08,150 --> 00:26:11,920 So the ground state of the harmonic oscillator saturates 395 00:26:11,920 --> 00:26:15,280 the uncertainty principal and the others don't. 396 00:26:15,280 --> 00:26:19,630 So this argument, I think, is just good for ground state 397 00:26:19,630 --> 00:26:21,750 energies. 398 00:26:21,750 --> 00:26:23,902 One more question. 399 00:26:23,902 --> 00:26:26,510 AUDIENCE: It appears that this method really works. 400 00:26:26,510 --> 00:26:29,638 Doesn't particularly work well if we have a potential that 401 00:26:29,638 --> 00:26:32,932 has an odd power, because we can't use [? the packet ?], 402 00:26:32,932 --> 00:26:35,430 like x [INAUDIBLE] expectation value x to the fourth 403 00:26:35,430 --> 00:26:37,670 is something, some power, expectation. 404 00:26:37,670 --> 00:26:40,820 PROFESSOR: Right, if it's an odd power, 405 00:26:40,820 --> 00:26:42,450 the method doesn't work well. 406 00:26:42,450 --> 00:26:45,000 But actually for an odd power, the physics 407 00:26:45,000 --> 00:26:50,010 doesn't work well either, because the system doesn't 408 00:26:50,010 --> 00:26:51,410 have ground states. 409 00:26:51,410 --> 00:26:56,020 And so let's say that if you had x 410 00:26:56,020 --> 00:26:59,470 to the fourth plus some x cubed, the physics could still 411 00:26:59,470 --> 00:27:00,740 make sense. 412 00:27:00,740 --> 00:27:04,590 But then it's not clear I can do the same. 413 00:27:04,590 --> 00:27:06,080 Actually you can do the same for x 414 00:27:06,080 --> 00:27:09,570 to the eighth and any sort of powers of this type. 415 00:27:09,570 --> 00:27:12,470 But I don't think it works for x to the sixth. 416 00:27:12,470 --> 00:27:14,780 You can try a few things. 417 00:27:14,780 --> 00:27:19,390 OK, so we leave the uncertainty principle 418 00:27:19,390 --> 00:27:28,480 and begin to understand more formerly the operators 419 00:27:28,480 --> 00:27:33,120 for which there's no uncertainty and you can simultaneously 420 00:27:33,120 --> 00:27:34,860 diagonalize them. 421 00:27:34,860 --> 00:27:39,150 So we're going to find operators like A and B, 422 00:27:39,150 --> 00:27:41,050 that they commute. 423 00:27:41,050 --> 00:27:43,070 And then sometimes you can simultaneously 424 00:27:43,070 --> 00:27:43,810 diagonalize them. 425 00:27:43,810 --> 00:27:44,310 Yes? 426 00:27:44,310 --> 00:27:45,645 You have a question. 427 00:27:45,645 --> 00:27:50,082 AUDIENCE: So part of [INAUDIBLE] we use here is [INAUDIBLE], 428 00:27:50,082 --> 00:27:50,582 right? 429 00:27:50,582 --> 00:27:51,290 PROFESSOR: Right. 430 00:27:54,558 --> 00:27:59,031 AUDIENCE: If we get an asset-- is there any way that we 431 00:27:59,031 --> 00:28:03,272 can better our [INAUDIBLE] principle based on the wave 432 00:28:03,272 --> 00:28:05,008 function with non saturated? 433 00:28:05,008 --> 00:28:07,988 Can we get an upper bound for just 434 00:28:07,988 --> 00:28:10,521 [INAUDIBLE] principle with an h bar over it? 435 00:28:10,521 --> 00:28:12,020 PROFESSOR: Can I get an upper bound? 436 00:28:12,020 --> 00:28:15,260 I'm not sure I understand your question. 437 00:28:15,260 --> 00:28:18,966 AUDIENCE: [INAUDIBLE] the fact that the [INAUDIBLE] principle 438 00:28:18,966 --> 00:28:20,250 will not be saturated. 439 00:28:20,250 --> 00:28:26,542 Can you put the bound for just taking [INAUDIBLE]? 440 00:28:33,240 --> 00:28:36,070 PROFESSOR: Yeah, certainly. 441 00:28:36,070 --> 00:28:38,460 You might have some systems in which 442 00:28:38,460 --> 00:28:40,270 you know that this uncertainty might 443 00:28:40,270 --> 00:28:43,730 be bigger than the one warranted by the uncertainty principles. 444 00:28:43,730 --> 00:28:45,620 And you use that information. 445 00:28:45,620 --> 00:28:49,420 But on general grounds, it's hard to know 446 00:28:49,420 --> 00:28:52,730 that a system might come very close to satisfy 447 00:28:52,730 --> 00:28:55,120 the uncertainty principle in its ground state. 448 00:28:55,120 --> 00:28:56,050 We don't know. 449 00:28:56,050 --> 00:28:59,220 There are systems that come very close in the ground state 450 00:28:59,220 --> 00:29:01,800 to satisfy this and some that are far. 451 00:29:01,800 --> 00:29:04,790 If they are far, you must have some reason 452 00:29:04,790 --> 00:29:06,710 to understand that to use it. 453 00:29:06,710 --> 00:29:10,320 So I don't know. 454 00:29:10,320 --> 00:29:15,080 So let me turn now to this issue of operators 455 00:29:15,080 --> 00:29:25,156 and diagonalization of operators. 456 00:29:30,090 --> 00:29:36,350 Now you might be irritated a little even by the title. 457 00:29:36,350 --> 00:29:37,800 Diagonalization of operation. 458 00:29:37,800 --> 00:29:42,970 You'll be talking about diagonalization of matrices. 459 00:29:42,970 --> 00:29:46,160 Well, there's a way to state what 460 00:29:46,160 --> 00:29:51,660 we mean by diagonalizing an operator in such a way 461 00:29:51,660 --> 00:29:55,990 that we can talk later about the matrix. 462 00:29:55,990 --> 00:29:58,670 So what is the point here? 463 00:29:58,670 --> 00:30:01,360 You have an operator, and it's presumably 464 00:30:01,360 --> 00:30:05,090 an important operator in your theory. 465 00:30:05,090 --> 00:30:09,250 You want to understand this operator better. 466 00:30:09,250 --> 00:30:12,280 So you really are faced with a dilemma. 467 00:30:12,280 --> 00:30:17,480 How do I get some insight into this operator? 468 00:30:17,480 --> 00:30:20,610 Perhaps the simplest thing you could do is to say, 469 00:30:20,610 --> 00:30:26,530 OK let me choose some ideal basis of the vector space, 470 00:30:26,530 --> 00:30:29,370 such as that operator is as simple as possible 471 00:30:29,370 --> 00:30:30,530 in that basis. 472 00:30:30,530 --> 00:30:32,870 So that's the origin of this thing. 473 00:30:32,870 --> 00:30:37,090 Find a basis in the state space, so the operator 474 00:30:37,090 --> 00:30:39,310 looks as simple as possible. 475 00:30:39,310 --> 00:30:43,800 So you say that you can diagonalize an operator 476 00:30:43,800 --> 00:30:51,950 if you can find the basis such that the operator has 477 00:30:51,950 --> 00:30:53,300 just diagonal entries. 478 00:30:56,420 --> 00:30:58,770 So let me just write it like this. 479 00:31:03,160 --> 00:31:20,990 So if you can find a basis in V where the matrix representing 480 00:31:20,990 --> 00:31:34,950 the operator is diagonal, the operator 481 00:31:34,950 --> 00:31:39,150 is said to be diagonalizable. 482 00:31:46,900 --> 00:31:49,870 So to be diagonalizable is just a statement 483 00:31:49,870 --> 00:31:54,370 that there is some basis where you look at the matrix 484 00:31:54,370 --> 00:31:56,410 representation operator, and you find 485 00:31:56,410 --> 00:31:59,300 that it takes form as a diagonal. 486 00:31:59,300 --> 00:32:04,770 So let's try to understand this conceptually 487 00:32:04,770 --> 00:32:07,220 and see what actually it's telling us. 488 00:32:07,220 --> 00:32:09,440 It tells us actually a lot. 489 00:32:09,440 --> 00:32:24,590 Suppose t is diagonal in some basis u1 up to un. 490 00:32:27,620 --> 00:32:31,460 So what does it mean for it to be diagonal? 491 00:32:31,460 --> 00:32:34,790 Well, you may remember all these definitions 492 00:32:34,790 --> 00:32:37,580 we had about matrix action. 493 00:32:37,580 --> 00:32:42,700 If T acting on a ui is supposed to be 494 00:32:42,700 --> 00:32:49,970 Tki uk in some basis sum over k. 495 00:32:49,970 --> 00:32:54,070 You act on ui, and you get a lot of u's. 496 00:32:54,070 --> 00:32:57,610 And these are the matrix elements of the operator. 497 00:32:57,610 --> 00:33:01,150 Now the fact that it's diagonalizable 498 00:33:01,150 --> 00:33:05,010 means that in some basis, the u basis, this is diagonal. 499 00:33:05,010 --> 00:33:10,465 So ki in this sum only happens to work out 500 00:33:10,465 --> 00:33:12,640 when k is equal to i. 501 00:33:12,640 --> 00:33:16,440 And that's one number and you get back to the vector ui. 502 00:33:16,440 --> 00:33:19,860 So if it's diagonal in this basis, 503 00:33:19,860 --> 00:33:28,060 you have the T on u1 is lambda a number times u1 T on u2 504 00:33:28,060 --> 00:33:31,140 is lambda 2 in u2. 