1 00:00:00,060 --> 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 --> 00:00:10,130 to offer high quality educational resources for free. 5 00:00:10,130 --> 00:00:12,700 To make a donation, or to view additional materials 6 00:00:12,700 --> 00:00:16,580 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,580 --> 00:00:17,210 at ocw.mit.edu. 8 00:00:22,560 --> 00:00:24,960 PROFESSOR: It's good to be back. 9 00:00:24,960 --> 00:00:28,930 I really want to thank both Aaron and Will who 10 00:00:28,930 --> 00:00:34,890 took my teaching duties over last week. 11 00:00:34,890 --> 00:00:39,340 You've been receiving updates of the lecture notes, 12 00:00:39,340 --> 00:00:42,700 and, in particular, as I don't want 13 00:00:42,700 --> 00:00:45,980 to go back over some things, I would 14 00:00:45,980 --> 00:00:50,980 like you to read some of the material you have there. 15 00:00:50,980 --> 00:00:54,500 In particular, the part on projectors 16 00:00:54,500 --> 00:00:57,460 has been developed further. 17 00:00:57,460 --> 00:01:02,670 We will meet projectors a lot in this space, 18 00:01:02,670 --> 00:01:07,420 because in quantum mechanics, whenever you do a measurement, 19 00:01:07,420 --> 00:01:11,392 the effect of a measurement is to act 20 00:01:11,392 --> 00:01:14,260 on a stage with a projector. 21 00:01:14,260 --> 00:01:17,530 So projectors are absolutely important. 22 00:01:17,530 --> 00:01:19,920 And orthogonal projectors are the ones 23 00:01:19,920 --> 00:01:21,990 that we're going to use-- are the ones that 24 00:01:21,990 --> 00:01:25,850 are relevant in quantum mechanics. 25 00:01:25,850 --> 00:01:28,470 There's a property, for example, of projectors 26 00:01:28,470 --> 00:01:33,250 that is quite neat that is used in maximization and fitting 27 00:01:33,250 --> 00:01:33,990 problems. 28 00:01:33,990 --> 00:01:37,800 And you will see that in the PSET. 29 00:01:37,800 --> 00:01:39,820 In the PSET, the last problem has 30 00:01:39,820 --> 00:01:44,330 to do with using a projector to find 31 00:01:44,330 --> 00:01:48,750 best approximations to some functions using polynomials. 32 00:01:48,750 --> 00:01:52,010 So there's lots of things to say about projectors, 33 00:01:52,010 --> 00:01:57,870 and we'll find them along when we go and do later stuff 34 00:01:57,870 --> 00:01:58,490 in the course. 35 00:01:58,490 --> 00:02:03,820 So please read that part on projectors. 36 00:02:03,820 --> 00:02:08,490 The other thing is that much of what we're going to do 37 00:02:08,490 --> 00:02:15,740 uses the notation that we have for describing inner products-- 38 00:02:15,740 --> 00:02:22,370 for example, u, v. And then, as we've mentioned, 39 00:02:22,370 --> 00:02:27,250 and this is in the notes-- this, in the bracket notation, 40 00:02:27,250 --> 00:02:29,920 becomes something like this. 41 00:02:29,920 --> 00:02:32,800 And the bracket notation of quantum mechanics 42 00:02:32,800 --> 00:02:36,610 is fairly nice for many things, and it's 43 00:02:36,610 --> 00:02:40,560 used sometimes for some applications. 44 00:02:40,560 --> 00:02:44,150 Everybody uses the bracket notation for some applications. 45 00:02:44,150 --> 00:02:47,910 I hope to get to one of those today. 46 00:02:47,910 --> 00:02:53,660 So much of the theory we've developed 47 00:02:53,660 --> 00:02:58,120 is done with this as the inner product. 48 00:02:58,120 --> 00:03:02,260 Nevertheless, the translation to the language of bras and kets 49 00:03:02,260 --> 00:03:04,190 is very quick. 50 00:03:04,190 --> 00:03:06,830 So the way the notes are going to be structured-- 51 00:03:06,830 --> 00:03:09,160 and we're still working on the notes, 52 00:03:09,160 --> 00:03:10,720 and they're going to chang a bit-- 53 00:03:10,720 --> 00:03:13,670 is that everything regarding the math 54 00:03:13,670 --> 00:03:17,070 is being developed more in this notation, 55 00:03:17,070 --> 00:03:19,480 but then we turn into bras and kets 56 00:03:19,480 --> 00:03:22,490 and just go quickly over all you've 57 00:03:22,490 --> 00:03:26,130 seen, just how it looks with bras and kets 58 00:03:26,130 --> 00:03:28,970 so that you're familiar. 59 00:03:28,970 --> 00:03:31,080 Then, in the later part of the course, 60 00:03:31,080 --> 00:03:34,820 we'll use sometimes bras and kets, and sometimes this. 61 00:03:34,820 --> 00:03:37,220 And sometimes some physicisists use 62 00:03:37,220 --> 00:03:39,223 this notation with parentheses. 63 00:03:42,110 --> 00:03:44,560 So for example, Weinberg's recent book 64 00:03:44,560 --> 00:03:46,735 on quantum mechanics uses this notation. 65 00:03:46,735 --> 00:03:49,905 It doesn't use bras and kets I think at all. 66 00:03:53,570 --> 00:03:56,890 So you have to be ready to work with any notation. 67 00:03:56,890 --> 00:04:01,290 The bra and ket notation has some nice properties 68 00:04:01,290 --> 00:04:05,140 that make it very fast to do things with it. 69 00:04:05,140 --> 00:04:06,740 It is very efficient. 70 00:04:06,740 --> 00:04:12,410 Nevertheless, in some ways this notation is a little clearer. 71 00:04:12,410 --> 00:04:17,160 So many of the things we'll develop is with this notation. 72 00:04:17,160 --> 00:04:19,350 So today I'm going to develop the idea 73 00:04:19,350 --> 00:04:22,190 of the Hermitian conjugator for an operator, 74 00:04:22,190 --> 00:04:24,440 or the adjoint of an operator. 75 00:04:24,440 --> 00:04:31,630 And this idea is generally a little subtle, a little hard 76 00:04:31,630 --> 00:04:33,140 to understand. 77 00:04:33,140 --> 00:04:38,200 But we'll just go at it slowly and try to make it very clear. 78 00:04:38,200 --> 00:04:44,100 So adjoints or Hermitian operators, or Hermitian 79 00:04:44,100 --> 00:04:54,260 conjugates-- adjoints or Hermitian conjugates. 80 00:04:59,680 --> 00:05:06,250 So the idea of Adjoints, or Hermition conjugates, 81 00:05:06,250 --> 00:05:14,220 really begins with some necessary background 82 00:05:14,220 --> 00:05:18,640 on what they're called-- linear functionals. 83 00:05:18,640 --> 00:05:20,800 It sounds complicated, but it's not. 84 00:05:20,800 --> 00:05:23,270 What is a linear functional? 85 00:05:23,270 --> 00:05:33,440 A linear functional on V-- on a vector 86 00:05:33,440 --> 00:05:44,250 space V-- is a linear map from V to the numbers 87 00:05:44,250 --> 00:05:48,870 F. We've always been calling F the numbers. 88 00:05:48,870 --> 00:05:53,060 So it's just that, something that, once you have a vector, 89 00:05:53,060 --> 00:05:55,120 you get a number and it's linear. 90 00:05:58,120 --> 00:06:02,900 So a linear function of Phi, if it's a linear functional, 91 00:06:02,900 --> 00:06:11,880 Phi on v belongs to F. Phi acts on a vector, v, 92 00:06:11,880 --> 00:06:15,720 that belongs to the vector space and gives you a number. 93 00:06:15,720 --> 00:06:25,710 So "linear" means Phi of v1 plus v2 is Phi of v1 plus Phi of v2. 94 00:06:25,710 --> 00:06:33,950 And Phi of av, for a is number, is a Phi of v. 95 00:06:33,950 --> 00:06:39,120 So seems simple, and indeed it is. 96 00:06:39,120 --> 00:06:43,430 And we can construct examples of linear functionals, 97 00:06:43,430 --> 00:06:45,515 some trivial ones, for example. 98 00:06:48,350 --> 00:06:58,030 Let Phi be a map that takes the vector space, reals in three 99 00:06:58,030 --> 00:07:00,840 dimensions, to the real numbers. 100 00:07:00,840 --> 00:07:02,950 So how does it act? 101 00:07:02,950 --> 00:07:09,000 Phi acts on a vector, which is x1, x2, and x3-- three 102 00:07:09,000 --> 00:07:10,190 components. 103 00:07:10,190 --> 00:07:13,680 And it must give a numbers, so it 104 00:07:13,680 --> 00:07:21,170 could be 3x1 minus x2 plus 7x3, as simple as that. 105 00:07:24,850 --> 00:07:26,480 It's linear. 106 00:07:26,480 --> 00:07:31,560 x1, x2, and x3 are the coordinates of a single vector. 107 00:07:31,560 --> 00:07:34,780 And whenever you have this vector, that is, 108 00:07:34,780 --> 00:07:37,530 this triplet-- now, I could have written it 109 00:07:37,530 --> 00:07:42,445 like this-- Phi of x1, x2, and x3, as a vector. 110 00:07:42,445 --> 00:07:43,770 It looks like that. 111 00:07:43,770 --> 00:07:47,600 But it's easier to use horizontal notation, 112 00:07:47,600 --> 00:07:51,785 so we'll write it like that. 113 00:07:51,785 --> 00:07:55,060 And, if you have an inner product 114 00:07:55,060 --> 00:08:01,170 on this space-- on this three dimensional vector space-- 115 00:08:01,170 --> 00:08:03,570 there's something you can say. 116 00:08:03,570 --> 00:08:07,810 Actually this Phi is equal-- and this we 117 00:08:07,810 --> 00:08:15,900 call the vector V-- is actually equal to u, 118 00:08:15,900 --> 00:08:22,020 inner product with v, where u is the vector that has components 119 00:08:22,020 --> 00:08:26,860 3, minus 1, and 7, because if you 120 00:08:26,860 --> 00:08:30,810 take the inner product of this vector with this vector, 121 00:08:30,810 --> 00:08:34,659 in three dimensions real vector spaces-- 122 00:08:34,659 --> 00:08:37,380 inner product is a dot product. 123 00:08:37,380 --> 00:08:42,970 And then we make the dot product of u with the vector V. 124 00:08:42,970 --> 00:08:46,270 Maybe I should have called it v1, v2, v3. 125 00:08:46,270 --> 00:08:53,360 I'll change that-- v1, v2, v3 here 126 00:08:53,360 --> 00:09:01,950 are components of the vector-- v1, v2, and v3, 127 00:09:01,950 --> 00:09:04,190 not to be confused with three vectors. 128 00:09:04,190 --> 00:09:10,890 This whole thing is a vector V. So this linear functional, 129 00:09:10,890 --> 00:09:14,650 that, given a vector gives me a number. 130 00:09:14,650 --> 00:09:18,050 The clever thing is that the inner product 131 00:09:18,050 --> 00:09:20,730 is this thing that gives you numbers out of vectors. 