资源描述：
《QFT Solutions - Srednicki》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。
1、QuantumFieldTheory:ProblemSolutionsMarkSrednickiUniversityofCalifornia,SantaBarbaramark@physics.ucsb.edu!c2007byM.SrednickiAllrightsreserved1MarkSrednickiQuantumFieldTheory:ProblemSolutions21Attemptsatrelativisticquantummechanics1.1)β2=1⇒eigenvalue-squared=1⇒eigenvalue=±1
2、.α21=1⇒Trβ=Trα21β.Cyclicpropertyofthetrace⇒Trα21β=Trα1βα1.Then{α1,β}=0⇒Trα1βα1=−Trα21β=−Trβ.ThusTrβequalsminusitself,andsomustbezero.Trαi=0followsfromthisanalysisbytakingβ→αiandα1→β.1.2)Fornotationalsimplicity,switchtoadiscretenotation:!!=d3x...d3x,1nδ=δ3(x−y),xya1=a(x1),
3、ψ=ψ(x1,...,xn;t).(1.40)Using[X,AB...C]=[X,A]B...C+A[X,B]...C+...+AB...[X,C],(1.41)whichfollowsfromwritingoutthetermsonbothsides,wehave[a†a,a†...a†]=[a†a,a†]a†...a†+a†[a†a,a†]a†...a†xy1nxy12n1xy23n+...+a†...a†[a†a,a†].(1.42)1n−1xynWehave[a†a,a†]=a†[a,a†]±[a†,a†]a,xyixyi∓xi
4、∓y=δa†(1.43)iyxwhere[A,B]∓=AB∓BA.Usingthisanday
5、0&=0,wefind"n(a†a)a†...a†
6、0&=(a†...a†)δ
7、0&.(1.44)xy1n1ni→xiyi=1Similarly,wehave"n(a†a†aa)a†...a†
8、0&=(a†...a†)xyyx1n1ni→x
9、0&(1.45)i,j=1j→yforbothbosonsandfermions.(Extraminussignswithfermionscancelwhenwemovea†xanda†intotheposi
10、tionsformerlyoccupiedbya†anda†.)yijNowconsider!"n!!d3xa†(x)∇2a(x)
11、ψ&=d3x(ψ∇2δ)(a†...a†)
12、0&.(1.46)xxxi1ni→xi=1Wehave∇2xδxi=∇2iδxi.Thenwecanintegratebypartstoput∇2iontoψ,andthenintegrateoverxusingthedeltafunctiontoget!"n!d3xa†(x)∇2a(x)
13、ψ&=(∇2ψ)a†...a†
14、0&.(1.47)xi1ni=1MarkSr
15、ednickiQuantumFieldTheory:ProblemSolutions3Similarly,!"n!d3xU(x)a†(x)a(x)
16、ψ&=U(x)
17、ψ&,(1.48)ii=1and!"nd3xd3yV(x−y)a†(x)a†(y)a(y)a(x)
18、ψ&=V(x−x)
19、ψ&,(1.49)iji,j=1whichyieldsthedesiredresult.#†††1.3)N=iaiai.Then[N,aj]=+ajand[N,aj]=−ajforbothbosonsandfermions.Thus,usingeq.(1.41
20、),wefind††††[N,ai1...ainaj1...ajm]=(n−m)ai1...ainaj1...ajm.(1.50)Thusifthenumberofa’sequalsthenumberofa†’s,theoperatorcommuteswithN.MarkSrednickiQuantumFieldTheory:ProblemSolutions42LorentzInvariance2.1)Startwitheq.(2.3)andletΛµρ+δµρ+δωµρ.WewillalwaysdroptermsthatareO(δω2)
21、orhigher.Thenwehaveg=g(δµ+δωµ)(δν+δων)ρσµνρρσσ=g(δµδν+δµδων+δωµδν)µνρσρσρσ=g+gδων+gδωµρσρνσµσρ=g