505 00:33:31,140 --> 00:33:35,883 And Tun equal lambda n un. 506 00:33:38,990 --> 00:33:43,740 So what you learn is that this basis vector-- 507 00:33:43,740 --> 00:33:46,010 so you learn something that maybe you 508 00:33:46,010 --> 00:33:47,910 thought it's tautological. 509 00:33:47,910 --> 00:33:49,350 It's not tautological. 510 00:33:49,350 --> 00:33:53,210 You learn that if you have a set of basis vectors 511 00:33:53,210 --> 00:33:56,140 in which the operator is diagonal, 512 00:33:56,140 --> 00:34:02,450 these basis vectors are eigenvectors of the operator. 513 00:34:02,450 --> 00:34:06,510 And then you learn something that is quite important, 514 00:34:06,510 --> 00:34:12,100 that an operator is diagonalizable if, 515 00:34:12,100 --> 00:34:16,179 and only if, it has a set of eigenvectors 516 00:34:16,179 --> 00:34:19,230 that span the space. 517 00:34:19,230 --> 00:34:21,830 So the statement is very important. 518 00:34:21,830 --> 00:34:29,830 An operator T is that diagonalizable 519 00:34:29,830 --> 00:34:43,717 if it has a set of eigenvectors that span the space. 520 00:34:43,717 --> 00:34:50,980 Span V. If and only if. 521 00:34:50,980 --> 00:34:53,210 If this double f. 522 00:34:53,210 --> 00:34:56,770 If and only if. 523 00:34:56,770 --> 00:35:00,490 So here it's diagonalizable, and we have a basis, 524 00:35:00,490 --> 00:35:03,720 and it has a set of these are eigenvectors. 525 00:35:03,720 --> 00:35:06,000 So diagonalizable realizable really 526 00:35:06,000 --> 00:35:08,560 means that it has a set of eigenvectors 527 00:35:08,560 --> 00:35:11,370 that span the space. 528 00:35:11,370 --> 00:35:14,590 On the other hand, if you have the set of eigenvectors that 529 00:35:14,590 --> 00:35:19,520 span the space, you have a set of u's that satisfy this, 530 00:35:19,520 --> 00:35:23,250 and then you read that, oh yeah, this matrix is diagonal, 531 00:35:23,250 --> 00:35:26,220 so it's diagonalizable. 532 00:35:26,220 --> 00:35:30,050 So a simple statement, but an important one, 533 00:35:30,050 --> 00:35:35,600 because there are examples of matrices that immediately you 534 00:35:35,600 --> 00:35:39,330 know you're never going to succeed to diagonalize. 535 00:35:39,330 --> 00:35:42,648 So here is one matrix, 0 0 1 0. 536 00:35:46,060 --> 00:35:50,450 This matrix has eigenvalues, so you 537 00:35:50,450 --> 00:35:54,560 do the characteristic equation lambda squared equals 0. 538 00:35:54,560 --> 00:35:58,790 So the only eigenvalue is lambda equals 0. 539 00:35:58,790 --> 00:36:01,350 And let's see how many eigenvectors 540 00:36:01,350 --> 00:36:04,050 you would have for lambda equals 0. 541 00:36:04,050 --> 00:36:09,320 Well, you would have if this is T, T on some vector 542 00:36:09,320 --> 00:36:13,560 a b must be equal to 0. 543 00:36:13,560 --> 00:36:25,750 So this is 0 1 0 0 on a b, which is b and 0, must be zero. 544 00:36:25,750 --> 00:36:28,990 So b is equal to 0. 545 00:36:28,990 --> 00:36:33,090 So the only eigenvector here-- I'll 546 00:36:33,090 --> 00:36:36,190 just write it here and then move to the other side. 547 00:36:36,190 --> 00:36:38,770 The only eigenvector for lambda equals 0, 548 00:36:38,770 --> 00:36:41,510 the only eigenvector is with b equals 0. 549 00:36:41,510 --> 00:36:43,610 So it's 1 0. 550 00:36:43,610 --> 00:36:45,610 One eigenvector only. 551 00:36:45,610 --> 00:36:47,620 No more eigenvectors. 552 00:36:47,620 --> 00:36:51,780 By the theorem, or by this claim, 553 00:36:51,780 --> 00:36:55,305 you know it's a two dimensional vector space you just 554 00:36:55,305 --> 00:36:57,310 can't diagonalize this matrix. 555 00:36:57,310 --> 00:36:58,560 It's impossible. 556 00:36:58,560 --> 00:37:01,700 Can't be done. 557 00:37:01,700 --> 00:37:05,760 OK, a couple more things that I wish to 558 00:37:05,760 --> 00:37:08,550 say about this process of diagonalization. 559 00:37:12,300 --> 00:37:17,820 Well, the statement that an operator is diagonal 560 00:37:17,820 --> 00:37:23,370 is a statement about the existence of some basis. 561 00:37:23,370 --> 00:37:26,820 Now you can try to figure out what that basis is, 562 00:37:26,820 --> 00:37:30,950 so typically what is the problem that you face? 563 00:37:30,950 --> 00:37:35,264 Typically you have a vector spaces V. Sorry? 564 00:37:35,264 --> 00:37:36,430 AUDIENCE: I have a question. 565 00:37:36,430 --> 00:37:37,055 PROFESSOR: Yes? 566 00:37:39,555 --> 00:37:43,190 If you had an infinite dimensional space 567 00:37:43,190 --> 00:37:49,340 and you had an operator whose eigenvectors do not 568 00:37:49,340 --> 00:37:53,175 span the space, can it still have eigenvectors, 569 00:37:53,175 --> 00:37:56,090 or does it not have any then? 570 00:37:56,090 --> 00:37:56,860 PROFESSOR: No. 571 00:37:56,860 --> 00:37:58,930 You said it has some eigenvectors, 572 00:37:58,930 --> 00:38:00,394 but they don't span the space. 573 00:38:00,394 --> 00:38:01,810 So it does have some eigenvectors. 574 00:38:01,810 --> 00:38:03,899 AUDIENCE: So my question is was what I just 575 00:38:03,899 --> 00:38:08,060 said a logical contradiction in an infinite dimensional space? 576 00:38:08,060 --> 00:38:10,450 PROFESSOR: To have just some eigenvectors? 577 00:38:10,450 --> 00:38:10,950 I think-- 578 00:38:10,950 --> 00:38:12,680 PROFESSOR: I'm looking more specifically 579 00:38:12,680 --> 00:38:14,825 at a dagger for instance. 580 00:38:14,825 --> 00:38:15,450 PROFESSOR: Yes. 581 00:38:15,450 --> 00:38:17,160 AUDIENCE: In the harmonic oscillator, 582 00:38:17,160 --> 00:38:21,660 you I think mentioned at some point that it does not have-- 583 00:38:21,660 --> 00:38:23,980 PROFESSOR: So the fact that you can' 584 00:38:23,980 --> 00:38:27,060 diagonalize this thing already implies 585 00:38:27,060 --> 00:38:29,450 that it's even worse in higher dimensions. 586 00:38:29,450 --> 00:38:32,720 So some operator may be pretty nice, 587 00:38:32,720 --> 00:38:36,370 and you might still be able to diagonalize it, 588 00:38:36,370 --> 00:38:39,154 so you're going to lack eigenvectors in general. 589 00:38:39,154 --> 00:38:40,570 You're going to lack lots of them. 590 00:38:40,570 --> 00:38:42,730 And there are going to be blocks of Jordan. 591 00:38:42,730 --> 00:38:46,420 Blocks are called things that are above the diagonal, things 592 00:38:46,420 --> 00:38:47,750 that you can't do much about. 593 00:38:51,610 --> 00:38:56,530 Let me then think concretely now that you have a vector space, 594 00:38:56,530 --> 00:39:03,380 and you've chosen some basis v1 vn. 595 00:39:03,380 --> 00:39:08,920 And then you look at this operator T, 596 00:39:08,920 --> 00:39:12,540 and of course, you chose an arbitrary basis. 597 00:39:12,540 --> 00:39:16,000 There's no reason why its matrix representation 598 00:39:16,000 --> 00:39:17,210 would be diagonal. 599 00:39:17,210 --> 00:39:23,540 So T on the basis v-- Tij. 600 00:39:23,540 --> 00:39:27,440 Sometimes to be very explicit we write Tij like that-- 601 00:39:27,440 --> 00:39:30,700 is not diagonal. 602 00:39:30,700 --> 00:39:34,310 Now if it's not diagonal, the question 603 00:39:34,310 --> 00:39:38,390 is whether you can find a basis where it is diagonal. 604 00:39:38,390 --> 00:39:42,050 And then you try, of course, changing basis. 605 00:39:42,050 --> 00:39:44,090 And you change basis-- you've discussed 606 00:39:44,090 --> 00:39:46,710 that in the homework-- with a linear operator. 607 00:39:46,710 --> 00:39:49,800 So you use a linear operator to produce 608 00:39:49,800 --> 00:39:53,490 another basis, an invertible in your operator. 609 00:39:53,490 --> 00:39:58,670 So that you get these vectors uk being equal to some operator 610 00:39:58,670 --> 00:40:01,280 A times vk. 611 00:40:01,280 --> 00:40:06,560 So this is going to be the u1's up to un's are 612 00:40:06,560 --> 00:40:09,690 going to be another basis. 