1 00:00:00,810 --> 00:00:03,525 PROFESSOR: All right, spin orbit. 2 00:00:07,810 --> 00:00:16,000 We have it still there, delta H spin orbit, orbit. 3 00:00:18,510 --> 00:00:20,280 Well, we know what V is. 4 00:00:20,280 --> 00:00:27,180 So that derivative, dv dr can be taken care of, the 1 over r. 5 00:00:27,180 --> 00:00:37,560 That gives you e squared over 2m c squared 1 over r cubed S, L. 6 00:00:37,560 --> 00:00:45,960 And you remember that S, L, from 805, J squared minus S 7 00:00:45,960 --> 00:00:47,880 squared minus L squared. 8 00:00:47,880 --> 00:00:51,990 The reason that's why you tend to do addition of angular 9 00:00:51,990 --> 00:00:59,470 momentum, because S and L, calculating the matrix 10 00:00:59,470 --> 00:01:02,140 elements of this thing is very easy, 11 00:01:02,140 --> 00:01:05,830 in that base is because every state there 12 00:01:05,830 --> 00:01:09,940 has a fixed value of S squared, a fixed value of L squared. 13 00:01:09,940 --> 00:01:14,750 And j squared plus a couple of possibilities. 14 00:01:14,750 --> 00:01:20,800 So we must work with the couple basis. 15 00:01:24,880 --> 00:01:25,810 Basis. 16 00:01:25,810 --> 00:01:34,500 And therefore, we can attempt to find E1 of a N, L, J, MJ, spin 17 00:01:34,500 --> 00:01:48,530 orbit, equal e squared over 2m squared c squared and LJ, MJ, 18 00:01:48,530 --> 00:01:52,160 S, L, R cubed. 19 00:01:52,160 --> 00:01:56,090 The R cube has to stay inside the expectation value, 20 00:01:56,090 --> 00:01:58,730 because the expectation value includes 21 00:01:58,730 --> 00:02:00,460 integration over space. 22 00:02:00,460 --> 00:02:03,290 So this is a very important. 23 00:02:03,290 --> 00:02:08,270 N, L, J, MJ. 24 00:02:08,270 --> 00:02:19,260 And again the useful question, the couple basis 25 00:02:19,260 --> 00:02:21,270 will have have degeneracies. 26 00:02:21,270 --> 00:02:24,120 All the states are degenerate there. 27 00:02:24,120 --> 00:02:30,540 So this time, we fixed n, because the degeneracies 28 00:02:30,540 --> 00:02:33,260 happen only when you fix n. 29 00:02:33,260 --> 00:02:36,740 So do we have the right to do this, 30 00:02:36,740 --> 00:02:40,250 to use the formula form perturbation, 31 00:02:40,250 --> 00:02:45,560 the non-degenerate perturbation theory to do this calculation? 32 00:02:45,560 --> 00:02:52,790 And the answer is yes, because the perturbation S.L over R 33 00:02:52,790 --> 00:03:05,570 cubed commutes with with L squared, with J squared, 34 00:03:05,570 --> 00:03:06,320 and with Jz. 35 00:03:09,760 --> 00:03:15,360 you need all that because you can have degeneracies 36 00:03:15,360 --> 00:03:18,280 by having different L values. 37 00:03:18,280 --> 00:03:21,330 And that would be taken care by this operator that 38 00:03:21,330 --> 00:03:24,610 has different eigenvalues when L is different. 39 00:03:24,610 --> 00:03:28,020 You can have the degeneracies involving different J values. 40 00:03:28,020 --> 00:03:31,220 This would be taken by this operator. 41 00:03:31,220 --> 00:03:33,120 And you can have degeneracies when 42 00:03:33,120 --> 00:03:38,160 m has different value so that involves the Jz operator. 43 00:03:38,160 --> 00:03:40,290 So you really need a perturbation 44 00:03:40,290 --> 00:03:44,340 that commutes with all of them. 45 00:03:44,340 --> 00:03:49,590 And why does it commute with all of them? 46 00:03:52,190 --> 00:03:54,090 You can see it in several ways. 47 00:03:54,090 --> 00:03:56,310 Let's do L squared. 