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where
j
µ
(x)=ψ
¯
(x)γ
µ
ψ(x)
(20.10)
Thiscurrentdiffers fromthething g ψ
¯
(x
1
)ψ(x
2
)inthe4-point functionbecausethecurrentis evaluateda
singlespace-timepoint. Wecanrepresentthiswithanoperatorinsertion
p
1
p
2
+
p
1
p
2
+k
p
1
−k
k
p
2
Ifyoufindthisoperatorlanguageconfusing,don’tworryaboutit. Intheend,wewillhavecalculatedthe
fulle
+
e
→µ
+
µ
. Thecurrentlanguagejustletsusbreaktheproblem intopieces,sowecanseewhat’s
importantateachstage,butifyouprefertothinkaboutdoingitallatonce,gorightahead!
Inadditiontotheloopcorrections tothe4-point function,tocanceltheinfrareddivergences,wewill
needtocalculatealsorealemissiongraphs.Thesearethetreelevelcontributionstoa5-pointfunction
G
5
=
0|T{ψ
¯
ψψ
¯
ψA
µ
}|0
(20.11)
Orwiththecurrent,simplythecurrentmatrixelementsG
J
withanextraphoton.Thegraphsare
+
We willdotheloops first,thentherealemissiongraphs,andthenshowthat we cantakem
γ
→0 after
thenarecombinedintothefullcrosssectiondσ(e
+
e
→µ
+
µ
(+γ)).
Wetakem
e
=m
µ
forsimplicity.
20.3 Loops
20.3.1 Treelevelgraph
Thetree-levelgraphthatweneedtointerferewithis
iM
0
=
p
1
p
2
=iu¯(p
2
µ
v(p
1
)
(20.12)
Thenthenormalizationissetby
pols
|M
0
|
2
=−Tr[p
2
γ
µ
p
1
γ
µ
]=8p
1
p
2
=4Q
2
(20.13)
Sowritingσ
0
=4Q
2
letsusrescaleourresultstomatchthefulle
+
e
→µ
+
µ
.
20.3 Loops
211
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20.3.2 VertexCorrection
Thevertexcorrectionis
iM
Γ
=
p
1
p
2
+k
p
1
−k
k
p
2
=iu¯(p
2
µ
v(p
1
)
(20.14)
with
Γ
µ
=(−i)(i)
2
(−ie)
2
d
4
k
(2π)4
g
αβ
−(1−ξ)
k
α
k
β
k2
k2
γ
α
p
2
+k
(p
2
+k)2
γ
µ
p
1
−k
(p
1
−k)2
γ
β
(20.15)
whichisthesameamplitudewe’vealreadyencountered.Wewroteitintermsofformfactors
Γ
µ
=F
1
(p
2
µ
+
µν
2m
e
p
ν
F
2
(p
2
)
(20.16)
where p=p
1
+p
2
isthemomentumcomingintothecurrentvertex. Recall,onlyF
1
isdivergent,andthe
divergenceisgivenbyF
1
(0)=−δ
1
+finite.
TheinterferencewithM
0
willgive
M
Γ
M
0
+M
0
M
Γ
=2Tr[p
2
Γ
µ
p
1
γ
µ
]+c.c.
(20.17)
=2F
1
(p2)Tr
p
2
γµp
1
γ
µ
+
1
m
e
p
ν
F
2
(p2)Tr[p
2
σµνp
1
γ
µ
(20.18)
Thefirsttermjust gives 2F
1
(Q
2
0
. Thesecondtermhasanoddnumberof γ matricesinthetrace,soit
vanishes(NB:thiswouldnot be trueif the fermionsweremassive,or ifwelookedat theNNLO,α
2
cor-
rection). Thusthetotalcorrectionatorderαisgivenby2F
1
(Q
2
0
.
WithaPauli-VillarsregulatorfortheUVandaphotonmassm
γ
fortheIR
F
1
(Q2)=
α
0
1
dxdydzδ(x+y+z−1)
log
2
zm
γ
2
−xyQ2
+
(1−x)(1−y)Q2
zm
γ
2
−xyQ2
(20.19)
Thisisactuallyconsiderablysimplerthanthegeneralexpression,becausewearetakingmasslessfermions.