132 00:09:20,730 --> 00:09:25,100 So you've reconstructed this linear functional 133 00:09:25,100 --> 00:09:30,120 as the inner product of some vector with the vector you're 134 00:09:30,120 --> 00:09:33,490 acting on, so, where u is given by that. 135 00:09:39,440 --> 00:09:42,770 The most important result about linear functionals 136 00:09:42,770 --> 00:09:45,610 is that this is not an accident. 137 00:09:45,610 --> 00:09:48,190 This kind be that very generally. 138 00:09:48,190 --> 00:09:51,960 So any time you give me a linear functional, 139 00:09:51,960 --> 00:09:57,930 I can find a vector that, using the inner product, 140 00:09:57,930 --> 00:10:01,620 acts on the vector you're acting on the same way 141 00:10:01,620 --> 00:10:05,480 as the linear function of thus. 142 00:10:05,480 --> 00:10:08,540 The most general linear functional 143 00:10:08,540 --> 00:10:13,320 is just some most general vector acting this way. 144 00:10:13,320 --> 00:10:16,800 So let's state that and prove it. 145 00:10:16,800 --> 00:10:20,190 So this is a theorem, it's not a definition or anything 146 00:10:20,190 --> 00:10:22,370 like that. 147 00:10:22,370 --> 00:10:34,820 Let Phi be a linear functional on v. Then 148 00:10:34,820 --> 00:10:48,800 there is a unique vector u belonging to the vector space 149 00:10:48,800 --> 00:11:10,370 such that Phi acting on v is equal to u, v. 150 00:11:10,370 --> 00:11:13,360 Since this is such a canonical thing, 151 00:11:13,360 --> 00:11:15,490 you could even invent a notation. 152 00:11:15,490 --> 00:11:20,760 Call this the linear functional created by u, 153 00:11:20,760 --> 00:11:25,250 acting on v. Everybody doesn't use this, 154 00:11:25,250 --> 00:11:27,860 but you could call it like that. 155 00:11:27,860 --> 00:11:30,870 This is a linear functional acting on v, 156 00:11:30,870 --> 00:11:34,180 but it's labeled by u, which is the vector that you've 157 00:11:34,180 --> 00:11:34,900 use there. 158 00:11:38,945 --> 00:11:43,210 This is important enough that we better understand why it works. 159 00:11:43,210 --> 00:11:44,930 So I'll prove it. 160 00:11:52,370 --> 00:11:56,070 We're going to use an orthonormal basis, 161 00:11:56,070 --> 00:12:07,582 say e1 up to en is an orthonormal, O-N, basis. 162 00:12:07,582 --> 00:12:09,040 AUDIENCE: That means we're assuming 163 00:12:09,040 --> 00:12:10,532 v is finite dimensional here? 164 00:12:10,532 --> 00:12:11,240 PROFESSOR: Sorry? 165 00:12:11,240 --> 00:12:13,656 AUDIENCE: We're assuming V is finite dimensional, correct? 166 00:12:16,459 --> 00:12:18,125 PROFESSOR: Yeah, it's finite dimensional 167 00:12:18,125 --> 00:12:22,776 I'm going to prove it using a finite basis like that. 168 00:12:22,776 --> 00:12:24,790 Is true finite dimensional? 169 00:12:24,790 --> 00:12:27,050 I presume yes. 170 00:12:27,050 --> 00:12:28,529 AUDIENCE: If it's not [INAUDIBLE]. 171 00:12:33,813 --> 00:12:34,938 PROFESSOR: What hypothesis? 172 00:12:34,938 --> 00:12:37,403 AUDIENCE: You say continuous when you're talking 173 00:12:37,403 --> 00:12:38,890 [INAUDIBLE]. 174 00:12:38,890 --> 00:12:41,745 PROFESSOR: OK, I'll check. 175 00:12:41,745 --> 00:12:46,280 But let's just prove this one finite dimensional like this. 176 00:12:50,990 --> 00:12:51,980 Let's take that. 177 00:12:51,980 --> 00:12:57,800 And now write the vector as a superposition of these vectors. 178 00:12:57,800 --> 00:12:59,900 Now we know how to do that. 179 00:12:59,900 --> 00:13:07,560 We just have the components of v along each basis vector. 180 00:13:07,560 --> 00:13:11,700 For example, the component of v along e1 181 00:13:11,700 --> 00:13:15,560 is precisely e1, v. So then you go on 182 00:13:15,560 --> 00:13:22,350 like that until you go en, v, en. 183 00:13:28,420 --> 00:13:31,150 I think you've derived this a couple of times already, 184 00:13:31,150 --> 00:13:33,920 but this is a statement you can review, 185 00:13:33,920 --> 00:13:38,410 and let's take it to be correct. 186 00:13:38,410 --> 00:13:44,540 Now let's consider what is Phi acting on a v like that. 187 00:13:44,540 --> 00:13:48,960 Well, it's a linear map, so it takes on a sum of vectors 188 00:13:48,960 --> 00:13:51,630 by acting on the vectors, each one. 189 00:13:51,630 --> 00:13:53,970 So it should act on from this plus that, plus that, 190 00:13:53,970 --> 00:13:54,800 plus that. 191 00:13:54,800 --> 00:13:57,610 Now, it acts on this vector. 192 00:13:57,610 --> 00:13:59,200 Well, this is a number. 193 00:13:59,200 --> 00:14:00,840 The number goes out. 194 00:14:00,840 --> 00:14:02,100 It's a linear function. 195 00:14:02,100 --> 00:14:19,190 So this is e1, v, Phi of e1, all the way up to en, v Phi of en. 196 00:14:24,260 --> 00:14:30,320 Now this is a number, so let's bring it 197 00:14:30,320 --> 00:14:32,060 into the inner product. 198 00:14:32,060 --> 00:14:35,820 Now, if you brought it in on the side of V as a number 199 00:14:35,820 --> 00:14:37,850 it would go in just like the number. 200 00:14:37,850 --> 00:14:40,230 If you bring it into the left side, 201 00:14:40,230 --> 00:14:43,200 remember it's conjugate homogeneous, 202 00:14:43,200 --> 00:14:45,750 so this enters as a complex number. 203 00:14:45,750 --> 00:15:05,730 So this would be e1, Phi of e1 star times V plus en, Phi 204 00:15:05,730 --> 00:15:23,240 of en star, v. And then we have our result that this Phi of v 205 00:15:23,240 --> 00:15:25,050 has been written now. 206 00:15:25,050 --> 00:15:29,690 The left input is different on each of these terms, 207 00:15:29,690 --> 00:15:31,630 but the right input is the same. 208 00:15:31,630 --> 00:15:36,120 So at this moment linearity on the first input 209 00:15:36,120 --> 00:15:46,240 says that you can put here e1, Phi of e1 star plus up to en, 210 00:15:46,240 --> 00:15:57,950 Phi of en star, v. And this is the vector you were looking 211 00:15:57,950 --> 00:16:05,030 for, the vector U. Kind of simple, at the end of the day 212 00:16:05,030 --> 00:16:09,300 you just used the basis and made it clearer. 213 00:16:09,300 --> 00:16:11,180 It can always be constructed. 214 00:16:11,180 --> 00:16:21,590 Basically, the vector you want is e1 times Phi of u1 star 215 00:16:21,590 --> 00:16:24,270 plus en up to Phi of en star. 216 00:16:24,270 --> 00:16:27,900 So if you know what the linear map does to the basis vectors, 217 00:16:27,900 --> 00:16:31,640 you construct the vector this way. 218 00:16:31,640 --> 00:16:33,780 Vector is done. 219 00:16:33,780 --> 00:16:37,760 The only thing to be proven is that it's unique. 220 00:16:37,760 --> 00:16:43,210 Uniqueness is rather easy to prove at this stage. 221 00:16:43,210 --> 00:16:48,260 Suppose you know that u with v works and gives you 222 00:16:48,260 --> 00:16:49,240 the right answer. 223 00:16:49,240 --> 00:16:51,660 Well, you ask, is there a u prime 224 00:16:51,660 --> 00:16:56,570 that also gives the right answer for all v? 225 00:16:56,570 --> 00:16:58,970 Well, pass it to the other side, and you 226 00:16:58,970 --> 00:17:04,550 would have u minus u prime, would have zero inner product 227 00:17:04,550 --> 00:17:11,740 with v for all v. Pass to the other side, 228 00:17:11,740 --> 00:17:14,010 take the difference, and it's that. 229 00:17:14,010 --> 00:17:16,410 So u minus u prime is a vector that 230 00:17:16,410 --> 00:17:19,720 has zero inner product with any vector. 231 00:17:19,720 --> 00:17:23,319 And any such thing as always zero. 232 00:17:23,319 --> 00:17:27,109 And perhaps the easiest way to show that, in case you 233 00:17:27,109 --> 00:17:30,730 haven't seen that before, if x with v 234 00:17:30,730 --> 00:17:39,053 equals 0 for all for all v. What can you say about x? 235 00:17:41,590 --> 00:17:46,880 Well, take v is the value for any v. So take v equal x. 236 00:17:46,880 --> 00:17:50,050 So you take x, x is equal to 0. 237 00:17:50,050 --> 00:17:52,840 And by the axioms of the inner product, 238 00:17:52,840 --> 00:17:59,130 if a vector has 0 inner product with itself, it's 0. 239 00:17:59,130 --> 00:18:04,290 So at this stage, you go u minus u prime equals 0, 240 00:18:04,290 --> 00:18:06,470 and u is equal to u prime. 241 00:18:06,470 --> 00:18:10,070 So it's definitely unique, you can't 242 00:18:10,070 --> 00:18:11,755 find another one that works. 243 00:18:16,720 --> 00:18:20,250 So we have this thing. 244 00:18:20,250 --> 00:18:21,875 This theorem is proven. 245 00:18:25,900 --> 00:18:29,220 And now let's use to define this the adjoint, 246 00:18:29,220 --> 00:18:32,310 which is a very interesting thing. 247 00:18:32,310 --> 00:18:36,700 So the adjoing, or Hermitian conjugate, 248 00:18:36,700 --> 00:18:42,130 sometimes called adjoint-- physicists 249 00:18:42,130 --> 00:18:48,320 use the name Hermitian conjugate, which 250 00:18:48,320 --> 00:18:49,487 is more appropriate. 251 00:18:49,487 --> 00:18:51,320 Well, I don't know if it's more appropriate. 252 00:18:51,320 --> 00:18:57,260 It's more pictorial if you have a complex vector space. 253 00:18:57,260 --> 00:19:00,460 And if you're accustomed with linear algebra 254 00:19:00,460 --> 00:19:03,490 about Hermition matrices, and what they are, 255 00:19:03,490 --> 00:19:06,085 and that will show up a little later, 256 00:19:06,085 --> 00:19:10,610 although with a very curious twist. 257 00:19:10,610 --> 00:19:18,170 So given an operator T belonging to the set of linear operators 258 00:19:18,170 --> 00:19:26,950 on a vector space, you can define T dagger, 259 00:19:26,950 --> 00:19:33,730 also belonging to l of v. So this 260 00:19:33,730 --> 00:19:38,420 is the aim-- constructing an operator called the Hermitian 261 00:19:38,420 --> 00:19:40,816 conjugate. 