c# display pdf in browser : Add page to pdf in preview software application dll winforms azure html web forms QFT-Schwartz14-part445

togive
x
j+1
e
−iHδt
x
j
=
dp
x
j+1
|pp|e
−i
p
2
2m
+V(x
j
,t
j
)
δt
|x
j
(14.11)
=e
−iV(x
j
,t
j
)δt
dp
e
−i
p
2
2m
δt
e
−ip[x
j+1
−x
j
]
(14.12)
NowwecanusetheGaussianintegralformulawecomputedabove.
−∞
dpe
1
2
ap2+Jp
=
a
e
J
2
2a
(14.13)
Forourintegral,a=i
δt
m
,J=−i(x
j+1
−x
j
)soweget:
x
j+1
e−iHδt
x
j
=Ne
−iV(x,t)δt
e
i
m
2
δt
(x
j+1
−x
j
)
2
(δt)
2
(14.14)
=Ne
i[
1
2
mx˙2−V(x,t)]δt
(14.15)
=Ne
iL(x,x˙)δt
(14.16)
where N N is s the normalization, , whichwe’ll ignore. . All l that happenedhere is that the Gaussianintegral
performedtheLegendretransformforus,togofromH(x,p)→L(x,x˙).
So,
f|i
=N
dx
n
dx
1
e
iL(x
n
,x˙
n
)δt
e
iL(x
1
,x˙
1
)δt
(14.17)
andtakingthelimitδt→0weget
f|i=N
Dx(t)e
iS(x,x˙)
(14.18)
wherewesumoverallpathsx(t)andtheactionisS=
dtL.
14.4 PathintegralinQFT
Thefieldtheoryderivationinfieldtheory,isverysimilar,butthesetofintermediatestatesis morecom-
plicated. Let’sstartwithjustthevacuummatrixelement
0,t
f
|0,t
i
∼
0
e
−iH(t
f
−t
i
)
0
(14.19)
InQMwebrokethisdownintointegralsover|xx|forintermediatetimes. Thestates|xareeigenstates
ofthexˆoperator. Infieldtheory,theequivalentof xˆis s φ
ˆ
(x),where,incaseyou’veforgotten,
φ
ˆ
(x
)=
d
3
k
(2π)3
(a
k
e
ik
x
+a
k
e
−ik
x
)
(14.20)
Thisisaninfinitenumberofoperators,oneateachpointx. Weputthehatonφtoremindyouthatit’s
anoperator.Thentheequivalentof|x
x|isacompletesetofeigenstatesof φ
ˆ
φ
ˆ
(x
)|Φ=Φ(x
)|Φ
(14.21)
TheeigenvaluesarefunctionsofspaceΦ(x
). Ofcourse,Φ(x
)willhavetosatisfysomekindofnormaliza-
tioncondition,sothatithasafixednumberofparticles,butwewillworryaboutthatlater.
Nowourbasisisoverstates|Φ
. Soourmatrixelementcanbechoppedupinto
0,t
f
|0,t
i
=
1
(x)
n
(x)
0
e
−iH(t
n
)δt
n
Φ
2
|e
−iH(t
2
)δt
1

Φ
1
|e
−iH(t
1
)δt
|0
14.4 Pathintegral l inQFT
141
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Noweachofthesepiecesbecomes
Φ
j+1
|e
iH(t
j
)δt
j
=e
i
d
3
xL[Φ,Φ
˙
]δt
(14.22)
Andtakingδt→0gives
0,t
f
|0,t
i
=
DΦ(x,t)e
iS[Φ]
(14.23)
whereS[Φ]=
d4xL.
So the path integral tells s us tointegrate over all classical fieldconfigurations Φ. . Note e that Φ is not
justtheone-particle states,itcanhave2-particle states,etc. . We e canrememberthis by drawingpictures
for the paths – – including disconnectedbubbles. . Actually, , we really sum over r all l kinds of discontinuous,
disconnectedrandomfluctuations,but the onesthatdominate thepathintegralwillhave anicephysical
interpretationaswecansee.
