1 00:00:00,599 --> 00:00:14,880 PROFESSOR: We'll begin by discussing the wave packets and uncertainty. 2 00:00:14,880 --> 00:00:22,500 So it's our first look into this Heisenberg uncertainty relationships. 3 00:00:22,500 --> 00:00:31,600 And to begin with, let's focus it as fixed time, t equals zero. 4 00:00:31,600 --> 00:00:34,809 So we'll work with packets at t equals zero. 5 00:00:34,809 --> 00:00:43,731 And I will write a particular wave function that you may have at t equals 0, and it's 6 00:00:43,731 --> 00:00:50,719 a superposition of plane waves. 7 00:00:50,719 --> 00:00:54,180 So it would be e to the ikx. 8 00:00:54,180 --> 00:01:02,039 You sum over many of them, so you're going to sum over k, but you're going to do it with 9 00:01:02,039 --> 00:01:07,990 a weight, and that's 5k. 10 00:01:07,990 --> 00:01:16,869 And there's a lot to learn about this, but the physics that is encoded here is that any 11 00:01:16,869 --> 00:01:28,649 wave at time equals 0, this psi of x at time equals 0, can be written as a superposition 12 00:01:28,649 --> 00:01:34,219 of states with momentum h bar k. 13 00:01:34,219 --> 00:01:44,179 You remember e to the ikx represents a particle or a wave that carries momentum h bar k. 14 00:01:44,179 --> 00:01:52,370 So this whole idea here of a general wave function being written in this way carries 15 00:01:52,370 --> 00:01:56,859 physical meaning for us. 16 00:01:56,859 --> 00:02:03,840 It's a quantum mechanical meaning, the fact that this kind of wave has momentum. 17 00:02:03,840 --> 00:02:15,610 But this phi of k, however, suppose you know this wave function at time equals 0. 18 00:02:15,610 --> 00:02:19,280 Phi of k is then calculable. 19 00:02:19,280 --> 00:02:25,910 Phi of k can be determined, and that's the foundation of what's called Fourier's theorem, 20 00:02:25,910 --> 00:02:30,310 that gives you a formula for phi of k. 21 00:02:30,310 --> 00:02:33,490 And it's a very similar formula. 22 00:02:33,490 --> 00:02:39,840 1 over 2 pi, this time an integral over x. 23 00:02:39,840 --> 00:02:50,110 So you take this of psi of x0 that you know and then multiply by e to the minus ikx. 24 00:02:50,110 --> 00:02:57,470 Integrate over x, and out comes this function of k. 25 00:02:57,470 --> 00:03:06,060 So if you know phi of x0, you know phi of k. 26 00:03:06,060 --> 00:03:13,270 You can calculate this interval and you can rewrite phi of x0 as a superposition of plane 27 00:03:13,270 --> 00:03:14,790 waves. 28 00:03:14,790 --> 00:03:18,790 So that's how you would do a Fourier representation. 29 00:03:18,790 --> 00:03:26,460 So somebody can give you an initial wave function, and maybe it's a sine function or a Gaussian 30 00:03:26,460 --> 00:03:34,400 or something, then what you would do if you wanted to rewrite it in this way, is calculate 31 00:03:34,400 --> 00:03:42,650 phi of k, because you know this psi, you can calculate this integral, at least with a computer. 32 00:03:42,650 --> 00:03:48,820 And once you know phi of k, you have a way of writing psi as a superposition of plane 33 00:03:48,820 --> 00:03:50,930 waves. 34 00:03:50,930 --> 00:03:58,329 So we've talked about this before, because we were doing wave packets before and we got 35 00:03:58,329 --> 00:04:04,550 some intuition about how you form a wave packet and how it moves. 36 00:04:04,550 --> 00:04:10,940 Now we didn't put the time dependence here, but that can wait. 37 00:04:10,940 --> 00:04:22,330 What I wish to explain now is how by looking at these expressions, you can understand the 38 00:04:22,330 --> 00:04:31,460 uncertainties that you find on the wave function, position, and momentum uncertainties, how 39 00:04:31,460 --> 00:04:32,650 they are related. 40 00:04:32,650 --> 00:04:39,670 So that is our real goal, understanding the role of uncertainties here. 41 00:04:39,670 --> 00:04:48,240 If phi of k has some uncertainty, how is the uncertainty in psi determined? 42 00:04:48,240 --> 00:04:52,150 So that's what we're looking for. 43 00:04:52,150 --> 00:04:55,350 So relationship of uncertainties. 44 00:04:55,350 --> 00:05:06,770 Now as before, we will take a phi of k, that we've usually be in writing, that depends 45 00:05:06,770 --> 00:05:13,640 on k and it's centered around some value k0. 46 00:05:13,640 --> 00:05:16,670 It's some sort of nice, centered function. 47 00:05:16,670 --> 00:05:26,410 And it has then, we say, some uncertainty in the value of the momentum. 48 00:05:26,410 --> 00:05:34,530 That is this signal, this phi of k that we're using to produce this packet. 49 00:05:34,530 --> 00:05:42,320 It has some uncertainty, it's not totally sharp, it's peaked around k0 but not fully 50 00:05:42,320 --> 00:05:43,420 sharp. 