Pictures Paths Particles Processes

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48 November 7, 2013 diagrams. With more external legs attached, this becomes ever so much more complicated : assigning a special rôle to one external line avoids this. In the last place, the above calculation is possible since all external lines are, so to speak, identical. In more dimensions, where external lines can carry momentum, this is no longer true. However, the effective potential, that is the effective action at zero momentum, does lend itself to such a calculation in higher dimensions 30 . We can extend this treatment to higher loop orders as well. Let us denote a vertex where at least n + 1 lines come together by n = n + n+1 + n+2 + · · · (1.99) and assign to this dressed vertex the Feynman rule n = − 1¯h S(n+1) (z) . (1.100) Now, we introduce the notion of a tadpole diagram : this is a connected diagram with precisely one external line and no source vertices. The effective action as given above then follows from writing out the 1PI tadpole diagrams, replacing propagators by dressed propagators and vertices by dressed vertices ; we can then simply read off the result. → 000 111 000 111 000 111 000 111 000 111 = − S(3) (z) 2S (2) (z) ⇒ Γ ′ 1(φ) = S(3) (φ) 2S (2) (φ) , (1.101) as before. In two loops, the 1PI tadpole is given by the diagrams + + + Dressing these tadpole diagrams gives us = ¯h 00000 11111 00000 11111 00000 11111 00 11 00 11 00 11 000000 111111 000000 111111 + 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 + 0000 1111 000 111 0000 1111 000 111 000 111 000 111 000 111 000 111 + 000 111 0000 1111 0000 1111 0000 1111 000 111 000 111 { S (3) (z) S (4) (z) − S(3) (z) 3 6 S (2) (z) 3 4 S (2) (z) + S(3) (z) S (4) } (z) − S(5) (z) 4 4 S (2) (z) 3 8 S (2) (z) 2 The two-loop contribution to the effective action is therefore (1.102) d dφ Γ 2(φ) = S(5) (φ) 8 S (2) (φ) − 5 S(3) (φ) S (4) (φ) + S(3) (φ) 3 2 12 S (2) (φ) 4 4 S (2) (φ) . (1.103) 4 30 For simple scalar theories. Of course external lines may carry more than just momentum information, that is, they can also carry spin/charge/colour· · · information. Then the calculation is again more difficult.
November 7, 2013 49 The effective action itself, the integral over the above experession, has no nice simple form as in Eq.(1.98), but is of course calculable as soon as S(φ) is explicitly given ; moreover, we see that it will becomes undefined where S ′′ (φ) vanishes. From our diagrammatic approach we see that this will persist in all loop orders 31 . 1.5.4 More fields So far, our main attention has been on theories of a single field. Suppose, for the sake of argument, that we have a theory of two fields instead: S(ϕ 1 , ϕ 2 ) = 1 2 µ 1ϕ 1 2 + 1 2 µ 2ϕ 2 2 + λ 4 ϕ 1 2 ϕ 2 2 . (1.104) This time, the coupling constant λ carries a factor 1/(2!)/(2!) since there are not four identical fields ‘meeting’ at the vertex, but rather two pairs of identical fields. We now need to distinguish between the two different fields, so we indicate the field type with either ‘1’ or ‘2’. The Feynman rules for this case are 1 1 ↔ ¯h µ 1 , ↔ J 1 ¯h , 2 2 ↔ ¯h µ 2 , 1 2 −λ ↔ 1 2 ¯h ↔ J 2 ¯h . (1.105) There are two coupled Schwinger-Dyson equations, one for each field : , 1 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 = 1 1 + 2 2 1 1 + 2 2 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 00000 11111 0000 1111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 2 + 1 2 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 + 1 000000 1 000000 111111 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 2 2 , 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 2 = 2 + 2 2 + 1 1 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 2 2 1 1 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 0000 1111 00000 11111 0000 1111 00000 11111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 2 + 1 1 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 2 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 + 2 000000 1 000000 111111 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 1 2 , (1.106) 31 Because in all 1PI diagrams we have to dress the propagators, which implies lots of S ′′ (φ) in the denominators.
