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where theseΛ
s haveto o do withthe explicit t waythe Lorentz groupacts,whichwe don’t careabout so
much. Thusanytheoryinvolvingamasslessspin2fieldshouldsatisfyaWardidentity:ifwereplaceeven
oneindexofthepolarizationtensorbyk
µ
thematrixelementsmuchvanish.
Whatdothe interactions look like? ? Well,we e havetocouplesomethingtoatensor fieldh
µν
. Wewill
beemittingonehsoitshouldbelinearinh
µν
. ByLorentzinvariance,weneedtwoderivatives s tofillup
theh
µν
indices. Weneed2morefieldstohaveaninteraction,sowecanhave
L=(charge)h
µν
µ
φ
1
ν
φ
2
(10.47)
or with h the e twoderivatives acting elsewhere. . Wecould d also have e more derivatives, but t then the e charge
would have e a different dimension n and we e can consider those interactions s separately. . If f there are any
derivativesactingonh
µν
Icanmove theover tothe φ
sby integratingby parts. . This s is totallygeneral,
andworksforanykindsoffields
Goingthroughthesameargumentasbefore,addingupdiagramsforsoftgravitonemission,wefind
M=M
0
incoming
(charge)
i
p
µ
i
p
i
q
ǫ
µν
p
ν
i
outgoing
(charge)
j
p
µ
j
p
j
q
ǫ
µν
p
ν
j
(10.48)
wherethesechargesarethegravitationalcharges of thethings. . Theextrafactorof f p
ν
i
hastobethereby
Lorentz invariance. . The only other r thing g it couldhave been is q
ν
, which could come from a derivative
actingonthegraviton,butwealreadymovedthoseawaybyintegrationbyparts.
ByLorentzinvariance,requiringamasslessspin2field,thisshouldvanishif ǫ
µν
=q
µ
Λ
ν
foranyΛ
ν
. So
M
0
Λ
ν
incoming
(charge)
i
p
ν
i
outgoing
(charge)
j
p
ν
j
=0
(10.49)
incoming
(charge)
i
p
ν
i
=
outgoing
(charge)
j
p
ν
j
(10.50)
Thissays thatthesumof(charge)
i
p
i
µ
isconserved. Butwealreadyknowthatthesumof p
i
µ
isconserved
–momentumconservation. So,forexample,wecansolvefor r p
1
µ
intermsoftheothers. Ifweandanother
constraintonthe p
i
µ
thentherewouldbeadifferent solutionfor r p
1
µ
,whichis impossibleunlessallthe p
i
µ
arezero. Theonlywaywecanhavenon-trivialscatteringisforallthechargestobethesame
(charge)
i
=(charge)
j
(10.51)
But that’sexactlywhat gravity does! Allparticles gravitatewiththesamestrength
1
M
P
. Gravityis s uni-
versal
Masslessspin2particlesimplygravityis universal
Wecankeepgoing.Formasslessspin3wewouldneed
incoming
(charge)
i
p
ν
i
p
µ
i
=
outgoing
(charge)
j
p
ν
j
p
µ
j
(10.52)
Forexample,theµ=ν=0componentofthissays
incoming
(charge)
i
E
i
2
=
outgoing
(charge)
j
E
j
2
(10.53)
that is s the sum m of the squares of f the energies s times s some e charges are e conserved. . That’s s way too con-
straining. The e only way out is if all the charges are 0,whichis aboring,non-interacting theory offree
masslessspin3field. So,
Therearenointeractingtheoriesofmasssless particleswithspin>2
10.7 Lorentz z invariance andcharge conservation
91
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Andinfact,nomasslessparticleswithspin>2haveeverbeenseen.
Nowyoucanseewhy Weinberg’s theorem is sofantastic.qIf wecanfindaniceway toignorethem,
thenperhapswewillgetsomethingfinite.
92
Scalar QED
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Chapter11
SpinorsandtheDiracEquation
11.1 Introduction
Fromnon-relativisticquantummechanics,wealreadyknowthattheelectronhasspin
1
2
. Weusuallywrite
itasadoublet
=

(11.1)
YoumightalsorememberthePaulimatrices
σ
1
=
0 1
1 0
, σ
2
=
0 −i
i 0
, σ
3
=
1 0
0 −1
(11.2)
whichsimplifythetreatmentofspin.
