c# display pdf in browser : Add page numbers to pdf document in preview application control cloud windows azure .net class QFT-Schwartz10-part441

IntheWeylrepresentation,theLorentzgeneratorsare
S
ij
=
1
2
ε
ijk
σ
k
σ
k
, K
i
=S
0i
=−
i
2
σ
i
−σ
i
(11.76)
Or,veryexplicitly
S
12
=
1
2
1
−1
1
−1
, S
13
=
i
2
0 1
−1 0
0 1
−1 0
, S
23
=
1
2
0 1
1 0
0 1
1 0
(11.77)
S
01
=
i
2
0 −1
−1 0
0 1
1 0
, S
02
=
1
2
0 −1
1 0
0 1
−1 0
, S
03
=
i
2
−1
1
1
−1
(11.78)
Theseareblockdiagonal. Thesearethesamegeneratorsweusedfor r the(
1
2
,0)and(0,
1
2
)representations
above. It t makes it clear that theDirac representation n is s reducible, , it is s thesum of aleft-handed d anda
right-handedspinorrepresentation.
AnotherimportantrepresentationistheMajoranarepresentation
γ0=
0 σ
2
σ
2
0
, γ1=
3
0
0 iσ
3
, γ2=
0 −σ
2
σ
2
0
γ3=
−iσ
1
0
0
−iσ
1
(11.79)
Inthisbasistheγ
sarepurelyimaginary.
TheWeylspinors,ψ
L
andψ
R
aremorefundamental,becausetheycorrespondtoirreduciblerepresen-
tationsoftheLorentzgroup.ButtheelectronisaDiracspinor.ThustodoQED,itiseasiestjusttostick
totheγ
sandtogetusedtomanipulatingthem. Eventually,whenyoudosupersymmetry,or r studythe
weakinteractions,youwillneedtousetheLandRrepresentationsagain.
11.6 Rotations
Nowlet’sseewhathappenswhenwerotatebyanangleθinthexyplane. Weuse
Λ=exp(iθ
xy
V
12
)
(11.80)
Howdoweexponentiateamatrix? Thestandardtrickistofirst t diagonalizingitwithaunitary transfor-
mation, do the rotation,then transform m back. This unitary transformationis like e choosing a direction,
exceptitispurelymathematicalasthedirectionmustbecomplex!
First,forthevectorrepresentation
V
12
=i
0
0 1
−1 0
0
=U
−1
0
−1
1
0
U
(11.81)
So,
Λ(θ
xy
)=exp(iθ
xy
V
12
)=U
−1
0
exp(−iθ
xy
)
exp(iθ
xy
)
0
U
(11.82)
Λ(2π)=1
(11.83)
Thatis,werotate360degreesandwe’rebacktowherewestarted.
11.6 Rotations
101
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Forthespinorrepresentation
Λ
s
=exp(iθ
µν
S
µν
)
(11.84)
the12rotationisalreadydiagonal:
S
12
=
1
2
1
−1
1
−1
(11.85)
So,
Λ(θ
xy
)=exp(iθ
xy
S
12
)=
exp(
i
2
θ
xy
)
−exp(
i
2
θ
xy
)
exp(
i
2
θ
xy
)
−exp(
i
2
θ
xy
)
(11.86)
Λ(2π)=
−1
1
−1
1
(11.87)
Thus a 2π π rotation n does not t bring g us s back k where e we started! ! If f we rotate by y 4π it t would. . So o we say
spinorsare spin
1
2
. What does s this meanphysically? ? I I havenoidea. There areplenty of physicalconse-
quencesof spinorsbeing spin
1
2
,but this business of rotatingby2π π is s not actually aphysicalthing g you
cando.
Asanaside,note thatthefactorof2isdeterminedby thenormalizationofthematrices,whichisset
by the Lie e Algebra. . For r each representation n by y itself, this s would be e arbitrary, , but it is s important t for
expressionswhichcombinetherepresentationstobeinvariant.
11.7 LorentzInvariants
TheγmatricesthemselvestransformnicelyundertheLorentzgroup.
