Pictures Paths Particles Processes

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116 November 7, 2013 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 = ⎛ ⎜ ⎝ 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 ⎞ ⎟ ⎠ ∗ If the diagram contains oriented lines, the time-reversal also inverts the orientation of those lines (if the orientation is indicated by an arrow, we reverse the arrow). We can write Eq.(4.26) diagrammatically as i f + 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 i 000000 111111 f + ∑ 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 k 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 i k k f = 0 . (4.34) It is possible to sharpen this equation to make it more useful. In the first place, Eq.(4.34) holds for whole matrix elements, evaluated to all orders in perturbation theory. This implies that it must also hold for each order separately 15 . However, even at some fixed order, M fi can contain very many diagrams. Consider a somewhat-complicated Feynman diagram in ϕ 3 theory : The corresponding Lagrangian reads (4.35) L = 1 2 (∂µ ϕ)(∂ µ ϕ) − 1 2 m2 ϕ 2 − 1 6 λϕ3 . (4.36) The unitarity structure of the above Feynman diagram is not immediately obvious since there are, at this order of perturbation theory, quite a few diagrams that contribute to this amplitude (58, in fact). We can, however, employ the following trick. Let us assign a different label to each line in the diagram, in an arbitrary manner, for instance 1 2 3 4 9 5 7 8 6 (4.37) and let us pretend that each line corresponds to a different field. This diagram can then be interpreted as coming from a theory with 9 distinct fields (with 15 If Eq.(4.34) were not to hold order-by-order, this would imply subtle relations between coupling constants, ¯h, and the like. We would then be in a position to actually compute coupling constants from first principles, which would be good — too good to be true, in fact.
November 7, 2013 117 identical mass) and Lagrangian L = 9∑ n=1 ( 1 2 (∂µ ϕ n )(∂ µ ϕ n ) − 1 2 m2 ϕ 2 n) − V , V = λ 123 ϕ 1 ϕ 2 ϕ 3 + λ 245 ϕ 2 ϕ 4 ϕ 5 + λ 349 ϕ 3 ϕ 4 ϕ 9 +λ 567 ϕ 5 ϕ 6 ϕ 7 + λ 789 ϕ 7 ϕ 8 ϕ 9 . (4.38) Nothing forbids us to assign to the various ϕϕϕ couplings precisely the value λ. Now, it is easily seen that, in order λ 123 λ 245 λ 349 λ 567 λ 789 , the diagram (4.37) is the only diagram that can contribute in this theory 16 ! We can do even more : by inspection of all possibilities, we can simply realize that the only final states k in the unitarity condition (4.34) must be precisely k = {2, 3}, {5, 9}, {2, 4, 9} or {3, 4, 5}, if we want to end up with the right order in perturbation theory 17 . In other words, + + + + + = 0 , (4.39) where we have omitted the line labellings : indeed, the same identity must hold for the original diagram (4.35)! This establishes the so-called cutting rules (also called the Cutkoski rules), which can be most simply expressed in words: take a diagram and move the cutting line through it from right to left in all possible manners, making sure that the two halves in which the diagram is cut remain connected and that neither the inital state or the final state is dissected. The particles described by internal lines through which the cut runs must be assumed to be on their mass shell 18 . The sum of all the possible contributions then vanishes 19 . 4.4.4 Infrared cancellations in QED As an illustration of how the cutting rules may be applied we shall make a slight jump ahead and consider quantum electrodynamics, that is the theory 16 The secret resides in the fact that in V the external fields 1,6 and 8 occur precisely once, and the other fields precisely twice. 17 Note that, for instance, the choice k = {5, 7, 8} would result in the right-hand half of the diagram being disconnected ; the choice k = {2, 4, 7} is inconsistent since both 6 and 8 are in the final state. 18 This may mean that the situation thus described fails to meet the restrictions of momentum/energy conservation ; then, that contribution vanishes. 19 You might object that in a theory with many different fields the symmetry factors of the diagrams will, in general, be different from those of a theory with only a single field, and this is true : however, in the summation over the ‘intermediate states’ k we must of course also include the ‘indentical-particle’ symmetry factor F symm, which precisely repairs the correspondence — another illustration of the crucial rôle of the symmetry factors !
