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202 November 7, 2013 7.4.4 The charged Klein-Gordon equation Just like the case of a Dirac particle in an e.m. field, that of a charged scalar in such a field allows us to write down a tree-level SDe for the scalar field, based on 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 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 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 (7.113) or, by explicitly use of the Fourier transforms of the fields : ∫ ∫ φ(x) = d 4 1 y (2π) 4 d 4 k e −ik·(x−y) i¯h k 2 − m 2 + iɛ (( i Q¯h ) ∫ 1 (2π) 8 d 4 pd 4 qe −ip·y−iq·y (p + q) µ A µ (q) φ(p) + 1 ) ) (2i Q2 A µ (y)A µ (y)φ(y) . (7.114) 2 ¯h Note the occurence of the symmetry factor 1/2 in the last line. We can therefore arrive at the following classical field equation, where we have used the Lorenz condition ∂ · A = 0 : ( − ∂ 2 − m 2) φ(x) = −iQA µ (x)∂ µ φ(x) − Q 2 A µ (x)A µ (x)φ(x) , (7.115) or ((i∂ + QA(x)) 2 − m 2 ) φ(x) = 0 . (7.116) This is de Klein-Gordon equation for charged scalar fields. We see that the same ‘minimal substitution rule’ p µ → p µ + QA µ as in the Dirac case is employed to account for the presence of the e.m. field ; and we see that the charge coupling constant Q is defined in the same way for both scalar and Dirac particles. 7.5 The Landau-Yang theorem 7.5.1 The photon polarisation revisited As stated above, any good amplitude for processes in which a photon is absorbed or produced must vanish under the handlebar operation. That means that, provided the amplitude is acceptable, we may add to any photon polarisation a piece of photon momentum. Let us consider a process with several photons present, with momenta q µ i and polarisation vectors ɛ µ i . We have, obviously, (q i · q i ) = (q i · ɛ i ) = 0 and (ɛ i · ɛ i ) = −1. From the above, we see that, if we wish, we may employ instead of ɛ i the more complicated object η i µ = ɛ i µ − (p · ɛ i) (p · q i ) q i µ , (7.117)
November 7, 2013 203 where p is any vector not proportional to q i . This has the properties (η i · q i ) = (η i · p) = 0 , η i 2 = −1 . (7.118) In numerous applications, η is actually more profitable to use than ɛ. But we should note that, in any amplitude described by more than one Feynman diagram, the shift from ɛ to η simply means that parts of some Feynman diagrams are ‘transferred’ to other diagrams : the total result must, of course, be the same. The most important difference between ɛ and η is in the handlebar, since η then vanishes : η i ⌋ ɛi→q i = 0 ; (7.119) therefore, any expression written in terms of η’s vanishes automatically under the handlebar. On the one hand this is, of course, nice ; on the other hand, it deprives us of a powerful check on the correctness of our diagrams, since almost any mistake made in writing them out will show up as a failure under the handlebar. 7.5.2 The Landau-Yang result Although this may seem to fall somewhat outside the province of QED, we can consider the decay of a spin-1 particle into photons. But even within QED this can be envisaged, since we may have a bound state of electron and positron (positronium) that may, of course, have some angular momentum. Such a positronium state can, unless we look really closely, be considered a single particle. In its ground state, positronium comes in two varieties : para-positronium in which the electron and positron’s spin are antiparallel and hence form total spin zero, and ortho-positronium in which the spins are parallel, leading to a total spin of one. Without knowing anything much about the bound-state structure of positronium, let us consider the amplitude for its decay into a pair of photons. Let us denote by P µ the positronium momentum (in its rest frame), and by q 1,2 and ɛ 1,2 the photon momenta and polarizations. We shall define q µ = (q µ 1 − qµ 2 )/2. In addition, the positronium being a spin-1 particle, we need its polarisation vector ɛ 0 . Any amplitude for the decay must necessarily be linear in ɛ 0 , ɛ 1 and ɛ 2 ; and to have current conservation we can, rather, take η 1,2 instead of ɛ 1,2 , where here η i = ɛ i − (P · η i )/(P · q i )q i . Since also (P · ɛ 0 ) = 0, the three polarisations (as well as the vector q) have no timelike component. Noting that, in this case, (q · η 1,2 ) = 0 as well, we see that to build an amplitude M we actually have but a very few structures that we can use 24 : M = A 1 (q · ɛ 0 )(η 1 · η 2 ) + A 2 ε(P, ɛ 0 , η 1 , η 2 ) + A 3 ε(P, q, η 1 , η 2 )(q · ɛ 0 ) . (7.120) The coefficients A are of course undetermined, but they can only depend on P 2 , q 2 and (P · q). This last product is zero, and P 2 = −4q 2 = M 2 where M is the 24 You might be tempted to write down a term like ε(q, ɛ 0 , η 1 , η 2 ) but since all these vectors have vanishing zeroth component, this Levi-Civita product is simply zero.