14.5 Classicallimit
Let’sputbackforamoment.Sincehasdimensionsofaction,itissimply
0,t
f
|0,t
i
=
DΦ(x,t)e
i
h
¯
S[Φ]
(14.24)
Let’sforgetabouttheiforamomentandsupposeweneedtocalculate
DΦ(x,t)e
1
h
¯
S[Φ]
(14.25)
Inthiscase,the integralwouldclearlybedominatedbyΦ
0
whichiswhereS[Φ]hasaminimum –every-
thingelsewouldgiveabiggerS[Φ]andbeinfinitelymoresuppressedas→0.
Now,whenweputtheibackin,thesamethinghappens,notbecausethenon-minimaltermsarezero,
butbecause away from theminimum youhave tosumover phasesswirlingaroundinfinitelyfast. . When
yousuminfinitelyswirlingphases,youalsogetsomethingwhichgoestozero. There’satheoremthatsays
this happens, related d to the method of stationary descent or sometimes method of steepest descent.
Anotherway tosee itis touse the more intuitive casewithe
−S[Φ]/
. Since e weexpect the answertobe
welldefined,it shouldbeananalyticfunctionof Φ
0
. So o we cantake →0inthe imaginary direction,
showingthattheintegralisstilldominatedbyS[Φ
0
].
Inanycase,theclassicallimitis dominatedby afieldconfigurationwhichminimizes theaction. . This
is the classical l paththat a particlewouldtake. . Sothepathintegral l has s anice intuitive classicalcorre-
spondence.
14.6 Time-orderedproducts
Supposeweinsertafieldatfixedxandtintothepathintegral
I=
DΦ(x,t)e
iS[Φ]
Φ(x
j
,t
j
)
(14.26)
Whatdoesthisrepresent?
Goingbackthroughourderivation,wehave
I=
1
(x)
n
(x)
0
e
iH(t
n
)δt
n
Φ
2
|e
iH(t
2
)δt
1

Φ
1
|e
iH(t
1
)δt
|0
Φ
j
(x
j
)
sincethesubscriptonΦisjustit’s pointintime,weknowwhichΦ
i
’sthesecorrespondto. Let’stakethe
partwithjustΦ
j
j
(x)
e
iH(t
n
)δt
j
Φ
j
(x
j
)Φ
j
|
ˆ
(x
j
)
j
(x)|Φ
j
Φ
j
|
(14.27)
142
PathIntegrals
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SowegettoreplaceΦ(x
j
)bytheoperator φ
ˆ
(x
j
)stuckinatthetimet
j
. Thenwecancollapseupallthe
integralstogive
DΦ(x,t)e
iS[Φ]
Φ(x
j
,t
j
)=
0
φ
ˆ
(x
j
,t
j
)
0
(14.28)
Ifyoufindthecollapsing-up-the-integrals confusing,justthinkaboutthederivationbackwards. . Aninser-
tionof φ
ˆ
(x
j
,t
j
)willendupby|Φ
j
Φ
j
|,andthenpulloutafactorofΦ(x
j
,t
j
).
Nowsayweinserttwofields
DΦ(x
,t)e
iS[Φ]
Φ(x
1
,t
1
)Φ(x
2
,t
2
)
(14.29)
Thefieldswillget insertedintheappropriatematrixelement. Butifyoucheck,youwillseethattheear-
lierfieldwillalwayscomeoutontherightofthelaterfield.Soweget
DΦ(x)e
iS[Φ]
Φ(x
1
)Φ(x
2
)=
0
T
φ
ˆ
(x
1
ˆ
(x
2
)
0
(14.30)
Wegettimeorderingforfreeinthepathintegral.Ingeneral
DΦ(x)e
iS[Φ]
Φ(x
1
)
Φ(x
n
)=
0
T
φ
ˆ
(x
1
)
φ
ˆ
(x
n
)
0
(14.31)
Whydoes this work? ? There e area fewcross checks you u cando. . As s aneasy one,suppose the answer
were
DΦ(x)e
iS[Φ]
Φ(x
1
)Φ(x
2
)=
0
φ
ˆ
(x
1
ˆ
(x
2
)
0
(14.32)
Well,theleft hand d side e doesn’t care whether Iwrite Φ(x
1
)Φ(x
2
) or Φ(x
2
)Φ(x
1
),since theseare classical
fields. SowhatwoulddeterminewhatorderIwritethe fieldsontheright? Weseeit mustbesomething
thatmakesthefieldseffectivelycommute,likethetime-orderingoperator. We’lldoanothercheckoncewe
getalittlemoreexperienceplayingwiththepathintegral.
