Peskin 量子场论答案.pdf-资料库

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Physics443Problemset1Fall2013DueOct9.ReadingPeskin&SchroederCh.2.Problems1.Peskin&Schroederproblem2.12.Peskin&Schroederproblem2.23.Peskin&Schroederproblem2.3

Physics443Homework1SolutionsProblem1(P&SProblem2.1)a)VaryingtheactionwithrespecttoAµweﬁndthatδS=−14d4xδ(FµνFµν)(1)=−12d4xδFµνFµν(2)=−12d4xδ(∂µAν−∂νAµ)Fµν(3)=−d4x∂µδAνFµν.(4)Sotheequationsofmotionaregivenby∂µFµν=0.Above,wehavebasicallyderivedthat∂L∂(∂µAν)=−Fµν,(5)sousingEuler-Lagrangeequationstheresultagainfollows.WewillnowuseEi=−F0iandijkBk=−Fijtorearrangetheseequations.Forν=iweobtain0=∂0F0i+∂jFji=−∂0Ei+ijk∂jBk(6)orequivalently∂E∂t=∇×B.(7)Forν=0weobtain0=∂0F00+∂jFj0=∂jEj(8)orequivalently∇.E=0.(9)ThelasttwoMaxwellequationsareBianchiidentitiesandtheyfollowfromthedeﬁnitionFµν=2∂[µAν]as3∂[µFνρ]=∂µFνρ+∂νFρµ+∂ρFµν=0.(10)Whenallindicesarespatialweget∂1F23+∂2F31+∂3F12=0(11)whichmeans∇.B=0.(12)Finally,whenoneoftheindexesisatimeindexonegetsijk(2∂jFk0+∂0Fjk)=0(13)giving∇×E+∂B∂t=0.(14)(NotethatBk=−12ijkFij.)1

b)Tµν=∂L∂(∂µAλ)(∂νAλ)−Lδµν.(15)Frompart(a)weknowthat∂L∂(∂µAλ)=−Fµλ,(16)andsoTµν=14FλρFλρgµν−Fµλ∂νAλ.(17)Althoughtheﬁrsttermissymmetricunderµ↔ν,thesecondtermisnot.Toremedythis,weaddaterm∂λ(FµλAν)toTµν.TheantisymmetryofFµλunderµ↔λgives∂µ∂λ(FµλAν)=0andhencethenewenergymomentumtensorisstillconserved.ˆTµν=Tµν+∂λ(FµλAν),(18)=14FλρFλρgµν−Fµλ∂νAλ+Fµλ∂λAν+∂λFµλAν.(19)Thelasttermvanishesonshell(i.e.whentheequationsofmotionaresatisﬁed)becauseofthe∂λFµλterm.WhatremainsisˆTµν=14FλρFλρgµν+FµλFλν,(20)whichissymmetricunderµ↔ν.1EnergydensityisgivenbytheˆT00term:ε=14FλρFλρg00+F0λFλ0(21)=−12F0iF0i+14FijFij+F0iF0i(22)=−12(−Ei)(−Ei)+14(−ijkBk)(−ijmBm)+(−Ei)(−Ei)(23)=12E2+12B2.(24)ThemomentumdensityisgivenbytheˆT0iterm:Si=F0λFλi=−F0jFji=−(−Ej)(−jikBk)=ijkEjBk,(25)whichgivesthefamiliarexpressionS=E×B.Problem2(P&SProblem2.2)InthisproblemwewillstudyafreecomlexscalarﬁeldwithactionS=d4x∂µφ∗∂µφ−m2φ∗φ,(26)andLagrangianL=∂µφ∗∂µφ−m2φ∗φ=˙φ∗˙φ−∇φ∗·∇φ−m2φ∗φ.(27)1Indeed,Lorentzsymmetryguaranteesthatonecanalwaysaddaterm∂λfλµν(withfλµν=−fµλν)andob-tainasymmetricenergy-momentumtensorwhichisequivalenttotheTµνobtainedfromNoetherprocedure.See’http://itf.fys.kuleuven.be/supergravity/index.php?id=1&type=ExtraCh1.html’forashortargumentonthisstatement.2