ThefirsttermisIRfinite,anditgives
F
1
(Q
2
)=
α
3
4
+
1
2
log
Λ
2
Q2
+
(20.20)
ThesecondtermisIRdivergentbutUVfinite. Theintegralsarehardtodo. Expandingnearm
γ
=0gives
F
1
(Q2)=
α
1
2
log
Λ2
m
γ
2
1
2
log2
m
γ
2
Q2
3
2
log
m
γ
2
Q2
7
4
+
π2
6
+O(m
γ
2
)
(20.21)
NoteatleastthatthelogΛ
2
termagreeswithwhatwehadabove.
Sotheresultofthisloop’scontributiontothetotalrateis
σ
Γ
=
α
σ
0
−log
2
m
γ
2
Q2
−3log
m
γ
2
Q2
7
2
+
π
2
3
+log
Λ
2
m
γ
2
(20.22)
Note the e double logs. . Double e logs are diagnostic of f infrared d divergences. . They y have e a a name: : Sudakov
doublelogarithms. Thereisalotoffunphysicsassociatedwiththem.
212
Infrared Divergences
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20.3.3 counterterms
Next,considerthecountertermcontribution.
Inbareperturbationtheory,wearetryingtocalculatetheGreen’sfunction
0|j
µ
R
ψ
¯
R
ψ
R
|0
=
Z
2
Z
j
0|j
µ
0
ψ
¯
0
ψ
0
|0
(20.23)
sowegetthecountertermsδ
2
−δ
j
. Wealreadyknowthatδ
j
=0bycurrentconservation,soallthat’sleft
isδ
2
.
RecallthatinPVwithamassivephotonandmasslesselectron
δ
2
=−
α
log
Λ
2
m
γ
2
(20.24)
Thusincludingtheinterferenceofthis counterterm withthetree-levelgraph,weget afactoroftwofrom
thecross-terming,andtherefore
σ
ct
0
α
log
Λ
2
m
γ
2
(20.25)
Thusthesumoftheradiativecorrectionsis
σ
V
ct
Γ
=
α
σ
0
−log
2
m
γ
2
Q2
−3log
m
γ
2
Q2
7
2
+
π
2
3
(20.26)
whichisUVfinite.
Thesamecalculationinrenormalizedperturbationtheorygetsafactorof δ
2
−δ
j
fromthevertex
L=
+
Z
2
Z
j
j
µ
ψ
¯
γ
µ
ψ
(20.27)
soit’sthesamething.
20.3.4 expandingouttheprocess
Let’s take in n moreof f thefulle
+
e
→ µ
+
µ
process to o see what’s goingona littlebetter. . Supposewe
workmoremorestepin.Thatis,wetrytocompute
p
1
p
2
Wherenowthecurrentsourceiscreatingaphoton. LetmecallthiscurrentZ
µ
. Sowearecalculating
0|T{Z
µ
ψ
¯
ψ
|0
(20.28)
Notethis operator Z
µ
s is adifferent operator fromtheleptonnumber current t j
µ
= ψ
¯
γ
µ
ψ wewereusing
before. Youcanthinkofthiscurrentas s aclassicalradiationsource. . Youcanalsothinkofis s simplyas an
external photon, however, we want t the photon to be off-shell, so that’s s not t quite e right. . Instead, , try
thinkingofitlikeaheavygaugeboson,saytheZ. Ok,justthinkofitaswhatever r youlike. . I’mgoingto
thinkofitasanoperator.
Thetreelevelcontributionisnow
spins,Zpols
|
0|T{Z
µ
ψ
¯
ψ
|0|
2
=
e
2
Q4
Tr[p
2
γ
µ
p
1
γ
ν
]=4
e
2
Q2
(20.29)
20.3 Loops
213
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It’sthesame,butwegetanextrafactorofefromthevertexandacoupleof
1
Q2
’sfromthephotonpropa-
gator. Thelooponthemuonsideisthesameasbefore
(20.30)
It’sdivergentpartisstill −δ
1
.
Since the current caps s the endof f the photon line, , so the photon should be considered internal. . So
thereareotherdiagramsandcountertermsaswell.First,thereistheinternalphotonloop
∼Π
2
(p2)
Thisisjustthevacuumpolarizationdiagram. Wecomputeditinthet-channel. Inthes-channel,it’sthe
sameuptoacrossingsymmetry(t→Q
2
),so
Π
2
(Q2)=−
π
0
1
dxx(1−x)log
Λ2
m
e
2
+Q2x(1−x)
(20.31)
Thisfunction’sIRdivergencesiscutoffnaturallybytheelectronmass. Thedivergenceis
Π
2
(0)=−
α
log
Λ
2
m
e
2
(20.32)
whichisjust −δ
3
.