Fromnowon,we’lljustuseφ(x)insteadofΦ(x),fortheclassicalfields.
14.6.1 Currentshorthand
There’sagreatwaytocalculate pathintegralsusingcurrents. . Let’s s addasourceterm toouraction. . So
define
Z[J]=
Dφexp
iS[φ]+i
d
4
xJ(x)φ(x)
(14.33)
Then,
Z[0]=
Dφe
i
d
4
xL[φ]
=0|0
(14.34)
Next,observethat
d
dJ(x
1
)
d
4
xJ(x)φ(x)=φ(x
1
)
(14.35)
So,
−i
dZ
dJ(x
1
)
=
Dφexp
iS[φ]+i
d
4
xJ(x)φ(x)
φ(x
1
)
(14.36)
Andthus
−i
dZ
dJ(x
1
)
|
J=0
=
Dφexp
{
iS[φ]
}
φ(x
1
)=
0|φ(x
1
)|0
(14.37)
Similarly,
(−i)
n
d
n
Z
dJ(x
1
)
dJ(x
n
)
|
J=0
=
0|T{φ(x
1
)
φ(x
n
)
}
|0
(14.38)
Sothisisanicewayofcalculatingtime-orderedproducts.
14.6 Time-orderedproducts
143
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14.7 SolvingthefreePathIntegral
First,let’slookatthefreetheory.
L=−
1
2
φ(+m
2
(14.39)
Then
Z
0
[J]=
Dφexp
i
d
4
x
1
2
φ(+m
2
)φ+J(x)φ(x)

(14.40)
Nowwecansolvethisexactly,sincewealreadyhaveourrelation
−∞
dp
e
1
2
p
Ap
+J
p
=
(2π)
n
detA
e
1
2
J
A−1J
(14.41)
Thepathintegralisjustaninfinitenumberof p
i
components.
Thisisexactlywhatwehave,withA=(+m
2
).Wehavealreadystudiedtheinverseof(+m
2
),
(+m
2
)φ(x)=J(x) ⇒
φ(x)=
d
4
xΠ(x−y)J(y)
(14.42)
whereΠistheGreen’sfunctionsatisfying
(
x
+m
2
)Π(x−y)=δ(x−y)
(14.43)
Explicitly,
Π(x−y)=
d
4
p
(2π)4
1
p2−m2
e
ip(x−y)
(14.44)
uptoboundaryconditions.Thus
Z
0
[J]=Nexp
i
d
4
x
d
4
y
1
2
J(x)Π(x−y)J(y)
(14.45)
Andso,
0|T{φ(x)φ(y)
}
|0
=(−i)2
d2Z
dJ(x)dJ(y)
|
J=0
=iΠ(x−y)=
d4p
(2π)4
i
p2−m2
e
ip(x−y)
(14.46)
Whathappenedtothetime-ordering?
Thestandardansweristhatweneedtoputaconvergencefactorinthepathintegral
Z
0
[J]=
Dφexp
i
d4x−
1
2
φ(+m2)φ+J(x)φ(x)
exp
−ε
d42)
(14.47)
=
Dφexp
i
d
4
x
1
2
φ(−−m
2
+iǫ)φ+J(x)φ(x)
(14.48)
=exp
1
2
d4x
d4yJ(x)D
F
(x−y)J(y)
(14.49)
WhereD
F
(x−y)istheFeynmanpropagator
0|T{φ(x)φ(y)}|0=(−i)
2
d
2
Z
0
dJ(x)dJ(y)
|
J=0
=
d
4
p
(2π)4
i
p2−m2+iε
e
ip(x−y)
=D
F
(x−y)
(14.50)
I’mnotsurehowlegitimatethisis. It’sreallyhardtodefinethepathintegralmathematically,butatleast
this convergence factor sounds believable. . Inanycase,we e knew wehadtoget thetime-orderedproduct
out,sothisisthecorrectanswer.
