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 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,305 at ocw.mit.edu. 8 00:00:21,275 --> 00:00:22,150 PROFESSOR: Very good. 9 00:00:22,150 --> 00:00:26,960 So today, we'll begin with a study 10 00:00:26,960 --> 00:00:31,360 of one dimensional potentials and the energy eigenstates 11 00:00:31,360 --> 00:00:33,150 or properties. 12 00:00:33,150 --> 00:00:39,190 And we'll discuss this for about half of the lecture. 13 00:00:39,190 --> 00:00:44,700 And then go into the variational principle. 14 00:00:44,700 --> 00:00:49,490 So let me go back then to our discussion of last time, 15 00:00:49,490 --> 00:00:51,960 where we were talking about energy eigenstates 16 00:00:51,960 --> 00:00:55,040 and the Schrodinger equation in one dimension. 17 00:00:55,040 --> 00:00:58,330 So there's a term that we use all the time. 18 00:00:58,330 --> 00:01:00,850 This term is called the bound state. 19 00:01:00,850 --> 00:01:06,160 And bound state seems something sort of very non-trivial. 20 00:01:06,160 --> 00:01:10,650 But mathematically, we can say quite clearly what it is. 21 00:01:10,650 --> 00:01:12,610 A bound state is something that is not 22 00:01:12,610 --> 00:01:16,500 spread all over space, basically. 23 00:01:16,500 --> 00:01:20,120 And the way we therefore use the terminology for bound 24 00:01:20,120 --> 00:01:23,010 state it that we only speak of bound states 25 00:01:23,010 --> 00:01:25,420 when there are energy eigenstates. 26 00:01:25,420 --> 00:01:28,500 So an energy eigenstate may or may not be a bound state, 27 00:01:28,500 --> 00:01:32,380 but any bound state is an energy eigenstate. 28 00:01:32,380 --> 00:01:35,330 So an energy eigenstate is a bound state 29 00:01:35,330 --> 00:01:38,710 if the wave function goes to zero 30 00:01:38,710 --> 00:01:42,400 when you go sufficiently far away. 31 00:01:42,400 --> 00:01:44,710 So it's a simple definition, but it 32 00:01:44,710 --> 00:01:47,710 helps us understand that the basic idea is 33 00:01:47,710 --> 00:01:52,480 at the state is just not spread all over the world. 34 00:01:52,480 --> 00:01:58,240 Now remember we're trying to find energy eigenstates, 35 00:01:58,240 --> 00:02:01,020 and that is to find wave functions, time 36 00:02:01,020 --> 00:02:04,920 independent wave functions, that solve the time 37 00:02:04,920 --> 00:02:07,350 independent Schrodinger equation. 38 00:02:07,350 --> 00:02:10,889 Which I have written there for convenience. 39 00:02:10,889 --> 00:02:15,110 This is the same equation we wrote last time. 40 00:02:15,110 --> 00:02:17,500 For a time independent potential, 41 00:02:17,500 --> 00:02:19,880 so that the full wave function can 42 00:02:19,880 --> 00:02:22,760 be written as a phase that contains 43 00:02:22,760 --> 00:02:27,590 the information of the energy times a function of x, 44 00:02:27,590 --> 00:02:29,280 sine of x. 45 00:02:29,280 --> 00:02:34,210 That still may be complex, but it doesn't have to be complex. 46 00:02:34,210 --> 00:02:36,355 As you can see, this is a real equation. 47 00:02:39,520 --> 00:02:45,010 Now in order to clean up some of the constants in this equation, 48 00:02:45,010 --> 00:02:47,146 it's-- yes? 49 00:02:47,146 --> 00:02:49,257 AUDIENCE: Why is it not V x minus e? 50 00:02:49,257 --> 00:02:51,590 PROFESSOR: Well you have to look back at the Schrodinger 51 00:02:51,590 --> 00:02:55,680 equation, and bring the things to the other side, 52 00:02:55,680 --> 00:02:58,330 and take care of the signs. 53 00:02:58,330 --> 00:03:00,250 It is correct as stated. 54 00:03:00,250 --> 00:03:00,965 Check it. 55 00:03:04,780 --> 00:03:10,580 Now we have, in order to do this and just 56 00:03:10,580 --> 00:03:15,930 to write things with a little more clarity, 57 00:03:15,930 --> 00:03:17,580 we scale the energy. 58 00:03:17,580 --> 00:03:20,590 So we define a calligraphic energy, 59 00:03:20,590 --> 00:03:33,140 which is 2m over h squared times energy E. And a calligraphic V 60 00:03:33,140 --> 00:03:35,100 with the same scaling factor. 61 00:03:40,890 --> 00:03:45,820 So that the Schrodinger equation then takes the form-- 62 00:03:45,820 --> 00:03:48,170 this same equation-- sine. 63 00:03:48,170 --> 00:03:53,200 And I'll use the double prime notation for two derivatives. 64 00:03:53,200 --> 00:04:03,250 Plus now E minus cal of x sine equals zero. 65 00:04:06,020 --> 00:04:08,890 So this is how our equation looks. 66 00:04:08,890 --> 00:04:11,030 It's relatively simple. 67 00:04:11,030 --> 00:04:14,920 And can be treated without carrying constants 68 00:04:14,920 --> 00:04:18,120 all over the place. 69 00:04:18,120 --> 00:04:23,200 Now we're going to discuss a few results that 70 00:04:23,200 --> 00:04:28,120 are extremely important that you keep in mind for the future. 71 00:04:28,120 --> 00:04:30,550 It helps you think clearly. 72 00:04:30,550 --> 00:04:35,360 So these results are little theorems that I'm 73 00:04:35,360 --> 00:04:37,080 going to state. 74 00:04:37,080 --> 00:04:39,980 And the proofs are given in the notes. 75 00:04:39,980 --> 00:04:42,500 I will skip some of the proofs, although I 76 00:04:42,500 --> 00:04:46,630 will tell you what the strategy is for the proofs 77 00:04:46,630 --> 00:04:50,950 and then show how the theorem implies, 78 00:04:50,950 --> 00:04:55,370 as a corollary, interesting things that 79 00:04:55,370 --> 00:04:58,110 could easily be misconstrued. 80 00:04:58,110 --> 00:05:00,890 So let me give you the first theorem. 81 00:05:00,890 --> 00:05:01,700 Theorem one. 82 00:05:05,520 --> 00:05:09,110 This is just the statement that if you 83 00:05:09,110 --> 00:05:13,070 are talking about bound states in one dimensional potentials, 84 00:05:13,070 --> 00:05:15,720 there are no degeneracies. 85 00:05:15,720 --> 00:05:18,080 So let's write it. 86 00:05:18,080 --> 00:05:33,190 There is no degeneracy for bound states of one 87 00:05:33,190 --> 00:05:35,840 dimensional potentials. 88 00:05:40,170 --> 00:05:44,030 So it's a fundamental result. 89 00:05:47,240 --> 00:05:52,080 We'll see some way to understand it a little later. 90 00:05:52,080 --> 00:05:55,650 And you have seen this in 804, I believe. 91 00:05:55,650 --> 00:05:59,610 So what is a strategy to a proof? 92 00:05:59,610 --> 00:06:01,760 So we want a rigorous proof. 93 00:06:01,760 --> 00:06:03,490 So what is a strategy? 94 00:06:03,490 --> 00:06:05,380 I will not go through the proof. 95 00:06:05,380 --> 00:06:10,280 The strategy is to assume there is a degeneracy. 96 00:06:10,280 --> 00:06:18,426 So strategy, assume a degeneracy. 97 00:06:21,940 --> 00:06:35,930 So sine 1 and sine 2 different with the same energy. 98 00:06:38,450 --> 00:06:43,180 And then you write this equation for sine 1 and sine 2, 99 00:06:43,180 --> 00:06:45,600 and they have the same energy. 100 00:06:45,600 --> 00:06:48,360 Then you take the two equations, and you multiply the first 101 00:06:48,360 --> 00:06:50,390 by something, the second by another thing. 102 00:06:50,390 --> 00:06:51,640 Subtract it. 103 00:06:51,640 --> 00:06:53,340 Do a bit of algebra. 104 00:06:53,340 --> 00:07:02,640 And you suddenly can prove that one solution 105 00:07:02,640 --> 00:07:06,490 is equal to a constant times the other solution. 106 00:07:11,980 --> 00:07:14,890 So you do the work, and that's what comes out. 107 00:07:14,890 --> 00:07:19,290 Once you've shown that one is proportional to the other, 108 00:07:19,290 --> 00:07:21,820 we've declared in quantum mechanics 109 00:07:21,820 --> 00:07:25,030 that two wave functions that differ by a constant 110 00:07:25,030 --> 00:07:27,580 have exactly the same physics. 111 00:07:27,580 --> 00:07:31,130 You can normalize it, and normalize them. 112 00:07:31,130 --> 00:07:33,030 And the constant can be just a phase, 113 00:07:33,030 --> 00:07:34,540 but the phase doesn't matter. 114 00:07:34,540 --> 00:07:38,130 So those two wave functions are really the same. 115 00:07:38,130 --> 00:07:40,610 So you get a contradiction. 116 00:07:40,610 --> 00:07:42,420 You thought they were different. 117 00:07:42,420 --> 00:07:46,770 But the difference is just too trivial, 118 00:07:46,770 --> 00:07:50,610 and therefore they are really the same wave function. 119 00:07:50,610 --> 00:07:52,950 Because we've said that two wave functions that 120 00:07:52,950 --> 00:07:56,190 differ by a constant should be identified. 121 00:07:56,190 --> 00:07:58,410 So that's this theorem. 122 00:07:58,410 --> 00:08:01,640 And it's a key theorem. 123 00:08:05,540 --> 00:08:10,470 Let's go to the next theorem that is also important. 124 00:08:10,470 --> 00:08:11,733 So theorem two. 125 00:08:16,070 --> 00:08:31,185 Energy eigenstates sine f of x can be chosen to be real. 126 00:08:38,650 --> 00:08:40,850 So here it is. 127 00:08:40,850 --> 00:08:43,650 We mentioned that this differential equation 128 00:08:43,650 --> 00:08:45,980 allows for real solutions. 129 00:08:45,980 --> 00:08:49,360 There's no contradiction, because there's no i in there. 130 00:08:49,360 --> 00:08:52,090 But now we say more. 131 00:08:52,090 --> 00:08:56,660 That even if you find a complex solution, 132 00:08:56,660 --> 00:08:59,070 you can work with real solutions. 133 00:08:59,070 --> 00:09:01,590 So what is the strategy here? 134 00:09:01,590 --> 00:09:02,990 Again strategy. 135 00:09:09,910 --> 00:09:13,005 To have you get a complex solution. 136 00:09:20,160 --> 00:09:22,160 Sine f of x. 137 00:09:22,160 --> 00:09:28,520 And then you go about to prove that actually 138 00:09:28,520 --> 00:09:32,240 this complex solution implies the existence of two 139 00:09:32,240 --> 00:09:34,160 real solutions. 140 00:09:34,160 --> 00:09:42,200 So this complex solution implies existence 141 00:09:42,200 --> 00:09:43,780 of two real solutions. 142 00:09:50,520 --> 00:09:54,960 And moreover, two real-- I should say it better here-- 143 00:09:54,960 --> 00:10:00,450 degenerate solutions. 144 00:10:00,450 --> 00:10:02,470 How could you prove that? 145 00:10:02,470 --> 00:10:05,010 Well, it probably would occur to you 146 00:10:05,010 --> 00:10:08,910 that you would take a solution and say, oh, this solves it. 147 00:10:08,910 --> 00:10:13,940 Then you would show that sine star also solves this equation. 148 00:10:13,940 --> 00:10:17,870 And then by linearity of the Schrodinger equation, 149 00:10:17,870 --> 00:10:23,270 you could form what we would call the real solution formed 150 00:10:23,270 --> 00:10:30,980 by taking sine plus sine star and adding them. 151 00:10:30,980 --> 00:10:37,450 Or the real solution that comes from the imaginary part. 152 00:10:37,450 --> 00:10:42,380 You see the real and imaginary parts of a complex number 153 00:10:42,380 --> 00:10:43,920 are real. 154 00:10:43,920 --> 00:10:46,490 So is the case here as well. 155 00:10:46,490 --> 00:10:51,180 That they can take sine and sine star that are different. 156 00:10:51,180 --> 00:10:53,930 And the sum is real by construction. 