Thecounterterms arealittledifferent. . Including g the fieldstrengthrenormalizationinthematrix ele-
ment(bareperturbationtheory)gives
0|T{Z
µ
R
ψ
¯
RψR
|0=Z
2
Z
Z
0|T{Z
µ
0
ψ
¯
0
ψ0
|0
(20.33)
(the
in
Z
Z
isjust aconvention). . Sowegetafactor r of f δ
2
1
2
δ
Z
forthecounterterms fromthefield
strengths. Wealsoget t afactorof f −δ
e
fromtherenormalizationof thebare electric charge intheQED
vertex:e
0
=e
R
−δ
e
.
Thus the e total UV divergences s are e δ
2
1
2
δ
Z
(Green
s function) − − δ
e
(electron) − − δ
3
(vac pol loop) −
δ
1
(vertexloop)=−
1
2
δ
Z
−δ
e
−δ
3
. Butsinceδ
e
=−
1
2
δ
3
wegetδ
Z
3
. Ifwehadincludedthis s current in
theLagrangian(renormalizedperturbationtheory),itwouldbe
L=
+
Z
3
Z
Z
Z
µ
A
µ
(20.34)
Sowegetafactorof
1
2
δ
3
1
2
δ
Z
fromthisterm.WealsohavetherealQEDvertexcounterterm,δ
1
andthe
vertexloop −δ
1
. Thenthevacuum m polarizationgives s −δ
3
andits counterterm δ
3
. Soweareleft t with
just
1
2
3
−δ
Z
). Thuswefindδ
Z
3
. Thefinitepartsoftheloopsareofcoursethesame. Wearequickly
confirmingthatbareandrenormalizedperturbationtheoryarequitetriviallythesamething.
Addthecountertermtothevacuumpolarizationgraphgives,forQ≫m
e
,
Π
2
(Q
2
)+δ
3
=
π
0
1
dxx(1−x)log
1+
Q
2
m
e
2
x(1−x)
(20.35)
α
log
Q
2
m
e
2
(20.36)
Thisgivesanadditionalcontributiontothetotalcrosssection. Notethatthechargeappearinghereisthe
renormalizedcharge,definedbywhatismeasuredintheCoulombforceat p=0.
214
Infrared Divergences
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Alsonotethat the electronmassappears,butwewantedtosetm
e
=0. Recallthatour r definitionof
theelectricchargewasthingthingmeasuredatp=m
e
∼0. So,them
e
in
α
log
Q2
m
e
isreallyjusttherenor-
malizationpoint.Thatis,moregenerally,wecanwrite
Π
2
(Q
2
)=
α
R
log
Q
2
p
0
2
(20.37)
whereα
R
isdefinedastheelectricchargemeasuredatp=p
0
. Forcomparisonwithotherexperiments,it’s
easierjusttoleavem
e
0forthiseffect.
First, note that t the e tree level cross s section becomes (stillnormalization the e phase space e to o 1) σ
0
=
e
2
4Q
2
. So
σ
tot
=4
e
R
2
Q2
+4
e
R
2
Q2
e
R
2
12π2
log
Q
2
m
e
2
(20.38)
Thisisnothingmorethantherunningoftheelectricchargeinthebarecrosssection
σ
tot
=16πQ
2
α
Q
(20.39)
with
α
Q
eff
(Q
2
)=α
R
1+
α
R
log
Q
2
m
e
(20.40)
Thisisthesamescale-dependenteffectivecouplingwewereusingforthecorrectionstoCoulomb’slaw.
20.3.5 fulle
+
e
→µ
+
µ
Nowwhathappenswhenwecalculatethefullmatrixelements
e
µ
e+
µ
+
=iM
0
Correspondingtomatrixelements
0|e+eµ+µ|0
(20.41)
withnoweirdoperatorsanymore(phew!). Thetotalcrosssectionis
σ
0
=
4πα
Q
2
3Q2
(20.42)
where I have included d the e vacuum-polarization n loop in writing α
Q
insteadof α
R
for this s cross s section.
Rememberα
Q
R
+O(α
R
2
).
