1 00:00:00,000 --> 00:00:02,520 The following content is provided under a Creative 2 00:00:02,520 --> 00:00:03,970 Commons license. 3 00:00:03,970 --> 00:00:06,360 Your support will help MIT OpenCourseWare 4 00:00:06,360 --> 00:00:10,660 continue to offer high quality educational resources for free. 5 00:00:10,660 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,190 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,190 --> 00:00:18,370 at ocw.mit.edu. 8 00:00:24,282 --> 00:00:25,145 PROFESSOR: Question. 9 00:00:25,145 --> 00:00:26,770 We know how to do matching and running. 10 00:00:26,770 --> 00:00:28,560 We've seen an example of that. 11 00:00:28,560 --> 00:00:31,620 In this area, there's this label v on the fields, 12 00:00:31,620 --> 00:00:33,900 and we want to figure out how that affects 13 00:00:33,900 --> 00:00:35,428 doing matching and running. 14 00:00:35,428 --> 00:00:37,470 So we started out talking about the wave function 15 00:00:37,470 --> 00:00:41,550 for normalization calculation. 16 00:00:41,550 --> 00:00:43,560 So we have these fields, which we 17 00:00:43,560 --> 00:00:46,410 can have a bare version of and a renormalized version 18 00:00:46,410 --> 00:00:49,510 of in the usual way. 19 00:00:49,510 --> 00:00:52,560 So this is renormalized, and this is bare. 20 00:00:52,560 --> 00:00:57,120 Some Z factor between them that I'll call Zh for Z heavy. 21 00:00:57,120 --> 00:01:05,570 And so we have to calculate at the lowest order, this diagram, 22 00:01:05,570 --> 00:01:12,200 send in a momentum P. This is Q plus p.? 23 00:01:12,200 --> 00:01:13,220 Use the Feynman rules. 24 00:01:19,330 --> 00:01:21,356 Use dimensional regularization. 25 00:01:30,600 --> 00:01:34,066 Use dimensional regularization with MS bar, 26 00:01:34,066 --> 00:01:38,115 so there's some extra factors. 27 00:01:41,220 --> 00:01:44,700 And use Feynman gauge, which is usually the simplest gauge 28 00:01:44,700 --> 00:01:45,720 choice. 29 00:01:45,720 --> 00:01:48,300 So each of these vertices here gets 30 00:01:48,300 --> 00:01:53,790 a v. There's a v mu from this vertex. 31 00:01:53,790 --> 00:01:56,490 A v mu from that vertex. 32 00:01:56,490 --> 00:01:59,440 And that's where this v squared comes from. 33 00:01:59,440 --> 00:02:03,090 And then there's one propagator here. 34 00:02:03,090 --> 00:02:05,490 It's very traditional to denote a heavy quark 35 00:02:05,490 --> 00:02:07,020 by two lines rather than one line, 36 00:02:07,020 --> 00:02:08,395 so you know which lines are heavy 37 00:02:08,395 --> 00:02:09,660 and which lines are light. 38 00:02:09,660 --> 00:02:11,970 So these two lines is a heavy quark, 39 00:02:11,970 --> 00:02:16,350 and that gives this propagator here, this v dot q plus P. 40 00:02:16,350 --> 00:02:18,900 And then there's a relativistic propagator for the gluon, 41 00:02:18,900 --> 00:02:20,970 and that's the q squared. 42 00:02:20,970 --> 00:02:23,890 I've taken into account all the i's when I put this minus sign, 43 00:02:23,890 --> 00:02:26,040 and this is just the fundamental Casimir 44 00:02:26,040 --> 00:02:28,815 from dotting two TAs together. 45 00:02:31,620 --> 00:02:32,970 OK. 46 00:02:32,970 --> 00:02:35,530 So when you have an integral like this-- 47 00:02:35,530 --> 00:02:36,490 so v squared is 1. 48 00:02:36,490 --> 00:02:39,372 That's one simplification. 49 00:02:39,372 --> 00:02:40,830 When you have an integral like that 50 00:02:40,830 --> 00:02:43,920 where you have a linear momentum in one propagator 51 00:02:43,920 --> 00:02:47,160 and a quadratic momentum in the other propagator, 52 00:02:47,160 --> 00:02:49,290 you don't want to use the standard Feynman trick. 53 00:02:52,050 --> 00:02:53,690 I call this trick the George I trick. 54 00:02:56,520 --> 00:03:00,030 So it's very similar to the Feynman trick 55 00:03:00,030 --> 00:03:03,880 but slightly different. 56 00:03:03,880 --> 00:03:09,720 So you use an integral that goes from 0 to infinity, 57 00:03:09,720 --> 00:03:15,180 and you can convince yourself that this is true. 58 00:03:15,180 --> 00:03:21,900 And so then set a equal the q squared and b equal to v dot 59 00:03:21,900 --> 00:03:22,770 q plus p. 60 00:03:25,668 --> 00:03:27,960 And the reason that you'd want to use this trick rather 61 00:03:27,960 --> 00:03:30,865 than the usual one is that if you use the usual one, 62 00:03:30,865 --> 00:03:32,490 you'd get an x multiplying this and a 1 63 00:03:32,490 --> 00:03:34,320 minus x multiplying that. 64 00:03:34,320 --> 00:03:37,300 But then you would have an x multiplying the q squared. 65 00:03:37,300 --> 00:03:39,687 So when you complete this, it would 66 00:03:39,687 --> 00:03:41,520 be-- when you'd want to complete the square, 67 00:03:41,520 --> 00:03:43,240 you'd like nothing to multiply the q squared. 68 00:03:43,240 --> 00:03:45,323 You'd like the q squared just to be bare by itself 69 00:03:45,323 --> 00:03:47,380 with no Feynman parameter multiplying it. 70 00:03:47,380 --> 00:03:49,110 And this trick does that, because a 71 00:03:49,110 --> 00:03:52,650 has no Feynman parameter multiplying it, 72 00:03:52,650 --> 00:03:55,920 or George I parameter in this case. 73 00:03:55,920 --> 00:04:05,070 OK, so we combine denominators in the usual way, 74 00:04:05,070 --> 00:04:09,130 and the denominator would become this. 75 00:04:09,130 --> 00:04:13,260 And if I kept the i epsilon it would 76 00:04:13,260 --> 00:04:17,100 be that if we combine these two denominators here. 77 00:04:17,100 --> 00:04:19,920 So this factor would be that. 78 00:04:23,650 --> 00:04:26,310 And we can then complete the square, right. 79 00:04:26,310 --> 00:04:28,020 This is some momentum squared minus 80 00:04:28,020 --> 00:04:37,800 whatever is left over where t is q plus lambda v, 81 00:04:37,800 --> 00:04:44,790 and then A is the rest of the stuff, which is this. 82 00:04:49,905 --> 00:04:59,040 OK, so then this guy, now we just 83 00:04:59,040 --> 00:05:02,310 have our usual quadratic integral 84 00:05:02,310 --> 00:05:05,880 that we can use the standard rules to do. 85 00:05:16,945 --> 00:05:18,570 And then instead of an integral over x, 86 00:05:18,570 --> 00:05:19,987 we have this integral over lambda. 87 00:05:33,280 --> 00:05:34,905 There's some factors that I'm dropping. 88 00:05:40,380 --> 00:05:44,100 So there's that e to the Euler gamma times epsilon and the 4 89 00:05:44,100 --> 00:05:47,020 pi minus epsilon. 90 00:05:47,020 --> 00:05:50,850 So write in everything with d equals 4 minus 3 epsilon. 91 00:05:56,100 --> 00:06:01,950 Do that integral, which is just giving some gamma functions. 92 00:06:17,990 --> 00:06:20,290 And if you think about the dimensions 93 00:06:20,290 --> 00:06:23,092 here, so we end up with something that's d minus 3. 94 00:06:23,092 --> 00:06:25,300 If you think-- if you want to look at the dimensions, 95 00:06:25,300 --> 00:06:29,380 d over 2 minus 2 is dimensionless, OK. 96 00:06:29,380 --> 00:06:30,970 The lambda had dimensions. 97 00:06:30,970 --> 00:06:32,620 If we go back here, that's obvious. 98 00:06:32,620 --> 00:06:34,960 q is dimensionful, v is dimensionless, 99 00:06:34,960 --> 00:06:37,150 so lambda has dimension 1. 100 00:06:37,150 --> 00:06:38,925 So the dimensions of the-- 101 00:06:38,925 --> 00:06:43,390 actually, the dimensions of the mu to the 2 epsilon 102 00:06:43,390 --> 00:06:45,950 are compensating the dimensions here, 103 00:06:45,950 --> 00:06:48,130 and then there's one power of dimension left. 104 00:06:48,130 --> 00:06:49,960 And that's why if I take d equals 4, 105 00:06:49,960 --> 00:06:53,560 I'm getting one power of momentum upstairs, which 106 00:06:53,560 --> 00:06:55,630 is what we would expect for an inverse propagator 107 00:06:55,630 --> 00:06:59,770 for a heavy quark, is 1 factor of v dot P. 108 00:06:59,770 --> 00:07:01,030 So expand this. 109 00:07:12,438 --> 00:07:17,350 And we get a divergence. 110 00:07:17,350 --> 00:07:25,288 So add a counter term for wave function renormalization. 111 00:07:39,940 --> 00:07:43,873 So if we did that in MS bar, then the Zh would be just this. 112 00:07:43,873 --> 00:07:45,790 And that would be the appropriate counter term 113 00:07:45,790 --> 00:07:48,947 to kill off the 1 over epsilon divergence. 114 00:07:48,947 --> 00:07:50,530 I'm going to carry out the calculation 115 00:07:50,530 --> 00:07:53,800 today, or this discussion of matching in MS bar 116 00:07:53,800 --> 00:07:54,760 for everything. 117 00:07:54,760 --> 00:07:57,010 And I'll show you some of the slight complication that 118 00:07:57,010 --> 00:07:59,020 shows up in that case. 119 00:07:59,020 --> 00:08:03,920 But we said that we could do matching-- 120 00:08:03,920 --> 00:08:06,350 basically, you could have two choices here. 121 00:08:06,350 --> 00:08:08,440 You could either use on-shell renormalization, 122 00:08:08,440 --> 00:08:10,000 or you could use MS bar. 123 00:08:10,000 --> 00:08:11,590 What's the difference? 124 00:08:11,590 --> 00:08:14,260 In MS bar, you just keep the divergence in the z. 125 00:08:14,260 --> 00:08:17,950 In on-shell renormalization, you keep some extra terms here, 126 00:08:17,950 --> 00:08:19,270 all right. 127 00:08:19,270 --> 00:08:21,098 And either way that we choose to do things, 128 00:08:21,098 --> 00:08:23,140 we should actually end up with the same matching, 129 00:08:23,140 --> 00:08:26,210 and I want to show you why that's true. 130 00:08:26,210 --> 00:08:29,050 So in order to show you why that's true, 131 00:08:29,050 --> 00:08:31,360 I'll pick to use MS bar at this point, 132 00:08:31,360 --> 00:08:35,700 and we'll see what complication that leads to later on. 133 00:08:35,700 --> 00:08:38,620 OK, so that's one thing that needs to be renormalized. 134 00:08:38,620 --> 00:08:43,870 And if we look at this guy and we compare it to z psi, 135 00:08:43,870 --> 00:08:47,980 this is not the same as z psi in QCD. 136 00:08:47,980 --> 00:08:50,138 So this is something different, and the reason 137 00:08:50,138 --> 00:08:52,180 it's different is because we did a different loop 138 00:08:52,180 --> 00:08:55,150 integral that had this heavy quark propagator, not 139 00:08:55,150 --> 00:08:56,701 the light quark propagator. 140 00:09:02,830 --> 00:09:06,040 So we have to renormalize also local operators. 141 00:09:06,040 --> 00:09:08,200 And so let's think about something that would make 142 00:09:08,200 --> 00:09:09,790 a heavy to light transition. 143 00:09:09,790 --> 00:09:14,430 So for example, if you looked at b goes to u, 144 00:09:14,430 --> 00:09:20,680 electron neutrino, then that would 145 00:09:20,680 --> 00:09:22,010 be a heavy to light transition. 146 00:09:22,010 --> 00:09:22,870 To b quark is heavy. 147 00:09:22,870 --> 00:09:24,520 The b quark is light. 148 00:09:24,520 --> 00:09:26,050 So to describe that, you would use 149 00:09:26,050 --> 00:09:29,830 some operator that has one heavy quark and one light quark. 150 00:09:29,830 --> 00:09:34,690 So let me call the light quark small q and the heavy quark 151 00:09:34,690 --> 00:09:40,780 big Q. And so you'd have some operator looks like that. 152 00:09:40,780 --> 00:09:52,830 And we could write down a renormalized operator 153 00:09:52,830 --> 00:09:57,370 with renormalized fields and then group all the z factors 154 00:09:57,370 --> 00:09:59,040 into a counter term. 155 00:10:06,506 --> 00:10:09,140 So let's think about doing the perturbation theory that way. 156 00:10:09,140 --> 00:10:10,700 So this is the renormalized operator. 