144
PathIntegrals
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So,forfor thefreefieldcase wehave solvedthe freepathintegralexactly. Weget the samethingas
justusingtheclassicalequationsofmotionandpluggingbackin. Thatiswhatwehavebeencallinginte-
gratingoutafield. Butnotethat wearenot integratingout thefieldwhenwedothe Gaussianintegral,
weareactuallycomputingthepathintegral.Thisonlyworksfor free fields,wherewecansolveitexactly.
So there isnodifference betweenaclassical andquantum systemof freefields. . Ifthere e areinteractions,
we canapproximate the integralby expandingaroundthe minimum andthenintegrating out the field.
Thisworksifthequantumcorrectionsaresmall,ortheinteractionsareweak.
14.7.1 4-pointfunction
Wecanalsocomputehigherorderproducts
0|T{φ
ˆ
(x
1
ˆ
(x
2
ˆ
(x
3
ˆ
(x
4
)
|0
=(−i)
4
d
4
Z
0
dJ(x
1
)
dJ(x
4
)
|
J=0
(14.51)
=
d
4
dJ(x
1
)
dJ(x
4
)
e
1
2
d
4
x
d
4
yJ(x)D
F
(x−y)J(y)
|
J=0
(14.52)
=
d
4
dJ(x
1
)dJ(x
2
)dJ(x
3
)
d
4
zD
F
(x
4
−z)J(z)
e
1
2
d4x
d4yJ(x)D
F
(x−y)J(y)
|
J=0
(14.53)
Beforewecontinue,let’ssimplifythenotation
Letusabbreviatethisas
0|T{φ
1
φ
2
φ
3
φ
4
}
|0
=
d4
dJ
1
dJ
2
dJ
3
dJ
4
e
1
2
J
x
D
xy
J
y
|
J=0
(14.54)
=
d
3
dJ
1
dJ
2
dJ
3
(−J
z
D
z4
)e
1
2
J
x
D
xy
J
y
|
J=0
(14.55)
=
d
2
dJ
1
dJ
2
(−D
34
+J
z
D
z3
J
w
D
w4
)e
1
2
J
x
D
xy
J
y
|
J=0
(14.56)
=
d
dJ
1
(D
34
J
z
D
z2
+D
23
J
w
D
w4
+J
z
D
z3
D
24
−J
z
D
z3
J
w
D
w4
J
r
D
r2
)e
1
2
J
x
D
xy
J
y
|
J=0
(14.57)
=D
34
D
12
+D
23
D
14
+D
13
D
24
(14.58)
Soweget
0|T{φ
ˆ
(x
1
ˆ
(x
2
ˆ
(x
3
ˆ
(x
4
)
|0
=
(14.59)
D
F
(x
3
−x
4
)D
F
(x
1
−x
2
)+D
F
(x
2
−x
3
)D
F
(x
1
−x
3
)+D
F
(x
1
−x
3
)D
F
(x
2
−x
4
)
(14.60)
These are the 3possible contractions. . This s is exactly what we found d from m time-dependent perturbation
theory.
14.8 Interactions
Nowsupposewehaveinteractions
L=−
1
2
φ(+m
2
)φ+λφ
4
(14.61)
Then,wecanwrite
Z[J]=
Dφe
i
d4x
1
2
φ(−−m2+iε)φ+J(x)φ(x)+λφ4
(14.62)
=
Dφe
i
d4x
1
2
φ(−−m2+iε)φ+J(x)φ(x)
e
i
d4x
λ
4!