Fortheloops,thereisthevertexcorrectionontheleftandthevacuumpolarizationgraph,plusanew
graph–thevertexcorrectiononthelefthandside.Thusthetotalloopdivergencesare
divergences=−δ
1
−δ
3
−δ
1
(20.43)
For counterterms,we get justthetwovertexcounterterms andthe photonpropagatorcounterterm. . The
externallinesareamputatedsotheydonotcontribute.Thuswehave
counterterms=δ
1
3
1
(20.44)
Soallthedivergencescancel.
Thusthetotalloopcontributionis
σ
V
0
α
2
−log
2
m
γ
2
Q2
−3log
m
γ
2
Q2
7
2
+
π
2
3
20.3 Loops
215
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andα
R
isdefinedintermsofthechargemeasuredatp→0. Sowejustgettwicewhatwehadforjustthe
looponthemuonside,plusthevacuumpolarizationeffect.
20.4 Real-emission:Finalstateradiation
Therealemissiongraphsareeasytocompute. However,thephasespaceintegralswithamassivephoton
can be e somewhat tedious. . This s analysis is taken from Rick k Field’s s book “Applications of f perturbative
QCD”.
First,takemasslessphotonemissionfromthemuonsideonly. Weneedtocalculatethesediagrams
p
2
p
4
p
3
+
Whichare
u¯(p
2
µ
i
p
4
+p
2
γ
ν
v(p
3
)+u¯(p
2
µ
i
p
4
+p
3
γ
ν
v(p
3
)
(20.45)
notep
4
+p
3
=p−p
2
.
LetusdefinetheMandelstamvariables
s=(p
3
+p
4
)
2
=Q
2
(1−x
2
)
(20.46)
t=(p
2
+p
4
)
2
=Q
2
(1−x
1
)
(20.47)
u=(p
2
+p
3
)
2
=Q
2
(1−x
3
)
(20.48)
Notethatnows+t+u=Q
2
sincethecurrentisoff-shell.
Thecalculationisstraightforwardandwefind
d2σ
dsdt
0
α
s
t
+
t
s
+
2u
st
(20.49)
orequivalently
d
2
σ
dx
1
dx
2
0
α
x
1
2
+x
2
2
(1−x
1
)(1−x
2
)
(20.50)
Withamassive photon,this gets modified. . There e arebothexplicit factorsof f m
γ
that comeintothe
matrixelements,butalsofactorsofm
γ
inthephasespaceintegralstomakeacrosssection. Theresultis
d
2
σ
dx
1
dx
2
0
α
1
(1−x
1
)(1−x
2
)
x
1
2
+x
2
2
2(x
1
+x
2
)+
(1−x
1
)
2
+(1−x
2
)
2
(1−x
1
)(1−x
2
)
+2β
(20.51)
withβ=
m
γ
2
Q2
. Now,s+t+u=Q2+Q2β2.
Tocalculatethetotalcrosssection,wehavetointegrateoverphasespace. Since
x
1
+x
2
+x
3
=2−β
(20.52)
0<x
i
<1
(20.53)
Wegetrestrictedphasespace. Forexample,if f x
3
=0 then x
1
<1−βorelsex
2
wouldhavetobelarger
than1.Thelimitsare
0x
i
1−β
(20.54)
1−β−x
1
x
2
1−x
1
−β
1−x
1
(20.55)
216
Infrared Divergences
Thus,
σ
R
=
0
1−β
dx
1
1−β−x
1
1−β−x
1
1−x
1
dx
2
d
2
σ
dx
1
dx
2
0
α
log
2
m
γ
2
Q2
+3log
m
γ
2
Q2
π
2
3
+5
(20.56)
Recalling
σ
V
ct
Γ
=
α
σ
0
−log2
m
g
2
Q2
−3log
m
g
2
Q2
7
2
+
π2
3
(20.57)
Weseethatallthelogm
γ
termspreciselycancel,andweareleftwith
σ
R
V
=
σ
0
(20.58)
Soweseethatifweincludethevirtualcontributionandtherealemission,theIRdivergencescancel.
Wehave,atthispoint,onlyincludedrealemissionfromthemuonside,andthevertexcorrectionloop
fromthemuonside. We e mustnowalsobeabletocancelthe infrareddivergence fromthevertexcorrec-
tionontheelectronside.
