c# display pdf in browser : Add a blank page to a pdf control application platform web page azure winforms web browser QFT-Schwartz11-part442

Using
p·σ
=
E−p
z
0
0
E−p
z
,
p·σ¯
=
E+p
z
0
0
E−p
z
(11.189)
Wecanwritemoregenerally
u
s
(p)=
p·σ
ξ
s
p·σ¯
ξ
s
, v
s
(p)=
p·σ
η
s
p·σ¯
η
s
(11.190)
wherethesquarerootofamatrixmeansdiagonalizeitandthentakethesquareroot. Inpractice,wewill
alwayspickpalongthez axis,sowedon’treallyneedtoknowhowtomakesenseoftheseformulas. But
theyareusefulbecausetheyareexplicitlyLorentzcovariantsowecanmanipulatethemcleanly.
11.12.2 normalizationandspinsums
Wehavechosenanormalizationhere,whichamountsto
ψ
¯
ψ
=u
s
(p)γ
0
u
s
(p)=
p·σ
ξ
s
p·σ¯
ξ
s
0 1
1 0
p·σ
ξ
s
p·σ¯
ξ
s
(11.191)
=2
E−p
z
0
0
E+p
z

E+p
z
0
0
E−p
z
ξ
s
ξ
s
(11.192)
=2mδ
ss
(11.193)
Wetakenξ
s
ξ
s
ss
byconvention. Thisisthenormalizationfortheinnerproduct.
Form=0thiswayofwritingtheconditionisnotuseful. Insteadwecanjustsetthenormalizationin
theframeoftheparticle;sdirectionbynormalizing
u
s
u
s
=
p·σ
ξ
s
p·σ¯
ξ
s
p·σ
ξ
s
p·σ¯
ξ
s
=2Eξ
s
ξ
s
=2Eδ
ss
(11.194)
whichhasthesame 2E
factorswe’vebeenusing.
Wecanalsocomputetheouterproduct
s=1
2
u
s
(p)u¯
s
(p)=p
+m
(11.195)
wherethesumisovers=1,2correspondingto ξ=
1
0
andξ=
0
1
.Fortheantiparticles,
s=1
2
v
s
(p)v¯
s
(p)=p
−m
(11.196)
Itisalsotruethat
s=1
2
u
s
s
=
s=1
2
s
v
s
=0
(11.197)
Youshouldtrytoverifytheserelationsonyourown.
Tokeepstraighttheinnerandouterproducts,itmaybehelpfultocomparetospin1particles.Inthat
casewehadformassivespin1
µ
i
(p)]
ǫ
µ
j
(p)=−δ
ij
s
(p)u
s
(p)=2mδ
ss
(innerproduct)
(11.198)
i=1
3
µ
i
(p)]
ǫ
ν
i
(p)=η
µν
p
µ
p
ν
m2
s=1
2
u
s
(p)u¯
s
(p)=p
+m
(outerproduct)
(11.199)
11.12 Solvingthe e Dirac equation
111
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So whenwe sum Lorentz indices s or internal l spinor r indices,we use aninner product and d get t a a number.
Whenwesumoverpolarizations/spins,wegetamatrix.
11.12.3 chargeconjugation
Intherestframe,
u
s
=
ξ
s
ξ
s
, v
s
=
η
s
−η
s
(11.200)
whereξandηareconstants,forspinupandspindown.
Letusseehowthespinstatestransformunderchargeconjugation
|↑
c
=
1
0
c
=−iσ
2
1
0
c
=
0 i
−i 0

1
0
=
0
−i
=−i|↓
(11.201)
|↓
c
=
0
1
c=−iσ
2
0
1
c=
0 i
−i 0

0
1
=
i
0
=i|↑
(11.202)
Sochargecongugationflipsthespin. Thus
(u
p
s
)
c
=
ξ
−s
−ξ
−s
(11.203)
Thiswillberelevanttostudyingcharge-conjugationinvariance.