157 00:10:10,700 --> 00:10:11,742 This is the counter term. 158 00:10:16,580 --> 00:10:18,955 So there's a wave function factor Zq for the light quark 159 00:10:18,955 --> 00:10:21,070 and Zh for the heavy quark, and then there's 160 00:10:21,070 --> 00:10:24,381 some Zo for the operator renormalization. 161 00:10:34,210 --> 00:10:38,150 OK, so to renormalize this operator at one loop, 162 00:10:38,150 --> 00:10:42,440 we insert it, and we do a one-loop diagram. 163 00:10:46,370 --> 00:10:49,160 I'm just going to tell you the answers, 164 00:10:49,160 --> 00:10:51,526 but let's draw the diagram. 165 00:10:51,526 --> 00:10:53,080 So here's the operator inserted. 166 00:10:53,080 --> 00:10:54,080 Here's your heavy quark. 167 00:10:54,080 --> 00:10:56,000 Here's your light quark. 168 00:10:56,000 --> 00:10:57,230 You have a diagram like that. 169 00:11:03,880 --> 00:11:10,030 And then you combine this calculation with the wave 170 00:11:10,030 --> 00:11:16,150 function renormalize Zh and Zq. 171 00:11:16,150 --> 00:11:18,530 Zq is the same as Z psi. 172 00:11:18,530 --> 00:11:24,620 We should've called this Zq for the light quark. 173 00:11:24,620 --> 00:11:26,908 Combine these things together, and then, 174 00:11:26,908 --> 00:11:29,450 because that graph is telling you to count this counter term, 175 00:11:29,450 --> 00:11:31,950 you need Zo. 176 00:11:31,950 --> 00:11:37,820 And that calculation, which you can look at in your reading 177 00:11:37,820 --> 00:11:40,700 if you want to look at more details, 178 00:11:40,700 --> 00:11:43,145 just gives you something about what you'd expect. 179 00:11:43,145 --> 00:11:48,390 It gives you a 1 over epsilon divergence factor of g squared. 180 00:11:48,390 --> 00:11:53,150 So the operator here has renormalization that's 181 00:11:53,150 --> 00:11:59,700 just minus alpha s over pi. 182 00:11:59,700 --> 00:12:01,385 That's the anomalous dimension. 183 00:12:04,000 --> 00:12:07,335 So what is this anomalous dimension doing? 184 00:12:07,335 --> 00:12:11,850 If you were to consider this kind of process in full QCD, 185 00:12:11,850 --> 00:12:15,000 then you'd have here gamma mu 1 minus 5, 186 00:12:15,000 --> 00:12:17,550 and that's a partially conserved current. 187 00:12:17,550 --> 00:12:19,830 So there would be actually no anomalous dimension 188 00:12:19,830 --> 00:12:20,970 to this operator. 189 00:12:20,970 --> 00:12:23,880 The vertex graph that we just drew over 190 00:12:23,880 --> 00:12:26,250 there would cancel the wave function graphs exactly. 191 00:12:26,250 --> 00:12:28,290 There'd be no anomalous dimension. 192 00:12:28,290 --> 00:12:30,420 But here we have one. 193 00:12:30,420 --> 00:12:33,580 And that's because these guys are not equivalent anymore. 194 00:12:33,580 --> 00:12:36,810 You saw that the Z for the heavy quark changed, 195 00:12:36,810 --> 00:12:39,240 and the vertex graph also changes. 196 00:12:39,240 --> 00:12:41,520 And we're left with something. 197 00:12:41,520 --> 00:12:45,150 And this remainder has to do with renormalization group 198 00:12:45,150 --> 00:12:47,933 evolution below the mass of the heavy quark. 199 00:12:47,933 --> 00:12:49,350 Above the mass of the heavy quark, 200 00:12:49,350 --> 00:12:52,140 there's no renormalize group evolution of this current. 201 00:12:52,140 --> 00:12:54,000 Below the mass of the heavy quark, there is. 202 00:13:22,577 --> 00:13:24,160 So there's additional logs, and that's 203 00:13:24,160 --> 00:13:27,580 because MQ is now being treated as infinite. 204 00:13:27,580 --> 00:13:31,690 So things that, from the point of view of QCD, were logs of MQ 205 00:13:31,690 --> 00:13:34,000 have now become UV divergences, and that gives 206 00:13:34,000 --> 00:13:35,691 an extra anomalous dimension. 207 00:13:39,380 --> 00:13:42,320 One thing you can note about this anomalous dimension 208 00:13:42,320 --> 00:13:47,150 is that I didn't really specify for you what the gamma was. 209 00:13:47,150 --> 00:13:49,790 I told you for this process it would be gamma mu 1 minus gamma 210 00:13:49,790 --> 00:13:50,720 5. 211 00:13:50,720 --> 00:13:56,250 And the results here, actually, if you carry out 212 00:13:56,250 --> 00:13:58,380 this calculation with arbitrary gamma, 213 00:13:58,380 --> 00:14:04,680 you find that it's independent of gamma. 214 00:14:04,680 --> 00:14:08,310 So you get the same universal anomalous convention 215 00:14:08,310 --> 00:14:11,520 for any spin structure. 216 00:14:11,520 --> 00:14:13,800 And that's partly related to the things 217 00:14:13,800 --> 00:14:17,670 that we talked about last time with the spin symmetry of HQET, 218 00:14:17,670 --> 00:14:21,360 which is telling you that certain couplings are not 219 00:14:21,360 --> 00:14:22,890 sensitive to the spin. 220 00:14:22,890 --> 00:14:26,030 And effectively, in this diagram, 221 00:14:26,030 --> 00:14:27,740 you're getting a v slash here. 222 00:14:27,740 --> 00:14:30,000 Well, let me not try to go through it 223 00:14:30,000 --> 00:14:32,427 but leave it for looking at in your reading, 224 00:14:32,427 --> 00:14:34,760 but we'll talk a little bit more about this in a minute. 225 00:14:38,970 --> 00:14:41,160 I won't try to explain where that 226 00:14:41,160 --> 00:14:43,960 comes from from the diagram. 227 00:14:43,960 --> 00:14:47,350 So let's look at another case. 228 00:14:47,350 --> 00:14:49,470 The only real interesting thing that happened here 229 00:14:49,470 --> 00:14:51,750 was that we got-- well, the fact that the answer was 230 00:14:51,750 --> 00:14:54,060 non-0 was interesting, and the fact that it 231 00:14:54,060 --> 00:14:57,060 was independent of gamma. 232 00:14:57,060 --> 00:14:58,610 But the v didn't show up. 233 00:14:58,610 --> 00:15:01,110 And the reason that the v didn't show up in this calculation 234 00:15:01,110 --> 00:15:03,870 here is because there was only one v, 235 00:15:03,870 --> 00:15:06,120 and v squared was equal to 1. 236 00:15:06,120 --> 00:15:07,830 So v couldn't really show up because we 237 00:15:07,830 --> 00:15:12,740 had to get a scalar answer, and since v squared is equal to 1, 238 00:15:12,740 --> 00:15:15,160 just, it's not showing up. 239 00:15:15,160 --> 00:15:18,360 So something more interesting is to look at, instead 240 00:15:18,360 --> 00:15:21,230 of a heavy to light transition like that, 241 00:15:21,230 --> 00:15:22,980 a heavy to heavy transition. 242 00:15:28,326 --> 00:15:32,450 So we'll spend a little bit more time on this one. 243 00:15:32,450 --> 00:15:36,560 So let's have two heavy fields. 244 00:15:36,560 --> 00:15:38,540 And I'm going to take them in a current 245 00:15:38,540 --> 00:15:43,168 where they have different velocities, v and v prime. 246 00:15:43,168 --> 00:15:45,710 So let me imagine I went through this procedure of separating 247 00:15:45,710 --> 00:15:55,370 out counter term and renormalized the operator just 248 00:15:55,370 --> 00:15:56,300 like I did over there. 249 00:15:59,650 --> 00:16:02,000 So I have these two terms, two types of structures. 250 00:16:02,000 --> 00:16:04,650 Now I don't have a Zq, but I have two heavy quarks. 251 00:16:04,650 --> 00:16:09,740 So I have Zh, root Zh root Zh, which is just Zh. 252 00:16:09,740 --> 00:16:12,620 So an example of this would be something 253 00:16:12,620 --> 00:16:19,160 like B meson changing to, say, a D star meson electron 254 00:16:19,160 --> 00:16:22,890 and a neutrino, so having a charm quark replace our up 255 00:16:22,890 --> 00:16:23,390 quark. 256 00:16:28,790 --> 00:16:30,590 OK, so now the charm quark and the B quark 257 00:16:30,590 --> 00:16:32,270 could both be thought of as heavy. 258 00:16:32,270 --> 00:16:34,880 They have different masses, but we take both of those masses 259 00:16:34,880 --> 00:16:37,310 to infinity, so we can use HQET for both of them. 260 00:16:42,210 --> 00:16:45,320 So Mb and Mc are going to infinity. 261 00:16:45,320 --> 00:16:49,190 And there's no reason why, in this process, 262 00:16:49,190 --> 00:16:51,290 that the b quark and the charm quarks 263 00:16:51,290 --> 00:16:54,143 should have the same velocity, and so we'll 264 00:16:54,143 --> 00:16:56,060 give them different velocities, v and v prime. 265 00:16:59,230 --> 00:17:02,800 OK, so we can go through the same thing. 266 00:17:02,800 --> 00:17:04,630 We already calculated Zh, so we just 267 00:17:04,630 --> 00:17:11,230 have to calculate a graph like this or two, 268 00:17:11,230 --> 00:17:14,710 where you have two heavy quarks. 269 00:17:14,710 --> 00:17:17,109 But the heavy quarks have different velocities. 270 00:17:21,222 --> 00:17:22,930 So what would that calculation look like? 271 00:17:27,619 --> 00:17:34,970 Again in Feynman gauge, and let me just take 272 00:17:34,970 --> 00:17:39,290 the external quarks to have zero momentum for simplicity, 273 00:17:39,290 --> 00:17:42,510 zero residual momentum. 274 00:17:42,510 --> 00:17:47,900 So this guy has k equals 0 and k equals 0. 275 00:17:47,900 --> 00:17:50,480 So the HQET Feynman rule for this guy 276 00:17:50,480 --> 00:17:54,380 has a v dot q if q is the loop momenta, 277 00:17:54,380 --> 00:17:57,890 and this guy has a v prime dot q, OK. 278 00:17:57,890 --> 00:18:01,550 So the integral that have to do is that integral. 279 00:18:01,550 --> 00:18:04,880 And you can do this again with one of these tricks where 280 00:18:04,880 --> 00:18:07,100 you use lambda parameters. 281 00:18:07,100 --> 00:18:10,220 And then one of the handouts on the web page, 282 00:18:10,220 --> 00:18:12,080 I've given you the appropriate trick 283 00:18:12,080 --> 00:18:14,090 that's two lambda parameters for this integral. 284 00:18:19,430 --> 00:18:21,710 This integral actually has both ultraviolet 285 00:18:21,710 --> 00:18:27,340 and infrared divergences. 286 00:18:27,340 --> 00:18:28,780 We're here in our discussion only 287 00:18:28,780 --> 00:18:33,235 interested in the ultraviolet ones, 288 00:18:33,235 --> 00:18:35,860 because we're worrying about the anomalous dimension right now. 289 00:18:39,760 --> 00:18:41,260 And again, I'm not going to drag you 290 00:18:41,260 --> 00:18:43,440 through the details of this calculation. 291 00:18:46,300 --> 00:18:48,910 It's essentially just, do the Feynman parameter, 292 00:18:48,910 --> 00:18:52,840 combine denominators with two Feynman parameters, 293 00:18:52,840 --> 00:18:55,240 complete the square, do the loop integral, 294 00:18:55,240 --> 00:18:59,290 do the Feynman parameter integrals, get an answer. 295 00:18:59,290 --> 00:19:01,840 Combine it together with the counter terms of the wave 296 00:19:01,840 --> 00:19:05,470 function renormalize, and see what type of counter term 297 00:19:05,470 --> 00:19:07,750 you get left over. 298 00:19:07,750 --> 00:19:14,560 And it's more interesting than the heavy to light case. 299 00:19:14,560 --> 00:19:29,400 So here's what it looks like, where w is v dot v prime, 300 00:19:29,400 --> 00:19:33,660 and this function, r of w, is the following. 301 00:19:33,660 --> 00:19:43,840 It's log w plus square root w squared minus 1 302 00:19:43,840 --> 00:19:45,940 over the square root of w squared minus 1. 303 00:19:52,080 --> 00:19:57,660 OK, so it's a non-trivial structure. 