48 00:03:56,310 --> 00:04:01,440 L squared is Casimir, it commutes with any LI. 49 00:04:01,440 --> 00:04:04,350 It doesn't even think about S because it doesn't 50 00:04:04,350 --> 00:04:06,660 know anything about S. So it commutes with S. 51 00:04:06,660 --> 00:04:11,640 So L squared with any LI and commutes with S. 52 00:04:11,640 --> 00:04:13,980 And L squared is an invariant, it 53 00:04:13,980 --> 00:04:18,540 commute with r squared because r squared 54 00:04:18,540 --> 00:04:20,100 is rotational invariant. 55 00:04:20,100 --> 00:04:22,740 So everything commute with that. 56 00:04:22,740 --> 00:04:25,440 In order to do the other ones, you 57 00:04:25,440 --> 00:04:35,530 can also think in terms of this matrix J squared over here 58 00:04:35,530 --> 00:04:39,160 and S squared and do all of them. 59 00:04:39,160 --> 00:04:44,430 You should do it and convince yourself that they all commute. 60 00:04:44,430 --> 00:04:45,870 So we can do this. 61 00:04:49,520 --> 00:04:52,040 If we can do it, it's good because then we 62 00:04:52,040 --> 00:04:55,700 can evaluate these quantities. 63 00:05:00,610 --> 00:05:05,306 So let's do a little of the the evaluation. 64 00:05:13,750 --> 00:05:23,740 So this E1nljmj is equal to. 65 00:05:23,740 --> 00:05:28,570 Let's evaluate the S dot L part by using 1/2 j squared 66 00:05:28,570 --> 00:05:30,560 s squared minus L squared. 67 00:05:30,560 --> 00:05:38,390 So that gives you a factor of h bar squared over 2, 68 00:05:38,390 --> 00:05:39,620 with this 2 over there. 69 00:05:39,620 --> 00:05:44,720 So you get e squared h squared over 4m 70 00:05:44,720 --> 00:05:51,230 squared c squared, J times J plus 1 minus l times l 71 00:05:51,230 --> 00:06:03,590 plus 1 minus 3/4, times nljmj, 1 over r cubed, mljmj. 72 00:06:14,090 --> 00:06:14,590 OK. 73 00:06:17,370 --> 00:06:20,310 That should be clear from the fact 74 00:06:20,310 --> 00:06:23,130 that you have a j squared minus s squared. 75 00:06:23,130 --> 00:06:27,090 That's a 3/4 spin is always 1/2, and l squared. 76 00:06:29,890 --> 00:06:33,940 This is known. 77 00:06:33,940 --> 00:06:43,500 It's equal, in fact, to nlml 1 over cubed nlml. 78 00:06:47,720 --> 00:06:51,140 Which is equal, let me discuss that again. 79 00:06:51,140 --> 00:06:59,140 It's a little-- a0l, l plus 1, l plus 1/2. 80 00:06:59,140 --> 00:06:59,960 OK. 81 00:06:59,960 --> 00:07:05,810 So this is a known result. It's one of those expectation values 82 00:07:05,810 --> 00:07:08,720 that you can get from Feynman, Hellman, 83 00:07:08,720 --> 00:07:12,500 or for other recursion relations. 84 00:07:12,500 --> 00:07:16,365 And this is always computed in the original uncoupled basis. 85 00:07:18,950 --> 00:07:22,575 But we seem to need it in the coupled basis. 86 00:07:22,575 --> 00:07:24,701 So again, are we in trouble? 87 00:07:24,701 --> 00:07:25,200 No. 88 00:07:25,200 --> 00:07:27,510 This is actually the same. 89 00:07:27,510 --> 00:07:30,780 And it is the same only because this answer 90 00:07:30,780 --> 00:07:33,240 doesn't depend on ml. 91 00:07:33,240 --> 00:07:37,050 Because this states involve various combinations of ml 92 00:07:37,050 --> 00:07:40,650 and ms. But doesn't depend on ml, 93 00:07:40,650 --> 00:07:42,610 so the answer is really the same. 94 00:07:42,610 --> 00:07:47,580 So these things are really the same. 95 00:07:47,580 --> 00:07:51,570 Happily, that simplifies our work. 96 00:07:51,570 --> 00:08:01,320 And now we have E1 nljm is equal to En 0-- 97 00:08:01,320 --> 00:08:06,140 this is yet another notation, this is the ground state energy 98 00:08:06,140 --> 00:08:06,760 here-- 99 00:08:06,760 --> 00:08:16,770 mc squared n, j, j plus 1 minus l, l plus 1 minus 3/4, 100 00:08:16,770 --> 00:08:23,790 over l, l plus 1/2, l plus 1. 101 00:08:23,790 --> 00:08:27,370 OK this is spin orbit.