a)Momentumdensitiesconjugatetoφandφ∗areπ(x)=∂L∂˙φ(x)=˙φ∗(x)andπ∗(x)=∂L∂˙φ∗(x)=˙φ(x).(28)HamiltonianassociatedwiththissystemisthengivenbyH=d3xπ(x,t)˙φ(x,t)+˙φ∗(x,t)π∗(x,t)−L(x,t),(29)=d3xππ∗+∇φ∗·∇φ+m2φ∗φ.(30)WewillnowstudythequantummechanicalsystemwiththisHamiltonianandhavingcanonicalequaltimecommutationrelations[φ(x,t),π(x,t)]=iδ3(x−x),(31)and[φ∗(x,t),π∗(x,t)]=iδ3(x−x),(32)withallotherequaltimecommutatorsbetweenφ,φ∗,π,π∗settozero.NotethatwewillbeworkinginHeisenbergpicturewhereoperatorsaretimedependentandstatesaretimeindependent.Moreover,theHamiltonianhasnoexplicittimedependenceandthereforeHeisenbergequationsofmotiongivedH/dt=0orinotherwordsH(t)=H(t)holdsasanoperatoridentity.UsingthesecommutationrelationsandHeisenbergequationsofmotionwecancompute∂φ/∂t.Notethathere∂/∂tdoesnotmeanderivativewithrespecttoexplicittimedependence.Itisthetotaltimederivativeoftheoperatorφ(x,t)whichalsodependsonthespacetimecoordinatexwhichwecanthinkofasanadditionalindexontheoperatorφ.∂φ∂t=−i[φ(x,t),H(t)](33)=−id3x[φ(x,t),π(x,t)]π∗(x,t)(34)=π∗(x,t).(35)Wecansimilarlycomputethetimederivativeoftheπ∗(x,t)operator.∂π∗∂t=−i[π∗(x,t),H(t)](36)=−id3x∇[π∗(x,t),φ∗(x,t)]·∇φ(x,t)+m2[π∗(x,t),φ∗(x,t)]φ(x,t)(37)=−d3x∇δ3(x−x)·∇φ(x,t)+m2δ3(x−x)φ(x,t)(38)=∇2φ(x,t)−m2φ(x,t).(39)Finally,wecancombinethesetwoequationsas∂2φ∂t2=∂π∗∂t=∇2φ(x,t)−m2φ(x,t),(40)whichcanberewrittenas(∂2+m2)φ=0.(41)Similarly,fortheHermiteanconjugateﬁeldoperatorwehave(∂2+m2)φ∗=0.(42)3

b)Theequationsofmotionaresatisﬁedbye±ip·xwherep0isconstrainedtobeEp=p2+m2Then,wewritetheﬁeldoperatoratt=0asaFourierexpansionandtheequationsofmotionfoundinpart(a)givestheﬁeldoperatorforalltasφ(x)=d3p(2π)312Epape−ip·x+b†peip·x,(43)whereapandbparetimeindependentoperators.Wealsoimposearealityconditionφ∗(x)=φ(x)†.Theequation∂φ∂t=π∗thengivesπ∗(x)=−id3p(2π)3Ep2ape−ip·x−b†peip·x.(44)Wecaninvertthesetwoexpressionsasap=eiEpx0d3xEp2φ(x)+i2Epπ∗(x)eip·x(45)andb†p=e−iEpx0d3xEp2φ(x)−i2Epπ∗(x)e−ip·x.(46)Then,usingthecommutationrelationswehavesofar,somealgebragivesap,a†p=12ei(Ep−Ep)x0d3xei(p−p)·xEpEp+EpEp(47)whichwecanintegrateasap,a†p=(2π)3δ3(p−p).(48)Similarly,bp,b†p=(2π)3δ3(p−p).(49)Atthispointweinterpretapandbpasannihilationoperatorsanddeﬁnethevacuumstatebyap|0=0andbp|0=0forallp.(50)OurﬁnaltaskistoputtheﬁeldexpressionswefoundintotheHamiltonianandﬁnditintermsofcreationandannihilationoperators.H=d3xd3p(2π)3d3p(2π)3EpEp4ape−iEpt−b†−peiEpta†peiEpt−b−pe−iEptei(p−p)·x+p·p+m24EpEpa†peiEpt+b−pe−iEptape−iEpt+b†−peiEpte−i(p−p)·x,(51)=d3p(2π)3Ep2apa†p+b†−pb−p+a†pap+b−pb†−p,(52)=d3p(2π)3Epa†pap+b†pbp+d3pEpδ3(0).(53)WeneglecttheconstantinﬁnitetermandweseethattheﬁrsttermgivesanenergyofEp=p2+m2forparticlescreatedbya†pandb†p.Inotherwords,therearetwokindsofparticleswhereeachhasmassm.4