112
Spinorsand theDirac c Equation
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Chapter12
QuantumElectrodynamics
12.1 Introduction
WehavefoundtheLagrangianforQED:
L=−
1
4
F
µν
2
+iψ
¯
D
ψ−mψ
¯
ψ
(12.1)
Inordertocomputeanythingwithitinthequantumtheory,wefirsthavetoquantizethespinors.
12.2 QuantizingSpinors
Forcomplexscalarthefieldiswehad
φ(x)=
d
3
p
(2π)3
1
p
a
p
e
ipx
+b
p
e
−ipx
(12.2)
φ
(x)=
d
3
p
(2π)3
1
p
b
p
e
ipx
+a
p
e
−ipx
(12.3)
whichimpelsthatb
p
createsparticlesoftheoppositechargeandsamemass,whichwecalledantiparticles.
FortheDiracequationtheLagrangianis
L=ψ
¯
(iD
−m)ψ
(12.4)
Andtheequationsofmotionare
(i∂
+eA
−m)ψ=0
(12.5)
ψ
¯
(−i∂
−eA
−m)=0
(12.6)
Thefreefieldsolutionscanbe labeledby constanttwo-componentspinors ξ andη. Theactualsolutions
dependonmomentainafunnyway. Labellingthemu
s
andv
s
wheres=1,2isthespin,wehave
u
p
1
=
p·σ
0
p·σ¯
0
,u
p
2
=
0
p·σ
0
p·σ
,v
p
1
=
p·σ
0
p·σ¯
0
,v
p
2
=
0
p·σ
0
p·σ
(12.7)
Theu
s
arethepositive frequencyelectrons, andthev
s
arenegative frequencyelectrons, or equivalently,
positivefrequencypositrons.
113
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Thuswetake
ψ(x)=
s
d
3
p
(2π)3
1
p
a
p
s
u
p
s
e
ipx
+b
p
s
v
p
s
e
−ipx
(12.8)
ψ
¯
(x)=
s
d
3
p
(2π)3
1
p
b
p
s†
p
s
e
ipx
+a
p
s
p
s
e
−ipx
(12.9)
Whereasalwaystheenergyisalwayspositiveanddeterminedbythe3-momentumω
p
=
p
2+m
2
.
Notetheconventionthat ψ(x) gets bothuandvandψ
¯
gets u¯ ¯ andv¯. . Itmighthavemade e sense to
useuandv¯ in n ψ, but t sincewhenwe foundthe free states,v v were thenegativefrequency y solution,this
wayofassigninguandvtoψandψ
¯
isconsistent.Thisisjustaconvention.
12.3 DiracPropagator
Next,let’scalculatethepropagatorforaDiracspinor.
0|T{ψ
¯
(x)ψ(y)}|0
(12.10)
Let’sfirstreviewwhathappenswithacomplexscalarwehad
φ(x)=
d
3
p
(2π)3
1
p
a
p
e
ipx
+b
p
e
−ipx
(12.11)
φ(x)=
d3p
(2π)3
1
p
b
p
eipx+a
p
e−ipx
(12.12)
Then
0|φ(x)φ(0)|0
=
d3p
(2π)3
d3q
(2π)3
1
p
1
q
0|
b
p
eipx+a
p
e−ipx
a
p
s
+b
p
s
)|0
=
d
3
k
(2π)3
1
k
e
−ikx
=
d
3
k
(2π)3
1
k
e
−iω
k
t+ik
x
(12.13)
0|φ(0)φ
(x)|0=
d
3
k
(2π)3
1
k
e
ikx
=
d
3
k
(2π)3
1
k
e
k
t−ik
x
(12.14)
Then,
0|T{φ(x)φ(0)}|0
=
d3k
(2π)3
1
k
eik
x
−iω
k
tθ(t)+e−ik
x
+iω
k
tθ(−t)
(12.15)
=
d3k
(2π)3
1
k
eik
x
e
−iω
k
|t|
(12.16)
Thenusing
e
−iω
k
|t|
=(2ω
k
)
−∞
(2π)
i
ω2−ω
k
2
+iε
e
iωt
(12.17)
Wefind
0|T{φ
(x)φ(0)}|0=
d
4
k
(2π)4
i
k2−m2+iε
e
ikx
(12.18)
Whichistheregularscalarpropagator.
