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138 November 7, 2013 5.3.3 The Dirac propagator, and a convention We have now arrived at a possible choice for the Dirac propagator. Since the two spin states described by uu should propagate in the same manner 21 , we shall use the projection operator T (p) = /p + m , (5.72) and adopt this choice also off the mass shell (where it is actually used). The Dirac propagator therefore takes the form i¯h /p + m p 2 − m 2 + iɛ . The fact that the numerator is linear in p means that the propagator is oriented, in contrast to what we have used so far. To indicate this we define the orientation with an arrow, and adhere to the convention that the momentum is counted in the direction of the arrow, irrespective of the sign of the energy component. The first Dirac Feynman rule therefore becomes k ↔ i¯h /k + m k · k − m 2 + iɛ internal lines Feynman rules, version 5.1 (5.73) In writing out Feynman diagrams containing Dirac particles, we of course have to keep track of the Dirac indices resident in propagator and vertices. This may lead to incredibly cumbersome notation, that may however be greatly simplified if we adopt the following writing convention : write out the Dirac-index carrying factors in order, moving against the orientation of the line. Then, all these factors are contracted together using the usual rules for matrix multiplication, and one hardly ever needs to write the Dirac indices explicitly. This convention is really to be urged on anyone contemplating any calculation involving Dirac particles 22 ! A final word on notation : since ( /p + m ) ( /p − m ) = p 2 − m 2 , (5.74) the Dirac propagator might be written as /p + m i¯h p 2 − m 2 + iɛ = i¯h /p − m + iɛ . (5.75) In instances were the iɛ can be neglected, this is certainly allowed ; however in more delicate situations (such as inside loops) the first alternative is probably to be preferred. Nevertheless we shall occasionally also use Eq.(5.75). 21 Otherwise we would not consider them to be states of the same particle 22 Try it out for yourself ; after at most ten minutes you will be convinced.
November 7, 2013 139 5.3.4 Truncating Dirac particles : external Dirac lines Let us now return to the truncation argument that gave us the Feynman rule for external lines in chapter 4. We shall redo this for Dirac particles moving between production and decay. As a first case, let the ‘p’-line connecting production and decay be oriented from production to decay, as indicated in the following diagram : } p p a b 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 A 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 k 1,2,... p B 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 } q 1,2,... According to the convention described above we then have for the amplitude i¯h(/p + m) M = [B] p 2 − m 2 [A] . (5.76) + imΓ Note that, in this amplitude, the factor [A] must carry the upper Dirac index of a spinor, and [B] the lower index of a conjugate spinor. p µ , obviously, carries positive energy. As we let Γ vanish and p µ approaches the mass shell, we may then write /p + m = ∑ u(p, s) u(p, s) , (5.77) s where the sum over s runs over two values, s µ and −s µ . Following the truncation argument, we readily see that the spinor u(p, s) must then be included in the decay amplitude, and u(p, s) in the production amplitude. In the alternative case, where the line is oriented against the flow of energy, the amplitude is given by } p p a b 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 A 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 k 1,2,... p B 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 } q 1,2,... and reads (again with our convention !) i¯h(−/p + m) M = [A] p 2 − m 2 [B] . (5.78) + imΓ Note that it is now [A] that is the conjugate spinor, and [B] the regular one. Of course, they describe a physical process different from the first case ! We are
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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
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.
Page 116 and 117: 116 November 7, 2013 0000000 111111
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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 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
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
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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
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220 November 7, 2013 We can therefo
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222 November 7, 2013 by p Q q k 1 k
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224 November 7, 2013 However, from
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226 November 7, 2013 If we were all
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228 November 7, 2013 with Z α as i
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230 November 7, 2013 ¯hQ U Q W M 2
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232 November 7, 2013 the process UU
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234 November 7, 2013 we have pretty
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236 November 7, 2013 where X µνα
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238 November 7, 2013 9.4 The Higgs
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240 November 7, 2013 in any dynamic
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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
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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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