φ4
(14.63)
14.8 Interactions
145
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=
Dφe
i
d4x
1
2
φ(−−m2+iε)φ+J(x)φ(x)
1+iλ
d44(z)+
(iλ)2
2!
d4z
d44(z)φ4(w)+
Then
0|T{φ
ˆ
(x
1
ˆ
(x
2
)
|0
=
0|T{φ
ˆ
I
(x
1
ˆ
I
(x
2
)
|0
+iλ
d
4
z
0|T{φ
ˆ
I
(x
1
ˆ
I
(x
2
ˆ
I
(z)
4
|0+
=
0|T{φ
ˆ
I
(x
1
ˆ
I
(x
2
)e
i
d
4
zλφ
ˆ
I
(z)
4
|0
(14.64)
IhaveaddedtheI subscript t toremindus thatthese arefields inthe freetheory,evolvingwiththefree
Hamiltonian(whatwecalledtheinteractionpicturebefore).Sowehavereproducedexactlytheexpression
fortheperturbationexpansionthe generatedtheFeynmanrules. . Thatis,havereproducedtheFeynman
rulesfromthepathintegral.
Anotheroccasionallyusefulnotationis
Z[J]=e
i
d4xV[
d
dJ(x)
]
Dφe
1
2
d
4
x
d
4
yJ(x)D
F
(x−y)J(y)
(14.65)
=e
i
d4xV[
d
dJ(x)
]
Z
0
[J]
|
J=0
(14.66)
ThisemphasizesthateverythingisdeterminedbyZ
0
.
Finally,rememberthatwhatwewereinterestedinforS-matrixelementswas
Ω|T{φ(x
1
)
φ(x
n
)}|Ω=
0|T{φ(x
1
)
φ(x
n
)e
i
d4zλφ
ˆ
I
(z)4
|0
0|e
i
d4zλφ
ˆ
I
(z)4
|0
(14.67)
=
1
Z[J]
d
n
Z[J]
dJn
|
J=0
(14.68)
SincethesameZ appearsinthenumeratorandthedenominator,thenormalizationofZ dropsout,which
iswhywehavebeenabletoignoreitsofar.
14.9 Thewardidentity
Oneofthekeythingsthatmakespathintegralsusefulisthatwecandofieldredefinitions.
Forexample,saywehavethepathintegralforsomegaugetheory
L[A
µ
i
]=−
1
4
F
µν
2
+|D
µ
φ
i
|
2
−m
2
i
|
2
+
(14.69)
Thenthepathintegralis
Z[0]=
i
i
DA
µ
e
i
d
4
xL[A,φ
i
]
(14.70)
Itisjusttheintegraloverallthefields.
Nowsupposewechangevariables
A
µ
=A
µ
(x)+V
µ
(x)
(14.71)
φ
i
i
(x)+∆
i
(x)
(14.72)
Sinceweareintegratingoverall fields,thepathintegralis automaticallyinvariant,foranyV
µ
(x)andany
i
(x).
i
i
DA
µ
e
i
d4xL[A
i
]
=
i
i
DA
µ
e
i
d4xL[A,φ
i
]
(14.73)
146
PathIntegrals
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Thisisnotveryimpressive.
Nowconsiderachangeofvariableswhichisagaugetransformation
A
µ
(x)=A
µ
(x)+∂
µ
α(x)
(14.74)
φ
i
(x)=e
iα(x)
φ
i
(14.75)
For some function α(x). . Then n we e canmake stronger r statements. . The e Lagrangianby itself is invariant,
becausethisisagaugetransformation.
e
i
d4xL[A
i
]
=e
i
d4xL[A,φ
i
]
(14.76)
The measure DA A is s invariant because this is s just t a a linear r shift. . The measure Dφ
i
changes by y a a factor
e
iα(x)
butDφ
i
changesby e
−iα(x)
andthesetwofactorscancel. Sincefor r everyfieldweneedtointegrate
overitanditsconjugate,thecancellationwillalwaysoccur. So
i
i
DA
µ
=
i
i
DA
µ
(14.77)
Stillnotveryimpressive.