262 00:19:40,816 --> 00:19:43,370 Now the way we're going to do it is 263 00:19:43,370 --> 00:19:47,940 going to be defining something that is a T star. 264 00:19:47,940 --> 00:19:51,130 Well, I said "T star" because mathematicians in fact 265 00:19:51,130 --> 00:19:52,810 call it star. 266 00:19:52,810 --> 00:19:55,320 And most mathematicians, they complex conjugate 267 00:19:55,320 --> 00:19:59,290 if a number is not z star but z bar. 268 00:19:59,290 --> 00:20:03,200 So that's why we call it T star and I 269 00:20:03,200 --> 00:20:06,060 may make this mistake a few times today. 270 00:20:06,060 --> 00:20:09,120 We're going to use dagger. 271 00:20:09,120 --> 00:20:12,150 And so I will make a definition that 272 00:20:12,150 --> 00:20:16,400 will tell you what T dagger is supposed to be, 273 00:20:16,400 --> 00:20:17,540 acting on things. 274 00:20:17,540 --> 00:20:20,860 But it might not be obvious, at least at first sight, 275 00:20:20,860 --> 00:20:23,070 that it's a linear operator. 276 00:20:23,070 --> 00:20:29,030 So let's see how does this go. 277 00:20:29,030 --> 00:20:30,390 Here is the claim. 278 00:20:30,390 --> 00:20:37,590 Consider the following thing-- u, 279 00:20:37,590 --> 00:20:51,380 T, v-- this inner product of u with T, v. And think of it 280 00:20:51,380 --> 00:20:53,510 as a linear functional. 281 00:20:53,510 --> 00:20:56,820 Well, it's certainly a linear functional 282 00:20:56,820 --> 00:21:05,380 of v. It's a linear functional because if you 283 00:21:05,380 --> 00:21:08,550 put a times v the a goes out. 284 00:21:08,550 --> 00:21:11,970 And if you put v1 plus v2 you get it's linear. 285 00:21:11,970 --> 00:21:15,150 So it's linear, but it's not the usual one's 286 00:21:15,150 --> 00:21:20,590 that we've been building, in which the linear functional 287 00:21:20,590 --> 00:21:26,700 looks like u with v. I just put an operator there. 288 00:21:26,700 --> 00:21:36,700 So by this theorem, there must be some vector 289 00:21:36,700 --> 00:21:40,210 that this can be represented as this acting 290 00:21:40,210 --> 00:21:46,180 with that vector inside here, because any linear operator is 291 00:21:46,180 --> 00:21:49,400 some vector acting on the vector-- on the vector 292 00:21:49,400 --> 00:21:54,720 v. Any linear functional, I'm sorry-- not linear operator. 293 00:21:54,720 --> 00:21:57,940 Any linear functional-- this is a linear functional. 294 00:21:57,940 --> 00:22:01,730 And every linear function can be written as some vector acting 295 00:22:01,730 --> 00:22:05,430 on v. So there must be a vector here. 296 00:22:05,430 --> 00:22:12,160 Now this vector surely will depend on what u is. 297 00:22:12,160 --> 00:22:13,760 So we'll give it a name. 298 00:22:19,080 --> 00:22:26,260 It's a vector that depends on U. I'll write it as T dagger u. 299 00:22:26,260 --> 00:22:37,360 At this moment, T dagger is just a map from v 300 00:22:37,360 --> 00:22:42,320 to v. We said that this thing that we must put here 301 00:22:42,320 --> 00:22:45,860 depends on u, and it must be a vector. 302 00:22:45,860 --> 00:22:49,310 So it's some thing that takes u and produces 303 00:22:49,310 --> 00:22:54,300 another vector called T dagger on u. 304 00:22:54,300 --> 00:22:56,440 But we don't know what T dagger is, 305 00:22:56,440 --> 00:22:58,225 and we don't even know that it's linear. 306 00:23:00,820 --> 00:23:05,530 So at this moment it's just a map, and it's a definition. 307 00:23:05,530 --> 00:23:13,020 This defines what T dagger u is, because some vector-- 308 00:23:13,020 --> 00:23:17,590 it could be calculated exactly the same way we calculated 309 00:23:17,590 --> 00:23:20,160 the other ones. 310 00:23:20,160 --> 00:23:27,270 So let's try to see why it is linear. 311 00:23:27,270 --> 00:23:38,280 Claim T dagger belongs to the linear operators in v. 312 00:23:38,280 --> 00:23:39,640 So how do we do that? 313 00:23:39,640 --> 00:23:41,120 Well, we can say the following. 314 00:23:41,120 --> 00:23:47,032 Consider u1 plus u1 acting on Tv. 315 00:23:49,960 --> 00:24:01,700 Well, by definition, this would be the T dagger of u1 plus u2, 316 00:24:01,700 --> 00:24:07,160 some function on u1 plus u2, because whatever is here 317 00:24:07,160 --> 00:24:12,570 gets acted by T dagger times v. On the other hand, 318 00:24:12,570 --> 00:24:20,970 this thing is equal to u1, Tv plus u2, 319 00:24:20,970 --> 00:24:39,200 Tv, which is equal to T dagger u1, v plus T dagger u2, v. 320 00:24:39,200 --> 00:24:48,700 And, by linearity, here you get equal to T dagger u1 plus T 321 00:24:48,700 --> 00:24:50,590 dagger on u2. 322 00:24:55,740 --> 00:24:59,680 And then comparing this too-- and this is true for arbitrary 323 00:24:59,680 --> 00:25:05,080 v-- you find that T dagger, acting on this sum of vectors, 324 00:25:05,080 --> 00:25:06,960 is the same as this thing. 325 00:25:10,930 --> 00:25:20,078 And similarly, how about au, Tv? 326 00:25:22,800 --> 00:25:41,060 Well, this is equal to T dagger on au, v. Now, T dagger on au, 327 00:25:41,060 --> 00:25:44,500 do you think the a goes out as a or as a bar? 328 00:25:47,830 --> 00:25:49,386 Sorry? 329 00:25:49,386 --> 00:25:51,110 a or a-bar? 330 00:25:51,110 --> 00:25:55,120 What do you think T dagger and au is supposed to be? 331 00:25:55,120 --> 00:26:01,730 a, because it's supposed to be a linear operator, so no dagger 332 00:26:01,730 --> 00:26:02,230 here. 333 00:26:02,230 --> 00:26:05,920 You see-- well, I didn't show it here. 334 00:26:05,920 --> 00:26:11,770 Any linear operator, T on av, is supposed to be a T of v. 335 00:26:11,770 --> 00:26:13,800 And we're saying T dagger is also 336 00:26:13,800 --> 00:26:17,360 a linear operator in the vector space. 337 00:26:17,360 --> 00:26:20,850 So this should be with an a. 338 00:26:20,850 --> 00:26:22,010 We'll see what we get. 339 00:26:22,010 --> 00:26:34,010 Well, the a can go out here, and it becomes a star u1, Tv, 340 00:26:34,010 --> 00:26:35,220 which is equal. 341 00:26:35,220 --> 00:26:37,250 I'm going through the left side. 342 00:26:37,250 --> 00:26:43,600 By definition, a bar T dagger of u, 343 00:26:43,600 --> 00:26:47,470 v. And now the constant can go in, 344 00:26:47,470 --> 00:26:54,330 and it goes back as a, T dagger u, v. So 345 00:26:54,330 --> 00:26:59,100 this must be equal to that, and you 346 00:26:59,100 --> 00:27:07,340 get what we're claiming here, which is T dagger on au, 347 00:27:07,340 --> 00:27:10,420 is equal to a T dagger of u. 348 00:27:13,010 --> 00:27:17,200 So the operator is linear. 349 00:27:17,200 --> 00:27:25,720 So we've defined something this way, and it's linear, 350 00:27:25,720 --> 00:27:30,030 and it's doing all the right things. 351 00:27:30,030 --> 00:27:33,470 Now, you really feel proud at this stage. 352 00:27:33,470 --> 00:27:36,470 This is still not all that intuitive. 353 00:27:36,470 --> 00:27:38,600 What does this all do? 354 00:27:38,600 --> 00:27:41,230 So we're going to do an example, and we're 355 00:27:41,230 --> 00:27:46,710 going to do one more property. 356 00:27:46,710 --> 00:27:49,660 Let me do one more property and then stop for a second. 357 00:27:49,660 --> 00:27:56,220 So here is one property-- ST dagger is supposed to be 358 00:27:56,220 --> 00:27:58,290 T dagger S dagger. 359 00:27:58,290 --> 00:28:00,210 So how do you get that? 360 00:28:00,210 --> 00:28:03,970 Not hard-- u, STv. 361 00:28:06,700 --> 00:28:12,300 Well, STv is really the same as S acting on Tv. 362 00:28:12,300 --> 00:28:15,330 Now the first S can be brought to the other side 363 00:28:15,330 --> 00:28:19,610 by the definition that you can bring something 364 00:28:19,610 --> 00:28:20,830 to the other side. 365 00:28:20,830 --> 00:28:22,220 Put in a dagger. 366 00:28:22,220 --> 00:28:24,550 So the S is brought there, and you 367 00:28:24,550 --> 00:28:33,060 get S dagger on u, T on v. And then the T can be brought here 368 00:28:33,060 --> 00:28:40,450 and act on this one, and you get T dagger S dagger u, 369 00:28:40,450 --> 00:28:48,670 v. So this thing is the dagger of this thing, 370 00:28:48,670 --> 00:28:50,230 and that's the statement here. 371 00:28:53,920 --> 00:28:57,540 There's yet one more simple property, 372 00:28:57,540 --> 00:29:09,320 that the dagger of S dagger is S. You take dagger twice 373 00:29:09,320 --> 00:29:11,550 and you're back to the same operator. 374 00:29:11,550 --> 00:29:13,570 Nothing has changed. 375 00:29:13,570 --> 00:29:17,062 So how do you do that? 376 00:29:17,062 --> 00:29:24,120 Take, for example, this-- take u, put S dagger here, and put 377 00:29:24,120 --> 00:29:33,740 v. Now, by definition, this is equal to-- you 378 00:29:33,740 --> 00:29:37,000 put the operator on the other side, adding a dagger. 379 00:29:37,000 --> 00:29:44,655 So that's why we put that one like this. 380 00:29:44,655 --> 00:29:47,540 The operator gets daggers, so now you've 381 00:29:47,540 --> 00:29:49,410 got the double dagger. 382 00:29:49,410 --> 00:29:52,540 So at this moment, however, you have 383 00:29:52,540 --> 00:29:57,170 to do something to simplify this. 384 00:29:57,170 --> 00:30:00,470 The easiest thing to do is probably the following-- 385 00:30:00,470 --> 00:30:02,760 to just flip these two, which you 386 00:30:02,760 --> 00:30:08,170 can do the order by putting a star. 387 00:30:08,170 --> 00:30:10,230 So this is equal. 388 00:30:10,230 --> 00:30:12,155 The left hand side is equal to this. 389 00:30:14,980 --> 00:30:22,406 And now this S dagger can be moved here and becomes an S. 390 00:30:22,406 --> 00:30:32,290 So this is u, Sv, and you still have the star. 391 00:30:32,290 --> 00:30:37,680 And now reverse this by eliminating the star, 392 00:30:37,680 --> 00:30:45,050 so you have S-- I'm sorry, I have 393 00:30:45,050 --> 00:30:46,640 this notation completely wrong. 