613 00:40:09,690 --> 00:40:12,350 The n vector here is the operator 614 00:40:12,350 --> 00:40:15,680 acting with the n vector on this thing. 615 00:40:15,680 --> 00:40:19,340 And then you prove, in the homework, a relationship 616 00:40:19,340 --> 00:40:27,010 between these matrix elements of T in the new basis, 617 00:40:27,010 --> 00:40:28,310 in the u basis. 618 00:40:31,300 --> 00:40:35,500 And the matrix elements of T in the v basis. 619 00:40:39,830 --> 00:40:42,650 You have a relationship like this, 620 00:40:42,650 --> 00:40:50,720 or you have more explicitly Tij in the basis u 621 00:40:50,720 --> 00:41:03,602 is equal to A minus 1 ik Tkp of v Apj. 622 00:41:09,370 --> 00:41:10,955 So this is what happens. 623 00:41:14,040 --> 00:41:16,550 This is the new operator in this basis. 624 00:41:16,550 --> 00:41:19,020 And typically what you're trying to do 625 00:41:19,020 --> 00:41:24,020 is find this matrix A that makes this thing 626 00:41:24,020 --> 00:41:26,090 into a diagonal matrix. 627 00:41:26,090 --> 00:41:30,020 Because we say in the u basis the operator is diagonal. 628 00:41:36,560 --> 00:41:41,740 I want to emphasize that there's a couple of ways in which you 629 00:41:41,740 --> 00:41:44,230 can think of diagonalization. 630 00:41:44,230 --> 00:41:47,490 Sort of a passive and an active way. 631 00:41:47,490 --> 00:41:51,050 You can imagine the operator, and you 632 00:41:51,050 --> 00:41:54,610 say look, this operator I just need 633 00:41:54,610 --> 00:41:58,860 to find some basis in which it is diagonal. 634 00:41:58,860 --> 00:42:02,280 So I'm looking for a basis. 635 00:42:02,280 --> 00:42:06,070 The other way of thinking of this operator 636 00:42:06,070 --> 00:42:17,940 is to think that A minus 1 TA is another operator, 637 00:42:17,940 --> 00:42:22,930 and it's diagonal in original basis. 638 00:42:22,930 --> 00:42:29,020 So it might have seem funny to you, but let's stop again 639 00:42:29,020 --> 00:42:31,170 and say this again. 640 00:42:31,170 --> 00:42:34,620 You have an operator, and the question of diagonalization 641 00:42:34,620 --> 00:42:39,540 is whether there is some basis in which it looks diagonal, 642 00:42:39,540 --> 00:42:41,060 its matrix is diagonal. 643 00:42:41,060 --> 00:42:44,650 But the equivalent question is whether there 644 00:42:44,650 --> 00:42:48,690 is an operator A such that this is 645 00:42:48,690 --> 00:42:52,800 diagonal in the original basis. 646 00:42:52,800 --> 00:42:57,680 To make sure that you see that, consider the following. 647 00:42:57,680 --> 00:43:03,170 So this is diagonal in the original basis. 648 00:43:08,120 --> 00:43:16,240 So in order to see that, think of Tui is equal to lambda i ui. 649 00:43:16,240 --> 00:43:19,010 We know that the u's are supposed 650 00:43:19,010 --> 00:43:23,350 to be this basis of eigenvectors where the matrix is diagonal, 651 00:43:23,350 --> 00:43:25,210 so here you got it. 652 00:43:25,210 --> 00:43:27,200 Here the i not summed. 653 00:43:30,160 --> 00:43:31,820 It's pretty important. 654 00:43:31,820 --> 00:43:34,300 There's a problem with this eigenvalue notation. 655 00:43:34,300 --> 00:43:36,580 I don't know how to do it better. 656 00:43:36,580 --> 00:43:40,340 If you have several eigenvalues, you want to write this, 657 00:43:40,340 --> 00:43:44,950 but you don't want this to think that you're acting on u1 658 00:43:44,950 --> 00:43:46,910 and you get lambda 1 u1. 659 00:43:46,910 --> 00:43:49,320 Not the sum right here. 660 00:43:49,320 --> 00:43:55,640 OK, but they ui is equal to A on vi. 661 00:44:00,130 --> 00:44:05,560 So therefore this is lambda i A on vi. 662 00:44:09,200 --> 00:44:12,080 And then you'll act with A minus 1. 663 00:44:12,080 --> 00:44:16,000 Act with A minus 1 from the left with the operator. 664 00:44:16,000 --> 00:44:25,610 So you get A minus 1 TA vi is equal to lambda i vi. 665 00:44:28,540 --> 00:44:30,090 So what do you see? 666 00:44:30,090 --> 00:44:33,270 You see an operator that is actually 667 00:44:33,270 --> 00:44:37,130 diagonal in the v basis. 668 00:44:37,130 --> 00:44:41,730 So this operator is diagonal in the original basis. 669 00:44:41,730 --> 00:44:43,950 That's another way of thinking of the process 670 00:44:43,950 --> 00:44:46,160 of diagonalization. 671 00:44:46,160 --> 00:44:53,700 There's one last remark, which is that the columns of A 672 00:44:53,700 --> 00:44:57,770 are the eigenvectors, in fact. 673 00:44:57,770 --> 00:45:02,343 Columns of A are the eigenvectors. 674 00:45:06,580 --> 00:45:09,670 Well, how do you see that? 675 00:45:09,670 --> 00:45:11,700 It's really very simple. 676 00:45:11,700 --> 00:45:14,030 You can convince yourself in many ways, 677 00:45:14,030 --> 00:45:19,760 but the uk are the eigenvectors. 678 00:45:19,760 --> 00:45:22,480 But what are uk's? 679 00:45:22,480 --> 00:45:23,470 I have it somewhere. 680 00:45:23,470 --> 00:45:24,290 There it is. 681 00:45:24,290 --> 00:45:25,510 A on vk. 682 00:45:31,800 --> 00:45:36,740 And A on vk is this matrix representation 683 00:45:36,740 --> 00:45:45,528 is sum over i Aik vi. 684 00:45:50,800 --> 00:45:54,320 So now if this is the original basis, 685 00:45:54,320 --> 00:45:57,560 the vi's are your original basis, 686 00:45:57,560 --> 00:46:04,240 then you have the following, that the vi's 687 00:46:04,240 --> 00:46:09,530 can be thought as the basis vectors and represented 688 00:46:09,530 --> 00:46:16,740 by columns with a 1 in the ith entry. 689 00:46:16,740 --> 00:46:24,590 So this equation is saying nothing more, or nothing less 690 00:46:24,590 --> 00:46:32,300 than uk, in terms of matrices or columns, 691 00:46:32,300 --> 00:46:49,230 is equal to A1k v1 plus Ank vn, which is just A1k A2k Ank. 692 00:46:51,910 --> 00:46:56,590 Because vi is the ith basis vector. 693 00:46:56,590 --> 00:47:01,680 So 1 0's only in the ith position. 694 00:47:07,110 --> 00:47:12,860 So these are the eigenvectors. 695 00:47:12,860 --> 00:47:16,070 And they're thought as linear combinations of the vi's. 696 00:47:16,070 --> 00:47:19,180 The vi's are the original basis vectors. 697 00:47:19,180 --> 00:47:23,950 So the eigenvectors are these numbers. 698 00:47:23,950 --> 00:47:33,130 OK, we've talked about diagonlization, 699 00:47:33,130 --> 00:47:38,890 but then there's a term that is more crucial for our operators 700 00:47:38,890 --> 00:47:40,870 that we're interested in. 701 00:47:40,870 --> 00:47:43,170 We're talking about Hermitian operators. 702 00:47:43,170 --> 00:47:46,420 So the term that is going to be important for us 703 00:47:46,420 --> 00:47:53,020 is unitarily diagonalizable. 704 00:47:58,240 --> 00:48:04,180 What is a unitarily diagonalizable operator? 705 00:48:04,180 --> 00:48:06,800 Two ways again of thinking about this. 706 00:48:06,800 --> 00:48:10,950 And perhaps the first way is the best. 707 00:48:10,950 --> 00:48:13,030 And I will say it. 708 00:48:13,030 --> 00:48:19,740 A matrix is set to be unitarily diagonalizable 709 00:48:19,740 --> 00:48:24,720 if you have an orthonormal basis of eigenvectors. 710 00:48:27,840 --> 00:48:34,390 Remember diagonalizable meant a basis of eigenvectors. 711 00:48:34,390 --> 00:48:35,470 That's all it means. 712 00:48:35,470 --> 00:48:40,420 Unitarily diagonalizable means orthonormal basis 713 00:48:40,420 --> 00:48:42,190 of eigenvectors. 714 00:48:42,190 --> 00:48:53,845 So T has an orthonormal basis of eigenvectors. 715 00:48:59,750 --> 00:49:03,826 Now that's a very clear statement. 716 00:49:06,860 --> 00:49:09,870 And it's a fabulous thing if you can achieve, 717 00:49:09,870 --> 00:49:14,820 because you basically have broken down the space 718 00:49:14,820 --> 00:49:20,270 into basis spaces, each one of them with a simple thing 719 00:49:20,270 --> 00:49:21,420 before your operators. 720 00:49:21,420 --> 00:49:23,860 And they're orthonormal, so it's the simplest 721 00:49:23,860 --> 00:49:26,510 possible calculational tool. 722 00:49:26,510 --> 00:49:29,710 So it's ideal if you can have this. 723 00:49:29,710 --> 00:49:33,730 Now the way we think of this is that you 724 00:49:33,730 --> 00:49:36,370 start with-- concretely, you start 725 00:49:36,370 --> 00:49:42,735 with a T of some basis v that is an orthonormal basis. 