Page 1: Pictures Paths Particles Processes
Page 4 and 5: 4 CONTENTS 1.5.1 The effective acti
Page 6 and 7: 6 CONTENTS 5.3.7 Massless Dirac par
Page 8 and 9: 8 CONTENTS 9.3.3 W, Z and γ four-p
Page 10 and 11: 10 November 7, 2013
Page 12 and 13: 12 November 7, 2013 the like. The m
Page 14 and 15: 14 November 7, 2013 cay widths. The
Page 16 and 17: 16 November 7, 2013 scenario of a s
Page 18 and 19: 18 November 7, 2013 which yields th
Page 20 and 21: 20 November 7, 2013
Page 22 and 23: 22 November 7, 2013 The function S(
Page 24 and 25: 24 November 7, 2013 For a single-fi
Page 26 and 27: 26 November 7, 2013 Computing the p
Page 28 and 29: 28 November 7, 2013 Using the serie
Page 30 and 31: 30 November 7, 2013 multiplied. The
Page 32 and 33: 32 November 7, 2013 carries a symme
Page 34 and 35: 34 November 7, 2013 therefore also
Page 36 and 37: 000000 111111 000000 111111 000000
Page 38 and 39: 38 November 7, 2013 1.4 Planck’s
Page 40 and 41: 0000000 1111111 0000000 1111111 000
Page 42 and 43: 42 November 7, 2013 1.4.3 The class
Page 44 and 45: 44 November 7, 2013 Such subdominan
Page 46 and 47: 000000 111111 000000 111111 000000
Page 50 and 51: 50 November 7, 2013 with the follow
Page 52 and 53: 52 November 7, 2013 1.6 Renormaliza
Page 54 and 55: 54 November 7, 2013 C naive 2 C nai
Page 56 and 57: 00000 11111 00000 11111 00000 11111
Page 58 and 59: 58 November 7, 2013 We may formulat
Page 60 and 61: 60 November 7, 2013 Using Eq.(1.136
Page 62 and 63: 62 November 7, 2013 action has been
Page 64 and 65: 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.
Page 116 and 117:
116 November 7, 2013 0000000 111111
Page 118 and 119:
000000000 111111111 000000000 11111
Page 120 and 121:
120 November 7, 2013 4.5.2 Two-body
Page 122 and 123:
122 November 7, 2013 which, to lowe
Page 124 and 125:
124 November 7, 2013
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
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
Page 174 and 175:
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 202 and 203:
202 November 7, 2013 7.4.4 The char
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
Page 222 and 223:
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
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
Page 244 and 245:
244 November 7, 2013 remain unaffec
Page 246 and 247:
246 November 7, 2013 Note that the
Page 248 and 249:
248 November 7, 2013 following type
Page 250 and 251:
250 November 7, 2013 + B 1 (1, 2, 5
Page 252 and 253:
Page 254 and 255:
254 November 7, 2013 constant λ 4
Page 256 and 257:
256 November 7, 2013 term is minima
Page 258 and 259:
258 November 7, 2013 10.2 More on s
Page 260 and 261:
260 November 7, 2013 The effective
Page 262 and 263:
262 November 7, 2013 10.4 Alternati
Page 264 and 265:
264 November 7, 2013 there are thre
Page 266 and 267:
266 November 7, 2013 10.5 Concavity
Page 268 and 269:
268 November 7, 2013 10.6 Diagram c
Page 270 and 271:
270 November 7, 2013 for Φ. If the
Page 272 and 273:
272 November 7, 2013 10.6.2 Countin
Page 274 and 275:
274 November 7, 2013 where the stra
Page 276 and 277:
276 November 7, 2013 In the table w
Page 278 and 279:
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
Page 284 and 285:
284 November 7, 2013 10.9 The funda
Page 286 and 287:
286 November 7, 2013 For this matri
Page 288 and 289:
288 November 7, 2013 where S, p µ
Page 290 and 291:
290 November 7, 2013 Second irregul
Page 292 and 293:
292 November 7, 2013 The next equiv
Page 294 and 295:
294 November 7, 2013 Since the spin
Page 296 and 297:
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,
Page 312 and 313:
312 November 7, 2013 which may help
Page 314 and 315:
314 November 7, 2013 Let now form t
Page 316:
316 November 7, 2013 and by inspect
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