394 00:30:46,640 --> 00:30:50,600 Sv-- this is u. 395 00:30:50,600 --> 00:30:52,960 The u's v's are easily confused. 396 00:30:52,960 --> 00:30:57,630 So this is v, and this is u. 397 00:31:00,340 --> 00:31:08,260 I move the S, and then finally I have Su, v without a star. 398 00:31:08,260 --> 00:31:12,240 I flipped it again. 399 00:31:12,240 --> 00:31:19,175 So then you compare these two, and you get the desired result. 400 00:31:23,560 --> 00:31:26,760 OK, so we've gone through this thing, which 401 00:31:26,760 --> 00:31:30,252 is the main result of daggers, and I 402 00:31:30,252 --> 00:31:31,960 would like to see if there are questions. 403 00:31:31,960 --> 00:31:36,730 Anything that has been unclear as we've gone along here? 404 00:31:36,730 --> 00:31:37,355 And question? 405 00:31:46,970 --> 00:31:48,316 OK. 406 00:31:48,316 --> 00:31:50,230 No questions. 407 00:31:50,230 --> 00:31:59,070 So let's do a simple example, and it's good 408 00:31:59,070 --> 00:32:04,500 because it's useful to practice with explicit things. 409 00:32:04,500 --> 00:32:05,630 So here's an example. 410 00:32:08,190 --> 00:32:16,290 There's a vector space V, which is three complex numbers, 411 00:32:16,290 --> 00:32:19,090 three component vectors-- complex vectors. 412 00:32:19,090 --> 00:32:29,706 So a v is equal to v1, v2, v3-- three numbers are all the vi. 413 00:32:29,706 --> 00:32:32,380 Each one belongs to the complex number. 414 00:32:32,380 --> 00:32:37,130 So three complex numbers makes a vector space like this. 415 00:32:37,130 --> 00:32:39,210 So somebody comes along and gives you 416 00:32:39,210 --> 00:32:47,400 the following linear map-- T on a vector, v1, v1, v3, 417 00:32:47,400 --> 00:32:48,870 gives you another vector. 418 00:32:48,870 --> 00:32:49,940 It's a linear map. 419 00:32:49,940 --> 00:32:51,740 So what is it? 420 00:32:51,740 --> 00:33:00,720 It's 0 times v1 plus 2v2 plus iv3 for the first component. 421 00:33:00,720 --> 00:33:02,780 The first component of the new vector-- I 422 00:33:02,780 --> 00:33:09,450 put the 0v1 just so you see that it just depends on v2 and v3. 423 00:33:09,450 --> 00:33:15,800 The second component is v1 minus iv2 plus 0v3. 424 00:33:19,500 --> 00:33:20,990 Those are not vectors. 425 00:33:20,990 --> 00:33:22,100 These are components. 426 00:33:22,100 --> 00:33:22,950 These are numbers. 427 00:33:22,950 --> 00:33:24,740 So this is just a complex number. 428 00:33:24,740 --> 00:33:27,980 This is another complex number, as it should be. 429 00:33:27,980 --> 00:33:30,090 Acting on three complex numbers gives 430 00:33:30,090 --> 00:33:33,400 you, linearly, three other ones. 431 00:33:33,400 --> 00:33:36,350 And then the third component-- they don't have space there, 432 00:33:36,350 --> 00:33:43,340 so I'll put it here-- 3iv1 plus v2 plus 7v3. 433 00:33:49,640 --> 00:33:54,090 And the question is two questions. 434 00:33:54,090 --> 00:34:06,990 Find T dagger, and write the matrix representations 435 00:34:06,990 --> 00:34:09,969 of T and T dagger. 436 00:34:09,969 --> 00:34:19,830 Write the matrices T and T dagger 437 00:34:19,830 --> 00:34:23,730 using the standard basis in which the three basis 438 00:34:23,730 --> 00:34:33,440 vectors are 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, and 0, 0, 1. 439 00:34:33,440 --> 00:34:38,920 These are the three basis vectors-- e1, e2, and e3. 440 00:34:38,920 --> 00:34:42,590 You know, to write the matrix you need the basis vectors. 441 00:34:42,590 --> 00:34:45,300 So that's a problem. 442 00:34:45,300 --> 00:34:49,080 It's a good problem in order to practice, 443 00:34:49,080 --> 00:34:52,739 to see that you understand how to turn 444 00:34:52,739 --> 00:34:54,690 an operator into a matrix. 445 00:34:54,690 --> 00:34:56,639 And you don't get confused. 446 00:34:56,639 --> 00:34:57,540 Is it a row? 447 00:34:57,540 --> 00:34:58,540 Is it a column? 448 00:34:58,540 --> 00:35:01,340 How does it go? 449 00:35:01,340 --> 00:35:05,700 So let's do this. 450 00:35:05,700 --> 00:35:10,820 So first we're going to try to find the rules for T dagger. 451 00:35:10,820 --> 00:35:14,230 So we have the following. 452 00:35:14,230 --> 00:35:16,130 You see, you use the basic property. 453 00:35:16,130 --> 00:35:29,312 u on Tv is equal to T dagger u on v. 454 00:35:29,312 --> 00:35:36,870 So let's try to compute the left hand side, and then look at it 455 00:35:36,870 --> 00:35:41,640 and try to see if we could derive the right hand side. 456 00:35:41,640 --> 00:35:45,350 So what is u supposed to be a three component vector? 457 00:35:45,350 --> 00:35:53,360 So for that use, u equals u1, u2, u3. 458 00:35:53,360 --> 00:36:00,520 OK, now implicit in all that is that when somebody tells you-- 459 00:36:00,520 --> 00:36:03,650 OK, you've got a three dimensional complex vector 460 00:36:03,650 --> 00:36:06,610 space what is the inner product? 461 00:36:06,610 --> 00:36:08,800 The inner product is complex conjugate 462 00:36:08,800 --> 00:36:09,880 of the first component. 463 00:36:09,880 --> 00:36:12,850 That's first component of the second, plus complex conjugate 464 00:36:12,850 --> 00:36:16,390 of the second times star, times star. 465 00:36:16,390 --> 00:36:19,770 So it's just a generalization of the dot product, 466 00:36:19,770 --> 00:36:21,865 but you complex conjugate the first entries. 467 00:36:25,280 --> 00:36:27,080 So what is this? 468 00:36:27,080 --> 00:36:30,430 I should take the complex conjugate of the first term 469 00:36:30,430 --> 00:36:34,350 here-- u1-- times the first one. 470 00:36:34,350 --> 00:36:38,940 So I have 2v2 plus iv3. 471 00:36:41,570 --> 00:36:45,660 This is the left hand side, plus the complex conjugate 472 00:36:45,660 --> 00:36:47,260 of the second component-- there's 473 00:36:47,260 --> 00:36:56,490 the second component-- so u2 times v1 minus iv2 474 00:36:56,490 --> 00:36:59,380 plus-- well, 0v3-- his time I won't write 475 00:36:59,380 --> 00:37:07,010 it-- plus u3 bar times the last vector, 476 00:37:07,010 --> 00:37:11,710 which is 3iv1 plus v2 plus 7v3. 477 00:37:15,460 --> 00:37:17,450 OK, that's the left hand side. 478 00:37:20,810 --> 00:37:24,320 I think I'm going to use this blackboard here, 479 00:37:24,320 --> 00:37:26,490 because otherwise the numbers are going 480 00:37:26,490 --> 00:37:29,160 to be hard to see from one side to the other. 481 00:37:29,160 --> 00:37:36,235 So this information, those two little proofs, 482 00:37:36,235 --> 00:37:36,985 are to be deleted. 483 00:37:40,150 --> 00:37:45,190 And now we have this left hand side. 484 00:37:45,190 --> 00:37:49,090 Now, somehow when you say, OK, now I'm 485 00:37:49,090 --> 00:37:54,940 going to try to figure out this right hand side your head goes 486 00:37:54,940 --> 00:37:56,330 and looks in there and says well, 487 00:37:56,330 --> 00:38:00,250 in the left hand side the u's are sort of the ones that 488 00:38:00,250 --> 00:38:05,000 are alone, and the v's are acted upon. 489 00:38:05,000 --> 00:38:07,230 Here the v's must be alone. 490 00:38:07,230 --> 00:38:12,760 So what I should do is collect along v. 491 00:38:12,760 --> 00:38:20,160 So let's collect along v. So let's put "something" times v1 492 00:38:20,160 --> 00:38:26,150 plus "something" like v2 plus "something" like v3. 493 00:38:26,150 --> 00:38:31,400 And then I will know what is the vector T star this. 494 00:38:31,400 --> 00:38:32,710 So let's do that. 495 00:38:32,710 --> 00:38:34,530 So v1, let's collect. 496 00:38:34,530 --> 00:38:41,750 So you get u2 bar for this v1, and 3iu3 bar. 497 00:38:46,530 --> 00:39:05,220 v2 will have 2u1 bar minus iu2 bar plus u3 bar. 498 00:39:11,180 --> 00:39:12,860 I think I got them right. 499 00:39:12,860 --> 00:39:14,430 OK. 500 00:39:14,430 --> 00:39:21,840 And then v3, let's collect-- iu1 bar, nothing here, 501 00:39:21,840 --> 00:39:28,300 and v3 7u3 bar. 502 00:39:28,300 --> 00:39:34,240 OK, and now I must say, OK, this is the inner product 503 00:39:34,240 --> 00:39:38,320 of T dagger u times v3. 504 00:39:38,320 --> 00:39:46,900 So actually, T dagger on u, which is u1, u2, u3, 505 00:39:46,900 --> 00:39:52,980 must be this vector with three components for which this thing 506 00:39:52,980 --> 00:40:02,120 is the inner product of this vector with the vector V. 507 00:40:02,120 --> 00:40:04,290 So I look at this I say, well, what 508 00:40:04,290 --> 00:40:06,600 was the formula for the inner product? 509 00:40:06,600 --> 00:40:10,960 Well, you complex conjugate the first entry of this 510 00:40:10,960 --> 00:40:12,840 and multiply by the first entry of that. 511 00:40:12,840 --> 00:40:14,900 Complex conjugate the second entry. 512 00:40:14,900 --> 00:40:23,970 So here I should put u2 minus 3iu3, 513 00:40:23,970 --> 00:40:25,940 because the complex conjugate of that 514 00:40:25,940 --> 00:40:28,340 is that as multiplied by v1. 515 00:40:28,340 --> 00:40:35,675 So here I continue-- 2u1 plus iu2 plus u3. 516 00:40:38,540 --> 00:40:43,680 And, finally, minus iu1 plus 7u3. 517 00:40:46,690 --> 00:40:48,960 And that's the answer for this operator. 518 00:41:00,260 --> 00:41:01,990 So the operator is there for you. 519 00:41:01,990 --> 00:41:04,300 The only thing we haven't done is the matrices. 520 00:41:07,870 --> 00:41:11,180 Let me do a little piece of one, and you 521 00:41:11,180 --> 00:41:12,710 try to compute the rest. 522 00:41:12,710 --> 00:41:14,890 Make sure you understand it. 523 00:41:14,890 --> 00:41:22,190 So suppose you get T on the basis vector e1. 524 00:41:22,190 --> 00:41:23,980 It's easier than what it looks. 