304 00:19:57,660 --> 00:20:00,870 So that counter term would lead to an anomalous dimension, 305 00:20:00,870 --> 00:20:02,670 which depends on this r of w. 306 00:20:13,320 --> 00:20:15,090 So the reason that that can happen 307 00:20:15,090 --> 00:20:18,960 is because v squared is 1, v prime squared is 1, 308 00:20:18,960 --> 00:20:20,880 but v dot v prime need not be 1. 309 00:20:24,220 --> 00:20:29,350 So v squared was 1, v prime squared was 1, 310 00:20:29,350 --> 00:20:32,755 but v dot v prime is not fixed. 311 00:20:36,890 --> 00:20:40,121 Well, it's not fixed simply by-- 312 00:20:40,121 --> 00:20:41,620 that's a poor choice of words. 313 00:20:45,763 --> 00:20:47,180 This is a parameter that can vary. 314 00:20:50,402 --> 00:20:52,340 So it will be fixed by kinematics, 315 00:20:52,340 --> 00:20:59,570 but it can depend on the kinematics. 316 00:20:59,570 --> 00:21:02,720 So let me go through this and organize it 317 00:21:02,720 --> 00:21:04,430 as a bunch of notes, comments. 318 00:21:07,140 --> 00:21:13,220 So this is what I just said. 319 00:21:13,220 --> 00:21:14,837 The answer depends on v dot v prime. 320 00:21:14,837 --> 00:21:16,670 And the way that you should think about this 321 00:21:16,670 --> 00:21:20,590 is that you have a current in the effective theory. 322 00:21:20,590 --> 00:21:23,870 So this is in the HQET current. 323 00:21:28,670 --> 00:21:33,320 And just like we label it by its Lorentz index mu, 324 00:21:33,320 --> 00:21:35,570 we should label it also by the v and v prime, 325 00:21:35,570 --> 00:21:38,940 because the fields involve v and v prime. 326 00:21:38,940 --> 00:21:43,610 So if we thought of this as some vector current or axial vector 327 00:21:43,610 --> 00:21:48,650 current labeled by mu, we would also label the current 328 00:21:48,650 --> 00:21:49,640 by v and v prime. 329 00:21:52,500 --> 00:21:55,082 So these are indices that are just indices on the current, 330 00:21:55,082 --> 00:21:56,540 and when you think about the Wilson 331 00:21:56,540 --> 00:21:57,980 coefficient or the anomalous dimension, 332 00:21:57,980 --> 00:21:59,272 it can depend on those indices. 333 00:22:06,770 --> 00:22:12,050 Now because the Wilson coefficient is a scalar, 334 00:22:12,050 --> 00:22:15,470 it really only depends on w. 335 00:22:15,470 --> 00:22:18,560 So if you think about Wilson coefficients, 336 00:22:18,560 --> 00:22:19,700 they depend on alpha s. 337 00:22:19,700 --> 00:22:22,730 They depend on scale mu. 338 00:22:22,730 --> 00:22:26,540 They depend on the hard scales in your problem, 339 00:22:26,540 --> 00:22:28,010 and the hard scales in the problem 340 00:22:28,010 --> 00:22:32,960 here for us are Mb times v and Mc times v prime, 341 00:22:32,960 --> 00:22:35,750 because those are the heavy scales of the heavy charm 342 00:22:35,750 --> 00:22:38,810 quark and the heavy b quark that we're removing. 343 00:22:38,810 --> 00:22:40,940 But since we know that this thing is a scalar, 344 00:22:40,940 --> 00:22:43,760 we can just square these, dot them into each other. 345 00:22:43,760 --> 00:22:50,990 So it's really a function of mu of alpha s mu Mb squared, 346 00:22:50,990 --> 00:22:54,890 from squaring the Mb term, M charm squared, and then 347 00:22:54,890 --> 00:22:55,865 this w factor. 348 00:22:59,088 --> 00:23:01,130 Those are the scalar quantities it can depend on. 349 00:23:11,580 --> 00:23:15,180 So what should I take for the w? 350 00:23:15,180 --> 00:23:17,340 Well, we could work that out for an example 351 00:23:17,340 --> 00:23:24,058 like the one I was saying that we're doing, B to D star L nu. 352 00:23:24,058 --> 00:23:25,600 So I want to show you how that works. 353 00:23:39,710 --> 00:23:43,720 So for the B meson, for momentum, 354 00:23:43,720 --> 00:23:48,280 we can take it to be M capital B meson times v mu. 355 00:23:48,280 --> 00:23:50,770 And by momentum conservation, that's 356 00:23:50,770 --> 00:23:53,680 going to be the D star momentum, which 357 00:23:53,680 --> 00:23:58,480 we can take to be MD Star times v prime mu 358 00:23:58,480 --> 00:24:02,260 then some momentum transfer, which I call q. 359 00:24:02,260 --> 00:24:05,170 This is a different q than my loop momentum q. 360 00:24:05,170 --> 00:24:06,040 Sorry about that. 361 00:24:10,998 --> 00:24:13,540 So then we can just take this relation, and we can square it, 362 00:24:13,540 --> 00:24:14,950 and we can get v dot v primes. 363 00:24:14,950 --> 00:24:36,660 So if you look at 2 squared, we can solve this for v dot v 364 00:24:36,660 --> 00:24:38,220 prime, and it's just fixed. 365 00:24:38,220 --> 00:24:40,770 You see that v dot v prime is fixed in terms of the meson 366 00:24:40,770 --> 00:24:43,380 masses and the momentum transfer, 367 00:24:43,380 --> 00:24:46,710 and that's the momentum transfer to the leptons here. 368 00:24:46,710 --> 00:24:48,665 How much momentum do they carry away? 369 00:24:48,665 --> 00:24:50,790 So you could think of all those things as external. 370 00:24:50,790 --> 00:24:53,940 You fixed how much momentum the leptons carry away. 371 00:24:53,940 --> 00:24:54,990 These are fixed numbers. 372 00:24:54,990 --> 00:24:57,960 You look them up in the PDG, and then what value of v 373 00:24:57,960 --> 00:25:00,390 dot v prime to plug-in here. 374 00:25:00,390 --> 00:25:01,770 Now it's a function of q squared. 375 00:25:01,770 --> 00:25:04,570 Q squared can vary in the process. 376 00:25:04,570 --> 00:25:08,430 So in that sense, it's a non-trivial function, not just 377 00:25:08,430 --> 00:25:10,230 a fixed number. 378 00:25:10,230 --> 00:25:12,750 But for any fixed kinematic configuration, 379 00:25:12,750 --> 00:25:17,380 any fixed q, then it would just be a number. 380 00:25:17,380 --> 00:25:19,230 So if you look at this in practice, 381 00:25:19,230 --> 00:25:23,070 you find that this guy for this particular case 382 00:25:23,070 --> 00:25:24,585 goes from 1 to 1.5. 383 00:25:24,585 --> 00:25:26,460 So that's the kinematic range that's allowed. 384 00:25:47,370 --> 00:25:52,020 So it is fixed by external kinematics, kinematics 385 00:25:52,020 --> 00:25:54,780 that is external to the dynamics inside the loop. 386 00:25:59,072 --> 00:26:00,780 And then that way, the Wilson coefficient 387 00:26:00,780 --> 00:26:04,150 here is more non-trivial than the ones we saw earlier, 388 00:26:04,150 --> 00:26:05,442 which just depended on masses. 389 00:26:05,442 --> 00:26:07,650 Now it's depending on masses as well as this function 390 00:26:07,650 --> 00:26:10,590 of v dot v prime, OK? 391 00:26:17,170 --> 00:26:21,220 Again, one finds that gamma T comes out 392 00:26:21,220 --> 00:26:24,580 to be exactly the same, independent of the choice 393 00:26:24,580 --> 00:26:27,850 of the spin structure. 394 00:26:27,850 --> 00:26:30,520 So we could do this calculation with any spin structure 395 00:26:30,520 --> 00:26:33,880 we like, and heavy quark symmetry in this case 396 00:26:33,880 --> 00:26:40,030 is all it takes to show that this gamma T is independent 397 00:26:40,030 --> 00:26:42,872 of the spin structure. 398 00:26:42,872 --> 00:26:44,830 So if you think about that from the loop graph, 399 00:26:44,830 --> 00:26:48,190 actually in this case, it's pretty easy to see, 400 00:26:48,190 --> 00:26:51,910 because remember that this vertex didn't have any spin 401 00:26:51,910 --> 00:26:52,420 structure. 402 00:26:52,420 --> 00:26:53,980 The propagator had no spin structure. 403 00:26:53,980 --> 00:26:56,478 So nothing in the calculation had spin structure. 404 00:26:56,478 --> 00:26:58,270 So the only thing that's had spin structure 405 00:26:58,270 --> 00:27:00,500 is the gamma you stuck in there. 406 00:27:00,500 --> 00:27:02,500 So it's just a scalar times gamma, 407 00:27:02,500 --> 00:27:04,660 so it doesn't care about the gamma. 408 00:27:04,660 --> 00:27:07,270 On whatever the tree level gamma is, it just goes through. 409 00:27:07,270 --> 00:27:10,960 It's not touched by heavy quark, by the HQET Lagrangian. 410 00:27:15,100 --> 00:27:17,530 So what is physically going on here, 411 00:27:17,530 --> 00:27:20,260 and what is HQET doing for you is 412 00:27:20,260 --> 00:27:25,420 that there's logs like this, MQ over lambda QCD and QCD. 413 00:27:25,420 --> 00:27:32,470 And by going over to HQET, this becomes a log 414 00:27:32,470 --> 00:27:37,810 of mu over lambda QCD, which is encoded 415 00:27:37,810 --> 00:27:43,990 in HQET operators like this current, 416 00:27:43,990 --> 00:27:49,750 as well as a log of mu over MQ or MQ 417 00:27:49,750 --> 00:27:56,410 over mu, which is in the HQET coefficient functions, HQET 418 00:27:56,410 --> 00:27:58,220 Wilson coefficients. 419 00:27:58,220 --> 00:28:00,400 So just how we-- much exactly the same 420 00:28:00,400 --> 00:28:02,650 as how we talked about it for integrating out 421 00:28:02,650 --> 00:28:06,520 a heavy particle, the logs get split up into pieces, 422 00:28:06,520 --> 00:28:08,950 and the Wilson coefficient into pieces, and the matrix 423 00:28:08,950 --> 00:28:11,800 elements, operators. 424 00:28:11,800 --> 00:28:15,220 Here we're separating MQ heavy quark mass, 425 00:28:15,220 --> 00:28:19,540 and it's both the charm and the bottom in the case of what 426 00:28:19,540 --> 00:28:22,370 we're talking about. 427 00:28:22,370 --> 00:28:25,370 And the anomalous dimension that we calculated sums up those 428 00:28:25,370 --> 00:28:30,645 logs, and summing up those logs involves a non-trivial function 429 00:28:30,645 --> 00:28:31,145 of this w. 430 00:28:44,840 --> 00:28:47,240 But we actually know exactly the non-trivial function, 431 00:28:47,240 --> 00:28:50,650 because we can calculate it. 432 00:28:50,650 --> 00:28:53,600 And it's just this guy here. 433 00:29:00,450 --> 00:29:03,540 OK, so the new wrinkle that can come in HQET 434 00:29:03,540 --> 00:29:06,210 is that the anomalous dimension can become a more 435 00:29:06,210 --> 00:29:07,040 non-trivial thing. 436 00:29:14,000 --> 00:29:17,260 So if you look at it at leading log order, 437 00:29:17,260 --> 00:29:19,250 the rest is pretty straightforward. 438 00:29:19,250 --> 00:29:23,698 So if you go through the leading log result, 439 00:29:23,698 --> 00:29:25,990 you would do the same type of thing that we did before. 440 00:29:25,990 --> 00:29:29,980 You would match it, some scale. 441 00:29:29,980 --> 00:29:32,230 And at that scale, you could normalize things 442 00:29:32,230 --> 00:29:37,420 so that the Wilson coefficient at mu equals MQ is just 1 443 00:29:37,420 --> 00:29:39,640 at tree level. 444 00:29:39,640 --> 00:29:43,600 And then the leading log result is the function 445 00:29:43,600 --> 00:29:47,980 of these various things, which in general is C of MQ 446 00:29:47,980 --> 00:29:52,240 times some evolution from MQ to mu, 447 00:29:52,240 --> 00:29:55,540 suppressing some of the dependencies. 448 00:29:55,540 --> 00:29:59,140 The leading log result is 1 for this. 449 00:29:59,140 --> 00:30:00,820 And then this guy at the lowest order 450 00:30:00,820 --> 00:30:02,500 is just a ratio of alphas again. 451 00:30:07,730 --> 00:30:10,810 And then there's the gamma. 452 00:30:10,810 --> 00:30:12,900 For the purpose of solving the RGE is just, 453 00:30:12,900 --> 00:30:14,650 the only thing that matters is it's alpha. 454 00:30:21,270 --> 00:30:26,450 So a gamma's a constant for heavy to light, just a number. 