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Pictures Paths Particles Processes
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4 CONTENTS 1.5.1 The effective acti
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6 CONTENTS 5.3.7 Massless Dirac par
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8 CONTENTS 9.3.3 W, Z and γ four-p
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10 November 7, 2013
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12 November 7, 2013 the like. The m
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14 November 7, 2013 cay widths. The
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16 November 7, 2013 scenario of a s
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18 November 7, 2013 which yields th
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20 November 7, 2013
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22 November 7, 2013 The function S(
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24 November 7, 2013 For a single-fi
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26 November 7, 2013 Computing the p
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28 November 7, 2013 Using the serie
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30 November 7, 2013 multiplied. The
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32 November 7, 2013 carries a symme
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34 November 7, 2013 therefore also
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000000 111111 000000 111111 000000
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38 November 7, 2013 1.4 Planck’s
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0000000 1111111 0000000 1111111 000
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42 November 7, 2013 1.4.3 The class
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44 November 7, 2013 Such subdominan
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000000 111111 000000 111111 000000
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48 November 7, 2013 diagrams. With
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50 November 7, 2013 with the follow
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52 November 7, 2013 1.6 Renormaliza
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54 November 7, 2013 C naive 2 C nai
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00000 11111 00000 11111 00000 11111
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58 November 7, 2013 We may formulat
Page 60 and 61:
60 November 7, 2013 Using Eq.(1.136
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62 November 7, 2013 action has been
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64 November 7, 2013 zero-dimensiona
Page 66 and 67: 66 November 7, 2013 The easiest way
Page 68 and 69: 68 November 7, 2013 are deemed to b
Page 70 and 71: 70 November 7, 2013 To check that t
Page 72 and 73: 72 November 7, 2013 Upon ‘discret
Page 74 and 75: 74 November 7, 2013 − λ 4 6 ( φ
Page 76 and 77: 76 November 7, 2013 Here we plot a
Page 78 and 79: 78 November 7, 2013 The continuum f
Page 80 and 81: 80 November 7, 2013 For large m|⃗
Page 82 and 83: 82 November 7, 2013 there is a law
Page 84 and 85: 84 November 7, 2013 k ↔ ¯h | ⃗
Page 86 and 87: 86 November 7, 2013 the integral is
Page 88 and 89: 88 November 7, 2013 by (x − y) 2
Page 90 and 91: 90 November 7, 2013 3.2.4 The iɛ p
Page 92 and 93: 92 November 7, 2013 Im k 4 Re k 4 I
Page 94 and 95: 94 November 7, 2013 The response of
Page 96 and 97: 96 November 7, 2013 = 1 (2π) 3 ∫
Page 98 and 99: 98 November 7, 2013 we find that th
Page 100 and 101: 100 November 7, 2013 x x B B E > 0
Page 102 and 103: 102 November 7, 2013 in which avail
Page 104 and 105: 104 November 7, 2013 the time-evolu
Page 106 and 107: 106 November 7, 2013 Now, the secon
Page 108 and 109: 108 November 7, 2013 Note that the
Page 110 and 111: 110 November 7, 2013 the story the
Page 112 and 113: 112 November 7, 2013 The n-particle
Page 114 and 115: 114 November 7, 2013 interactions.