726 00:49:48,340 --> 00:49:52,700 Start with an orthonormal basis, and then 727 00:49:52,700 --> 00:49:57,940 pass to another orthonormal basis u. 728 00:49:57,940 --> 00:50:01,110 So you're going to pass to another orthonormal basis 729 00:50:01,110 --> 00:50:04,390 u with some operator. 730 00:50:04,390 --> 00:50:06,570 But what you have learned is that if you 731 00:50:06,570 --> 00:50:12,600 want to pass from v orthonormal to another basis 732 00:50:12,600 --> 00:50:17,040 u, a vector that is also orthonormal, 733 00:50:17,040 --> 00:50:19,330 the way to go from one to the other 734 00:50:19,330 --> 00:50:21,520 is through a unitary operator. 735 00:50:21,520 --> 00:50:25,560 Only unitary operators pass you from orthonormal to orthonormal 736 00:50:25,560 --> 00:50:27,530 basis. 737 00:50:27,530 --> 00:50:31,860 Therefore really, when you start with your matrix 738 00:50:31,860 --> 00:50:35,320 in an orthonormal basis that is not diagonal, 739 00:50:35,320 --> 00:50:41,600 the only thing you can hope is that T of u 740 00:50:41,600 --> 00:50:52,900 will be equal to sum u dagger, or u minus 1, T of v u. 741 00:50:52,900 --> 00:51:02,500 Remember, for a unitary operator, where u is unitary, 742 00:51:02,500 --> 00:51:05,420 the inverse is the dagger. 743 00:51:05,420 --> 00:51:08,040 So you're doing a unitary transformation, 744 00:51:08,040 --> 00:51:12,580 and you find the matrix that is presumably then diagonal. 745 00:51:12,580 --> 00:51:16,720 So basically, unitarily diagonalizable 746 00:51:16,720 --> 00:51:18,390 is the statement that if you start 747 00:51:18,390 --> 00:51:23,820 with the operator in an arbitrary orthonormal basis, 748 00:51:23,820 --> 00:51:26,840 then there's some unitary operator that takes you 749 00:51:26,840 --> 00:51:31,080 to the privilege basis in which your operator is diagonal, 750 00:51:31,080 --> 00:51:32,900 is still orthonormal. 751 00:51:32,900 --> 00:51:39,430 But maybe in a more simple way, unitarily diagonalizable 752 00:51:39,430 --> 00:51:41,830 is just a statement that you can find 753 00:51:41,830 --> 00:51:46,050 an orthonormal basis of eigenvectors. 754 00:51:46,050 --> 00:51:52,010 Now the main theorem of this subject, 755 00:51:52,010 --> 00:51:54,110 perhaps one of the most important theorems 756 00:51:54,110 --> 00:51:57,540 of linear algebra, is the characterization 757 00:51:57,540 --> 00:52:02,020 of which operators have such a wonderful representation. 758 00:52:07,500 --> 00:52:09,870 What is the most general operator T 759 00:52:09,870 --> 00:52:14,320 that will have an orthonormal basis of eigenvectors? 760 00:52:14,320 --> 00:52:20,390 Now we probably have heard that Hermitian operators do the job. 761 00:52:20,390 --> 00:52:22,720 Hermitian operators have that. 762 00:52:22,720 --> 00:52:24,830 But that's not the most general ones. 763 00:52:24,830 --> 00:52:29,280 And given that you want the complete result, 764 00:52:29,280 --> 00:52:33,400 let's give you the complete result. 765 00:52:33,400 --> 00:52:36,740 The operators that have this wonderful properties 766 00:52:36,740 --> 00:52:40,030 are called normal operators, and they 767 00:52:40,030 --> 00:52:41,930 satisfy the following property. 768 00:52:41,930 --> 00:52:56,400 M is normal if M dagger, the adjoint of it, commutes with M. 769 00:52:56,400 --> 00:52:59,520 So Hermitian operators are normal, 770 00:52:59,520 --> 00:53:04,750 because M dagger is equal to M, and they commute. 771 00:53:04,750 --> 00:53:07,940 Anti Hermitian mission operators are also normal, 772 00:53:07,940 --> 00:53:10,360 because anti Hermitian means that dagger 773 00:53:10,360 --> 00:53:14,920 is equal to minus M, and it still commutes with M. 774 00:53:14,920 --> 00:53:21,180 Unitary operators have U dagger U equal to U U 775 00:53:21,180 --> 00:53:23,260 dagger equals to 1. 776 00:53:23,260 --> 00:53:27,750 So U and U dagger actually commute as well. 777 00:53:27,750 --> 00:53:32,370 So Hermitian, anti Hermitian, unitary, they're 778 00:53:32,370 --> 00:53:33,706 all normal operators. 779 00:53:37,470 --> 00:53:39,590 What do we know about normal operators? 780 00:53:42,190 --> 00:53:45,390 There's one important result about normal operators, 781 00:53:45,390 --> 00:53:46,060 a lemma. 782 00:53:48,970 --> 00:54:07,970 If M is normal and W is an eigenvector, 783 00:54:07,970 --> 00:54:17,110 such that MW is equal to lambda W. 784 00:54:17,110 --> 00:54:22,090 Now normal operators need not have real eigenvalues, 785 00:54:22,090 --> 00:54:25,200 because they include unitary operators. 786 00:54:25,200 --> 00:54:32,042 So here I should write Hermitian, anti Hermitian, 787 00:54:32,042 --> 00:54:35,715 and unitary are normal. 788 00:54:41,050 --> 00:54:45,460 So here is what a normal operator is doing. 789 00:54:45,460 --> 00:54:46,810 You have a normal operator. 790 00:54:46,810 --> 00:54:49,760 It has an eigenvector with some eigenvalue. 791 00:54:49,760 --> 00:54:54,640 Lambda is a complex number in principle. 792 00:54:54,640 --> 00:54:57,190 Then the following result is true, 793 00:54:57,190 --> 00:55:06,300 then M dagger omega is also an eigenvector of M dagger. 794 00:55:06,300 --> 00:55:08,920 And it has eigenvalue lambda star. 795 00:55:17,380 --> 00:55:20,830 Now this is not all that easy to show. 796 00:55:20,830 --> 00:55:23,320 It's a few lines, and it's done in the notes. 797 00:55:23,320 --> 00:55:24,170 I ask you to see. 798 00:55:24,170 --> 00:55:25,585 It's actually very elegant. 799 00:55:28,702 --> 00:55:31,740 What is the usual strategy to prove things 800 00:55:31,740 --> 00:55:37,210 like that Is to say oh, I want to show this is equal to that, 801 00:55:37,210 --> 00:55:40,540 so I want to show that this binds that is 0. 802 00:55:40,540 --> 00:55:42,760 So I have a vector that is zero what 803 00:55:42,760 --> 00:55:45,400 is the easiest way to show that it's 0? 804 00:55:45,400 --> 00:55:47,780 If I can show it's norm is 0. 805 00:55:47,780 --> 00:55:49,900 So that's a typical strategy that you 806 00:55:49,900 --> 00:55:52,350 use to prove equalities. 807 00:55:52,350 --> 00:55:56,430 You say, oh, it's a vector that must be 0 as my equality. 808 00:55:56,430 --> 00:55:57,740 Let's see if it's 0. 809 00:55:57,740 --> 00:56:00,130 Let's find its norm, and you get it. 810 00:56:00,130 --> 00:56:02,280 So that's a result. 811 00:56:02,280 --> 00:56:11,695 So with this stated, we finally have the main result 812 00:56:11,695 --> 00:56:13,740 that we wanted to get to. 813 00:56:16,680 --> 00:56:20,660 And I will be very sketchy on this. 814 00:56:24,700 --> 00:56:28,200 The notes are complete, but I will be sketchy here. 815 00:56:28,200 --> 00:56:30,005 It's called the spectral theorem. 816 00:56:35,770 --> 00:56:48,105 Let M be an operator in a complex vector space. 817 00:56:55,820 --> 00:57:12,420 The vector space has an orthonormal basis 818 00:57:12,420 --> 00:57:31,780 of eigenvectors of M if and only if M is normal. 819 00:57:31,780 --> 00:57:37,370 So the normal operators are it. 820 00:57:37,370 --> 00:57:43,270 You want to have a complete set of orthonormal eigenvectors. 821 00:57:43,270 --> 00:57:46,900 Well, this will only happen if your operator 822 00:57:46,900 --> 00:57:50,580 is normal, end of story. 823 00:57:50,580 --> 00:57:59,200 Now there's two things about this theorem 824 00:57:59,200 --> 00:58:06,650 is to show that if it's diagonalizable, it is normal, 825 00:58:06,650 --> 00:58:11,110 and the other thing is to show, that if it's normal, 826 00:58:11,110 --> 00:58:12,780 it can be diagonalized. 827 00:58:12,780 --> 00:58:17,270 Of course, you can imagine the second one 828 00:58:17,270 --> 00:58:19,620 is harder than the first. 829 00:58:19,620 --> 00:58:23,050 Let me do the first one for a few minutes. 830 00:58:23,050 --> 00:58:28,010 And then say a couple of words about the second. 