525 00:41:23,980 --> 00:41:26,620 I'm going to have to write some things in order 526 00:41:26,620 --> 00:41:28,950 to give you a few components, but then 527 00:41:28,950 --> 00:41:32,450 once you get a little practice, or you look what it means, 528 00:41:32,450 --> 00:41:33,670 it will become clear. 529 00:41:33,670 --> 00:41:35,520 So what is T on e1? 530 00:41:35,520 --> 00:41:40,990 Well, it's T on the vector 1, 0, 0. 531 00:41:40,990 --> 00:41:45,750 T on the vector 1, 0, 0-- look at the top formula ther3-- 532 00:41:45,750 --> 00:41:50,870 is equal to 0, 1, and 3i. 533 00:41:56,710 --> 00:42:02,660 Top formula-- the v1 is 1, and all others are 0. 534 00:42:02,660 --> 00:42:06,995 And this is e2 plus 3ie3. 535 00:42:10,330 --> 00:42:14,150 So how do you read, now, matrix elements? 536 00:42:14,150 --> 00:42:20,230 You remember the formula that T on ei 537 00:42:20,230 --> 00:42:27,560 is supposed to be Tkiek-- sum over k. 538 00:42:27,560 --> 00:42:36,770 So this thing is supposed to be equal to T11e1 539 00:42:36,770 --> 00:42:41,840 plus T21e2 plus T31e3. 540 00:42:45,340 --> 00:42:51,350 Your sum over the first index, T of e1, is there for that. 541 00:42:51,350 --> 00:42:58,280 So then I read this, and I see that T21 is equal to 1. 542 00:42:58,280 --> 00:43:01,050 This is equal to 3i. 543 00:43:01,050 --> 00:43:03,580 And this is equal to 0. 544 00:43:03,580 --> 00:43:07,140 So you've got a piece of the matrix, 545 00:43:07,140 --> 00:43:13,140 and the rest I will just tell you how you see it. 546 00:43:13,140 --> 00:43:17,240 But you should check it. 547 00:43:17,240 --> 00:43:19,790 You don't have to write that much after you 548 00:43:19,790 --> 00:43:23,130 have a little practice with this. 549 00:43:23,130 --> 00:43:27,370 But, the matrix T-- what you've learned 550 00:43:27,370 --> 00:43:29,640 is that you have 0, 1, and 3i. 551 00:43:33,070 --> 00:43:38,890 So 0, 1, and 3i are these numbers, 552 00:43:38,890 --> 00:43:42,280 in fact-- 0, 1, and 3i. 553 00:43:42,280 --> 00:43:43,460 And they go vertical. 554 00:43:43,460 --> 00:43:48,580 So 2, minus i, and 1 is the next column. 555 00:43:52,140 --> 00:43:56,070 2, minus i, and 1 is the next column, 556 00:43:56,070 --> 00:44:00,580 and the third one would be i-- look at the v3 there. 557 00:44:00,580 --> 00:44:06,500 It has an i for the first entry, a 0 for the second, and a 7. 558 00:44:06,500 --> 00:44:07,620 So this is the matrix. 559 00:44:13,790 --> 00:44:16,360 How about the matrix T dagger? 560 00:44:19,760 --> 00:44:23,470 Same thing-- once you've done one, don't worry. 561 00:44:23,470 --> 00:44:24,370 Don't do the one. 562 00:44:24,370 --> 00:44:28,040 So this you look for the first column. 563 00:44:28,040 --> 00:44:32,505 It's going to be a 0-- no u1 here-- a 2, and a minus i. 564 00:44:38,070 --> 00:44:48,390 0, 2, and a minus i, then 1, i, and 0, minus 3i, 1, and 7. 565 00:44:48,390 --> 00:44:51,190 And those are it. 566 00:44:51,190 --> 00:44:54,150 And look how nice. 567 00:44:54,150 --> 00:44:58,430 The second one is in fact the Hermitian conjugate 568 00:44:58,430 --> 00:44:59,120 of the other. 569 00:44:59,120 --> 00:45:01,890 Transpose and complex conjugate gives it to you. 570 00:45:04,750 --> 00:45:09,090 So that example suggests that that, of course, 571 00:45:09,090 --> 00:45:11,830 is not an accident. 572 00:45:11,830 --> 00:45:14,225 So what do you need for that to happen? 573 00:45:16,810 --> 00:45:21,020 Nobody said that what you're supposed to do to find T dagger 574 00:45:21,020 --> 00:45:23,260 is transpose some complex conjugate, 575 00:45:23,260 --> 00:45:27,840 but somehow that's what you do once you have the matrix, 576 00:45:27,840 --> 00:45:29,550 or at least what it seems that you 577 00:45:29,550 --> 00:45:30,940 do when you have the matrix. 578 00:45:30,940 --> 00:45:35,040 So let's see if we can get that more generally. 579 00:45:35,040 --> 00:45:37,100 So end of example. 580 00:45:37,100 --> 00:45:48,170 Look at T dagger u, v is equal to u, Tv. 581 00:45:48,170 --> 00:45:50,230 We know this is the key equation. 582 00:45:50,230 --> 00:45:53,130 Everything comes from this. 583 00:45:53,130 --> 00:45:56,650 Now take u and v to be orthonormal vectors, 584 00:45:56,650 --> 00:46:03,930 so u equal ei, and v equal ej. 585 00:46:03,930 --> 00:46:05,990 And these are orthonormal. 586 00:46:05,990 --> 00:46:08,730 The e's are going to be orthonormal each time 587 00:46:08,730 --> 00:46:12,300 we say basis vectors-- e, orthonormal. 588 00:46:12,300 --> 00:46:21,942 So put them here, so you get T dagger on ei times ej 589 00:46:21,942 --> 00:46:25,810 is equal to ei, Tej. 590 00:46:29,480 --> 00:46:36,590 Now use the matrix action on these operators. 591 00:46:36,590 --> 00:46:43,848 So T dagger on ei is supposed to be T dagger kiek. 592 00:46:48,700 --> 00:46:52,870 The equation is something worth knowing by heart. 593 00:46:52,870 --> 00:46:55,330 What is the matrix representation? 594 00:46:55,330 --> 00:46:58,530 If the index of the vector goes here, 595 00:46:58,530 --> 00:47:00,610 the sum index goes like that. 596 00:47:00,610 --> 00:47:06,910 So then you have ej here, and here you have ei, 597 00:47:06,910 --> 00:47:08,345 and you have Tkjek. 598 00:47:33,530 --> 00:47:36,470 So now this basis orthonormal. 599 00:47:36,470 --> 00:47:41,570 This is a number, and this is the basis. 600 00:47:41,570 --> 00:47:43,860 The number goes out. 601 00:47:43,860 --> 00:47:50,520 T dagger ki-- remember, it's on the left side, 602 00:47:50,520 --> 00:47:53,540 so it should go out with a star. 603 00:47:53,540 --> 00:47:56,970 And then you have ekej. 604 00:47:56,970 --> 00:48:00,880 That's orthonormal, so it's delta kej. 605 00:48:00,880 --> 00:48:03,910 The number here goes out as well, 606 00:48:03,910 --> 00:48:06,220 and the inner product gives delta ik. 607 00:48:09,360 --> 00:48:10,585 So what do we get? 608 00:48:14,470 --> 00:48:24,140 T dagger ji star is equal to Tij. 609 00:48:29,090 --> 00:48:33,495 First, change i for j, so it looks more familiar. 610 00:48:33,495 --> 00:48:39,500 So then you have T dagger ij star is equal to Tji. 611 00:48:42,710 --> 00:48:46,730 And then take complex conjugate, so that finally you 612 00:48:46,730 --> 00:48:53,843 have T dagger ij is equal to Tji star. 613 00:48:56,860 --> 00:49:04,050 And that shows that, as long as you have an orthonormal basis 614 00:49:04,050 --> 00:49:10,750 you can see the Hermitian conjugate of the operator 615 00:49:10,750 --> 00:49:13,600 by taking the matrix, and then what you usually 616 00:49:13,600 --> 00:49:16,730 call the Hermitian conjugate of the matrix. 617 00:49:16,730 --> 00:49:20,360 But I want to emphasize that, if you 618 00:49:20,360 --> 00:49:23,880 didn't have an orthonormal basis-- if you 619 00:49:23,880 --> 00:49:30,640 have your operator, and you want to calculate the dagger of it, 620 00:49:30,640 --> 00:49:32,600 and you find its matrix representation. 621 00:49:32,600 --> 00:49:35,790 You take the Hermitian conjugate of the matrix. 622 00:49:35,790 --> 00:49:41,910 It would be wrong if your basis vectors are not orthonormal. 623 00:49:41,910 --> 00:49:42,880 It just fails. 624 00:49:42,880 --> 00:49:46,050 So what would happen if the basis vectors are not 625 00:49:46,050 --> 00:49:47,390 orthonormal? 626 00:49:47,390 --> 00:49:57,460 Instead of having ei with ej giving you delta iej, 627 00:49:57,460 --> 00:50:02,930 you have that ei with ej is some number. 628 00:50:02,930 --> 00:50:07,830 And you can call it aij, or alpha iej, or gij, I think, 629 00:50:07,830 --> 00:50:09,510 is maybe a better name. 630 00:50:09,510 --> 00:50:21,550 So if the basis is not orthonormal, then ei with ej 631 00:50:21,550 --> 00:50:25,540 is some sort of gij. 632 00:50:28,190 --> 00:50:30,020 And then you go back here. 633 00:50:30,020 --> 00:50:36,970 And, instead of having deltas here, you would have g's. 634 00:50:36,970 --> 00:50:48,240 So you would have the T dagger star ki with gkj 635 00:50:48,240 --> 00:50:53,808 is equal to Tkj, gik. 636 00:50:58,380 --> 00:51:01,480 And there's no such simple thing as saying, 637 00:51:01,480 --> 00:51:04,310 oh, well you just take the matrix and complex 638 00:51:04,310 --> 00:51:06,260 conjugate and transpose. 639 00:51:06,260 --> 00:51:08,250 That's not the dagger. 640 00:51:08,250 --> 00:51:11,920 It's more complicated than that. 641 00:51:11,920 --> 00:51:15,180 If this matrix should be invertible, 642 00:51:15,180 --> 00:51:17,320 you could pass this to the other side 643 00:51:17,320 --> 00:51:19,590 using the inverse of this matrix. 644 00:51:19,590 --> 00:51:23,450 And you can find a formula for the dagger 645 00:51:23,450 --> 00:51:27,970 in terms of the g matrix, its inverses and multiplications. 646 00:51:27,970 --> 00:51:29,935 So what do you learn from here? 647 00:51:29,935 --> 00:51:35,730 You learn a fundamental fact, that the statement 648 00:51:35,730 --> 00:51:39,010 that an operator-- for example, you have T. 649 00:51:39,010 --> 00:51:43,510 And you can find T dagger as the adjoint. 650 00:51:43,510 --> 00:51:48,360 The adjoint operator, or the Hermitian conjugate operator, 651 00:51:48,360 --> 00:51:52,420 has a basis independent definition. 652 00:51:52,420 --> 00:51:56,350 It just needs that statement that we've 653 00:51:56,350 --> 00:52:02,050 written many times now, that T dagger u, 654 00:52:02,050 --> 00:52:06,106 v is defined via this relation. 655 00:52:06,106 --> 00:52:08,680 And it has nothing to do with a basis. 656 00:52:08,680 --> 00:52:11,030 It's true for arbitrary vectors. 