LAMBDA PHI CUBED QED

Ap?ind=6 | | $4ind=4 QED Overall comments madewhenthisbinder in#wasfinished, april22,1979 ThefirstthingIlearnedwashowyourenormalize afieldtheoryandgeta e finite result. Basically, you show finiteness byexamining integral equations. “The trick istouse counterterms (ie, rescalings) which contain the cutoff. Ittook mealong time tounderstand howyoudoallthis, butIthink I've nowgotit.Cantake QED, eg,and 7 compute something and get aclear finite answer. Iunderstand these things without having understood Hepp. Theoverlap analysis wascrucial toshowing hawacertain 2, factor was removed, similar toBDdeal. Alltheabove work wasdone without thinking about “renormalization poitn" or anything fancy. Ichoose torenormalize atthephysical mass m:ie,Isetupmy "renormalization conditions" atp*=m’, Basically, thishadtodowithsetting thé séale ofthe vertex and propagactr. Iused mass cointerterm method inthe above, never had abare mass. Inthe next section ofnotes Iconéidered the idea ofusing some other point todefine the renormalization conditions. Ithink that each "renormalization part” in a‘field theory needs renorm conditions, maybe more than one ifD.GT.0. Qbviously when youhaveadivergent n-point function youaregoing tohavetoeffect subtraction ofthe divergence with counterterms, orrescaling ofbasic objects liké the chargé. Ifyou have to'subtract somefunction,youlosecontroloveritsfinitéconstantpart.Thus,you OD: mst specity the velue ofthe function somewhere. This iswhat arenortalization condition does. Thevalue’ ofp*where youdothesubtracticn orspeficiation orwhatever iscalled | therenorm point. SoIdid&againwithanarbitrary renorm pointu,.TheZ'syouget coming outofthe original bare theory then depend onuj, sotherefore does the charge.| Mymass m,happened nottodepend onu,,butAbarb's didsoIkept this possible dependence. Then Iasked: suppose you choose some other renorm poitn up.Then youget adifferent but finite theory, your Z's are alittle different. These theories 1and 2 are then connected byafinite GRmultiplication rule, Its not véry complicated: ifyou change your renorm point u,touj,youwant toimpose thesame renorm condition oneach renorm part, butyoudosoatdifferent value ofp’. Same asletting keep samerenorm point and allowing differnt scale sets, eg, differn "residue" etc. These scale sets arenotobservables; youcanalways scale upyourpropagator andscale downyourvertex and have same result. Only the Z's have changed, etc. Physical finite theory iwthe same. This leads tothe idea ofthe invariant vertex function orcharge d.You find + acombination which isthe same despite rescalings: this isthe object ofphysical interest. Youaskhowitbehaves atlarge andsmall q”andgetasymptotic freedom perhaps, 1 orjustafiniteinfrared chargeinthecaseofQEDR.This“running coupler" isthe 6 object ofespecial interest: you want toknow all about it. Some information canbe obtained from the BETA function. p?ind=6 Contents of this Section. : ©)3)Summary ofPindub,ie,summary ofnotesinlatersection ofthisbinder,. Thiswork was done using usual mass renorm point m.Main idea was toshow the finiteness ofthe renormed theory due tointegral equations. 2.Modifications when you gotoarbitrary renorm point u.(summary page) 3. Dimensionalities. 4.Howtorenormalize awayfrom theusual m*point. (still massmview) 5.The bare mass viewpoint. 6.Renorm atarbitrary mu, bare mass view. Showing what depends onwhat variables. 7.Connection with Abarbanel: our m's do-not agree. 8.Emgnation from the tare theory tothe 1-theory orthe 2-theory. Idea that the theory you arrive atdepends onthe renorm point. Little 2's connect these two finite theories. Construction ofaninvariant function dinthis theory. Renorm group transformations. 9.Properties ofthe invariant charge function d. (taken from older notes) a)construction offunctions that look like they are oftwo variables but are not. b)the RGdifferential equation for d;massless cases. 10.Show that BETA =BETA(e) only inmassless theory. Also, how tomove from onerenorm fe) pointtoanother, dependencies. 11. Another emanation map showing three theories. 12, THE RGE for the propagator. 13.RGE for general 1PIvertex function (including the inverse propagator) 14. Concept ofrenorm dimension, general RGE. 15.Calculation ofBETA in@,using earlier perturbation results. Massless case. 16,Connection between theinvariant dfunction and "running coupling constent 2"of Gross and Wilezek. Mention ofBDsolution, zeros inBETA, etc. 17. AFEW residual QED notes, tobekilled soon. 1 se) oe 7a - °D=DD ped T=yeep 3= $:Thom dag “Ka K Ay o hom2s" ' *§=Bg (2)gowcenorwalige ds 'D= \ = i -eReeeac) (Pew)[1°25] BC) =emt) 4ewe"BG °F) =0aalbedlad. “O(Pant)=2 >om=\-oFGut)SalsasZsPom Qe)Senormalagid ¢ “ D= \ = \ =&Dp pow -2@9 Cem L-BEY] QB >B= GH) (2) +,LH) > owt) =By°S7 Od) ZOM= Cw) wE@ =2,(Ze)- Foo] 6)Remon. Condiduoor 2 AEGR ©(edlySbfoaletueda0.) >poleinakemoae M. a.Be)20Gaonlommiad given d=Rd SS wewideadono GosiaBLO)Sonata GowBoateLo) (@)LovettOrtonWaresSDgangs le) 2 o) Ned a _ “CACY =Ste Fe]AbatBoat)+Gant[Baaecgst| Soe EO OCrvmomnatinghs sgnfetantABQ») TERED =AL=segrnatne Per. Teta =PG)=By,SalmanByapt LyCenonmmed s Terr Ret >pep=2x, Fa%eF Res=she e)_remswen oonduhue 2Top\=. >»d-Br Toy =4 a (aayeckDeaLO.beomateehan), (d)lowedhonderworkerUwenown TED=Ye OG[Rate Setaay] ara 3.Youn+gprwolts CoyOKRA RATSBeRHD=ATLETiesTIABE uaa. “= Mc+oe ++ NU+k. K=Way kK Ww fo) ToOz)°C wae Tr=prope N-powk NeeQvar2EJedotttheQuerinLSD,=NooneNereequ” © WKawhTecomeeWritten oaskiebm oxpanracone , 0 K=SAG» Yo oderwm. Kdhbe divogmer— suasPadDooNgoeomDa—2SetNeagonoleadSuleqnag Saint SkeDaeeee SNe SfSaray DyerSoyabson (2)Whe,warensnmadvanSifn , Sey =ewQvvra ke=4Eel 4ae \ ~ >=&By[Bot\oohhsSat *fzexhoot _JoSF8am SEe— @) 75 ()Os EDD =EGt)E GOVE) =2pCee , > Bates =EVpoi-¢\fookonF =+Ce -4Oe aavil\ebenD Pop KooD Leese Tug,aasdOak!{ee)=RYEG)+Wel=e eras” “aksSw D2awiea ‘S. ZR)=X23EQ) (BySemovmad vars 2a eon iB@)=SFLEGV-E@| =YakQ Zo) =0sma Wee —SDEGE). 6 Ya=145EG) —B®=_AGwe) [I= 4EG] . 5.TheVerges Spucheos . |Oowigennsdvas;vaAFeV=143)0K Tes YFELVoBK =Keg “oe b)gemonwrad wersior: Ren=Be+Shook =&++Gee /sSivwinate. LsSomWadegkEdoll.° mee’ PGK) =eRook =[rans Feel ¥ Thon. Dy=\~Flys)=|lott +See) ¥ =Regn 2OFGX=srioty Gente. OS.SrordOrd,Gieutatons @EG)=EAQK =Eatsdean)+GhodsgLemrsci)Ett) E(r) > P.ASNEMG)=14NE[Crateenti Res Lacey D@Y = Ge) [\-4 Sac’) Sot VLLat LARabaon) VO2\-2Lens Sos) Ress\aedated fe) pes --sth.ye 3PMyO Qeerde ACH Ws) =ALE BPW?) =-£F [Gersrpan +(Bees)+Wan) | Renormalized Lagrangian and graphical interpretation ofthe renormali: 0 Itisclearthattheprogramoutlines above,basically thatdescribed inBD, isequivalent tosimultaneously rescaling thefield fromanunrenromed field $, toarenormed field ,andatthe same time rescaling the charge. Ie,wetake the bare lagrangian, ° aye CS opt RBM =EAMHFE- okPY+We’44kG andmakethese substitutions, aK ‘=\25% do=BsBsd toget this new "renormalized lagrangian": 3 a a“gt Ka LEW) =EBs(Qaang wd!) +2s)&xetCRB)& 2y here Z,andZ,arepowerseries in). Onecanthenusethe"Feynmen rules" implied bythis lagrangian, a —=-&Ss=(Wed)—x-—=(dx) fo)ne and onefinds that the result ofanycAlculation isfinite, ifyou compute toafinite order in%.Infact, thisis@waytosetthevarious Z'sanddm’.Ususlly, onecomputes things using theBDframework however, though this inprinciple works aswell. ‘The idea ofhaving unrenormed and renormed fields issimply thist 3, 2g ~Y= KITAPI D= LATA =WY, 8.Couterterm philosophy, Onecanobtain theabove "renormed Lagrangian" inanother way. Start with theoriginal Lbutinterms ofg,then addcounterterms togetnewlagrangian. Inthis view, there isnofield $,.Theoriginal Lhasdivergences, butbyadding counterterms you get rid ofthose pieces. Since the counterterms arise from rescaling, itisobvious that the counterterms are ofthe form ofterms already present. Once we choose aparticular mand, the Z'saredetermined, sothecounterterms arefixed. Changing mand tonewvalues changes the Z'sand changes the counterterms. Thefreedom to addarbitrary counterterms (ofsaid form) reflects the freedom tohave mand) bewhatever values you like: they cannot becomputed inafield theory! Modifications ofthepreceding results whenyouusearbitrary renormpointmu._ 01,Propagator: Zoe=o apdachon|\ Ey=o OGY=GRO” >ofWWHY oRety=x=\~“2Ge) Ee Bea EN|(ACMaeBBak)+(GewtyTBdeRtC(gtof XNBaeanestbedalentpS, Qua, a= HAN Lames CARD]. Bak, army, Wweh)acmvataedawinnddifferently maspaDOs,“K) ¥aa‘the Conseilspanking, a(usdante,X)=StOSToBI+CGE>*) fo)2.Vertex: y_PoiweYsyGrades. u 4 Dan, Bg=ATRASH]. asRae SO -FED\ 3.New subtraction rule for self-energy: an 2 °a 1uyo ct tZep wlHo-Zyl 2“Aty’~ ca'[cq-cigs). le) Dimensionalities inf°indab fo)Lookatthelagrangian. Willbeintegrated overd°xsomustbeB°.Ifmassismass, thendim(g) =E*.Thus, coupling constant isdimensionless! Ie, Lghastheentire 6dimensions intheP.Then,eg,thepropagator inx-spacehasdin(D(x))= 4.But then youfourier transform togetD(p) =dxD(x) =-6+4 =-2which isstill thepole. Howtodof°using arenorm point different fromthemassms 01.Startoffbylooking at"propagator" section ofnotesandseewhichequations change: Det few @ SF) =LH) +Get) BG) 7 E(w) 0TWhoxgdeWkrwfe,soDitbaneapoteRee ° *. " »aedaa DCgepT)=Ba mooouta Da,s°y=(PEG). Um) —_ Veogodspemomme) cee \ ee ee a | ee-oU) GU Tle% fo)oFweBe) iYsnev/. w. ZBP= Maw FU) - oD ye°,”(2=%LEO LEG csmaosattepent| Awe, ZQM=o ok Deepy= —L.= pow 2.Inowwishtoarguethatourrequirements that27(m2)=OandthataM)=0 cause both these pieces oftherenormalized self-energy todepend implicitly onweIfthis istrue, then ofcourse the renormed propagator depends onue. Lets see how this goes: on Sous lyse) BQMY=0 5WHi—FG ase) aWeU(ph NN’), one S> YMGpew y=Beds DesBQalSW), Jhon v fe) §X=GBA =FG, HY)dd. ~we doSoa dsiwNN) Due Joea nadappeane.. BD)DueyereonseyYa=Uo( yw5NY)am : %, ogyAytye 2)Romans Bote BeFOEAyOFHoyts hye 0 =CRN) OZhw,NGEwa08)05) _ aiave ge=FGhx)NY. 2, QueGe)deguade. mpSepale~ah-w samy: ar eneL(geots XY) ° - L . ‘Te,wacinseacerhaunomacceaunbelem —SmutCopx)gowwaleeMuleJe. . a =:euu XYSoyon\sapVeinsomeCone—VonpeVdd Mamd |Sie),YerWELohromge mt| ae av ue‘a FT ayt 3)Sensaity 5SOM=SCSHSN), W)Nowlodecknarhere 3h. ao. v weRR cL Y=BrWPCesSw,RS)>T=PCspyyyMY). .4mtwee Ou Ragone) aeQAPym woGe yttye on)R(t tt)=Oh First, thepoint oftheabove istoshow that really allrenormalized objects aregoing tobefunctions ofybecause oftwosources: theZ's, andconverting ),to.Itwould beatruly stunning coicidence ifsomehow arenormed object (like self-energy, vertex, propagator, kernel, etc) didnotdepend ontherenorm point u. So, this being the case, consider the two boxed equations above: these are ‘theconditions that putthepole atmass m,andthecharge atvalue \.Suppose youb start offwith some particulat choice ofuandA,Youalso chooise mand), .Youthen construct the above functions ofallthose argyments. Ie, the "theory" determines these functions. Now, suppose youvayy ua little, butyou insist that the above conditions stall betrue. Theclaim isthat both your mass andcharge will move alittle. Ofcourse you could make the mass and charge not move bychanging your comterterm functions like dn?ortheZ's.Buttheideaistokeep’themthesamefunctionsthatthey.were. A Thus wegettheimportant point: achange ofrenorm point changes your mass and charge, ifyou hold the form ofyour counterterm functions fixed. This isalso what the differential renormalization group equation will tell us: that achagge intherenormalization point will beaccompanied by@change inthe mass and 0thecharge! 3.About finite renormalizations in theory. G woe ‘ =DsaD Weah Xeacan TheBare MAss Viewpoint. . (@)1,Motivation: Ithinkthisviewpoint ismoreusefulwhenyouaretaliingaboutrenorm group transformations. Idont yet know how, but that ismymotivation. 