c# display pdf in window : Add page to pdf without acrobat SDK control project winforms azure .net UWP QFT-Schwartz5-part460

ThisbeautifulLorentzinvariantobjectiscalledtheFeynmanpropagator.
Keepinmind:
• k
0
k
2
+m
2
anymore.Itisaseparateintegrationvariable. Thepropagatingfieldcanbeoff-shell!
• Theicomesfromacontourintegral. . Wewillalways s get factors of f iintwo-pointfunctionsofreal
fields.
• Theεisjustatrickfor r representingthetimeorderingasimpleway. . Wewillalwaystakeε→0at
theend,andoftenleaveitimplicit. Youalwaysneeda a +iεfortimeorderedproducts,butisreally
justshorthandforapole-prescriptioninthecontourintegral,whichisexactlyequivalenttoadding
variousθ(t)factors.
• Forε=0this s looks justliketheclassicalGreen’s functionfor theKleinGordonequationφ=J.
That’sbecauseitisthesamething.Wearejustcomputingclassicalpropagationinareallycompli-
catedway.
7.2 Feynmanpropagator
51
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Chapter8
FeynmanRules
8.1 Time-DependentPerturbationTheory
FromtheLSZtheorem,weneedobjectslike
f|i∼0|T{φ(x
1
)
φ(x
n
)}|0
(8.1)
wheretheseφ
sarenotfreebutthefieldswhichevolveduetothefullHamiltonian.
First,weseparate,asusual,theHamiltonianintoH
0
andtheperturbationV.
H=H
0
+V
(8.2)
Wealreadyknowhowthefreefieldsevolve,duetoH
0
φ
I
(x,t)=e
iH
0
t
φ
0
(x)e
−iH
0
t
=
d
3
k
(2π)3
1
k
a
k
e
ikx
+a
k
e
−ikx
(8.3)
The subscript I I stands s for “interaction”, , since we are now in the interactionpicture. . Note e that thefree
fields aredefinedatareferencetimet=t
0
=0andω
k
=
k
2
+m2
. Sogoingtotheinteractionpictureis
easyinQFT,andparticularlynicebecausethephasefactorsbecomeLorentzInvariant.
SoreallywhatwecomputedtheFeynmanpropagatorlasttime,wewerecomputing
0|T{φ
I
(x)φ
I
(y)}|0=
d
4
k
(2π)4
1
k2−m2+iε
e
ik(x−y)
(8.4)
Nowlet’sdefine anoperatorU U whichexpressesthefullfieldφ(x,t)intermsoftheevolvingfreefield
φ
I
(x,t) where0 0 isthe reference timeat t whichthefreefieldsare equalto o the interactingfields. . We e are
relatingφtoφ
I
atthesametimet,buttheevolutionhasstartedfromtimet
0
.
Weknow
φ(x,t)=e
iHt
φ
0
(x,0)e
−iHt
=e
iHt
e
−iH
0
t
(e
iH
0
t
φ
0
(x,0)e
−iH
0
t
)e
iH
0
t
e
−iHt
(8.5)
Then
U(t,0)=e
iH
0
t
e
−iHt
(8.6)
Itfollowsthat
i∂
t
U(t,0)=eiH
0
t(−H
0
+H)e−iHt
(8.7)
=
e
iH
0
t
Ve
−iH
0
t
e
iH
0
t
e
−iHt
(8.8)
So
i∂
t
U(t,0)=V
I
(t)U(t,0)
(8.9)
where
V
I
(t)=e
iH
0
t
Ve
−iH
0
t
(8.10)
53
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All we’ve done is conjugate everything with e
iH
0
t
and we are e left t with the e Schrodinger r equation with
bunchof I’s.
Nowweneedtosolve
i∂
t
U(t,t
0
)=V
I
(t)U(t,t
0
)
(8.11)
Ifeverythingcommuted,thesolutionwouldjustbeU
I
(t,t
0
)=exp(−i
t
0
t
V
I
(t
)dt
). ButV
I
(t
1
)doesnot
necessarilycommutewithV
I
(t
2
),sothisisnottherightanswer. Itturnsouttherightansweris s verysim-
ilar
U(t,t
0
)=T
exp[−i
t
0
t
dt
V
I
(t
)]
(8.12)
whereT isthetimeorderingoperator. Thissolutionworksbecausetimeorderingeffectivelymakesevery-
thinginsidecommute:
T{AB}=T{BA}
(8.13)
Whichiswhywegettheexponentialsolution.