Forexample,theinteractionoftheelectronwithamagneticfieldappears inwhatissometimescalled
theSchrodinger-Pauliequation:
i∂
t
ψ=Hψ=

p
2
2m
+V(r)−µ
B
B
·L
1
2×2
−2µ
B
B
·σ
ψ
(11.3)
whereµ
B
=
e
2m
e
isthe“Bohrmagneton”(thesizeoftheelectron’sorbitalmagneticmoment)andL
=x
×
p
istheangular momentumoperator. . Wehavealsowrittenσ
=(σ
1
2
3
) tocallattentiontothe fact
that theseσ
i
s transform as thecomponents of avector,justlikethemagneticfieldB
i
. Thus(σ
·B
)ψ is
rotationallyinvariant.Thisisnon-trivial,andonlyworksbecause
i
j
]=2iε
ijk
σ
k
(11.4)
whicharethesamealgebraicrelationssatisfiedbyinfinitesimalrotations(wewillreviewthisshortly).
Keepinmindthatσ
i
donotchangeunderrotations–they arealways givenbyEq. (2)inanyframe.
ψischangingandB
i
ischanging,andthesechangescancelwhenwewrite(σ
·B
)ψ.
Wecouldalsohavewrittendownarotationallyinvariantequationofmotionforψ
1·∂
t
ψ−∂
i
σ
i
ψ=0
(11.5)
Since∂
i
transforms likea3-vectorandsodoes σ
i
ψ,this equationis rotationallyinvariant. . It t turnsout it
isLorentzinvarianttoo. Infact,thisisjusttheDiracequation!
Ifwewrite
σ
µ
=(1
2×2
1
2
3
)
(11.6)
Thenitis
σµ
µ
ψ=0
(11.7)
which is nice e and d simple looking. . Actually, , this is the e Dirac equationfor a a Weyl spinor, , which h is s not
exactlythesameastheequationcommonlycalledtheDiracequation.
By theway,itdoes notfollowthat thisequationis Lorentzinvariantjustbecausewe’vewrittenit as
σµ
µ
. Forexample,
µ
µ
+m)ψ=0
(11.8)
93
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isnot Lorentzinvariant. Tounderstandtheseenigmatictransformationproperties,wehavetoknowhow
toconstruct the Lorentzgroupout of the σ
µ
’sandsee howthose act on n ψ. . But t insteadofjustguessing
theanswer(orlookingitup),we’regoingtodosomethingalittlemoregeneral,andmotivatewhatkinds
oftransformationpropertiesarepossible.
11.2 RepresentationsoftheLorentzGroup
Wehavealreadyarguedthatparticlesinour worldshouldtransformunder unitaryrepresentationsofthe
Poincare group. . These e are characterizedby mass m andspin n j. . Wemotivatedthis s inour study of the
spin1fields. MassisLorentzinvariant,soitisanobviousquantumnumber.Momentumisalsoconserved,
but it is Lorentz covariant. If f choose e a a frame e in which the e momentum has some e canonical l form, for
example p
µ
= (m,0,0,0) for m m >0, then the particles s are characterizedby the groupthat t holds s this
momentumfixed. Thatisthelittlegroup,inthiscasethegroupof3Drotations,SO(3). Thegroupof3D
rotations provides s the e second d quantum m number, , j. The way the polarizations s transform under the full
LorentzgroupistheninducedbythetransformationsunderSO(3)andthewaythemomentumtransforms
underboosts.
We alsosaw that todofieldtheory,wehavetowritedownLargrangians involvingfields. . These e are
things like V
µ
or φ or T
µν
whichdon’thaveto involve particles per se,althoughobviously inaphysical
theorytheywill. Aswesawforspin1,there’s s alotatrouble thatcomesfromhavingtoembedparticles
offixedmassandspinintofieldslikeV
µ
. Forexample,V
µ
has4degreesoffreedomwhichdescribes spin0
andspin1, sothe Lagrangian has s tobe carefully chosentomakesure thephysical l theory never r excites
thespin0component. Inaddition,whenwewanttheparticletohavem=0andspin1,weneededtogo
furtherandmakesurethelongitudinalpolarizationis neverproduced. . Thisleddirectlytochargeconser-
vation. The next t logical step p to make these embeddings s a a bit more systematic c is to o see e what kinds of
Lorentz-invariant fields wecanwrite downat all. This willrevealtheexistenceofthespin
1
2
states,and
helpuscharacterizetheirembeddingsintofields.