2.Consider theunrenormalized,bare propagator. Ithasapole atm,-Question: where isthepole oftheunrenormalized but fully dressed propagator? Answer: itieat aposition which you cannot fix but which isfixed bythe "theory". Te: °. MDO sw) = Pore—SCBSdott)At 2. We, PCs NarnM)=AD bee dake es GoanDeargalegather MeQuy|Wien"ot(BEMdemo,L)= dslewuvu Ae. SoMo=Me(doyne, fe) ie]3.Nowgototherenormalized, dressedpropagator. Itissupposedtoberelatedto ‘theunréfformed byaZ3factor andithasthedform D=ates awa Bw =0Poe FO pee Ke ae : Ay < > x =~ at attra 2 ‘ D=BwDys=U) as-wt- FEV-~|pw (Fd : , aioe tee RewereodadienQearshins O=mame—(peas dome,A). OT ose et are : On:EL(parm Yo,omto,K)=owewit 1,PCr,do,NY=O. Thus,ifyouhadchosendefinite valuesofmy,yandL,thiswouldforceyourmass mtogeataparticular value. Usually, however, you think ofmand Aas being given , then this renorm condition defines acurve through thespace m,»do- 3.Nowlets worry about Zs. Here ishowyouwant toexpand theself-energies: _ “rt 1 ot ae CBE wehot)=PLGrerede Cpa) "2.CBwade) 0Te,imaginethatoncewehavecomputédthismassmintermsofmdolasdiscussedabove, wethenprocede toexpand theselfenergy around thatvalue andwetherebyeffect awell-defined definition forthetwiddled nought function. Similarly: wt Lots ZEY= Ow)+Gwe)ZOQY a a. at : Que: Same [|Lewy+owe)TE]= .: Za a et xt. DBP goat LE) +Guy? Cay eres : =BIGOT _ =Gee ll-Z@)| : ; ) DLS =1-ZE ° as .® ° DelLITBewyy=\ ameBO Cuntacautte), st a : >BamVn"2Pawsheoh) “JaDy=2aCormte,N) . =v,Qe, KY, a)UMLMH(MeV,N), So,4f2,isgiven thismmerical velue, thenthepoleresidue willbeunity. ‘aNext, whatabout Z,?Compute’ theunrenormed vertex, thengotosympoint andyou gotitsths\ ° 2Wy=TQkerds hye)/e| ==DeQos,0). lo 5.Nowwhat about therenormalized charge? Both 2'saredetermined,so italso is determined, you cannot set it. Infact wehave, from the xmmuirmment definition of 0renormed charge, that: - ~ Yoo 3 d= Popy= 2Ter)=bs : She q|WaDERaeSACen) | 6.Wecould have introduced aZtodescribe the mass renormalization, had wechosen to: a L ae a 0t=Bwy = (FeYd5osMe,K)=cms(2\-1) yor co soondogikB=Qdots, NY|Regacea amen (hemoyd). a) 7.Solets sumerize: youstart offwith anunrenorm lagrangian with m,andA». You calculate thet thedressed, unrenorm prop hasapoleatsoneqierdplaceM,-By examining acertain equation involving thefunction 2°,youdeduce thatitispossible toconstruct arenormed propagator which hasapolea atm,residue 1,Thevalue ofm ispredetermined, andsoistheZ,thatmakes theresidue 1.Finally, yougoonto compute Zsand).Thefollowing items arecompltetly determined bym,andyt ZysZ,ymy « irksx 8.Lets nowwrite outthemass-set equation andthe charge-set- equationst QO) ameomrere, N) Ma YW BWOsw A). re)GAK=AQore,A) CYYs=%WComea)Presumably wecaninverte equations (1)and (4)togetvalues ofthebare mass and charge which will produce any desired renormed mass and charge. Then wegett Mo= Weyer, NY 24=BOwm,0) Yo=doQyrm, A) Ye=UAsw, de Nowwesimply pick anymassandcharge wewant, mand),then.we cancompute myhor andZ,andZsthatmakethetheory work: ie,mwillbethemass, )willbethe charge; residue will beunity. Again: ifsomeone hands you mand ),youwill have toadjust m,anddotospecific values tomake thetheory work. Renorm with (a) arbitrary renorm point and(b)bare mass viewpoint. 1.Asinthefirst"baremass"sheet,wecomputethepoleM,oftheunrenprop.Noone ‘e)really cares what thiswalue is.Then wedefine therenormed prop bysome Z;andwe want ittohave pole atm,notatrenorm point jp.Thus, wegett \ pe te Bet ae)Femmes SBFRewW\ 0SBpteKom some GO)oeede9[a=rme(dove,A) yg Although wehave chosen. asour“renorm point", wehave notused ityet. Theidea ofmass" requires ustolook atp“=m~ andtherenorm point isnotyetinvobked. 2.Now weinvoke the renonm point torequire that: DEER) =~\ =_ = i==.. . Su we ayFwZH) eel-EG9) Bowe em Inother words, weare requiring that, atthe renorm point, the propagator beequal tosome specific number (which number weknow) (because weknow mbythe above). There fo)arepropablyotherwaystoimposethisrenormcondition. Asjlapproaches m,this condition duplicates theusual "mass shell"renorm condition; forgeneral pnotice that weare not atapole. Itisprobably agood idea tonot impose this condition atapole! WhatZ,willcause theabove tobetrue? Lookatpage(2)oflastsection and duplicate all definitions there regrding self-energies. Then you will find that tosatisfyHeo oSRe1Fear may| =o =~°%ao Estee) =>Bs= \~°Z Gah) neh) =pw= sYowee,NL) Now you see clearly that, besdies being dependent onthe bare mass and charge and cutoff, 2depends forsureontherenorm point because uappears inthedefinition ofZgitttr —_\ an= (pr ye. 3.NowdoZs y= Pte pe,wayh) [ds ; Dy=WsGure wo,A). \ 0Sotosummarize: Z,isdetermined byacondition ontherenormed self-energy (2-poitn fetn) ~ attherenorm point; Zs;isdetermined byacondition ontheunrenormed vertex (3-pnt) atthe renorm point. ace “~hL- 0d=ValdonaD(adoranydsa 5.Guawanaarye + Qaeckme+ oAFeKoyo, N) Wage \ D=_NGadorme, RY boyadh Bs=Dw(jp,do,Wo,A) Pa=Vp,do,me,) 6,Tawardogah+ Ma=Mle(prydom,NY Nosdolpyharm, fe) Ys=LeQa,dsm,d) 25=Cudym,d) 6: we —Gonnection withthenotesofAbarbanel: _ 0.“AbarbEq.(17)reads,whenconverted tomynotation: Oe ze :Ws ghwt2%) . : woe 2oe) a teaBMS\—aetyy=\=>acice| =o e ers Se eR y= tw) at SF(ghhy= ©em)ZG Loe. aeee(phww)epimt)eo. SoWidonigta: =AABPAC GSeeal2, we ype weBEX|=FGR)+ Grew22l St(pt)| aro G+ ew)Se. *ZR! a 2, 2Be Jvedeyand Y= SG GHEE) . . Br, MenWe sey, Zp) =o. Thus, Aberbanels "residue condition" isnot the same asmine. Hehas acertain equation shatxkimxamkfemnexgx thetnyself-energyfunctionmustsatisfy.Buthisequationis (o)notthesameasmine. Thus,myZ,willbedifferent fromhisZ.HisZwilldepend onthesane paramters asmyZ,,butthese Z'swill bedifferent functions oftheir paramters. . Ithink the lesson here issimple: any "residue type condition" will dothe trick, aslong asitisaspecific condition onthe self-energy and thus itdetermines the function Z. . Similarly, look atAbarb equation (20). Inmynotation itreads: x u & .~~ 2(e)=- = % a \ °ZOC)=-w =seya0 |VGH)=ax Obviously the symbol msointroduced byabarbanel has nothing todowith the location ofthe pole ofthe renormalized propagator! Abarb even states this onpage 40h. Good. 2.SohowshouldIcompare my"conditions" tohis"conditions". Ithinkthereason for having "conditions" isthis: the conditions determined the Z's, and the Z's then determined the counterterms. Sohaving conditions just teels you what counterterms to add tothe theory toget afinite theory. Our conditions are different, soour counterterms aredifferent,butbothourtheorieswillbefinite.Mymhasmasspolesignificance, his fe)mhas nosignicicance except asnoted above, . en? (Abarb con'd) fe) So,eithersetofconditions (hisormine)leadt6specificfunctions, which are the Z's. One interesting difference isthis: mycondition leads to aneution m=m(m)sh) andsimply doesnotinvolve therenorm point p..This isreally.an accident because Z,should have appeared inthemequation butitdid not. Thus, the derivative hecalle BTA onpage 407 always vanishes for me: if Ichange nyrenorm point pwhile holding m,andhefixed, mymdoes notmove!!! 6 8 voor Howto"emanate" from thebare theory intwodifferent ways: .1.Startoff‘iththebaretheoyrwithm,andNo.Choosearenormpointcallitjy. Compute upobjects mandd,,eachwillbeefunction ofpyy-meyroy A+(Except, ifmds really themass, itwont beafunctin ofpy,but lets keep general anyway and pretend itisafunctin ofph.) The "transformation" which takés usffom the bare theory tothe 1-theory isaBigEnchilada andischaracterized bysome Z's. Ie, wedomultiplicative renorm bythe Z's that are releavnt. 2.Alternatively, youcould choose same m,andAobutnowrenorm point is2. Gofor pnother bigenchilada andarrive atthe2-theory. Claim: thenthefinite 1-theory and the finite 2-theory will beconnected byafinite multiplicative renorm ofobjects!!! 2,Maye . (yy4 ' $ oe Ne ay ‘e)“/ Ds‘y Aono —784,)Xb.QO82a. :By fe) ————__ : ory b= %=2yD\ SS3 HYT oe Ore aay>Zabywa Psy» oyQs 3 33 ‘ 3» Ww @ oo= oy 3 "_3sNMjle 2s~\Rel-TRe| ee_Theclaimisthat‘'#)z,,forexample,isfinite,has ReSeBS\y |nologsofcutoffinit.Letscheckthisinperturbation th, =m. omg”. theory tomake sure:teWey Speywm) Oyo ay ‘|=VEER[Bah+eCmt)] MWB u fo)MyaLeNL[RaatCCndy] Suresaoshe | S40). = Syste oe) a RBy&(VEfBenrcods (FerWwansetn’sl)ceFENELect)<oGnbs\ Oat)=pote -2- Saythisinworlis:thefiniteR-grouptransformation isinducedsimplybychanging fe) therenormalization point, holding Lsanddoandfixed. Thistransformation is simply amultiplicative rescaling ofobjects bysome finite Z's, just asinQED. 3.Construction ofapossible “invariant charge" object: hich=0y ebay 2 ay duimQ=27 X-Lo@d|)=(Law 2weo ~ wewae2 UReee] Chaeee) Qa,deedwnYoDaromudeatksquat)Vepoe beyeytok aay? OFAQ) =(Abe) hina=6Sz otot~%Thcheey QhonwaWontenstuched onunvenoak obeysch 6 SM By XC weetAGEteh) =AGB) Qn:acemut‘)byes~ye2\z}yrs = \ = oeen aame Gea GeLees! Uses d= 4 AGf= ; Teak/SeWIph= vee Doan| . iN afad 2 AlsBoke) =wfROMCISINNC?)ARC) oun: o/s aos pot ann’ F Kets We)=tsWY.(Gi)-Loy A: theryy f / oom -3- Nodadfins.Sia.ered,dawenst aaa.Songs: , auy & ve \ QOeX pe =Re\oE-| x4((aw Q-Y EN Qen 2 a mtgh a gtalsyae)=$aars Te, this thing isafunction ofthree variables, now Iamchanging the variables from one set to another set. Now notice that: u_ uN MSA, BVH ACNE BS Now sustale: a 2 eb: ee ARC Gee? |b 3. ~ a RngweOTNem/AR(gesodt 9AR) =ofS Cool /Heset | . Qa walund Wok a ghowt a etoMa RY=AGE) Ss a : ae, 4FEV= et x | Alea,ME)= {alee 4Se t v Thus, Ihave been able tocome upwith adefinition ofadimensionless charge eand aninvariant function dwhichobeysthesamerealesasIfoundbefore intheQEDlike case. toe —Yy—- Corse) ) 2.Propertiesoftheinvariantchargefunction. ©ayauscatespreset, (MARE TH "onecanrepeatallthesbohechoosingpyinsteadofgi’.Onewillthenendupwithachargee,instead ofe,.”Thepropagators Dy‘andDyarethenconnected by sonefinite reriormalization whichwewillcallZandwhichipunrelated ‘tothe2,ugedinthelast,section.Thusi ee ;MR eeahaa 77whthee&& &~ew?\B-ze] ow) Ba, AH =D. Thur ae L AS) =AR) Oe EY Sposh wareA(15)on: : at hatsgape>=AFERE) re)wok . aout nkv (6) qea = er=ae, BF) ubwrveearealJoa(AYomd CvDrak Me) =. v (a) Me,WIR)= QaWwideGunes Cpe)ondGye)&guesby Man 2yp.2G) it , «ey 2 at we‘ 2) Rogue OF(6)boy(Ibb)begahOiasyplen . 2 tant eu te ok we)fo}A(R) =aCaso, feFE)<Fett th ASABE) =eh Randy CodieomAQ2)vo )) -5- en _. Le oe ee leita 20. 0 b)nomass scale present wet . Heré,thinking ofpureYMT,wesetmp0anddropthethirdargument ofa(x,y,z)andwerewrite’ alltheprevious equations asfollows: © a gt 2 gt8,FE)=ACB) () cs 2yteealee,3) oma(492) (2) 2 ag \=e onk (\49.2) (2) . qu ayyB=&o= ay) = er- (23) a e aceEs) ' 2 gt ae yh auiCA(d5k) =4(aca), ge) Va @x) Xt 0.aleyvy =ef / These equations apply either topure YMT ortoQED inanappropriate high energy Limit according toWeinberg's Theorem which allows you toneglect the electron mass asascale. Onewould like tosolve the functional equation ford(x,y) with the boundary condition shown .This will bedone below after other stuff. oe oe Le)St=yinC5)teXt ©et Crs . aCe8aE)=A(er,xae,wh) (as) Noodo2/3Xowbowsite GunathE1degah Ly ont 2 dn98) =AEa(S, aeswh) es) Replece CTimlotrbtclonesquodiae waing(16)bogat(det) eer wd 2Swab Say zy wkWFR) =AACR RI,Bee) =ACHE) 2 ute) gt 2ykwe) ak wwe 2 >WAG ee)=weaLCaeeB), eet) =AEE) an These functions HyandHyareexamples offunctions which appear todepend on fo) threearguments. Duetotheinvariance ofd,bothHyandHyareinfactindependent.ofthesecond argument, asonesees bylooking onthe extreme right. Ifwenow letm0sonomassscale intheory, these equations becoue: ” Weve)=d(alee,#7)=act, r|(a) teA WGP =Ma,CACO) =d208\) FS (365 Inthiscase, thefunctions Happear todepend onthevariable r,buttheyreally don'tdepend onthevariable r.Obviously thesearetheIdadofStatements thatwillallow ustoshowwhytheBETAfunction isafunction onlyofoneargument. -7- a)RGEforfunctioa di {gs[AQF] =ObyStyJ(tins Oa aie atesyy (3\) Qrusfes.,o=pm{aevasmye) (3) Oye ey >0.|a3e+(Ma, Ja+Aes2heeds) Ca ~ ope Pees! Sey Vee faa)mal PAE =i =% Inequation (33)thepartial wrtp,issupposed topickouttheexplicit or dependence which sppears inthesecond andthird arguments ofd(x,y,z). The implicitdependence throughthefirstargumentistakenbyd/de,with,ofcourse, fo)thesecond twoarguments heldfixedasimplied bythepartial derivnotation.Below wewilldistuss further theobject "(b", andthepossibility ofd/dm,, b)-RGE’forfunction f. ahsx. ory~ + Senet. a Aat RottAGLBB) -SHAT) Rinadria wi?(33)do Sond: y .San,Quinhe(38)doSoak y / ‘ 2m2 .2gtBEY [saz+PSq+73+2fag ° = (ea) P=MsSep [o) =+,(wel =4Fe wim7Ee Te EEson SlumsaskB=We)incmmontnns Wawetye 2 Oo lasteddeSonata” Jakke couadaGaActheyhe gh Oe OCCOeOO TOO OU ——Men\esk_ck Oreo2-Day: ee.