157 00:10:53,930 --> 00:10:57,700 You put the star, and recalling that star of star 158 00:10:57,700 --> 00:11:02,400 gives nothing, you see that this function is equal to its star. 159 00:11:02,400 --> 00:11:03,880 So it's real. 160 00:11:03,880 --> 00:11:05,370 So is this one. 161 00:11:12,940 --> 00:11:17,750 This function, which is the imaginary part of the solution 162 00:11:17,750 --> 00:11:20,760 that you had, is also real. 163 00:11:20,760 --> 00:11:23,270 So you get two real solutions. 164 00:11:23,270 --> 00:11:27,850 Moreover, you can show that sine and sine star 165 00:11:27,850 --> 00:11:29,100 have the same energy. 166 00:11:29,100 --> 00:11:32,780 Therefore sine r and sine m have the same energy. 167 00:11:32,780 --> 00:11:35,770 So I'm really skipping very little. 168 00:11:35,770 --> 00:11:38,940 But I don't want to clutter the blackboard 169 00:11:38,940 --> 00:11:42,340 with the whole derivation at this moment. 170 00:11:42,340 --> 00:11:43,825 So here is the strategy. 171 00:11:43,825 --> 00:11:47,790 A complex solution implies two real solutions. 172 00:11:47,790 --> 00:11:51,480 And therefore, since you have real solutions, 173 00:11:51,480 --> 00:11:55,160 you can always choose them to be real. 174 00:11:55,160 --> 00:11:56,740 No loss of generality. 175 00:11:56,740 --> 00:11:58,560 You had a complex solution. 176 00:11:58,560 --> 00:12:00,870 You say, I like just real solutions. 177 00:12:00,870 --> 00:12:02,700 And you chose your two real solutions. 178 00:12:02,700 --> 00:12:03,600 You give them. 179 00:12:03,600 --> 00:12:06,850 You've given the whole thing. 180 00:12:06,850 --> 00:12:10,570 But then there is a nice corollary 181 00:12:10,570 --> 00:12:14,370 for the case of one dimension bound states. 182 00:12:14,370 --> 00:12:20,590 So this is the part that is perhaps somewhat interesting. 183 00:12:20,590 --> 00:12:24,070 And the corollary is that if you're 184 00:12:24,070 --> 00:12:29,530 talking bound states of one dimension, 185 00:12:29,530 --> 00:12:37,970 any solution is equal to a real solution up to a constant. 186 00:12:37,970 --> 00:12:39,980 You may say, well what's the difference? 187 00:12:39,980 --> 00:12:45,970 Let's write it down and look at it. 188 00:12:45,970 --> 00:12:58,480 Corollary for bound states of one dimensional potentials. 189 00:13:05,700 --> 00:13:23,940 Any solution is, up to a phase, equal to a real solution. 190 00:13:30,120 --> 00:13:37,840 So how does that go? 191 00:13:37,840 --> 00:13:43,580 Maybe we can use more of this blackboard. 192 00:13:43,580 --> 00:13:46,710 Why would any solution be, up to a phase, 193 00:13:46,710 --> 00:13:48,020 equal to a real solution? 194 00:13:48,020 --> 00:13:50,340 OK. 195 00:13:50,340 --> 00:13:52,870 Suppose you have these two solutions here. 196 00:13:52,870 --> 00:13:55,592 Well this is real, so that's trivially true. 197 00:13:55,592 --> 00:13:56,300 And this is real. 198 00:13:56,300 --> 00:14:00,270 But what if I put a linear combination of them? 199 00:14:00,270 --> 00:14:02,620 Or take the original complex solution? 200 00:14:02,620 --> 00:14:06,880 Why would it be, up to a phase, equal to a real solution? 201 00:14:06,880 --> 00:14:08,800 The answer is the following. 202 00:14:08,800 --> 00:14:13,470 Because of theorem and, that there's no degeneracy, 203 00:14:13,470 --> 00:14:17,180 you've got two solutions that claim 204 00:14:17,180 --> 00:14:20,840 to be different and degenerate. 205 00:14:20,840 --> 00:14:22,890 But that can't be. 206 00:14:22,890 --> 00:14:31,550 So by theorem one, you must have that-- in the notations, 207 00:14:31,550 --> 00:14:36,310 I have here-- sine imaginary of x 208 00:14:36,310 --> 00:14:41,245 must be equal to a constant times sine real of x. 209 00:14:48,580 --> 00:14:54,960 Now if these function's anyway are real, 210 00:14:54,960 --> 00:14:57,120 the constant here must be real. 211 00:15:01,070 --> 00:15:05,630 Therefore, even if you form the original complex solution, 212 00:15:05,630 --> 00:15:14,501 sine, which is sine real plus i sine imaginary. 213 00:15:14,501 --> 00:15:16,470 You can check that. 214 00:15:16,470 --> 00:15:19,940 It's clear by definition that that's the way it should be. 215 00:15:19,940 --> 00:15:25,350 If you think this, this is equal to 1 plus i times 216 00:15:25,350 --> 00:15:30,860 c times sine real of x. 217 00:15:30,860 --> 00:15:35,940 Therefore this is equal to a constant times 218 00:15:35,940 --> 00:15:39,250 the phase of-- this is a complex number. 219 00:15:39,250 --> 00:15:43,070 So this is yet another constant that we 220 00:15:43,070 --> 00:15:52,000 could say, the norm of 1 plus i c times some phase times sine 221 00:15:52,000 --> 00:15:53,460 r of x. 222 00:15:53,460 --> 00:15:58,000 So you see that any solution, any linear combination, 223 00:15:58,000 --> 00:16:02,080 in fact, with any numbers of sine r and sine imaginary, 224 00:16:02,080 --> 00:16:07,760 will be a constant times a phase times this thing. 225 00:16:07,760 --> 00:16:13,020 So it is really just real anyway. 226 00:16:13,020 --> 00:16:17,455 So you can just have one solution. 227 00:16:17,455 --> 00:16:19,950 And any solution that you may write, 228 00:16:19,950 --> 00:16:23,960 that represents a bound state, is, up to a phase, 229 00:16:23,960 --> 00:16:25,115 equal to a real solution. 230 00:16:30,110 --> 00:16:30,780 OK. 231 00:16:30,780 --> 00:16:34,690 So that's our theorem two. 232 00:16:34,690 --> 00:16:35,760 Theorem three. 233 00:16:43,210 --> 00:16:47,980 It's also very famous. 234 00:16:47,980 --> 00:17:01,640 If the potential V of x is even-- that is, V of minus x 235 00:17:01,640 --> 00:17:14,089 is equal to V of x-- the eigenstate can be chosen 236 00:17:14,089 --> 00:17:26,730 to be even or odd under x2 minus x. 237 00:17:31,720 --> 00:17:32,220 OK. 238 00:17:32,220 --> 00:17:39,040 So here is another claim. 239 00:17:39,040 --> 00:17:41,480 So the potential is even. 240 00:17:41,480 --> 00:17:44,210 There's no theorem for all the potentials. 241 00:17:44,210 --> 00:17:47,170 No clear theorems, no simple theorems for all the-- 242 00:17:47,170 --> 00:17:50,580 but if the potential is even, the claim 243 00:17:50,580 --> 00:17:55,280 is that you can choose eigenstates to be even or odd. 244 00:17:55,280 --> 00:17:58,460 The word chosen is very important here. 245 00:17:58,460 --> 00:18:00,800 Otherwise, this would not be a precise statement. 246 00:18:00,800 --> 00:18:04,160 You cannot say the eigenstates are even or odd. 247 00:18:04,160 --> 00:18:06,400 You can choose them to be. 248 00:18:06,400 --> 00:18:08,460 So how does the proof go? 249 00:18:08,460 --> 00:18:18,200 Strategy begin with a wave function, 250 00:18:18,200 --> 00:18:22,840 sine of x, that is neither even nor odd. 251 00:18:29,720 --> 00:18:33,580 And then you do a little work with the Schrodinger equation. 252 00:18:33,580 --> 00:18:36,630 Take the Schrodinger equation, change all x's 253 00:18:36,630 --> 00:18:42,180 to minus x's and show that, in fact, not only is sine 254 00:18:42,180 --> 00:18:48,640 of x a solution, but sine of minus x is also a solution. 255 00:18:48,640 --> 00:18:50,720 With the same energy. 256 00:18:50,720 --> 00:18:58,250 So prove that sine of minus x is a solution 257 00:18:58,250 --> 00:18:59,220 with the same energy. 258 00:19:07,930 --> 00:19:12,550 And in this case, of course, we can already 259 00:19:12,550 --> 00:19:15,090 have shown that we can choose these wave 260 00:19:15,090 --> 00:19:16,470 functions to be real. 261 00:19:16,470 --> 00:19:21,520 So we can choose all of these wave functions to be real. 262 00:19:21,520 --> 00:19:24,040 And what do we do next? 263 00:19:24,040 --> 00:19:28,200 If we have these two solutions with the same energy, 264 00:19:28,200 --> 00:19:31,710 then you can build of sine s, which 265 00:19:31,710 --> 00:19:36,800 is 1/2 of sine of x plus sine of minus x. 266 00:19:40,530 --> 00:19:43,200 And of sine a. 267 00:19:43,200 --> 00:19:46,500 s for symmetric, and a for anti-symmetric. 268 00:19:46,500 --> 00:19:50,750 And of sine a, that is 1/2 of sine 269 00:19:50,750 --> 00:19:57,900 of x minus sine of minus x. 270 00:19:57,900 --> 00:20:02,930 And this tool would be this even, 271 00:20:02,930 --> 00:20:06,260 under the exchange of x for minus x. 272 00:20:06,260 --> 00:20:12,350 This one odd, under the exchange of x for minus x. 273 00:20:12,350 --> 00:20:15,700 And both would be solutions by superposition. 274 00:20:15,700 --> 00:20:17,805 And both would have the same energy. 275 00:20:21,130 --> 00:20:29,400 So that's the end of the theorem because then these things are 276 00:20:29,400 --> 00:20:32,640 even or odd and have the same energy. 277 00:20:32,640 --> 00:20:39,300 So the solutions can be chosen to be even or odd under x. 278 00:20:39,300 --> 00:20:43,120 So if you've proven this, you've got it already. 279 00:20:43,120 --> 00:20:44,805 But now we get the corollary. 280 00:20:49,490 --> 00:21:01,660 For bound states in one dimension, 281 00:21:01,660 --> 00:21:05,900 the solutions not anymore the word chosen. 282 00:21:05,900 --> 00:21:08,710 We can delete the word chosen. 283 00:21:08,710 --> 00:21:13,465 The solutions are either odd or even. 284 00:21:28,020 --> 00:21:29,810 So it's a lot stronger. 285 00:21:29,810 --> 00:21:32,380 It's not anymore, you can choose them to be, 286 00:21:32,380 --> 00:21:35,981 but a general one is neither odd nor even. 287 00:21:35,981 --> 00:21:36,480 No. 288 00:21:36,480 --> 00:21:39,660 You try to find the solution that is neither odd nor even, 289 00:21:39,660 --> 00:21:41,800 and you can't find it. 290 00:21:41,800 --> 00:21:42,930 So it's very strong. 291 00:21:42,930 --> 00:21:44,216 Yes? 292 00:21:44,216 --> 00:21:46,070 AUDIENCE: Is this for even potentials? 293 00:21:46,070 --> 00:21:47,195 PROFESSOR: Even potentials. 294 00:21:47,195 --> 00:21:47,970 Yes. 295 00:21:47,970 --> 00:21:52,980 For bound states in one dimension with even potentials. 296 00:21:52,980 --> 00:21:54,255 V of x. 297 00:21:54,255 --> 00:21:56,770 V of minus x equal V of x. 298 00:21:56,770 --> 00:21:58,940 Yes. 299 00:21:58,940 --> 00:22:01,720 So how do we show that? 300 00:22:01,720 --> 00:22:07,500 Well, again, you've got two solutions here 301 00:22:07,500 --> 00:22:10,750 that are degenerate that have the same energy. 302 00:22:10,750 --> 00:22:13,730 Sine of x and sine of minus x. 303 00:22:13,730 --> 00:22:17,780 So given that there's no degeneracy in one 304 00:22:17,780 --> 00:22:21,620 dimensional problems, this thing that they have the same energy, 305 00:22:21,620 --> 00:22:23,950 the only thing that can be happening 306 00:22:23,950 --> 00:22:30,400 is that sine of minus x is equal to a constant times sine of x. 307 00:22:30,400 --> 00:22:33,670 Where this constant is real. 308 00:22:33,670 --> 00:22:34,630 Why real? 309 00:22:34,630 --> 00:22:37,590 Because we said already by the previously theorem, 310 00:22:37,590 --> 00:22:40,590 wave functions can be chosen to be real. 311 00:22:40,590 --> 00:22:45,930 So you've got this thing already that this is true. 312 00:22:45,930 --> 00:22:51,670 Sine of minus x is equal to c times sine of x. 313 00:22:51,670 --> 00:22:55,040 Which, if you use this property again by saying, 314 00:22:55,040 --> 00:23:03,400 oh, but sine of x is equal to c times sine of minus x. 315 00:23:03,400 --> 00:23:06,100 Basically, what this says is that you can change 316 00:23:06,100 --> 00:23:09,480 the sign of the argument by putting an extra c. 317 00:23:09,480 --> 00:23:13,530 So you do it again, sine of x is equal to this. 318 00:23:13,530 --> 00:23:18,820 So this is c squared times sine of minus x. 319 00:23:18,820 --> 00:23:23,580 So c squared must be equal to 1. 