657 00:52:11,030 --> 00:52:14,170 Nevertheless, how you construct T dagger, 658 00:52:14,170 --> 00:52:19,360 if you have a basis-- well, sometimes 659 00:52:19,360 --> 00:52:22,360 it's a Hermitian conjugate matrix, 660 00:52:22,360 --> 00:52:25,110 if your basis is orthonormal. 661 00:52:25,110 --> 00:52:28,700 But that statement, that the dagger is the Hermitian 662 00:52:28,700 --> 00:52:33,090 conjugate basis, is a little basis dependent, 663 00:52:33,090 --> 00:52:38,030 is not a universal fact about the adjoint. 664 00:52:38,030 --> 00:52:40,330 It's not always constructed that way. 665 00:52:40,330 --> 00:52:44,596 And there will be examples where you will see that. 666 00:52:44,596 --> 00:52:45,096 Questions? 667 00:52:51,806 --> 00:52:52,802 No questions? 668 00:53:00,290 --> 00:53:04,500 Well, let's do brackets for a few minutes 669 00:53:04,500 --> 00:53:08,490 so that you see a few properties of them. 670 00:53:08,490 --> 00:53:13,520 With the same language, I'll write formulas that we've-- OK, 671 00:53:13,520 --> 00:53:16,190 I wrote a formula here, in fact. 672 00:53:16,190 --> 00:53:21,160 So for example, this formula-- if I 673 00:53:21,160 --> 00:53:28,110 want to write it with bras and kets, I would write u Tv. 674 00:53:28,110 --> 00:53:34,360 And I could also write it as u T v, 675 00:53:34,360 --> 00:53:38,620 because remember this means-- the bra and the ket-- 676 00:53:38,620 --> 00:53:43,270 just says a way to make clear that this object is a vector. 677 00:53:43,270 --> 00:53:47,060 But this vector is obtained by acting T on the vector 678 00:53:47,060 --> 00:53:52,770 v. So it's T on the vector v, because a vector v is just 679 00:53:52,770 --> 00:53:55,770 something, and when you put it like that that's still 680 00:53:55,770 --> 00:54:01,850 the vector v. The kit doesn't do much to it. 681 00:54:01,850 --> 00:54:03,670 It's almost like putting an arrow, 682 00:54:03,670 --> 00:54:08,560 so that's why this thing is really this thing as well. 683 00:54:08,560 --> 00:54:10,500 Now, on the other hand, this thing-- 684 00:54:10,500 --> 00:54:19,350 let's say that this is equal to v, T dagger u star. 685 00:54:19,350 --> 00:54:26,295 So then you would put here that this is v T dagger u star. 686 00:54:31,470 --> 00:54:35,230 So this formula is something that most people remember 687 00:54:35,230 --> 00:54:39,750 in physics, written perhaps a little differently. 688 00:54:39,750 --> 00:54:44,460 Change v and u so that this left hand side now 689 00:54:44,460 --> 00:54:52,520 reads u T dagger v. And it has a star, 690 00:54:52,520 --> 00:54:58,780 and the right hand side would become v T u. 691 00:54:58,780 --> 00:55:01,170 And just complex conjugated it. 692 00:55:01,170 --> 00:55:16,550 So u T dagger v is equal to v T u star-- a nice formula that 693 00:55:16,550 --> 00:55:22,530 says how do you get to understand what T dagger is. 694 00:55:22,530 --> 00:55:25,870 Well, if you know T dagger's value 695 00:55:25,870 --> 00:55:29,014 in between any set of states, then you 696 00:55:29,014 --> 00:55:34,300 know-- well, if you know T between any set of states 697 00:55:34,300 --> 00:55:36,400 u and v, then you can figure out what 698 00:55:36,400 --> 00:55:43,100 T dagger is between any same two states by using this formula. 699 00:55:43,100 --> 00:55:45,210 What you have to do is that this thing 700 00:55:45,210 --> 00:55:47,230 is equal to the reverse thing. 701 00:55:47,230 --> 00:55:50,600 So you go from right to left and reverse it here. 702 00:55:50,600 --> 00:55:54,990 So you go v, then T, then u, and you put a star, 703 00:55:54,990 --> 00:55:58,595 and that gives you that object. 704 00:56:01,400 --> 00:56:04,380 Another thing that we've been doing all the time when 705 00:56:04,380 --> 00:56:11,605 we calculate, for example, ei, T on ej. 706 00:56:14,400 --> 00:56:17,900 What is this? 707 00:56:17,900 --> 00:56:20,050 Well, you know what this is. 708 00:56:20,050 --> 00:56:23,160 Let's write it like that-- ei. 709 00:56:23,160 --> 00:56:32,368 Now T on ej is the matrix T kjek. 710 00:56:32,368 --> 00:56:39,460 If this is an orthonormal basis, here is a delta iek. 711 00:56:39,460 --> 00:56:43,930 So this is nothing else but Tij. 712 00:56:43,930 --> 00:56:49,200 So another way of writing that matrix element, ij, of a matrix 713 00:56:49,200 --> 00:56:54,840 is to put an ei, an ej here, and a T here. 714 00:56:54,840 --> 00:57:07,240 So people write it like that-- Tij is ei comma Tej. 715 00:57:07,240 --> 00:57:13,960 Or, in bracket language, they put ei T ej. 716 00:57:21,330 --> 00:57:24,860 So I need it to be flexible and just 717 00:57:24,860 --> 00:57:27,720 be able to pass from one notation to the other, 718 00:57:27,720 --> 00:57:31,890 because it helps you. 719 00:57:31,890 --> 00:57:39,900 One of the most helpful things in this object 720 00:57:39,900 --> 00:57:44,960 is to understand, for example, in bra and ket notation, what 721 00:57:44,960 --> 00:57:48,070 is the following object? 722 00:57:48,070 --> 00:57:52,910 What is ei ei? 723 00:57:57,370 --> 00:58:00,950 This seems like the wrong kind of thing, 724 00:58:00,950 --> 00:58:05,880 because you were supposed to have bras acting on vectors. 725 00:58:05,880 --> 00:58:07,980 So this would be on the left of that, 726 00:58:07,980 --> 00:58:10,000 but otherwise it would be too trivial. 727 00:58:10,000 --> 00:58:13,140 If it would be on the left of it, it would give you a number. 728 00:58:13,140 --> 00:58:18,920 But think of this thing as a object that stands there. 729 00:58:18,920 --> 00:58:23,470 And it's repeated endlessly, so it's summed. 730 00:58:23,470 --> 00:58:25,680 So what is this object? 731 00:58:25,680 --> 00:58:29,210 Well, this object is a sum of things like that, 732 00:58:29,210 --> 00:58:41,200 so this is really e1 e1 plus e2 e2, and it goes on like that. 733 00:58:43,790 --> 00:58:47,310 Well, let it act on a vector. 734 00:58:47,310 --> 00:58:50,160 This kind of object is an operator. 735 00:58:50,160 --> 00:58:52,210 Whenever you have the bra and the ket 736 00:58:52,210 --> 00:58:54,930 sort of in this wrong position-- the ket first, 737 00:58:54,930 --> 00:58:59,210 and the bra afterwards-- this is, in Dirac's notation, 738 00:58:59,210 --> 00:59:03,160 an operator, a particular operator. 739 00:59:03,160 --> 00:59:06,610 And you will see in general how it is the general operator 740 00:59:06,610 --> 00:59:07,660 very soon. 741 00:59:07,660 --> 00:59:09,140 So look at this. 742 00:59:09,140 --> 00:59:12,120 You have something like that, and why do we 743 00:59:12,120 --> 00:59:14,210 call it an operator? 744 00:59:14,210 --> 00:59:16,130 We call it an operator business if it 745 00:59:16,130 --> 00:59:19,355 acts on a vector-- you put a vector here, 746 00:59:19,355 --> 00:59:23,760 a bra-- this becomes a number, and there's still 747 00:59:23,760 --> 00:59:25,310 a vector left. 748 00:59:25,310 --> 00:59:34,480 So this kind of structure, acting on something like that, 749 00:59:34,480 --> 00:59:38,660 gives a vector, because this thing goes in here, 750 00:59:38,660 --> 00:59:41,840 produces a number, and the vector is left there. 751 00:59:41,840 --> 00:59:53,041 So for example, if you act with this thing on the vector a-- 752 00:59:53,041 --> 01:00:00,720 an arbitrary vector a-- what do you get? 753 01:00:03,810 --> 01:00:07,226 Whatever this operator is is acted on a. 754 01:00:07,226 --> 01:00:12,230 Well, you remember that these thing are the components of a, 755 01:00:12,230 --> 01:00:13,870 and these are the basis vectors. 756 01:00:13,870 --> 01:00:17,795 So this is nothing else but the vector a again. 757 01:00:22,710 --> 01:00:29,530 You see, you can start with a equals some alpha i's with 758 01:00:29,530 --> 01:00:33,540 ei's, and then you calculate what are the alpha i's. 759 01:00:33,540 --> 01:00:41,610 You put an ej a, and this ej on that gives you alpha j. 760 01:00:41,610 --> 01:00:44,430 So alpha j-- these numbers are nothing else 761 01:00:44,430 --> 01:00:47,190 but these things, these numbers. 762 01:00:47,190 --> 01:00:51,050 So here you have the number times the vector. 763 01:00:51,050 --> 01:00:55,470 The only difference is that this is like ei alpha i. 764 01:00:55,470 --> 01:00:58,650 The number has been to the right. 765 01:00:58,650 --> 01:01:02,680 So this thing acting on any vector is the vector itself. 766 01:01:02,680 --> 01:01:07,280 So this is perhaps the most fundamental relation 767 01:01:07,280 --> 01:01:15,270 in bracket notation, is that the identity operator is this. 768 01:01:19,664 --> 01:01:20,163 Yes. 769 01:01:20,163 --> 01:01:23,040 AUDIENCE: Is that just 1 e of i, or sum over all e of i? 770 01:01:23,040 --> 01:01:24,475 PROFESSOR: It's sum of over all. 771 01:01:24,475 --> 01:01:33,568 So here implicit sum is the sum of all up to en en. 772 01:01:33,568 --> 01:01:36,430 You will see, if you take just one of them, 773 01:01:36,430 --> 01:01:41,050 you will get what is an orthogonal projector. 774 01:01:41,050 --> 01:01:43,590 Now this allows you to do another piece 775 01:01:43,590 --> 01:01:49,860 of very nice Dirac notation. 776 01:01:49,860 --> 01:01:52,005 So let's do that. 777 01:01:59,240 --> 01:02:09,610 Suppose you have an operator T. You put a 1 778 01:02:09,610 --> 01:02:15,140 in front of it-- a T and a 1 in front of it. 779 01:02:15,140 --> 01:02:22,070 And then you say, OK, this 1, I'll put ei ei. 780 01:02:22,070 --> 01:02:31,650 Then comes the T, and then comes the ej ej-- another 1. 781 01:02:35,460 --> 01:02:37,850 And then you look at that and you suddenly 782 01:02:37,850 --> 01:02:40,851 see a number lying there. 783 01:02:40,851 --> 01:02:41,350 Why? 784 01:02:41,350 --> 01:02:43,165 Because this thing is some number. 