Timeorderingofanexponentialisdefinedintheobviouswaythroughitsexpansion
U(t,0)=1−i
0
t
dt
V
I
(t
)−
1
2
0
t
dt
0
t
dt
′′
T{V
I
(t
)V
I
(t
′′
)}+
(8.14)
8.1.1 perturbativesolutionforDyson’sseries
Ifyoudon’tfindthe previous argument compelling,wecansolvefor U(t) the hardway, usingperturba-
tiontheory.
i∂
t
U(t)=V
I
(t)U(t)
(8.15)
Atzerothorder
U(t)=1
(8.16)
TofirstorderinV
I
U(t)=1+
0
t
dt
V
I
(t
)U(t
)=1+
0
t
dt
V
I
(t
)
(8.17)
Tosecondorder
U(t)=1−
0
t
dt
V
I
(t
)
1−
0
t
dt
′′
V
I
(t
′′
)U(t
′′
)
(8.18)
=1−
0
t
dt
V
I
(t
)−
0
t
dt
0
t
dt
′′
V
I
(t
)(t
′′
)
(8.19)
Notethatt
>t
′′
sowecanwritethisas
U(t)=1−
0
t
dt(t)−
0
t
dt
0
t
dt′′T
{
V
I
(t)V
I
(t′′)
}
(8.20)
whereT meansthethingsinbracketsaretimeordered. Sothisis
U(t)=1−
0
t
dtV
I
(t)−
1
2
0
t
dt
0
t
dt′′T
{
V
I
(t)V
I
(t′′)
}
+
(8.21)
Andinfact
U(t)=T
exp[−i
0
t
dt
V
I
(t
)]
(8.22)
Moregenerally
U(t,t
0
)=T
exp[−i
t
0
t
dt
V
I
(t
)]
(8.23)
54
FeynmanRules
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8.1.2 Urelations
ItisconvenienttoabbreviateU with
U
21
≡U(t
2
,t
1
)=T
exp[−i
t
1
t
2
dtV
I
(t)]
(8.24)
RememberthatinQFTwealwayshavelatertimesontheLEFT. Itfollowsthat
U
21
U
12
=1 ⇒
U
21
−1
=U
21
=U
12
(8.25)
andfor t
1
<t
2
<t
3
U
32
U
21
=U
31
(8.26)
So,
φ(x,t)=U
(t,0)φ
I
(x,t)U(t,0)
(8.27)
φ(x,t
1
)=U
(t
1
,0)φ
I
(x,t
1
)U(t
1
,0)=U
10
φ(x,t
1
)U
10
(8.28)
=U
01
φ(x,t
1
)U
10
(8.29)
8.1.3 GroundState
Thelastthingweneedtoknow is s what the groundstateisinthe interactingtheory. . Callit |Ω. . Inthe
freetheory
|0
I
=e
iH
0
t
|0=|0
(8.30)
Then
|Ω(t)=U(t,−∞)|0=U
t−∞
|0
(8.31)
Thatis,thereferencetime,forwhichthe|Ω=|0ist=−∞.
Thuswecanevolveeverythingtoacommontimet
0
. Recalling
φ(x,t
1
)=U
01
φ(x,t
1
)U
10
(8.32)
wefind
Ω|φ(x
1
)φ(x
2
)|Ω=0|U
∞0
U
01
φ
I
(x
1
)U
10
U
02
φ
I
(x
2
)U
20
U
0−∞
|0
(8.33)
Andtime-orderedproductshaveaparticularlynicestructure,becausewecancommutethingsaroundand
combinetheU
s:
Ω|T
{
φ(x
1
)φ(x
2
)
}
|Ω
=
0|T{φ
I
(x
1
I
(x
2
)U
∞0
U
01
U
10
U
02
U
20
U
0−∞
}
|0
(8.34)
=
0|T{φ
I
(x
1
I
(x
2
)U
∞−∞
}
|0
(8.35)
=
0
T
φ
I
(x
1
I
(x
2
)exp[−i
−∞
dtV
I
(t)]

0
(8.36)
Whichisaremarkablysimplyfinalresult.