320 00:23:23,580 --> 00:23:28,290 And therefore c is equal to either plus or minus 1. 321 00:23:28,290 --> 00:23:30,030 No other option. 322 00:23:30,030 --> 00:23:35,060 So the functions are either even or odd, 323 00:23:35,060 --> 00:23:38,750 but they can't be arbitrary. 324 00:23:38,750 --> 00:23:41,270 This point is sufficiently settled 325 00:23:41,270 --> 00:23:44,760 that part two general exam at MIT 326 00:23:44,760 --> 00:23:48,910 10 years ago had a question like that. 327 00:23:48,910 --> 00:23:51,700 And the person that invented the problem 328 00:23:51,700 --> 00:23:53,990 claimed that there would be a solution that 329 00:23:53,990 --> 00:23:57,580 could be neither even nor odd. 330 00:23:57,580 --> 00:24:01,530 So even faculty members at MIT sometimes get this wrong. 331 00:24:01,530 --> 00:24:06,260 It's not as weak as this, that can be chosen. 332 00:24:06,260 --> 00:24:11,247 But it's really either or other in the case 333 00:24:11,247 --> 00:24:12,205 you have one dimension. 334 00:24:16,040 --> 00:24:16,630 OK. 335 00:24:16,630 --> 00:24:20,730 So these are our main theorems. 336 00:24:20,730 --> 00:24:23,670 And we're going to proceed now by clarifying 337 00:24:23,670 --> 00:24:26,090 a little more the nature of the spectrum. 338 00:24:26,090 --> 00:24:27,225 So are there questions? 339 00:24:32,680 --> 00:24:33,180 Yes? 340 00:24:33,180 --> 00:24:34,930 AUDIENCE: Can you give an example of state 341 00:24:34,930 --> 00:24:36,857 that's not bound? 342 00:24:36,857 --> 00:24:37,440 PROFESSOR: OK. 343 00:24:37,440 --> 00:24:40,220 The question is can I give an example 344 00:24:40,220 --> 00:24:42,870 of a state that is not bound? 345 00:24:42,870 --> 00:24:43,700 Yes. 346 00:24:43,700 --> 00:24:45,310 We can give such state. 347 00:24:45,310 --> 00:24:47,990 You have a potential like this. 348 00:24:47,990 --> 00:24:51,350 And you have an energy like that. 349 00:24:51,350 --> 00:24:53,920 And then the wave function could look like this. 350 00:24:53,920 --> 00:24:55,940 Then do something like that. 351 00:24:55,940 --> 00:24:59,260 And then look like that. 352 00:24:59,260 --> 00:25:02,637 It just doesn't vanish when you go to infinity. 353 00:25:02,637 --> 00:25:04,220 AUDIENCE: So that can't be normalized? 354 00:25:04,220 --> 00:25:05,640 PROFESSOR: Can't be normalized. 355 00:25:05,640 --> 00:25:06,230 That's right. 356 00:25:06,230 --> 00:25:07,980 If it's not bound, it can't be normalized. 357 00:25:14,200 --> 00:25:16,212 Other questions? 358 00:25:16,212 --> 00:25:17,156 Yes? 359 00:25:17,156 --> 00:25:20,932 AUDIENCE: So you can't really represent-- it's doesn't really 360 00:25:20,932 --> 00:25:24,250 represent single particles, more like a stream of particles? 361 00:25:24,250 --> 00:25:24,900 PROFESSOR: Yes. 362 00:25:24,900 --> 00:25:27,680 So it doesn't represent a single particle. 363 00:25:27,680 --> 00:25:30,550 Now trying to interpret it as a stream of particles 364 00:25:30,550 --> 00:25:32,990 is a little delicate. 365 00:25:32,990 --> 00:25:37,320 So what we usually do is we build superpositions 366 00:25:37,320 --> 00:25:42,160 of those states that can represent a localized thing. 367 00:25:42,160 --> 00:25:45,130 But it's true that, morally speaking, 368 00:25:45,130 --> 00:25:49,370 it seems to represent more than one particle. 369 00:25:49,370 --> 00:25:50,510 OK. 370 00:25:50,510 --> 00:25:53,950 So now we talk a little about the nature of the spectrum. 371 00:26:05,020 --> 00:26:12,040 So what do we want to say here? 372 00:26:12,040 --> 00:26:16,210 We want to go back to the Schrodinger equation here. 373 00:26:16,210 --> 00:26:21,050 And just move one term to the right hand side. 374 00:26:21,050 --> 00:26:24,160 And just see what can happen in terms 375 00:26:24,160 --> 00:26:26,860 of singularities and discontinuities. 376 00:26:33,720 --> 00:26:40,040 So first of all, we always begin with the idea 377 00:26:40,040 --> 00:26:44,840 that sine must be continuous. 378 00:26:44,840 --> 00:26:48,060 And the reason sine must be continuous is 379 00:26:48,060 --> 00:26:53,930 that we don't want singularities worse 380 00:26:53,930 --> 00:26:57,110 than delta functions in potentials. 381 00:26:57,110 --> 00:27:03,010 If a function is continuous, the derivative 382 00:27:03,010 --> 00:27:06,170 might be discontinuous, and the second derivative 383 00:27:06,170 --> 00:27:08,760 would have a delta function. 384 00:27:08,760 --> 00:27:11,540 So the second derivative would have a delta function. 385 00:27:11,540 --> 00:27:14,760 But if the function is not even continuous, 386 00:27:14,760 --> 00:27:18,480 the second derivative would have derivatives of delta functions. 387 00:27:18,480 --> 00:27:21,230 So somehow the potential would have to have that. 388 00:27:21,230 --> 00:27:22,570 And we don't want it. 389 00:27:22,570 --> 00:27:27,206 So to simplify our life, we say that sine must be continuous. 390 00:27:32,910 --> 00:27:39,060 Now if sine is continuous, we will 391 00:27:39,060 --> 00:27:44,850 consider several possibilities, possibilities for V, 392 00:27:44,850 --> 00:27:51,370 possibilities for V of x. 393 00:27:51,370 --> 00:27:55,715 So first possibility-- one, V is continuous. 394 00:27:59,900 --> 00:28:03,910 So psi is continuous, and V is continuous. 395 00:28:03,910 --> 00:28:07,200 If psi is continuous and V is continuous, 396 00:28:07,200 --> 00:28:08,610 this product is continuous. 397 00:28:08,610 --> 00:28:11,230 Psi double prime is continuous. 398 00:28:11,230 --> 00:28:13,230 And psi prime is continuous. 399 00:28:19,360 --> 00:28:22,810 Two, V has finite jumps. 400 00:28:28,860 --> 00:28:35,830 Well, if V has finite jumps, and psi is continuous, 401 00:28:35,830 --> 00:28:39,590 this product has finite jumps. 402 00:28:39,590 --> 00:28:44,760 So psi double prime has finite jumps. 403 00:28:44,760 --> 00:28:49,360 If psi prime has finite jumps, the worst 404 00:28:49,360 --> 00:28:53,715 is that psi prime still must be continuous. 405 00:28:57,280 --> 00:28:59,410 But it changes. 406 00:28:59,410 --> 00:29:02,520 Psi prime could look like that, could have a corner. 407 00:29:02,520 --> 00:29:04,620 But it cannot be worse than that. 408 00:29:04,620 --> 00:29:10,970 Because if V has finite jumps, if psi double 409 00:29:10,970 --> 00:29:15,025 prime has finite jumps, and if psi prime is not continuous, 410 00:29:15,025 --> 00:29:16,880 it would have delta functions. 411 00:29:16,880 --> 00:29:22,110 So for these two conditions, continuous or even 412 00:29:22,110 --> 00:29:27,760 finite jumps, psi prime is still continuous. 413 00:29:27,760 --> 00:29:32,250 Things change qualitatively if three, V has delta functions. 414 00:29:39,500 --> 00:29:44,050 If V has a delta function, then psi 415 00:29:44,050 --> 00:29:47,670 double prime has a delta function. 416 00:29:47,670 --> 00:29:52,410 And psi prime therefore jumps. 417 00:29:52,410 --> 00:29:54,400 Psi prime is not continuous. 418 00:29:54,400 --> 00:29:56,860 Psi prime jumps. 419 00:29:56,860 --> 00:30:00,090 This may be reminiscent to you whenever 420 00:30:00,090 --> 00:30:03,290 you had to solve the problem of a bound state 421 00:30:03,290 --> 00:30:05,510 of a delta function, you got a wave function 422 00:30:05,510 --> 00:30:10,220 that looked like this in which psi prime jumps. 423 00:30:10,220 --> 00:30:11,520 And it has to jump. 424 00:30:11,520 --> 00:30:15,970 Because psi double prime has a delta function. 425 00:30:15,970 --> 00:30:19,940 Another case in which psi prime jumps 426 00:30:19,940 --> 00:30:22,450 is for if V has a hard wall. 427 00:30:26,530 --> 00:30:30,835 That is, the potential suddenly at one point becomes infinite, 428 00:30:30,835 --> 00:30:34,970 and it prevents the particle from moving across. 429 00:30:34,970 --> 00:30:38,470 A hard wall is a place, as you remember 430 00:30:38,470 --> 00:30:41,350 in the infinite square well, in which the wave 431 00:30:41,350 --> 00:30:44,890 function vanishes, but the derivative doesn't vanish. 432 00:30:44,890 --> 00:30:49,370 So you could say that psi is 0 here outside. 433 00:30:49,370 --> 00:30:55,160 Psi prime, then, jumps in that it's non-0 here 434 00:30:55,160 --> 00:30:56,960 and it's 0 afterwards. 435 00:30:56,960 --> 00:31:02,850 So now, you could object to what I'm 436 00:31:02,850 --> 00:31:06,730 saying by explaining that, well, we shouldn't talk 437 00:31:06,730 --> 00:31:09,950 about the wave function beyond the hard wall. 438 00:31:09,950 --> 00:31:12,010 And in some sense, you're right. 439 00:31:12,010 --> 00:31:14,850 But suppose we do and we say the wave function is just 440 00:31:14,850 --> 00:31:18,390 0 everywhere, we see that psi prime jumps. 441 00:31:18,390 --> 00:31:23,640 So really, psi prime jumps. 442 00:31:23,640 --> 00:31:27,250 So this is as bad as things can get. 443 00:31:27,250 --> 00:31:33,570 So we can summarize this in just one sentence. 444 00:31:33,570 --> 00:31:39,390 And the sentence reads, psi and psi prime 445 00:31:39,390 --> 00:31:57,510 are continuous unless V has delta functions or hard walls, 446 00:31:57,510 --> 00:32:07,120 in which case psi prime can have finite jumps. 447 00:32:14,560 --> 00:32:19,770 So basically, psi and psi prime, continuous. 448 00:32:19,770 --> 00:32:25,850 Exceptions-- delta functions and hard walls, and psi prime 449 00:32:25,850 --> 00:32:28,200 can change. 450 00:32:28,200 --> 00:32:30,860 So questions. 451 00:32:30,860 --> 00:32:32,292 There was a question before. 452 00:32:32,292 --> 00:32:33,250 Maybe it's still there. 453 00:32:33,250 --> 00:32:33,860 Yes. 454 00:32:33,860 --> 00:32:37,668 AUDIENCE: Do you absolutely need that function [INAUDIBLE]? 455 00:32:37,668 --> 00:32:40,980 Or do we just assume that [INAUDIBLE]? 456 00:32:40,980 --> 00:32:43,150 PROFESSOR: We assume it-- so the question is, do I 457 00:32:43,150 --> 00:32:46,960 need that in absolute generality to do quantum mechanics? 458 00:32:46,960 --> 00:32:49,440 I don't think so. 459 00:32:49,440 --> 00:32:53,185 I presume you could discuss some potentials that 460 00:32:53,185 --> 00:32:56,610 lead to psi that are not continuous 461 00:32:56,610 --> 00:32:58,640 and still make sense. 462 00:32:58,640 --> 00:33:00,320 But we will not discuss them. 463 00:33:00,320 --> 00:33:03,495 And actually, I don't think I've encountered any of those. 