455 00:30:30,120 --> 00:30:33,300 So for that current, it was a constant. 456 00:30:33,300 --> 00:30:36,760 This solution is actually valid for both of them. 457 00:30:36,760 --> 00:30:47,190 And then it's a function of this w for the current, where 458 00:30:47,190 --> 00:30:48,720 we have two heavy quarks. 459 00:30:48,720 --> 00:30:50,640 So the w dependence just goes along 460 00:30:50,640 --> 00:30:52,973 for the ride when you're solving the anomalous dimension 461 00:30:52,973 --> 00:30:54,130 equation. 462 00:30:54,130 --> 00:30:54,630 OK? 463 00:30:54,630 --> 00:30:58,920 So that's what re-summing the logs would look like. 464 00:30:58,920 --> 00:31:01,890 Essentially, each log is getting extra powers 465 00:31:01,890 --> 00:31:03,330 of this factor of gamma. 466 00:31:06,697 --> 00:31:11,360 So with number four, is there any questions about that? 467 00:31:14,776 --> 00:31:17,009 OK, pretty straightforward. 468 00:31:21,320 --> 00:31:23,690 So much of this, essentially all of the story, 469 00:31:23,690 --> 00:31:26,590 except for this one wrinkle, is very similar to integrating out 470 00:31:26,590 --> 00:31:28,970 a massive particle. 471 00:31:28,970 --> 00:31:33,130 And the other part of the story that's similar 472 00:31:33,130 --> 00:31:37,060 is that the HQET matrix elements depend on mu as well. 473 00:31:46,090 --> 00:31:49,050 And so in our example that we talked about last time, 474 00:31:49,050 --> 00:31:50,350 we had a matrix element. 475 00:31:50,350 --> 00:31:55,810 So let me give it to you in the context of that example. 476 00:31:55,810 --> 00:31:57,750 So last time we were talking about something 477 00:31:57,750 --> 00:32:02,708 which was a decay constant, and that's one example 478 00:32:02,708 --> 00:32:04,000 of this heavy to light current. 479 00:32:07,970 --> 00:32:10,310 So we had our current which had one heavy quark, one 480 00:32:10,310 --> 00:32:13,280 light quark, and then a heavy meson. 481 00:32:13,280 --> 00:32:18,920 And we figured out that this was giving some a times v mu, 482 00:32:18,920 --> 00:32:21,380 and now I'm telling you that you should think of the a 483 00:32:21,380 --> 00:32:23,330 as a function of mu. 484 00:32:23,330 --> 00:32:25,190 The matrix element here is a function of mu. 485 00:32:30,415 --> 00:32:33,440 OK, so that's just a slight modification to what 486 00:32:33,440 --> 00:32:36,870 we talked about last time. 487 00:32:36,870 --> 00:32:42,470 And again, if you want, for this matrix element, 488 00:32:42,470 --> 00:32:47,690 you'd want mu to be, say, a GeV or some scale that's 489 00:32:47,690 --> 00:32:51,788 greater than lambda QCD, and so what you would do 490 00:32:51,788 --> 00:32:53,330 is you would evaluate matrix elements 491 00:32:53,330 --> 00:32:56,180 and define that parameter at some scale like a GeV 492 00:32:56,180 --> 00:32:59,360 and do renormalization group evolution from the heavy quark 493 00:32:59,360 --> 00:33:02,480 mass down to a GeV. 494 00:33:02,480 --> 00:33:03,020 OK? 495 00:33:03,020 --> 00:33:12,250 So that's how the renormalization group evolution 496 00:33:12,250 --> 00:33:13,030 story would be. 497 00:33:16,292 --> 00:33:18,500 You don't want to run all the way down to lambda QCD, 498 00:33:18,500 --> 00:33:21,500 because the anomalous dimension has to remain perturbative. 499 00:33:21,500 --> 00:33:24,715 So you would decide what your cutoff 500 00:33:24,715 --> 00:33:26,840 is for where you think perturbation theory is still 501 00:33:26,840 --> 00:33:27,950 valid. 502 00:33:27,950 --> 00:33:32,270 Often people pick something like 1 GeV or 1 1/2 GeV for these 503 00:33:32,270 --> 00:33:33,050 types of problems. 504 00:33:44,910 --> 00:33:47,810 And again, this is just separating out all the MQs, 505 00:33:47,810 --> 00:33:51,080 making sure that your matrix element here has no MQs, 506 00:33:51,080 --> 00:33:54,230 it does have an extra cutoff mu. 507 00:33:54,230 --> 00:33:55,850 OK? 508 00:33:55,850 --> 00:33:58,580 So that's the RGE story. 509 00:33:58,580 --> 00:34:01,010 Let's also talk a little bit about the matching story. 510 00:34:05,210 --> 00:34:13,000 So these are the perturbative corrections at the scale MQ, 511 00:34:13,000 --> 00:34:16,239 or alpha s at MQ. 512 00:34:16,239 --> 00:34:18,172 We had perturbative corrections at the w scale 513 00:34:18,172 --> 00:34:19,630 when we integrated out the w boson. 514 00:34:19,630 --> 00:34:21,310 Now we have another set of perturbative corrections 515 00:34:21,310 --> 00:34:23,409 at the heavy quark mass scale when we integrated 516 00:34:23,409 --> 00:34:25,000 out the heavy quark mass. 517 00:34:25,000 --> 00:34:27,670 It's different because we're now passing from something 518 00:34:27,670 --> 00:34:29,679 that looked like a full QCD theory 519 00:34:29,679 --> 00:34:31,330 with some external operators. 520 00:34:31,330 --> 00:34:34,120 We're now passing to this HQET theory for the heavy quark. 521 00:34:40,159 --> 00:34:43,940 So if you like, we previously had Mw. 522 00:34:43,940 --> 00:34:45,710 We knew how to do renormalization group 523 00:34:45,710 --> 00:34:50,929 evolution there, and we had a Hw theory here. 524 00:34:50,929 --> 00:34:53,630 Now we integrate out the heavy quark mass, 525 00:34:53,630 --> 00:34:57,620 and we pass to an HQET theory below that scale. 526 00:35:03,065 --> 00:35:09,870 So if you go back to the Hw theory, if we can call it that, 527 00:35:09,870 --> 00:35:13,140 and we want to match that onto HQET, 528 00:35:13,140 --> 00:35:17,230 then we do it with a calculation like this. 529 00:35:17,230 --> 00:35:20,400 And I'll just use this heavy, light example still. 530 00:35:25,482 --> 00:35:27,190 So here's a matrix element that you could 531 00:35:27,190 --> 00:35:31,110 consider for the matching. 532 00:35:31,110 --> 00:35:33,328 And let me write it with a bunch of schematic objects 533 00:35:33,328 --> 00:35:34,620 and then explain what they are. 534 00:35:55,460 --> 00:35:57,480 So we use our spinners. 535 00:35:57,480 --> 00:36:01,980 I'm taking zero momentum here, just for simplicity. 536 00:36:05,010 --> 00:36:08,430 These Rs are residue factors that come in. 537 00:36:08,430 --> 00:36:14,550 So so far when we calculated the wave function renormalization 538 00:36:14,550 --> 00:36:19,050 graphs, we just took the divergent piece. 539 00:36:19,050 --> 00:36:21,960 And if you do that when you do the matching computation, 540 00:36:21,960 --> 00:36:23,580 you have to use LSZ. 541 00:36:23,580 --> 00:36:25,770 And so the finite pieces come back in, 542 00:36:25,770 --> 00:36:28,020 and you call them residues. 543 00:36:28,020 --> 00:36:37,460 So these are finite residues that you have to take account 544 00:36:37,460 --> 00:36:40,670 of, finite in the UV sense. 545 00:36:40,670 --> 00:36:45,780 So UV finite residues you have to take into account if you're 546 00:36:45,780 --> 00:36:48,390 just using MS bar, and this here would 547 00:36:48,390 --> 00:36:51,150 be the vertex renormalization graph. 548 00:36:54,782 --> 00:36:57,340 OK, both diagrams look like this. 549 00:36:57,340 --> 00:37:03,590 In QCD, they're both the same type of structure. 550 00:37:03,590 --> 00:37:07,390 And then in HQET, it's a similar thing. 551 00:37:07,390 --> 00:37:09,310 We can write down a formula for the s matrix 552 00:37:09,310 --> 00:37:14,600 element, the same states, now with our effective theory 553 00:37:14,600 --> 00:37:15,100 current. 554 00:37:17,830 --> 00:37:22,000 And we know how to transition from effective theory 555 00:37:22,000 --> 00:37:26,350 and full theory states. 556 00:37:26,350 --> 00:37:29,130 We talked about that last time. 557 00:37:29,130 --> 00:37:32,240 And so there again would be some residue factors. 558 00:37:32,240 --> 00:37:33,682 And if the residue factors are not 559 00:37:33,682 --> 00:37:36,140 the same in the two theories, you have to account for that. 560 00:37:36,140 --> 00:37:38,890 And they won't be, because one of the heavy quark residue 561 00:37:38,890 --> 00:37:41,200 will be different than the heavy quark residue here. 562 00:38:02,260 --> 00:38:07,720 So this guy here is this finite piece of this graph. 563 00:38:07,720 --> 00:38:10,460 This guy here will be the same as above. 564 00:38:10,460 --> 00:38:19,240 And this guy here is the heavy quark vertex 565 00:38:19,240 --> 00:38:22,270 graph, which is independent of the spin structure. 566 00:38:22,270 --> 00:38:24,640 It's not independent of the spin structure up there. 567 00:38:34,713 --> 00:38:36,880 So we could carry out the calculations of those loop 568 00:38:36,880 --> 00:38:39,403 diagrams, and then we could subtract, 569 00:38:39,403 --> 00:38:40,820 and we could see what's left over. 570 00:38:40,820 --> 00:38:44,290 And whatever's left over is the Wilson coefficient. 571 00:38:44,290 --> 00:38:47,320 OK, so very similar to what we did before. 572 00:38:50,400 --> 00:38:53,430 Calculate, subtract. 573 00:38:59,410 --> 00:39:02,140 What you find when you do that is that there's actually 574 00:39:02,140 --> 00:39:02,950 two currents. 575 00:39:02,950 --> 00:39:11,430 If you consider a vector current where gamma is gamma mu, 576 00:39:11,430 --> 00:39:15,600 then the effective field theory, which is HQET, 577 00:39:15,600 --> 00:39:21,450 has two effective theory currents that are vector. 578 00:39:24,140 --> 00:39:27,145 So you have C1 and C2. 579 00:39:32,278 --> 00:39:34,820 The reason that there's two is because we have another vector 580 00:39:34,820 --> 00:39:37,670 to play with, which is v mu. 581 00:39:37,670 --> 00:39:43,140 So that can have q bar v mu replacing the gamma mu. 582 00:39:46,420 --> 00:39:48,880 So v mu wasn't an external thing in QCD. 583 00:39:48,880 --> 00:39:50,920 It was part of the dynamics. 584 00:39:50,920 --> 00:39:52,420 Here it's an external thing, so it's 585 00:39:52,420 --> 00:39:55,900 allowed to replace gamma mu as one of the structures. 586 00:39:58,430 --> 00:40:00,253 And if you go through the calculation, 587 00:40:00,253 --> 00:40:02,920 this is the result, just to show you what the result looks like. 588 00:40:14,915 --> 00:40:16,290 Remember, the heavy to light case 589 00:40:16,290 --> 00:40:18,520 is the case where you're not getting a non-trivial function 590 00:40:18,520 --> 00:40:19,478 on the right hand side. 591 00:40:24,370 --> 00:40:26,660 So you'd get a non-zero result of order alpha 592 00:40:26,660 --> 00:40:32,490 s for both of those coefficients, OK? 593 00:40:32,490 --> 00:40:35,640 So the reading also goes through this whole thing 594 00:40:35,640 --> 00:40:38,040 for the heavy to heavy case, which is more interesting. 595 00:40:38,040 --> 00:40:40,890 But it's not really more non-trivial than what 596 00:40:40,890 --> 00:40:42,368 we've talked about so far already 597 00:40:42,368 --> 00:40:43,410 with anomalous dimension. 598 00:40:43,410 --> 00:40:45,830 You get results that are functions of w. 599 00:40:45,830 --> 00:40:50,300 Wilson coefficient would have functions of w showing up here, 600 00:40:50,300 --> 00:40:51,590 OK. 601 00:40:51,590 --> 00:40:55,360 So I won't go through that. 602 00:40:55,360 --> 00:40:57,970 Now if you wanted to carry out this calculation, 603 00:40:57,970 --> 00:41:00,560 it looks like it's kind of involved, 604 00:41:00,560 --> 00:41:02,470 this graph, this graph, this graph, 605 00:41:02,470 --> 00:41:05,510 all these diagrams to consider. 