1 00:00:00,920 --> 00:00:03,040 Welcome back everyone. I'm Charles Clark. 2 00:00:03,040 --> 00:00:06,381 This is Exploring Quantum Physics. This part we're going to start studying 3 00:00:06,381 --> 00:00:09,687 the quantum theory of angular momentum. And in particular we'll begin by 4 00:00:09,687 --> 00:00:13,573 introducing and practicing a few basic techniques that are generally useful in 5 00:00:13,573 --> 00:00:17,459 computing the commutation relations associated with operators in quantum 6 00:00:17,459 --> 00:00:21,730 mechanics. They're particularly useful for 7 00:00:21,730 --> 00:00:26,410 understanding level of detail, the theory of angular momentum. 8 00:00:26,410 --> 00:00:30,762 So, in the next few parts of the lecture, we're going to, derive a number of 9 00:00:30,762 --> 00:00:35,114 foundational properties of the angular momentum operator and then apply to 10 00:00:35,114 --> 00:00:40,988 calculations. First we're going to start by reviewing 11 00:00:40,988 --> 00:00:45,680 some standard procedures, the, these may be familiar to some of you but I think a 12 00:00:45,680 --> 00:00:50,678 review here is appropriate. And, of interest to those who haven't 13 00:00:50,678 --> 00:00:54,740 seen them before. They're really quite elementary ideas, 14 00:00:54,740 --> 00:01:01,420 but very powerful in application. And the, first thing we discuss is the, 15 00:01:01,420 --> 00:01:05,146 Einstein's summation con, convention was first introduced by Albert Einstein in 16 00:01:05,146 --> 00:01:12,360 the context of, general relativity. But it's broadly useful and we will, see 17 00:01:12,360 --> 00:01:17,321 how it applies too. Now in the discussion we're going to be, 18 00:01:17,321 --> 00:01:22,008 continue to use vector notation. But, the symbols that we are, really are 19 00:01:22,008 --> 00:01:26,034 using, are all operators so, in correspondence to classical mechanics, 20 00:01:26,034 --> 00:01:30,258 the angle of momentum vector has three separate compliments, one along each 21 00:01:30,258 --> 00:01:34,702 axis. so we'll write it symbolically as a 22 00:01:34,702 --> 00:01:38,734 vector but just keep in mind that the copy, the components of that vector are 23 00:01:38,734 --> 00:01:43,586 in fact operators given by this familiar expression. 24 00:01:43,586 --> 00:01:49,340 Now, there are three useful tools that we're going to use hereafter. 25 00:01:50,580 --> 00:01:54,012 And, in fact, you'll find them used quite widely throughout physics and I think 26 00:01:54,012 --> 00:01:57,730 you'll see the advantage of that in a moment. 27 00:01:57,730 --> 00:02:02,827 So, the first is a notational convention. Probably familiar to some of you, and 28 00:02:02,827 --> 00:02:07,070 that is to replace the standard XYZ Cartesian coordinates. 29 00:02:07,070 --> 00:02:12,270 By index at x1, x2, x3, which are just, I mean it doesn't matter what you choose. 30 00:02:12,270 --> 00:02:18,322 There's just but it's, a convenience instead of having x, y, and z to have an 31 00:02:18,322 --> 00:02:24,374 index set, because in running a, running over sums over all coordinates it just is 32 00:02:24,374 --> 00:02:34,131 convenient to introduce them, by indices rather than different letters. 33 00:02:34,131 --> 00:02:39,331 Then corespondingly we define an, appropriate set of unit vectors, which 34 00:02:39,331 --> 00:02:46,439 are just related the old x, y, and z unit vectors call them e1, e2, and e3. 35 00:02:46,439 --> 00:02:53,190 Okay, now the second aspect is the Einstein summation convention. 36 00:02:53,190 --> 00:02:57,990 So in his work on general relativity, Einstein was evaluating many multiple 37 00:02:57,990 --> 00:03:02,542 sums over coordinates and he made a observation. 38 00:03:02,542 --> 00:03:07,031 And let's describe his observations, in terms of the familiar example, which is 39 00:03:07,031 --> 00:03:15,144 the scalar product of two vectors, a.b. And in our, index notation we would write 40 00:03:15,144 --> 00:03:20,890 that as a sum of I equal 1 to 3 of ai times bi. 41 00:03:22,530 --> 00:03:29,592 Now Einstein noticed the following thing and that is; When you have a summation 42 00:03:29,592 --> 00:03:36,654 sign and the sum is over indices that are repeated in the sum, you may just as well 43 00:03:36,654 --> 00:03:47,292 eliminate the, the summation sign and write the expression thusly. 