464 00:33:06,587 --> 00:33:08,326 Yes. 465 00:33:08,326 --> 00:33:09,700 AUDIENCE: Can you give an example 466 00:33:09,700 --> 00:33:12,015 of a physical system whose potential is well approximated 467 00:33:12,015 --> 00:33:13,348 by a delta function [INAUDIBLE]? 468 00:33:16,150 --> 00:33:24,540 PROFESSOR: Yes, there are systems like that. 469 00:33:24,540 --> 00:33:28,450 For example, this one that is somewhat well 470 00:33:28,450 --> 00:33:30,790 approximated by a delta function. 471 00:33:30,790 --> 00:33:34,000 For example, a nucleus sometimes is 472 00:33:34,000 --> 00:33:38,990 considered to be like a spherical cavity in which 473 00:33:38,990 --> 00:33:42,470 particles are bound by a deep potential and don't escape. 474 00:33:42,470 --> 00:33:46,250 So the nuclei are moving like that and don't escape. 475 00:33:46,250 --> 00:33:49,030 In that case, it would be a three dimensional delta 476 00:33:49,030 --> 00:33:53,410 function that vanishes at the origin. 477 00:33:53,410 --> 00:33:56,070 I presume there are many examples. 478 00:33:56,070 --> 00:33:59,560 Any potential that sort of begins 479 00:33:59,560 --> 00:34:02,850 like a finite square well that is sufficiently deep 480 00:34:02,850 --> 00:34:07,235 will start to look like a delta function after awhile. 481 00:34:07,235 --> 00:34:08,110 AUDIENCE: [INAUDIBLE] 482 00:34:12,530 --> 00:34:17,929 PROFESSOR: Well, it's again neither-- yeah, I guess so. 483 00:34:17,929 --> 00:34:21,719 But it depends how big is this psi. 484 00:34:21,719 --> 00:34:26,050 So yeah, probably you're right. 485 00:34:26,050 --> 00:34:29,960 This looks a little more like an analog of a hard wall. 486 00:34:29,960 --> 00:34:33,690 But if a hard wall is very narrow and very deep, 487 00:34:33,690 --> 00:34:35,300 it looks like a delta function. 488 00:34:35,300 --> 00:34:38,090 So it's idealizations for sure. 489 00:34:38,090 --> 00:34:42,310 But I'm sure we could get a better example. 490 00:34:42,310 --> 00:34:45,580 And I'll try to find one. 491 00:34:45,580 --> 00:34:48,310 Now, the next thing we want to do 492 00:34:48,310 --> 00:34:52,800 is give you intuition for this incredible result 493 00:34:52,800 --> 00:34:56,380 that there's no degeneratives in one dimensional potentials. 494 00:34:56,380 --> 00:35:02,450 That is not to say that the proof is not good enough. 495 00:35:02,450 --> 00:35:07,440 It is just to say that we can illustrate that 496 00:35:07,440 --> 00:35:09,920 without going into a mathematical proof that 497 00:35:09,920 --> 00:35:12,010 is more complicated. 498 00:35:12,010 --> 00:35:14,970 So how do we do that? 499 00:35:14,970 --> 00:35:17,980 We'll consider the following case, a simple case, 500 00:35:17,980 --> 00:35:23,670 an example of a potential of this form. 501 00:35:23,670 --> 00:35:26,260 V of x-- this is x. 502 00:35:26,260 --> 00:35:27,800 And here is V of x. 503 00:35:31,390 --> 00:35:36,410 And we will try to find a solution with some energy that 504 00:35:36,410 --> 00:35:42,780 is like that, an energy that is right there below the barrier. 505 00:35:45,740 --> 00:35:47,720 So this would be a bound state. 506 00:35:47,720 --> 00:35:48,520 Why? 507 00:35:48,520 --> 00:35:52,332 Because solutions here are exponentials 508 00:35:52,332 --> 00:35:54,820 that decay, exponentials that decay. 509 00:35:54,820 --> 00:35:57,500 And here, the wave function would 510 00:35:57,500 --> 00:35:59,970 be oscillating presumably. 511 00:35:59,970 --> 00:36:02,900 So the wave functions go to 0 and infinity. 512 00:36:02,900 --> 00:36:04,190 You could get a bound state. 513 00:36:04,190 --> 00:36:07,150 So let's see how we get a bound state. 514 00:36:07,150 --> 00:36:08,970 Now, the argument I'm going to follow 515 00:36:08,970 --> 00:36:11,110 is just an elaboration of something 516 00:36:11,110 --> 00:36:13,450 you can read in Shankar. 517 00:36:13,450 --> 00:36:16,900 And it's a nice and simple argument. 518 00:36:16,900 --> 00:36:24,000 So we want to understand why we would get here no degeneracies. 519 00:36:24,000 --> 00:36:28,440 Or even more-- in fact not just no degeneracies, 520 00:36:28,440 --> 00:36:32,050 but the spectrum is quantized. 521 00:36:32,050 --> 00:36:35,710 That is, you find one energy, and then another energy maybe, 522 00:36:35,710 --> 00:36:37,730 and another energy. 523 00:36:37,730 --> 00:36:39,590 So how do we see that? 524 00:36:39,590 --> 00:36:43,690 Well, you look at the way you can write solutions 525 00:36:43,690 --> 00:36:46,750 and count the parameters of the solutions 526 00:36:46,750 --> 00:36:50,980 and try to see how many conditions you have to satisfy. 527 00:36:50,980 --> 00:36:55,980 So here, the wave function would be a decay in exponential. 528 00:36:55,980 --> 00:36:59,160 A decay in exponential is of the form 529 00:36:59,160 --> 00:37:06,040 alpha e to the K, kappa, x. 530 00:37:06,040 --> 00:37:07,730 Because x here is negative. 531 00:37:07,730 --> 00:37:11,260 So this decays as x goes to minus infinity 532 00:37:11,260 --> 00:37:12,810 if kappa is positive. 533 00:37:12,810 --> 00:37:15,170 And that's how a solution looks. 534 00:37:15,170 --> 00:37:19,140 You need one coefficient here to determine this solutions. 535 00:37:19,140 --> 00:37:20,604 So I'll put a 1 here. 536 00:37:23,330 --> 00:37:25,650 Now, in here, the solution is oscillatory. 537 00:37:25,650 --> 00:37:29,012 So it's a sine plus cosine. 538 00:37:29,012 --> 00:37:30,220 So you need two coefficients. 539 00:37:33,720 --> 00:37:37,930 In here, the solution must again be decaying. 540 00:37:37,930 --> 00:37:40,810 And therefore, you just need one coefficient. 541 00:37:40,810 --> 00:37:45,150 Again, this time it would be a beta e to the minus Kx. 542 00:37:45,150 --> 00:37:46,870 The fact that this potential looks 543 00:37:46,870 --> 00:37:50,350 symmetric-- I'm not assuming it is. 544 00:37:50,350 --> 00:37:50,920 Yes. 545 00:37:50,920 --> 00:37:52,544 AUDIENCE: Won't one of the coefficients 546 00:37:52,544 --> 00:37:53,960 be unconstrained by normalization? 547 00:37:53,960 --> 00:37:56,160 Isn't one just the normalization factor? 548 00:37:56,160 --> 00:37:58,720 PROFESSOR: OK, how about normalization? 549 00:37:58,720 --> 00:38:04,600 Indeed, we have one, two, and two and one, so 550 00:38:04,600 --> 00:38:08,230 a total of four coefficients, four parameters. 551 00:38:13,400 --> 00:38:17,710 But indeed, suppose you wrote your whole solution. 552 00:38:17,710 --> 00:38:21,800 You could say, look, let me divide this solution by 3. 553 00:38:21,800 --> 00:38:24,460 That's an equivalent solution. 554 00:38:24,460 --> 00:38:27,350 I'm just checking if it solves the Schrodinger equation. 555 00:38:27,350 --> 00:38:28,520 That's all I have to check. 556 00:38:28,520 --> 00:38:30,910 I don't have to check normalization. 557 00:38:30,910 --> 00:38:34,870 Normalization, in fact, is sort of irrelevant here. 558 00:38:34,870 --> 00:38:37,040 You just need to know if a bound state exists. 559 00:38:37,040 --> 00:38:42,200 So indeed, even though you have these four parameters, given 560 00:38:42,200 --> 00:38:45,670 that you can multiply the solution by a constant, 561 00:38:45,670 --> 00:38:51,455 there's just three constants to fix. 562 00:38:54,950 --> 00:38:58,660 Four parameters via normalization 563 00:38:58,660 --> 00:39:02,540 or the multiplication by any constant-- just three 564 00:39:02,540 --> 00:39:04,050 constants to fix. 565 00:39:04,050 --> 00:39:07,910 But this potential is nice enough 566 00:39:07,910 --> 00:39:10,760 that psi and psi prime must be continuous. 567 00:39:10,760 --> 00:39:15,690 So you get two conditions here and two conditions here. 568 00:39:15,690 --> 00:39:19,680 So four conditions-- continuity of psi and psi prime, 569 00:39:19,680 --> 00:39:22,650 continuity of psi and psi prime, four conditions. 570 00:39:26,260 --> 00:39:27,730 So what did we get? 571 00:39:27,730 --> 00:39:30,120 We got in trouble. 572 00:39:30,120 --> 00:39:34,700 We've shown that this is unsolvable in general. 573 00:39:34,700 --> 00:39:38,990 Because there are more conditions than parameters. 574 00:39:38,990 --> 00:39:41,350 Now, this equation could to be a little peculiar. 575 00:39:41,350 --> 00:39:43,330 Maybe this is not completely general. 576 00:39:43,330 --> 00:39:47,810 But you seem to have more conditions than parameters. 577 00:39:47,810 --> 00:39:51,210 But here comes the catch. 578 00:39:51,210 --> 00:39:55,830 The solution really is determined. 579 00:39:55,830 --> 00:39:58,670 Like kappa-- do you know kappa? 580 00:39:58,670 --> 00:40:02,960 Well, you know kappa if you know the energy of the solution. 581 00:40:02,960 --> 00:40:05,210 Kappa is determined by the energy of the solution. 582 00:40:05,210 --> 00:40:08,890 So some parameters in the solution depend on the energy. 583 00:40:08,890 --> 00:40:12,150 So the way we have to think of this 584 00:40:12,150 --> 00:40:16,220 is that in fact three constants to fix, but four conditions. 585 00:40:16,220 --> 00:40:18,510 So we really need four constants to fix. 586 00:40:18,510 --> 00:40:22,090 And the fourth constant is the energy. 587 00:40:22,090 --> 00:40:24,980 So the energy is the fourth constant to fix. 588 00:40:24,980 --> 00:40:28,040 And with four conditions, these three constants 589 00:40:28,040 --> 00:40:32,450 that we had there and the energy, they can just be fixed. 590 00:40:32,450 --> 00:40:35,670 So the solution should fix the energy 591 00:40:35,670 --> 00:40:39,080 and should fix this coefficient. 592 00:40:39,080 --> 00:40:50,820 So the solution exists for some energy, or possibly 593 00:40:50,820 --> 00:40:54,560 some values of the energies, but not all values of the energy. 594 00:40:54,560 --> 00:40:59,030 So this shows, or at least very clearly illustrates, 595 00:40:59,030 --> 00:41:04,310 that you are going to find sets of energies for which you have 596 00:41:04,310 --> 00:41:08,030 solutions depending on how the equations look, 597 00:41:08,030 --> 00:41:09,920 and one solution each time. 598 00:41:09,920 --> 00:41:17,466 So you get what is called a discrete non-degenerate 599 00:41:17,466 --> 00:41:17,965 spectrum. 600 00:41:24,160 --> 00:41:29,110 Now, there are more cases to discuss, 601 00:41:29,110 --> 00:41:31,630 the case in which you have just the step, 602 00:41:31,630 --> 00:41:35,540 or the case in which you have three bound states. 603 00:41:35,540 --> 00:41:43,340 And I will not do them but state the results. 604 00:41:43,340 --> 00:41:46,940 Again, all that I don't do explicitly 605 00:41:46,940 --> 00:41:49,180 can be found in the notes. 606 00:41:49,180 --> 00:41:55,200 So you would look at them later. 607 00:41:55,200 --> 00:42:00,940 And so here is the second case, a potential like this 608 00:42:00,940 --> 00:42:06,460 and an energy like that, energy level. 609 00:42:06,460 --> 00:42:10,590 And what you get here is that in fact, doing the counting 610 00:42:10,590 --> 00:42:12,700 and analyzing the boundary conditions, 611 00:42:12,700 --> 00:42:15,010 you should do it by yourselves. 