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Page 120 and 121: 120 November 7, 2013 4.5.2 Two-body
Page 122 and 123: 122 November 7, 2013 which, to lowe
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Page 126 and 127: 126 November 7, 2013 Therefore, T (
Page 128 and 129: 128 November 7, 2013 where Q is som
Page 130 and 131: 130 November 7, 2013 (P ) coefficie
Page 132 and 133: 132 November 7, 2013 Obviously, the
Page 134 and 135: 134 November 7, 2013 Now, we also k
Page 136 and 137: 136 November 7, 2013 p 2 = m 2 we c
Page 138 and 139: 138 November 7, 2013 5.3.3 The Dira
Page 140 and 141: 140 November 7, 2013 now forced by
Page 142 and 143: 142 November 7, 2013 under Lorentz
Page 144 and 145: 144 November 7, 2013 = 1 ( 1 + 1 (
Page 146 and 147: 146 November 7, 2013 Dirac propagat
Page 148 and 149: 148 November 7, 2013 The first diag
Page 150 and 151: 150 November 7, 2013 5.5 The Dirac
Page 152 and 153: 152 November 7, 2013 gives us preci
Page 154 and 155: 154 November 7, 2013 this gives the
Page 156 and 157: 156 November 7, 2013 and is picture
Page 158 and 159: 158 November 7, 2013 5.7.3 The muon
Page 160 and 161: 160 November 7, 2013 µ p ν ↔ i
Page 162 and 163: 162 November 7, 2013 (T z ) µ ν =
Page 164 and 165: 164 November 7, 2013 The SDe for a
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166 November 7, 2013 operator is th
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168 November 7, 2013 However, a pro
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170 November 7, 2013 transverse one
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172 November 7, 2013 in contrast to
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174 November 7, 2013
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176 November 7, 2013 µ ↔ iQ¯h
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178 November 7, 2013 the correspond
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180 November 7, 2013 The handlebar
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182 November 7, 2013 The Dirac equa
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184 November 7, 2013 7.3 Some QED p
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186 November 7, 2013 The cross sect
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188 November 7, 2013 We therefore h
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190 November 7, 2013 so that up to
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192 November 7, 2013 The radiative
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194 November 7, 2013 Hard Bremsstra
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196 November 7, 2013 Double-pole te
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198 November 7, 2013 with constants
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200 November 7, 2013 with the trivi
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202 November 7, 2013 7.4.4 The char
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204 November 7, 2013 positronium ma
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206 November 7, 2013 We now postula
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208 November 7, 2013 For the QED di
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210 November 7, 2013 8.3 The three-
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212 November 7, 2013 be renormaliza
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214 November 7, 2013 Also, in the a
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216 November 7, 2013 8.4 Four-gluon
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218 November 7, 2013 so that A 34 t
Page 220 and 221:
220 November 7, 2013 We can therefo
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222 November 7, 2013 by p Q q k 1 k
Page 224 and 225:
224 November 7, 2013 However, from
Page 226 and 227:
226 November 7, 2013 If we were all
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228 November 7, 2013 with Z α as i
Page 230 and 231:
230 November 7, 2013 ¯hQ U Q W M 2
Page 232 and 233:
232 November 7, 2013 the process UU
Page 234 and 235:
234 November 7, 2013 we have pretty
Page 236 and 237:
236 November 7, 2013 where X µνα
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238 November 7, 2013 9.4 The Higgs
Page 240 and 241:
240 November 7, 2013 in any dynamic
Page 242 and 243:
242 November 7, 2013 with the contr
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244 November 7, 2013 remain unaffec
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246 November 7, 2013 Note that the
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248 November 7, 2013 following type
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250 November 7, 2013 + B 1 (1, 2, 5
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252 November 7, 2013
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254 November 7, 2013 constant λ 4
Page 256 and 257:
256 November 7, 2013 term is minima
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258 November 7, 2013 10.2 More on s
Page 260 and 261:
260 November 7, 2013 The effective
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262 November 7, 2013 10.4 Alternati
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264 November 7, 2013 there are thre
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266 November 7, 2013 10.5 Concavity
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268 November 7, 2013 10.6 Diagram c
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270 November 7, 2013 for Φ. If the
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272 November 7, 2013 10.6.2 Countin
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274 November 7, 2013 where the stra
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276 November 7, 2013 In the table w
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278 November 7, 2013 resulting cont
Page 280 and 281:
280 November 7, 2013 The continuum
Page 282 and 283:
282 November 7, 2013 in the spirit
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284 November 7, 2013 10.9 The funda
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286 November 7, 2013 For this matri
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288 November 7, 2013 where S, p µ
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290 November 7, 2013 Second irregul
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292 November 7, 2013 The next equiv
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294 November 7, 2013 Since the spin
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296 November 7, 2013 the operator f
Page 298 and 299:
298 November 7, 2013 |2, 1〉 = |+0
Page 300 and 301:
300 November 7, 2013 − 1 4 (∆
Page 302 and 303:
302 November 7, 2013 our candidate
Page 304 and 305:
304 November 7, 2013 × ∆ µα∆
Page 306 and 307:
306 November 7, 2013 where m and m
Page 308 and 309:
308 November 7, 2013 cross section
Page 310 and 311:
310 November 7, 2013 As we can see,
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312 November 7, 2013 which may help
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314 November 7, 2013 Let now form t
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316 November 7, 2013 and by inspect
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