Λ
s
−1
γ
µ
Λ
s
=(Λ
V
)
µν
γ
ν
(11.88)
wheretheΛ
s
aretheLorentz transformationsactingoneachγ
µ
individually,as amatrix,andtheΛ
V
is
thevector representationwhichmixesuptheLorentzindices. . Thatis,writingoutthematrixindices s γ
µ
αβ
,
thismeans
s
−1
)
δα
γ
µ
αβ
s
)
βγ
=(Λ
V
)
µν
γ
ν
αβ
(11.89)
whereµreferstowhichγmatrix,andαandβindextheelementsofthatmatrix.
Thentheequation
µ
ν
}=2η
µν
(11.90)
reallymeans
γ
µ
αγ
γ
ν
γβ
ν
αγ
γ
µ
γβ
=2η
µν
δ
αβ
(11.91)
Andtheequation
S
µν
=
i
4
µ
ν
]
(11.92)
Shouldreallybewrittenas
S
µν
αβ
=
i
4
γ
µ
αγ
γ
ν
γβ
−γ
ν
αγ
γ
µ
γβ
(11.93)
Foranexpressionlike
V
µ
V
µ
=V
µ
η
µν
V
ν
=V
µ
µ
ν
}V
ν
(11.94)
102
Spinorsand theDirac c Equation
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Tobeinvariant,itmustbethatµtransformsinthevectorrepresentation.
Nextconsider
ψ
ψ→
ψ
Λ
s
s
ψ)
(11.95)
For thistobe invariant,wewouldneedΛ
s
s
−1
,thatis,fortherepresentationof the Lorentzgroupto
beunitary. Thespinorrepresentation,likeanyotherrepresentationisnot unitary,becausetheboostgen-
eratorsareanti-Hermetian.
It is s useful to study the properties s of the Lorentz generators from m the e Dirac algebra a itself, , without
needingtochooseaparticularbasisfortheγ
µ
.Firstnotethat
µ
ν
}=2η
µν
γ
0
2
=1, γ
i
2
=−1
(11.96)
Sotheeigenvalues of γ
0
are ±1andthe e eigenvaluesof f γ
i
are ±i. . Thus s if wediagonalize γ
0
,wewould
finditHermetian,andifwediagonalize γ
1
γ
2
orγ
3
wewouldfindtheyareanti-Hermetian. Thisistruein
general.
γ
0
0
, γ
i
=−γ
i
(11.97)
Then,
(
S
µν
)
=
i
4
[
γ
µ
ν
]
=−
i
4
γ
ν
µ
=
i
4
γ
µ
ν
(11.98)
Whichimplies
S
ij
=S
ij
S
0i
=−S
0i
(11.99)
Again, we see that therotations s are unitaryand d the e boosts are not. . You u can seethis s from the explicit
representationsabove.Butbecauseweshoweditalgebraically,itistrueinANYrepresentation.
Now,observethatone of theDiracmatricesisHermetian,γ
0
0
isthe only HermetianDiracmatrix
becausethemetricsignatureis(1,-1,-1-,1)).Moreover
γ
0
γ
i
γ
0
=−γ
i
i
, γ
0
γ
0
γ
0
0
0
(11.100)
γ
µ
0
γ
µ
γ
0
(11.101)
⇒γ
0
S
µν
γ
0
0
i
4
γ
µ
ν
γ
0
=
i
4
γ
0
γ
µ
γ
0
0
γ
ν
γ
0
=
i
4
µ
ν
]=S
µν
(11.102)
0
Λ
s
γ
0
)
0
exp(iθ
µν
S
µν
)
γ
0
=exp(−iθ
µν
γ
0
S
µν
γ
0
)=exp(−iθ
µν
S
µν
)=Λ
s
−1
(11.103)
Then,finally,
ψγ
0
ψ→
ψΛ
s
γ
0
(
Λ
s
ψ
)
=
ψγ
0
Λ
s
−1
Λ
s
ψ
γ
0
ψ
(11.104)
whichisLorentzinvariant.
We have just beenrederivingfrom the Diracalgebrapoint ofviewwhat we foundby handfrom the
Weylpointofview. Theconjugateof ψisnot ψ
but
ψ
¯
≡ψ
γ
0
(11.105)
Thepointisthatψ
¯
transformsaccordingtoΛ
s
−1
. Thus s ψ
¯
ψisLorentzinvariant.