612 00:42:15,010 --> 00:42:17,810 But you will see the answers in the notes. 613 00:42:17,810 --> 00:42:26,000 You get here continuous spectrum, non-degenerate. 614 00:42:31,320 --> 00:42:35,930 So you will get a solution for every value 615 00:42:35,930 --> 00:42:40,850 of the energy-- that's to mean, continuous spectrum-- and one 616 00:42:40,850 --> 00:42:43,820 solution each time. 617 00:42:43,820 --> 00:42:52,770 Finally, this case-- if you have an energy like this, 618 00:42:52,770 --> 00:43:11,350 e, you get continuous spectrum and doubly degenerate, 619 00:43:11,350 --> 00:43:12,320 so two solutions. 620 00:43:17,360 --> 00:43:25,190 Now, after this, there's one more result 621 00:43:25,190 --> 00:43:30,480 that qualifies as a theorem. 622 00:43:30,480 --> 00:43:35,030 And it's hard to prove rigorously. 623 00:43:35,030 --> 00:43:39,650 I will not attempt to prove it here nor even in the notes. 624 00:43:39,650 --> 00:43:40,700 It's hard enough. 625 00:43:40,700 --> 00:43:45,110 So this theorem has to do with nodes. 626 00:43:45,110 --> 00:43:58,930 Theorem-- so if you have the discrete bound state 627 00:43:58,930 --> 00:44:11,560 spectrum of a one dimensional potential, 628 00:44:11,560 --> 00:44:22,620 and you list the energies E1 less than E2 less than E3 629 00:44:22,620 --> 00:44:26,080 like that, E1 is the ground state energy. 630 00:44:31,940 --> 00:44:34,380 Remember, this spectrum is discrete. 631 00:44:34,380 --> 00:44:36,740 So this is less than E2, less than E3, 632 00:44:36,740 --> 00:44:40,430 and it goes on like that, the ground state energy. 633 00:44:40,430 --> 00:44:45,680 Then, you have associated wave functions, energy eigenstates 634 00:44:45,680 --> 00:44:51,470 psi 1 of x, psi 2 of x, psi 3 of x. 635 00:44:55,440 --> 00:44:56,890 Here is the theorem. 636 00:44:56,890 --> 00:45:01,220 The theorem tells you something about the vanishing of the wave 637 00:45:01,220 --> 00:45:01,990 function. 638 00:45:01,990 --> 00:45:09,530 It says that psi 1 has no nodes. 639 00:45:09,530 --> 00:45:14,930 Psi 2 has one node. 640 00:45:14,930 --> 00:45:19,570 Psi 3 has two nodes. 641 00:45:19,570 --> 00:45:27,565 And so it goes so that psi n has n minus one node. 642 00:45:27,565 --> 00:45:34,500 So psi n is greater than-- well, it's correct. 643 00:45:34,500 --> 00:45:41,070 Any n greater or equal to 1 has n minus 1 nodes. 644 00:45:41,070 --> 00:45:46,330 Now, there are several ways people show this. 645 00:45:46,330 --> 00:45:52,090 Mathematicians show it in a rather delicate analysis. 646 00:45:52,090 --> 00:45:55,210 Physicists have an argument as well 647 00:45:55,210 --> 00:46:02,800 for this, which is based on approximating any potential 648 00:46:02,800 --> 00:46:05,950 by infinite square wells to begin with. 649 00:46:05,950 --> 00:46:09,960 So suppose you have a potential like that. 650 00:46:09,960 --> 00:46:14,700 Well, think of it as first being a little potential like that, 651 00:46:14,700 --> 00:46:16,700 an infinite square well. 652 00:46:16,700 --> 00:46:21,310 And you start making the window of the square well bigger. 653 00:46:21,310 --> 00:46:25,060 The argument-- it's a neat argument. 654 00:46:25,060 --> 00:46:28,650 Maybe you can discuss it in recitation. 655 00:46:28,650 --> 00:46:30,470 I would suggest that. 656 00:46:30,470 --> 00:46:31,690 It's a good argument. 657 00:46:31,690 --> 00:46:33,730 But it's not rigorous. 658 00:46:33,730 --> 00:46:37,040 But still, one can do something like that. 659 00:46:37,040 --> 00:46:38,140 You make it grow. 660 00:46:38,140 --> 00:46:41,330 And what you know is that the infinite square well, 661 00:46:41,330 --> 00:46:43,620 the first wave function has no node. 662 00:46:43,620 --> 00:46:48,660 And as you change the screen to make the potential really what 663 00:46:48,660 --> 00:46:52,230 it's supposed to be and not just that of a square well, 664 00:46:52,230 --> 00:46:57,080 the wave function cannot gain a node. 665 00:46:57,080 --> 00:47:00,100 On the other hand, what you will show in the homework 666 00:47:00,100 --> 00:47:04,230 is something that is a partial result which 667 00:47:04,230 --> 00:47:10,890 says that the solution with n plus 1 668 00:47:10,890 --> 00:47:17,200 has at least one more node than the solution with n. 669 00:47:17,200 --> 00:47:20,350 So it's part of what you can show. 670 00:47:20,350 --> 00:47:22,130 And it doesn't take too much effort. 671 00:47:22,130 --> 00:47:23,540 And you can prove it rigorously. 672 00:47:23,540 --> 00:47:25,740 So we will assign that eventually 673 00:47:25,740 --> 00:47:26,940 for a homework to do. 674 00:47:31,950 --> 00:47:34,630 In the homework that you have for the first homework, 675 00:47:34,630 --> 00:47:39,360 you also have a problem with delta functions. 676 00:47:39,360 --> 00:47:43,660 And I suggest that you read the notes that 677 00:47:43,660 --> 00:47:46,110 will be posted today. 678 00:47:46,110 --> 00:47:49,030 Because there's an example there with delta functions. 679 00:47:49,030 --> 00:47:51,490 If you study that example, you'll 680 00:47:51,490 --> 00:47:54,980 find the problem quite easy to solve. 681 00:47:54,980 --> 00:47:56,540 You may have solved already. 682 00:47:56,540 --> 00:48:00,620 Some of you are very eager to get going with the homework. 683 00:48:00,620 --> 00:48:07,590 So it's something you can study first, and then 684 00:48:07,590 --> 00:48:10,920 make your life a little easier. 685 00:48:10,920 --> 00:48:17,710 So what we're going to do now for the rest of the lecture 686 00:48:17,710 --> 00:48:20,330 is consider the variational problem, which 687 00:48:20,330 --> 00:48:24,100 is something you probably haven't seen before, 688 00:48:24,100 --> 00:48:25,475 the variational problem. 689 00:48:38,450 --> 00:48:43,410 This problem has to do with calculus of variations. 690 00:48:43,410 --> 00:48:45,680 Now, calculus of variations is something 691 00:48:45,680 --> 00:48:49,200 considered fairly advanced. 692 00:48:49,200 --> 00:48:52,535 And therefore, as you will see, we 693 00:48:52,535 --> 00:48:57,390 will avoid some of the major difficulties of calculus 694 00:48:57,390 --> 00:49:01,930 of variations in our discussion of the variational principle. 695 00:49:01,930 --> 00:49:11,860 But I wanted to mention a little story about this problem. 696 00:49:11,860 --> 00:49:18,590 So this calculus of variations is a more complicated version 697 00:49:18,590 --> 00:49:26,140 of maxima and minima in which in maxima and minima of functions 698 00:49:26,140 --> 00:49:27,800 you look at the function. 699 00:49:27,800 --> 00:49:30,580 And if you could plot it, you could say, here's a maximum, 700 00:49:30,580 --> 00:49:32,360 here's a minimum. 701 00:49:32,360 --> 00:49:35,427 If you want to figure out where they are, you know. 702 00:49:35,427 --> 00:49:38,050 You take a derivative, set it equal to 0, 703 00:49:38,050 --> 00:49:41,300 you find the maxima and minima. 704 00:49:41,300 --> 00:49:44,030 So the typical calculus problem is 705 00:49:44,030 --> 00:49:45,635 one in which you have a function, 706 00:49:45,635 --> 00:49:47,930 and you want the maxima and minima. 707 00:49:47,930 --> 00:49:51,920 The variational problem is a problem 708 00:49:51,920 --> 00:49:56,190 in which you want to maximize or minimize something. 709 00:49:56,190 --> 00:50:00,890 But what you don't know is not where the maximum or minimum 710 00:50:00,890 --> 00:50:04,440 occurs, but which kind of function 711 00:50:04,440 --> 00:50:07,090 will give you this maximum or minimum. 712 00:50:07,090 --> 00:50:09,890 So your unknown is not a point where 713 00:50:09,890 --> 00:50:13,790 there's a maximum or a minimum but a function where 714 00:50:13,790 --> 00:50:16,310 there is a maximum and a minimum. 715 00:50:16,310 --> 00:50:20,660 So it's slightly more complicated. 716 00:50:20,660 --> 00:50:22,940 So this is the calculus of variations. 717 00:50:22,940 --> 00:50:28,290 And people wonder when did it start. 718 00:50:28,290 --> 00:50:31,530 It actually seems to have first been discussed by Newton. 719 00:50:31,530 --> 00:50:36,080 And it's quite an interesting story. 720 00:50:36,080 --> 00:50:39,060 Newton was trying to understand apparently 721 00:50:39,060 --> 00:50:40,470 the following problem. 722 00:50:40,470 --> 00:50:49,610 If you would have a cross sectional area like this, 723 00:50:49,610 --> 00:50:53,700 he asked the question, how should you 724 00:50:53,700 --> 00:50:59,640 make a solid out of this by tapering it and ending 725 00:50:59,640 --> 00:51:02,810 with this, tapering it, in such a way 726 00:51:02,810 --> 00:51:06,070 that as it moves in a viscous fluid, 727 00:51:06,070 --> 00:51:10,980 the resistance is the minimum possible-- very complicated 728 00:51:10,980 --> 00:51:12,950 problem. 729 00:51:12,950 --> 00:51:16,180 And as you can imagine, this is a complicated problem 730 00:51:16,180 --> 00:51:19,590 because you're trying to find a shape-- not just a maximum 731 00:51:19,590 --> 00:51:22,060 or a minimum of a function but what 732 00:51:22,060 --> 00:51:25,930 shape maximizes or minimizes this. 733 00:51:25,930 --> 00:51:28,600 So apparently, he solved the problem 734 00:51:28,600 --> 00:51:33,950 and wrote it in Principia but didn't explain his solution. 735 00:51:33,950 --> 00:51:36,650 And people for years were trying to figure it out. 736 00:51:36,650 --> 00:51:41,130 And nobody could figure out how he did it. 737 00:51:41,130 --> 00:51:47,710 Then, the story goes that this mathematician Johann Bernoulli 738 00:51:47,710 --> 00:51:53,870 in 1696 came up with a challenge to all mathematicians. 739 00:51:53,870 --> 00:51:56,380 At that time, people would announce a problem 740 00:51:56,380 --> 00:51:59,110 and challenge to see who's smart, 741 00:51:59,110 --> 00:52:01,580 who can solve this problem. 742 00:52:01,580 --> 00:52:13,380 So Johann Bernoulli in around 1696 743 00:52:13,380 --> 00:52:18,410 poses a problem of, you're given two points in the plane, 744 00:52:18,410 --> 00:52:23,440 in the vertical plane like this blackboard, point this, A, 745 00:52:23,440 --> 00:52:28,560 and point B in here. 746 00:52:28,560 --> 00:52:32,260 You must design the curve of shortest time 747 00:52:32,260 --> 00:52:35,430 for fall, so some curve here. 748 00:52:35,430 --> 00:52:37,465 If you put an object and let it fall, 749 00:52:37,465 --> 00:52:42,270 it will get the fastest to that point, so maybe 750 00:52:42,270 --> 00:52:44,950 something that looks like this. 751 00:52:44,950 --> 00:52:49,860 It's a complicated curve, or at least not all that simple. 752 00:52:49,860 --> 00:52:56,330 And he asked all the people to solve it, 753 00:52:56,330 --> 00:53:00,740 gave them one year to solve it. 754 00:53:00,740 --> 00:53:03,920 So who was around at that time? 755 00:53:03,920 --> 00:53:08,060 Well, one person that got the letter was Leibniz. 756 00:53:08,060 --> 00:53:12,800 He got it on the 9th of June of that year, 1696. 