Agroupisasetofelements{g
i
}andaruleg
i
⊗g
j
→g
k
whichtellshoweachpairofelementsismulti-
pliedtogetathird. Theruledefinesthegroup,independentofanyparticularwaytowritethegroupele-
ments downas matrices. . A representation is particular embeddingof these g
i
s intomatrices. . Oftenwe
talkaboutthevectors onwhichthe matrices act as beingtherepresentation,but technically the matrix
embeddingistherepresentation. Anygrouphasthetrivialrepresentations s r:g
i
→1. Butmoregenerally
wecareaboutembeddingswhichare faithful,forwhicheachelementgetsit’sownmatrix.
RecallthattheLorentzgroupisthesetofrotationsandboostswhichpreservetheMinkoswkimetric
Λ
T
ηΛ=η
(11.9)
Lorentztransformationcanactonspace-time4-vectorsas
V
µ
→Λ
µν
V
ν
(11.10)
WhereΛ is acombinationofrotationsandboosts. . Wesawthat wecouldwriteaLorentztransformation
astheproductof3rotationsand3boosts:
Λ=
1
0
0
0
0 cosθ
xy
−sinθ
xy
0 sinθ
xy
cosθ
xy
0
0
0
0
1
1
cosθ
xz
−sinθ
xz
1
sinθ
xz
cosθ
xz
1
1
cosθ
xz
−sinθ
xz
sinθ
xz
cosθ
xz
(11.11)
×
coshβ
x
sinhβ
x
sinhβ
x
coshβ
x
1
1
coshβ
y
sinhβ
y
1
sinhβ
y
coshβ
y
1
coshβ
z
sinhβ
z
1
1
sinhβ
z
coshβ
z
(11.12)
94
Spinorsand theDirac c Equation
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Thisisaparticularrepresentation. oftheLorentzgroup. . Thatis,itisoneembeddingofthegroupintoa
setofmatrices.
Thegroupitselfis amathematicalobjectindependentofanyparticularrepresentation. . Toextractthe
groupawayfrom itsrepresentations, it iseasiest tolook at infinitesimaltransformations. . Wecanalways
writethesegroupelementsas
Λ=exp(iθ
i
λ
i
)=1+iθ
i
λ
i
+
(11.13)
whereθ
i
are6numbers,correspondingtothe6anglesθ
xy
xz
yz
xy
xz
yz
andΛ
i
arecalledgenera-
tors. Fortheregular,vector,representationabove,thegeneratorsare
λ
1
=V
12
=i
0
0 −1
1 0
0
, λ
2
=V
13
=i
0
0
−1
0
1
0
, λ
3
=V
23
=i
0
0
0 −1
1 0
λ
4
=V
01
=i
0 1
1 0
0
0
, λ
5
=V
02
=i
0
1
0
1
0
0
, λ
6
=V
03
=i
0
1
0
0
1
0
(11.14)
Theλ
sandtheV
sarejustdifferentnotationsforthesameobjects. Thegeneratorssatisfy:
[V
µν
,V
ρσ
]=i(η
νρ
V
µσ
−η
µρ
V
νσ
−η
νσ
V
µρ
µσ
V
νρ
)
(11.15)
whichmeans
[V
01
,V
12
]=iV
02
, [V
12
,V
23
]=iV
13
,
(11.16)
SonowwecandefinetheLorentzgroupasthesetoftransformationsgeneratedbythesegenerators.
AnotherrepresentationortheLorentzgeneratorsisgiven
J
µν
=i(x
µ
ν
−x
ν
µ
)
(11.17)
Theseare theclassical generators ofangularmomentumgeneralizedtoincludetime. Youcancheckthat
J
µν
satisfythecommutationrelationsoftheLorentzalgebra.