Wecanalsoconstructobjectslikelike
ψ
¯
γ
µ
ψ, ψ
¯
γ
µ
γ
ν
ψ, ψ
¯
µ
y
(11.106)
alltransformnicelyundertheLorentzgroup.Also
L=ψ
¯
(iγ
µ
µ
−m)ψ
(11.107)
isLorentzinvariant.Weabbreviatethiswith
L=ψ
¯
(i∂
−m)ψ
(11.108)
11.7 Lorentz z Invariants
103
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ThisistheDiracLagrangian.
TheDiracequationfollowsfromthisLagrangianbytheequationsofmotion
(i∂
−m)ψ=0
(11.109)
Tobeexplicit,thisisshorthandfor
(iγ
µ
αβ
µ
−mδ
αβ
β
=0
(11.110)
Bymultiplyingontheright,wefind
0=(i∂
+m)(i∂
−m)ψ=(−
1
2
µ
ν
µ
ν
}−
1
2
µ
ν
µ
ν
]−m
2
(11.111)
=−(∂
2
+m
2
(11.112)
SoψsatisfiestheKlein-Gordonequation
(+m
2
)ψ=0
(11.113)
It is inthis sense that t people e sometimes say the Dirac equation n is the e square-root of the Klein-Gordon
equation.
WecanintegratetheLagrangianbypartstoderivetheequationsofmotionforψ
¯
:
L=ψ
¯
(i∂
−m)ψ=−i∂
µ
ψ
¯
µ
−m)ψ
(11.114)
So,
−i∂
µ
ψ
¯
γ
µ
−mψ
¯
=0
(11.115)
Thisγ
µ
ontherightisalittleannoying,soweoftenhideitbywriting
ψ
¯
(−i∂
−m)=0
(11.116)
wherethederivativeactstotheleft. Thismakes s theconjugateequationlookmoreliketheoriginalDirac
equation.
11.8 Couplingtothephoton
Underagaugetransformψtransformsjustlikeascalar,sincethegaugeinvariancehasnothingtodowith
spin.So,
ψ→e
ψ
(11.117)
Thenweusethesamecovariantderivative∂
µ
−ieA
µ
asforascalar. So
D
µ
ψ=(∂
µ
−ieA
µ
(11.118)
ThentheDiracequationbecomes
(i∂
+eA
−m)ψ=0
(11.119)
NowwetrytoreproducetheKleinGordonequationforascalarfieldcoupledtoA
µ
.
[(i∂
µ
+eA
µ
)2−m2
φ
(11.120)
Followingthesamerouteasbefore
0=(i∂
+eA
+m)(i∂
+eA
−m)ψ
(11.121)
=
(i∂
µ
+eA
µ
)(i∂
ν
+eA
ν
µ
γ
ν
−m
2
ψ
(11.122)
=
1
4
{i∂
µ
+eA
µ
,i∂
ν
+eA
ν
}{γ
µ
ν
}+
1
4
[i∂
µ
+eA
µ
,i∂
ν
+eA
ν
][γ
µ
ν
]−m
2
ψ
(11.123)
104
Spinorsand theDirac c Equation
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Beforetheantisymmetriccombinationdroppedout,butnowwefind
1
4
[i∂
µ
+eA
µ
,i∂
ν
+eA
ν
]=e[i∂
µ
A
ν
−i∂
ν
A
µ
]=eiF
µν
(11.124)
Soweget
(i∂
µ
+eA
µ
)
2
+
ie
4
F
µν
µ
ν
]−m
2
ψ
(11.125)
WhichcontainsanextratermcomparedtotheKlein-Gordonequation.
Whatisthisterm? Well,recallthat
i
4
µν]=Sµν,our Lorentzgenerators. . Thesehavetheform(in
theWeylrepresentation)
S
ij
=
1
2
ε
ijk
σ
k
σ
k
, S
0i
=−
i
2
σ
i
−σ
i
(11.126)
Andsince
F
0i
=E
i
, F
ij
ijk
B
k
(11.127)
Weget
(∂
µ
−ieA
µ
)
2
+m
2
e
2
(B
+iE
(B
−iE

ψ=0
(11.128)
Thiscorrespondstoamagneticdipolemoment.
