1 00:00:03,130 --> 00:00:06,280 There are a certain number of key commutator identities that are useful for 2 00:00:06,280 --> 00:00:10,480 working out the properties of the quantum theory of angular momentum. 3 00:00:10,480 --> 00:00:15,163 And they can be derived using the methods that we introduced in Part 3. 4 00:00:15,163 --> 00:00:19,131 And I strongly encourage you to work these all out for yourself, but I'm going 5 00:00:19,131 --> 00:00:24,335 to go through and outline the steps. Work out a few examples, and then give 6 00:00:24,335 --> 00:00:31,300 you some pointers on how to proceed. Now, we'll keep using the standard 7 00:00:31,300 --> 00:00:35,380 definition of the angular momentum operator and then the Einstein summation 8 00:00:35,380 --> 00:00:41,585 convention to designate it. And in fact, everything that we were 9 00:00:41,585 --> 00:00:48,515 doing in this this part of the lecture derives from this one uncertainty 10 00:00:48,515 --> 00:00:58,720 principal commutator pj xk, this is h bar over i, delta jk. 11 00:00:58,720 --> 00:01:03,040 In fact, you can get all the results by going back to this fundamental form. 12 00:01:03,040 --> 00:01:07,115 Nothing else is needed. However, it's very helpful to derive 13 00:01:07,115 --> 00:01:12,980 some, to build up on this and derive some other fundamental identities, which will 14 00:01:12,980 --> 00:01:19,874 then save you a lot of algebra. So let's try a simple application of 15 00:01:19,874 --> 00:01:26,000 applying this identity to a slightly more complicated expression. 16 00:01:26,000 --> 00:01:30,837 That's an in video quiz. So I hope that was clear. 17 00:01:30,837 --> 00:01:36,465 in other words, by simple application of this, you can derive this and it just 18 00:01:36,465 --> 00:01:42,261 involves an, an explicit calculation of the commutator and then substitution of, 19 00:01:42,261 --> 00:01:50,900 of this, of this relationship whenever there is a product of x and p. 20 00:01:54,440 --> 00:01:59,603 Now, we're going to get the first of the really significant results. 21 00:01:59,603 --> 00:02:04,427 And again, everything precedes from the definition, and this identity, though 22 00:02:04,427 --> 00:02:09,035 one, you know, here, we'll find a case where the use of its this subsidiary 23 00:02:09,035 --> 00:02:17,200 relationship is also important. So our task is calculate the commutator 24 00:02:17,200 --> 00:02:21,680 of the one component of the angular momentum with another component, the 25 00:02:21,680 --> 00:02:27,560 position operator, okay? So to be explicit, this commutator, just 26 00:02:27,560 --> 00:02:32,540 remember, is just computed La xb is just La xb minus xb La. 27 00:02:32,540 --> 00:02:38,478 That's the definition. So now you see, we start using the 28 00:02:38,478 --> 00:02:45,930 specific representation of L in terms of x and p. 29 00:02:45,930 --> 00:02:53,825 So we have here that Lu is epsilon uvw, xp, pw. 30 00:02:53,825 --> 00:03:00,854 xv, pw, [INAUDIBLE] is La is a epsilon acd, cx, pd. 31 00:03:00,854 --> 00:03:09,130 And you see, we use c and d here, because b has already been reserved. 32 00:03:09,130 --> 00:03:12,610 One could use any choice of letters, it's, it's good to, to keep a set that 33 00:03:12,610 --> 00:03:16,400 are handy in the mind. By the way, I, I avoid using i in these 34 00:03:16,400 --> 00:03:19,572 expressions, because that's a conventional description for the square 35 00:03:19,572 --> 00:03:24,400 root of minus one, and one doesn't want to have any confusion about that. 36 00:03:25,730 --> 00:03:31,050 Anyway, okay, so, epsilon acd, xcpd computed with with xb. 37 00:03:31,050 --> 00:03:39,593 Well, then we just apply this and we get epsilon acd, h bar of i, xc delta bd. 38 00:03:39,593 --> 00:03:48,540 And now if you want to go through this yourself, but here's what happens. 