— _ . at teComa) pd a Tia Aaa kgk Se $C BypagNY RkKai fadeoeaepe OeceA6 Eas 1pataomBL9BOD _. = Cong,Ayfayfa) ars ees pt edge Some RT RG ee ee Meet re eeaSGyl OER eeceee en = ra men ~b- SBE SCE) EY —2Q-+ a _—-=|7E TE cena SeaySeg REREN —Qaundander CeEen en eee ————(No deanteSebodsopaatoin) Dueangeeech soandqrsanth| oe ge eg 6 Se = a = = 3 fo —Se Se be pgatt eg eee ee=(para = ~ 2ESBS Reg ~=fllee) =damSe ____Thus,theBBTAdefined inthiswayhasthesameproperty astheotherBETA:itis _ oeafunction onlyofthearguements shown, andisdimensionless. ee ~~6.ncnjotamgslecsHoyaldeestietaste @ mens &Qioa oe eontim AjDrennan Rennfhspace fo) uh hees 4ey ts wh —otpsecrmabigeds4 “De Ss AyZ ee v @) , ws ey 23= [3 lo 2 Oy Ge* 2=esdel onasadSoaee\ OUSt en HymVddWo,do,1qrred+LockeokDMuery, yorseg® @),Za=2s(primero, N) Ng=We(PrMe,do,) Yam PeCprwe,do,9). Boh,Awebak yeVedpramsdrSocadgoroF - @ y> aYa272s(preom,yp) om,=MeCMe;amtApe) 2 MeRms, AY) : Bijouts865pop, GDM©Arraggmt Drckrm,PoypyaanWahJurad “eal can Nee Lib ok qdfhedrearann’, Jens TrPoSun[DUms0)]=fe) Dbed \a 0 -42@b,DChimdye)=\emaway}DOtimbn) t Quesaeat: ny0[algTabemnl) “ewig|Deimay)=o 2ySe =nn%, =pe OhmXepaBed= SNR). Ss tS MUR AAALD Fmd) =O : en amwae: b= (2stI))| =&ha(HespwdAme 0 %=Be(22bane)|.A. L2po[Zelers rpg) QueRCEfogunnol APTgree, ©lacey =GCtay 2 GMa) WG.0 )er ayTne pytek, » . Qawou, a°D=Z,d. he.G=le)=BG *4=Bd Ss. aE(w) ~ *Boh p=Boo poo[Pv] means =>one) 5Pm y= 27Py GySetree Gs daeAAreal,on: fren) a-% als)\(omdr)=“a, iy(mde) ah . . OtyrangeBoPde: pa(=)=BQNgeWSs=-2@)% Rentd whGrenhe: (x Foam pdaswede thyTe;may=o wtObSs)6(fone 8C=(B)G>Qevenndu=1. TH=(B)P fem dee=om *D=(Qe)? Sfsxdr_=42, 5aD f 54RD:By,+pasarmS p39Y(mdyJ=o . fe) , \ \ \ ‘o) : 2h nt foBerean =(aly: pies) ne v=BY =B=WBE). =2,\ := a ©Ww,, Dy(podeteK=(=922(FEYdomo) Baky= SDEGY+ba” wp aytsn)=SME) baGet) ECM S, B= -Fpy= XVEW DB nMe KEW) ee DL aLanai ~\ ° 1 “+‘ Ye~PYfe=AESTEBAS AGO], a . Ber ; >Wevroe=[ibs(Baud(pt)|Tes(ria+$¥p:)) meFURENS(Blane ORE)[LEM CRRA HI)| 0.=\aeLagBeffacieSte|Of). , uu Que AW 2Bar) 2(g0,8t)\e , 1ME=8EBD+pai(Reaewe+ =®e=2%(perros)oft . fete, oy 3% > ay|+ -3g- 2pby>) =\\SB-F+&M5e,IpsSRN 3 3 Smee SS NTOL Hoever, IdontthinktheCandftermsarereallypresent. Thesearedimensionless functions andtherefore should have been difiensioned tothe renorm point. No.Inamassive theory (m,/0),, thefunctions CandF*canexist and these terms will thus contribute toBETA. Inamassless theory, bydimensional argument ©andfcannot depend onwy,80those derivatives mst vanish, inwhich case the entire answer ib as shown. ) 3 oe) &--> (28+F). eM =O. ee VU Qn.theconnectin. between ourinvariant, function d,andtheso-called “running coupler, ..~---we_know thefollowing equation stobetrues 0 eee BobsesARIAS) eA age @ANSE @®—S= aleAig). oe pyr af aSy Sn Oe Abed seach wonade 9_\-§+ (OSAGS350 mR reared—@ BQ= 2-42 AGS an @_ A@a nwLaat ee .ot =Aes}. a at ~s— OS RG) SiVat)towagalecegy a ae i Ca Oe Ba,yecastSanenalosaul Miesoehay eS aea 3GQeajeg fa TR --MeaaaapgctaBye agg a eerXccarpaplekic. rssicon! SeSpaad,posta —_OST ent teSF “Piper ebcpda \.@ ~hoe2), ~| (ASIA) geet Ae .. __ava =-’@) ~=Ae =\hJieok\ s soe * Sen. Theconnection between global anddifferential renormalization group equations, why(>=@'(é)only,andrelated facts. 1,Renormalization oftheGauge Propagator, definition offunctions F,f,d. .+28 2 p aA@Q~=T(E Rd)WrapFeqnqy|Adakome Gace: a = {7 52) > owe=id k|=(3we)>,ancuanesgayel=0. . NS/ ~Qo =aDEF, Na8) [agpraan. po?tom,oud60. © Heremisthephysical electron mass,¢,isthebareelectron charge, A=acutoff. Pet, TtALimplemented ingauge invariant way,there areno"don't care” terms intheoe saugeself-energy. Oneshouldimagine theobjects“{fandDasbeingcomputed tosomedefinite order ine,2,saytoorder’ n,4e,toorder (¢2)". Thelowest order term inD4szeroth order, thelowest-order term inWVisfirst-order [forthis reason, BDusee217" =WT].Thefunction TTtoanyfinite orderisInk divergent. From the eelf-energy Dyson equation weknow that: a BARE fe)PETES =|— \° a) \a(ed, of,Syne) a7 BeepikaahinkMy Letusnowchooseanarbitrarypointq”=BToneshouldimaginethistobeanegaitve realnumber soyoustay offthenormal thresholds eteofvarious functions] and aimply define aZ,inthis way: —\ews u 2,=V4WS Sam ~Qa(h) divaguk ey Nowusing this number 2,wewake amultiplicative rennomalization ofthecharge andgauge propagator [theelectron propagator goeswith2pythevertex withZ)=Z, byWard, but weignore those functions here] asfollows: .SO = a (3) D=zZo v= Bet Maintaining accuracy toorder n(ie, making anerror inthenext order n+1) we canincorporate thismultiplicative renormalization asasubtraction inthe © cvnosinator: BARE= 4 0bee tee |VET (ESN oo)TT(9dph,re) poe -_— Now define asubtracted, scalar self-energy function like sot a _ae a Let le)TEipod)=ee,|WES4Rt—(es,phKo)| (s) Presumably theInA. divergence isremoved inmaking this subtraction, sothe limit {seo canbetaken andthefunction sodefined isindependent ofA. lethen replace thevariable e,withe; asdefined in( 3)onboth sides ofthe equation (4)toget thefollowing propagator which is"renormalized" in thatit.1sindependent ofthecutoff, andwhich is"neraalized” tobeequal tothebarepropagator atthemomentum pointq*=rteos at BAREDsiahphes) =Sraaceaces D () Parry Vee Cttwrymt) swee Tehhyd) =0. a) Wemight nowdefine afunction Fsothat: 0 rr [i+TECeattnt Feet) phwt)= 3C4)BM) AlaaFRR) =4 ) BARE weeay8 at DO] peg)=Fg pty) D Thus,bydoingthismultiplicative renormalization bythelogarithmically divergent2,[a'procedure éneimplements byaddingcounterterms tothelagrangian, which areofthesameformasterasalréady present], weerupwithafinttepropagator andafinitechargee,,afiniteelectron'adss atromthestarts,andafintte remorgialistionpolatjy”.Paranetets e,andALaregone.WeShellreferbelow tothistypeofinfinite renormalizaticn a3TheBigEnchilada, yang,haephysiadmeeWereonetochoosé pt=0inGEDonewouldfindthat@,-e,,the measuredphysical electron charge. Fora=-10Gov?onegetssomecharge ake, e;=finite. Onecould compute e,intersié“of egiven aknowledgé ofthe function d(‘x,y,2) tobédefined below. [bho cancncpale omy°t] e Sincethéfunction Fisdimensionless wereplaceitfimktwith function f+ -3- (ee Soh we virED =FAS, at) @) ty my) _RICE, UY S, Lond Baneers xakg : Q&ig wes FS HE) DY (10) Instead ofdealing withf,itismoreconvenient toreplace itwithafunction 4 inthis way: ezgtWw = ye iMeFoet) =sfFe,QM) ay . 2n (22.8 at dow(SDE tol =aseah) DB (2) Theobvious convenience ofthefungtica disthatitisinvariant underfurthermultiplicative renormalizations ofthechargeandthepropagator ,sinceitis equal totheir product, modulo D(bare). Forthis reason dissometimes called "theinvariantcharge"(BS).Belowwewillseethatdistherunning’coupling fe) constant" ofTit,withits“arguments slightly redefined. “ note 1.Electron mass renormalization istaken care ofbymass counterterm dmwhich ‘oneadjusts inanyordersothatA(n,du)-0; thenZpcanbewritten intermsof3 which {sInAdivergent. —— note2.Forthegaugeparticle thereisnomassrenormalization unless hasapoleorsingularity ‘atq?=0.InQEDthereisnosigularity sincethelowest.singularityisateeanInQCDthisthreshold andall‘othersfallintotheoriginaiidone’doesnot,knowtheFedult,Also,2,isgiveninterns"ofatsomepoint,notw/. blank tab Comments: thiswasjustatesting ground formetostudy theideaofrenormalization. §~..Yes,this’theory is*notusefulbecauseifzsunstabledtietothe‘shapeofthe‘potential, fo)**gnd'soon.Butitisthesimplest “coupling-you canimagine, thereisnospinology to ““get, corifused about, sot feel“it 43theright place tobegin oties study ofrenormalization Index _toNotes: Ca ee rn ee 1. Brelim sttempt to compute: sthorder séifcenerfy. ,., bate oe "~~Thiswaswherg1,SekarTcomputedioioncra ‘lized ‘objects intertis ofk,. TogettheAthorderself-energy Iusedtheasymmetric ‘Dygdnéqiatiion.. Mysictation sutlias“and"S°ete“was getihariother ‘setofnotesstored ifRenorm Binder. The_result ofthis calculation was‘atfirst that. tlierenordied 4thorder delf energy”*seemgato still“havealoginit,butitalmost,cantelled. My,mistake herewasthatIwasforgetting toexpose both divergent ‘subintegratitris in“thé 4th ofder, loop graph. _Ineffect, this ‘modification causes“a2toappear”(red)"which,when¢arried‘through, causés 4finite self-energy infoutth order! Most’other restlts ofthis,section aré 2,The overlapping divergence in®- graphe.. ’ "Here Iexioséd forthefirsttimebothdivergent, subintegration logs.This yieldedfinally anatcurate, expression forthe“integral “ofD°D°f“where£was‘the: +“finite‘part’ofthevertexihthird’oder’“Rattier”thanbeingustguadratic pluslogaéIfirst thought,Ifoundthat,thisexpression reallyhad,(1og)*terms.due.to the overlap loginteracting withtheoverall log. .'. . aon . 3.Newcalculation of4thorder self-energy accounting forthe‘overlap. "_ ThistimeIwrotearenormedversionoftheasymmetric Dygonequation. I [e)inserted the,correct form for,thegraph discussed above andafter, doing the‘varioussubtractions’ etc,Icameupwithawell-defined andfiniteresult,for‘the4thorderself-energy. Thiswassatisfying because up.tothatpdintIhadthéfeeling thatthisobject,reallywasnot,finiteexcentonshell.This,Section. ends‘withasummarypage ‘showing allresults tothis point. -1 . li,Reinterpret, results using’ acharge’ counterterm. in Mponae st|.4,Thisturned, outtobe'io‘big,deal, Iwanted to‘have,dand&\Appear,inthe Lagratigian,buttheir-sumcan’stiltbecalled\s.’Theresultsof,working’inthis way are so'that you see only the symbol Xahd’ not ),. But this isthe same asjust dealing with renormalized objects, andequations asI.started todo.later, - ‘*"5,Simple self-energy algorithm. _ re Le \"Here Icameupwitha,veryshort,formulafortlierenoriied selfenergyusing therenormed ‘asymm8tric Dyson. The result wasZ,times deItach. Atthis point.it was“realmysteryhowthemissedoverlapsingularities ofthe,asymmetric integralwere ‘finetoexactlycancelthe!logspresent; in,Z,1,1wasreally stuck, 0,. . 6.Vertex Intégral. Eavation. . . eo oye‘After anabortive attempt, acame upwith anintegrai equation forthe vertex which hed nooverlaps. Byexciuding two-particle states from the ‘horizontal ,>\chatinel ofKyou remove theoverlap problem, atleast naively. ‘Infact Bjorken and+.“prellanalyse thisinmoredetail, and,doindeed,éonclude that,thereis.nooyerlap+” GntheQED‘arialog, Soin,a redboxIwas able tocorrectly state the,"vertex equation" “An,termsofrenormedobjectsand’forthefirsttimeIyasreallyawarethatasingle fe)subtraction removes allproblems; andIhadaway’tocémpute’ Zs.Itis"this vertex equation that provides iteration used inlater analyses. ne : - sn 4 "7" "Here T"wrote"doy‘the‘basiciterative, idéa..usitig”the”thie.fidin’ingredients, HIsk@letion expakisiohé ,‘VertaxDyson, ‘and’sélf-chergy Dyson.The,Programatthisstagel its Wasscomplate “except Ididnt‘knowjow£0Handle the“overlap inthegeneral sefl-ener]Case. a ss / 8.Howyoudevelop powers oflogs incomputing [. -- whenTfirstconsidered the.oyerlap problem, myapproach, was,toworkwith bareobjects and,tetrytoNexpose™ alltheoverlap siriguldritiés etc.LaterIfound *abetter waytddo'thia, butLtmotivated aninvéstigaticn ofwhat Zxand{l°really _100kLike’iinténtisofftheir’16g’strugtte, Basically this‘is‘Howitforkss,youknow“thatin'termsofX*Zs“Can“have‘opily,atie,ower“ofLogaripingfrom,théoyérall .divergence.”But’thisiPmgtiplied bya"eetseriesin")"andwheiiyoushiftto “Ag "yout tint that.2,Isapolyriowial in(IRA)..This“polyriomial is‘edeytsinterpret *"Anteiof‘simplg Foprings graghs.IYealized fitthe,Paytitan rhileétondoing°"Pend¥matized caléiationg ‘Yaeludethigrule;:Sigpesces‘sHouldbesubtracted. Actually tlerulé“is touse’renorited ‘Vertéces which a¥é finite. Ie,youshould work through the integral equations. Generally Iput solid boxes around avertex that issubtracted. 