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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
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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
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66 November 7, 2013 The easiest way
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68 November 7, 2013 are deemed to b
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70 November 7, 2013 To check that t
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72 November 7, 2013 Upon ‘discret
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74 November 7, 2013 − λ 4 6 ( φ
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76 November 7, 2013 Here we plot a
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78 November 7, 2013 The continuum f
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80 November 7, 2013 For large m|⃗
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82 November 7, 2013 there is a law
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84 November 7, 2013 k ↔ ¯h | ⃗
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86 November 7, 2013 the integral is
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88 November 7, 2013 by (x − y) 2
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90 November 7, 2013 3.2.4 The iɛ p
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92 November 7, 2013 Im k 4 Re k 4 I
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94 November 7, 2013 The response of
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96 November 7, 2013 = 1 (2π) 3 ∫
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98 November 7, 2013 we find that th
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100 November 7, 2013 x x B B E > 0
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102 November 7, 2013 in which avail
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104 November 7, 2013 the time-evolu
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106 November 7, 2013 Now, the secon
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108 November 7, 2013 Note that the
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110 November 7, 2013 the story the
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112 November 7, 2013 The n-particle
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114 November 7, 2013 interactions.
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116 November 7, 2013 0000000 111111
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000000000 111111111 000000000 11111
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120 November 7, 2013 4.5.2 Two-body
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122 November 7, 2013 which, to lowe
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124 November 7, 2013
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126 November 7, 2013 Therefore, T (
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128 November 7, 2013 where Q is som
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130 November 7, 2013 (P ) coefficie
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132 November 7, 2013 Obviously, the
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134 November 7, 2013 Now, we also k
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136 November 7, 2013 p 2 = m 2 we c
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138 November 7, 2013 5.3.3 The Dira
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140 November 7, 2013 now forced by
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142 November 7, 2013 under Lorentz
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144 November 7, 2013 = 1 ( 1 + 1 (
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146 November 7, 2013 Dirac propagat
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148 November 7, 2013 The first diag
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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
Page 166 and 167: 166 November 7, 2013 operator is th
Page 168 and 169: 168 November 7, 2013 However, a pro
Page 170 and 171: 170 November 7, 2013 transverse one
Page 172 and 173: 172 November 7, 2013 in contrast to
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Page 176 and 177: 176 November 7, 2013 µ ↔ iQ¯h
Page 178 and 179: 178 November 7, 2013 the correspond
Page 180 and 181: 180 November 7, 2013 The handlebar
Page 182 and 183: 182 November 7, 2013 The Dirac equa
Page 184 and 185: 184 November 7, 2013 7.3 Some QED p
Page 186 and 187: 186 November 7, 2013 The cross sect
Page 188 and 189: 188 November 7, 2013 We therefore h
Page 190 and 191: 190 November 7, 2013 so that up to
Page 192 and 193: 192 November 7, 2013 The radiative
Page 194 and 195: 194 November 7, 2013 Hard Bremsstra
Page 196 and 197: 196 November 7, 2013 Double-pole te
Page 198 and 199: 198 November 7, 2013 with constants
Page 200 and 201: 200 November 7, 2013 with the trivi
Page 204 and 205: 204 November 7, 2013 positronium ma
Page 206 and 207: 206 November 7, 2013 We now postula
Page 208 and 209: 208 November 7, 2013 For the QED di
Page 210 and 211: 210 November 7, 2013 8.3 The three-
Page 212 and 213: 212 November 7, 2013 be renormaliza
Page 214 and 215: 214 November 7, 2013 Also, in the a
Page 216 and 217: 216 November 7, 2013 8.4 Four-gluon
Page 218 and 219: 218 November 7, 2013 so that A 34 t
Page 220 and 221: 220 November 7, 2013 We can therefo
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Page 224 and 225: 224 November 7, 2013 However, from
Page 226 and 227: 226 November 7, 2013 If we were all
Page 228 and 229: 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
Page 238 and 239: 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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Page 246 and 247: 246 November 7, 2013 Note that the
Page 248 and 249: 248 November 7, 2013 following type
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252 November 7, 2013
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254 November 7, 2013 constant λ 4
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256 November 7, 2013 term is minima
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258 November 7, 2013 10.2 More on s
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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
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280 November 7, 2013 The continuum
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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
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298 November 7, 2013 |2, 1〉 = |+0
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300 November 7, 2013 − 1 4 (∆
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302 November 7, 2013 our candidate
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304 November 7, 2013 × ∆ µα∆
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306 November 7, 2013 where m and m
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308 November 7, 2013 cross section
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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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