785 01:02:45,790 --> 01:02:47,901 So this is the magic of the Dirac notation. 786 01:02:47,901 --> 01:02:49,650 You write all this thing, and suddenly you 787 01:02:49,650 --> 01:02:53,750 see numbers have been created in between. 788 01:02:53,750 --> 01:02:59,770 This number is nothing else but this matrix representation 789 01:02:59,770 --> 01:03:01,140 of the operator. 790 01:03:01,140 --> 01:03:04,550 T, between this, is Tij. 791 01:03:04,550 --> 01:03:12,445 So this is ei Tij ej. 792 01:03:16,990 --> 01:03:22,610 So this formula is very fundamental. 793 01:03:22,610 --> 01:03:27,210 It shows that the most general operator that you can ever 794 01:03:27,210 --> 01:03:33,770 invent is some sort of ket before a bra, 795 01:03:33,770 --> 01:03:37,470 and then you superimpose them with these numbers which 796 01:03:37,470 --> 01:03:43,280 actually happen to be the matrix representation of the operator. 797 01:03:43,280 --> 01:03:45,860 So the operator can be written as a sum 798 01:03:45,860 --> 01:03:51,180 of, if this is an n by n matrix n squared thinks of this form-- 799 01:03:51,180 --> 01:03:55,300 1 with 1, 1 with 2, 1 with 3, and all of them. 800 01:03:55,300 --> 01:03:58,340 Bu then, you know this formula is so important 801 01:03:58,340 --> 01:04:00,770 that people make sure that you realize 802 01:04:00,770 --> 01:04:03,420 that you're summing over i and j. 803 01:04:03,420 --> 01:04:05,190 So just put it there. 804 01:04:08,010 --> 01:04:12,730 Given an operator, these are its matrix elements. 805 01:04:12,730 --> 01:04:19,340 And this is the operator written back in abstract notation. 806 01:04:19,340 --> 01:04:21,370 The whole operator is back there for you. 807 01:04:24,450 --> 01:04:27,440 I want to use the last part of the lecture 808 01:04:27,440 --> 01:04:35,790 to discuss a theorem that is pretty interesting, 809 01:04:35,790 --> 01:04:39,230 that allows you to understand things 810 01:04:39,230 --> 01:04:44,580 about all these Hermitian operators and unitary operators 811 01:04:44,580 --> 01:04:46,620 much more clearly. 812 01:04:46,620 --> 01:04:51,400 And it's a little mysterious, this theorem, 813 01:04:51,400 --> 01:04:53,720 and let's see how it goes. 814 01:04:58,140 --> 01:05:01,650 So any questions about this Dirac notation at this moment, 815 01:05:01,650 --> 01:05:03,730 anything that I wrote there? 816 01:05:03,730 --> 01:05:05,720 It takes a while to get accustomed 817 01:05:05,720 --> 01:05:06,940 to the Dirac notation. 818 01:05:06,940 --> 01:05:10,540 But once you get the hang of it, it's 819 01:05:10,540 --> 01:05:13,425 sort of fun and easy to manipulate. 820 01:05:16,070 --> 01:05:17,480 No questions? 821 01:05:17,480 --> 01:05:18,458 Can't be. 822 01:05:22,860 --> 01:05:24,670 You can prove all kinds of things 823 01:05:24,670 --> 01:05:29,230 with this matrix representation of the identity. 824 01:05:29,230 --> 01:05:32,140 For example, you can prove easily 825 01:05:32,140 --> 01:05:35,420 something you proved already, that when you multiply 826 01:05:35,420 --> 01:05:39,341 two operators the matrices multiply. 827 01:05:39,341 --> 01:05:41,120 You can prove all kinds of things. 828 01:05:41,120 --> 01:05:43,180 Pretty much everything we've done 829 01:05:43,180 --> 01:05:45,530 can also be proven this way. 830 01:05:45,530 --> 01:05:49,500 OK, so here comes the theorem I want to ask you about. 831 01:05:49,500 --> 01:05:54,870 Suppose somebody comes along, and they tell you, 832 01:05:54,870 --> 01:05:59,160 well, you know, here's a vector v, 833 01:05:59,160 --> 01:06:04,520 and I'm going to have a linear operator acting on this space. 834 01:06:04,520 --> 01:06:07,300 So the operator's going to be T, and I'm 835 01:06:07,300 --> 01:06:12,770 going act with the vector v. 836 01:06:12,770 --> 01:06:15,560 And moreover, I find that this is 837 01:06:15,560 --> 01:06:26,150 0 for all vectors v belonging to the vector space. 838 01:06:26,150 --> 01:06:30,280 And the question is-- what can we say about this operator? 839 01:06:35,950 --> 01:06:39,050 From all vectors it's just 0. 840 01:06:39,050 --> 01:06:43,290 So is this operator 0, maybe? 841 01:06:43,290 --> 01:06:44,680 Does it have to be 0? 842 01:06:44,680 --> 01:06:46,245 Can it be something else? 843 01:06:48,800 --> 01:06:53,120 OK, we've been talking about real and complex vector spaces. 844 01:06:53,120 --> 01:06:55,720 And we've seen that it's different. 845 01:06:55,720 --> 01:06:59,000 The inner product is a little different. 846 01:06:59,000 --> 01:07:01,410 But let's think about this. 847 01:07:01,410 --> 01:07:05,530 Take two dimensions, real vector space. 848 01:07:05,530 --> 01:07:07,760 The operator that takes any vector 849 01:07:07,760 --> 01:07:11,440 and rotates it by 90 degrees, that's a linear operator. 850 01:07:14,300 --> 01:07:20,570 And that is a non-trivial linear operator, and it gives you 0. 851 01:07:20,570 --> 01:07:26,710 So case settled-- there's no theorem here, nothing 852 01:07:26,710 --> 01:07:28,500 you can say about this operator. 853 01:07:28,500 --> 01:07:31,140 It may be non-zero. 854 01:07:31,140 --> 01:07:33,440 But here comes the catch. 855 01:07:33,440 --> 01:07:40,010 If you're talking complex vector spaces, T is 0. 856 01:07:40,010 --> 01:07:42,350 It just is 0, can't be anything else. 857 01:07:42,350 --> 01:07:44,460 Complex vector spaces are different. 858 01:07:44,460 --> 01:07:48,540 You can't quite do that thing-- rotate all vectors by something 859 01:07:48,540 --> 01:07:50,420 and do things. 860 01:07:50,420 --> 01:07:53,020 So that's a theorem we want to understand. 861 01:07:53,020 --> 01:08:05,830 Theorem-- let v be a complex inner product space. 862 01:08:10,270 --> 01:08:13,810 By that is a complex vector space with an inner product. 863 01:08:13,810 --> 01:08:25,890 Then v, Tv equals 0 for all v implies 864 01:08:25,890 --> 01:08:28,170 that the operator is just 0. 865 01:08:35,580 --> 01:08:39,930 I traced a lot of my confusions in quantum mechanics 866 01:08:39,930 --> 01:08:42,390 to not knowing about this theorem, 867 01:08:42,390 --> 01:08:46,109 that somehow it must be true. 868 01:08:46,109 --> 01:08:48,520 I don't know why it should be true, 869 01:08:48,520 --> 01:08:56,250 but somehow it's not, because it really has exceptions. 870 01:08:56,250 --> 01:08:57,300 So here it is. 871 01:08:57,300 --> 01:09:01,340 We tried to prove that. 872 01:09:01,340 --> 01:09:05,060 It's so important, I think, that it should be proven. 873 01:09:05,060 --> 01:09:08,590 And how could you prove that? 874 01:09:08,590 --> 01:09:12,330 And at first sight it seems it's going to be difficult, 875 01:09:12,330 --> 01:09:16,490 because, if I do just a formal proof, 876 01:09:16,490 --> 01:09:18,760 how is it going to know that I'm not 877 01:09:18,760 --> 01:09:21,069 talking real or complex vector spaces. 878 01:09:21,069 --> 01:09:24,430 So it must make a crucial difference in the proof 879 01:09:24,430 --> 01:09:27,359 whether it's real or complex. 880 01:09:27,359 --> 01:09:31,939 So this property really sets the complex vector spaces 881 01:09:31,939 --> 01:09:33,970 quite apart from the real ones. 882 01:09:33,970 --> 01:09:38,899 So let's see what you would need to do. 883 01:09:38,899 --> 01:10:01,630 Well, here's a strategy-- if I could prove that u, 884 01:10:01,630 --> 01:10:10,800 Tv is equal to 0 for all u and all v. You see, 885 01:10:10,800 --> 01:10:14,010 the problem here is that these two are the same vector. 886 01:10:14,010 --> 01:10:17,320 They're all vectors, but they're the same vector. 887 01:10:17,320 --> 01:10:23,230 If I could prove that this is 0 for all u and v, then 888 01:10:23,230 --> 01:10:24,380 what would I say? 889 01:10:24,380 --> 01:10:27,900 I would say, oh, if this is 0 for all u and v, 890 01:10:27,900 --> 01:10:34,040 then pick u equal to Tv. 891 01:10:34,040 --> 01:10:39,410 And then you find that Tv, Tv is 0, 892 01:10:39,410 --> 01:10:43,600 therefore Tv is the 0 vector. 893 01:10:43,600 --> 01:10:45,660 By the axiom of the inner product, 894 01:10:45,660 --> 01:10:50,580 for all v is a 0 vector, so T kills all vectors, 895 01:10:50,580 --> 01:10:53,630 therefore T is 0. 896 01:10:53,630 --> 01:11:00,060 So if I could prove this is true, I would be done. 897 01:11:03,150 --> 01:11:05,510 Now, of course, that's the difficulty. 898 01:11:05,510 --> 01:11:09,470 Well, I wouldn't say of course. 899 01:11:09,470 --> 01:11:13,120 This takes a leap of faith to believe 900 01:11:13,120 --> 01:11:16,060 that this is the way you're going to prove that. 901 01:11:16,060 --> 01:11:20,050 You could try to prove this, and then it would follow. 902 01:11:20,050 --> 01:11:21,890 But maybe that's difficult to prove. 903 01:11:21,890 --> 01:11:24,490 But actually that's possible to prove. 904 01:11:24,490 --> 01:11:28,780 But how could you ever prove that this is true? 905 01:11:28,780 --> 01:11:31,690 You could prove it if you could somehow 906 01:11:31,690 --> 01:11:40,626 rewrite u and Tv as some sort of something with a T 907 01:11:40,626 --> 01:11:47,530 and something plus some other thing with a T, 908 01:11:47,530 --> 01:11:52,450 and that other thing plus some-- all kinds of things like that. 909 01:11:52,450 --> 01:11:57,100 Because the things in which this is the same as that are 0. 910 01:11:59,710 --> 01:12:03,660 So if you can do that-- if you could re-express this left hand 911 01:12:03,660 --> 01:12:07,840 side as a sum of things of that kind-- that would be 0. 