20.5 Initialstateradiation
That’s allwellandgoodfor the muonside,whereit makes sensetosum over thepossibilityof emitting
muonsandphotonsincalculatingthetotalcrosssection.
Ifweintegrateoverallinitialstatephotonsaswell,wewouldfind
σ
tot
[e
+
e
(+γ)→µ
+
µ
(+γ)]=σ
0
1+
eff
(20.59)
whichisperfectlyfinite.However,shouldn’twebeabletocollidethee
+
e
withoutsummingalsooverini-
tialstatephotons?
Actually, no. . While e it does not make e sense e to o integrate e over all possible initial-state photons, , it t is
impossibletomakesureyoujusthaveanelectronandnotanelectronandabunchofphotons. Soatleast
youmustsum overthesoft stuff,whichwill be enoughtocut-offthedivergence. . Onereasonthis s might
seemconfusingis becauseour languagehas beenimprecise. . Allalongwe e havepretendedwe’vebeencol-
lidingmomentum eigenstates,i.e planewaves. Infact, wearecollidingwavepackets. . Theuncertainty y in
the wave-packet has to do with h not t knowinghow w much energy thereis, andalso not t knownhow w much
energyisintheelectronorinitssoupofsurroundingsoftphotons,whicharealsointhewave-packet.
Forexample,supposethewavepackethassomeenergyuncertaintyE
i
≪Q. E
i
canbeverysmall,even
E
i
≪m
e
,butitmustbegreaterthanzero. ThenintegratingthephotonenergyuptoE
i
willgive
σ
R
0
α
log
2
m
γ
2
E
i
2
+3log
m
γ
2
E
i
+c
(20.60)
forsomeconstantc. Thenaddingthistoourcorrectionwillcancelthelogsof m
γ
leaving
σ
tot
[e
+
e
(+γ)→µ
+
µ
(+γ)]=σ
0
1+
log
2
Q
2
E
i
2
+3log
Q
2
E
i
2
+c

(20.61)
whichisfinite–itdoesn’tdependonthefictitiousphotonmass.
How do o we deal l with h this s in practice? ? First t of f all, we e could just choose E
i
of some e scale e so o that
αlog
2
Q2
E
i
2
≪1. Butdowehavetoreallyknowwhattheuncertaintyisonourwavepacketsandincludethat
inthecalculation? Nowedon’t. . Infact,therearemany y uncertainties whicharemuchlarger thanwave-
packet width, , suchas s the impact parameter of the e
+
ande
we are colliding. . Averaging g over impact
parameters completelyoverwhelms any uncertainty-relation n boundedwave-packet t uncertainty. . The e prac-
ticalresolutionisthatwe shouldjuststick toobservables that arenot sensitivetotheelectronside. Let
megivesomeexamples.
20.5 Initialstateradiation
217
20.6 Applicationsofinfrareddivergences
Aprocesslikee
+
e
→µ
+
µ
isnotwell-definedinperturbationtheorybecauseoftheinfrareddivergence.
Wefoundthatweneedtosumoveradditionalphotonemissionstogetafinitequantity. Quantitieswhich
areinsensitivetotheinfraredregulator arecalledinfrared safe. Thereisatheorem m due toKinoshita-Lee
andNauenberg(KLNtheorem)whichsaysthatanyphysicallyobservablequantityisinfraredsafe.
Hereareafewexamplesoftheimportanceofinfraredsafetyinpractice:
• Decaywidths
For situations s where e the e cross s sectionis important, such as s Z Z bosonproduction, , it’s usually resonance
production. Infact, , the soft photons s help p bring the e Z Z to o theresonance peak, aprocess s calledradiative
return. Inthis s case, , it’s just the decay widthyoumeasure,so you only needthe finalstateloops. . The
decaywidthiscalculable,finite,anddoesnotdependonwhetheritwas e
+
e
or somethingelsethatpro-
ducedtheZ. Thusthecorrectiontothewidthisawaytotestforthe
correction.
For example, , suppose the Z Z decays s to o quarks s and d to muons. . Quarks s havecharge
2
3
or −
1
3
,so the
photonloopcorrectionwillbesmallerforthewidthtoquarksthanthewidthtomuons. Thustheratioof
partialwidths directlymeasures αandthechargesoftheparticles inthedecay. . The electronside drops
outcompletely. Sowedon’tneedtobotherwiththeE
i
cutofforthevertexcorrectionloop.