757 00:53:12,800 --> 00:53:17,340 And he answered it, sent an email back, 758 00:53:17,340 --> 00:53:20,000 by the 16th of June with a letter 759 00:53:20,000 --> 00:53:21,285 with a complete solution. 760 00:53:24,470 --> 00:53:29,640 It's a funny thing that actually apparently Newton was very busy 761 00:53:29,640 --> 00:53:32,540 and didn't receive this letter. 762 00:53:32,540 --> 00:53:35,510 Or something happened, and he got it one day, 763 00:53:35,510 --> 00:53:39,750 and he actually solved the problem in one night. 764 00:53:39,750 --> 00:53:43,170 It took him one full night to solve it. 765 00:53:43,170 --> 00:53:45,090 Now, you say, well, how brilliant. 766 00:53:45,090 --> 00:53:49,670 And true, but given that he had solved this problem, 767 00:53:49,670 --> 00:53:54,590 he was criticized as being really slow and-- how come 768 00:53:54,590 --> 00:53:56,680 you took 12 hours to solve this problem? 769 00:54:00,250 --> 00:54:02,210 So it's quite amazing. 770 00:54:02,210 --> 00:54:03,980 There's a lot of Bernoullis. 771 00:54:03,980 --> 00:54:08,970 And apparently, this question by Jacob Bernoulli, 772 00:54:08,970 --> 00:54:11,990 the main purpose of this question 773 00:54:11,990 --> 00:54:16,820 was to demonstrate to everybody that his older brother, 774 00:54:16,820 --> 00:54:20,640 Jacob Bernoulli, who had invented the Bernoulli numbers, 775 00:54:20,640 --> 00:54:22,790 was actually an incompetent person that 776 00:54:22,790 --> 00:54:25,990 could not solve this problem. 777 00:54:25,990 --> 00:54:29,110 So that was apparently what he wanted to do. 778 00:54:29,110 --> 00:54:31,650 It's a rather famous family. 779 00:54:31,650 --> 00:54:34,000 But they obviously didn't get along. 780 00:54:34,000 --> 00:54:37,330 But apparently, Jacob did manage to solve the problem. 781 00:54:37,330 --> 00:54:41,515 So Jacob Bernoulli, Leibniz, and Newton all solved the problem. 782 00:54:44,130 --> 00:54:47,670 Johann Bernoulli, the one that started this problem-- and I 783 00:54:47,670 --> 00:54:50,220 think it's maybe with a double N, 784 00:54:50,220 --> 00:54:54,510 I'm sorry-- his son is Daniel Bernoulli. 785 00:54:54,510 --> 00:54:56,470 And engineers know him, because that's 786 00:54:56,470 --> 00:55:01,160 the Bernoulli of the Bernoulli fluid dynamics stuff. 787 00:55:01,160 --> 00:55:05,775 So the problem is not all that easy. 788 00:55:05,775 --> 00:55:09,490 And calculus of variations determines this shape. 789 00:55:09,490 --> 00:55:14,580 So the calculus of variation applied to quantum mechanics 790 00:55:14,580 --> 00:55:21,000 asks, here, this function is determined by the principle 791 00:55:21,000 --> 00:55:24,250 that it minimizes time. 792 00:55:24,250 --> 00:55:26,730 So you have the Schrodinger Equation. 793 00:55:26,730 --> 00:55:31,130 And you could ask, you have all these Eigenfunctions. 794 00:55:31,130 --> 00:55:33,870 What do they minimize? 795 00:55:33,870 --> 00:55:36,610 Is there something they're minimize? 796 00:55:36,610 --> 00:55:38,230 And the answer is yes. 797 00:55:38,230 --> 00:55:40,510 And this is what you'll understand 798 00:55:40,510 --> 00:55:43,330 in the next few minutes. 799 00:55:43,330 --> 00:55:46,060 So what is the problem we want to solve? 800 00:55:46,060 --> 00:55:52,280 We want to solve the problem h psi equal e psi. 801 00:55:52,280 --> 00:55:54,670 So some Hamiltonian. 802 00:55:54,670 --> 00:56:00,390 Now my notation will be such that it 803 00:56:00,390 --> 00:56:02,740 applies to three dimensions as well. 804 00:56:02,740 --> 00:56:04,990 So I'll put just arrows on top of it, 805 00:56:04,990 --> 00:56:07,850 and you would have to write the proper Hamiltonian. 806 00:56:07,850 --> 00:56:11,690 It will not be necessary for you to know the Hamiltonian. 807 00:56:11,690 --> 00:56:15,890 So I'll put psi of x here, meaning 808 00:56:15,890 --> 00:56:21,360 that this is equally valid for more than one dimension. 809 00:56:21,360 --> 00:56:28,200 Now we want to find solutions of this equation. 810 00:56:28,200 --> 00:56:30,870 And you can say, what do they maximize or minimize? 811 00:56:30,870 --> 00:56:33,600 Well we won't get to it until 15 minutes. 812 00:56:33,600 --> 00:56:37,110 First let's try something simpler. 813 00:56:37,110 --> 00:56:39,740 How about, can we learn something 814 00:56:39,740 --> 00:56:43,950 about the ground state energy of the system? 815 00:56:43,950 --> 00:56:48,360 So let's try to think about the ground state energy. 816 00:56:50,950 --> 00:56:53,240 State energy. 817 00:57:00,670 --> 00:57:05,440 Now consider ground state energy and we'll 818 00:57:05,440 --> 00:57:14,390 consider an arbitrary-- arbitrary 819 00:57:14,390 --> 00:57:17,960 is the most important word here-- 820 00:57:17,960 --> 00:57:20,103 psi of x that is normalized. 821 00:57:23,070 --> 00:57:23,722 Is normalized. 822 00:57:26,600 --> 00:57:34,020 So integral the x of psi squared is equal to 1. 823 00:57:39,480 --> 00:57:42,800 And here comes the claim. 824 00:57:42,800 --> 00:57:49,030 The first claim that we can make. 825 00:57:49,030 --> 00:57:51,470 You see, this wave function doesn't 826 00:57:51,470 --> 00:57:53,450 solve the Schrodinger equation. 827 00:57:53,450 --> 00:57:56,390 That's what we mean by arbitrary. 828 00:57:56,390 --> 00:58:02,950 It's just any function of space that is normalizable. 829 00:58:02,950 --> 00:58:04,980 Doesn't solve the Schrodinger equation. 830 00:58:04,980 --> 00:58:07,110 Never the less, you are commanded 831 00:58:07,110 --> 00:58:09,070 to compute the following quantity. 832 00:58:19,000 --> 00:58:22,090 This quantity is also by definition 833 00:58:22,090 --> 00:58:24,480 what we call the expectation value 834 00:58:24,480 --> 00:58:29,090 of the Hamiltonian in the state psi. 835 00:58:29,090 --> 00:58:33,850 I love the fact, the remarkable fact that we're 836 00:58:33,850 --> 00:58:38,000 going to show now, is that this thing provides 837 00:58:38,000 --> 00:58:46,615 an upper bound for the ground state energy for all psi. 838 00:58:50,230 --> 00:58:54,550 So let me try to make sure we understand 839 00:58:54,550 --> 00:58:58,160 what's happening here. 840 00:58:58,160 --> 00:59:00,990 Here it says you don't know the ground state energy 841 00:59:00,990 --> 00:59:03,880 but you're going to learn something about it. 842 00:59:03,880 --> 00:59:06,880 Something that's interesting is if you 843 00:59:06,880 --> 00:59:10,000 know that it has an upper bound, so the ground state energy 844 00:59:10,000 --> 00:59:13,190 is definitely not higher than this one, 845 00:59:13,190 --> 00:59:14,230 so you learn something. 846 00:59:14,230 --> 00:59:17,080 Would be ideal if you had also lower bound 847 00:59:17,080 --> 00:59:19,500 so you knew it's in this range. 848 00:59:19,500 --> 00:59:24,040 But an upper bound is a nice thing to have. 849 00:59:24,040 --> 00:59:28,250 And the claim here is that each time you try an arbitrary 850 00:59:28,250 --> 00:59:32,090 function, you put anything here, you ever write, 851 00:59:32,090 --> 00:59:34,840 you've got an upper bound. 852 00:59:34,840 --> 00:59:37,020 So how is this used? 853 00:59:37,020 --> 00:59:42,450 You try arbitrary functions that you think look possibly 854 00:59:42,450 --> 00:59:45,560 like the wave function of a bound state. 855 00:59:45,560 --> 00:59:47,730 And you get numbers and you already 856 00:59:47,730 --> 00:59:49,880 know that the ground state energy smaller 857 00:59:49,880 --> 00:59:52,470 than some number. 858 00:59:52,470 --> 00:59:57,240 So it's a rather nice way of getting some information 859 00:59:57,240 --> 00:59:59,350 about the ground state energy. 860 00:59:59,350 --> 01:00:04,120 So this psi effects is called a trial wave function. 861 01:00:06,870 --> 01:00:10,320 Is a trial wave function. 862 01:00:17,330 --> 01:00:22,040 So is the statement of this theorem clear? 863 01:00:22,040 --> 01:00:24,250 Not the proof, but the statement. 864 01:00:24,250 --> 01:00:25,260 Do we have questions? 865 01:00:30,955 --> 01:00:31,455 Yes. 866 01:00:34,125 --> 01:00:36,368 AUDIENCE: Is there any statement about how 867 01:00:36,368 --> 01:00:38,808 using the wave function will give us 868 01:00:38,808 --> 01:00:43,200 how accurate an estimate [INAUDIBLE] 869 01:00:43,200 --> 01:00:44,870 PROFESSOR: No. 870 01:00:44,870 --> 01:00:47,655 We're going to become good and figure out 871 01:00:47,655 --> 01:00:52,280 some nice way of choosing wave functions, but no. 872 01:00:52,280 --> 01:00:55,380 Once you tried you got some information. 873 01:00:55,380 --> 01:00:58,420 And you may not know so easily whether you 874 01:00:58,420 --> 01:01:00,806 could go much lower. 875 01:01:00,806 --> 01:01:05,210 You can try a little, but there's no clear way to know. 876 01:01:05,210 --> 01:01:10,170 This is just some partial information. 877 01:01:10,170 --> 01:01:10,670 OK. 878 01:01:10,670 --> 01:01:14,270 So let me first prove this. 879 01:01:14,270 --> 01:01:21,180 We'll prove it and then explain a little more what it all 880 01:01:21,180 --> 01:01:23,430 means. 881 01:01:23,430 --> 01:01:25,350 So the proof. 882 01:01:28,690 --> 01:01:32,480 Now it's a proof under quotation marks. 883 01:01:32,480 --> 01:01:36,360 I will make a few assumptions. 884 01:01:36,360 --> 01:01:41,030 Basically, that I don't have a continuous spectrum. 885 01:01:41,030 --> 01:01:45,740 Now that assumption is done for me to write a simpler proof, 886 01:01:45,740 --> 01:01:47,780 not because the result doesn't hold. 887 01:01:47,780 --> 01:01:51,830 So the proof is good, but I will just 888 01:01:51,830 --> 01:01:55,750 consider for notational purposes no continuous spectrum. 889 01:01:55,750 --> 01:01:58,190 So we'll have a ground state energy 890 01:01:58,190 --> 01:02:03,610 which is e1 that is maybe less than or equal to e2 less than 891 01:02:03,610 --> 01:02:07,550 or equal e3 like that. 892 01:02:07,550 --> 01:02:13,730 So you even may consider the [? genorisies. ?] 893 01:02:13,730 --> 01:02:19,220 And we have h psi n is equal to en psi n. 894 01:02:23,580 --> 01:02:26,890 So what do we have? 895 01:02:26,890 --> 01:02:29,130 We have a trial wave function. 896 01:02:29,130 --> 01:02:33,810 So your trial wave function since it's 897 01:02:33,810 --> 01:02:36,780 an arbitrary function of x should 898 01:02:36,780 --> 01:02:42,775 be expandable by completeness as a serious or a superb 899 01:02:42,775 --> 01:02:46,670 position of the energy eigenstates. 900 01:02:49,440 --> 01:02:52,210 Let me clarify this point. 901 01:02:52,210 --> 01:02:54,550 This is a trial wave function. 902 01:02:54,550 --> 01:02:57,910 Doesn't solve the Schrodinger equation. 903 01:02:57,910 --> 01:03:04,180 So this Doesn't solve this energy eigenstate equation. 904 01:03:04,180 --> 01:03:06,970 So in fact, it doesn't solve it because this 905 01:03:06,970 --> 01:03:09,690 is a superb position of many in here. 906 01:03:09,690 --> 01:03:13,110 So that's consistent with this, and the fact 907 01:03:13,110 --> 01:03:17,970 that this wave function as given in here 908 01:03:17,970 --> 01:03:22,200 just can be represented using the energy eigenstates. 