9.Khas no,divergent subintegrations. ..1... . “ ._Her®T‘showusingthe,'skeléton -exparision exactly whyitisthattherenormed, KemnelKGanhavpno,divergerit subintegrations insideitself, eventhoughK°does. “20."Théevéms 1¢youignore overlép, delf-energy Will notbe,finite. - -Thiswasawarmuptomotivate mysearchTortheself-energy problem's solution. Ijustshowed,that4fyou,Asnorsovex}aphiddendivergences, youdon’t,standa change ofgetting "afinite "gerenergy! ° : Atthis,point,Iwentand“studied‘BYorken.aiid,Drell,Beforedoing,that.Isqnchow O2Lighrerontthat,youRagba,doesmlti-kernel giippsion ofthe,vertex,inorderto \-Sxatithe,thedverlag, problem;then.Iwasgoirig£6.Pushrikernels to’théleft;and fyKertiels ‘totHe right’ orstimething. Inoticed. that, BD.used asimilar expansion and’ finally Iread that famous section 19,11 onwhy‘rétiorm isfinite. ° Themajor deal here was.the useofasymmetri, Dyson equation rather than,theasymmetric one.Inthissympetric Dyson. themumxtapax subdivergences always’°gancel“hetyiéen’ thepyo,terms.Mynotel."oy thie,réading re.filed’iri BDwithother« BarligFnotes; ‘onthattastchapter. ~ we fa 11.Discussion offinite 4th’order \ising BDtechniGue. “Here Ifound outexactly howthedangerous. subintegrations cencle. inthis simple,£48¢;.Whenyouputtogether allgraphs whichappear in’the'symmettic self-energy ~“2faysony,yon,Findthet.youhateprecisely. pulledoythothsybdivergenced sothe.restWas:only’ovetallAivergsnce. Thisled.'toity.precise,formulatién ofhow you“otipute “selr-energies: *colistrit ‘objectBesubtzact’ and,divide. bypole, thensubtract ‘again. Thefirststbtractich killsoverall constant, thesecond kills thelogmittiplying (p*-m2) andtheresult isthen finite. Divergences from subintegratio|always‘santel Afyouinclude all.terms.Theself-etiérgy., problem was,finally solved! 12.6thorder. self-energy, -.~ 1” . - ay«Here Iwrote-déjmBadplaesifiel’ Someofthegraphs’astheyappear-in E(p?). raYTbealiged‘thatyet‘really’want.toclassify” térms”bythéirkeyné]-nuiiber, as“inBD. “Ishowddthatfor'ns0,1’ “there.arenddivefgent, sibintegratiins, je,nooverlapPrObIih.“Forna2Ihaftositdownatidponder. ‘ aantog fe)13.‘TheSubgraph Theorém* ~‘ - . This just says that the divergence ofasubintegration cannot beremoved by anoverall subtraction unlésé @11 éxternal mometita flow bhrough your subgraph. 14. The Finiteness proof. ~"—"HereTdevelopthekennel-number expansion forthevertexasinBD.Then Iconsider" explicitly the2-kernel sector andIexamine théVarious subintegrations to"showthattheyarefinite.Thisalwaysinvolvesatrickyéonbination ofterms. 0 Thus,"you canseethatyouaregetting cancellation ofbadsubintégrativvis betweenVarious ofYour°2-kernel terms. The general idea isthat divergént subintegrations cacel ineach order ofkernel count. BDdoitfor n-kernel piece, Idid itfor 2-kernel piece. Idont really have itdown rigorously ,but Iget the main ideas. 15.Sumary ofthe&renorm program. HereIredowhatIdidin7above. Therearereally onlythreeingreidénts calledA,B,C.Thereisthenageneralalgorithm forcomputing everything in the theory toany order you ant; remormed andbare objects, Z's, dm@,everything. 16.HereIapplied thealgorithm inlowest orderandwasabletoduplicate myearlierresults. Ithink its inthe bag now. 17. Discussion ofrenormed graph expansion and counterterm philosophy. fe)Comment: Inthefollowing discussion, allquantities willbeun-renormalized, withtheexception ofthe4thorderself-energy itselfwhichIwill"Fsinitsrenormed form. There will bethree different constants 2;there isnoWard entity tereduce this number of constants. 1._Definition ofthe_three constants: z\ oe aT=4-ZiGe) hk“DG =Be on 5. asy Gr)=Zee Yoo Ded Inorder for the theory torenormalize correctly, Ianticipate this relation between the various Z's: areal eat fe)2-_Compute_vertex throughthird-order, get25~ ° is] aP@rtatsy =do+YoLRM Fp] =<+ i=dol)4%(FAK EG))] s\ia=Ye=\4%CRah +Hep) 3-_Compute_second-order self-enerey_get_Z3+ is) ict OCC : BhSs AR(pewt)LRoneG 3“AG)=£%[Bardscq] : Seo) . Os %=\-4%[bet+eot] Here, second-order mass counterterm was adjusted tokill the quadradic divergence A -2- 4._Compute_renromalized second-order_self enerey 7 ~ mo QB PeOrVOHve) =ree) +c JCcE) =EFI(8) A=3Cr) s a taaa Dr,a ~ woh = w=rt08) a ZEGI= 4264) This shows that the second-order renormalized self-energy isfinite when expressed in terms oftherenormalized coupling constant, because function C(p*) isfinite. 5«Compute_unrenormalized fourth-order_self energy. Basically there are three terms here: you can feed back the second-order correction into either propagator, and you can feed-back the vertex correction. The propagator insecond-order isgiven byt a)V@= _1 eeEw)PZ] Gave) (\-ERLBaatee@)) OF a on=°D@)+ ErThad+eq) "DQ. fe)Correction ont:xo) on), mea, i“yr oyFMayaEAH TODJEReonsOIE) AGomer) Onn ahs 3 “ ® ;+Age VET IDSbrane tggoly 2 >=Lod saan)[ARsFotos oFKt 4ew ea TARE BAGGY 7 a. als} +NEAR GO BEM AI +Oe These last twointegrals arewhat you getincluding first C(k'2) andsecond f(.,.,.)- Iamassuming that the functions Cand fdon't change the UVbehavior ofthese integrals sothey look roughly the same, inUV, asthe raw: integral. Weshould comment onthe various divergences. The second-order propagator correction which wedre feeding back already includes the effect ofthe second-order masscounterterm. Thatiswhythepropagator correction haspolealsoatpeat. Thenthe overallloopintegration inthefirsttermgenerates newquadradic andlogdivergences. fo)Inparticular, wegetalog” term arising from thefeedback logendtheoverall log. Now weadjust the“jth order mass counterterm tokill all divergent constants. Thus: my . at a 4 if 2 7 ew .BH=eMwarronLovinsceyy &(GBe)ah+GOEDCR') . won 3. «Wehisseshe‘alformul: ore aEGimeficaRenoverwehavetoLeciinletheinteyectionofthesecond-order denomwiththesecond-ordernumeratorwhichgivesextra4thorder: fo) atw 4 27 ,SO)=2»qBRFYInd.DCC)+ACG+pete}+i%ZAsceDE a LEN PBmnxc(wyl > mwAGy= 4%5QB+QF)ManDEE)+2OCH2OQ, #(Bon+eom ace>f . =+»t(BER)tateAc(g)+QC)+LOCC)+abot)onie)t 2 w 2 aBSBm =ghec)+bhGear ccepy+NSxQpincte) Now doyour charge renormalization toget rid ofdivergences: wl abed a 4 sty 4 ‘w= BX=Xe Qe) Yo=WN=N+le) fo) DRAEN se DYepre. ts) ~1 >Bey=EDtecet) sENGaew)tatacys) +NMh\+oO) a@ .=2Xaccety +|xDyPact)+pGarAonece) |4pte oe a0) a uv a =o(3842F) WM dS ‘ >Bo Boerne) oe -\Ti] ese )AK u ) =DyaloSe EB +OQu So,ifvinfact Z,hasadivergent part asshown here, youwill achieve your goal ofobtaining afinitt 4-thorderselfenergy!!! 7._But_what is2,tosecondorderbased_on relation tootherZ's2? -\3% (2) () a) @) {eu OR=BV aHOH (37s. Gary) =-an—eare? =3544X[Behre(neNK+aj-X%[FansPopyle : he . 11 ._— _ L +aby=1-4(38PPPatm%(462)42Hom) fe)Ttagrees, except forthefactor-of-2 discrepancy intheFterm asindicated bythearrows. * 8.Possible sourceofdiscrepancy. Iamnotsureofthissection,buthaveanidea. Ithink ithas todowith the overlapping singularity. (a) the term Finthe third correction term for the 4th order self energy represents thédivergent subintegration onthe right side. Somehow I‘havé neglected todisplay thedivergent subintegration ontheleft sides This issomehow still contained in the fintegration. . Inother words, Iamguessing now that myform for the fintegral iswrong, and in fact looks like this: (ops60 \e*20ayBPC) PGKte9) a re’=FALARS Gm TaN ecegyg aAnis SDAnt Gg) Ie, £has some pinch orsomething initwhich generates this extra term which ineffect doubles theweight ofFinthe self energy. Then ifyou foElow the redink 2'sIhave putin,things work outconsistently andIhave finally been able toproduce afinite Ith order self-energy. y loIneera eae eb @Med cond, ee wp . ee ee _SE 2DRATODO. [easPane — oex——-.—__—-- ee es oebhioutdoateumenact supitanien 80OninTipeatangpg.= eae ee Oe higanastoidegalsallyA Oo8tetAiadcipalokSRT [SRDEA- a ee ee Nieto Qeagai aide=bespade.wee a 8.=ae bosYORE) “yo— ee —ELA ReGoadLOQEee AKoaimal, JdDW?=ANGit)LOegy c=bwyaDTU) |Fons Seay} (|RAR GEDRUSei)F a AGot)BLK) a Ws LANE ARE aN Gem)JABEQIAAFCATEI +wet mceconeoe MLcc@)AFM +XC) a @Que was phoShe VentomdocdadeDapepe ORES oprkts Bereget 92m, Sak}acig')-[Ita+dante]+araloo ——énQnDeveroaedeldo ©Yop TF Sertog ne)B69) =Carateae) 7 Seth daeanagynefidebingpannedSega ——. POGN=. ByDGD coe RG gy oo. a ©co0)Nd2>| ne pow. pe 6 aris ina Tay Se G-2OHase ssWA eat. ons mae . o° ©@Wed.Qala.tiskofOrrcordacdanhitee edeS a SeDERE REG we tONeLEee WOT LL RS Aete) =bxQL) es w/e GES LEESt)ee To RaeVEST) AS Ce RhsGOY 6SS88fe=AyGt)REISraw—Vaajr, 2 OO ee(Bon COVAAN|Bok — —--N)--——qy- - ee aES SeOFSESGeuw)!Esha). — a ef--———= Lt NAY [Rena cigh\t f. &XTanéeceaty Rohe gok 3 — _ ow fo a FREde AREAS SOTACSG. f=. -atOLMAb) RGPESAYEANSee (—SES AY AN a ee . YB = NACA ehAY AKG- -—+. By DaCa(Ad)etByDal gyy aw ef _y- ~—Als,~ Seger te geoeOn 2)NaessDYANAGem[RaeOQ ne oy SatePWatsteedPPS Sae we Re)LANemTBAGOL RA CgoPBELBELeLRAMMAGY} Ss LARA AR) Be ia)TREGtLEcee)2y=Boe oy SakDTAche)=ARNE Gem)[Bahk Ce(pjs Ee= _LG)=St.preoieubindaieat EN RARADAD)4ENCE) |OELAVATEcet8,BSGbnEp=.Rese)Nighy}_ ee TANSSEQTeakey}} oe ee \are ea a Ce SEENON Pearls aee —a.QAR 2A, +2AQOF ooGent) {heres Got)385]]8SahSpE) GF aRED28Oe)+2St of ~3- Owe, gerSeedobgeosee ee oOBae aes=ABSuleCnt)BR)+WAAL)E a nn oo —Hala(sun =OC.(BOK)+2aBAOBF) i ae __ EESSNSCESCra)+21Sihacgle— cee pt 8).ep ee[Bon =SENS aescp)Lenaace] _. _ a Latacetdee) T 0. Nae ~ee eeeTA se) 2) 52 SYeeeee Ze MPLS OED, De gy| anitCiacbatin)aangheenate etc) 2eee sey Snweas\-4Roye)2+a2sgh+a:acces.— =| 2a)wrtfoc nace=2.86).aerg* ceeeee oh ae +Suvmmouy adQawhe a OS. Saag Ges oS RGR” =aay Date Ee PAR GSeee) Oey SO TS =SesERE Ganam Pa) SFOS psGustepGnn ©ak eia-i-8) SON aaa Fae =netafests Ssi ——$~Recindnpih. nome,saiseg.osShrosgecnienbeaanit ee oe (Ooh Be2Gatagassaatatemg gewe Sea pheohege. Bye errs WOT =HB Pe) USS SSS SiGGe) Ea @bk Basbh =OME) a aSSECCeSTSWSea aoT peconngede Qe WSTT=OER) GPE) ) TS Tags=xay 7 ©.Wow)Atarsvin Qeaacegealdapeobo i.weeS —— .26) =eetO LS)ViRODE PEL). @ Brome ateet oo RAL ASsARR SAYDeesameacknnethascmallned ——Aeatentacksanmnjynik,aagaciieage an,eotowtedhins of,dafer So,this.Schargecounterterm method*worksfine,finally! Inthismethod,everything4sexpressedinpowersof},ratherthan),rightfromthe.beginning. Thatwasthe Sa right PatheSAA Rela“ROLL et ERP sOSrcenhdeisir aadvantage,TwaeSriginglty lspleingfor,Still.42,49very.convenient to.maleute ~ef-the.symbol \ydefined as\}+$).+ This calculation simply duplicates:the earlierme no etre calculation.en See a ne ———_-_—--- -—-———- -a“Whatwasmyearliermistakethatkeptwrecking thiscalculation? Ikept Pa AI Jeysclera ntfatsnncmsintoarenes aceed___getting &renormalized Ithordetself-entrgy thatdepended onthecutoff.Mymain -misteke waswiththeoverlap subtlety; Iwasfailing tofullyexposeadivergence, ~~" “always,cancel. AnothermistakeImadewasleavingouttheDyson1/2,butIdon'tthing)thatwasmuchofan.error, couldhave:beenabsorbed intothedksymbol, butnowIam _____thet_was muchofan.error, couldhave!beey-absorbedEntethedksymbol,butnewTam always shoyingtheDyson1/2‘factorexplicitly. oe Suite tandShewnaaon oo QWeRkpeng | TESSEntaeaSDOM TRF MEGS4eB@ Sale DGG) ROMe)_=byA):Gawd) ea we ee -=)ROY). -. me ER) =$ADRONA | Oo.Tn Sige TR i) SAR OsoRRRSDRTIA |ETE SACS La ea RGR) SELBEanyLeak eBBHLABEGG. —a?) sence akan fas)Fat*AGEDSANRI=FEVACGy t__i keys OSETAB AAA St) JTF exeBOR TT ghLacGEbraGGT6AG) ~FOALEDE a EEBaSOYCacegsyyf.ee Ps) —Qrinomtaodaromamendai,ginesaciaaen —-Doves)onsalevtheaponhSawant — OWatkins SeeagindeRemQabeng, TWA omidn —Qaeayn sepDeDaralentdoitaeolyaeaallduvegyenen BakaDakaencecagtaontGayfll] OatAnte,yore aeoi.Cee ©@Sarwea faleandar KoolaHietechveBBP YD a SR ee ee ee a Wo .. ONGat seISG RYDODE) KER A) RG =A alPGER)BODO KEae Qo eeded -@ViodainoctnaSeeincanda Sasage ew ce ce Me TTBeSVQTEAWSOMT MELE FY—Q |Pres dedak kSDW DK)KON,ge) _ |f@edeTha PakAdwnmkage [ TERA KK) DUO RAMAY OL ue t Oa atoMSO Ki) =GROLaEAT SO a onaFePORE ~-®Voouses oneah ee —_N@nrs 1ahsLenS) —ELA) =Fagele ar aie) “Wow that.ve_have slinitatedthehidden singularities whichwerepresentinthat.otherintegral_equationa, thesingularitige tritiellicy cancel.Theonlysingulerities (omisfiniteandthereis.na_oxextap probes. ce as aE-cn ERE OSES Re ee ey —— aeBeal IRS DeAS Pe "tuts isendwportant stepinthechain.Fronthevertexintegralequationwecan -MistiRusomatinedin dfYuendeo oTAigTRDRS eeange)Opnpetanenaalngaywah DaneckteJed KEKaeneeaie a Vk hike AO Mowat, FOS TSPy ee —————dhobataday evesManat an OS eS oS ae = o=3es5EESE=RieGee) TS as La ag CO Raswaiiy= REGS. Gea feet Se RO ~oEDe eG) Os ay a) QiagenRk SSSEe (Renate ER YSESDENFM Oia Oe Soesepa =oPLSBPSapgEY hey STE SSR Tt geSE ey ae yeagfSyPees geSSSOSsHeke PSS SSL “Tad SoSeay elGade oo Comments:asfarasIcantell,theonly.missingingredientinthisproofofthe a)Fenortalizeabitity ‘of0?theory. in.da6,AsthefacbithotTnayg.notyetiptovén,that tidbeltiencigy-utiin-Aterated doesnot.contadn. divergences Recall,that.ydy:até*supposed,tofind,nodivergencesineach,ordershaveexplicttlyshownsdidisauth. order,and’thitdemonstration, inrdlved,the,overlapproblewéritlay. Thotherwords, itremains formetoshowthat thelogs in2smagically cancel against thelogs indeltah,astheydidinthe4thordercase.wf Lets suppose Ican prove this. Then amIdone? Conveniently, itisrelatively easytoseehowthedivergences goawaywhen"youiterate upyourvertex function. ‘There wasnonooverlap inthat equation, that's whyitwaseasier. Look atthe overall procedure. Yes, the only missing piece istoshow that the self- energy isfinite toany order. ce) o) j ae . WAAAY -—~Howyougetpowersoflog'sinthevertexBUCKOS_- we CO Hasse,wemustnecaretul shout thenotationwdedforFencraalizing objects. . © wemustbecareful aboutthenotation udedforrenormalizingobjects. ---__Lpropose this a = = ~©. 