8.1.4 meaningof subscript
Wesawthatφ
I
isjustthefreefieldφ
0
withexplicittime-dependenceinthephasefactors
φ
I
(x,t)=e
iH
0
t
φ
0
(x)e
−iH
0
t
=
d
3
k
(2π)3
1
k
a
k
e
ikx
+a
k
e
−ikx
(8.37)
8.1 Time-Dependent t PerturbationTheory
55
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Similarly the interactionpotential is just the potentialwritteninterms of interacting φ
s. That is,sup-
poseweknowtheinteractionintermsofthefieldsatsometimet
0
:
V=
d
3
x
λφ
3
=
d
3
x
λφ
0
3
(8.38)
We use e φ
0
insteadof the interactingfield d φ φ not because the fields are not interacting,but because our
basisoffieldsatthetimet=t
0
isthebasisoffreefields. Then
V
I
=e
iH
0
t
d
3
x
λφ
0
3
e
−iH
0
t
=
d
3
x
λφ
I
3
(8.39)
We will usuallyjust drop p the e I subscripts onV V and d φ, becausegoingto the e interactionpicture just
meansaddingit
0
ω
k
tothephasefactors,whichisunambiguous.Forthislecture,however,wewillkeepI’s
intomakeanexplicitdistinctionbetweenfreeandinteractingfields.
8.1.5 Summary
Insummary, the I I subscript t refers s to the e free theory evolving with h the e free Hamiltonian. . No o subscript
meansthefullinteractingfields.Matrixelementsarerelatedby
Ω|φ(x
1
)φ(x
2
)|Ω=0|U
∞0
U
01
φ
I
(x
1
)U
10
U
02
φ
I
(x
2
)U
20
U
0−∞
|0
(8.40)
where
U
ij
=T
exp[−i
t
j
t
i
dt
V
I
(t
)]
(8.41)
and|Ω
isthegroundstateintheinteractingtheory.
Forthespecialcaseoftimeorderedproducts,suchaswhatweneedforS-matrixelements,thissimpli-
fiesto
Ω|T
{
φ(x
1
)φ(x
2
)
}
|Ω
=
0
T
φ
I
(x
1
I
(x
2
)exp[−i
−∞
dtV
I
(t)]
0
(8.42)
8.2 Timeorderedproductsandcontractions
Let’susethisformulatocalculatesomething. Let’stakeourinteractiontobeoftheform φ
3
. Then
V
I
d
3
xφ(x)
3
(8.43)
Tonextorder,weusetheformula
Ω|T{φ(x
1
)φ(x
2
)}|Ω=
0
T
φ
I
(x
1
I
(x
2
)exp
−i
−∞
dtV
I
(t)

0
(8.44)
=
0
|
T{φ
I
(x
1
I
(x
2
)
|
0
−iλ
d4x
0
T{φ
I
(x
1
I
(x
2
I
(x)3
0
(8.45)
−λ
2
d
4
x
d
4
y
0
T{φ
I
(x
1
I
(x
2
I
(x)
3
φ
I
(y)
3
0
+
(8.46)
Notice howthe
−∞
dtcombinedwiththe
d3xinV togive e aLorentzinvariant integral. Sonowwe
needtobeabletocomputeexpressionslike
0
T
φ
I
(x
1
I
(x
2
I
(x)
3
φ
I
(y)
3
0
(8.47)
56
FeynmanRules
Tobegin,recallthateachφ
I
(x)hasacreationandanannihilationoperatorandsomeexponentials
φ
I
(x)=
d
4
p
(2π)4
1
p
a
p
e
ipx
+a
p
e
−ipx
(8.48)
So the expressionbecomes s abunchof a
p
s anda
p
†′
s andexponentials. . Inour r example,wegetsomething
like
(a
p
1
e
ip
1
x
1
+h.c)(a
p
2
e
ip
2
x
2
+h.c)(a
p
3
e
ip
3
x
+h.c.)(a
p
4
e
ip
3
x
+h.c.)
(a
p
8
e
ip
8
x
+h.c.)
(8.49)
integratedoverx,yandthe8p
s.
Forthistogiveanon-zeromatrixelementonthevacuum,allthea
p
i
’swillhavetobepairedoffwith
all the a
p
j
’s. This s is really easy todo– – for r each h field, from m right to left, either create afieldfrom the
vacuum,or annihilateonethat’s alreadybeencreated. . e just sum overallways s ofdoingthis. . Theneach
pairingwillgive
0|T{φ
I
(x)φ
I
(y)}|0=
d
4
k
(2π)4
i
k2−m2+iε
e
ik(x−y)
≡D
F
(x,y)
(8.50)
Ifthere’smorea
p
sthana
p
†′
stheanswerwillbezero. Iftwop
sarethesame,thentherecouldbesome2-
particlestates,or more generally n-particle states,whichmight give a a factors s of f n. So o we’llhave to be
carefultodivideoutbysymmetryfactorsforidenticalparticles.