Thisisprettyremarkable.Forafreespinor,wereproducetheequationofmotionofascalarfield. But
when the e spinor is s coupledto the e photon, we find d anadditional interaction corresponds to a magnetic
dipolemoment.Wecanreadoffthattheelectronhasspin
1
2
. Note:thecouplingtotheelectricfieldisnot
anelectricdipolemoment–thatwouldnothaveani,butissimplytheeffectofamagneticmomentina
boostedframe.
11.9 Weylspinors–Helicity
Diracspinors,whatwehavebeenusing,are4componentcomplexfieldsinthe (
1
2
,0)⊕(0,
1
2
)representa-
tionoftheLorentzgroup.Let’sreturnforamomenttothinkingaboutthe2-componentfields.
IntheWeylbasis,theγmatriceshavetheform
γ
µ
=
0 σ
µ
σ¯
µ
0
(11.129)
andLorentzgeneratorsareblockdiagonal
δψ=
(iθ
i
i
i
(iθ
i
−β
i
i
ψ
(11.130)
Wecanwriteourspinorastheleftandrighthelicityparts
ψ=
ψ
L
ψ
R
(11.131)
TheDiracequationis
−m
µ
D
µ
iσ¯
µ
D
µ
−m

ψ
L
ψ
R
(11.132)
meaning
(iσ¯
µ
µ
+eσ¯
µ
A
µ
R
=mψ
L
(11.133)
(iσ
µ
µ
+eσ
µ
A
µ
L
=mψ
R
(11.134)
Sotheelectronmassmixestheleftandrighthandedstates.
11.9 Weyl l spinors s –Helicity
105
Intheabsenceofamass,thisimplies
0=iσ
µ
µ
ψ
R
=(E+σ
·p
R
(11.135)
0=iσ¯
µ
µ
ψ
L
=(E−σ
·p
L
(11.136)
So the left t andright handed d states s are eigenstates of theoperator σ
· p
withoppositeeigenvalue. This
operatorprojectsthespinonthemomentum direction. Wecallspinprojectedonthedirectionofmotion
thehelicity,sotheleftandrighthandedstateshaveoppositehelicity. Sincethemassmixesleftandright
handedstates, massive fermions s arenot eigenstates of helicity. . This s makes ssene– – inthe electron’s s rest
frame,thereisnodirectionofmotionsohelicityisnotwelldefinedlinthemasslesslimit,thereisnorest
frame,andaleft-handedelectroncanneverturnintoaright-handedelectron,andaright-handedelectron
canneverturnintoaleft-handedelectron,eveninthepresenceofelectromagneticfields.
Thefactthatprojectionofspinonthedirectionofmomentumisagoodquantumnumberformassless
particlesworksformasslessparticlesofanyspin. Foranyspin,wewillalwaysfind(E±sJ
p
s
=0,where
J
are the rotationgenerators of spin n s. . For r spin1/2, J
=
σ
2
. For r photons,the e rotation n generators are
listedinsection2. Forexample,J
z
=V
23
has eigenvalues s ±1witheigenstates(0,i,1,0)and(0,−i,1,0).
Thesearethestatesofcircularlypolarizedlightinthezdirection. Theyarehelicityeigenstates. Somass-
lessparticles alwayshavetwohelicitystates. . Itistrueforspin1/2andspin1,aswehaveseen, , itistrue
for gravitons (spin2),Rarita-Schwing g fields s (spin3/2) andspins s>2(although,as wehaveseen,it is
impossibletohaveinteractingtheorieswithmasslessfieldsofspins>2).
Wehaveseenthattheψ
L
andψ
R
states
• donotmixunderLorentzTransformation
• ψ
L
and ψ
R
eachhavetwocomponentsonwhichthe σ
sact. Thesearethetwospinstates s ofthe
electron–bothleftandrighthandedspinorshave2spinstates.
• ψ
L
andψ
R
haveoppositehelicity.
Using
ψ
¯
=
ψ
L
ψ
R
γ
0
=
ψ
R
ψ
L
(11.137)
TheLagrangian
L=ψ
¯
(iγ
µ
µ
+eγ
µ
A
µ
−m)ψ
(11.138)
becomes
L=iψ
L
σ¯µD
µ
ψ
L
+iψ
R
σµD
µ
ψ
R
−m(ψ
L
ψ
R
R
ψ
L
)
(11.139)
Whichis what we derivedin n the e beginning. . Notethat t ψ
L
andψ
R
bythemselves mustbe massless . . To
rightdownamassterm,weneedbothaψ
L
andaψ
R
.