Technically, we say the generators s make uptheLorentz Algebraso(1,3)whichgenerates theLorentz
GroupSO(1,3). Acommonconventionistouselowercaselettersforthenamesofalgebrasanduppercase
lettersforgroups. Asanothertechnicality,wenotethatitispossiblefortwodifferentgroupstohavethe
same algebra. . For r example, , the Proper r Orthonchronous s Lorentz z groupandthe Lorentz Grouphave the
same algebra, , but t the e Lorentz group p has in addition the discrete e symmetries s time e reversal l and d parity
reversal. It’sasmalldifference,butit’sadifferenceandthegroupsarenotidentical.
11.3 Generalrepresentations
ThereisaverynicewaytostudytherepresentationsoftheLorentzgroup. Startwiththerotationgener-
atorsJ
i
andtheboostgeneratorsK
j
. Forexample,youcouldtakeJ
1
=V
23
,K
1
=V
01
,etc.asaparticular
representationofthisbasis. Thesesatisfy
[J
i
,J
j
]=iǫ
ijk
J
k
(11.18)
[J
i
,K
j
]=iǫ
ijk
K
k
(11.19)
[K
i
,K
j
]=−iǫ
ijk
J
k
(11.20)
whereǫ
ijk
isdefinedbyε
123
=1andbeingantisymmetricinallofitsindices(itiscalledthetotallyanti-
symmetrictensor.)Asisprobablycleartoyoualready,[J
i
,J
j
]=iǫ
ijk
J
k
isthealgebraforrotations,so(3)
sotheJ
i
generatethesubgroupofrotationsin3D.
11.3 General l representations
95
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Nowtakethelinearcombinations
J
i
+
=
1
2
(J
i
+iK
i
)
(11.21)
J
i
=
1
2
(J
i
−iK
i
)
(11.22)
whichsatisfy
[J
i
+
,J
j
+
]=iǫ
ijk
J
k
+
(11.23)
[J
i
,J
j
]=iǫ
ijk
J
k
(11.24)
[J
i
+
,J
j
]=0
(11.25)
Sowehavefound2commutingsubgroupsoftheLorentzgroup. ThealgebraoftheJ
sisthe3Drotation
algebra,so(3),morecommonlycalledso(2).Sowehaveshownthat
so(1,3)=su(2)×su(2)
(11.26)
(technically,it shouldreallybesl(2,R)=so(1,1) notsu(2),butthesealgebrasarethesameuptosome
i’s,andphysicistsusuallyjustsaysu(2).).
Thenames so(n) andsu(n)standfor specialorthogonalalgebraandspecialunitary algebra. . Orthog-
onalmeans it preserves anorm definedwithtranspose:V
T
V is s invariant under SO(n). . Unitarymeans s it
preservesanormdefinedwithadjointV
V isinvariantunder r SU(n). . Specialmeansthedeterminantis1,
whichjustnormalizestheelementsandgetsridofphases.
Thedecompositionso(3,1)=su(2)×su(2) makes studyingthe representations veryeasy. Wealready
knowwhattherepresentationsareofsu(2),sincethisisthealgebraofPaulimatrices,whichgeneratesthe
3DrotationgroupSO(3) (so(3)=su(2)). . Therepresentations s arecharacterizedbyaquantum number r j,
andhave2j+1elements,labeledbym=j
z
,SorepresentationsoftheLorentzgrouparecharacterizedby
twonumbersAandB. The(A,B)representationhas(2A+1)(2B+1)degreesoffreedom.
TheregularrotationgeneratorsareJ
=J
+
+J
,whereInowusethevectorsuperscripttocallatten-
tion the fact t that the e spins s must be e added vectorially (remember Clebsch-Gordon n coefficients?). . The
bottomlineisthateach(A,B)representationcontainsparticlesofspinJ=A+B,A+B−1,
,|A−B|.
Forexample,
repofsu(2)×su(2)
repof3Drotationgroup
(A,B)=(0,0)
J=0
(A,B)=(
1
2
,0)
J=
1
2
(A,B)=(0,
1
2
)
J=
1
2
(A,B)=(
1
2
,
1
2
)
J=1,0
(A,B)=(1,0)
J=1
(A,B)=(1,1)
J=2,1,0
(11.27)
Soweseethat theregular 4Dvector representationA
µ
containsspins 1,0soit corresponds tothe (
1
2
,
1
2
)
representationoftheLorentzgroup.