831 00:58:28,010 --> 00:58:31,660 And you may discuss this in recitation. 832 00:58:31,660 --> 00:58:35,150 It's a little mathematical, but it's all 833 00:58:35,150 --> 00:58:37,350 within the kind of things that we do. 834 00:58:37,350 --> 00:58:40,970 And really it's fairly physical in a sense. 835 00:58:40,970 --> 00:58:43,860 We're accustomed to do such kinds of arguments. 836 00:58:43,860 --> 00:58:54,270 So suppose it's unitarily diagonalizable, 837 00:58:54,270 --> 00:59:02,880 which means that M-- so if you have U dagger, 838 00:59:02,880 --> 00:59:09,490 MU is equal to a diagonal matrix, DM. 839 00:59:09,490 --> 00:59:11,140 I'm talking now matrices. 840 00:59:11,140 --> 00:59:14,550 So these are all matrices, a diagonal matrix. 841 00:59:17,370 --> 00:59:23,130 There's no basis to the notion of a diagonal operator, 842 00:59:23,130 --> 00:59:26,380 because if you have a diagonal operator, 843 00:59:26,380 --> 00:59:29,120 it may not look diagonal in another basis. 844 00:59:29,120 --> 00:59:34,570 Only the identity operator is diagonal in all basis, but not 845 00:59:34,570 --> 00:59:36,330 the typical diagonal operator. 846 00:59:36,330 --> 00:59:40,440 So unitarily diagonalizable, as we said, 847 00:59:40,440 --> 00:59:43,850 you make it-- it's gone somewhere. 848 00:59:43,850 --> 00:59:44,540 Here. 849 00:59:44,540 --> 00:59:46,900 You act with an inverse matrices, 850 00:59:46,900 --> 00:59:48,640 and you get the diagonal matrix. 851 00:59:48,640 --> 01:00:01,190 So from this, you find that M is equal to U DM U dagger 852 01:00:01,190 --> 01:00:05,610 by acting with U on the left and with U dagger from the right, 853 01:00:05,610 --> 01:00:09,830 you solve for M, and it's this. 854 01:00:09,830 --> 01:00:15,670 And then M dagger is the dagger of these things. 855 01:00:15,670 --> 01:00:22,100 So it's U to DM dagger U dagger. 856 01:00:22,100 --> 01:00:26,490 The U's sort of remain the same way, 857 01:00:26,490 --> 01:00:30,100 but the diagonal matrix is not necessarily real, 858 01:00:30,100 --> 01:00:34,200 so you must put the dagger in there. 859 01:00:34,200 --> 01:00:40,540 And now M dagger M. To check that the matrix is normal 860 01:00:40,540 --> 01:00:42,360 that commutator should be 0. 861 01:00:42,360 --> 01:00:46,380 So M dagger M. You do this times that. 862 01:00:46,380 --> 01:00:51,030 You get U DM dagger. 863 01:00:51,030 --> 01:00:53,790 U dagger U. That's one. 864 01:00:53,790 --> 01:00:57,416 DM U dagger. 865 01:00:57,416 --> 01:01:02,110 And M M dagger you multiply the other direction 866 01:01:02,110 --> 01:01:08,640 you get U DM DM dagger U dagger. 867 01:01:08,640 --> 01:01:22,750 So the commutator of M dagger M is equal to U DM dagger DM 868 01:01:22,750 --> 01:01:29,880 minus DM Dm dagger U dagger. 869 01:01:29,880 --> 01:01:34,290 But any two diagonal matrices commute. 870 01:01:34,290 --> 01:01:36,430 They may not be that simple. 871 01:01:36,430 --> 01:01:38,970 Diagonal matrices are not the identity matrices, 872 01:01:38,970 --> 01:01:40,770 but for sure they commute. 873 01:01:40,770 --> 01:01:45,570 You multiply elements along with diagonal so this is 0. 874 01:01:45,570 --> 01:01:53,260 So certainly any unitarily diagonalizable matrix 875 01:01:53,260 --> 01:01:54,970 is normal. 876 01:01:54,970 --> 01:01:56,910 Now the other part of the proof, which 877 01:01:56,910 --> 01:02:03,170 I'm not going to speak about, it's actually quite simple. 878 01:02:03,170 --> 01:02:08,440 And it's based on the fact that any matrix in a complex vector 879 01:02:08,440 --> 01:02:11,500 space has at least one eigenvalue. 880 01:02:11,500 --> 01:02:16,610 So what you do is you pick out that eigenvalue and it's 881 01:02:16,610 --> 01:02:19,110 eigenvector, and change the basis 882 01:02:19,110 --> 01:02:22,510 to use that eigenvector instead of your other vectors. 883 01:02:22,510 --> 01:02:24,730 And then you look at the matrix. 884 01:02:24,730 --> 01:02:27,012 And after you use that eigenvector, 885 01:02:27,012 --> 01:02:31,490 the matrix has a lot of 0's here and a lot of 0's here. 886 01:02:31,490 --> 01:02:35,920 And then the matrix has been reduced in dimension mansion, 887 01:02:35,920 --> 01:02:38,130 and then you go step by step. 888 01:02:38,130 --> 01:02:41,590 So basically, it's the fact that any operator 889 01:02:41,590 --> 01:02:45,190 has at least one eigenvalue and at least one eigenvector. 890 01:02:45,190 --> 01:02:47,150 It allows you to go down. 891 01:02:47,150 --> 01:02:55,530 And normality is analogous to Hermiticity in some sense. 892 01:02:55,530 --> 01:02:59,130 And the statement that you have an eigenvector 893 01:02:59,130 --> 01:03:03,800 generally tells you that this thing is full of 0's, but then 894 01:03:03,800 --> 01:03:06,170 you don't know that there are 0's here. 895 01:03:06,170 --> 01:03:09,990 And either normality or Hermiticity 896 01:03:09,990 --> 01:03:11,830 shows that there are 0's here, and then you 897 01:03:11,830 --> 01:03:13,490 can proceed at lower dimensions. 898 01:03:13,490 --> 01:03:16,580 So you should look at the proof because it 899 01:03:16,580 --> 01:03:22,160 will make clear to you that you understand what's going on. 900 01:03:22,160 --> 01:03:26,480 OK but let's take it for granted now you have these operators 901 01:03:26,480 --> 01:03:32,750 and can be diagonalized. 902 01:03:32,750 --> 01:03:37,787 Then we have the next thing, which 903 01:03:37,787 --> 01:03:39,120 is simultaneous diagonalization. 904 01:03:42,700 --> 01:03:45,550 What is simultaneous diagonalization? 905 01:03:45,550 --> 01:03:47,670 It's an awfully important thing. 906 01:03:47,670 --> 01:03:51,690 So we will now focus on simultaneous diagonalization 907 01:03:51,690 --> 01:03:53,380 of Hermitian operators. 908 01:03:53,380 --> 01:04:02,150 So simultaneous diagonalization of Hermitian ops. 909 01:04:06,210 --> 01:04:11,990 Now as we will emphasize towards the end, 910 01:04:11,990 --> 01:04:16,250 this is perhaps one of the most important ideas 911 01:04:16,250 --> 01:04:18,210 in quantum mechanics. 912 01:04:18,210 --> 01:04:22,540 It's this stuff that allows you to label and understand 913 01:04:22,540 --> 01:04:24,240 your state system. 914 01:04:24,240 --> 01:04:28,000 Basically you need to diagonalize 915 01:04:28,000 --> 01:04:30,340 more than one operator most of the time. 916 01:04:30,340 --> 01:04:33,340 You can say OK, you found the energy eigenstates. 917 01:04:33,340 --> 01:04:34,930 You're done. 918 01:04:34,930 --> 01:04:37,270 But if you find your energy eigenstates 919 01:04:37,270 --> 01:04:39,980 and you think you're done, maybe you 920 01:04:39,980 --> 01:04:44,250 are if you have all these energy eigenstates tabulated. 921 01:04:44,250 --> 01:04:46,390 But if you have a degeneracy, you 922 01:04:46,390 --> 01:04:51,640 have a lot of states that have the same energy. 923 01:04:51,640 --> 01:04:53,241 And what's different about them? 924 01:04:53,241 --> 01:04:54,990 They're certainly different because you've 925 01:04:54,990 --> 01:04:57,770 got several states, but what's different about them? 926 01:04:57,770 --> 01:05:00,600 You may not know, unless you figure out 927 01:05:00,600 --> 01:05:04,010 that they have different physical properties. 928 01:05:04,010 --> 01:05:07,250 If they're different, something must be different about them. 929 01:05:07,250 --> 01:05:09,550 So you need more than one operator, 930 01:05:09,550 --> 01:05:13,220 and your facing the problem of simultaneously diagonalizing 931 01:05:13,220 --> 01:05:17,500 things, because states cannot be characterized just by one 932 01:05:17,500 --> 01:05:19,440 property, one observable. 933 01:05:19,440 --> 01:05:24,160 Would be simple if you could, but life is not that simple. 934 01:05:24,160 --> 01:05:27,520 So you need more than one observable, 935 01:05:27,520 --> 01:05:31,250 and then you ask when can they be 936 01:05:31,250 --> 01:05:34,090 simultaneously diagonalizable. 937 01:05:34,090 --> 01:05:35,710 Well, the statement is clear. 