Forourexampleinteraction
0
T
φ
I
(x
1
I
(x
2
I
(x)3φ
I
(y)3
0
(8.51)
Thereare3possiblecontractions
1. D
F
(x
1
,x
2
)D
F
(x,y)
3
2. D
F
(x
1
,x)D
F
(x
2
,y)D
F
(x,y)D
F
(x,y)
3. D
F
(x
1
,x)D
F
(x
2
,x)D
F
(x,y)D
F
(y,y)
Sotheresultisthesumofthese,plusthephases,withappropriatesymmetryfactors.
8.2.1 Wick’stheorem
This procedure we described d follow from m a more general theorem m calledWick’s theorem. . I I don’t really
knowwhyyouneedthemoregeneralform,butwecan’treallycoverfieldtheorywithoutmentioningit.
RecalltheFeynmanpropagator
0|T
{
φ
I
(x)φ
I
(y)
}
|0
=
d4k
(2π)4
i
k2−m2+iε
e
ik(x−y)
≡D
F
(x,y)
(8.52)
Thiscamefrom
0|(a
p
1
e
ip
1
x
+a
p
1
e
−ip
1
x
)(a
p
2
e
ip
2
y
+a
p
2
e
−ip
2
y
|0
(8.53)
We only usedthea
p
1
a
p
2
termintheexpansion. Infactreally,whatweusedwas[a
p
1
,a
p
2
],sincethis is a
deltafunction. Ifweexpandtheoperatorwefindinadditiontothe[a
p
1
,a
p
2
]term usedfortheFeynman
propagator,termslike
a
p
1
a
p
2
+a
p
1
a
p
2
+a
p
1
a
p
2
(8.54)
All of these have a
p
’s onthe e left and d a
p
’s on the e right,so they have vanishingmatrix x elements in the
vacuum.Wecallsuchtermsnormalordered,andindicateitwitha
.
• normal l ordering: all the a
p
’s are on n the e right, and d all the e a
p
s are on n the e left. . Represented d with
colons.
Thusforthetwopointfunction:
T{φ
I
(x)φ
I
(y)}=D
F
(x,y)+:φ
I
(x)φ
I
(y):
(8.55)
8.2 Timeorderedproductsandcontractions
57
Wick’stheoremsaysthat
T{φ
I
(x
1
)
φ
I
(x
n
)}=:φ
I
(x
1
)
φ
I
(x
n
)+allpossiblecontractions:
(8.56)
whereacontractionmeans taketwofields φ(x
i
)and φ(x
j
)from anywhere intheseriesandreplacethem
withafactorof D
F
(x
i
,x
j
) for eachpair offields. . Allpossible e contractions includes onecontraction,two
contractions, etc, involving any of the fields. . But t each h field can only y be contracted d once. . Since e normal
orderedproductsvanishunlessallthefieldsarecontractedthisimpliesthat
Theproofisbyinduction. Foronefield,
T{φ
I
(x)}=:φ
I
(x):
(8.57)
Now suppose the e theorem holds for n fields. . Since e we can commute the e fields s within a a time e ordering
symbol,let’sjustputthemallintimeorder,andreliablethexsappropriately(wecouldhavecalledthem
ysifyoulike). Then
T
{
φ
I
(x
1
I
(x
2
)
φ
I
(x
n
)
}
I
(x
1
)T{φ
I
(x
2
)
φ
I
(x
n
)}
(8.58)
=(a
p
1
e
ip
1
x
+a
p
1
e
−ip
1
x
)[:φ
I
(x
2
)
φ
I
(x
n
)+contractionsof{x
2
,
,x
n
}:]
(8.59)
Nowweneedtogetthea
p
1
anda
p
1
insidethenormalordering. Fora
p
1
it’s easysinceitisalreadyonthe
left,whereitneedstobe. Thetermwitha
p
1
hastobecommutedpassedeverybody.Eachtimewegeta
a
p
1
φ(x
j
)=
a
p
1
(a
p
j
+a
p
j
)
=
φ(x
j
)a
p
1
+
a
p
1
,a
p
j
(8.60)
Butthiscommutatoris justadelta a function,whichis s the samethingweusedtocalculate theFeynman
propagator.Sothisgivesafactorof D
F
(x
1
,x
j
). Aswemovea
p
along,wepickupallthepossiblecontrac-
tions,whichisallthereistotheproof.