11.3.1 unitaryrepresentations
RepresentationsoftheLorentzgrouparenotalwaysunitary. UnitaritymeansΛ
Λ=1,whichisnecessary
tohaveLorentzinvariantmatrixelements
ψ|ψ→
ψ|Λ
Λ|ψ
(11.28)
96
Spinorsand theDirac c Equation
SincethegroupelementistheexponentialofthegeneratorΛ=e
,unitarityrequiresthatλ
=λ,thatis,
that λ be Hermetian. . Looking g backat t our explicit 4D representationfor the Lorentz generators, wesee
thattherotations
J
1
=i
0
0 −1
1 0
0
, J
2
=i
0
0
−1
0
1
0
, J
3
=i
0
0
0 −1
1 0
areHermetian. Whiletheboosts
K
1
=i
0 1
1 0
0
0
, K
2
=i
0
1
0
1
0
0
, K
3
=i
0
1
0
0
1
0
(11.29)
Arenot.(Theyareanti-Hermetianλ=−λ.)
Non-unitaryrepresentationsoftheLorentzgroupareokforfields,butnotforparticles.Particlesmust
transformunderunitary representations of the Poincare group,sothat we canconstruct physical ampli-
tudeswiththem. AsWignershowed,andwediscussedwhenweintroducedgaugeinvariance,unitaryrep-
resentations andarecharacterizedbymass andspin. . This s spinindicatesarepresentationof the3Drota-
tiongroup,whichisunitary.
Now,thegeneratorsforthesu(2)×su(2)decompositionJ
±
=J
±iK
areHermetian. Thisisbecause
su(2) is thespecialunitarygroup,soofcourseithas unitaryrepresentations. But thisdoesn’tmeanthat
thecorrespondingrepresentationsoftheLorentzgroupareunitary. Youhavetobeverycarefulaboutthe
factorsof iandhowthesegroupsaredefined.WesaidthataLorentzgroupelementis
Λ=exp(iθ
i
J
i
+iβ
i
K
i
)
(11.30)
where the θ
i
arethe rotationangles and d β
i
theboosts “angles”. . Theseare real numbers. . If f we took the
basisλ
i
=J±,thenthecorrespondingθ
i
swouldhavetobecomplex.SotherepresentationsoftheLorentz
grouparedeterminedbythealgebrasu(2)×su(2),butsincetheanglesθ arereal,thecorrespondingrep-
resentationsoftheLorentzgrouparenotunitary.
11.3.2 summary
Insummary,the irreduciblerepresentations of theLorentz groupare characterizedbytwospins (A,B).
Physical particles are e characterizedby mass, , m andspin n j. Fields s generally describe particles s of f many
spins,andonehastomakesurethataphysicaltheoryonlyhascausallypropagatingparticlesofthespins
wewant.
11.4 Spin
1
2
representation
Now that we e know w that the e irreducible representations of the Lorentz group are e characterized d by y two
spins,(A,B),howdowefindtheexplicitrepresentations? Inparticular,whatdothe(
1
2
,0)and(0,
1
2
)rep-
resentationslooklike? Weneedtofind2x2matricesthatsatisfy
[J
i
+
,J
j
+
]=iǫ
ijk
J
k
+
(11.31)
[J
i
,J
j
]=iǫ
ijk
J
k
(11.32)
[J
i
+
,J
j
]=0
(11.33)
11.4 Spin
1
2
representation
97
Butwealreadyknowsuchmatrices:thePaulimatrices
i
j
]=2iε
ijk
σ
k
(11.34)
rescalingwefind
[
σ
i
2
,
σ
j
2
]=iε
ijk
σ
k
2
(11.35)
Anotherusefulfactisthat
i
j
}=σ
i
σ
j
j
σ
i
=2δ
ij
(11.36)
ThuswecansetJ
i
+
=
σi
2
.Thisisthe(
1
2
,).WhataboutJ
i
?Thisshouldbethe(,0),sotheobviousthing
todoisjusttakethetrivialrepresentationJ
i
=0. Sothe(
1
2
,0)representationis
(
1
2
,0): J
i
+
=
σ
i
2
, J
i
=0
(11.37)
Similarly,the(0,
1
2
)representationis
(0,
1
2
): J
i
+
=0, J
i
=
σ
i
2
(11.38)
What does this meanfor actualLorentz transformations? ? Well,the e rotations are J
=J
+
+J
andthe
boostsareK
=i(J
−J
+
)so
(
1
2
,0): J
i
=
σ
i
2
, K
i
=−i
σ
i
2
(11.39)
(0,
1
2
): J
i
=
σ
i
2
, K
i
=i
σ
i
2
(11.40)
SincethePaulimatricesareHermetianσ
i
i
weseethattherotationsareHermetianbuttheboostsare
anti-Hermetian. Thisis s the sameas what we foundfor the vector representation. . Also noticethat these
tworepresentationsarecomplexconjugatesofeachother.Incontrast,thevectorrepresentationwasreal.