Forafermion,westartthesameway
ψ(x)=
s
d
3
p
(2π)3
1
p
a
p
s†
u
p
s
e
ipx
+b
p
s
v
p
s
e
−ipx
(12.19)
ψ
¯
(x)=
s
d
3
p
(2π)3
1
p
b
p
s†
p
s
e
ipx
+a
p
s
p
s
e
−ipx
(12.20)
114
QuantumElectrodynamics
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Then,
0|ψ(0)ψ
¯
(x)|0
=
d
3
p
(2π)3
d
3
q
(2π)3
1
p
1
q
×
s,s
0|
a
p
s†
u
p
s
+b
p
s
v
p
s
b
q
s
q
s
e
ipx
+a
q
s
q
s
e
−ipx
|0
(12.21)
=
d3p
(2π)3
d3q
(2π)3
1
p
1
q
s,s
v
p
s
q
s
0|b
p
s
b
q
s
|0
eipx
=
d
3
k
(2π)3
1
k
s
v
k
s
k
s
e
ikx
Nowwesumoverpolarizationsusingtheouterproduct
s=1
2
u
s
(p)u¯
s
(p)=p
+m,
s=1
2
v
s
(p)v¯
s
(p)=p
−m
(12.22)
Giving
0|ψ(0)ψ
¯
(x)|0
=
d3k
(2π)3
1
k
eikx(k
+m)
Similarly
0|ψ
¯
(x)ψ(0)|0
=
d
3
p
(2π)3
d
3
q
(2π)3
1
p
1
q
0|
b
p
s
p
s
e
−ipx
+a
p
s†
p
s
e
ipx

a
q
s
u
q
s
+b
q
s†
v
q
s
|0
=
d
3
k
(2π)3
1
k
e
−ikx
(k
−m)
(12.23)
So,
0|ψ(0)ψ
¯
(x)|0
=
d3k
(2π)3
1
k
eikx(k
+m)=(−i∂
+m)
d3k
(2π)3
1
k
e−ik
x
+iω
k
t
(12.24)
0|ψ
¯
(x)ψ(0)|0
=
d
3
k
(2π)3
1
k
e
−ikx
k
−m
=(i∂
−m)
d
3
k
(2π)3
1
k
e
ik
x
−iω
k
t
(12.25)
Then,
0|T{ψ
¯
(x)ψ(0)}|0
=(−i∂
+m)
d
3
k
(2π)3
1
k
e
ik
x
−iω
k
t
θ(t)−e
−ik
x
+iω
k
t
θ(−t)
(12.26)
=(−i∂
+m)
d3k
(2π)3
1
k
eik
x
e
−iω
k
|t|
[
θ(t)−θ(−t)
]
(12.27)
=(−i∂
+m)
d
4
k
(2π)4
i
ω2−ω
k
2
+iε
e
ikx
(−1)
|t|
t
(12.28)
Waitaminute–thisshouldhavesimplified. Whathappened?
Wereallywantthatminussigntobeaplus. Wecangetthatbydefiningthetimeorderedpropagator
as
T{ψ
¯
(x)ψ(0)}=ψ
¯
(x)ψ(0)θ(t)−ψ(0)ψ
¯
(x)θ(−t)
(12.29)
Then,
0|T{ψ
¯
(x)ψ(0)}|0
=(−i∂
+m)
d4k
(2π)4
i
ω2−ω
k
2
+iε
eikx
(12.30)
0|T{ψ
¯
(x)ψ(0)}|0
=
d
4
k
(2π)4
i(k
+m)
k2−m2+iε
(12.31)
ThisisbeautifullyLorentzinvariant.