909 01:03:22,200 --> 01:03:25,050 But being a superb position, it's 910 01:03:25,050 --> 01:03:28,490 not an energy eigenstate which is true because a trial wave 911 01:03:28,490 --> 01:03:31,100 function is something that you invent out of your head. 912 01:03:31,100 --> 01:03:32,670 It's not a solution. 913 01:03:32,670 --> 01:03:36,090 If you had a solution, you wouldn't need this. 914 01:03:36,090 --> 01:03:37,550 So you don't have a solution. 915 01:03:37,550 --> 01:03:40,080 You invent the trial wave function, and you have this. 916 01:03:44,130 --> 01:03:46,160 A couple of things. 917 01:03:46,160 --> 01:03:51,395 The psi squared integral being one. 918 01:03:53,960 --> 01:03:58,960 You can do that integral, that condition, 919 01:03:58,960 --> 01:04:05,410 is the sum of their bn-2nd is equal to 1. 920 01:04:10,640 --> 01:04:12,880 This I use the orthonormality. 921 01:04:12,880 --> 01:04:15,190 You can elevate this. 922 01:04:15,190 --> 01:04:18,430 It's sort of the kinds of things we're doing last time. 923 01:04:18,430 --> 01:04:21,910 Please make sure you know how to do that. 924 01:04:21,910 --> 01:04:23,960 Then there's the other computation 925 01:04:23,960 --> 01:04:28,020 that we also have sketched last time, which 926 01:04:28,020 --> 01:04:32,535 is that the integral of psi star h 927 01:04:32,535 --> 01:04:37,570 psi which is what we want to compute, 928 01:04:37,570 --> 01:04:45,760 h hat psi is actually bn-2nd en. 929 01:04:51,170 --> 01:04:53,920 So that was something we're doing 930 01:04:53,920 --> 01:04:56,650 towards the end of last lecture. 931 01:04:56,650 --> 01:05:00,920 And this computation takes a few lines, but it was there. 932 01:05:00,920 --> 01:05:03,330 It's in the notes. 933 01:05:03,330 --> 01:05:09,860 And now comes the kind of thing we want to say. 934 01:05:09,860 --> 01:05:11,170 Look at this sum. 935 01:05:11,170 --> 01:05:17,360 It has b1, e1, b2, e2, b3, e3, but e2, e3, e4, 936 01:05:17,360 --> 01:05:23,130 all those are bigger, or at most, equal to e1. 937 01:05:23,130 --> 01:05:27,440 So if I did, here, the following bad joke 938 01:05:27,440 --> 01:05:35,140 of substituting en for e1, which is not the same, if I put here 939 01:05:35,140 --> 01:05:39,450 bn squared n equals 1 to infinity. 940 01:05:39,450 --> 01:05:44,500 I put here e1, well this is bigger 941 01:05:44,500 --> 01:05:47,480 than that because e2 is possibly bigger than e1, 942 01:05:47,480 --> 01:05:49,450 e3 is bigger than e1. 943 01:05:49,450 --> 01:05:52,610 But it may be equal. 944 01:05:52,610 --> 01:05:55,940 But at this moment, e1 can go out of the sum. 945 01:05:55,940 --> 01:05:59,860 So this is e1 times this sum which 946 01:05:59,860 --> 01:06:06,180 is 1 because bn is equal to 1. 947 01:06:06,180 --> 01:06:10,790 And e1 is the ground state energy by definition. 948 01:06:10,790 --> 01:06:13,040 So the ground state energy is less 949 01:06:13,040 --> 01:06:15,970 than this which is the expectation 950 01:06:15,970 --> 01:06:17,760 value of the Hamiltonian. 951 01:06:17,760 --> 01:06:19,670 Pretty simple, in fact, the proof 952 01:06:19,670 --> 01:06:22,700 is really a little too simple. 953 01:06:27,530 --> 01:06:30,790 Where do we go from now? 954 01:06:30,790 --> 01:06:36,540 Well let's make a more general statement 955 01:06:36,540 --> 01:06:38,590 of the variational principal. 956 01:06:38,590 --> 01:06:42,230 Again, sometimes it's not all that 957 01:06:42,230 --> 01:06:45,380 convenient to have normalized wave functions. 958 01:06:45,380 --> 01:06:58,890 So recall that if psi of x is not normalized, 959 01:06:58,890 --> 01:07:06,810 psi of x over the square root of integral of psi-2nd dx is. 960 01:07:12,210 --> 01:07:18,510 Therefore if you hand me an arbitrary 961 01:07:18,510 --> 01:07:20,950 psi of x that is really arbitrary. 962 01:07:20,950 --> 01:07:24,250 You don't even bother to normalize it. 963 01:07:24,250 --> 01:07:28,200 Then when I plug here in this formula 964 01:07:28,200 --> 01:07:30,190 it is supposed to be normalized. 965 01:07:30,190 --> 01:07:33,530 So I plug the second expression there. 966 01:07:33,530 --> 01:07:40,670 So therefore I get that egs less than 967 01:07:40,670 --> 01:07:46,320 or equal to the integral of psi star h 968 01:07:46,320 --> 01:07:54,650 psi the x over integral of psi star psi the x. 969 01:08:03,890 --> 01:08:08,860 This is actually nicer in one way 970 01:08:08,860 --> 01:08:12,770 because you don't have to work with normalized wave functions. 971 01:08:12,770 --> 01:08:15,550 And that result must be true still. 972 01:08:15,550 --> 01:08:16,430 Yes? 973 01:08:16,430 --> 01:08:17,305 AUDIENCE: [INAUDIBLE] 974 01:08:21,063 --> 01:08:21,729 PROFESSOR: Sure. 975 01:08:21,729 --> 01:08:23,779 It cannot be completely arbitrary, 976 01:08:23,779 --> 01:08:27,569 the function should we normalizable. 977 01:08:27,569 --> 01:08:29,654 Doesn't have to be normalized but normalizable. 978 01:08:32,470 --> 01:08:34,160 So here you got. 979 01:08:34,160 --> 01:08:36,890 And here let me introduce just a little name. 980 01:08:39,630 --> 01:08:41,564 f of psi. 981 01:08:45,710 --> 01:08:49,586 f of psi is what is called a functional. 982 01:08:52,810 --> 01:08:54,084 f of is a functional. 983 01:08:58,100 --> 01:09:00,240 What is a functional? 984 01:09:00,240 --> 01:09:03,899 A functional is a machine or an expression 985 01:09:03,899 --> 01:09:10,750 whose input is a function and whose output is a number. 986 01:09:10,750 --> 01:09:16,390 So here f of psi is a functional. 987 01:09:16,390 --> 01:09:19,390 And maybe I should use brackets. 988 01:09:19,390 --> 01:09:23,529 Many times people with brackets denote that, watch out, 989 01:09:23,529 --> 01:09:26,710 this is not a function, it's a functional. 990 01:09:26,710 --> 01:09:29,680 And here it is. 991 01:09:29,680 --> 01:09:33,670 No dash there. 992 01:09:33,670 --> 01:09:38,250 You give me a psi of x, which is a function, 993 01:09:38,250 --> 01:09:40,760 and then this is a number because you've 994 01:09:40,760 --> 01:09:42,640 done the integrals. 995 01:09:42,640 --> 01:09:47,600 So that is like the Brachistochrone problem, that 996 01:09:47,600 --> 01:09:50,590 a funny name for it. 997 01:09:50,590 --> 01:09:51,359 Here it is. 998 01:09:51,359 --> 01:09:54,790 There is a function now which is the time 999 01:09:54,790 --> 01:09:56,660 that it takes to go here. 1000 01:09:56,660 --> 01:10:00,030 You give me a path, a function, and I 1001 01:10:00,030 --> 01:10:03,570 can calculate the time it will take the mass to get here. 1002 01:10:03,570 --> 01:10:08,460 So this was the issue of finding a critical point 1003 01:10:08,460 --> 01:10:09,840 of a functional. 1004 01:10:09,840 --> 01:10:14,740 So actually we start to see, it seems that the ground state 1005 01:10:14,740 --> 01:10:17,835 energy is the minimum of this functional. 1006 01:10:22,300 --> 01:10:25,080 And what this interesting as well 1007 01:10:25,080 --> 01:10:31,150 is that when you get the minimum you will have gotten 1008 01:10:31,150 --> 01:10:33,270 a ground state wave function. 1009 01:10:33,270 --> 01:10:36,580 So the ground state wave function actually 1010 01:10:36,580 --> 01:10:40,830 is the thing that minimizes this functional 1011 01:10:40,830 --> 01:10:45,250 and gives you some value, the ground state energy. 1012 01:10:45,250 --> 01:10:50,080 A little more in a couple of minutes. 1013 01:10:50,080 --> 01:10:52,990 Let's do an example. 1014 01:10:52,990 --> 01:11:08,350 How do we use this so now if you think about this carefully 1015 01:11:08,350 --> 01:11:14,220 it's kind of dizzying because what is sell functional, 1016 01:11:14,220 --> 01:11:15,670 really? 1017 01:11:15,670 --> 01:11:19,720 It's, in some sense, a function in an infinite dimensional 1018 01:11:19,720 --> 01:11:24,770 space because a function itself is specified 1019 01:11:24,770 --> 01:11:27,670 by infinitely many numbers that you can change. 1020 01:11:27,670 --> 01:11:33,750 So how many dimensions you have is how many ways 1021 01:11:33,750 --> 01:11:38,660 you can move your hands, but they are linearly independent. 1022 01:11:38,660 --> 01:11:41,250 But if you have a function, and you can change it, 1023 01:11:41,250 --> 01:11:43,490 you can change it here, you can change it there, 1024 01:11:43,490 --> 01:11:45,760 or you can change it there, and those are all 1025 01:11:45,760 --> 01:11:48,170 orthogonal directions. 1026 01:11:48,170 --> 01:11:50,180 You're finding a critical point. 1027 01:11:50,180 --> 01:11:52,190 When you find the critical point, 1028 01:11:52,190 --> 01:11:56,070 you should imagine that you're plotting a function that is not 1029 01:11:56,070 --> 01:12:00,040 one dimensional function or two dimensional function, 1030 01:12:00,040 --> 01:12:02,740 but it's infinitely dimensional function. 1031 01:12:02,740 --> 01:12:09,060 Direction one, direction two, infinitely many directions. 1032 01:12:09,060 --> 01:12:12,010 And suddenly in this infinitely many directions 1033 01:12:12,010 --> 01:12:14,770 you find the critical point. 1034 01:12:14,770 --> 01:12:19,720 It's really incredible that one can do these things. 1035 01:12:19,720 --> 01:12:21,620 So you're at that critical point, 1036 01:12:21,620 --> 01:12:25,250 and you can deform the energy eigenstate 1037 01:12:25,250 --> 01:12:28,230 by making it a little fatter here or thinner 1038 01:12:28,230 --> 01:12:29,460 here or up there. 1039 01:12:29,460 --> 01:12:31,120 And those are the infinite directions 1040 01:12:31,120 --> 01:12:33,550 in any direction that you will the energy 1041 01:12:33,550 --> 01:12:37,837 goes up because you're at the global minimum 1042 01:12:37,837 --> 01:12:38,670 it's pretty amazing. 1043 01:12:41,540 --> 01:12:45,850 Something that you will prove in the homework 1044 01:12:45,850 --> 01:12:50,550 is that actually it's even more. 1045 01:12:50,550 --> 01:12:58,560 Every single eigenstate is a critical point 1046 01:12:58,560 --> 01:12:59,580 of this functional. 1047 01:12:59,580 --> 01:13:02,050 So you've got the lowest energy state 1048 01:13:02,050 --> 01:13:05,850 and in that infinite dimensional space in every direction 1049 01:13:05,850 --> 01:13:08,700 that you move you go up. 1050 01:13:08,700 --> 01:13:12,960 The first excited state is another critical point. 1051 01:13:12,960 --> 01:13:17,750 But it will not be an absolute minimum. 1052 01:13:17,750 --> 01:13:20,530 It will be a [? saddle ?] an infinite dimensional 1053 01:13:20,530 --> 01:13:22,560 [? saddle ?] which are infinitely 1054 01:13:22,560 --> 01:13:25,010 many directions in which you go up. 1055 01:13:25,010 --> 01:13:27,560 There's one direction in which you go down 1056 01:13:27,560 --> 01:13:31,600 because you could flow towards the ground state. 1057 01:13:31,600 --> 01:13:35,820 So the first excited state is the [? saddle ?] but these 1058 01:13:35,820 --> 01:13:38,640 are all stationary points of this functional. 