It is helpfulto o be ableto projectout t theleft or righthandedWeylspinors fromaDirac spinor. . We
candothatwiththeγ
5
matrix
γ
5
=iγ
0
γ
1
γ
2
γ
3
(11.140)
intheWeylrepresentation
γ
5
=
−1
1
,
(11.141)
So
P
R
=
1+γ
5
2
=
0
1
, P
L
=
1−γ
5
2
=
1
0
(11.142)
TheseareprojectionoperatorssinceP
R
2
=P
R
andP
L
2
=P
L
and
P
R
ψ
L
ψ
R
=
0
ψ
R
, P
L
ψ
L
ψ
R
=
ψ
L
0
(11.143)
Theseprojectorsareusefulbecausetheyarebasis-independent.
106
Spinorsand theDirac c Equation
11.10 Spinandstatistics
Recallthatwecouldn’twritedownamasstermψ
L
ψ
L
forjustalefthandedspinorbecause
δψ
L
=
1
2
(iθ
i
i
i
ψ
L
(11.144)
δψ
L
=
1
2
(−iθ
i
i
L
σ
i
(11.145)
δψ
L
ψ
L
i
ψ
L
ψ
L
0
(11.146)
Thisjustindicatesthattheboostsarenotunitarytransformations.
Togetsomethingnew,recallthatforthePaulimatrices,σ
1
andσ
3
arereal,andσ
2
isimaginary.
σ
1
=
0 1
1 0
, σ
2
=
0 −i
i 0
, σ
3
=
1 0
0 −1
(11.147)
Wecanseethat
σ
1
1
, σ
2
=−σ
2
, σ
3
3
(11.148)
σ
1
T
1
, σ
2
T
=−σ
2
, σ
3
T
3
(11.149)
Forthei=1andi=3,σ
i
T
σ
2
i
σ
2
=−σ
2
σ
1
. Fori=2,σ
i
T
σ
2
=−σ
2
σ
2
=−σ
2
σ
i
. So
σ
i
T
σ
2
=−σ
2
σ
i
(11.150)
Then
δ(ψ
L
T
σ
2
)=
1
2
(iθ
i
i
L
T
σ
i
T
σ
2
=
1
2
(−iθ
i
−β
i
)
ψ
L
T
σ
2
σ
i
(11.151)
Whichcancelsthetransformationpropertyof ψ
L
. Thus
ψ
L
T
σ
2
ψ
L
(11.152)
isLorentzinvariant.
Theonlyproblemisthatσ
2
=
−i
i
sothisisjust
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
)
(11.153)
Thisistrueforanyspinorcomponents ψ
1
andψ
2
.So,
ψ
1
ψ
2
−ψ
2
ψ
1
isLorentzinvariant
(11.154)
This is anexampleof the spin-statistics theorem. I thinkit’s s the simplest way to o seetheconnection
betweenanti-symmetrizationandLorentzinvariance.
Youknow it t from m quantum m mechanics. . Fermions s come in anti-symmetric states. . It t may look more
familiarifweusearrowsfortheψ
1
andψ
2
states:
=
1
2
(
|↑|↓−|↓|↑
)
(11.155)
We see that wavefunctions are e antisymmetric c to o the exchange of f particles. . This s is the Pauli i exclusion
principle. HereweseethatitfollowssimplyfromLorentzinvariance.
Obviously if the fermioncomponents commutethisis just zero. Sofermioncomponents can’tbereg-
ularnumbers,theymustbeanticommutingnumbers. Suchthingsare e calledGrassman numbersandsat-
isfy a Grassman algebra. . We e will see that it’s easy to make the quantum fields s anti-commute, by
demandingthatthecreationandannihilationoperatorsanti-commute,a
p
a
q
=−a
q
a
p
,. Andwe’llrederive
spin-statisticsfrompropertiesoftheS-matrix. Buthereweareseeingthatifwearetomakeanysenseof
spinorsclassically,orinanywayatall,theymustanticommute.