TheresultofWick’s theorem is that time-orderedproducts aregivenbyabunchof contractions plus
normal orderedproducts. . Since e thenormal-orderedproducts vanishinvacuum matrix elements, we get
the specialcaseweusedabove. Thebottom lineisthatallof theS-matrixelementswewillcomputecan
bereducedtoabunchof D
F
sandabunchofphasefactors.
8.3 Feynmandiagrams
There’saveryimportantwaytorewritethesumofcontractionscalledFeynmandiagrams. Itsaysdrawa
point for eachx
i
orxor yintheexpression. TheneachD
F
(x,y)isalinebetweenxandy. Thenthere
areasmany linescomingout of the xs asthereare insertions. . Thefinalamplitude e willbeasumof all
thewaysofcontractingeverythingtogether.
Let’sworkthroughourexample. Westartedfrom
V
I
=λφ(x)
3
(8.61)
andwereleadto
0
T
φ(x
1
)φ(x
2
)φ(x)
3
φ(y)
3
0
2
d
4
x
d
4
yD
F
(x
1
,x)D
F
(x
2
,y)D
F
(x,y)D
F
(x,y)+
ThistermcorrespondstooneparticularFeynmandiagram:
x
1
x
2
x
y
Toevaluatethisdiagram,firstwriteeveryD
F
inmomentumspace
D
F
(x,y)=
d
4
p
(2π)4
i
p2−m2+iε
e
ip(x−y)
(8.62)
58
FeynmanRules
Thusallthex
sappearonlyinexponentials. Therewere8 8 p
s tobeginwith,andwepairedthemoff,so
thereare4p’sleftfromtheFouriertransformsofthe4D
F
’s. Wehave
M=λ
2
d
4
x
d
4
y
d
4
p
(2π)4
4
e
ip
1
(x
1
−x)
e
ip
2
(x
2
−y)
e
ip
3
(x−y)
e
ip
4
(x−y)
i
p
1
2
+iε
i
p
2
2
+iε
i
p
3
2
+iε
i
p
4
2
+iε
Nowwecandothexandyintegrals,whichproduceδ
(4)
(p
1
+p
3
+p
4
) andδ
(4)
(p
2
+p
3
+p
4
). This s says
that momentum is s conservedat t eachvertex intheFeynmandiagram. . If f we integrate over r p
3
usingthe
firstδ-function,p
3
=−p
1
−p
4
thesecondδfunctionbecomesδ
(4)
(p
1
+p
2
).Thenwehave,relabelling p
4
=
k:
M=−λ
2
d
4
k
(2π)4
d
4
p
1
(2π)4
d
4
p
2
(2π)4
e
ip
1
x
1
e
ip
2
x
2
i
p
1
2
+iε
i
p
2
2
+iε
i
(p
1
+k)2+iε
i
k2+iε
δ
(4)
(p
1
+p
2
)
Next,rememberthatweareinterestedinanS-matrixelement. FromtheLSZtheorem,
f|S|i
=
i
d
4
x
1
e
−ip
i
x
1
(
1
2
−m
2
)

i
d
4
x
2
e
ip
f
x
2
(
2
2
−m
2
)
Ω|T
{
φ(x
1
)φ(x
2
)
}
|Ω
(8.63)
2
d4x
1
e−ip
i
x
1
(−p
i
)2
d4x
2
e
ip
f
x
2
(−p
f
2
)
M
(8.64)
Nowwenotethat thex
1
integralgivesδ
4
(p
1
−p
i
)andthex
2
integralgivesaδ
4
(p
2
+p
f
). Sowecannow
dothep
1
andp
2
integrals,giving
f|S|i=λ
2
d
4
k
i
(p
i
+k)2+iε
i
k2+iε
δ
(4)
(p
i
−p
f
)
(8.65)
Notehowthetwopropagatorfactorsinthebeginninggetcanceled. Thisalwayshappensforexternallegs
–rememberthepointofLSZwastoforcetheexternallinestobeon-shell.
Thisintegralis obviouslyinfinite. . Wewillspendalargepart of theremainderofthecoursetryingto
makesenseoutofthesethings. Finally,the δ
4
(p
1
−p
2
)termintheanswerforces overallmomentumcon-
servation,andwillalwaysbepresentinanycalculation. Butwewillalwaysfactoritout,likewedidwhen
werelateddifferentialscatteringamplitudestoS-matrixelements.
WecansummarizethisprocedurewiththeFeynmanrules
• Internallinesgetpropagators
i
p2−m2+iε
• Verticescomefrom m interactionsintheLagrangian. Theygetfactorsofthecouplingconstanttimes
ifromtheexp(iV).