ButnowconsiderthecorrelationfunctionofsomethinginvolvinganA
µ
field
0|A
µ
(x
1
)|0=
i
i
DA
µ
e
i
d4xL[A,φ
i
]
A
µ
(x
1
)
(14.78)
Nowmakethesamereplacement.Thenanychangeofvariableswillnotaffectthepathintegral,
i
i
DA
µ
e
i
d4xL[A,φ
i
]
A
µ
(x
1
)=
i
i
DA
µ
e
i
d4xL[A
i
]
A
µ
(x)
(14.79)
Butalso,sincethemeasureandactionareseparatelyinvariant,thissimplifiesto
=
i
i
DA
µ
e
i
d4xL[A,φ
i
]
[A
µ
(x
1
)+∂
µ
α(x
1
)]
(14.80)
whichmeans
i
i
DA
µ
e
i
d
4
xL[A,φ
i
]
µ
α(x
1
)=0
(14.81)
ThisisanexampleoftheWardidentity.
Itfollowsmoregenerallyaswell.Supposeweaddinsomemorefields
0|A
µ
(x
1
i
(x
2
)
φ
j
(x
n
)|0=
i
i
DA
µ
e
i
d4xL[A,φ
i
]
A
µ
(x
1
i
(x
2
)
φ
j
(x
n
)
(14.82)
=
i
i
DA
µ
e
i
d4xL[A,φ
i
]
[A
µ
(x
1
)+∂
µ
α(x
1
)]e
iα(x
2
)
φ
i
(x
2
)
e
iα(x
n
)
φ
j
(x
n
)
(14.83)
Thesephasefactorsarejustnumbers,whichdon’tmatter.Forexample,justtakingtheabsolutevalue,we
find
|0|A
µ
(x
1
i
(x
2
)
φ
j
(x
n
)|0|=|0|A
µ
(x
1
i
(x
2
)
φ
j
(x
n
)|0
+0|∂
µ
α(x
1
i
(x
2
)
φ
j
(x
n
)|0
|
Thisimpliesthat0|∂
µ
α(x
1
i
(x
2
)
φ
j
(x
n
)|0
=0whichisthewardidentitywithoneA
µ
.
Explicitly,ifwewrite
0|A(x
1
i
(x
2
)
φ
j
(x
n
)|0=
d
4
k
(2π)4
ε
µ
(k)M
µ
(k,x
2
,
x
n
)e
ikx
1
(14.84)
wehaveshown
d
4
k
(2π)4
µ
α(x
1
)M
µ
(k,x
2
,
x
n
)e
ikx
1
=−α(x
1
)
d
4
k
(2π)4
k
µ
M
µ
(k,x
2
,
x
n
)e
ikx
1
=0
(14.85)
⇒ k
µ
M
µ
(k,x
2
,
x
n
)=0
(14.86)
14.9 The e wardidentity
147
whichistheWardidentityinmomentumspace.
Finally,supposetherearelotsof A
µ
’s.Wefind
0|A
µ
(x
1
)
A
µ
(x
n
)
|0=0|(A
µ
(x
1
)+∂
µ
α(x
1
))
(A
µ
(x
n
)+∂
µ
α(x
n
)
|0
(14.87)
Forαinfinitesimal,thisimplies
0|∂
µ
(x
1
)
A
µ
(x
n
)|0
+
0|A
µ
(x
1
)∂
µ
(x
2
)
|0
+
0|A
µ
µ
α(x
n
)
|0=0
(14.88)
whichisthestatement that wehavetosumover allpossibleways that aphotoncanbereplacedwitha
k
µ
. Because e the photons are identical particles,we wouldhave tosum over r all these graphs s to o get t the
amplitudeforthepolarizationǫ
µ
=k
µ
,sothisisexactlywhatwewouldcalculate.
Thus,thepathintegralmakes thederivationoftheWardidentityeasy. . Itisalsocompletelynon-per-
turbative.
14.10 Gaugeinvariance:Fadyeev-Popov
Anotherthingthat’seasytodowithpathintegralsistoprovegaugeinvariance,meaningξdependenceof
theamplitudes.