39 00:03:48,540 --> 00:03:58,069 La xb is i h bar here, let's start here. La xb is i h bar epsilon abc, xc. 40 00:04:00,120 --> 00:04:04,034 Now, here's a, possibly a helpful mnemonic for you. 41 00:04:04,034 --> 00:04:08,935 You want to check this. You can rewrite this expression in the 42 00:04:08,935 --> 00:04:16,790 following symbolic form, 1 half L plus x minus x cross L is i h bar x. 43 00:04:17,820 --> 00:04:21,541 I consider this to be a really quantum expression, because of course, for 44 00:04:21,541 --> 00:04:25,628 ordinary vectors and classical mechanics, L cross x is necessary, necessarily 45 00:04:25,628 --> 00:04:32,218 perpendicular to x. So this expression says that for the 46 00:04:32,218 --> 00:04:37,310 angular momentum operator, when you apply the cross product to a vector, you get 47 00:04:37,310 --> 00:04:44,484 the vector back itself in a certain sense, with an imaginary prefactor. 48 00:04:44,484 --> 00:04:48,839 And you can think of an other way of looking at this some of you may have seen 49 00:04:48,839 --> 00:04:55,833 this before. And that is for example, if we take, L3 50 00:04:55,833 --> 00:05:08,199 applied to x1, 1, it gives it gives x2. And this is a, is a fact, the action of 51 00:05:08,199 --> 00:05:17,698 L3 rotates, you might say x1 and x2. And then, and in this sense this is way 52 00:05:17,698 --> 00:05:21,986 of displaying the fact that the angle momentum operator is a, is a generative 53 00:05:21,986 --> 00:05:27,435 rotation. So, it's my recommendation now it's my 54 00:05:27,435 --> 00:05:33,555 recommendation as your friend, it's, it's not mandatory that, this is a good time 55 00:05:33,555 --> 00:05:38,760 to pause. We're, the, we're concluding this part. 56 00:05:38,760 --> 00:05:42,980 I'll give you some time to do a few exercises. 57 00:05:42,980 --> 00:05:47,670 which basically, the, the, more and I suggest to do them in this order. 58 00:05:47,670 --> 00:05:54,300 So, this one is just a repetition of what we did for La xb, but now, applying it to 59 00:05:54,300 --> 00:06:02,012 the momentum operator rather than the position operator. 60 00:06:02,012 --> 00:06:09,220 And you should find that you get exactly the same, exactly the same identity if 61 00:06:09,220 --> 00:06:16,644 you exchange x and p in here. Furthermore you get the same identity if 62 00:06:16,644 --> 00:06:21,054 you, you, if you operate on L, so this gives, you know, even nicer expression 63 00:06:21,054 --> 00:06:27,114 with the cross prod of the angular momentum factor with itself. 64 00:06:27,114 --> 00:06:32,169 It's just the angular momentum operator again times an imaginary number. 65 00:06:34,170 --> 00:06:45,024 And finally this important identity which shows that all, any given, so L1, i 66 00:06:45,024 --> 00:06:57,160 squared equal L2, i squared equal L3, L squared equal zero. 67 00:06:57,160 --> 00:07:05,970 Each component of the angular momentum commutes with the sum of the squares. 68 00:07:05,970 --> 00:07:09,420 And this, this square angular momentum operator is something that enters into 69 00:07:09,420 --> 00:07:13,880 the equations of motion. That's actually going to be the subject 70 00:07:13,880 --> 00:07:17,802 of the next part. But meantime, I suggest that you just 71 00:07:17,802 --> 00:07:23,550 pause and work these things about, so that you have some experience with them. 72 00:07:23,550 --> 00:07:25,960 We'll be using them frequently in the next part.