938 01:05:35,710 --> 01:05:40,720 If you have two operators, S and T that 939 01:05:40,720 --> 01:05:46,050 belong to the linear operators in a vector space, 940 01:05:46,050 --> 01:05:49,650 they can be simultaneously diagonalized 941 01:05:49,650 --> 01:05:58,330 if there is a basis for which every basis vector is 942 01:05:58,330 --> 01:06:02,210 eigenstate of this and an eigenstate of that. 943 01:06:02,210 --> 01:06:04,420 Common set of eigenstates. 944 01:06:04,420 --> 01:06:07,330 So they can be simultaneously diagonalized. 945 01:06:07,330 --> 01:06:12,920 Diagonalizable is that there is a basis where this basis is 946 01:06:12,920 --> 01:06:16,410 comprised of the eigenvectors of the operator. 947 01:06:16,410 --> 01:06:19,760 So this time you require more, that that basis 948 01:06:19,760 --> 01:06:24,690 be at the same time a basis set of eigenvectors of this 949 01:06:24,690 --> 01:06:28,320 and a set of eigenvectors of the second one. 950 01:06:28,320 --> 01:06:36,450 So a necessary condition for simultaneous diagonalization 951 01:06:36,450 --> 01:06:38,570 is that they commute. 952 01:06:38,570 --> 01:06:40,760 Why is that? 953 01:06:40,760 --> 01:06:44,240 The fact that two operators commute or they don't commute 954 01:06:44,240 --> 01:06:48,130 is an issue that is basis independent. 955 01:06:48,130 --> 01:06:51,010 If they don't commute, the order gives something 956 01:06:51,010 --> 01:06:54,730 different, and that you can see in every basis. 957 01:06:54,730 --> 01:06:59,850 So if they don't commute and they're simultaneously 958 01:06:59,850 --> 01:07:04,680 diagonalizable, there would be a basis in which both 959 01:07:04,680 --> 01:07:07,300 are diagonal and they still wouldn't commute. 960 01:07:07,300 --> 01:07:09,960 But you know that diagonal matrices commute. 961 01:07:09,960 --> 01:07:12,890 So if two operators don't commute, 962 01:07:12,890 --> 01:07:15,130 they must not commute in any base, 963 01:07:15,130 --> 01:07:18,260 therefore there can't be a basis in which both 964 01:07:18,260 --> 01:07:21,350 are at the same time diagonal. 965 01:07:21,350 --> 01:07:25,500 So you need, for simultaneous diagonalizable, 966 01:07:25,500 --> 01:07:29,420 you need that S and P commute. 967 01:07:29,420 --> 01:07:34,660 Now that may not be enough, because not all operators 968 01:07:34,660 --> 01:07:35,820 can be diagonalized. 969 01:07:35,820 --> 01:07:38,900 So the fact that they commute is necessary, but not everything 970 01:07:38,900 --> 01:07:40,560 can be diagonalizable. 971 01:07:40,560 --> 01:07:44,530 Well, you've learned that every normal operator, 972 01:07:44,530 --> 01:07:47,800 every Hermitian operator is diagonalizable. 973 01:07:47,800 --> 01:07:51,820 And then you got now a claim of something 974 01:07:51,820 --> 01:07:54,160 that could possibly be true. 975 01:07:54,160 --> 01:07:58,590 Is the fact that whenever you have two Hermitian operators, 976 01:07:58,590 --> 01:08:01,330 each one can be diagonalized by themselves. 977 01:08:01,330 --> 01:08:03,230 And they commute. 978 01:08:03,230 --> 01:08:08,030 There is a simultaneous set of eigenvectors of 1 979 01:08:08,030 --> 01:08:11,040 that are eigenvectors of the first and eigenvectors 980 01:08:11,040 --> 01:08:12,240 of the second. 981 01:08:12,240 --> 01:08:25,710 So the statement is that if S and T are commuting 982 01:08:25,710 --> 01:08:34,990 Hermitian operators, they can be simultaneously diagonalized. 983 01:08:49,569 --> 01:08:57,420 So this theorem would be quite easy to show 984 01:08:57,420 --> 01:08:59,294 if there would be no degeneracies, 985 01:08:59,294 --> 01:09:01,680 and that's what we're going to do first. 986 01:09:01,680 --> 01:09:04,510 But then we'll consider the case of degeneracies. 987 01:09:08,430 --> 01:09:13,100 So I'm going to consider the following possibilities. 988 01:09:13,100 --> 01:09:16,700 Perhaps neither one has a degenerate spectrum. 989 01:09:16,700 --> 01:09:18,790 What does it mean a degenerate spectrum? 990 01:09:18,790 --> 01:09:22,229 Same eigenvalue repeated many times. 991 01:09:22,229 --> 01:09:26,120 But that is a wishful thinking situation. 992 01:09:26,120 --> 01:09:28,960 So either both are non degenerate, 993 01:09:28,960 --> 01:09:32,890 either one is non degenerate and the other is degenerate, 994 01:09:32,890 --> 01:09:35,939 or both are degenerate. 995 01:09:35,939 --> 01:09:40,130 And that causes a very interesting complication. 996 01:09:40,130 --> 01:09:43,439 So let's say there's going to be two cases. 997 01:09:43,439 --> 01:09:44,670 It will suffice. 998 01:09:44,670 --> 01:09:46,630 In fact, it seems that there are three, 999 01:09:46,630 --> 01:09:49,510 but two is enough to consider. 1000 01:09:49,510 --> 01:10:01,020 There is no degeneracy in T. So suppose 1001 01:10:01,020 --> 01:10:05,600 one operator has no degeneracy, and let's call 1002 01:10:05,600 --> 01:10:07,990 it T. So that's one possibility. 1003 01:10:07,990 --> 01:10:12,300 And then S may be degenerate, or it may not be degenerate. 1004 01:10:12,300 --> 01:10:19,500 And the second possibility is that both S and T 1005 01:10:19,500 --> 01:10:20,125 are degenerate. 1006 01:10:25,780 --> 01:10:30,150 So I'll take care of case one first. 1007 01:10:30,150 --> 01:10:33,530 And then we'll discuss case two, and that 1008 01:10:33,530 --> 01:10:36,730 will complete our discussion. 1009 01:10:36,730 --> 01:10:42,380 So suppose there's no the degeneracy 1010 01:10:42,380 --> 01:10:46,860 in the spectrum of T. So case one. 1011 01:10:49,450 --> 01:10:52,200 So what does that mean? 1012 01:10:52,200 --> 01:10:56,290 It means that T is non degenerate. 1013 01:10:56,290 --> 01:11:00,280 There's a basis U1 can be diagonalized 1014 01:11:00,280 --> 01:11:10,690 to UM, orthonormal by the spectral theorem. 1015 01:11:10,690 --> 01:11:15,170 And there's eigenvectors T U-- these are eigenvectors. 1016 01:11:15,170 --> 01:11:17,480 Lambda I Ui. 1017 01:11:17,480 --> 01:11:22,010 And lambda I is different to lambda j 1018 01:11:22,010 --> 01:11:25,606 for i different from j. 1019 01:11:25,606 --> 01:11:30,540 So all the eigenvalues, again, it's not summed here. 1020 01:11:30,540 --> 01:11:36,300 All the eigenvalues are different. 1021 01:11:36,300 --> 01:11:39,640 So what do we have? 1022 01:11:39,640 --> 01:11:42,490 Well, each of those eigenvectors, 1023 01:11:42,490 --> 01:11:46,630 each of the Ui's that are eigenvectors, 1024 01:11:46,630 --> 01:11:49,905 generate invariant subspaces. 1025 01:11:49,905 --> 01:11:52,570 There are T invariant subspaces. 1026 01:11:52,570 --> 01:11:57,580 So each one, each vector U1 you can 1027 01:11:57,580 --> 01:12:00,020 imagine multiplying by all possible numbers, 1028 01:12:00,020 --> 01:12:01,330 positive and negative. 1029 01:12:01,330 --> 01:12:04,910 And that's an invariant one dimensional subspace, 1030 01:12:04,910 --> 01:12:08,700 because if you act with T, it's a T invariant space, 1031 01:12:08,700 --> 01:12:12,760 you get the number times a vector there. 1032 01:12:12,760 --> 01:12:15,930 So the question that you must ask now 1033 01:12:15,930 --> 01:12:20,770 is you want to know if these are simultaneous eigenvectors. 1034 01:12:20,770 --> 01:12:26,350 So you want to figure out what about S. 1035 01:12:26,350 --> 01:12:30,010 How does S work with this thing? 1036 01:12:30,010 --> 01:12:34,640 So you can act with S from the left. 1037 01:12:34,640 --> 01:12:43,340 So you get STUi is equal to lambda i SUi. 1038 01:12:47,081 --> 01:12:57,960 So here each Ui generates an invariant subspace Ui T 1039 01:12:57,960 --> 01:12:58,460 invariant. 1040 01:13:02,540 --> 01:13:13,796 But S and T commute, so you have T SUi is equal to lambda i SUi. 1041 01:13:13,796 --> 01:13:16,170 And look at that equation again. 1042 01:13:16,170 --> 01:13:20,160 This says that this vector belongs 1043 01:13:20,160 --> 01:13:31,370 to the invariant subspace Ui, because it satisfies exactly 1044 01:13:31,370 --> 01:13:36,610 the property that T acting on it is equal to lambda i Ui. 1045 01:13:36,610 --> 01:13:40,560 And it couldn't belong to any other of the subspaces, 1046 01:13:40,560 --> 01:13:44,700 because all the eigenvalues are different. 