7 neae eee PGS BPG) we aig pha he28DeSootherxonds,thefunctional formof.all renormalized...-—--objects_is defined toautomatically beafunctign ofthenewcharges Asfor2, - you-can-write themin either oftwoways: ee I a= BaQuyBat) =By(doSa) —_——~Ie, -when-I_am_thinkcing_of .a_Z.as_heing-expanded’ inthebarecharge, J_will_put_a bar. .~--over_it-because it_will beadifferent functional form. 2, NowTetereviewoutanalysts ofthé-vertex integral equation. Wefoundy—-————— ee nneaneee anna iia ecineneneneeneneows 2s+A FOLD DKA) ~ 26=12QRTODADOKGW Ce.“Youassume‘youknowfiniterenormalized objects intheintegrand asseriesin\,so:theonlysingularity arising fromtheintegral istheoveralllog.Thuswrite: wen2oe a i ee —STFOLIAGAL=1¥=rato) _— N=. 1eeXofCoN). =A stoh)ise oo“Wehavealready notedhowthisrelation letsyouiterate tocompute therenormed_______vertex toonehigherorderin}andnodivergences areintroduced. 3s However, inordertotalkabouttheoverlap problem andotherthings, Iwantto : _....__Testate theaboveequations intermsofthebarecharge. LetsstudyZ,@irst: a ee eo met __Notice thisfact:when2;isexpressed intermsof),theonlypowerof(InA)youever_ _._stt isfirstpowersnonabtertowhatorderyoucompute, ButwhendealingwithZ5(be)-OY_0, 321_get_apolynomtal in(1n))ifyoucomputefoafiniteorder.“Iets doit;_ | 20) ==NCEA) | Z ee Ra EEO =25822See OS nha GS:Gotan MSSERIO Fn havaanaytointerpre’orunderstandwhorethsepalynoniais ia(Ia)are a TkLP0kasOO LOS ED iw a jhny — oe ~.TBO OS Phemae = BMA) QO =OR) TSS REE hw) -h- gmThepointhereisthatAfyouinvertZy,ithastheamegeneralformagaseries, Oo xceptthepolynomisls inln\arenowdifferent polynomials. Theirdegreeisthe_-—_-sane_as_before, Socontinue wththediscusgion startedontopoflastpagera TE PS ROWERS RGN ELBwTX —— 5 oe ooTe FeO 2Vela sneee apoNt ee ——$$$$$_3a—_——-POSS asRGA44)ay,pag,Cad)dovinve. CY_EFUaeWictingowtheSoy?“tatsase,Awubintegrationbas@TogwachThen|” multipliestheoverallloopwiththesubintegration shrunktoapoint.Theoverall ee -O) wk i oe PN. @te SS 6. “ve\\yt "ALT graphsofthistypemakepolynomial inIn— cone Sst bes — --we ee we oe.Oa Qi=acy ___Lan finallystartingtounterstand howthisrenormalization stuffworks.__.Lhevenotyetworkedthroughtheself-energy stuffsinceit1smorecomplicated, but ______ there 4smichthetIcanalreadysayjustfromthe#rtexanalysis, ___in termsofthebarecoupling constant isveryeasy.Youjustwritedomallthe5th ______allways cancelled bythemasscounterterm insersion, soyounevergetthese.You,just._____..._g0aheadandcompute thingsandyougotapolynomial ofdggree2inlog,Thereis_ ____. __...0the‘otherhand,suppose, you,wanttodothesamethingintermsofthe_eee renormalized charge,ie,compute2;tofifthorder,say.Nowtheclaimisyouwill __--draw_sl1_Sth onderfeymmangraphs,nowputting\abeachvertéxinplaceofky_-Ten.you drawLinesisolating youryertex.insersions, overlapping ordisjoint or._ ______and direct subtraction which killsthelogandtakes ftoAt.Inturn,thisobject, _.._.HO)"yhichhasbeenthussubtracted appears ontheright’sideinthegenerationanna. of1),teidestheniasimpler. foreach vertexinsertion youdoonesubtraction —_ _....-.ahthesymmetric point. Thus ateachnested sublevel thelogisremoved. ----—--__Actually, thediscussion aboveonlyhandles thefollowing graphs: 4 Xv_Thave notyetlearned howtodotheself-energy right,nor have| |__I shownthattherenormalized Khasnologs.Bg,whathappens eee __withthtsgraphs: oO 7 ; ~~ a Seed Mig Nal a aaees -6- "xm SowhereamI? a lol Thavenotfinished yetbatTcanseothattherewillemerge@setof __Mrenormalization” ruleswhichmustaccompany theFeynman rules.Oneoftoserulesseems tobe:anytime youseeasimple vertex insertion, subtractthatsubimtegration, ~_vand'do 50atthesymetric point,thiebeingthepointwere{1.1toailorders. 7 When accompanied withtherestoftherenormrules(whichIhaveyettoderive), you"ean computeanyamplitude toanyfiniteorderandgetauniqueandfiniteresult/i? _. Forexample, suppose youwerecomputing theanomoloas magaetic momentof —___ the“electron” [email protected],"dey therenormalized vertex. Youdosoangyougetafiniteresult. Notethatyou____certainly cannotignorethenonentgraph?! verygraphstilicontributes something_tothefiniterestr s—~—“—sSSSCSCSC“C“‘(‘SNNNSNNNSCSCS™S ——-a re en CQ... — co eee _ Howtoshow that Khasnodivergences using theskeleton expansion, —_ ,(Similarly forthetriangle). ‘_ee : "1 Thnavenoticedthat’kdoesinfactcontainsubdivergences. However, thesearealways “__ duetovertex andself-energy insertions andKdoesnothavethemasTwillnowshow. ____Propbably thisargunent couldbephrased intermsofanintegral equation, butIhave eenunable towrite anintegral equation forK.However, Idoknowtheskeleton ____been.unabletowriteanintegralequationforKyHowevertdolmowtheskeleton. expansion: RE one aei ceeeS aae ==sBQemed\ — = EGPQOMDOS) =FLSETOL DAP "New ouJustTookatthe“shape” ofanyterminthe skeletonexpansion andfactoroff ae ce wee we ;(, a — "Here yougeethattherenormalized YernelcalledKisaseriesinrenormalized [and “D, Since these objects havenodivergences, theonlydivergences inKarethose thet _____might arisefromoverall loopintegrations, ButpowercountshowsthatKhasnoeuchdivergences. Thus,usingtheskeleton expansion we.caninprinciple compute Ktoanyiain eea a order ingivenexpressions for1endDe "2, ‘Thisallowsustoimproveourformerargunent concerning thevertex. Theonlymissing linkishowyouiteratetoshowthatself-energyisdivergencefree.Wehave‘already ______ shownhowtoiterate thevertexJ'andaloohowtogdtKandbotharpfinitetoany _. 3.Thiesameskeleton argument canbeusedforthevertexinplaceottheintegral --Sgquatdon argument esfollows ee ae ooOc eeEP Ses ECT)=BATHE 29) wee - ~ Sen —4TilabadrnaltoastaBae) EODap~ SPENT DoeETS we eda. | De PQSARFO)FOAPSE ed .Sos: —-(Nocioni yomohtoaccoficuslag,fie,SistasAnangual cous fncuslop JiaTratue Aamo Oath 0CopyingSensackiaacuee vetboek weEGS SetaDeeeTSakDODGY HOM) Wa Jafvaderpsnsumsang:Miamtgdormagaaseas inQaQikiaslaopoh SassBRRLATA+Gat)ShesSoap —_ee ee _ dee2AOaefenagnnssiwee aOO GSS RESAESESoa pO ea PN |My ©RASDaenseMangonuredQegh ne ga tnt bg - oeoo TS a (Rie thingSaviowg etclacineTE OD Oeen eres. psec ____—_ ve ee ee ”ThusIhave atleast shown‘theTollowingr ifyouignoretheoverlapsubdivergepees whiten occur attheTetevertex,thentherenormalized Selrenergymustnotbefinite,whatIwould “Bowketoshow19Wit17Youincludetheeffectofthese’subdivergences, ‘therenormed— ~pelfcenergy UsInfaefinite en - Disoussion-ef finiteness ofself-energy tofourth-order using—BD bechnig 1o.-— CY aosietemaethecomteitatng eos 7 OT ___1,First,here2 3: —, .—_—— hp. fi ee _______The circled vertices inthesegraphsareofcourserenormalized vertices, sonot___Just Feynman graphs. Ie,they havebeen internally subtracted soonly finite part a remains, The kernel inthethird graph issosimple that subtraction makes no _ _..difference. Thesegraphsallcomeintheone-kernelclassification catagory. __2.Letslookforpossible divergent subintegrations whichcannotbetamedbyoverall —.._ -Subtraction because thesearetheonlythings that,cangetus.Thecombination integral —-. .Alpobviously canbeoverall subtraction-tamed, soweneedonlyconsider 1,and1p - ..Separately. Butsymmetry thensaysweneedonlyexamine the1,integration. __ . -—..3+ Thefirstoftheabovethreetermsobviously hasnodanger withregard tothe1, a soweconsider theothér twoterms. Basic: whathappens isthis: dis —@)___at tneroteendcancelsbetweenthesetuotermsleavingyouOK.Write: —-Sorts Spnyoe-ee Lege Sepp s.Zn}Yipnkn=\omwee Ob) ey ____. __Hereyoucanseeexactlywhathappens. Thesubtraction terminsideterm2,called —... .here(2a),isnoproblem because itislogtimes theBornloopandcanthusbe - cleared byanoverall subtraction. Itistheotherpieceofthesecond termwhich -—.. contains theproblem.Thisotherpieceisexactly thebargraphwhichdoescontain —____the_undenired divergent subintegration, Butthisgraphandthusitsdangerous __-—_--..ivergent_subintegration, isexactly cancelled bythethirdterm) @)4ctaritication: tnetern22vitorisZstimesBornloop_is. not.just_a_constant_that. ~—__—_can be.removed byoverall subtraction. .The_point_is_that anoverall subtraction would —-~~——Remave thelog-arising-in-the-2y~integration-and-thatis allyonneed_to.show.— le) 1.Theinstruction istoaddthreegraphsasdiscussed onthelastsheet: teAhamepole. le>- +e -oO be’ v e[talods 0}vdasbod+ -o{-§= 2%ko|+o- Because ofthewaythat|isdefined, theterms inthisequation areactually second- order inthecoupling constant. Aswillbeshown below, thesecond-order partof25 isnegative. The25xBornLoop sitting there will exactly cancel thetwodivergent subintegrations onthetwosides andyouarethenleftwithjusttheoverall divergence which then isremoved byyour overall subtraction! Infact: va nes @ ~ Be=\-2\akssk 9Be=-\e \=-4[Rae Yer] 4 eacDe =2[Dr LO]aAe Got(BRK AGED) | BDmekrod . beh.0Que,Wan = atasrhe+S\CO-)+28rene) a WEVA) =~FFFew)[BELLCES| FANECnty(BySal4G,() Well, Isee how the divergent subintegrations cancel out inthe result for the unsubtracted self-energy. Butwhat isthis object inmynotation? Ishould goback and construct the self-energy equation inmyown notation: Pa ha | 2 e .CEE)=BukPHWeHeHOF [9,0ey,romenmadages R=ar hu Fook —\ ~ rey ye 23Lae +WNgacFoot-Vkdte, FoowoF| Leen enet4Cgve)[shel we = wa\eoNe WE) =>SNP) =DsTse] = CYR a oe = 3) =ANGYs Os, asin abt[eater 1GaFooeont | Bo. =Tema. _24Sowhenyousubtractthisobjectyoudon'texpecttogetsomethingfinite.Its fe)onlyafter youfactor offthep’-m? factor andthensubtract thatresult should befinite. InBDthisfactor istheq°a" tensor thing, andthere isnoconstant toworry about because gauge invariance blocks mass renormalization. Sohere is away Ican write mything: Be [EMA =SQ=Kxety AQ, Mes ABGD =Cr-w)HG) = SSC =WE) Go) So Now ia Oe ane: Ney=[A —Wod| Bey=[WG-Weed) ae UG) = SGN Eee) poet a| i ee 2o |ubeSQ)={AeFok-uaForest] ed =4-002 -4-<eSe- 2=Bepov® =Bex|e-| 3.Thisisnowverynice. Inowunderstand thattheexpression E(p”) =Eis devoid ofthose divergent subintegrations. Thus, you see bywriting itinthe other waywith 2;that thedivergences arising when thelines ontheright of the asymmetric loop are slid tothe left, these divergences must beexactly cancelled bytheleading 2;factor. Thus, thequantity Econtains only "overall" divergences. These overall divergences are ofthe quandratic and log variety. When youdoyour first subtraction tomakeE(p*)-E(m?), thequadratic divergences go away because they only appear inconstants. Then you divide through bythe pole, and finally you resubtract. Inthis second subtraction the log divergences goaway! The result is finite!!!!! alculation oftheSixth Order SelfEnergy in¢theory withdbisBDmethod. fe) 1.ThefirsttaskistocomputewhatIhavecalledtheexpression B(p*).Sor JO)+ see 4 wininpudoaday Guncansck baN= \\aeFov—\ataal,Fok’vat| MWateaSick|emlutouvors (Wathtok| 7Speak gage 52\ROr” tg> »yee»»x qc oR ee ae ° ca ya\ye - - 4So” <a> )LPweerys@ aft >— gyQeserrssa) -&D>— — a)yo?Kv?v® ~<ab— \-Ax>— 0 (sixthorder,page2) ao,YostoatlemeottohQedSn coma! 2.Once you accept that Ehas nodivergent subintegratitns, you see why the calculation givesafiniteresult:namely,bydirectpowercounting bachintegralgivesaterm fo) oftheusual form: const L7+(p®-m")[ Blogl +C(p®)]~ since De2. Theconstant goes away inthe first subtraction, then the inside log goes away inthe second subtraction, and you get afinite result. Remember that each insertion (sith box around itinfigure) contains nologs becuase itisarenormalized object. Ifa vertex insertion, then itisdirectly subtracted before itgoes into the graph. If its alower order self-energy insertion, the thing inserted isthe finite result of the lower order caitcultion. 