Explicitly, if f ψ
L
is a (
1
2
, 0) ) spinor, , known also o as s a left-handed Weyl spinor, then under rotations
anglesθ
i
andboostanglesβ
i
ψ
L
→e
1
2
(iθ
i
σ
i
i
σ
i
)
ψ
L
=(1+
1
2
i
σ
i
+
1
2
β
i
σ
i
+
L
(11.41)
Similarly,
ψ
R
→e
i
σ
i
−β
i
σ
i
ψ
R
=(1+
1
2
i
σ
i
1
2
β
i
σ
i
+
R
(11.42)
Infinitesimally,
δψ
L
=
1
2
(iθ
i
i
i
ψ
L
(11.43)
δψ
R
=
1
2
(iθ
i
−β
i
i
ψ
R
(11.44)
Noteagaintheanglesθ
i
andβ
i
arerealnumbers. AlthoughwemappedJ
i
orJ
i
+
to0,westillhavenon-
trivialactionofalltheLorentzgenerators. Sotheseare faithful irreduciblerepresentationsoftheLorentz
group. Similarly
δψ
L
=
1
2
(−iθ
i
i
L
σ
i
(11.45)
δψ
R
=
1
2
(−iθ
i
−β
i
L
σ
i
(11.46)
11.4.1 Lorentzinvariants
NowthesimplestthingtodowouldbetowritedownaLagrangianwithtermslike
L
)
ψ
L
+m
2
L
)
ψ
L
(11.47)
98
Spinorsand theDirac c Equation
However,wecanseefromtheabovethatthisisnotLorentzinvariant:
δψ
L
ψ
L
=
1
2
L
)
[(iθ
i
i
i
ψ
L
]+
1
2
[(ψ
L
)
(−iθ
i
i
i
L
(11.48)
=
β
i
2
ψ
L
σ
i
ψ
L
0
(11.49)
Thisisjustthemanifestationofthefactthattherepresentationisnotunitarybecausetheboostgenera-
torsareanti-Hermetian. Ifweallowourselvestwofields,ψ
L
andψ
R
,wecanwritedowntermslikeψ
R
ψ
L
.
UnderinfinitesimalLorentztransformations,
δ(ψ
R
ψ
L
)=ψ
R
1
2
(−iθ
i
−β
i
i
ψ
L
R
1
2
(iθ
i
i
i
ψ
L
=0
(11.50)
Whichisgreat. However,thistermisnotreal. ButthenwecanjustaddtheHermetianconjugate,so
m
ψ
R
ψ
L
L
ψ
R
(11.51)
isok.
Whataboutkineticterms? Wecouldtry
ψ
R
ψ
L
L
ψ
R
(11.52)
whichisbothLorentz invariantandreal. . But t this isactuallynot averyinterestingLagrangian. . We e can
alwayssplitupourfieldintocomponents ψ
L
=
ψ
1
ψ
2
,whereψ
1
andψ
2
arejustregularfields. Thenwesee
thatthis isjusttheLagrangianforacoupleofscalars. . Soit’snotenoughtodeclaretheLorentztransfor-
mationpropertiesofsomething,theLagrangianhastoforce thosetransformationproperties. Inthesame
way,avectorfieldisjust4scalarsuntilwecontractitwith∂
µ
intheLagrangian.