1047 01:13:44,700 --> 01:13:53,180 So spaces that are in Ui are the spaces-- vectors that are in Ui 1048 01:13:53,180 --> 01:13:56,700 are precisely all those vectors that are left invariant 1049 01:13:56,700 --> 01:13:59,800 by the action of T. They're scaled only. 1050 01:13:59,800 --> 01:14:03,590 So this vector is also in Ui. 1051 01:14:03,590 --> 01:14:07,480 If this vector is in Ui, SUi must 1052 01:14:07,480 --> 01:14:12,640 be some number Wi times Ui. 1053 01:14:12,640 --> 01:14:18,940 And therefore you've shown that Ui is also an eigenvector of S, 1054 01:14:18,940 --> 01:14:22,836 possibly with a different eigenvalue of course. 1055 01:14:22,836 --> 01:14:24,335 Because the only thing that you know 1056 01:14:24,335 --> 01:14:26,340 is that SUi is in this space. 1057 01:14:26,340 --> 01:14:29,250 You don't know how big it is. 1058 01:14:29,250 --> 01:14:34,850 So then you've shown that, indeed, these Ui's 1059 01:14:34,850 --> 01:14:45,280 that were eigenstates of T are also eigenstates of S. 1060 01:14:45,280 --> 01:14:47,910 And therefore that's the statement 1061 01:14:47,910 --> 01:14:51,290 of simultaneously diagonalizable. 1062 01:14:51,290 --> 01:14:55,340 They have the common set of eigenvectors. 1063 01:14:55,340 --> 01:14:58,770 So that's this part. 1064 01:14:58,770 --> 01:15:04,020 And it's relatively straight forward. 1065 01:15:04,020 --> 01:15:07,380 Now we have to do case two. 1066 01:15:07,380 --> 01:15:09,280 Case two is the interesting one. 1067 01:15:12,540 --> 01:15:16,240 This time you're going to have degeneracy. 1068 01:15:16,240 --> 01:15:19,410 We have to have a notation that is good for degeneracy. 1069 01:15:19,410 --> 01:15:28,610 So if S is degeneracies, has degeneracies, 1070 01:15:28,610 --> 01:15:31,820 what happens with this operator? 1071 01:15:31,820 --> 01:15:38,050 It will have-- remember, a degenerate operator 1072 01:15:38,050 --> 01:15:43,440 has eigenstates that form higher than one dimensional spaces. 1073 01:15:43,440 --> 01:15:45,840 If you have different eigenvalues, 1074 01:15:45,840 --> 01:15:52,360 each one generates a one dimensional operator invariant 1075 01:15:52,360 --> 01:15:53,670 subspace. 1076 01:15:53,670 --> 01:15:55,480 But if you have degeneracies, there 1077 01:15:55,480 --> 01:15:58,960 are operators-- there are spaces of higher dimensions that 1078 01:15:58,960 --> 01:16:00,330 are left invariant. 1079 01:16:00,330 --> 01:16:09,330 So for example, let Uk denote the S invariant 1080 01:16:09,330 --> 01:16:16,680 subspace of some dimension Dk, which 1081 01:16:16,680 --> 01:16:18,790 is greater or equal than 1. 1082 01:16:22,190 --> 01:16:29,160 I will go here first. 1083 01:16:29,160 --> 01:16:39,380 We're going to define Uk to be the set of all vectors 1084 01:16:39,380 --> 01:16:51,510 so that SU is equal to lambda k U. 1085 01:16:51,510 --> 01:16:59,150 And this will have dimension of Uk is going to be Dk. 1086 01:16:59,150 --> 01:17:01,570 So look what's happening. 1087 01:17:01,570 --> 01:17:05,240 Basically the fact is that for some eigenvalues, 1088 01:17:05,240 --> 01:17:09,870 say the kth eigenvalue, you just get several eigenvectors. 1089 01:17:09,870 --> 01:17:13,510 So if you get several eigenvectors not just scaled 1090 01:17:13,510 --> 01:17:15,690 off each other, these eigenvectors 1091 01:17:15,690 --> 01:17:19,200 correspond to that eigenvalue span of space. 1092 01:17:19,200 --> 01:17:21,750 It's a degenerate subspace. 1093 01:17:21,750 --> 01:17:25,140 So you must imagine that as having 1094 01:17:25,140 --> 01:17:29,280 a subspace of some dimensionality with some basis 1095 01:17:29,280 --> 01:17:32,840 vectors that span this thing. 1096 01:17:32,840 --> 01:17:35,460 And they all have the same eigenvector. 1097 01:17:35,460 --> 01:17:38,940 Now you should really have visualized 1098 01:17:38,940 --> 01:17:40,800 this in a simple way. 1099 01:17:40,800 --> 01:17:45,520 You have this subspace like a cone or something like that, 1100 01:17:45,520 --> 01:17:50,250 in which every vector is an eigenvector. 1101 01:17:50,250 --> 01:17:54,410 So every vector, when it's acted by S is just scaled up. 1102 01:17:54,410 --> 01:17:57,720 And all of them are scaled by the same amount. 1103 01:17:57,720 --> 01:18:01,320 That is what this statement says. 1104 01:18:01,320 --> 01:18:12,310 And corresponding to this thing, you have a basis of vectors, 1105 01:18:12,310 --> 01:18:14,790 of these eigenvectors, and we'll call 1106 01:18:14,790 --> 01:18:28,700 them Uk1, the first one, the second, up to U Dk1, 1107 01:18:28,700 --> 01:18:31,280 because it's a subspace that we say 1108 01:18:31,280 --> 01:18:35,270 it has the dimensionality Dk. 1109 01:18:35,270 --> 01:18:37,540 So look at this thing. 1110 01:18:37,540 --> 01:18:39,240 Somebody tells you there's an operator. 1111 01:18:39,240 --> 01:18:40,810 It has degenerate spectrum. 1112 01:18:40,810 --> 01:18:46,120 You should start imagining all kind of invariant subspaces 1113 01:18:46,120 --> 01:18:48,540 of some dimensionality. 1114 01:18:48,540 --> 01:18:51,970 If it has degeneracy, it's a degeneracy each time 1115 01:18:51,970 --> 01:18:54,870 the Dk is greater than 1, because if it's 1116 01:18:54,870 --> 01:18:58,740 one dimensional, it's just one basis vector one 1117 01:18:58,740 --> 01:19:01,250 eigenvector, end of the story. 1118 01:19:01,250 --> 01:19:04,250 Now this thing, by the spectral theorem, 1119 01:19:04,250 --> 01:19:06,490 this is an orthonormal basis. 1120 01:19:06,490 --> 01:19:10,010 There's no problem, when you have a degenerate subspace, 1121 01:19:10,010 --> 01:19:11,810 to find an orthonormal basis. 1122 01:19:11,810 --> 01:19:17,080 The theorem guarantees it, so these are all orthonormal. 1123 01:19:17,080 --> 01:19:20,300 So at the end of the day, you have a decomposition 1124 01:19:20,300 --> 01:19:28,470 of the vector space, V as U1 plus U2 plus maybe up to UM. 1125 01:19:31,660 --> 01:19:36,330 And all of these vector spaces like U's here, they 1126 01:19:36,330 --> 01:19:39,340 may have some with just no degeneracy, and some 1127 01:19:39,340 --> 01:19:42,990 with degeneracy 2, degeneracy 3, degeneracy 4. 1128 01:19:42,990 --> 01:19:44,700 I don't know how much degeneracy, 1129 01:19:44,700 --> 01:19:48,420 but they might have different degeneracy. 1130 01:19:48,420 --> 01:19:53,350 Now what do we say next? 1131 01:19:53,350 --> 01:19:58,870 Well, the fact that S is a Hermitian operator 1132 01:19:58,870 --> 01:20:01,210 says it just can be diagonalized, 1133 01:20:01,210 --> 01:20:03,550 and we're can find all these spaces, 1134 01:20:03,550 --> 01:20:06,020 and the basis for the whole thing. 1135 01:20:06,020 --> 01:20:15,110 So the basis would look U1 of the first base up to U d1 1136 01:20:15,110 --> 01:20:17,430 of the first base. 1137 01:20:17,430 --> 01:20:19,659 These are the basis vectors of the first 1138 01:20:19,659 --> 01:20:21,200 plus the basis vectors of the second. 1139 01:20:21,200 --> 01:20:28,155 All the basis vectors U1 up to Udm of the mth space. 1140 01:20:31,460 --> 01:20:32,970 All this is the list. 1141 01:20:32,970 --> 01:20:43,650 This is the basis of V. So I've listed 1142 01:20:43,650 --> 01:20:48,130 the basis of V, which a basis for U1, all these vectors. 1143 01:20:48,130 --> 01:20:50,540 U2, all of this. 1144 01:20:50,540 --> 01:20:52,720 So you see, we're not calculating anything. 1145 01:20:52,720 --> 01:20:55,740 We're just trying to understand the picture. 1146 01:20:55,740 --> 01:21:03,940 And why is this operator, S, diagonal in this basis? 1147 01:21:03,940 --> 01:21:04,720 It's clear. 1148 01:21:04,720 --> 01:21:11,940 Because every vector here, every vector is an eigenvector of S. 1149 01:21:11,940 --> 01:21:14,940 So when you act with S on any vector, 1150 01:21:14,940 --> 01:21:17,270 you get that vector times a number. 