44 00:03:47,292 --> 00:03:53,140 So this save space, especially if you're doing multiple sums. 45 00:03:53,140 --> 00:04:00,680 So the message now is when indices are repeated in the expression, then the sum 46 00:04:00,680 --> 00:04:07,154 is implied. And we'll, go through that, let's see if 47 00:04:07,154 --> 00:04:14,749 you can apply it now to a simple problem. Okay, I hope that elementary construction 48 00:04:14,749 --> 00:04:18,394 was clear to you. It's just a way of writing a vector and 49 00:04:18,394 --> 00:04:21,850 just summing over all the cartesian coordinates of the projection on to that 50 00:04:21,850 --> 00:04:25,576 coordinate times the unit vector on that coordinate, and you just use the Einstein 51 00:04:25,576 --> 00:04:33,680 summation. Okay the third tool that we'll be using 52 00:04:33,680 --> 00:04:41,340 is the Levi-Civita symbol. initially introduced by Tullio 53 00:04:41,340 --> 00:04:49,979 Levi-Civita and it's a function with three arguments indices i, j and k. 54 00:04:49,979 --> 00:04:56,410 I, J and K are, are, are taken from the values one, two and three. 55 00:04:56,410 --> 00:05:03,230 This is simply used for vector manipulation and has the following 56 00:05:03,230 --> 00:05:11,042 properties so when I, J, K is 123, 231 or 312 in that order the value of epsilon 57 00:05:11,042 --> 00:05:17,770 ijk is one. Those are the so called cyclic 58 00:05:17,770 --> 00:05:23,870 permutations, so you can see you can get from here to here just by wrapping. 59 00:05:23,870 --> 00:05:34,094 Wrapping, it's like having the the arguments on a ring let's say, and now by 60 00:05:34,094 --> 00:05:46,390 rotating the ring, you just take 1 to 2, 2 to 3, and 3 to 1. 61 00:05:46,390 --> 00:05:51,348 So this, cyclic for cyclic permutations of the arguments from the natural order 62 00:05:51,348 --> 00:05:56,084 the symbol takes the value plus 1 for odd permutations, which are just the 63 00:05:56,084 --> 00:06:00,746 permutations you get by transposing two of these and then executing cyclic 64 00:06:00,746 --> 00:06:08,233 permutation those. The symbol is minus 1 and otherwise it's 65 00:06:08,233 --> 00:06:13,822 0, and this is important in the sense that it, vanishes when, any two, any two 66 00:06:13,822 --> 00:06:22,740 of these indices are the same. So let's, try a simple example of 67 00:06:22,740 --> 00:06:29,840 applying this to describe the conventional vector cross product. 68 00:06:32,370 --> 00:06:36,792 Now, the way I phrased that previous example within, in the video quiz was to 69 00:06:36,792 --> 00:06:41,013 indicate the in factors, there's no fundamental significance to the use of 70 00:06:41,013 --> 00:06:46,819 any symbol for an index. Very often we have, a large number of, 71 00:06:46,819 --> 00:06:51,442 of, index symbols that need to be kept independent so we just keep throwing them 72 00:06:51,442 --> 00:06:57,072 in. So here is, important contraction 73 00:06:57,072 --> 00:07:05,061 identity for the Levi-Civita symbol. And again I'm writing this using Einstien 74 00:07:05,061 --> 00:07:13,650 summation convention. So this, this identity here implicitly 75 00:07:13,650 --> 00:07:21,560 refers to a summation over the repeated index I. 76 00:07:22,940 --> 00:07:28,088 So, it says it when you have a product of two Levi-Civita symbols, and you're 77 00:07:28,088 --> 00:07:33,440 selling over one index, then you get a contraction. 78 00:07:33,440 --> 00:07:39,537 And it's so if, again it has this, expresses the symmetry of the system, so 79 00:07:39,537 --> 00:07:49,386 this is, the, the value of this is delta js, delta kt minus delta jt, delta ks. 80 00:07:49,386 --> 00:07:55,734 where the delta functions the Kronecker delta which is 1 if the arguments agree 81 00:07:55,734 --> 00:08:02,622 and 0 otherwise. Now, you can actually work this out, it's 82 00:08:02,622 --> 00:08:07,800 not a deep identity you can just look at the several cases. 83 00:08:07,800 --> 00:08:12,216 And it might be helpful for you to pause here and just try a few cases for 84 00:08:12,216 --> 00:08:16,632 yourself and persuade, persuade yourself that this is true because it, it is 85 00:08:16,632 --> 00:08:22,494 something that's broadly useful. And it's a good idea for you to develop 86 00:08:22,494 --> 00:08:25,650 some intuition about how and when it can be used correctly. 87 00:08:28,630 --> 00:08:33,859 Okay, we're going to conclude this part with two examples of the use of this 88 00:08:33,859 --> 00:08:41,872 formalism from, with applied to the standard problems of vector calculus. 89 00:08:41,872 --> 00:08:48,144 So, let's first look at this, this product, A.B cross C, a tripe vector 90 00:08:48,144 --> 00:08:55,120 product. Okay, so how do we write this in the 91 00:08:55,120 --> 00:09:03,576 Einstein summation convention notation? Well this is just that, because we're 92 00:09:03,576 --> 00:09:07,894 now. We're, we're, we're going to go down to 93 00:09:07,894 --> 00:09:15,000 specific indexing of the, the various components and see what we get. 94 00:09:15,000 --> 00:09:19,698 So, this is, this is a scalar process. A sub i times the i'th component of the 95 00:09:19,698 --> 00:09:24,992 vector B cross C. Well, from the definition of the cross 96 00:09:24,992 --> 00:09:30,788 product that you, uh,uh, look, you, you reviewed in the, in the M video quiz, the 97 00:09:30,788 --> 00:09:36,940 i'th component of E cross C is epsilon ijkBjCk. 