Toproceed,let’slookat
ψ
L
σ
j
ψ
L
(11.53)
Thistransformsas
δψ
L
σ
j
ψ
L
=
1
2
ψ
L
σ
j
[(iθ
i
i
i
ψ
L
]+
1
2
L
(−iθ
i
i
i
j
ψ
L
(11.54)
=
i
2
ψ
L
(
σ
j
σ
i
−σ
j
σ
i
)
ψ
L
+
β
i
2
ψ
L
i
σ
j
j
σ
i
L
(11.55)
i
ε
ijk
ψ
L
σ
k
ψ
L
j
ψ
L
ψ
L
(11.56)
Thuswehavefoundthat
δψ
L
ψ
L
=
β
i
2
ψ
L
σ
i
ψ
L
(11.57)
δψ
L
σ
j
ψ
L
i
ε
ijk
ψ
L
σ
k
ψ
L
j
ψ
L
ψ
L
(11.58)
Similarly,
δψ
R
ψ
R
=−
β
i
2
ψ
L
σ
i
ψ
L
(11.59)
δψ
R
σ
j
ψ
R
i
ε
ijk
ψ
L
σ
k
ψ
L
+−β
j
ψ
L
ψ
L
(11.60)
Now,recallhowavectortransforms
δV
0
i
V
0
(11.61)
δV
j
j
V
0
i
ε
ijk
V
k
(11.62)
ThisisjustthetransformationofthevectorV
µ
L
=(ψ
L
ψ
L
L
σ
j
ψ
L
).Thus
ψ
L
t
ψ
L
L
j
σ
j
ψ
L
(11.63)
11.4 Spin
1
2
representation
99
isLorentzinvariant.SimilarlyV
µ
R
=(ψ
R
ψ
R
R
σ
j
ψ
R
)transformslikeavector
Defining
σ
µ
=(1,σ
), σ¯
µ
=(1,−σ
)
(11.64)
WecanwritealltheLorentz-invarianttermswehavefoundas
L=iψ
L
σ
µ
µ
ψ
L
+iψ
R
σ¯
µ
µ
ψ
R
+m
ψ
R
ψ
L
L
ψ
R
(11.65)
Ihaveaddedafactorof ionthekinetictermsforlaterconvenience.
There’sanevenshorter-handwaytowritethis. Ifwewriteourspinorasadoublet
ψ=
ψ
L
ψ
R
(11.66)
Then,define
ψ
¯
=
ψ
R
ψ
L
(11.67)
Andusingthe4x4matrices
γ
µ
=
σ
µ
σ¯
µ
(11.68)
OurLagrangianbecomes
L=ψ
¯
µ
µ
−m)ψ
(11.69)
WhichistheconventionalformoftheDiracequation.
11.5 Diracmatrices
Expandingthemout,theDiracmatricesare
γ
0
=
1
1
γ
i
=
0
σ
i
−σ
i
0
(11.70)
Or,
γ
0
=
0
1
0
1
1
0
1
0
, γ
1
=
0
1
0 1
−1 0
−1
0
(11.71)
γ
2
=
0
−i
0 i
i 0
−i
0
, γ
3
=
0
1
0
−1
−1
0
1
0
(11.72)
Theysatisfy
µ
ν
}=2η
µν
(11.73)
InthesamewaythatthealgebraoftheLorentzgroupismorefundamentalthananyparticularrepresen-
tation,thealgebraofthe γ
sismorefundamentalthananyparticularrepresentationofthem. Wesaythe
γ
sformtheDiracalgebra,whichisaspecialcaseofaCliffordalgebra. ThisparticularformoftheDirac
matricesisknownastheWeylrepresentation.
TheLorentzgeneratorsare
S
µν
=
i
4
µ
ν
]
(11.74)
Theysatisfy theLorentzalgebrafor any γ
ssatisfyingtheCliffordalgebra. Thatis,youcanderive from
µ
ν
}=2η
µν
that
[S
µν
,S
ρσ
]=i(η
νρ
S
µσ
−η
µρ
S
νσ
−η
νσ
S
µρ
µσ
S
νρ
)
(11.75)
100
Spinorsand theDirac c Equation
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