RecallthatvectorfieldLagrangianis
L=−
1
4
F
µν
=
1
2
A
µ
(−k
2
η
µν
+k
µ
k
ν
)A
ν
+J
µ
A
µ
(14.89)
Andtheequationsofmotionare
(−k
2
η
µν
+k
µ
k
ν
)A
ν
=J
µ
(14.90)
whichisnotinvertiblebecausethismatrixhas0determinant(ithasaneigenvectork
µ
witheigenvalue0).
Thephysicalreasonit’snotinvertibleis becausewecan’tuniquelysolveforA
µ
intermsof J
µ
becauseof
gaugeinvariance:
A
µ
→A
µ
+∂
µ
α(x)
(14.91)
That is, many currents correspond d to the e same vector field. . Our r previous solution n was s to o gauge fix x by
adding the e term
1
ξ
(∂
µ
A
µ
)
2
to the Lagrangian. . Now w we will l justify y that prescription, and prove e gauge
invarianceingeneral,throughtheFadyeev-Popovprocedure.
Withageneralsetoffieldsφ
i
andinteractionsweareinterestedincomputing
Z
O
=
DA
µ
i
e
i
d
4
xL[A,φ
i
]
O
(14.92)
where Oreferstowhateverwe’retakingthecorrelationfunctionof(forexample,O=φ(x)φ(y))
Ω|T{O}|Ω=
Z
O
Z
1
(14.93)
Recall that under r a a gauge transformation n ∂
µ
A
µ
= α. . Since e we can always go to a a gauge e where
µ
A
µ
=0,thismeanswecanfindafunctionα=
1
µ
A
µ
.Nowconsiderthefollowingfunction
f(ξ)=
Dπe
i
d
4
x
1
ξ
(π)
2
(14.94)
Thisisjustsomefunctionof ξ,probablyinfinite.Nowlet’sshift
π(x)→π(x)−α(x)=π(x)−
1
µ
A
µ
(14.95)
148
PathIntegrals
Thisisjustashift,sotheintegrationmeasuredoesn’tchange. Then
f(ξ)=
Dπe
i
d4x
1
ξ
(π−∂
µ
A
µ
)2
(14.96)
Thisisstilljustthesamefunctionof ξ,whichdespiteappearances,isindependentof f A
µ
.So,
Z
O
=
1
f(ξ)
DπDA
µ
i
e
i
d4xL[A,φ
i
]+
1
ξ
(π−∂
µ
A
µ
)2
O
(14.97)
Nowlet’sdothegaugetransformationshift,withπ(x)asourgaugeparameter.
A
µ
=A
µ
+∂
µ
π
(14.98)
φ
i
=e
φ
i
(14.99)
AgainthemeasureDπDA
µ
i
andtheactionL[A,φ
i
]don’tchange. Omaychange. . Forexample,ifit’s
O=A
µ
(x
1
)
A
µ
(x
n
i
(x
n+1
)
φ
j
(x
m
)
TheA
µ
transformtoA
µ
+k
µ
andthek
µ
partsvanishbytheWardidentity. Thefieldsφ
i
willgivephases
e
iπ(x
i
)
,whichdon’tdon’tdependonξ. Thenweget
Dπe
iπ(x
n+1
)
e
iπ(x
m
)
=
(14.100)
where π
(x) = = e
iπ(x
n+1
)
e
iπ(x
m
)
π(x). So o this factor r is actually independent of all l the x
s. Therefore,
overall,wegetsomeconstantnormalizationtimesthegauge-fixedpathintegral
Z
O
=N
ξ
DA
µ
i
e
i
d
4
xL[A,φ
i
]+
1
ξ
(∂
µ
A
µ
)
2
O
(14.101)
whereN
ξ
=
1
f(ξ)
. Butsincewearealwayscomputing
Ω|T{O}|Ω=
Z
O
Z
1
(14.102)
The N
ξ
drops out. . The e point is that Z
O
onlydepends on n ξ throughit’s s normalization. . Thus s thetime-
orderedproductsareξindependent.