• ProtonsinQCD
For QCD, , where you u don’t have e free quarks, , the wave-packets s have e a a natural size e – – the proton. . The
protonisabundleofquarksandgluons,wherethegluonsaremostlysoft,andcanbethoughtoftoafirst
approximation like e the photons s we e have been using in this discussion. . These e virtual and d soft t gluons
stronglyinfluencethedistributionofenergies ofquarks insidetheproton,inacalculableway. . Thisisthe
physicalbasisof partondistributionfunctions. Sothere’s s alotofphysics ininitialstateradiation,it just
doesn’thavemuchtodowithtotalcrosssectionsinQED.
• Finalstateenergycuts.
Swappingalogm
γ
foralogE worksforfinalstateradiationtoo,andthereitisimportant.Wedon’thave
tointegrateoverallphotonsonthe µ
+
µ
side. Wecandosomethinglessinclusive (moreexclusive),and
integrateonlyoverphotonsuptoanenergyE
f
. Ifweputacutonthephotonenergy,wewillgetlogsof
thatenergydividedby Qinthecrosssection.
tot
dE
f
tot
α
2
log
2
E
f
2
Q2
+
(20.62)
This E
f
is often n a a realistic c quantity related d to o actual parameters of f an experiment. . For r example, the
experimentBABARat SLACmeasures the decays of B mesons tokaons andphotons(B→Kγ). . They
areonlysensitive tophotons harder than1.6 6 GeV. . Softer r photonsarebelowthethresholdsensitivityof
their detector. . Thus s this logis avery important t quantitative e feature of the cross sectiontheymeasure,
andmustbeincludedinattemptstoconstrainnewphysics(eg.supersymmetry)usingB→K+γdecays.
• Soft/Collinearsingularities
NotethatforsmallE
f
≪Qtherewillbelargelogarithmswhichmayinvalidateyourperturbationexpan-
sion. Infact, , it seems s that ifwelookat dσ/dE
f
it willbe completely dominatedby small E
f
. Another
wayofsayingthisisthatthedominantradiationwillbesoft. WecanseethisfromtheFeynmandiagrams
directly
p
2
p
4
p
3
218
Infrared Divergences
Thepropagatoris ∼
1
(p
4
+p
2
)2
1
E
γ
E
2
. Sothedivergence e is comingfromintegrating
0
E
f
1
E
γ
dE
γ
. More
generally,thispropagatorwillvanishwhen
p
4
·p
2
=0
(20.63)
Writingp
4
=E
γ
(1,0,0,1),thishappenswheneitherE
γ
→0,whichiscalledasoftdivergence,orwhenp
3
isalsointhez direction,that is p
3
·p¯
4
=0. That t iscalledacollinear singularity. . Sothecrosssectionis
dominatedbysoftandcollinearsingularities.
• Jets
Itisnot hardtoseethat whentheradiationis softenough,theenhancementinthecross sectionwillbe
substantial. Thisiscompensatedforbythefactthatthesoft radiationishardtosee. . Butwhatitmeans
isthattheoutgoingparticle,saythemuon,willbeaccompaniedbylotsofsoftorcollinear photons. . This
broadenstheoutgoingparticlefrom aplanewave intosomethinglikeawave-packet. . Thiswave-packet t is
notaGaussianwave-packet,andnoteasytodescribetheoretically,butithasacalculableshape.
While thesefinal-statemuon-photonwave-packetsarearehardtoseeinQED,theyareeasytoseein
QCD. InQCD,the e muonis replacedby aquark andthephotonreplacedbya gluon. The quark itself,
and the e additional soft t gluons s turninto o separate observable particles,suchas s pions andkaons. . Thus s a
quarkinQCDturnsintoajetofhadrons.
These jets s are e a a very y real and d characteristic c phenomenon n of f all high-energy y collisions. . We e have
explainedtheirexistencebystudyingtheinfraredsingularitystructureofFeynmandiagramsinquantum
fieldtheory.
• Lambshift
The dominant contribution n tothe e Lambshift is given n by y an n infrared divergent t effect. . The e loop p gives s a
log
m
γ
2
m
e
2
andtherealemissioncontributionreplacesthism
γ
withthebindingenergyoftheorbitalE
B
. The
resultis∆E∼log
m
γ
2
E
B
2
∼1000MHz.Wewilldiscussthisindetailinthenextlecture.