3.The fourth order vertex belongs either tothe one ortwo kernel expansion terms. Ithink itis correct tocompletely cancel that overall factor of$and change Kto K"because atwist can always beundone togive aterm already present. So, lets now construct the second and fourth order kemels firstt 1@Ko. (Busey Aeedd wdade Ae 5Oe Ee Pana ag--Chel ~ap+8+etFae=ik x ae—Zay ~ig 1S) 7 bn.Bos > >“dt ees co = -Fay=a me oy 4.Nowlets examine theoverlepping divergence business andtrytoseewhythere are nodivergent subintegrations floating around here. First, the basic instruction is toclassify all6thorder contributions according tothenumber ofkernels involved. Tothis order, that number can be0,1 or2and nomore. First lets deal with the single kernel contributions: -~>— +—§*>- =lypedampirlernl qugracofoddo OO <=\toovee =de|-S>| +\pvoernnZ oe. Tyeypor) Gp.+~~ (sixcth order, page 3) This extra term wejust got will then cancel against the single-kernel part ofthe le) secondterm,leavinguswiththisoverellresultinthesingle-kernel sector:2,times adressed Born Loop. But this object obviously has nooverlap problems and becomes finite after the instructions: subtract, divide bypole, subtract again. The overlap proplematic part ofthefirst termcompletely cancelled with thesecond term, just ashappened inthesimpler 4thorder case. Somuch for the single-kernel contributions. The zero kernel ones are at once noproblem because they are already inthe dressed Born loop form, sotospeak. The real test ix to deal with the 2-kernel contributions. fe) lo) Theorem: Thedivergence arising fromasubintegration cannotberemoved byan ; overall subtraction onyour Feynman graph unless thesubintegration ties toall fe) theexternal lines. oS so _theexternal Lines 4 re apegetena “'"thereasonforthisisverysimpleandcanbeillustratedwithafewexamples. Basically theideaisthis: whenyouspeakofanoverall subtraction ontheFeynman aaically,theridea tFoeeee ee aeOnene,Ee graph,youarereferring tosubtracting adivergent constant. Ie,inanoveréll sibtractiion youcanonlyrémove something whichisIndependent oftheexternal momenta. Thecharacteristic ofadivergent subintegration which is'detached from someoftheexternal momerita isthatitscontribution totheoverall graph is dependent onthe momenta which were not touched. For example, consider this graph: s e wa ° dy \ ‘ ty is 4 =(dwemt)x41; 'yp 2 NW, RS > HereIhavepickedoutacertainsubintegration whichweshallpresumeisdivergent. ro)Lets take most optimistic case, ie,that thecums divergence ofthis subintegration isjust aconstant (ashasoften been thecase, maybe isalways thewase). Obviously this divergent constant isgoing toget multiplied bythe "reduced" graph shown in thebracket andclearly thisthingisgoingtodepende onthosemomenta! Thus,such adivergent term could notberemoved byanoverall subtraction. Here isanother example": x \ \ .Ss y=(dum) * 1 ed i>. ySO) Ifyour selected divergent subgraph misses even one external momenum, asinthis example, the function multiplying the constant will beafunction ofthat momentum andthus cannot beremoved byanoverall constant. Onthe other hand, ifyour subgraph touches all the external moemta, the “reduced” graph inthat case isjust adot with all lines touching, which istosay itisaconstant; such adivergente then could beremoved byoverall subtraction. Relevance ofthistheorem . ; GntheproofsoffinitenessofvarousobjectsinBD,_vouoftenaredealingrey with objects having D-0 which can bereduced toD1 byanoverall subtraction. Theproofalyaysrequires youtoshowthattherearenosubintegrations whichdivergewhichfailtomake,contactwithalltheexternal particles. IftherewereaD=0 subintegration ofthis type, youwould beintrouble because itwould bealog thatcouldnotberemoved byanoverall subtraction, andyourprofof thatPI,eg,wasfinite, would fail. | ~ . . . Seefootnote onpage329forconfirmation ofthis theorem, « t*) Pooofthattheself-energy isfinite. 1.FirstIconsidered thefourthorderslef-energy andthenstartedconsidering the fe) sixth order, but now this does not seem the right way togo. The significant quantity isnot the "order" but rather the "kernel number" ofthe self-energy. The idea isto prove finiteness for each kernel-number sector, regradless ofthe order. Ofcourse any order, such asthe 6th order, can bedecomposed into sum of0,1 and 2kernel contributions. Ihave already verified the finiteness ofthe 0and 1st kemel number sectors. Here Iwill show how itworks for 2-kernels. The arguments here could ofcourse bespecialized tothat portion ofthe sixth order slef-energy, but noneed todoso. 2.Lets stick with the notation ofBD, except Iplan tochange afew signs. The notation Iamreferring toisonpage 337 ofBDand refers tothe finite part ofintegrals. The net result isaniteration for the finite functions ofthe vertex by"kernel-number" rather than byspecific order. With mychoice ofsigns, here iswhat you have: SmovK =Lag+NooGQeeie) 0 faa\*ZaBe\-Bau Thefirstlinejustsays:duetopowercounting, Liatwillbetheconstant divergent term inthe integral, quandratic plus log; and Awill beafunction that vanishes atthe symmetric point, due tonorm condition asshown below. Thru this iteraction wegenerate anexpression for25andforthefinite vertex toorder ninkernel count. Ifyou are interested inacertain finite order ofperturbation theory, then you will only have togouptoafinite number ofkernels, sonoproblem. 3.Lets derive the above equations tomake sure weknow what weare doing: 3fl=By+\iook [Soundond varlerDyou. 4 0) ao a =y=Yeu eek =GPA LYSAG =FcR Ly+Ay dO JAor)#0,OwnenceLAG-Abs\*Ags)=y+Alp) fe)Soosama Lyhydered 3Ai(syn) =o.wa ~ a) RoGmy= 1a Welly (page 3,finiteness proof) Now you seeexactly when the n-kernel contribution tothevertex (renormed) is: Oengoreed.yepore POF DT i_ iN = = As. You can now see precisely the operational rule for computing this finite object. Start atthe innernmost nested level, dothe innermost integration and throw away thedivergent part, keepthefinite partcalled A,(throw outL,).Thenmoveup tothenextnestanddothenextintegral. Throw outeybutkeepthefinite part A,,Finally, gototheoutermost level. Throw outL,butkeepthefinite part. Thus, Iknow exactly howtocompute then-kernel finite vertex function.as acertain operation onthe kernel. Ofcourse toknow the answer, Ihave toknow the kernel, and the propagator, since that isinthere also. Ie, Ihave tohave anexpression for the finite kernel remermed and finite propagator renormed. 4.Nowletsreturntoourtaskofexaminingthe“expression E"toseeifthere fo)are any divergent subintegrations. The claim isthat there are not any. dsBy Se EG) =AOD +SPA +A\SBA - ~~~ aeee es ~ASOKOD ~DoDDA, ©~6oe YaarontofDreQkeweeh eve. Sywabsapauan., WudakSomeukAraAyonLaffsumerDearaeok,Yun a» ce exe=DBDAL— BEKDDBA, Raeb=-+ dadKODA, 2 4ep=~L:dD+dDOKMOA\— onysony =-l,\dd =-uo-|. Hereweereexaminingthe1,stbintegration toseeifthereareanydangerous eo)divergences which cannot betamed bythe routine proceess ofsubtractions. Thediscussion ofthisresult inBDappears onpage340below equation (19.79). Ourresult hereisLy (page 4,finiteness proof) times 2Born loop deal which contains the1,integration. Ourquestion is:isthis (o) aworrisome divergent subintegration? Theanswerisno:itstarts.outasDe2,butlowerstdD=Owhenwefactoroutthe(p?-n*) factor, andfinally dropstoD=-1when wesubtract. ThefactthatIhisadivergent constant doesnotmatter because L2 arose fromintegrating overthose other variables lyand1. Soasfarasthe 1,integration isconcerned, itdoes notrepresent adangerous divergent subintegration inE. You can see, however, that inthose two terms which cancelled, ineach ofthose terms aywas abadsubgraph causing adivergence. But they cancelled. This iks what usually happens, there isacancellation between the two major terms ofE. Analysis ofthe4integration issimilar bysymmetry soforget it. io_integration. This is different and somst bed considered: be Ae de EGY =ABA, —ADYKOD —POKDDA, QsokA=Ly+DOKby Bs 0Sy=(-uWA,+vOYDDA,) =AooeDD=veeson,re =-\deUtKeo)—Cutpo)pokn® =40-00 -4ypkoo +LSEkDo—PHKEDKODapte ~ok 2=dd—poxoroend =ULo-|-\<ara>-| Consider thefirstterm.Thedivergent constant 1?comesfromthe1and13int's and thus does not count. The born loop object then has D=-1 and isthus OK. As forthesecond term, theAloop there has6-2-2-2-2 =-2,just asforKitself, andisthus noproblem. Thus, there axgxmm isnoproblem associated with the1+ Alysubintegration. Herewegoonceagain: eHp=Vz +A\POA\— APOEDD —POKPDA, 1 Ade Bde =~LadD+NODA~ADHKDID =—EE-DH CNDD)L, SY re) Adv(A=KDD)ate Ak TMBS RR yt dydts te=-dBLL -ADDL =—--OL- oY - (page 5,finiteness proof) Asdiscussed onthe top ofpage 341, each ofthese terms isnoproblem. The second termisreallyanoverallsubtraction typeterm,youcanignorelysinceitcomesfrom fe) 1y- Thefirst term istrickier since part oftheintegration isintheconstant. Obviously, Icould reformulate these ideas inaclearer way, because ina subintegration search youarereally supposed tohold certain vairalbes fixed. This means youreally should gothrough andundo parts ofintegrations etctoseewhatthe result is. Such anatalysis would then yield the naive couting rules wehave been using here. SoIcanthen consider the1s integration andgetasimilar finite result. Potential problems always cancel between the two main terms! 6 0 a? . A-sumary oftherenormaligetion programof@theory. fo) 1.Theclaimisthatyoucancomputeanyofthebasicrenormalized objectstoany finite order and you will get something that isfinite, ifyou follow through with the simple "mltiplicative" renorm program. Lets review how this works. First, there arecertein important integral equations which describe thetheory. They are: . Paar as) A)velvisgate FL 8gar |—|apn . ot LT .Tot)26}atLtdwebed]ow] A-Ys agers UROLOSKL ERD2 B)tomar dagerwa. te roy twZe =FLAH-He} ata roy ) wes WEY = EGS -EC) Oo w i~ Gedgetgoth Soyol)VS Bee, oe) a ry DgTLEEAG) darnedlastedbaeeaeBL Sot =SEC) GE) Gad7eame oa) wel spdas FPOKT LEH TD) Be,Set Sabine: age=OE +HA +. aed: ieowks KeicBeal aWageickquaome=HSE.-..lL ‘. -2- 2.Herethenistheexactalgorithm forcomputing thistheory toanyorder: Oa sadubi Raa K®. =¥aayw=Gwe 4 KP=0 Wey Jom Abgt FA,v5 mp s . jun Bagh DW, dw __povengiaies, wedkanede a nckBBA. Sun Coo gk K ~~ Ww a Qo AagkPRD yon Brogd DOWDp chnneknwbotsale. Nua GhokKo oO evdl ©Maieabisia ohGeeDagsedSonbested ‘Osteo naOSeRScOTE © aAOC 2 Ae ©Mae ee nS 2 Qe ie Fe RGN ae — 0 Wielete ~~ |Set. -YSAM YO Working paper ont aproblem onthe counterterm interpeetation ofrenormalization. 9 1.1mowthatthepropagator Dcanberendered finite bysimply miltiplying itby the proper constant. Ialso know, according toVEVIOP idea, that this same operation can becaused byasimple rescaling ofthe field. But Ialso know that arescaling ofthe field inanylagrangian (including couplings) canbeaccomplished byadding the right counterterms tothe lagrangian. There fore, one concludes that the renormalizaticy ofthe propagator can be"impletmented” byadding counterterms tothe lagrangina. Morevover, this argument applies toany n-point Greens Function, and Ithink Ican extend ittoany IPI proper function too. 2.Myproblem isthis: toshow inlow order perturbation theory exactly how this works for the propagator. This means Ihave toshow something about the self-energy. Here iswhere Iamconfused: Idont know quite what Iwant toshow. This iswhere I teke upthe problem. 3.Comment: consider therelation D(h)=23"(h)D°(de)+ Suppose youareinterested inthe"second-order" part ofD,ie,thepart which isorder SY.There aretwocontributions tothis quantity: first, the2-vertex graph inD°contributes, butalso thezero-vertex QF contributes venyouincludetheorderXterminZ3.Similarly, therenormed second-order self emergy gets acontribution from the obvious 2-vertex graph, but italso gets acontribution via Zfrom the zero-vertex graph. Here isanother way to say itIthink: the residue ofthe renormalized, bare propagator isnot1butisz54whichisapolynomial inh. Thus,the"barepropagator" dethe zero-vertex "graph" actually con tributes inall orders!!! Ofcourse the residue ofthe dressed, renormed propagator is1. Thus, ifyou are computing the "second-order self energy", you should include contributions from the 2-vertex graph and the O-vertex graph. Ingeneral, n-th order self-energy will show contributions from all lower graphs because ofthe Z's!!!!1 6 Hypothesis: howtodo"renormalized feynman graphs" 61.Weknowthatthevarious GreensFunctions canberendered finitebyrescaling the fields andbyalsoredefining thecoupling constant. Ie,replace #°byBit g then define your greens functions asVEVTOP's inthenew fields: they will befinite whenexpressed intermsof),soyoualsogetridof\,byreplacing itwith2529!2\, Thus: = 3aD=£88) =DO)=AC) Da(hoe WsBw). 