1151 01:21:17,270 --> 01:21:20,190 But that vector is orthogonal to all the rest. 1152 01:21:20,190 --> 01:21:25,270 So when you have some U and S and another U, 1153 01:21:25,270 --> 01:21:29,430 this gives you a vector proportional to U. 1154 01:21:29,430 --> 01:21:31,030 And this is another vector. 1155 01:21:31,030 --> 01:21:34,900 The matrix element is 0, because they're all orthogonal. 1156 01:21:34,900 --> 01:21:39,190 So it should be obvious why this list produces something 1157 01:21:39,190 --> 01:21:43,130 that is completely orthogonal-- a diagonal matrix. 1158 01:21:43,130 --> 01:21:48,890 So S, in this basis, looks like the diagonal matrix 1159 01:21:48,890 --> 01:22:04,700 in which you have lambda 1 d1 times up to lambda m dm times. 1160 01:22:07,260 --> 01:22:11,900 Now I'll have to go until 2:00 to get the punchline. 1161 01:22:11,900 --> 01:22:15,340 I apologize, but we can't stop right now. 1162 01:22:15,340 --> 01:22:21,010 We're almost there, believe it or not. 1163 01:22:21,010 --> 01:22:22,705 Two more things. 1164 01:22:26,720 --> 01:22:33,890 This basis is good, but actually another basis 1165 01:22:33,890 --> 01:22:35,570 would also be good. 1166 01:22:35,570 --> 01:22:41,450 I'll write this other basis would be a V1 acting on the U1 1167 01:22:41,450 --> 01:22:50,960 up to V1 acting on that U1 up to a Vm acting on this U1 up to Vm 1168 01:22:50,960 --> 01:22:53,040 acting on that U1. 1169 01:22:53,040 --> 01:22:55,240 This is m. 1170 01:22:55,240 --> 01:22:55,920 m. 1171 01:22:55,920 --> 01:22:58,420 This is dm. 1172 01:22:58,420 --> 01:23:00,210 And here it's not U1. 1173 01:23:00,210 --> 01:23:00,830 It's Ud1. 1174 01:23:04,280 --> 01:23:08,660 You see, in the first collection of vectors, 1175 01:23:08,660 --> 01:23:15,420 I act with an operator V1 up to here with an operator Vm. 1176 01:23:15,420 --> 01:23:33,730 All of them with Vk being a unitary operator in Uk. 1177 01:23:33,730 --> 01:23:38,660 In every subspace, there are unitary operators. 1178 01:23:38,660 --> 01:23:41,930 So you can have these bases and act 1179 01:23:41,930 --> 01:23:46,140 with a unitary operator of the space U1 here. 1180 01:23:46,140 --> 01:23:48,350 A unitary operator with a space U2 here. 1181 01:23:48,350 --> 01:23:52,980 A unitary operator of the space Un here. 1182 01:23:52,980 --> 01:23:55,720 Hope you're following. 1183 01:23:55,720 --> 01:23:59,230 And what happens if this operator is unitary, 1184 01:23:59,230 --> 01:24:02,320 this is still an orthonormal basis in U1. 1185 01:24:02,320 --> 01:24:07,680 These are still orthonormal basis in Um. 1186 01:24:07,680 --> 01:24:10,870 And therefore this is an orthonormal basis 1187 01:24:10,870 --> 01:24:14,810 for the whole thing, because anyway those different spaces 1188 01:24:14,810 --> 01:24:17,320 are orthogonal to each other. 1189 01:24:17,320 --> 01:24:19,270 It's an orthogonal decomposition. 1190 01:24:19,270 --> 01:24:21,700 Everything is orthogonal to everything. 1191 01:24:21,700 --> 01:24:26,817 So this basis would be equally good to represent the operator. 1192 01:24:26,817 --> 01:24:27,316 Yes? 1193 01:24:27,316 --> 01:24:31,150 AUDIENCE: [INAUDIBLE] arbitrary unitary operators? 1194 01:24:31,150 --> 01:24:34,710 PROFESSOR: Arbitrary unitary operators at this moment. 1195 01:24:34,710 --> 01:24:37,170 Arbitrary. 1196 01:24:37,170 --> 01:24:42,300 So here comes the catch as to the main property 1197 01:24:42,300 --> 01:24:53,220 that now you want to establish is that the spaces Uk are also 1198 01:24:53,220 --> 01:24:54,043 T invariant. 1199 01:24:56,990 --> 01:24:59,870 You see, the spaces Uk were defined 1200 01:24:59,870 --> 01:25:03,370 to be S invariant subspaces. 1201 01:25:03,370 --> 01:25:07,950 And now the main important thing is that they are also T 1202 01:25:07,950 --> 01:25:12,660 invariant because they commute with that. 1203 01:25:12,660 --> 01:25:15,310 So let's see why that is the case. 1204 01:25:15,310 --> 01:25:18,950 Suppose U belongs to Uk. 1205 01:25:21,710 --> 01:25:26,970 And then let's look at the vector-- examine the vector Tu. 1206 01:25:26,970 --> 01:25:29,270 What happens to Tu? 1207 01:25:29,270 --> 01:25:35,740 Well, you want to act on S on Tu to understand it. 1208 01:25:35,740 --> 01:25:40,880 But S and T commute, so this is T SU. 1209 01:25:40,880 --> 01:25:44,260 But since U belongs to Uk, that's 1210 01:25:44,260 --> 01:25:46,890 the space with eigenvalue lambda k. 1211 01:25:46,890 --> 01:25:55,110 So this is lambda k times u, so you have Tu here. 1212 01:25:55,110 --> 01:26:00,260 So Tu acted with S gives you lambda k Tu. 1213 01:26:00,260 --> 01:26:05,785 So Tu is in the invariant subspace Uk. 1214 01:26:14,320 --> 01:26:20,570 What's happening here is now something very straightforward. 1215 01:26:20,570 --> 01:26:29,680 You try to imagine how does the matrix T look in the basis 1216 01:26:29,680 --> 01:26:31,660 that we have here. 1217 01:26:31,660 --> 01:26:34,240 Here is this basis. 1218 01:26:34,240 --> 01:26:38,340 how does this matrix T look? 1219 01:26:38,340 --> 01:26:46,910 Well, this matrix keeps the invariant subspaces. 1220 01:26:46,910 --> 01:26:50,070 So you have to think of it blocked diagonally. 1221 01:26:53,850 --> 01:26:56,990 If it acts on it-- here are the first vectors 1222 01:26:56,990 --> 01:26:59,100 that you're considering, the U1. 1223 01:26:59,100 --> 01:27:03,190 Well if you act on it with T of the U1 subspace, 1224 01:27:03,190 --> 01:27:05,140 you stay in the U1 subspace. 1225 01:27:05,140 --> 01:27:06,580 So you don't get anything else. 1226 01:27:06,580 --> 01:27:09,490 So you must have 0's all over here. 1227 01:27:09,490 --> 01:27:12,320 And you can have a matrix here. 1228 01:27:12,320 --> 01:27:17,720 And if you act on the second Uk U2, you get a vector in U2, 1229 01:27:17,720 --> 01:27:20,150 so it's orthogonal to all the other vectors. 1230 01:27:20,150 --> 01:27:22,330 So you get a matrix here. 1231 01:27:22,330 --> 01:27:24,070 And you get a matrix here. 1232 01:27:24,070 --> 01:27:30,150 So actually you get a blocked diagonal matrix 1233 01:27:30,150 --> 01:27:35,350 in which the blocks correspond to the degeneracy. 1234 01:27:35,350 --> 01:27:41,540 So if there's a degeneracy d1 here, it's a d1 times d1. 1235 01:27:41,540 --> 01:27:45,690 And d2 times d2. 1236 01:27:45,690 --> 01:27:48,710 So actually you haven't simultaneously 1237 01:27:48,710 --> 01:27:50,300 diagonalized them. 1238 01:27:50,300 --> 01:27:52,160 That's the problem of degeneracy. 1239 01:27:52,160 --> 01:27:55,560 You haven't, but you now have the tools, 1240 01:27:55,560 --> 01:27:59,020 because this operator is Hermitian, therefore it's 1241 01:27:59,020 --> 01:28:03,060 Hermitian here, and here, and here, and here. 1242 01:28:03,060 --> 01:28:05,840 So you can diagonalize here. 1243 01:28:05,840 --> 01:28:08,880 But what do you need for diagonalizing here? 1244 01:28:08,880 --> 01:28:10,950 You need a unitary matrix. 1245 01:28:10,950 --> 01:28:13,400 Call it V1. 1246 01:28:13,400 --> 01:28:16,140 For here you need another unitary matrix. 1247 01:28:16,140 --> 01:28:18,430 Call it V2. 1248 01:28:18,430 --> 01:28:19,910 Vn. 1249 01:28:19,910 --> 01:28:24,140 And then this matrix becomes diagonal. 1250 01:28:24,140 --> 01:28:26,990 But then what about the old matrix? 1251 01:28:26,990 --> 01:28:31,400 Well, we just explained here that if you change the basis 1252 01:28:31,400 --> 01:28:37,330 by unitary matrices, you don't change the first matrix. 1253 01:28:37,330 --> 01:28:39,900 So actually you succeeded. 1254 01:28:39,900 --> 01:28:45,030 You now can diagonalize this without destroying 1255 01:28:45,030 --> 01:28:46,130 your earlier result. 1256 01:28:46,130 --> 01:28:49,430 And you managed to diagonalize the whole thing. 1257 01:28:49,430 --> 01:28:52,180 So this is for two operators in the notes. 1258 01:28:52,180 --> 01:28:56,060 You'll see why it simply extends for three, four, and five, 1259 01:28:56,060 --> 01:28:58,410 or arbitrary number of operators. 1260 01:28:58,410 --> 01:29:00,860 See you next time.