20.7 e
+
e
→µ
+
µ
indimreg
Thecalculationofthetotalcrosssectioncanbedonealsoindimensionalregularizationwithminimalsub-
tractionor MS
. RepeatingthecalculationthiswayhelpsillustrateregulatorindependencenotjustofUV
quantities,butalsoofIRquantities. This s partis takenfromMuta“FoundationsofQuantum Chromody-
namics”,whichyoushouldlookattogetmoredetails.
20.7.1 loops
TheimportantloopindimensionalregularizationistheF
1
form factor. Wecomputeditalreadywithan
electronandphotonmass
F
1
(p
2
)=
e
R
2
(4π)
d/2
µ
4−d
0
1
dxdydzδ(x+y+z−1)
Γ(2−
d
2
)(2−ε)2
2−d/2
(20.64)
+
Γ(3−
d
2
)
3−d/2
p
2
[2(1−x)(1−y)−εxy]+m
e
2
[2(1−4z+z
2
)−ε(1−z)
2
]
(20.65)
where
∆=(1−z)
2
m
e
2
+zm
γ
2
−xyp
2
(20.66)
20.7 e
+
e
→µ
+
µ
indimreg
219
Anicethingaboutdimensionalregularizationisthat canbeusedtoregulateinfraredaswellasultra-
violetdivergences.Thisisbecause
ddk
(2π)4
1
k4
(20.67)
isfiniteinatsmallkford>4. Thatmeansshouldexpandford=4+ε. However,aslongaswekeepdas
avariable,wewill alwaysbeabletodotheintegrals s andexpandinε,not worryingabout whether then
signof εispositiveornegative. . So,wecanjustsetm
e
=m
γ
=0before doingtheintegrals,thenexpand
neard=4.
Form
e
=m
γ
=0indimregtheformfactorat p
2
=Q
2
becomes
F
1
(Q
2
)=
e
2
(4π)
d/2
µ
2
−Q2
4−d
0
1
dxdydzδ(x+y+z−1)
(20.68)
×
Γ(2−
d
2
)
(2−ε)
2
2
1
(xy)
2−
d
2
+Γ(3−
d
2
)
2(1−x)(1−y)−εxy
(xy)
3−d/2
(20.69)
=
α
4πµ2
−Q2
ε
Γ(3−
d
2
)
Γ(
d
2
−1)Γ(
d
2
)
Γ(d−1)
8
ε2
+
2
ε
−2
(20.70)
Expandingoutthe
1
ε
(−Q
2
)
ε
giveslog(−Q
2
)=logQ
2
+πi. AtorderαweonlyneedRe[M
Γ
M
0
]andM
0
is
real,sowecandroptheseimaginaryparts.However,notethat
log
2
(−Q
2
)=log
2
Q
2
+2πilogQ
2
−π
2
(20.71)
Sowegetveryimportantπ
2
termswhichaffectthefinalanswer.
Then,
Re[F
1
(Q
2
)]=
α
8
ε2
6+4log
µ˜2
Q2
ε
−2log
2
µ˜2
Q2
−log
µ˜2
Q2
−8+
2
6
(20.72)
Now, recall that t with the e massive electron n and d photon, , these masses s regulated the infrared diver-
gences. There,wefound
F
1
(p2)=
α
2
ε
+
(20.73)
SincetheUVdivergenceshavetodowithhighenergy,they areindependent ofanysmallmasstheelec-
tronor photonmight have. . Thus s wecanfigure outwhichoftheε’sinF
1
are UVdivergencesandwhich
areIRdivergences
Re[F
1
(Q
2
)]=
α
2
ε
UV
8
ε
IR
2
8+4log
µ˜2
Q2
ε
IR
−2log
2
µ˜2
Q2
−log
µ˜2
Q2
−8+
2
6
(20.74)
WealreadyknowthatthisUVdivergencewillgetcanceledbythecounterterm.Butinordertoappreciate
the IRcontentofthecounterterm,weneedtounderstandhowtocomputedimensionlessintegralsindim
reg.
20.7.2 dimensionless s integrals indim-reg
Thecountertermweneedisδ
1
. Recallthatδ
1
cancels thedivergenceofthevertexcorrectionandδ
2
can-
celsthedivergenceoftheselfenergydiagram. Sinceδ
1
2
,itiseasierjusttolookattheself-energydia-
gramwhichissimpler.
220
Infrared Divergences
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