606 00:41:05,510 --> 00:41:07,578 You'd like to make your life as easy as possible, 607 00:41:07,578 --> 00:41:09,370 and there's actually a very nice trick here 608 00:41:09,370 --> 00:41:11,650 for doing that I have to mention to you, 609 00:41:11,650 --> 00:41:14,887 because it's kind of magical. 610 00:41:14,887 --> 00:41:17,220 So what is the fastest way that I could get this result? 611 00:41:28,442 --> 00:41:30,400 So this is a nice trick to remember if you ever 612 00:41:30,400 --> 00:41:33,580 have to do a calculation like that, 613 00:41:33,580 --> 00:41:35,470 because it's not specific to HQET. 614 00:41:38,060 --> 00:41:43,070 So let's pick our infrared regulator 615 00:41:43,070 --> 00:41:45,960 to make the effective theory as simple as possible. 616 00:41:45,960 --> 00:41:48,860 We have some choice in how to pick the infrared regulator. 617 00:41:48,860 --> 00:41:50,480 The result for Wilson coefficients 618 00:41:50,480 --> 00:41:54,120 and anomalous dimensions will not depend on that choice. 619 00:41:54,120 --> 00:41:56,180 So let's use that freedom and make 620 00:41:56,180 --> 00:41:58,540 things as simple as we can. 621 00:42:03,010 --> 00:42:05,770 And the choice that does that here 622 00:42:05,770 --> 00:42:13,540 is to use dimensional regularization for the UV, 623 00:42:13,540 --> 00:42:19,010 as we've been discussing, but also for the infrared. 624 00:42:19,010 --> 00:42:23,350 So let's use dimensional regularization for both. 625 00:42:23,350 --> 00:42:28,690 If you do that, you can convince yourself that all heavy quark 626 00:42:28,690 --> 00:42:32,575 effective theory graphs with on-shell external momentum-- 627 00:42:32,575 --> 00:42:34,450 so I can take the external momentum on-shell. 628 00:42:34,450 --> 00:42:36,280 I don't need it to regulate divergences, 629 00:42:36,280 --> 00:42:40,428 because I'm going to use div reg to do that. 630 00:42:40,428 --> 00:43:00,880 So all the integrals are scaleless, 631 00:43:00,880 --> 00:43:06,520 and that means that they come out to be something that, 632 00:43:06,520 --> 00:43:10,990 if you think about it, is either zero or zero in a way where you 633 00:43:10,990 --> 00:43:13,240 have the ultraviolet divergence cancelling an infrared 634 00:43:13,240 --> 00:43:16,720 divergence, which is still zero. 635 00:43:16,720 --> 00:43:18,640 Now you have to think about the fact 636 00:43:18,640 --> 00:43:20,140 that there's both of these going on, 637 00:43:20,140 --> 00:43:21,973 because you still have to think about adding 638 00:43:21,973 --> 00:43:24,520 counter terms to HQET to cancel the UV divergences. 639 00:43:31,890 --> 00:43:34,758 But the answers are very simple, because you can throw away 640 00:43:34,758 --> 00:43:35,675 all the finite pieces. 641 00:43:35,675 --> 00:43:37,545 If you have 1 over epsilon minus 1 642 00:43:37,545 --> 00:43:42,930 over epsilon, if you multiply by epsilon, that term's not there. 643 00:43:42,930 --> 00:43:50,160 So 1 over epsilon gets removed by counter terms, 644 00:43:50,160 --> 00:43:54,100 and there's no finite pieces left over. 645 00:43:54,100 --> 00:44:00,095 So just use MS bar, so you just strip off that exactly, 646 00:44:00,095 --> 00:44:01,720 and you're left with 1 over epsilon IR. 647 00:44:06,480 --> 00:44:13,250 Now if you-- so the effective theory diagrams are just 648 00:44:13,250 --> 00:44:16,310 simply all 1 over epsilon IR. 649 00:44:16,310 --> 00:44:18,920 Now the reason that this is making things simple 650 00:44:18,920 --> 00:44:21,050 is because you also know a fact, which 651 00:44:21,050 --> 00:44:23,090 is that the IR divergences in the full theory 652 00:44:23,090 --> 00:44:25,700 and the effective theory have to match up. 653 00:44:25,700 --> 00:44:28,160 So these 1 over epsilon IRs have to match up 654 00:44:28,160 --> 00:44:29,610 with your full theory calculation. 655 00:44:32,430 --> 00:44:35,510 So if you renormalize the QCD calculation, which 656 00:44:35,510 --> 00:44:37,817 you can't really get around doing, 657 00:44:37,817 --> 00:44:39,150 you have to do that calculation. 658 00:44:39,150 --> 00:44:47,540 So you do that calculation in pure dim reg, same IR 659 00:44:47,540 --> 00:44:52,340 regulator, which is a nice regulator to use for QCD. 660 00:44:52,340 --> 00:44:54,860 You do the UV renormalization using the standard counter 661 00:44:54,860 --> 00:44:58,405 terms, and what will you get? 662 00:44:58,405 --> 00:44:59,780 You will get something that looks 663 00:44:59,780 --> 00:45:03,410 like a number over epsilon IR, and then you'll 664 00:45:03,410 --> 00:45:12,890 get numbers times logs mu over MQ plus other things. 665 00:45:12,890 --> 00:45:17,600 This thing here just cancels with the-- if I subtract HQET, 666 00:45:17,600 --> 00:45:21,200 this is just canceling this. 667 00:45:21,200 --> 00:45:30,275 So this guy here cancels when we subtract HQET. 668 00:45:34,190 --> 00:45:36,010 And so the matching is then just this. 669 00:45:40,790 --> 00:45:43,750 So I don't actually even have to consider calculating the heavy. 670 00:45:43,750 --> 00:45:46,910 If I use all these facts that I know, if I trust them, 671 00:45:46,910 --> 00:45:50,630 then I don't even have to calculate the HQET graphs. 672 00:45:50,630 --> 00:45:52,983 I just say, let me imagine that I calculate them. 673 00:45:52,983 --> 00:45:53,900 They're all scaleless. 674 00:45:53,900 --> 00:45:55,277 They look like that. 675 00:45:55,277 --> 00:45:57,110 Let me imagine that I renormalized them all. 676 00:45:57,110 --> 00:45:59,120 The 1 over epsilon UVs are gone. 677 00:45:59,120 --> 00:46:01,700 I'm left with 1 over epsilon IRs. 678 00:46:01,700 --> 00:46:02,790 I do this calculation. 679 00:46:02,790 --> 00:46:06,060 I say, let me imagine that these cancel each other. 680 00:46:06,060 --> 00:46:08,090 And then I have the matching. 681 00:46:08,090 --> 00:46:09,980 So that's exploiting all the facts 682 00:46:09,980 --> 00:46:12,740 that we know about effective theories and full theories 683 00:46:12,740 --> 00:46:15,235 to get the matching as quickly as possible by just 684 00:46:15,235 --> 00:46:16,610 doing the full theory calculation 685 00:46:16,610 --> 00:46:19,130 with a particular regulator. 686 00:46:19,130 --> 00:46:20,300 It's not checking anything. 687 00:46:20,300 --> 00:46:22,925 It's not checking that the full theory and the effective theory 688 00:46:22,925 --> 00:46:27,930 have the IR divergences matching up, et cetera, et cetera. 689 00:46:27,930 --> 00:46:29,840 But if you know that that's true, 690 00:46:29,840 --> 00:46:31,520 if you trust that it's true, then this 691 00:46:31,520 --> 00:46:33,145 is the fastest way to get the matching. 692 00:46:39,470 --> 00:46:40,490 Seems like magic, right? 693 00:46:45,580 --> 00:46:47,560 OK, so sometimes you can exploit what 694 00:46:47,560 --> 00:46:50,080 you know about the effective theory 695 00:46:50,080 --> 00:46:52,840 to get things more quickly. 696 00:46:52,840 --> 00:46:54,070 So questions about that? 697 00:46:58,830 --> 00:47:00,734 AUDIENCE: What is [INAUDIBLE]? 698 00:47:03,600 --> 00:47:05,110 PROFESSOR: Yeah. 699 00:47:05,110 --> 00:47:09,990 So if you think about the loop integral, 700 00:47:09,990 --> 00:47:12,240 then the d here, right, and 4 minus 2 701 00:47:12,240 --> 00:47:15,630 epsilon have an epsilon greater than 0, 702 00:47:15,630 --> 00:47:19,050 decreasing the powers of q in the numerator is making it more 703 00:47:19,050 --> 00:47:21,720 UV convergent. 704 00:47:21,720 --> 00:47:23,490 So to regulate the IR, you want to have 705 00:47:23,490 --> 00:47:26,140 epsilon on the other side. 706 00:47:26,140 --> 00:47:28,500 So this is what you need for regulating IR, 707 00:47:28,500 --> 00:47:30,910 and this is what you need for regulating UV. 708 00:47:30,910 --> 00:47:32,910 So it may seem contradictory that you could even 709 00:47:32,910 --> 00:47:34,618 do both of these things at the same time, 710 00:47:34,618 --> 00:47:38,160 because greater than zero and less than zero. 711 00:47:38,160 --> 00:47:41,010 But you could always think of splitting up this integral 712 00:47:41,010 --> 00:47:43,770 with a hard cutoff somewhere in between and then 713 00:47:43,770 --> 00:47:46,053 just using this above and this below that cutoff. 714 00:47:46,053 --> 00:47:47,970 And the cutoff dependence will cancel when you 715 00:47:47,970 --> 00:47:49,810 put the pieces back together. 716 00:47:49,810 --> 00:47:53,190 So it's actually valid to just do calculations. 717 00:47:53,190 --> 00:47:56,280 And for the most part, you can just close your eyes, 718 00:47:56,280 --> 00:47:58,140 and you'll get some gamma of epsilons 719 00:47:58,140 --> 00:47:59,700 and some gamma of minus epsilons, 720 00:47:59,700 --> 00:48:02,448 and those will be separately regulating the divergences. 721 00:48:02,448 --> 00:48:04,740 And if you ever worry about it, you can do what I said. 722 00:48:04,740 --> 00:48:06,900 You could put a cutoff in and check that you're not 723 00:48:06,900 --> 00:48:09,780 making mistakes, but for the most part, 724 00:48:09,780 --> 00:48:14,120 it just works automatically. 725 00:48:14,120 --> 00:48:14,740 Any other-- 726 00:48:14,740 --> 00:48:17,710 AUDIENCE: [INAUDIBLE] all of the epsilon UV [INAUDIBLE] epsilon 727 00:48:17,710 --> 00:48:18,210 IR. 728 00:48:18,210 --> 00:48:18,877 PROFESSOR: Yeah. 729 00:48:18,877 --> 00:48:21,280 AUDIENCE: [INAUDIBLE] minus epsilon UV [INAUDIBLE] zero 730 00:48:21,280 --> 00:48:21,780 [INAUDIBLE]. 731 00:48:21,780 --> 00:48:22,080 PROFESSOR: Yeah. 732 00:48:22,080 --> 00:48:23,033 AUDIENCE: --negative. 733 00:48:23,033 --> 00:48:23,700 PROFESSOR: Yeah. 734 00:48:23,700 --> 00:48:25,658 AUDIENCE: But formally, you can set the epsilon 735 00:48:25,658 --> 00:48:26,840 UV plus an epsilon IR. 736 00:48:26,840 --> 00:48:29,030 PROFESSOR: That's right, yeah. 737 00:48:29,030 --> 00:48:32,490 So formally, this is zero. 738 00:48:32,490 --> 00:48:34,490 And the reason that you have to worry about zero 739 00:48:34,490 --> 00:48:38,510 is because you have to add a counter term to cancel this. 740 00:48:38,510 --> 00:48:40,640 Your UV counter term, you have to still add it. 741 00:48:40,640 --> 00:48:42,223 And then it cancels this, and then you 742 00:48:42,223 --> 00:48:43,610 get something non-zero. 743 00:48:43,610 --> 00:48:46,280 So the bare graph is zero. 744 00:48:46,280 --> 00:48:48,080 The counter term's non-zero, and there 745 00:48:48,080 --> 00:48:51,310 renormalized graph is non-zero. 746 00:48:51,310 --> 00:48:57,890 This is a subtlety that's worth remembering if you ever want 747 00:48:57,890 --> 00:48:59,170 to do calculations this way. 748 00:49:07,296 --> 00:49:10,900 OK, so that's some of the complications 749 00:49:10,900 --> 00:49:15,535 and fascinating facts about HQET in the perturbative sector. 750 00:49:19,360 --> 00:49:30,030 Let's come back and talk about power corrections, which are-- 751 00:49:30,030 --> 00:49:32,050 I'll go under the title of-- well, 752 00:49:32,050 --> 00:49:35,150 maybe I should just call them power corrections. 753 00:49:35,150 --> 00:49:35,880 Better title. 754 00:49:41,640 --> 00:49:44,670 So we have an effective theory. 755 00:49:44,670 --> 00:49:46,830 We've so far talked about it at lowest order. 756 00:49:46,830 --> 00:49:48,120 We stopped at lowest order. 