1059 01:13:41,340 --> 01:13:44,356 So we'll conclude by doing this example. 1060 01:13:51,650 --> 01:13:52,971 Sorry, what is the question? 1061 01:13:52,971 --> 01:13:53,846 AUDIENCE: [INAUDIBLE] 1062 01:13:56,650 --> 01:13:59,380 PROFESSOR: I didn't assume it's non-degenerate. 1063 01:13:59,380 --> 01:14:02,960 So if you have two things that have the same ground state, 1064 01:14:02,960 --> 01:14:07,630 the functional will have, in fact, the degeneracy there 1065 01:14:07,630 --> 01:14:13,040 will be two solutions that have the same energy. 1066 01:14:13,040 --> 01:14:17,640 And any linear combination of them will have the same energy. 1067 01:14:17,640 --> 01:14:21,080 The proof that I did here doesn't assume non-degeneracy, 1068 01:14:21,080 --> 01:14:25,080 it's even true with degenerate things. 1069 01:14:25,080 --> 01:14:30,910 So the example is an example for illustration, 1070 01:14:30,910 --> 01:14:35,880 not for solving something that you can't do otherwise. 1071 01:14:35,880 --> 01:14:38,240 So it's a delta function potential. 1072 01:14:38,240 --> 01:14:44,580 v of x is minus alpha delta of x with alpha positive. 1073 01:14:44,580 --> 01:14:47,740 And the ground state energy is well known. 1074 01:14:47,740 --> 01:14:52,450 It's minus m alpha squared over 2h-2nd. 1075 01:14:52,450 --> 01:14:57,520 You've solved this problem many times in 804. 1076 01:14:57,520 --> 01:14:59,390 So trial wave function. 1077 01:14:59,390 --> 01:15:01,130 Well you know how it should look, 1078 01:15:01,130 --> 01:15:03,550 but let's assume you don't. 1079 01:15:03,550 --> 01:15:07,010 And you say, it's some sort of [INAUDIBLE]. 1080 01:15:07,010 --> 01:15:09,610 So trial. 1081 01:15:09,610 --> 01:15:15,630 It would be psi of x equals e to the minus x squared. 1082 01:15:15,630 --> 01:15:20,930 While this would do, you're going to work hard 1083 01:15:20,930 --> 01:15:23,330 and you're not going to reap all the benefits 1084 01:15:23,330 --> 01:15:25,310 of this calculation. 1085 01:15:25,310 --> 01:15:29,320 So what you should do at this moment is put a constant here. 1086 01:15:29,320 --> 01:15:35,190 Minus beta squared x squared and I'll put the minus one half. 1087 01:15:35,190 --> 01:15:36,815 This is our trial wave function. 1088 01:15:39,400 --> 01:15:45,330 You see, by this, you're going to get an expression. 1089 01:15:45,330 --> 01:15:47,460 You calculate this number, and you're 1090 01:15:47,460 --> 01:15:49,410 going to get the function of beta. 1091 01:15:49,410 --> 01:15:50,800 Beta is not going to disappear. 1092 01:15:53,460 --> 01:15:56,040 And therefore, you're going to know that the ground state 1093 01:15:56,040 --> 01:15:59,360 energy is less than this function of beta. 1094 01:15:59,360 --> 01:16:04,120 And then you can adjust beta to get the best bound. 1095 01:16:04,120 --> 01:16:07,020 So beta is put as a parameter to begin with, 1096 01:16:07,020 --> 01:16:09,260 and we hope to use it. 1097 01:16:09,260 --> 01:16:19,400 So note that integral of psi squared dx in this case 1098 01:16:19,400 --> 01:16:23,960 is equal to square root of pi over beta. 1099 01:16:23,960 --> 01:16:29,320 So we have to calculate this whole functional. 1100 01:16:29,320 --> 01:16:33,656 So this integral of psi star-- well 1101 01:16:33,656 --> 01:16:36,420 I don't have to bother with psi star because it's real. 1102 01:16:36,420 --> 01:16:42,920 h psi over psi psi, and what do we get? 1103 01:16:42,920 --> 01:16:45,070 Well the denominator is easy. 1104 01:16:45,070 --> 01:16:48,840 So we beta over square root of pi, 1105 01:16:48,840 --> 01:16:51,005 and let me write the whole thing here. 1106 01:16:51,005 --> 01:16:54,970 dx the psi would have e to the minus one half 1107 01:16:54,970 --> 01:16:58,750 beta squared x squared minus h squared 1108 01:16:58,750 --> 01:17:07,100 over 2m d-2nd dx2nd minus alpha delta of x. 1109 01:17:07,100 --> 01:17:09,910 And another wave function, e to the minus one 1110 01:17:09,910 --> 01:17:13,980 half beta squared x-2nd. 1111 01:17:13,980 --> 01:17:14,650 OK. 1112 01:17:14,650 --> 01:17:16,390 So you have to evaluate that. 1113 01:17:16,390 --> 01:17:20,096 And that's the part that is not so much fun. 1114 01:17:20,096 --> 01:17:23,760 For any integral that you have in 805, 1115 01:17:23,760 --> 01:17:29,340 we believe that you can use Mathematica or Maple or MATLAB 1116 01:17:29,340 --> 01:17:31,220 or whatever and do it. 1117 01:17:31,220 --> 01:17:34,550 The only reason not to use any of these things 1118 01:17:34,550 --> 01:17:37,240 is if you think you could not do the integral. 1119 01:17:37,240 --> 01:17:39,490 But once you realize, oh, this is an integral, 1120 01:17:39,490 --> 01:17:42,720 I know how to do, don't waste time. 1121 01:17:42,720 --> 01:17:44,340 Use any of those programs. 1122 01:17:44,340 --> 01:17:48,400 Now, this part of the integral is kind of easy 1123 01:17:48,400 --> 01:17:51,610 because the delta function just picks the value of 0. 1124 01:17:51,610 --> 01:17:56,020 So this part of the integral gives you 1125 01:17:56,020 --> 01:18:00,780 minus beta alpha over square root of pi. 1126 01:18:00,780 --> 01:18:03,530 The other part of the integral, however, 1127 01:18:03,530 --> 01:18:06,120 is a little more complicated. 1128 01:18:06,120 --> 01:18:10,240 Believe it or not, it makes a big difference 1129 01:18:10,240 --> 01:18:12,110 whether you take the two derivatives 1130 01:18:12,110 --> 01:18:14,710 of this function or you integrate by parts. 1131 01:18:14,710 --> 01:18:18,500 If you integrate by parts, you save a lot of work. 1132 01:18:18,500 --> 01:18:20,910 So let me integrate by parts. 1133 01:18:20,910 --> 01:18:26,120 This becomes, plus beta over square root of h, 1134 01:18:26,120 --> 01:18:32,320 h-2nd over 2m, integral dx of ddx of e 1135 01:18:32,320 --> 01:18:38,590 to the minus one half beta squared x-2nd squared. 1136 01:18:38,590 --> 01:18:41,850 So you integrate by parts one of the ddx 1137 01:18:41,850 --> 01:18:46,020 and then you have the other ddx, so it's the thing squared. 1138 01:18:46,020 --> 01:18:51,250 And that's an easier integral to do. 1139 01:18:51,250 --> 01:18:53,850 We don't want to bother with this, 1140 01:18:53,850 --> 01:18:58,280 but this whole thing then becomes 1141 01:18:58,280 --> 01:19:04,750 minus beta over square root of pi, alpha plus beta squared h 1142 01:19:04,750 --> 01:19:06,450 squared over 4m. 1143 01:19:06,450 --> 01:19:11,740 That's the whole answer And that's 1144 01:19:11,740 --> 01:19:16,330 the valuation of this whole thing. 1145 01:19:16,330 --> 01:19:18,490 So look what you get. 1146 01:19:18,490 --> 01:19:21,770 You got a function of beta indeed. 1147 01:19:21,770 --> 01:19:24,800 So how does that function of beta look? 1148 01:19:24,800 --> 01:19:29,850 It's 0 for beta equals 0 is 0 for some other possible beta, 1149 01:19:29,850 --> 01:19:31,860 it's going to look like this. 1150 01:19:31,860 --> 01:19:34,320 So there's a point at which this function is 1151 01:19:34,320 --> 01:19:37,800 going to have a minimum, and we should choose that point 1152 01:19:37,800 --> 01:19:41,545 to get the best upper bound on the function. 1153 01:19:45,270 --> 01:19:53,040 Our claim is that following from the variational theorem 1154 01:19:53,040 --> 01:19:55,690 that we've proven is that the e ground 1155 01:19:55,690 --> 01:20:03,220 state is less than or equal than the minimum value over beta 1156 01:20:03,220 --> 01:20:09,050 of this beta squared h squared over 4m 1157 01:20:09,050 --> 01:20:12,875 minus beta square root of pie alpha. 1158 01:20:12,875 --> 01:20:17,150 So you minimize over beta, and yet still the ground state 1159 01:20:17,150 --> 01:20:21,440 energy must be a little smaller than that. 1160 01:20:21,440 --> 01:20:22,860 Well what do you get? 1161 01:20:25,790 --> 01:20:29,910 You do this, this minimization gives beta equal 2m 1162 01:20:29,910 --> 01:20:34,660 alpha over h squared square root of pi. 1163 01:20:34,660 --> 01:20:37,190 It's a little messy but not terrible. 1164 01:20:37,190 --> 01:20:41,230 And you substitute, and you get that e ground state is 1165 01:20:41,230 --> 01:20:49,140 less than or equal than m alpha squared over pi h squared. 1166 01:20:49,140 --> 01:20:56,400 And to better write it as 2 over pi times minus m alpha 1167 01:20:56,400 --> 01:21:02,300 squared over 2h-2nd which is the true ground state energy. 1168 01:21:02,300 --> 01:21:06,660 So let's just make sure we understand what has happened. 1169 01:21:06,660 --> 01:21:08,640 Here is the energy. 1170 01:21:08,640 --> 01:21:10,880 Here is zero. 1171 01:21:10,880 --> 01:21:15,360 Energy as just a vertical plot here is zero. 1172 01:21:15,360 --> 01:21:20,350 The true ground state energy of this problem 1173 01:21:20,350 --> 01:21:25,280 is this one, let's call it egs, is negative. 1174 01:21:25,280 --> 01:21:27,970 And we go 2 over pi times that. 1175 01:21:27,970 --> 01:21:33,200 2 over pi is about 0.64. 1176 01:21:33,200 --> 01:21:40,560 So the bound says that here is 0.64 egs. 1177 01:21:40,560 --> 01:21:44,650 So that's what the bound told you. 1178 01:21:44,650 --> 01:21:49,180 The bound state energy must be lower than this quantity. 1179 01:21:49,180 --> 01:21:50,490 We got close. 1180 01:21:50,490 --> 01:21:53,590 Not impressively close, but the work 1181 01:21:53,590 --> 01:21:55,850 functional was not all that bad either. 1182 01:21:58,800 --> 01:21:59,966 Question? 1183 01:21:59,966 --> 01:22:01,830 AUDIENCE: [INAUDIBLE] always going 1184 01:22:01,830 --> 01:22:04,160 to be a constant times incorrect energy value 1185 01:22:04,160 --> 01:22:06,750 or is it just the closest approximation? 1186 01:22:06,750 --> 01:22:08,360 PROFESSOR: Is it going to be what? 1187 01:22:08,360 --> 01:22:10,176 AUDIENCE: Is it always going to be 1188 01:22:10,176 --> 01:22:12,936 a constant times the correct energy value or is it just-- 1189 01:22:12,936 --> 01:22:14,890 PROFESSOR: Well it typically is like that 1190 01:22:14,890 --> 01:22:19,000 because of dimensional units. 1191 01:22:19,000 --> 01:22:21,900 You're looking for a constant because you're not 1192 01:22:21,900 --> 01:22:23,150 looking for the function. 1193 01:22:23,150 --> 01:22:28,590 So you will get a number times a correct value, yes, indeed. 1194 01:22:28,590 --> 01:22:33,646 That's an illustration of the problem of wave functions. 1195 01:22:33,646 --> 01:22:36,310 You know the variational principle tells you 1196 01:22:36,310 --> 01:22:38,310 things about the ground state but allows 1197 01:22:38,310 --> 01:22:41,620 you to find the first excited state as well 1198 01:22:41,620 --> 01:22:45,570 if the potential is symmetric, will allow you to prove 1199 01:22:45,570 --> 01:22:49,060 that any attractive potential has a bound state. 1200 01:22:49,060 --> 01:22:51,750 And you will prove in the homework 1201 01:22:51,750 --> 01:22:56,040 that stationary points of these things are the eigenfunctions. 1202 01:22:56,040 --> 01:22:58,210 See you next Monday.