ThusallS-matrixelements ofgaugeinvariantoperatorsareguaranteedtobeindependentof f ξ,which
isthestatementofgaugeinvariance.Thisproofiscompletelynon-perturbative.
14.11 Fermionicpathintegral
For a a path h integral over fermions, , it’s s basically y the e same, , but t we have to o allow w for the e fact that the
fermions anticommute. . In n the e quantum theory, we saw that the quantum fields s must anticommute e to
haveaLorentz-invariantS-matrix. Butevenclassically,aLorentzinvariantspinorinner r productmustbe
antisymmetric. Forexample,theonlyLorentzinvariantconstructedfromasingleWeylspinoris
ψ
L
T
σ
2
ψ
L
=
ψ
1
ψ
2
i
−i

ψ
1
ψ
2
=i(ψ
1
ψ
2
−ψ
2
ψ
1
)=|↑|↓−|↓|↑
(14.103)
From this we concludedthat {ψ
1
2
}=0whichmeans ψ(x)is s valuedintheanti-commutingGrassman
algebra.
TheGrassmannumbersissimply asetofobjects whichaddcommutativelyandcanby multipliedby
realnumbers. Considerageneralfunctionf(θ)ofGrassmannumbersθ. Sinceθ
2
=
θ,θ}=0,theTaylor
expansionof f(θ)aroundθ=0stopsatfirstorder
f(θ)=a+bθ
(14.104)
14.11 Fermionic c pathintegral
149
ThebasicthingwewanttodowiththesefunctionsisintegratethemoverallGrassmannumbers togeta
realnumberout.Sincewe’remakinguptherulesforGrassmannumbersaswegoalong,let’sjustdefine
dθ(a+bθ)=b
(14.105)
Thatis,
dθ=0,
dθθ=1
(14.106)
Wealsoneedaconventionformultipleintegrals,sowesay
1
2
θ
2
θ
1
=
1
θ
1
=1
(14.107)
Thus,
2
1
θ
2
θ
1
=−1
(14.108)
Andsoon. Theserules s areinternally consistent,andhavesome propermathematicaljustification,which
istotallyuninteresting.
Now,let’stryourGaussianintegral
I=
dθe
1
2
2
=
dθ=0
(14.109)
Sinceθ
2
=0. Let’strywithtwoθ
s. Letψ=(θ
1
θ
2
)
T
andconsider ψ
T
Aψ. Then
ψ
T
Aψ=(A
12
−A
21
1
θ
2
=det(A)θ
1
θ
2
(14.110)
Where we have e taken A A antisymmetric c to o write it t as s a determinant. . The e symmetric part t of A A simply
doesn’tcontribute. Thenthisis
1
2
e
ψ
T
=
1
2
(1+det(A)θ
1
θ
2
)=det(A)
(14.111)
Thegeneralizationtonvariablesissimply
e
θ
=det(A)
(14.112)
Thisisverydifferentfromthebosonicintegral
dp
e
1
2
p
Ap
1
detA
=(detA)
−1/2
(14.113)
Butthat’swhatit is. This det factor comesupfromtimetotime,butmostly it isjust the infinitenor-
malizationofthepathintegral,whichwecanignoreintakingratios.
TocalculatethegeneratingfunctionalZ[J]weneedafermionicsourceη. Let’sjusttake2fieldsψand
ψ
¯
,andcompletethesquare
ψ
¯
Aψ+η¯ψ+ψ
¯
η=(ψ
¯
+η¯A−1)A(ψ+A−1η)−η¯A−1η
(14.114)
Then
dψdψ
¯
eψ
¯
Aψ+η¯ψ+ψ
¯
η=e−η¯A
−1
η
dψdψ
¯
e
¯
+η¯A−1)A(ψ+A−1η)
(14.115)
=e
−η¯A−1η
dψdψ
¯
e
ψ
¯
=det(A)e
−η¯A−1η
(14.116)
=Ne
−η¯A−1η
(14.117)
forsomenormalizationN.
150
PathIntegrals
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