757 00:49:48,120 --> 00:49:50,512 We had this HQET Lagrangian, and we 758 00:49:50,512 --> 00:49:52,470 talked about using that Lagrangian to carry out 759 00:49:52,470 --> 00:49:53,827 some perturbative calculations. 760 00:49:53,827 --> 00:49:56,160 What if we went to higher order in the power [INAUDIBLE] 761 00:49:56,160 --> 00:49:59,580 expansion, which is 1 over MQ? 762 00:49:59,580 --> 00:50:02,338 OK, so power corrections here means higher order 763 00:50:02,338 --> 00:50:02,880 in 1 over MQ. 764 00:50:07,419 --> 00:50:14,500 So let me show you how you can construct those terms. 765 00:50:14,500 --> 00:50:17,850 So let me go back to a representation of the full QCD 766 00:50:17,850 --> 00:50:23,110 Lagrangian, which we had in terms of this B field and the Q 767 00:50:23,110 --> 00:50:23,610 field. 768 00:50:26,340 --> 00:50:27,930 And when we first talked about this, 769 00:50:27,930 --> 00:50:30,382 we just dropped all the terms of the B, 770 00:50:30,382 --> 00:50:32,340 but now I'm going to do something a little more 771 00:50:32,340 --> 00:50:38,100 sophisticated with them and really just integrate them out. 772 00:50:41,740 --> 00:50:43,460 So we had this, and this was really 773 00:50:43,460 --> 00:50:47,030 just us writing QCD in a fancy way that was 774 00:50:47,030 --> 00:50:48,560 convenient for this discussion. 775 00:50:51,710 --> 00:50:53,960 So this is really just QCD written in a fancy way. 776 00:51:00,080 --> 00:51:02,590 So if we want to take this Lagrangian at tree level, 777 00:51:02,590 --> 00:51:06,040 we can just integrate out Bv. 778 00:51:06,040 --> 00:51:10,720 This is a Lagrangian that has quadratic dependence on Bv. 779 00:51:10,720 --> 00:51:13,790 So you could think that the path interval in this formula 780 00:51:13,790 --> 00:51:15,790 here would be quadratic path interval, and those 781 00:51:15,790 --> 00:51:16,832 we can always just solve. 782 00:51:19,500 --> 00:51:23,760 And what effectively integrating out Bv amounts 783 00:51:23,760 --> 00:51:27,270 to is solving for the equations of motion of Bv 784 00:51:27,270 --> 00:51:28,560 and plugging that back in. 785 00:51:57,038 --> 00:51:58,580 So the type of diagrams I was drawing 786 00:51:58,580 --> 00:52:00,038 before where I had this wiggly line 787 00:52:00,038 --> 00:52:02,720 and it was a Bv propagator, we can integrate out that 788 00:52:02,720 --> 00:52:05,320 by solving for the equation of motion. 789 00:52:05,320 --> 00:52:08,810 So we look for variation with respect 790 00:52:08,810 --> 00:52:12,890 to Bv bar and set it to 0. 791 00:52:12,890 --> 00:52:25,370 And that gives this equation, and then 792 00:52:25,370 --> 00:52:44,075 we solve this formally for Bv by just inverting this operator 793 00:52:44,075 --> 00:52:46,915 to get that equation. 794 00:52:46,915 --> 00:52:48,540 And then we can plug that equation back 795 00:52:48,540 --> 00:53:15,120 into this equation, which is still a QCD 796 00:53:15,120 --> 00:53:17,250 Lagrangian, actually, but-- 797 00:53:17,250 --> 00:53:18,360 and then we expand. 798 00:53:18,360 --> 00:53:21,810 And once we expand, we match onto the HQET Lagrangian order 799 00:53:21,810 --> 00:53:26,110 by order, tree level. 800 00:53:26,110 --> 00:53:29,670 So the first term is the term we've been discussing. 801 00:53:29,670 --> 00:53:34,770 The next term, we drop the v dot D here, because that's small. 802 00:53:34,770 --> 00:53:43,852 We just have 1 over 2 Q. There's two D transverses, 803 00:53:43,852 --> 00:53:46,060 and they'd be higher order terms as well, 804 00:53:46,060 --> 00:53:49,130 but we'll stop in that order. 805 00:53:49,130 --> 00:53:56,107 So this is L0, first order term, and this would be L1. 806 00:53:56,107 --> 00:53:57,604 They'd be higher order terms. 807 00:54:01,600 --> 00:54:04,373 So what is this guy? 808 00:54:04,373 --> 00:54:06,040 It's useful to write that guy, actually, 809 00:54:06,040 --> 00:54:07,415 in terms of two different things, 810 00:54:07,415 --> 00:54:10,348 and you'll see why momentarily. 811 00:54:10,348 --> 00:54:12,640 So it's got two covariant derivatives, and both of them 812 00:54:12,640 --> 00:54:14,680 are dotted into gamma matrices. 813 00:54:14,680 --> 00:54:18,550 It involves this thing, D transverse, which, remember, 814 00:54:18,550 --> 00:54:22,180 is the full D minus a projection onto 815 00:54:22,180 --> 00:54:26,200 v. So it's something that's transverse to v. So 816 00:54:26,200 --> 00:54:28,780 what I want to do to simplify this guy here 817 00:54:28,780 --> 00:54:31,040 is I want to do the following. 818 00:54:31,040 --> 00:54:32,240 I'm going to use the fact-- 819 00:54:32,240 --> 00:54:34,407 I'm going to write it in terms of the field strength 820 00:54:34,407 --> 00:54:38,230 by using the fact that the commutator of two Ds 821 00:54:38,230 --> 00:54:45,440 gives me a G. And the commutator of two sigmas, I'll write-- 822 00:54:45,440 --> 00:54:45,940 sorry. 823 00:54:45,940 --> 00:54:47,800 The commutator of two gammas gives me 824 00:54:47,800 --> 00:54:48,925 something I can call sigma. 825 00:54:56,000 --> 00:54:58,600 So let me write this as the symmetric piece and then 826 00:54:58,600 --> 00:54:59,800 the anti-symmetric piece. 827 00:55:03,660 --> 00:55:07,530 And I do a symmetrization in both the fields 828 00:55:07,530 --> 00:55:10,167 and the Dirac structures. 829 00:55:14,937 --> 00:55:23,700 Doing DT slash DT slash anti-commutator. 830 00:55:38,170 --> 00:55:40,020 So then since it's anti-symmetric, 831 00:55:40,020 --> 00:55:42,990 it automatically forces that anti-symmetric. 832 00:55:42,990 --> 00:55:48,970 This guy is a GB nu, so for this piece, we just get DT squared. 833 00:55:48,970 --> 00:55:55,350 And this piece, once we track all the 2s, 834 00:55:55,350 --> 00:56:02,180 the i's give sigma dot G. 835 00:56:02,180 --> 00:56:06,650 And so the usual way of writing L1 is as follows. 836 00:56:06,650 --> 00:56:12,250 You say L1, using this formula, plugging it in, has two terms. 837 00:56:15,200 --> 00:56:18,367 And you'll see why when I write them down 838 00:56:18,367 --> 00:56:19,450 that we wanted to do this. 839 00:56:32,836 --> 00:56:36,560 OK, so that's L1 is after plugging that in. 840 00:56:36,560 --> 00:56:39,590 Now the reason to do this is that if we ask about symmetry 841 00:56:39,590 --> 00:56:41,120 breaking, that's something that can 842 00:56:41,120 --> 00:56:42,940 happen from sub-leading terms. 843 00:56:42,940 --> 00:56:46,070 Lowest order, we had a symmetry, heavy quark symmetry, 844 00:56:46,070 --> 00:56:48,730 and that is broken by these interactions. 845 00:56:48,730 --> 00:56:50,480 But it's actually broken in different ways 846 00:56:50,480 --> 00:56:52,400 by these two terms. 847 00:56:52,400 --> 00:56:54,620 This term here doesn't have any spin structure, 848 00:56:54,620 --> 00:56:56,690 so it doesn't break the spin symmetry. 849 00:56:56,690 --> 00:56:59,030 It does have flavor structure because of the MQ, 850 00:56:59,030 --> 00:57:02,370 so it breaks the flavor symmetry. 851 00:57:02,370 --> 00:57:08,690 So this is a kinetic energy type correction, 852 00:57:08,690 --> 00:57:12,050 and it breaks flavor symmetry because of the dependence on Q 853 00:57:12,050 --> 00:57:12,950 in the MQ. 854 00:57:18,980 --> 00:57:21,050 And this guy breaks both, because it 855 00:57:21,050 --> 00:57:24,180 has a spin structure now, and it's a magnetic moment type 856 00:57:24,180 --> 00:57:24,680 term. 857 00:57:39,680 --> 00:57:49,186 So it's got the sigma dot B field type interaction. 858 00:57:49,186 --> 00:57:51,760 This is what I mean by the magnetic moment type term. 859 00:57:54,260 --> 00:57:54,760 OK? 860 00:57:54,760 --> 00:57:57,218 So that's what the sub-leading power corrections look like, 861 00:57:57,218 --> 00:57:59,560 and if we wanted to use the effective theory to talk 862 00:57:59,560 --> 00:58:02,440 about power corrections, we could do that. 863 00:58:02,440 --> 00:58:05,050 We're constructing them here by knowing the full theory, 864 00:58:05,050 --> 00:58:09,870 just integrating out explicitly the fields, OK, 865 00:58:09,870 --> 00:58:11,810 which is a very nice thing if you can do it. 866 00:58:14,770 --> 00:58:16,420 Now you could do it the other way, 867 00:58:16,420 --> 00:58:19,750 which would be to think about just writing them 868 00:58:19,750 --> 00:58:22,420 from the bottom up. 869 00:58:22,420 --> 00:58:23,920 And there is one way in which that's 870 00:58:23,920 --> 00:58:26,710 more general than what we've talked about, 871 00:58:26,710 --> 00:58:29,080 and that's because what we talked about was tree level. 872 00:58:34,480 --> 00:58:37,150 And if you wanted to include loop corrections, 873 00:58:37,150 --> 00:58:39,580 how do we know that there's not some other operator here 874 00:58:39,580 --> 00:58:42,310 that we missed because it just vanished at tree level, 875 00:58:42,310 --> 00:58:44,230 for example? 876 00:58:44,230 --> 00:58:46,042 We've seen examples where that happens. 877 00:58:46,042 --> 00:58:48,250 There's an operator that only shows up at loop level. 878 00:58:51,193 --> 00:58:53,110 So we could think about it from the bottom up, 879 00:58:53,110 --> 00:58:55,600 even though this is a top-down effective theory, in order 880 00:58:55,600 --> 00:58:57,730 to make sure we're not missing anything. 881 00:58:57,730 --> 00:58:59,320 And if we wanted to do that, we should 882 00:58:59,320 --> 00:59:02,650 enumerate all the possible things, the symmetries and all 883 00:59:02,650 --> 00:59:06,470 that we can use to constrain the form of the operators. 884 00:59:06,470 --> 00:59:09,042 So let's enumerate. 885 00:59:09,042 --> 00:59:10,750 So there's the power counting, of course. 886 00:59:14,735 --> 00:59:15,610 That's pretty simple. 887 00:59:15,610 --> 00:59:19,280 Here all the powers of 1 over MQ are being made explicit, 888 00:59:19,280 --> 00:59:21,280 and they just tell us what dimension of operator 889 00:59:21,280 --> 00:59:30,200 to look for, just as in our integrating 890 00:59:30,200 --> 00:59:33,488 out heavy particles. 891 00:59:33,488 --> 00:59:35,280 So we just know the dimension of the fields 892 00:59:35,280 --> 00:59:36,780 that we have to put in the numerator 893 00:59:36,780 --> 00:59:38,580 from how many MQs we're talking about. 894 00:59:45,180 --> 00:59:46,770 There's gait symmetry, of course. 895 00:59:46,770 --> 00:59:48,330 So use covariant derivatives. 896 00:59:51,930 --> 00:59:53,490 Very easy to take into account. 897 00:59:59,620 --> 01:00:02,710 There's discreet symmetries, charge conjugation, 898 01:00:02,710 --> 01:00:05,240 parity at time reversal, which are symmetries of QCD 899 01:00:05,240 --> 01:00:07,270 if we drop the theta term. 900 01:00:07,270 --> 01:00:12,628 And we can impose them as well, and again, that's easy. 901 01:00:15,227 --> 01:00:17,810 I wouldn't be making a list if there wasn't at least one thing 902 01:00:17,810 --> 01:00:25,280 that was hard and non-trivial. 903 01:00:25,280 --> 01:00:26,860 But discreet symmetries are easy. 904 01:00:33,570 --> 01:00:36,540 The thing that actually is the hardest is Lorentz symmetry. 905 01:00:41,000 --> 01:00:44,590 Oh, you say, just dot Lorentz indices into Lorentz indices. 906 01:00:47,303 --> 01:00:49,970 But you have to ask the question whether we even have Lorentzian 907 01:00:49,970 --> 01:00:53,060 variance in this theory. 