98 00:09:36,940 --> 00:09:42,682 So now, we can just bring the A inside and we have, you see we now have 99 00:09:42,682 --> 00:09:49,513 something that can be evaluated because we have implied summations over all three 100 00:09:49,513 --> 00:09:56,767 indicies. So this, this expresses this A.B cross C 101 00:09:56,767 --> 00:10:02,525 in a highly symmetric form, and now by, just. 102 00:10:02,525 --> 00:10:08,285 The, these identities here and here are trivial, they just involve cycling, doing 103 00:10:08,285 --> 00:10:16,120 cycling permutations on the indices here, and rearranging the order of a, b and c. 104 00:10:16,120 --> 00:10:20,740 Now here we're talking about classical vectors, so there's no issue computation. 105 00:10:20,740 --> 00:10:27,507 And so, what you can see is, that this original, expression A.B cross C, is the 106 00:10:27,507 --> 00:10:34,210 same as B.C cross A, and the same as C.A cross B. 107 00:10:34,210 --> 00:10:40,972 That's a, that's a an identity that I think is made quite transparent in this, 108 00:10:40,972 --> 00:10:48,940 in this particular notation. let's take another example with vector 109 00:10:48,940 --> 00:10:54,879 identity A cross B cross C. So this will give us, show us how 110 00:10:54,879 --> 00:11:02,010 contraction identity works and give us use of multiple Levi-Civita symbols. 111 00:11:02,010 --> 00:11:05,223 So, we start just by expanding in compliments, and this is the, the whole 112 00:11:05,223 --> 00:11:08,334 idea of this whole approach, is to take an expression, expand it out in the 113 00:11:08,334 --> 00:11:13,805 components, and then run contractions over all indices that are possible. 114 00:11:13,805 --> 00:11:18,868 So, we start by saying, okay, here's the cross product of, of, we put in the 115 00:11:18,868 --> 00:11:27,090 vector, A, epsilon ijk, epsilon iaj, then B cross C, the K complement of that. 116 00:11:28,200 --> 00:11:30,850 Now, what's the K complement of B cross C? 117 00:11:30,850 --> 00:11:36,874 Well we just throw in some more indices. That's epsilon KBSCT minus BTCS epsilon 118 00:11:36,874 --> 00:11:42,389 KSTBSCT. So this is now, expressed with the 119 00:11:42,389 --> 00:11:50,782 cross-product, and, let's see. Well, I've real, I've run a cyclic, 120 00:11:50,782 --> 00:11:58,250 permutation of the index ijk to get that. That's the same, the same thing. 121 00:11:58,250 --> 00:12:03,494 I'll, just pass this epsilon through to get a highly symmetric looking product 122 00:12:03,494 --> 00:12:07,855 here. So, we have the product of two 123 00:12:07,855 --> 00:12:14,950 Levi-Civita symbols with a repeat, one repeated in index k. 124 00:12:16,590 --> 00:12:22,510 So then we take this identity here, and perform the contraction. 125 00:12:22,510 --> 00:12:26,463 Now the notation's somewhat different, the, the, the specific symbols used for 126 00:12:26,463 --> 00:12:30,562 the indices, differ. But I think you can see that the whole 127 00:12:30,562 --> 00:12:34,531 point is, you take the common, the repeated index K and then the contraction 128 00:12:34,531 --> 00:12:38,815 is delta is delta jt minus delta js, delta it, that in other words, plus 1 for 129 00:12:38,815 --> 00:12:45,080 the same order of n to z minus 1 to the different order. 130 00:12:45,080 --> 00:13:01,360 So, now that gives us this expression. And let's see what do we have? 131 00:13:01,360 --> 00:13:07,870 Well we have a contraction on, we have a contraction on both indices so this gives 132 00:13:07,870 --> 00:13:14,880 us, for example, the delta is, the delta jt gives us eibi. 133 00:13:15,920 --> 00:13:23,812 So, that's that one there and then AjCj. And then, correspondingly, are it gives 134 00:13:23,812 --> 00:13:32,425 us, eiCi and js gives us Aj Bj. So, now we see that this here is the 135 00:13:32,425 --> 00:13:38,910 vector B and this here is the .product A.C and similarly. 136 00:13:38,910 --> 00:13:45,860 So we get this, we get this spectral identity out in a fairly transparent way. 137 00:13:45,860 --> 00:13:49,143 Okay, so I think it might be a good idea for you to review this and make sure that 138 00:13:49,143 --> 00:13:53,220 you're you're comfortable with the innervations. 139 00:13:53,220 --> 00:13:56,640 And we'll next, in the next part we'll start applying them to the angular 140 00:13:56,640 --> 00:13:58,507 momentum operator.