912 01:12:07,840 --> 01:12:11,700 So let's try. 913 01:12:11,700 --> 01:12:13,840 So what can you try? 914 01:12:13,840 --> 01:12:20,490 You can put u plus v here, and T of u plus v. That 915 01:12:20,490 --> 01:12:25,290 would be 0, because that's a vector, same vector here. 916 01:12:25,290 --> 01:12:29,450 But that's not equal to this, because it has the u, Tu, 917 01:12:29,450 --> 01:12:31,650 and it has the v Tv. 918 01:12:31,650 --> 01:12:34,130 And it has this in a different order. 919 01:12:34,130 --> 01:12:49,500 So maybe we can subtract u minus v, T of u minus v. Well, 920 01:12:49,500 --> 01:12:52,400 we're getting there, but all this 921 01:12:52,400 --> 01:12:59,570 is question marks-- u, Tu, v, Tv-- these cancel-- u, Tu, v, 922 01:12:59,570 --> 01:13:01,460 Tv. 923 01:13:01,460 --> 01:13:04,410 But, the cross-products, what are they? 924 01:13:04,410 --> 01:13:06,114 Well here you have a u, Tv. 925 01:13:08,904 --> 01:13:13,210 And here you have a v, Tu. 926 01:13:13,210 --> 01:13:16,660 And do they cancel? 927 01:13:16,660 --> 01:13:17,630 No. 928 01:13:17,630 --> 01:13:18,540 Let's see. 929 01:13:18,540 --> 01:13:22,160 u, Tv, and up here is u minus Tv about. 930 01:13:22,160 --> 01:13:25,000 But there's another minus, so there's another one there. 931 01:13:25,000 --> 01:13:28,050 And v, Tu has a minus, minus is a plus. 932 01:13:28,050 --> 01:13:32,010 So actually this gives me two of this plus two of that. 933 01:13:32,010 --> 01:13:34,290 OK, it shouldn't have been so easy anyway. 934 01:13:40,090 --> 01:13:43,673 So here is where you have to have the small inspiration. 935 01:13:48,070 --> 01:13:50,170 Somehow it shouldn't have worked, you know. 936 01:13:50,170 --> 01:13:53,860 If this had worked, the theorem would read different. 937 01:13:53,860 --> 01:13:57,240 You could use a real vector space. 938 01:13:57,240 --> 01:13:59,900 Nothing is imaginary there. 939 01:13:59,900 --> 01:14:02,600 So the fact that you have a complex vector space 940 01:14:02,600 --> 01:14:03,290 might help. 941 01:14:03,290 --> 01:14:06,150 So somehow you have to put i's there. 942 01:14:06,150 --> 01:14:08,260 So let's try i's here. 943 01:14:08,260 --> 01:14:16,000 So you put u plus iv and T of u plus iv. 944 01:14:19,090 --> 01:14:23,380 Well, then you probably have to subtract things as well, so 945 01:14:23,380 --> 01:14:29,740 u minus iv, T of u minus iv. 946 01:14:29,740 --> 01:14:34,030 These things will be 0 because of the general structure-- 947 01:14:34,030 --> 01:14:38,020 the same operator here as here. 948 01:14:38,020 --> 01:14:39,620 And let's see what they are. 949 01:14:39,620 --> 01:14:43,700 Well, there's u, Tu, and here's minus u, Tu, 950 01:14:43,700 --> 01:14:49,480 so the diagonal things go away-- the minus iv, minus iv, iv, 951 01:14:49,480 --> 01:14:53,940 and a T. You have minus iv, minus iv subtracted, 952 01:14:53,940 --> 01:14:55,800 so that also cancels. 953 01:14:55,800 --> 01:14:58,790 So there's the cross-products. 954 01:14:58,790 --> 01:15:05,790 Now you will say, well, just like the minus signs, 955 01:15:05,790 --> 01:15:09,050 you're not going to get anything because you're 956 01:15:09,050 --> 01:15:10,300 going to get 2 and 2. 957 01:15:10,300 --> 01:15:13,492 Let's see. 958 01:15:13,492 --> 01:15:15,640 Let's see what we get with this one. 959 01:15:15,640 --> 01:15:23,090 You get u with Tiv, so you get i u, Tv. 960 01:15:26,930 --> 01:15:32,870 But look, this i on the left, however, when you take it out, 961 01:15:32,870 --> 01:15:41,440 becomes a minus i, so you get minus i v, Tu. 962 01:15:46,780 --> 01:15:49,490 And the other products [INAUDIBLE]. 963 01:15:49,490 --> 01:15:56,050 So let's look what you get here-- a u with a minus iv 964 01:15:56,050 --> 01:16:00,366 and a minus here gives you a 2 here. 965 01:16:00,366 --> 01:16:03,450 And the other term, v, Tu-- well, 966 01:16:03,450 --> 01:16:06,570 this goes out as a plus i. 967 01:16:06,570 --> 01:16:11,020 But with a minus, it becomes a minus i, so v, Tu is this. 968 01:16:11,020 --> 01:16:12,185 So there's a 2 here. 969 01:16:17,850 --> 01:16:21,500 So that's what these terms give you. 970 01:16:21,500 --> 01:16:23,700 And now you've succeeded. 971 01:16:23,700 --> 01:16:24,220 Why? 972 01:16:24,220 --> 01:16:27,110 Because the relative sign is negative. 973 01:16:27,110 --> 01:16:28,830 So who cares? 974 01:16:28,830 --> 01:16:32,370 You can divide by i, and divide this by i. 975 01:16:32,370 --> 01:16:34,270 You are constructing something. 976 01:16:34,270 --> 01:16:37,070 So let me put here what you get. 977 01:16:42,230 --> 01:16:43,695 I can erase this blackboard. 978 01:16:50,020 --> 01:16:51,285 So what do we get? 979 01:16:55,860 --> 01:17:03,280 I claim that if you put one quarter of u plus v, 980 01:17:03,280 --> 01:17:15,230 T u plus v minus u minus v, T of u minus v, then, 981 01:17:15,230 --> 01:17:17,680 let's see, what do we need to keep? 982 01:17:17,680 --> 01:17:21,450 We need to keep u and Tv. 983 01:17:21,450 --> 01:17:27,250 So divide this by i plus 1 over i 984 01:17:27,250 --> 01:17:36,400 u plus iv, T of u plus iv minus 1 985 01:17:36,400 --> 01:17:42,730 over i, u minus iv, T of u minus iv. 986 01:17:46,027 --> 01:17:47,490 And close it. 987 01:17:50,700 --> 01:17:52,550 You've divided by i. 988 01:17:52,550 --> 01:17:57,180 You get here four of these ones, zero of these ones, 989 01:17:57,180 --> 01:18:00,820 and you got the answer you wanted. 990 01:18:00,820 --> 01:18:04,180 So this whole thing is written like that, 991 01:18:04,180 --> 01:18:11,270 and now, since this is equal to u with Tv, 992 01:18:11,270 --> 01:18:14,170 by the conditions of the theorem, 993 01:18:14,170 --> 01:18:20,500 any vector-- any vector here-- these are all 0. 994 01:18:20,500 --> 01:18:26,000 You've shown that this is 0, and therefore the operator is 0. 995 01:18:26,000 --> 01:18:28,030 And you should be very satisfied, 996 01:18:28,030 --> 01:18:30,940 because the proof made use of the fact 997 01:18:30,940 --> 01:18:32,840 that it was a complex vector space. 998 01:18:32,840 --> 01:18:36,790 Otherwise you could not add vectors 999 01:18:36,790 --> 01:18:38,250 with an imaginary number. 1000 01:18:38,250 --> 01:18:41,880 And the imaginary number made it all work. 1001 01:18:41,880 --> 01:18:45,010 So the theorem is there. 1002 01:18:45,010 --> 01:18:48,370 It's a pretty useful theorem, so let's use it 1003 01:18:48,370 --> 01:18:50,940 for the most obvious application. 1004 01:18:54,890 --> 01:19:01,170 People say that, whenever you find that v, 1005 01:19:01,170 --> 01:19:13,210 Tv is real for all v, then this operator 1006 01:19:13,210 --> 01:19:16,050 is Hermitian, or self-adjoint. 1007 01:19:16,050 --> 01:19:23,800 That is, then, it implies T dagger equals T. 1008 01:19:23,800 --> 01:19:25,950 So let's show that. 1009 01:19:28,590 --> 01:19:32,620 So let's take v, Tv. 1010 01:19:36,730 --> 01:19:38,962 Proof. 1011 01:19:38,962 --> 01:19:48,180 You take v, Tv, and now this thing is real. 1012 01:19:48,180 --> 01:19:55,795 So since this is real, you can say it's equal to v, Tv star. 1013 01:19:59,270 --> 01:20:02,750 Now, because it's real-- that's the assumption. 1014 01:20:02,750 --> 01:20:03,990 The number is real. 1015 01:20:03,990 --> 01:20:16,520 Now, the star off an inner product is Tv, v. 1016 01:20:16,520 --> 01:20:19,530 But on the other hand, this operator, 1017 01:20:19,530 --> 01:20:24,820 by the definition of adjoint, can be moved here. 1018 01:20:24,820 --> 01:20:30,120 And this is equal to T dagger v, v. So 1019 01:20:30,120 --> 01:20:34,530 now you have done this is equal to this. 1020 01:20:34,530 --> 01:20:37,090 So if you put it to one side, you 1021 01:20:37,090 --> 01:20:46,510 get that T dagger minus T on v times v is equal to 0. 1022 01:20:46,510 --> 01:20:53,560 Or, since any inner product that is 00-- 1023 01:20:53,560 --> 01:20:59,140 it's complex conjugate is 0-- you can write it as v, 1024 01:20:59,140 --> 01:21:08,180 T dagger minus v is 0 for all v. 1025 01:21:08,180 --> 01:21:11,430 And so this is an actually well known statement, 1026 01:21:11,430 --> 01:21:17,460 that any operator that gives you real things must be Hermitian. 1027 01:21:17,460 --> 01:21:23,110 But it's not obvious, because that theorem is not obvious. 1028 01:21:23,110 --> 01:21:26,660 And now you can use a theorem and say, well, 1029 01:21:26,660 --> 01:21:31,910 since this is true for all v, T dagger minus T is 0, 1030 01:21:31,910 --> 01:21:37,010 and T dagger is equal to T. Then you can also show, of course, 1031 01:21:37,010 --> 01:21:41,730 if T dagger is equal to T, this thing is real. 1032 01:21:41,730 --> 01:21:47,100 So in fact, this arrow is both ways. 1033 01:21:47,100 --> 01:21:53,320 And this way is very easy, but this way uses this theorem. 1034 01:21:53,320 --> 01:21:56,445 There's another kind of operators 1035 01:21:56,445 --> 01:21:58,630 that are called unitary operators. 1036 01:21:58,630 --> 01:22:01,060 We'll talk a little more about them next time. 1037 01:22:01,060 --> 01:22:04,940 And they preserve the norm of vectors. 1038 01:22:04,940 --> 01:22:06,480 People define them from you, and you 1039 01:22:06,480 --> 01:22:09,080 see that they preserve the norm of vectors. 1040 01:22:09,080 --> 01:22:11,375 On the other hand, you sometimes find an operator 1041 01:22:11,375 --> 01:22:14,460 that preserves every norm. 1042 01:22:14,460 --> 01:22:16,200 Is it unitary? 1043 01:22:16,200 --> 01:22:17,667 You will say, yes, must be. 1044 01:22:17,667 --> 01:22:18,500 How do you prove it? 1045 01:22:18,500 --> 01:22:20,550 You need again that theorem. 1046 01:22:20,550 --> 01:22:23,120 So this theorem is really quite fundamental 1047 01:22:23,120 --> 01:22:25,950 to understand the properties of operators. 1048 01:22:25,950 --> 01:22:27,750 And we'll continue that next time. 1049 01:22:27,750 --> 01:22:29,600 All right.