• External l lines s don’t get propagators,because these arecanceledby terms fromtheLSZ reduction
formula.
• Momentumisconservedateachvertex
• Integrateoverundeterminedmomenta.
8.3.1 symmetryfactors
Wehaven’tkepttrackofoverallnormalization. Therearelotsoffactorsof2,3,4,
that showup,which
havetobecarefullykepttrackof. Inpractice,wewoulddraweachuniquediagram,thentrytofigureout
theappropriatesymmetryfactor.Intheend,therulesarequitesimple.
• Thereisafactorof
1
m!
fromtheexpansionofexp(iV)=
1
m!
(iV)
m
. But ifweexpandtoorderm
there will be m identical vertices s in the e same diagram. . We e canalso swapthese vertices around,
leavingthediagramlookingthesame. Soifwecalculatethediagramonce,them!frompermuting
thediagramswillcancelthe
1
m!
fromtheexponential. Thus,wecanforgetaboutthispart.
8.3 Feynmandiagrams
59
• Consideraφ
n
(x)interaction. Itleadstoavertexatthepointxwithnlinesstickingout. Whenwe
connect thoselinesuptotherestofthediagram,weget D
F
(x,y
i
)
n
. The y
i
scanalsobex’s,but
first letssuppose they’re not. . Ifweswaparoundwhichline e connects to which h y
i
,wewill get the
sameexpression,correspondingtoa differentcontractionwiththe sameFeynmandiagram. . Sowe
cancalculate theFeynmandiagramonce andjustmultiply byfactorof n!foreveryvertexwithn
linesstickingout.Insteadofdoingthis,itisconventionaltorescalethevertexbyafactorofn!
λ
4!
φ
4
,
g
3!
φ
3
,
(8.66)
• Nowsupposesomeofthe e y
i
sarexs. Forexample,ifwehaveφ(x)2→D
F
(x,x)2. Thislookslikea
bubblestartingatxandendingatx. Inthis s case,thereisonlyonecontraction,sotherearefewer
diagrams than we thought t and d we have to o divide by 2! to compensate. . Similarly, , if f we had
φ(x)
2
φ(y)
2
, the diagram withtwo lines connecting x and y corresponds s to o only one contraction,
D
F
(x,y)2,sowe needto put back a2!heretoo. . Ingeneral, , weneedtoput backa factor r corre-
sponding to o the number of geometric symmetries s of the diagram. . See e chapter 4 4 of f Peskin and
Schroederforsomemoreexamples.
• Identicalparticles. . Identicalparticles s arealready takencare of inWick’stheorem. . Movingaround
the a
p
’s anda
p
’s has thealgebraofidenticalparticles inthem. . Theonly y timeit matters is when
the identicalparticlesareexternalparticles,thatis,whenwecan’t distinguishtheparticles weare
scattering. Andthenit’s s onlyforfinalstates,sincewedistinguishourinitialstatesbythesetupof
the experiment. . Thuswhennofthesameparticles s are produced,wehavetodividethe crosssec-
tionbyn!.
8.3.2 signsofmomenta
There is a standard d convention about t how w to choose the direction n that the momenta a are e going. For
externalmomentaitmakes sensetoassignthemtheir physicalvalues,whichshouldhavepositiveenergy.
Thenmomentumconservationbecomes
p
i
=
p
f
(8.67)
whichappearsinaδ-functionas
δ
(4)
(p
i
−p
f
)
(8.68)
Thisisnotthesameasδ
4
(Σp)whichwesometimeswriteasshorthand,butwouldobviouslyrequiresome
ofthemomentatohavenegativeenergy
For internallines, , weintegrateover r the momenta,sotheresultis symmetrytok
µ
→−k
µ
. Butit t is
important to o keeptrack of whichway the momentum is s goingso o that t all the δ-functions at the vertices
arealsoΣ(p
in
−p
out
).Sowedrawarrowsonthelinestoindicatethis
p
2
p
1
p
4
p
3
Actually, the arrows s willeventually come to o be associatedwiththedirection the charge is s flowing,and
thenwe willsometimesaddadditionalarrows torefer tojust themomentum. . Butfor r now,let’sstick to
thearrowsrepresentingthedirectionmomentumflow.
Also,sometimesFeynmandiagramsaredrawnwithtimegoingupwards.Iprefertohavetimegoingto
theright,butagain,thisisjustamatteroftaste.Youcouldhavetimegoingdiagonallynortheastandthe
diagramwouldhavethesamemeaning.
60
FeynmanRules
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