11.10 Spinandstatistics
107
11.11 Majoranafermions
If we allowfermions s tobe e Grassmannumbers, then n we e can n write e downa Lagrangianfor a single Weyl
spinorwithamassterm
L=iψ
L
σ
µ
µ
ψ
L
+i
m
2
L
σ
2
ψ
L
−ψ
L
T
σ
2
ψ
L
)
(11.156)
These kinds of mass terms arecalled d Majorana masses. . Note e that this s mass s term breaks the symmetry
underψ→eψ,since
ψ
L
T
σ
2
ψ
L
→ψ
L
T
e
σ
2
e
ψ
L
=e
2iα
ψ
L
T
σ
2
ψ
L
(11.157)
SoaparticlewithaMajoranamasscannotbecharged.
TheequationofmotionfortheMajoranafermionis
σ
µ
µ
ψ
L
+mσ
2
ψ
L
=0
(11.158)
whichfollowsfromtheLagrangianabove.
IfwehaveaMajoranafermion,wecanstillusetheDiracalgebratodescribeit,butwehavetoputit
intoa4-componentspinor
ψ=
ψ
L
−iσ
2
ψ
L
(11.159)
ThistransformslikeaDiracspinorbecauseσ
2
ψ
L
transformslike ψ
R
.
Since(intheWeylbasis)
−iγ
2
ψ=−i
0
σ
2
−σ
2
0

ψ
L
ψ
R
=
2
ψ
R
−iσ
2
ψ
L
(11.160)
WecanseethataMajoranafermionisonesatisfying
ψ=ψ
c
≡−iγ
2
ψ
(11.161)
We call l ψ
c
the charge conjugate fermion. . A A Majorana fermionis it’s owncharge conjugate. . Since e it t is
real,itisalsitsownantiparticle.
Finally,notethatintheWeylbasisγ
2
isimaginaryandγ
0
1
andγ
3
arereal. Ofcourse,wewecould
justaswellhavetakenγ
3
imaginaryandγ
2
real,butit’sconventionaltopickoutγ
2
. Wecanalsodefinea
newrepresentationofthe γ matrices by γˆ
µ
2
γ
µ
γ
2
. ThissatisfiestheDiracAlgebrabecause γ
2
2
=−1.
Nowdefine
ψ
c
=−iγ
2
ψ⋆ 
ψ=−iγ
2
ψ
c
(11.162)
IfwetaketheDiracequation
(i∂
+eA
−m)ψ=0
(11.163)
andtakethecomplexconjugateweget
(−iγ
µ
µ
+eγ
µ
A
µ
−m)ψ
=0
(11.164)
γ
2
(−iγ
µ
µ
+eγ
µ
A
µ
−m)γ
2
ψ
c
=0
(11.165)
(iγˆ
µ
µ
−eγˆ
µ
A
µ
−m)ψ
c
=0
(11.166)
Soψ
c
hastheoppisitechargefromψ,whichisanotherreasonthatMajoranafermionscan’tbecharged.
11.11.1 summary
Insummary,
Wehaveseenthreetypesofspinors
• DiracSpinors:massive,leftANDrighthanded
• WeylSpinors:massless,leftORrighthanded
108
Spinorsand theDirac c Equation
• Majoranaspinors:realconstrainedDiracspinors.
We arenot sureif theneutrinoisaMajoranaoraDiracspinor. Weylspinorsplayaveryimportantrole
insupersymmetry,andinthetheoryoftheWeakinteractions. ButforQED,wecanjuststickwithDirac
spinors.
11.12 SolvingtheDiracequation
Let’s take abreak from the Dirac equation n for r amoment,andrecallhow w wediscoveredantiparticles s for
complexscalarfields.TheLagrangianwas
L=
1
2
[(∂
µ
+ieA
µ
][(∂
µ
−ieA
µ
)φ]+m2φφ
(11.167)
L=φ
(−∂
µ
+ieA
µ
)(∂
µ
−ieA
µ
)φ]+m
2
φ
φ
(11.168)
Theequationsofmotionare
(∂
µ
−ieA
µ
)2φ+m2φ=0
(11.169)
(∂
µ
+ieA
µ
)
2
φ
+m
2
φ
=0
(11.170)
Sowe seethat φand d φ
haveoppositecharge,andweinterpret themas particle andantiparticle. Recall
thatwhenwequantizedthefieldφ,itcreatedaparticleanddestroyedandantiparticle,andvice-versafor
φ
. Butattheclassicallevel,wecanjustthinkof φasparticleandφ
asantiparticle.