51 00:05:43,420 --> 00:05:51,980 So the uncertainty is called delta k and it's some typical width over here. 52 00:05:51,980 --> 00:06:00,970 Delta k is then uncertainty. 53 00:06:00,970 --> 00:06:06,700 Now it's not the purpose of today's lecture to make a precise definition of what the uncertainty 54 00:06:06,700 --> 00:06:07,700 is. 55 00:06:07,700 --> 00:06:09,080 This will come later. 56 00:06:09,080 --> 00:06:15,650 At this moment, you just want to get the picture and the intuition of what's going on. 57 00:06:15,650 --> 00:06:21,500 And there is some uncertainty here, perhaps you would say, look at those points where 58 00:06:21,500 --> 00:06:29,150 the wave goes from peak value to half value and see what is the width. 59 00:06:29,150 --> 00:06:31,620 That's a typical uncertainty. 60 00:06:31,620 --> 00:06:37,909 So all what we're going to do in these arguments is get for you the intuition. 61 00:06:37,909 --> 00:06:43,670 Therefore, the factors of 2 are not trustable. 62 00:06:43,670 --> 00:06:49,120 If you're trying to make a precise statement, you must do precise definitions. 63 00:06:49,120 --> 00:06:54,650 And that will come later, probably in about one or two lectures. 64 00:06:54,650 --> 00:06:57,490 So at this moment, that's the uncertainty, delta k. 65 00:06:57,490 --> 00:07:11,419 And let's assume that this phi of k is real. 66 00:07:11,419 --> 00:07:18,540 And its peaked around k0 uncertainty delta k. 67 00:07:18,540 --> 00:07:25,010 Now what happens with psi of x? 68 00:07:25,010 --> 00:07:33,480 Well, we had our statements about the stationary phase that you already are practicing with 69 00:07:33,480 --> 00:07:38,050 them for this homework. 70 00:07:38,050 --> 00:07:46,050 If you want to know where this function peaks, you must look where the phase, this phi-- 71 00:07:46,050 --> 00:07:51,830 we say it's real, so it doesn't contribute to the phase-- where the phase, which is here, 72 00:07:51,830 --> 00:07:59,820 is stationary, given the condition that it should happen at k0. 73 00:07:59,820 --> 00:08:04,180 The only contribution to the integral is basically around k0. 74 00:08:04,180 --> 00:08:09,870 So in order to get something, you must have a stationary phase, and the phase must be 75 00:08:09,870 --> 00:08:14,320 stationary as a function of k, because you're integrating over k. 76 00:08:14,320 --> 00:08:27,640 And the phase is kx, the derivative with respect to k of the face is just x, and that must 77 00:08:27,640 --> 00:08:36,459 vanish, therefore, so you expect this to be peaked around x equals zero. 78 00:08:36,459 --> 00:08:52,170 So the x situation, so psi of x0 peaks at x equals 0. 79 00:08:52,170 --> 00:08:56,600 And so you have a picture here. 80 00:08:56,600 --> 00:09:03,980 And if I have a picture, I would say, well it peaks around the x equals 0. 81 00:09:03,980 --> 00:09:07,839 So OK, it's like that. 82 00:09:07,839 --> 00:09:10,360 And here we're going to have some uncertainty. 83 00:09:10,360 --> 00:09:19,660 Here is psi of x and 0, and here is x. 84 00:09:19,660 --> 00:09:28,269 And let me mention, I've already become fairly imprecise here. 85 00:09:28,269 --> 00:09:33,319 If you were doing this, you probably would run into trouble. 86 00:09:33,319 --> 00:09:40,519 I've sort of glossed over a small complication here. 87 00:09:40,519 --> 00:09:48,029 The complication is that this, when I talk about the peaking of psi, and you probably 88 00:09:48,029 --> 00:09:57,680 have seen it already, you have to worry whether psi is real or psi is complex. 89 00:09:57,680 --> 00:09:59,920 So what is this psi here? 90 00:09:59,920 --> 00:10:02,709 Should it be real? 91 00:10:02,709 --> 00:10:08,699 Well actually, it's not real. 92 00:10:08,699 --> 00:10:15,540 You've done, perhaps, in the homework already these integrals, and you see that psi is not 93 00:10:15,540 --> 00:10:16,540 real. 94 00:10:16,540 --> 00:10:23,029 So when we say it peaks at x equals 0, how am I supposed to plot psi? 95 00:10:23,029 --> 00:10:27,720 Am I plotting the real part, the imaginary part, the absolute value? 96 00:10:27,720 --> 00:10:38,230 So it's reasonable to plot the absolute value and to say that psi absolute value peaks at 97 00:10:38,230 --> 00:10:42,010 x equals 0. 98 00:10:42,010 --> 00:10:52,420 And there will be some width again here, delta x, width. 99 00:10:52,420 --> 00:10:58,410 And that's the uncertainty in psi of x. 100 00:10:58,410 --> 00:11:06,709 So the whole point of our discussion for the next 10 minutes is to just try to determine 101 00:11:06,709 --> 00:11:12,429 the relation between delta k and delta x and understand it intuitively.