12.3 Dirac c Propagator
115
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Sowewanttohavethetime-orderedproductofspinors haveanextraminus sign. . But thenforarbi-
traryspinors
T{ψ(x)ψ(y)}=ψ(x)ψ(y)θ(x
0
−y
0
)−ψ(y)ψ(x)θ(y
0
−x
0
)=−T{ψ(y)ψ(x)}
(12.32)
Thusunlesswewantthistovanish,wemustassumethat ψ(y)andψ(x)anticommute.
{ψ(x),ψ(y)
}
≡ψ(x)ψ(y)+ψ(y)ψ(x)=0
(12.33)
Similarly,fieldsmustanticommutewiththeiranti-fields,uptoadeltafunction
{ψ(x),ψ
¯
(y)
3
(x−y)
(12.34)
Thisisequivalentto
a
p
r
,a
q
s
=
b
p
r
,b
q
s
=(2π)3δ
(3)
(p−q)δrs
(12.35)
SoweseethatinordertohaveLorentzinvarianttime-orderedproducts,fermionsmustanti-commute.
Nowconsiderwhathappensifwetrytoconstructastatewithtwoofthesamefieldinit:
a
p
a
p
|0=−a
p
a
p
|0=0
(12.36)
Sowe can’t! ThisistheFermi-exclusionprinciple,andit follows directlyfromtheanticommutationrela-
tions.
Thisisanexampleofthespin-statisticstheorem. Let’stracebacktowhathappened.Wefound
0|φ
(x)φ(0)|0=
d
3
k
(2π)3
1
k
e
ikx
(12.37)
0|φ(0)φ
(x)|0=
d
3
k
(2π)3
1
k
e
−ikx
(12.38)
Ascomparedto(form=0):
0|ψ
¯
(x)ψ(0)|0
=
d
3
k
(2π)3
k
k
e
ikx
(12.39)
0|ψ(0)ψ
¯
(x)|0
=
d3k
(2π)3
k
k
e−ikx
(12.40)
Nowwe cansee that theproblem is that k
is oddunder the rotationwhich h takes s k→− k, , sothat t an
extra−1isgeneratedwhenwetrytocombinethetime-orderedsumfortheFermion. Rotatingk→−kis
arotationbyπ. Wesawthatthisgivesafactorof iinthefermioncase. Soherewehave2fermions,and
wegeta −1. Soit’sdirectlyrelatedtothespin
1
2
. Thiswillhappenforanyhalfintegerspin,whichgets
anextra −1intherotation.
Anotherwaytolookat itis thattheyk
factorcomesfromthepolarizationsum,whichinturncomes
fromtherequirementthat thefreesolutionssatisfyk
u
s
(k)=k
v
s
(k)=0. Infact,wecannowseedirectly
thatthesameproblemwillhappenforanyparticleofhalf-integerspin. Aparticleofspinn+
1
2
forinteger
n willhavea fieldwith h nvector r indices anda a spinor r index, , ψ
µ
1
µ
n
. Sopolarizationsummust t have a
factorof γ
µ
andtheonlythingaroundtocontract γ
µ
withisit’smomentumk
µ
. Thuswealwaysgetak
,
and the time-ordered product t can n never be e Lorentz invariant t unless s the fields s anticommute. . These e are
fermions. TheyobeyFermi-Diracstatistics.