908 01:00:53,060 --> 01:00:55,505 And it turns out that part of the Lorentz group 909 01:00:55,505 --> 01:00:58,430 was actually broken by having this heavy quark 910 01:00:58,430 --> 01:01:03,518 and doing this type of expansion that we've been talking about. 911 01:01:03,518 --> 01:01:05,810 So if you think about the six generators of the Lorentz 912 01:01:05,810 --> 01:01:08,690 group, the boosts and the rotations, 913 01:01:08,690 --> 01:01:11,600 there's a part that I could call the transverse part, which 914 01:01:11,600 --> 01:01:13,890 is transverse to the velocity. 915 01:01:13,890 --> 01:01:18,620 So in the rest frame, that would be M12, M23, and M13. 916 01:01:21,490 --> 01:01:24,820 And those are the rotations. 917 01:01:24,820 --> 01:01:28,600 So this i-- v is like this. 918 01:01:28,600 --> 01:01:30,340 So no matter what v you pick, there's 919 01:01:30,340 --> 01:01:34,570 always three generators that are rotations. 920 01:01:34,570 --> 01:01:37,098 And then there's the boosts. 921 01:01:37,098 --> 01:01:38,515 And you should think of the boosts 922 01:01:38,515 --> 01:01:43,090 as taking v mu and then M dotted into v 923 01:01:43,090 --> 01:01:45,070 and then making the other guy transverse, 924 01:01:45,070 --> 01:01:47,260 so when we denote it like that. 925 01:01:47,260 --> 01:01:50,020 So the new index is transverse, and in the rest from, 926 01:01:50,020 --> 01:01:54,008 that's M01, M02, and M03. 927 01:01:59,490 --> 01:02:07,320 And so introducing v mu actually breaks the boosts' symmetry. 928 01:02:26,970 --> 01:02:29,370 And if you like, you could think the reason 929 01:02:29,370 --> 01:02:31,290 it breaks the symmetry is because it gives 930 01:02:31,290 --> 01:02:34,340 a preferred frame, which is the rest frame of the heavy quark. 931 01:02:34,340 --> 01:02:36,090 If you have a preferred frame, then you've 932 01:02:36,090 --> 01:02:37,250 broken Lorentzian frames. 933 01:02:41,300 --> 01:02:43,945 So that's bad. 934 01:02:43,945 --> 01:02:45,320 And it turns out there's actually 935 01:02:45,320 --> 01:02:47,695 a hidden symmetry of this effective theory that partially 936 01:02:47,695 --> 01:02:48,695 restores this breaking. 937 01:02:51,830 --> 01:02:53,930 And it restores it in exactly the amount 938 01:02:53,930 --> 01:02:55,120 that it needs to restore it. 939 01:03:04,060 --> 01:03:07,170 That is, it restores it at low energies. 940 01:03:10,410 --> 01:03:16,170 And that's called reparameterization invariance, 941 01:03:16,170 --> 01:03:18,420 which I will write once, and then forever more, 942 01:03:18,420 --> 01:03:23,790 we talk about it as RPI, Reparameterization Invariance. 943 01:03:23,790 --> 01:03:27,465 And it's an additional symmetry that we have on v mu itself. 944 01:03:38,082 --> 01:03:40,165 So let's go back and think about how we introduced 945 01:03:40,165 --> 01:03:41,890 v mu in the first place. 946 01:03:41,890 --> 01:03:44,140 So we're saying that v mu breaks part of the cemetery, 947 01:03:44,140 --> 01:03:48,183 but how did we decide on what v mu was? 948 01:03:48,183 --> 01:03:49,600 How much freedom was there when we 949 01:03:49,600 --> 01:03:51,422 defined v mu at the beginning? 950 01:03:58,815 --> 01:04:00,440 If we're saying that it breaks, then we 951 01:04:00,440 --> 01:04:03,200 should know how much freedom there 952 01:04:03,200 --> 01:04:05,990 was, because a freedom to define different vs 953 01:04:05,990 --> 01:04:09,670 could restore symmetry, just realized in a different way. 954 01:04:09,670 --> 01:04:10,670 And that's what happens. 955 01:04:14,870 --> 01:04:16,860 So where did it come from? 956 01:04:16,860 --> 01:04:19,540 We had P, have a heavy quark, and we split that 957 01:04:19,540 --> 01:04:25,330 into two pieces, MQv plus k. 958 01:04:25,330 --> 01:04:29,800 But this split into two pieces is arbitrary by some amount. 959 01:04:29,800 --> 01:04:33,160 We could move pieces back and forth between here and here, 960 01:04:33,160 --> 01:04:35,170 and we would still have the same theory. 961 01:04:41,017 --> 01:04:43,600 We have to be careful that we're moving back pieces that don't 962 01:04:43,600 --> 01:04:45,610 violate the power counting. 963 01:04:45,610 --> 01:04:48,850 And that's what I mean by somewhat arbitrary here, 964 01:04:48,850 --> 01:04:50,530 not completely arbitrary. 965 01:04:50,530 --> 01:04:52,360 There was a point to doing this, because we 966 01:04:52,360 --> 01:04:54,820 wanted to separate out the big piece and the small piece. 967 01:04:54,820 --> 01:04:57,070 But we could always move a small piece back over here, 968 01:04:57,070 --> 01:05:01,450 and that wouldn't change this decomposition. 969 01:05:01,450 --> 01:05:06,680 So the invariance that you have is the following. 970 01:05:06,680 --> 01:05:10,390 You can take v mu and send it to v mu 971 01:05:10,390 --> 01:05:14,680 plus some epsilon mu over MQ. 972 01:05:14,680 --> 01:05:22,330 And k mu comes to k mu minus epsilon mu. 973 01:05:22,330 --> 01:05:24,500 That moves the piece back and forth between them, 974 01:05:24,500 --> 01:05:30,150 and as long as I think that epsilon mu is some parameter 975 01:05:30,150 --> 01:05:32,220 that doesn't have a power counting in it, i.e. 976 01:05:32,220 --> 01:05:35,490 it's order doesn't have any MQs in it-- 977 01:05:35,490 --> 01:05:39,810 it's just something of order lambda QCD, say-- 978 01:05:42,900 --> 01:05:45,720 then that makes the power counting still true. 979 01:05:45,720 --> 01:05:48,150 That was the point of this decomposition. 980 01:05:48,150 --> 01:05:50,400 And it allows us to move a small piece back and forth. 981 01:05:50,400 --> 01:05:52,950 The small piece is this epsilon. 982 01:05:52,950 --> 01:05:54,090 So that's a symmetry. 983 01:05:54,090 --> 01:05:56,465 That's called reparameterization invariance. 984 01:05:56,465 --> 01:05:57,840 So we have to make sure that when 985 01:05:57,840 --> 01:05:59,730 we construct our effective theory that it 986 01:05:59,730 --> 01:06:01,140 satisfies the symmetry if we want 987 01:06:01,140 --> 01:06:03,150 it to be a boost invariant-- 988 01:06:03,150 --> 01:06:07,360 if we want to restore boost invariance to the theory. 989 01:06:07,360 --> 01:06:10,140 So this parameter epsilon you can think of as-- 990 01:06:10,140 --> 01:06:13,110 you could consider it to be a finite reparameterization 991 01:06:13,110 --> 01:06:14,580 symmetry, but you don't really have 992 01:06:14,580 --> 01:06:16,163 to worry about finite transformations. 993 01:06:16,163 --> 01:06:18,090 You can just do the infinitesimal. 994 01:06:18,090 --> 01:06:21,237 So we'll think about epsilon mu as an infinitesimal. 995 01:06:29,840 --> 01:06:34,160 And it has that counting that I put over there. 996 01:06:34,160 --> 01:06:37,490 Now v squared was equal to 1, and that's also something 997 01:06:37,490 --> 01:06:39,120 we don't want to spoil. 998 01:06:39,120 --> 01:06:41,070 But that's easy. 999 01:06:41,070 --> 01:06:44,790 We just say that epsilon dot v is equal to 0. 1000 01:06:44,790 --> 01:06:46,712 That maintains this condition. 1001 01:06:46,712 --> 01:06:48,920 So that means that there's three different components 1002 01:06:48,920 --> 01:06:57,410 of epsilon, non-trivial components to epsilon. 1003 01:06:57,410 --> 01:06:59,120 And those three components of epsilon 1004 01:06:59,120 --> 01:07:02,144 are exactly related to the three boosts here. 1005 01:07:06,884 --> 01:07:08,350 OK, we have a three family-- 1006 01:07:08,350 --> 01:07:16,632 three-parameter family of transformations, 1007 01:07:16,632 --> 01:07:18,840 which are the three components of the epsilon, which, 1008 01:07:18,840 --> 01:07:21,382 in the rest frame, would just be the one, the two, and three. 1009 01:07:23,460 --> 01:07:25,260 What did we do-- what about the fields? 1010 01:07:25,260 --> 01:07:27,810 How does the field, the Qv change under this type 1011 01:07:27,810 --> 01:07:29,114 of transformation? 1012 01:07:33,760 --> 01:07:36,640 Let me take the field that x equals 0 for now. 1013 01:07:40,850 --> 01:07:44,530 So v slash on Qv, it was equal to 0. 1014 01:07:44,530 --> 01:07:48,140 And if I do the transformation, then this v slash changes. 1015 01:07:48,140 --> 01:07:53,020 It becomes v slash plus epsilon slash over MQ. 1016 01:07:53,020 --> 01:07:54,325 Let me imagine Qv changes. 1017 01:07:57,820 --> 01:08:00,785 It goes to Qv plus delta Qv, and I 1018 01:08:00,785 --> 01:08:02,035 have to do that on both sides. 1019 01:08:06,740 --> 01:08:08,530 Then I can take this-- so this thing here 1020 01:08:08,530 --> 01:08:11,380 is some order epsilon change. 1021 01:08:11,380 --> 01:08:14,770 Then I can take this equation, and I can just solve. 1022 01:08:14,770 --> 01:08:16,359 So the piece that's order epsilon 1023 01:08:16,359 --> 01:08:18,670 to the 0, just satisfied. 1024 01:08:18,670 --> 01:08:20,920 Solve for the piece that's order epsilon, 1025 01:08:20,920 --> 01:08:24,750 and that gives me an equation for delta Qv. 1026 01:08:24,750 --> 01:08:28,750 So rearranging this equation, I find 1027 01:08:28,750 --> 01:08:39,250 that 1 minus v slash delta Qv is epsilon slash over MQ times Qv. 1028 01:08:39,250 --> 01:08:46,729 And this equation has the solution, 1029 01:08:46,729 --> 01:08:53,410 but delta QV is epsilon slash over 2MQ. 1030 01:08:53,410 --> 01:08:57,470 Remember that epsilon is transverse to v, 1031 01:08:57,470 --> 01:09:04,040 so if I push the v slash-- so if I plug that solution in here 1032 01:09:04,040 --> 01:09:07,010 and I push a v slash through the epsilon, 1033 01:09:07,010 --> 01:09:10,279 well, then I can push the v slash through the epsilon, 1034 01:09:10,279 --> 01:09:11,359 let it hit the QV. 1035 01:09:11,359 --> 01:09:14,180 That's giving a factor of 2, because it's anti-commutes. 1036 01:09:14,180 --> 01:09:16,279 That's this 2 here. 1037 01:09:16,279 --> 01:09:20,130 And then I would get what I wrote on the right hand side. 1038 01:09:20,130 --> 01:09:23,420 So if it's not obvious, check for yourself 1039 01:09:23,420 --> 01:09:26,550 that that's a solution. 1040 01:09:26,550 --> 01:09:28,729 So that's how you derive the change to the field 1041 01:09:28,729 --> 01:09:33,649 under this reparameterization. 1042 01:09:33,649 --> 01:09:35,240 And so when we talk about operators 1043 01:09:35,240 --> 01:09:37,520 and the effective theory, we have 1044 01:09:37,520 --> 01:09:39,700 to worry, how does the symmetry act on them? 1045 01:09:47,279 --> 01:09:50,362 And it's a kind of a non-trivial symmetry. 1046 01:09:50,362 --> 01:09:52,279 Was not apparent to us when we started, right. 1047 01:09:52,279 --> 01:09:55,790 [CHUCKLES] 1048 01:09:55,790 --> 01:09:58,400 OK. 1049 01:09:58,400 --> 01:10:13,060 So the full reparameterization is v mu 1050 01:10:13,060 --> 01:10:18,250 goes to v plus epsilon over MQ. 1051 01:10:18,250 --> 01:10:26,662 And then if I take Qv of x, then that is what I said. 1052 01:10:26,662 --> 01:10:31,395 There's this epsilon slash over 2MQ piece. 1053 01:10:31,395 --> 01:10:32,770 This is the transformation, so it 1054 01:10:32,770 --> 01:10:35,290 goes to itself plus this extra piece. 1055 01:10:35,290 --> 01:10:38,980 And the fact that I take it at x adds one little slight wrinkle, 1056 01:10:38,980 --> 01:10:41,260 and it just gives this extra phase factor. 