c)ThechargeQcanbeexapndedasQ=i2xd3p(2π)3d3p(2π)3−i2EpEpa†peiEpt+b−pe−iEptape−iEpt−b†−peiEptei(p−p)·x−i2EpEpa†peiEpt−b−pe−iEptape−iEpt+b†−peiEptei(p−p)·x,(54)=14d3p(2π)32a†pap−2b−pb†−p,(55)=12d3p(2π)3a†pap−b†pbp−12d3pδ3(0).(56)Weneglecttheconstantinﬁnitetermandweseethata†pparticleshavecharge+1/2andb†pparticleshavecharge−1/2.NotethatthislastexpressionshowsmanifestlythatQisconservedsincenotimedependenceisleft.d)Letφ=φ1φ2whereφ1andφ2aretwofreecomplexKlein-Gordonﬁeldswithmassm.TheactionisthengivenbyS=d4x∂µφ†∂µφ−m2φ†φ.(57)Wecanimmediatelyseethatthetransformationφ→Aφ,φ†→φ†A†leavestheLagrangian(andhencetheaction)invariantforA∈U(2).LetAbeanelementclosetotheidentitymatrix.ThatisA=I−iBfor1.ThenAA†=IgivesB=B†tolinearorderin.Inotherwords,thegeneratorsofU(2)are2×2Hermiteanmatrices.ThesearegeneratedbythematricesI/2andσi/2(i=1,2,3).I/2generatestheU(1)partandσi/2generateSU(2).LetusdenotethegeneratorscollectivelyasTJ=(I/2,σi/2)forJ=0,1,2,3.Inﬁnitesimalversionsoftheﬁeldtransformationsareφa→φa−i(TJ)abφb(58)andφ∗a→φ∗a+iφ∗b(TJ)ba.(59)ByNoether’stheorem(andusingthefactthattheLagrangianremainsinvariant)foreachJwegetaconservedcurrentasjJµ=∂L∂(∂µφa)−i(TJ)abφb+iφ∗b(TJ)ba∂L∂(∂µφ∗a)(60)ThenjJ0=−iπa(TJ)abφb−φ∗a(TJ)abπ∗bandQJ=d3xjJ0isaconservedcharge.QJ=id3xφ∗a(TJ)abπ∗b−πa(TJ)abφb(61)andnotethatQ0isthesumoftwochargesasdeﬁnedinpart(c)andQifori=1,2,3aretheconservedchargesintheproblem.5

QI,QJ=−d3xd3y(TI)ab(TJ)cd[φ∗a(x,t)π∗b(x,t)−πa(x,t)φb(x,t),φ∗c(y,t)π∗d(y,t)−πc(y,t)φd(y,t)]=−d3xd3y(TI)abφ∗a(x,t)[π∗b(x,t),φ∗c(y,t)]π∗d(y,t)+φ∗c(y,t)[φ∗a(x,t),π∗d(y,t)]π∗b(x,t)+πa(x,t)[φb(x,t),πc(y,t)]φd(y,t)+πc(y,t)[πa(x,t),φd(y,t)]φb(x,t)(TJ)cd(62)=−i(TI)ab(TJ)cdd3x−φ∗aδbcπ∗d+φ∗cδadπ∗b+πaδbcφd−πcδadφb(63)=i(TI)ab(TJ)bd−(TI)bd(TJ)abd3x(φ∗aπ∗d−πaφd)(64)=iTI,TJadd3x(φ∗aπ∗d−πaφd).(65)Therefore,writingTI,TJ=ifIJKTKwegetQI,QJ=ifIJKQK.(66)Inparticular,forSU(2)partQi,Qj=iijkQk.(67)Ournotationgeneralizesinastraightforwardwaytothecasewhereonehasnidenticalcomplexﬁelds.Wehaveshownthattherearen2conservedchargeswhichsatisfyu(n)Liealgebra.Indeed,ncomplexﬁeldactioncanberewrittenasanactionfor2nidenticalrealscalarﬁelds,Φa,wherea=1,2,···,2n.S=d4x∂µΦa∂µΦa−m2ΦaΦa,(68)Then,atransformationbyA∈SO(2n),φ→Aφisasymmetry.Thisleadston(2n−1)conservedcharges,oneforeachgeneratorofso(2n)Liealgebra.Problem3(P&SProblem2.3)ALorentztransformationactingon(x−y)canbecompensatedbyaLorentztransformationonpandgiventheinvarianceofintegralmeasureunderLorentztransformationsD(x−y)=d3p(2π)312Epe−ip·(x−y)(69)isaLorentzinvariantfunctionof(x−y).Inparticular,foraspacelike(x−y)where(x−y)2=−r2wecanﬁndtheresultbyevaluatingtheintegralat(x−y)µ=(0,0,0,r).Usingsphericalcoordinatesforpintegrationweget:D(x−y)=12(2π)3∞0dpp2π0dθsinθ2π0dφ1p2+m2eiprcosθ,(70)=14π2r∞0dppp2+m2sin(pr),(71)=m4π2r∞0dusinh(u)sin(mrsinh(u)),(72)6

whereinthelastlinewechangedtheintegrationvariableasp=msinh(u).Finally,usingthefollowingintegralexpression2formodiﬁedBesselfunction,K1(t):K1(t)=∞0dusinh(u)sin(tsinh(u)),(73)wegetD(x−y)=m4π2rK1(mr).(74)2Ausefulwebsitethatincludessuchexpressionsis’http://functions.wolfram.com/’.Theparticularexpressionusedherecanbefoundat’http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/01/01/’whereonealsoneedstotakethelimitofKν(t)asν→1andusetheanalyticityinν.7

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