2.Suppose wemake exactly these replacements starting from thebare lagrangiant ae seth Re as ha FQRMF-vEGK)+BE+rok4 u ue 3 \u =£Rade’) +GyGisFoe 28 31 Here Ihave simply rewritten thelagrangin, making the above substitutions. Now lets read off the feynman rules ofthis new lagrangian: Ps —-=3=hawae fe)a mens a=BA)=ladey, Ofcourse Iknow that the dressed propagator isfinite, asisthe dressed vertex. lets compute these things inlowest few orders and see ifyou really doget finite results: \ enolond + o—-@9 =a= = ENoefontyGs=O odb- =r =pate -_roy @) §BYenim|—@-—=fey —O- +[-*-) <1) vt 2 7 L=BeGeet) +AD BeSlAmeGesy Gareat}aye a .=EXabetlied)gateAteGee yim}-ead =$84e(—)- Ce) it,(% to)=FO? Weget the right answer! Notice that ibiscrucial topick upthe second-order term o)fromthe"bare"propagator graph,sinceithasresidueZ3”'.Thesingularities ofthisgraph cancel those arising from the loop integration inthe Born Loop graph. why isthis the renormalized self-energy? Because that issupposed tobeeverything except theterm(pram). Infact, Ithink Icanargue thatthiswould beamethod tocalculate Zssecond order: youputinthelogtocancel theotherlog,andyouputintheC(m?) togetunit residue ontherenormed D.Then Z,mst beasIhave ithere. Ifyouwent tothorder, youwould adjust thefourth order part ofZ,sothat itcancelled all logs created bythehigher graphs, andsoresidue isreset to1inthat order too. Infact I'mpretty surethis will work exactly. Lets looknext atthevertex: a) . -- ahsant LesansSe) x x =YLFRA4FGry)\ +?TR Sly) =¥.of(2) ge teain wegetthecorrect answer: infact, wecould usethis tocompute Z,.Therequirementsle)on.2,arethatyoucancelthelogandthatthethird-order termvanishesatthesympoint. So, byplaying back and forth between these two equations you should be able tocompute thefinite vertex orself energy toanyorder youwant, andnowitis adirect graphical expansion! Ihave played with the fourth -order self energy and itseemstowork.Youthenhavetoadjust2,(4)sothat2(m2) =0,de,unitresidue. This time all the fancy double logs cancel, etc. Certainly this isnow the way togo toprove finiteness etc, but its a reasonable way tointerpret things . 3.So, ifthe above iscorrect, here iswhat Ihave shown: (a) ifyou rescale your fields and charge and think interms ofthe new fields, then the theory computed toany order isfinite, all logs cancel. (b) Another way tosay itisthis: start with abare lagrangian but write inintith g.Observe that propagator computed from this lagrangian haslogs. Nowaddcounterte: (kinetic, charge andmass) togetthe seconline lagrangian onprevious page, still g. Bythus changing the lagrangian, the new propagator isfinite! Thus, byadding count CO) termsoftheformofthosealready present, yourendered itfinite, Inthisview,you dont speak ofanyfield $,.Sothis isthefamous "counter-term implementation". liMore oncounterterms. Notice that wehave three counterterms toplay with. Basically,weadjustthemasscountertermtogetpoletobeatchosenmn.Wechoose Ce) the2,kinetic counterterm togetthepoletohaveunitresidue. Wechoose the 2,counterterm togetthecharge tobethechosen value\.Thefreedom toplay with the three independent counterterms isthe freedom toset mass, residuem a and charge towhatever you please. Ofcourseyoushouldrealize that,eg,theAthorder2;willdependonthe ssecond-order Zsand2,andmaybe masscounterterm also. Thestmucture interveaves all lower constnats into the higher order ones, asyou would expect. 6 6 o*ind=4 ~@— v=2 —@— 24, 3dmadinfrbAnalysis ofgitheoryindal. . AAT lo)‘Thistheoryisalmostthesameas@ina6whichIhavealreadydone.Butletsgothrueach step tomake sure nothing ishiding. i.Propagator: Defined insameway.Theonlydifference between thisandgisthe appearance ofthe self energy graph: the functional form isjust the same. Thus: ECE) =o=ome=ren (Sacto “ ost >Was1-72(PeptSwA) aeye oot oeZe=%-LZ Z| 2 cacy eg(at)=Sut+a+o ‘:a cost+(ewe)[Ben eG) =est Kw Notice, bytheway, that whether ornotwestart with thebare mass Moywealways include amass counterterm. Ifyou are going toallow rescaling D(ie, allow counterterms) then © youmightaswellallowamasscounterterm nomatter what.Theleatgraphisaconstant socanbeignored. Theother graph hassameformas#loopind=6,. ZiVertex: 7%isdefined inthesame way, butthere will beadifference inthe . charge relationship: T 2eee ° ~ \Ys=TG) P=&P \ de wilt) = ter r=TeBiMpy=BU eats] Vs ((A=\ 0=a =Vy=2,"MS)=23Xs a = |dr= 2s re \\ +ana V4Samefonoesod. es ee YS[Baas $(e-y) ro)Bytheway,oddn-pointfunctionsdonotexistinthistheory,easytoshow. 3.Kernel _:rule out 1particle horizonal; two ruled out since cant exist; also rule out three! So: ° =6Kttermmaiode) =ES SB 4.semeshallng.Lv 3 = an 6—e= . = KeK =3= He&CReay.= K) 3 : 4.Self-Energy Dyson equations: By-@=—*+2.aoda oe [ne3h FO) =whe WYD +AWYWETel 3.wea asl ~=ewe+Dy72s#ypvott <_—eS ge =EC). Again, thisresult isveryclose to-theory, justanextra Dinthere. Thisfact iscompensated bythedifferent charge condition ofcourse. Once reduced tofunction B(p2),thisequationisexactlysameas(,ooalltheremainingself-energyequations ()there apply here, inparticular, thedouble subtraction idea togetrenormed SE. 5.Vertex Equations: T=d.+YpoeorK » IR ee ~|R=2,+H\Foook, [soma adn Ur Ade LRP? >E@)=oN -any GI= LUBE gargk pea! =SB —STE =reweg. Since theequations aresonearly thesameas-ind=6,Ifeelcertain thatthe rs)iterativeproofoffinitenessworksinthesameway.Thereisreallynotmuchnewhere. . Qowmds aboo wwDelermminne) eG @Sepaging busrmee’ dowwelreddod: Qegrey5eondR payne TOM [ite geety Adcot YeaTESedfoe PY IA ene ~\ o -\ 3) 3atoMia tenwe ro)4YXLeOsYe. guj) senayyreneehe vt Xe y ie} © TABLE VII(continued) a Depth =3200hg/en? XO Theta, énthecenter ofthebin,ranges from46.25° to66.75° in2bdegree binds . Multiplicity =1 . “ Intensity +231-03' ,.247-05 264-03 .266-03 .288-03 .308-03 ,520-03 .332-93 .353-03 Error 106-04 971-05 .875-05 .677-05 .597-05 .521-05 491-05, .766-05 129-04 Multiplicity =2, Intensity +507-05 .421-05 433-05 .480-0$ 461-05 437-05 .422-05 .418-05 .313-05 : Error +895-06 .642-06 .§27-06 .S71-06 .330-06 .390-06 234-06 .467-06 .746-06 Multiplicity =3 . Intensity 216-06 460-06 544-06 .632-06 441-06 514-06 1423-06 +529-0§ 422-66 Error 142-06 156-06 116-06 .126-06 101-06 .881-07 .753-07 .796-07 125-06 Multiplicity =4, Intensity +133-06 163-06 .227-06 .242-06 201-06 .168-06 235-06 °.245-06. .187-06 Error +730-07 .881-07 645-07 .753-07 .633-07 466-07 .417-07 685-07 .107-06 Multiplicity 5 Average overallangles =0.5x1077+.1x10°77 Multiplicity 6 Average over allangles =0.4x1077 +.2x1077 Multiplicity 7Average overallangles =0.3x10°?+.2x1077 o Average angle =57.5° “ © © © Datta danteadage ZO ee 6sua ae Nate BOW) feet ostomy tanSalapades Bayh oe a = Sn a Nea peckBaseOuW).=Ge —.~(riede.CassiGees!FCressesmes) . Ne ae nee oe@Grandes SOW) = __a aCD Pe ©Yue cantalent’INpakgewe MatameganVinsecagt.(poke. _ ns or anaGewyme | oO2 =asa et VeryaaChakDaxNeNenetaeoddvale fey ——Yate (veystjo 7»Ye “olDeparture Point: Suriday March 11, 1979 Ihavebeentryingtounderstand exactlyhowyourenormaliz/e atheory.IhaVE fe) beenworking withtheGf‘thoory because itistheeasiest, oneeld,onecoupling, doderivatvies, etc. Myproblem isasfollows: when Icompute the renormalizg self-energy inthis theory, itstill seems todepend onthe cutoff. The renorm'ed self energy isgiven bythe difference ofthe unrenormed objects, divided bysométhing. Even ifthe log cancels inthe numerator, when you divide you aré stuck with its presence. For example, you are supposed toable compute the propagator off shell to any finite order inthe coupling’ and you are supposed toget afinite, iecutoff independent answer. But, suppose Iwere tocompute tosay"4th order. The denominator is1+second érder, and this second order interacts with the second order inthe numerator toyield afourth order term that islinear inthe log ofcutoff. Thus, you will bestuck with alog cutoff inyour prapagator. Idont atthe moment see any way around this thing. Iwant apropagator that is finite tofourth order evenawayfrommassshell. yu» InQED the Ward identity causes myrenorm condition tobe Z3=%1sothat youdogetjustasubtracted photonselfenergy.ButthenPI(0)=0snyfar.Thisisaspecialcase, Iwould like tounderstand howthis works in94without anyWard identities to add confusion. Inthe+theory ifyoucompute theunrenormed vertex tosome order youwill get, say, Ze where eislambda onthis typewriter. Therenorm condition will then bethet EX: 2. Z*tese 6‘Theother condition isthatnomassrenorm. Thus, twoequations inda”andde.Since there isnoWard identity, Iseenowaytogetarelation between ZandZ,. Ie,2, isrelated toFeynman graphs which compuse the vertex, whereas Zisrelated to‘feynman graphs which compse the self-energy. What Ihavenotmade-use ofin$*theory aretheintegral equations. Idont even know what they are! oaTiphoole faa genes me -- TS =TAS Baac TATE oeotstent yingAes -on =» \ ah as a J Ue pte) YL Ae 4. ~ oe lone ee ee—. © oo. se Qed Ae a 1 aeeT ee chad ermteBaseAifet)bokdaktonneoastecKACelasa Sibanr yokbeSpadeabandebewslemeds. BikasaomeMr a-=SieasedDoctA=dentsoodsnadagptyMereabeoAomanA=acastaub. Mat WeottaAGB)(BDthiegestae rennetDWboopasKamien, aintdhenhdane2 ee ee a@OteekaCOPEL DAYn)JohSeDY AN Rffat)Nan SOARBeDeDet) at 2.Waowash oaddo_ongaunr) fo EAR) = +Gt)ER) PS ate) =SS ae PE TSOES) Co Qn Ri NY a AfBQ)=26)| 85[eGo +fGoan|oe Rd 2 ;ATte= Tet= wag) Ge Eee Me ~~©CnodesSah_fLeerSremenBGI)SOVegsata):Teywk epgand, Ldn yreoan Qa! saat Taypet at) 2(Se TRO) SORE)Qo se eee) eed A) OEE) (7) of ey oe i : =LKR) S204) 300Le - BoD duncan ansdepeaed a ee eeON oeadDEaeJetissoSiosbpmchepeda. “Qian, Cs)Zo.SasoeSollee? Sested =\i SOPERG. SG Qt ye) a \teatas [20{agee(Mpkepks get) a| - Eee sen CEAO™—~—“CSOSCSSNSC“‘(SSCSC‘CY AvaSoh byeeekeee aee cA Se eETTEen—Annee styppa WodedeR PE TEAGRE Qa TSSRevesaASS Tae aayae ~,baliicin,9° Ve wasRegula? Sage =o |masegs Theorem forJt,Theory: Statement oftheorem: theself-energy hasDe2xcThe leadingterm4sa.cqnstant - fe)times AY.Thelinear term inN} vanjshes identically. Thecoefficient ofthe log: divergence isconstant times p*.And this isall the divergences you have to-worry about. lyFirst, we-camweite anintegral forthe‘selt-energy intermsof,thefull,vertex. This“shows the:Basicformoftheself-energy: -itis.adouble’ integral only,andithas,acertain functional form: “a Aad aah yea Beaten. Lent neo Cooseers mrt x f a +eRe ‘@--;(0B 6SE REGhee .:@ . vatredien Wake =P .tae aSy \ =SotaeGaal FCvewyPyRete) +eyeeae Eachpropagator depends onitsownscalar, andthevertexcandepend hate=2pxonthefour "masses" andtwoMandlestam inyarianst asthbsen in . ~ek oneparticular way-above. Ileft outmassmasargument off.Ie, ~GW +thisis’therinostgéneral functional depehdenté you'tah have’forany ***Feynmann graph inthe: self-energy. ad 74 a 2%Now,dim(f)=-6 bectuse D=2.Sowe‘can.usescaling argument torewrite aboveas: -.Odea ae Ge gh Ca ee ER ,0.SAARG pe igs Ae) Are 1 This isstill the exact integral. Now ifweare interested inUVparts, wecan set two arguments tozero and terebey define: ww) TV(jeBhomke= nr aPhePYGEE et) =F(a a55 0) © ®2® © This thing isdimensionless andindependent ofmass m.Wenow expand for afewterms insmall p,ofcourse wereally mean small p/k, since that iswhat issmall. Thus: ° “A esawe — i> —~.Vata bresPY, 1(ee)BY B19) BES Goo) +2CR)fyGis80) a — y = sow ate HEB Griese) +Qanemettan) Ignoring thesimension ofp,Ihave labelled thedimension ofeach term. Youseethat theremainder will yiled afinite integral, soallUV:pieces arénowéxposed. From thisformallresultsareobvious:(1)thejeading,termwillintegrate toconstant fo) times. \~ where const. isindependent of.p*.and m2.[upper endpointuevaluation onlyisimplied here]. The second term integrates tozero ‘because odd.under negation ofallk's.Sinceeachtermist(o)5obvious thatthirdandfourth termswillyiledp2timesconstant times lox/\”. Then lower endpoints: plus ‘remainder will yiald finite result. a JovetAcoe ~“Thins,ourresultLodksLike‘this:.