Incontrastforintegerspintherecanonlybeanevennumberof k
µ
2
inthepolarizationsum(therehad
betterbenok
µ
ssincek2=0!).SothesefieldsmustcommutetohaveLorentzinvarianttimeorderedprod-
ucts.Thesearebosons. TheyobeyBose-Einsteinstatistics
12.4 Spinstatistics
We saw w that t to o have a Lorentz invariant t time-orderedproduct for fermions, we needthem to o antisym-
metric. Thatis
ψ(x)ψ(y)=−ψ(y)ψ(x)
(12.41)
116
QuantumElectrodynamics
orequivalently
a
p
r
,a
q
s
=(2π)
3
δ
(3)
(p−q)δ
rs
(12.42)
Actingonthevacuum,this hasthatmultiparticlestatesareantisymmetricontheexchangeoftwoparti-
cles
ψ
s
1
p
1
s
2
p
2
=
a
s
1
(p
1
)
a
s
2
(p
2
)
|0
=−ψ
s
2
p
2
s
1
p
2
(12.43)
Ingeneral,wecouldaskwhatkindsofstatisticsarepossible. (This s discussionistakenfromsection4.1of
Weinberg).
Supposewehaveastatewithmanyidenticalparticles.Byidentical,wemeanthatifweswaphowthe
twoparticlesarerepresented,westillhavethesamephysicalstate.Onlythenormalizationcanchange. So
ψ
s
1
p
1
s
2
p
2
=αψ
s
2
p
2
s
1
p
2
(12.44)
whereαisanumber. Thisisthedefinitionof identicalparticles.
Ifwechageback,wefindthat
ψ
s
1
p
1
s
2
p
2
=αψ
s
2
p
2
s
1
p
2
2
ψ
s
1
p
1
s
2
p
2
(12.45)
Thusα
2
=1. Whatcouldαdependon? Sinceitisjustanumber,itcannotdependonthemomentapor
thespinss,astherearenonon-trivlialone-dimensionalrepresentationsoftheLorentzgroup. Thusα=±
1. Wecall α=1bosons s andα=−1fermions. . Aswehave e shownabove,byLorentzinvariance,integer
spinparticlesmustbebosonsandhalf-integerspinparticlesmustbefermions.
Thereisoneexception,thefactorαcandependontheway p
1
andp
2
areinterchanged. Forexample,
intheAharonov-Bohmeffect,youcanpickupaphasebyrotatingaparticlearoundamagneticflux. This
onlyworksin2+1dimensions(2spaceandone1dimension)wherewecanhavethingslikefluxesorvor-
tices,andthestatecanrememberhowmanytimesithas beenrotatedaroundit. . Thinkofastringwrap-
pingaroundapointin2D. Thenyoucangetαtodependonthepath,thatis,thenumberoftimesithas
gonearoundthepoint(thewindingnumber)whichleadstoparticlescalledanyonswithparastatics. But
in3+1 dimensions,youcan’t t wrapa a string arounda point t – – you can always s unwindsuch h path-depen-
dence. SoonlyFermiandBosestatisticsarepossible.
12.5 Causality
Let’sgobacktoarealscalarfield,
φ(x)=
d
3
q
(2π)3
a
q
e
iqx
+a
q
e
−iqx
(12.46)
wherewesaw
φ(x)φ(y)=
d
3
p
(2π)3
1
p
e
−ip(x−y)
(12.47)
[φ(x),φ(y)]=
d3p
(2π)3
1
p
e
−ip(x−y)
−e
ip(x−y)
(12.48)
Let’sworkin2dimensions,wherecausalityissimple.Thenfory−x=(t,z)weget
[φ(x),φ(y)]=
dp
1
p
e
i(ω
p
t−pz)
−e
−i(ω
p
t−pz)
(12.49)
=
−∞
dp
i
p2+m2
sin( p
2
+m
2
t−pz)
(12.50)
12.5 Causality
117
Nowsupposetheintervalisspacelike,forexampley−x=(0,z). Then
[φ(x),φ(y)]=
dp
−i
p2+m2
sin(pz)=0
(12.51)
Sincesin(pz)isanoddfunctionof p,thisintegralvanishes.