1057 01:10:41,260 --> 01:10:43,360 And that extra phase factor is exactly what 1058 01:10:43,360 --> 01:10:45,820 encodes the change of k. 1059 01:10:45,820 --> 01:10:49,390 So this, if you like, encodes that derivatives. 1060 01:10:49,390 --> 01:10:53,260 Should go to derivatives minus epsilon or in momentum space, 1061 01:10:53,260 --> 01:10:57,420 that k should go to k minus epsilon. 1062 01:10:57,420 --> 01:10:59,273 OK, so previously we had a rule for k, 1063 01:10:59,273 --> 01:11:00,940 but now I've encoded that in this phase. 1064 01:11:05,220 --> 01:11:05,720 OK? 1065 01:11:05,720 --> 01:11:10,220 So that's the symmetry that we should look into. 1066 01:11:15,740 --> 01:11:18,540 So what does it do? 1067 01:11:18,540 --> 01:11:21,460 So what this does is it restores invariance under boosts, 1068 01:11:21,460 --> 01:11:22,445 but only small boosts. 1069 01:11:34,897 --> 01:11:36,605 The reason that I call them a small boost 1070 01:11:36,605 --> 01:11:39,453 is because epsilon here had to be of order lambda QCD. 1071 01:11:39,453 --> 01:11:40,745 It couldn't have been order MQ. 1072 01:11:47,440 --> 01:11:49,410 That's what I mean by small. 1073 01:11:49,410 --> 01:11:51,160 And from the point of view of this theory, 1074 01:11:51,160 --> 01:11:52,243 this is all we care about. 1075 01:12:00,460 --> 01:12:02,440 Because we want to remain within the region 1076 01:12:02,440 --> 01:12:04,260 where the effective theory was valid, 1077 01:12:04,260 --> 01:12:06,340 the whole setup of the effective theory 1078 01:12:06,340 --> 01:12:08,800 involved dividing out a large piece and a small piece. 1079 01:12:08,800 --> 01:12:12,490 If we allow back large pieces, then the game is over, 1080 01:12:12,490 --> 01:12:15,070 and you wouldn't be formulating correctly 1081 01:12:15,070 --> 01:12:19,660 the effective theory, because you'd spoil the power counting. 1082 01:12:19,660 --> 01:12:21,940 OK, so this is this hidden symmetry 1083 01:12:21,940 --> 01:12:23,690 we're calling reparameterization variance. 1084 01:12:23,690 --> 01:12:25,270 And it's not special to HQET. 1085 01:12:25,270 --> 01:12:28,060 Any time you have fields that are labeled by something, 1086 01:12:28,060 --> 01:12:30,560 you should think about whether there's a symmetry like this. 1087 01:12:32,650 --> 01:12:35,230 OK, so that's the entire list of the symmetries 1088 01:12:35,230 --> 01:12:37,360 that you should consider in order 1089 01:12:37,360 --> 01:12:40,630 to think about doing a bottom-up approach to HQET. 1090 01:12:40,630 --> 01:12:42,670 Simple ones, and then there's this one that's 1091 01:12:42,670 --> 01:12:44,069 a little more complicated. 1092 01:12:48,290 --> 01:12:51,670 So let's go back and now consider the 1 1093 01:12:51,670 --> 01:12:58,895 over MQ operators in general. 1094 01:13:03,900 --> 01:13:07,380 And it turns out that there's not any missing operators, 1095 01:13:07,380 --> 01:13:11,220 that the two operators we have are actually 1096 01:13:11,220 --> 01:13:14,760 the complete set that you can write down 1097 01:13:14,760 --> 01:13:20,345 at this dimension using all the properties of the field. 1098 01:13:20,345 --> 01:13:22,470 So we didn't miss anything from that point of view. 1099 01:13:26,630 --> 01:13:29,800 So let me write them down again, and let me write them 1100 01:13:29,800 --> 01:13:33,010 down in a way where I imagine that radiative corrections have 1101 01:13:33,010 --> 01:13:34,387 come in as well. 1102 01:13:34,387 --> 01:13:36,220 And I'll give them some Wilson coefficients, 1103 01:13:36,220 --> 01:13:38,620 which are generically called Ck and Cf. 1104 01:13:38,620 --> 01:13:41,210 That's the standard notation. 1105 01:13:41,210 --> 01:13:42,936 So this is a Wilson coefficient. 1106 01:13:46,810 --> 01:13:48,280 That's a Wilson coefficient. 1107 01:13:48,280 --> 01:13:49,330 This is not 4/3. 1108 01:13:49,330 --> 01:13:52,710 it's Wilson coefficient. 1109 01:13:52,710 --> 01:13:55,490 The name is the same as the Cf that is 4/3, 1110 01:13:55,490 --> 01:13:58,240 but this is a little cf, not a big CF. 1111 01:14:09,820 --> 01:14:11,230 So if we want to-- 1112 01:14:11,230 --> 01:14:12,540 so it's gauge invariant. 1113 01:14:12,540 --> 01:14:15,250 It has the right parity, et cetera, et cetera. 1114 01:14:15,250 --> 01:14:17,746 We should worry about the reparameterization invariance. 1115 01:14:23,580 --> 01:14:25,320 So let's do that. 1116 01:14:25,320 --> 01:14:33,810 So at lowest order, the phase is what changes. 1117 01:14:33,810 --> 01:14:36,090 And the leading order Lagrangian is invariant, 1118 01:14:36,090 --> 01:14:37,630 because v dot epsilon is 0. 1119 01:14:44,550 --> 01:14:49,920 So at order MQ to the 0, since v dot epsilon is 0, 1120 01:14:49,920 --> 01:14:51,480 you don't get a leading order change. 1121 01:14:51,480 --> 01:14:54,240 So our Lagrangian was variant. 1122 01:14:54,240 --> 01:14:56,490 This invariance, this reparameterization invariance 1123 01:14:56,490 --> 01:14:58,830 mixes orders. 1124 01:14:58,830 --> 01:15:03,370 It connects orders in the expansion. 1125 01:15:03,370 --> 01:15:06,220 There was a term that was order 1, which is this piece, 1126 01:15:06,220 --> 01:15:08,020 and there's a term that's order 1 over MQ. 1127 01:15:08,020 --> 01:15:09,520 So the symmetry is actually making 1128 01:15:09,520 --> 01:15:11,590 a connection between leading order and sub-leading order 1129 01:15:11,590 --> 01:15:12,250 operators. 1130 01:15:22,560 --> 01:15:24,770 So we could ask about this delta L 0, 1131 01:15:24,770 --> 01:15:27,980 and there will be a piece at order 1 over MQ. 1132 01:15:27,980 --> 01:15:30,350 So let's just write out all these things. 1133 01:15:42,170 --> 01:15:43,810 Transforming everything. 1134 01:15:49,310 --> 01:15:53,210 This is our leading order Lagrangian. 1135 01:15:53,210 --> 01:15:57,620 After imposing the field change as well as v change, 1136 01:15:57,620 --> 01:16:00,530 the v dot D becomes this, and the field becomes that. 1137 01:16:04,140 --> 01:16:08,990 So there's three things here that are being changed. 1138 01:16:08,990 --> 01:16:11,010 Expand this out. 1139 01:16:11,010 --> 01:16:14,780 Use things like 1 plus v slash over 2, 1140 01:16:14,780 --> 01:16:19,700 epsilon slash 1 plus v slash over 2. 1141 01:16:19,700 --> 01:16:22,325 C equal to epsilon dot v is equal to 0. 1142 01:16:25,130 --> 01:16:29,810 Simplify, do some Dirac algebra, and you can boil this 1143 01:16:29,810 --> 01:16:32,390 down to something simple, which is 1144 01:16:32,390 --> 01:16:35,870 that the entire change is just an epsilon dot 1145 01:16:35,870 --> 01:16:41,540 D over MQ times QV. 1146 01:16:41,540 --> 01:16:44,420 And if you look at this, it has to cancel against something 1147 01:16:44,420 --> 01:16:46,856 that's order 1 over MQ. 1148 01:16:46,856 --> 01:16:50,270 And if you look at the terms that we had at order 1, 1149 01:16:50,270 --> 01:16:53,120 which are 1 over MQ, there was a kinetic piece. 1150 01:16:53,120 --> 01:16:56,670 We called it kinetic energy piece. 1151 01:16:56,670 --> 01:16:58,430 And if we do the change there, there 1152 01:16:58,430 --> 01:17:01,880 is a contribution from the phase in this case, 1153 01:17:01,880 --> 01:17:06,230 because we had transverse derivatives. 1154 01:17:06,230 --> 01:17:07,963 So we can add epsilon dot D transverse. 1155 01:17:07,963 --> 01:17:08,630 That's non-zero. 1156 01:17:15,005 --> 01:17:16,630 And if you go through the leading order 1157 01:17:16,630 --> 01:17:18,940 change to this guy, as well as the guy that's 1158 01:17:18,940 --> 01:17:24,580 the magnetic guy, you find that the magnetic guy 1159 01:17:24,580 --> 01:17:28,756 is 0 at this order. 1160 01:17:28,756 --> 01:17:32,320 It's non-zero at higher orders, but at this order, it's zero. 1161 01:17:32,320 --> 01:17:35,170 And the kinetic guy does have a transformation. 1162 01:17:35,170 --> 01:17:39,040 It has exactly the same form as this guy here. 1163 01:17:39,040 --> 01:17:43,340 And if epsilon's dotted into the D, then it's a D transverse. 1164 01:17:43,340 --> 01:17:45,080 But this guy has a Wilson coefficient. 1165 01:17:45,080 --> 01:17:46,715 This guy doesn't. 1166 01:17:46,715 --> 01:17:48,590 So in order for these to cancel, you actually 1167 01:17:48,590 --> 01:17:51,060 learn something non-trivial. 1168 01:17:51,060 --> 01:17:53,060 The symmetry teaches you something non-trivial 1169 01:17:53,060 --> 01:17:54,950 about the sub-leading Lagrangian. 1170 01:17:54,950 --> 01:17:56,330 That Wilson coefficient has to be 1171 01:17:56,330 --> 01:18:02,040 1 to [INAUDIBLE] perturbation theory in order 1172 01:18:02,040 --> 01:18:03,540 for the symmetry not to be violated. 1173 01:18:09,700 --> 01:18:12,300 AUDIENCE: But you are enforcing the symmetry? 1174 01:18:12,300 --> 01:18:13,350 PROFESSOR: Yeah. 1175 01:18:13,350 --> 01:18:15,150 So the symmetry is boost invariance, 1176 01:18:15,150 --> 01:18:17,850 and it seems like a reasonable symmetry to impose. 1177 01:18:17,850 --> 01:18:18,350 Yeah. 1178 01:18:25,550 --> 01:18:26,850 It'd be small boosts. 1179 01:18:43,470 --> 01:18:46,170 So as long as your scheme and your regulator 1180 01:18:46,170 --> 01:18:49,866 don't break the symmetry, which is always something 1181 01:18:49,866 --> 01:18:54,900 that you have to worry about in general, 1182 01:18:54,900 --> 01:18:58,640 then this guy is 1 to all orders. 1183 01:18:58,640 --> 01:18:59,140 OK? 1184 01:18:59,140 --> 01:19:00,960 So if you did something like dimensional regularization 1185 01:19:00,960 --> 01:19:02,490 and you thought you should calculate this guy, 1186 01:19:02,490 --> 01:19:04,050 you'd just find that it would be 1. 1187 01:19:04,050 --> 01:19:06,290 And you'd wonder, well, why is that? 1188 01:19:06,290 --> 01:19:09,694 It's the symmetry that tells you it's 1. 1189 01:19:09,694 --> 01:19:12,270 OK, so you don't have to figure out that guy. 1190 01:19:12,270 --> 01:19:14,760 The other guy, which we've called Cf, 1191 01:19:14,760 --> 01:19:18,000 you do have to figure out, because it wasn't constrained. 1192 01:19:18,000 --> 01:19:27,060 And at lowest order, the other coefficient 1193 01:19:27,060 --> 01:19:29,920 is not constrained in this way. 1194 01:19:29,920 --> 01:19:33,030 And so it does get an anomalous dimension. 1195 01:19:33,030 --> 01:19:36,420 And so we could calculate it. 1196 01:19:36,420 --> 01:19:38,280 It's a good homework problem. 1197 01:19:38,280 --> 01:19:40,140 You may see it on a future homework set. 1198 01:19:49,860 --> 01:19:51,360 And there is an anomalous dimension, 1199 01:19:51,360 --> 01:19:52,920 and when you solve that anomalous dimension, 1200 01:19:52,920 --> 01:19:54,462 you're again getting something that's 1201 01:19:54,462 --> 01:19:56,190 the ratio of alphas to some power. 1202 01:19:56,190 --> 01:19:59,475 In this case, it's a non Abelian power, so the adjoint Casimir. 1203 01:20:06,522 --> 01:20:10,140 OK, so that guy does have an anomalous dimension. 1204 01:20:10,140 --> 01:20:12,080 I think we'll stop there today. 1205 01:20:12,080 --> 01:20:15,260 And we'll talk more about power corrections 1206 01:20:15,260 --> 01:20:17,960 and the phenomenology of them, how 1207 01:20:17,960 --> 01:20:20,990 we can make non-trivial predictions of from them 1208 01:20:20,990 --> 01:20:22,840 next time.