Howdo weknowifwehavea particle or r anantiparticle? ? There e is aneasy wayto answer this. . The
freeequationsforφandφ
arethesame:
(+m2)φ=(+m2=0
(11.171)
Thesehaveplanewavesolutions
φ=φ
p
e
ip
µ
xµ
(11.172)
Intherestframe, p
0
2
=m2,so p
0
=±m. Thesolutionwithp
0
=−misconfusing. Itis s alegitimatesolu-
tiontotheequationofmotion,butitsaysthattheseparticlesaregoingbackwardintime!Butnotethat
φ=φ
p
e
ip
0
t
φ
p
e
−ip
0
t
(11.173)
So we can just t as easily interpret thesesolutions as anti-particles going g forwardintime. . Obviously y this
interpretationis easier toswallow, , but Feynmanspent some time showingthat there really is s no o physi-
callydistinguishabledifference.
Nowbacktospinors. TheDiracequationis
(i∂
+eA
−m)ψ=0
(11.174)
ψ
¯
(−i∂
−eA
−m)=0
(11.175)
Soψ
¯
isaparticlewithmassmandoppositechargetoψ:thepositron.
11.12.1 freefieldsolutions
Let’sexpandtheDiracspinorinplanewavesolutions
ψ
p
(x)=u
β
(p)e
ipx
(11.176)
u
β
andv
β
arethepolarizations.WiththisAnsatz,theDiracequationbecomes
−m
p
µ
σ
µ
p
µ
σ¯
µ
−m
u
β
(p)=0
(11.177)
11.12 Solvingthe e Dirac equation
109
Sincethespinoralsosatisfies(+m2)ψ=0,itfollows that p2=m2. Wealreadyknowthatthenegative
frequencysolutions p
0
=−marenotaproblemsincetheantiparticles
ψ
¯
=u¯e
−ipx
(11.178)
havepositivefrequency. Still,itis s somewhatconvenienttokeeptalkingabout p
0
<0fornow. For r p
0
<0
wecallthepolarizationsv
β
(p). Thenthesolutionsare
ψ
p,s
(x)=u
s
(p)e
ipx
, p
0
>0
(11.179)
ψ
p,s
(x)=v
s
(p)e
−ipx
, p
0
<0
(11.180)
u
s
andv
s
arethepolarizationsands=1,2labelsthespin.
Intherestframe,p=(m,0,0,0),theequationsofmotionreduceto
−1 1
1 −1
u
s
=0 and
−1 −1
−1 −1
v
s
=0
(11.181)
Sosolutionsareconstants
u
s
=
ξ
s
ξ
s
(11.182)
and
v
s
=
η
s
−η
s
(11.183)
Foranytwo-componentspinors ξandη. . Thusfors=1wecantakethebasis s spinortobeξ
1
=
1
0
and
s=2maybeξ
2
=
0
1
.
Nowlet’sboostinthez-direction. PeskinandSchroederdotheactualboost. Butwe’lljust solvethe
equationsagainintheboostedframeandmatchthenormalization. If f p=(E,0,0,p
z
)then
p
µ
σ
µ
=
E−p
z
0
0
E+p
z
, p
µ
σ¯
µ
=
E+p
z
0
0
E−p
z
(11.184)
Letx= E−p
z
andy= E+p
z
,thenm
2
=(E−p
z
)(E+p
z
)=xy.
ThentheDiracequationis
(i∂
−m)ψ=
−xy
0
x
2
0
0
−xy
0
y
2
y
2
0
−xy
0
0
x
2
0
−xy
u
β
(p)=0
(11.185)
Thesolutionsare
u
s
=u
ξ
(p)=
1
m
1
2
1
2
=
1
m
x 0
0 y
ξ
y 0
0 x
ξ
(11.186)
Check
−xy+x
2
y
2
=0,
(11.187)
Notealsothatintherestframep
z
=0intherestframe,x
2
=y
2
=msowearebacktotheξstates.
Sothesolutionsinthep
z
frameare
u
s
(p)=
E−p
z
0
0
E+p
z
ξ
s
E+p
z
0
0
E−p
z
ξ
s
, v
s
(p)=
E−p
z
0
0
E+p
z
η
s
E+p
z
0
0
E−p
z
η
s
(11.188)
110
Spinorsand theDirac c Equation
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