Next,takey−x=(t,0)=timelike.Then,
[φ(x),φ(y)]=
−∞
dp
i
p2+m2
sin( p
2
−m
2
t)
(12.52)
Nowtheintegrandis aneven functionof f p,soingeneraltheintegralwillnot t vanish. It’snot soeasyto
evaluatetheintegral,butthepointissimplythatitisnotzero. Thenwecanusethefactthat[φ(x),φ(y)]
is Lorentz covariant. . So o we findthat t ingeneralcommutators vanishat spacelike separations but not at
timelikeseparation.
[φ(x),φ(y)]
=0, x−yspacelike
0, x−ytimelike
(12.53)
Thisistrueinanynumberofdimensions.
What does this commutator have to do with physics? ? Remember, , we e are just doing quantum
mechanicshere. Sowhentooperatorscommutetheyaresimultaneously y observable. . Another r way tosay
that is that t if f the operators s commutethey y are uncorrelatedand d can’t t influence eachother. . So o herewe
findthatifthewemeasurethefield(rememberφmeasuresthefield)atx=0itdoesn’tinfluencethemea-
surementatxatthesametimet=0. Butifwemeasurethefieldatt=0itdoesaffectthefieldatt=tat
thesamepositionx.Thisisaclearstatementofcausality.
If youtry todothe samething in n non-relativistic c quantum mechanics, it willfail. . For r example,the
wavefunction is s spreadout inxbut fixedwithrespect to o t. Soif we collapse the wavefunction bymea-
suringitatx=0att=0itwillimmediatelyaffectwhatwecanseeatx=xandt=0.
We cantrytodothesamecalculationforspinors. . We e findthatitis theanticommutator whichvan-
ishes
{
ψ(x),ψ(0)
}
=0
(12.54)
Infact,
[ψ(x),ψ(0)]
0
(12.55)
Whatdowemakeofthis? Itimpliesifwecouldevermeasureafermionicwavefunction,wecouldviolate
causality. Thus s fermions are not observable. . Only y bosons are observable. . After r all, , all measurements in
theendarenumberswhicharebosons. YouwouldhavetomeasureaGrassmannumbertoseeaFermion.
Socausalityimpliesthatonlyfermionbilinearshave
¯
(x)ψ(x),ψ(y)ψ(y)]
=0, x−yspacelike
0, x−ytimelike
(12.56)
Thatissimplybecauseψ
¯
(x)ψ(x)isaboson.
12.6 QEDFeynmanrules
NowwehaveeverythingweneedforQED. TheLagrangianis
L=−
1
4
F
µν
2
+iψ
¯
D
ψ−mψ
¯
ψ
(12.57)
=−
1
4
F
µν
2
¯
(iγµ
µ
+eγµA
µ
−m)ψ
(12.58)
118
QuantumElectrodynamics
Weknowthepropagators. Aphotonissquigglyline
=
−i
k2+iε
η
µν
−(1−ξ)
k
µ
k
ν
k2
=
−iη
µν
k2+iε
(Feynmangauge)
WewillusuallyjustchooseFeynmangauge,inlessweareexplicitlycheckinggaugeinvariance.
Afermionisasolidlinewithanarrowindicatingparticleflow(momentumflowtimescharge)
=
i(p
+m)
p2−m2+iε
Externalphotonlinesgetpolarizationvectors
⊗=ε
µ
(p) (incoming)
(12.59)
µ
(p) (outgoing)
(12.60)
Heretheblobmeanstherestofthediagram.
Externalfermionlinesgetspinors,withusforelectronsandvsforpositrons.
⊗=us(p)
=u¯
s
(p)
⊗=v
s
(p)
=v¯
s
(p)
Theinteractionscanbereaddirectlyoffthevertex.
V=eψ
¯
γ
µ
ψA
µ
(12.61)
Sincethereisnofactorofmomentum,thesignisthesamenomatterwhatisgoingonatthevertex
e+
e+
=
e
+
e
=
e
e
+
=
e
e
=
−ieγ
µ
Theµ onthe γ
µ
willgetcontractedwiththeµofthephoton. Thisiseitherinthe η
µν
ofthephoton
propagator,ifthephotonisinternal,ortheǫ
µ
polarizationtensorifthephotonisexternal.
Theγ
µ
αβ
µ
asamatrixwillalsogetsandwichedbythetwofermionlines,suchas
u¯γ
µ
u=u¯
α
γ
αβ
µ
u
β
(12.62)
ifit’s ee− scattering g or r v¯γµv ifit’s s e+e− annihilation. . Thebarredspinor r always goesontheleft,since
the interaction is ψ
¯
A
µ
γ
µ
ψ. If f there is an internal fermion line between the e ends, , its s propagator goes
betweentheendspinors:
p
1
p
2
p
3
=(−ie)
2
u¯(p
3
µ
i(γ
α
p
2
α
+m)
p
2
2
−m2+iε
γ
ν
u(p
1
µ
ε
ν
(12.63)
Inthisexample,the3γmatricesgetmultipliedandthensandwichedbetweenthespinors.
If the fermion n is s a a closed loop, then the the spinor indices s of the multiplied d product t get contracted
witheachother.Ifweclosedtheloopinthediagramabove,itwouldbe
ε
µ
ε
ν
γµ
i(γαp
2
α
+m)
p
2
2
−m2+iε
γν
αα
=Tr
γµ
i(γαp
2
α
+m)
p
2
2
−m2+iε
γν
ε
µ
ε
ν
(12.64)
OftencomputingFeynmandiagramsinQEDwillinvolvetakingthetraceofproductsof γmatrices.
12.6 QED D Feynmanrules
119
Let us finallybecarefulaboutminus signs. A t-channeldiagramwillhave,interms of positionspace
propogators
x
2
x
1
x
4
x
3
y
x
=D(x
1
,x)D(x,x
3
)D(x
2
,y)D(y,x
4
A
(x,y)
Thiscorrespondstoaparticularcontractionfromthetime-orderedproduct
=0|T{Π
A
(x,y)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
3
)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
4
)||0
Theu-channeldiagramgivesthesamethingwithx
3
andx
4
interchanged
x
1
x
x
2
y
x
4
x
3
=D(x
1
,x)D(x,x
4
)D(x
2
,y)D(y,x
3
)
whichgives
=0|T{Π
A
(x,y)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
4
)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
3
)}|0
Thisissupposedtocorrespondtoadifferentcontractionofthesametime-orderedproduct. So,usingthe
anti-commutation,wefind
ψ
¯
(x
4
)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
3
)=
ψ
¯
(x
3
¯
(x
4
)ψ(x
1
¯
(x)ψ(x)
=−ψ
¯
(x
3
)ψ(x
1
¯
(x)ψ(x)ψ
¯
(x
4
)
So we get t a a minus s signinexchanging g x
3
andx
4
. This s is just Fermi i statistics. . Thus s we e get t an relative
minus signfor exchangingtwofermions. . The overallsignisunimportant,asmatrix elements s willalways
besquared.
Forloops,wehavesomethinglike(ignoringthephotons)
x
1
x
2
x
y
=D(x,x
4
)D(x
2
,y)
=
0|T{ψ
¯
(x)ψ(y)ψ
¯
(x)ψ(y)}|0
(12.65)
Togetthesignweneedtocompareittotheproduct withouttheloop,whichcontributestothevacuum
matrixelement,butcomesfromthesametime-orderedproduct
x
y
x
1
x
y
x
2
=D(x,x)D(y,y)
=
0|T{ψ
¯
(x)ψ(x)ψ
¯
(y)ψ(y)}|0
(12.66)
=−
0|T{ψ
¯
(x)ψ(y)ψ
¯
(y)ψ(x)}|0
(12.67)
Sowegetanoverallminus signfortheloop. . Thiswillalways s happenfor any loopwithanynumber of
fermions.
Thus
• -1forinterchangeofexternalfermions
120
QuantumElectrodynamics
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