\:.4° —@— =Ww AK&gBAN +oneECRPe)oeA HereAand"BHustbeSimple riumerical conétants. Thescalar, dilensivnaless‘ function fmayincltde ‘anypowef Ofp*in'itsexpahsioriy notjustpowers:beginning withp>.This isbecause we‘havetoinclude lower enpoint evaluations fromthe‘first’ téting. Te,~theUVtermsshownarejusttheUVupperendpoint-evaluations. Clearly, you could insert anminthelogandcorrectthefinitetermaccordingly. Youcould” algodothis: ’ 7 ‘ 4=SateAeBanChee)+atFey»‘ a‘ s. a a wy alos ARs8&1 +BOS) +antLGR)wee Nowyouadjustthemasseounterterm tg,cancel boththe.quadradic andJog'inbracket.Then you see that the remaining log divergence only enters Z..: - : . 1 vIfwetredgard. this:self-energy as-computed to.some finite ordefin‘hs,, thenAandBandhencean?willbelittle:dimensionless powerseriesin,andsowillf. Oneorechange:wewanttoadjustmasscountertermsothatself-energy vanishes (*) atp*sm, ‘Thus wecould include m2f(1) inthe counterterm‘to gét anew! finite -functiol which vanishes on shell: 1 «ey . &, »de ARceoBaehay)+Ghewyb.(RY.swtBY4) aMEC)42 renee ap Ge Erie) Ber. 2 ar Oe oO,a LGR)=ety[SG] +CH] ee un CO)=Joke B= wmircel omst. Onelast change. Write C=C(1) +D(p*/m2). Then: Cae ee TO Eee & @) aAes Go] (7 FH+dee]. oa,“G0 a 5 ne ~) ve powFE) Gee LL~Bans)-C0)+0(8)] Problem: cantheGjpropagator bemadefinite (cutoff independent) bymultiplicative renormalization, even-when itisoffshell? ,~. eran 1.Onpreceding page entitled “theorem” Ihave shown the ,computed toanyorder, the self-energy has this form: ot, et, a oneZA@) =CACO +Fea, n ae a ae etek aaron 5roat ote 4aeAtay sedy -Se ree Bye Mea. ae ed ow af) =LBIRGS) +CEI ae ‘ woe CFE =RC Fo)R/ GE, seCG) Qt - . ‘The function Cdeviates from zero starting insecond-order inthe coupling. 2.Asanaside,theleafgraphisjustaconstant andiscancelled directly byam”. . Ie,ithasnomomentum dependence andthushasnoeffect onan =tkDUD=CateFe)=Lae meFoe)+ah Dyn veFaeree WePrasat rte+93] Here wehave three little numerical constants. Bach begins infirst order; that iswhy the masscounterterm hasfirst-order termsinit.Allotherstuffbeginsinsecondorder. 3.Ifwedefinetherenormaliged propagator intheusualway,wefinds D@= 1s \ Pe LHS— GFaady[iFen} a a gt Oe Grane TeM= FBH-Bery =CH)-cHO AL-Bewty A=[Bout] \“ > Z. a) Ie) Se Be +BH=1PCRGE)<clo} Van: \2e0=ZDca) ThisiswhatIcalltherenormalizedself-energy.Tolowestorder,itisorder”and fe)iscutoff-independent, asisthe resultant propagator tosecond-order. Ie, weget: tA —P= Gap Teri] 3.Sowehavemadesomeprogress. Wehayecompyted inprinciple thepropagatorrenormalized through second order andhave-found it‘is‘ctitoff“indepéndent. Theconfusion nowariseswhenwetrytocomputeittohigherorder: [#) bad” . on tea badd : . :Sey= BZ" oO Ifinfact Dis &cutoff-independent toany order, you see that the second-order ofZ combines with the second-order ofDtoyield aterm offourth-order inthe self- energy. which does-depende onthecutoff, because thatsecond-order correction to Zhas log cutoff init. Thus, itseems that the4-th order self-energy will becutoff dependent! Howare wegoing toget around this? Bythe wey, onshell this function Dvanishes toall orders sothere isnever any problem there. ButIamtalking about offshell, because propagators appear inFeynman graphsintheiroffshellsenseft| . : Resolution of aParadox. —— re fe) Herewastheparadox. Lookbackattheintegral equation fortheselfegergy in$furththeory. There Iargued thatanygraph which contributes tothemainthing mest have the result that: rs 2 aAM= (r*)[Bate ACD)) Q)Xanu Ie, ifyou add upall graphs to some order, this must bethe form ofthe result, and the constant Band the function Cbegin insecond order. Ifthis had been true, then you arrive at this result: Pin a 2AQ =Hw)|cifre) ~0) @)~ \— Boat ~eQ) The claim was then made afollows: since function Ciscutoff independent, and since Bbegins insecond order, you are stuck infourth order with 1ncutoff terms inyour renormalized self energy and thus inyour renormalized propagator. Result was finite insecond order, but then diverges infourth order even after you have renormalized. Sotheproblemwesthet,unldssnumberBhapvenstobezeroinanyorder,you re)will sooner orlater get uncelnceléed logs. : Resolution: the form assumed above ineyaation (1) isonly correct insecond order. When you iterate the second order correction tothe propagator toget the fourth order correction tothe self energy, for example, ifyou were towrite the 4,thorder correction asshown in(1), you would find thet the function Cinfact. depends onthe cutoff, The reason isthis: ifdoes not depend onthe cutoff from the loop integration (superficial thing). That iselwasy the first log. Its just that the second order propagator correction itself conteins log, and when that is fed back through, botl/ terms in(1) will bemultiplied bythis log, sothe first term islog squared. Thus itisetleast conceivable that the fourth order term which islog dependent andwhich you generate in(2)bythe interaction ofthe quadradic term inZwith thequadretic terminC,this4th/order logdependent termcouldbecancelled byalog dependent fourth order teyin inC. Although Ihave not proved that this must happen, Ihave atleast gottedn- rid ofthe paradox which said tht the 4thorder renormalized self energy had todepend onlog cutoff. 1e)1.Considerthestartinglagrangian withbarefieldsJ,andm,and\p.Supposeyou areactually given values ofm,andhy,andsuppose youwantthephysical massend charge tocomecutbeing mand. Onehastogiveupthenotation whatdn*=mm. andthat dh=)-. Theactual values ofthese “counterterms” must besetsothat themass andcharge come outstvalues mand. Onecannot pick avalue mandm, and expect that dmhappens tocome out astheir difference. For example, suppose you start out with lagrangian in‘™. and ho«You could then add the above counterterms toshift the thing tomand). But you have not really added any counterterms yet, you have just rewritten the lagrangian. Now you add arbitrary counterterms and combine them with the other ones and you call these new, undetermined objects dmand dh. Ifyou want, you can now dothe following: specify your mand \,then compute dmanddhtosomeorder, thendefine m=m+dmandsimilarly forthecharge. Then toanygiven order, m,and), arecomputed objects interms ofmand}.Inusual theoreis these will always beinfinite. fe)2.InoYtheoryitturnsoutthat)and),asdefinedbyhy=»+dhalwaysdifferbysomething that issecond order. Either can bethought ofasthe expansion paratmer. 3.Itisperhapsmoreconvenient touse\,thanbecause‘phenyoudon'thavetopepperincounterterm vertex M™everytime youhaveavertex X%’. Ie,these always appear inanadditive way. Thus, for example, instead ofwriting: Av K+Hs Qea) » ON Gi hoWecandefineyasnoted,andthenwehaveonlytwographs: x+do ik.Ontheotherhand,onemayprefertohaveanexpansion % directly interms oftheplyk physical charge ).Itisnice, though, that bygoing from ),3) toh» youreplace mxmtwosynbols with one, 5.You can dothe same thing with mass. But whichever mass you use, you must have the same counterterm. Note that masses don't add quite sosimply and in effect reagragne thetheory alittle, butthat isOK.6.Ijustfeelalittleuncomfortableexpandinginheand“m,sincethesefereallyboth le)computedquantitities. I'dratherexpand@rectlyinthephysicalobjects:Thecostisthat youhave toinclude counterterms inyour feynman diagrams. But youhave to woryy about them anyway because youhaved tountangle eventually! 7.Forexample, suppose yougoahead andusem,andho,thereby avoiding allcounter- term feynman graphs except thebasic oneinthéself-energy. Then when youcompute to some order yougetsomething like this: 2 Oy oat eKAlnt)+BAGS) ou awot 2)WAM OES RL =[LEN bel] 41 DettXsAire) Yougetverysimple expressions formamt dmandd\asfunctions ofm,anddo.Butreally these arefunctional equations fordmanddhasfunctions offhefixed, known objects mand}. Inother words, itisrédiculous toexpand inm,because youdon't even know what m,is!Anditisnotthemass that wesinthebare Lagrangian! This isanimpértant point. What the above equations really asy isthis: U, Y= Bate Cea)FH(ahaBt)4eda)Rab+Su) =au ueaee QoDETR KWH, (weduE) =A[1+BED LMtin} Although thecharges always appear inasimple way,toactually finddu”youhaveto goinandexpand thefunctions f,andfo,etcetc, allofwhich work isnoharder thangoing theother route endexplicitly showing thecounterterms. elle. OoOBbAYondBowLagann: —GoKEG nh)=eh oo.he 2£G,g38 EB)DG++Gh O99 BO OLQUaAdonecoeMenebaalcausNeatomdSRERSbee aabihneyWate= ._xes(4.3.99, AG)=2GaksSantQexEarnoo tan BadeIKAseVeouting ogc 1@DiganQoeisoaion1ltcasegy) sanefowksaBP=BeyceeBO). Oh10080.40.sy apdpeStach nonWeaadadWyHSiSMat a oot Q=ast <fe ao-: :en a TENA GaU2@) hieadsegquavidehwenedit.—EGO©Daimeliasaunigeana)-these ta Sand,PRYspe ee we\ oe _. . 2 4 ~the22go'gantgh)=Dens Paar VR LL. |Naeasate fusdossLo£QOQwt)Sage SoG ©OXeapayin begE @_d=Bd=Bo.=5 1-9 _wn CgawST20D). ea(e _..Shaka passin DS=oy 0Rivalsoreged ate OS er oer ee———SSSO? ess ueayeP= aged)=BNC =EGat) ~ —— <> a @Satysronseget sass5Beg amend, wo enSo condWocheEN=o(Commaahwe) ©__Selook whatthismethodAisdoing,Thefirstthingyoudoisinsist that.thebare mass andbare coupling beequal toyour final physical values. Youmove these things|-~-pight-at~bhe-beginning~by-adding-some-counterterms. “Thenwhen-you-"dress" thetheory, ‘your conditions:vhich determined thecounterterns vslues.ere thattherebeno.GD further mass.renormalization, and rio-renormalization’ ofthe-coupling. ©-oseeMfo-any-given*order “inthe~coupling x~jrand-for-fixed-value“of the-cutoff, you~. Can_computethe.coordinates oftheinterzection ofthe'above curves, 000200 2. a -. Lewatnoh Ayod . ee no}@Own Be2 {pirhze i. Soe Ce De ZsgD =tk DeRD _Tei egies hike zigg) Sheasngand ate cagcleineemtway SQaSeat)=ctw)Be RM RTT ENOat reiyZe,rOVDeseo sega 2 Ew)TQ=GwAG~2)+LAW) RGIS) yalfee MEARS. +EAS or pt 2 ae Sayrahaupenn) SefdWokgonseanenactdeeouAbs oOc=4[a ===CoetiadAcota].fp Deismade Quer peadilSno Dewa eS a a ON oe Quantum Electrodynamics FF yaaa 3 D=4- £EF-Ee ® Renormalization ofQED —Y— Orabs. (o)Iplan todothis first without putting inthe i's, then later goback and do repairs. Comparison toBDnotation madesomewhere below.A\\\acm mstetan, 1.Propagator: Lobk this upingeneral renorm notes. Start with bare propagator inthe general a-gauge, then add upseries. Here isthe result: fon 2 Dyo=‘ee-*qnav une=Lag-Fg “Ow=1 Ww-Pac ons Woty=GoTQ’).Dewa TT Here wesee the familiar result that the longitudinal part ofthe propagator remains ‘unrenormalized", ie, undressed. The reason isthat the self-energy isentirely transverse. Toshow this, one uses the self-energy Dyson equation and the Ward identity: one assumes that the cutoff isimplemented inagauge-invariant manner. See page 302, Eq. 19.22 onthis point. 0 Ishallusethestandardrenormalization pointq°-0ratherthananarbitrarypoint u*,Thus, oneseeswhat2,willbe: -\ weaZa=\+W@=0; wee, S) Lets now write the renormalized propagator asfollows: — Bw= Tw =2ye =2D,\ewa) * 4 Quer: RA: W@0)=0 Trakasain ~ teem): =2ao [ontdase)« 3 According ththese definitions, wemay relate the renormed and unrenormed self-energy scalar functions inthis way: a ra e°. reTWS)= 4-\S WG) =2%)We)—"Te} 28 -2- Our assumption ofthe transversal self-energy yields the above form forthe dressedpropagator. Thissimplydoesnotallowanymassrenormalization unless fe)PIhas apole, which unitarity says itdoes not. Thus, gauge invariance inthe form ofthe Ward Identity blocks mass renorm, soinasense weavoid one problem there. Hence, wenever use any kind ofmass counterterm for the photon. Not needed. Perturbation theory: clearly thescalar function hasD=0. Using agauge invariant cutoff, BD find in vol 1that: . \ Ky + + &Wey =1,[mE -H\teeaae[h Gewal \toe 68 NS awe ON[a>Uddin,Time @ e .mw\-Ske & 3YWa\- SHO. ; A2.Thevertex.Themomentaaredefinedlikeso: 2s 2Pigvertex,Th Tguessthegeneral kinetic formis:mydi.wt ° ': Can ay DEReee) =Agee BGA, +COND Butifyousandwich between i(p) andu(p) andtake limit that q-0allcomponents, you find: 3 ~ ~ 6 We)FeGreyacy=e20>(fre) =e)RAHM) AndthenZ;issomecombination ofAandCabove,evaluated withtheirargumentsonmaceshell p*<n”. Notice thatthisvertex, unlike the&vertex, isafunction ofpq”andp'?andalsoofp,andp',,duetothespinclogy. Thus,inorderto define arenormalization point, youcannot justsetthing onmassshell. Justsetting tomass shell still leaves twopossibly differerd momenta p"andp!™. Sohere, the thingthatcorresponds tothesymmetric pointin#org*istotakep'y=p, so that g,=0 asafour vector, andalso putpandp!onmass shell. When this isdone, you see that the matrix element offthe vertex must be amiltiple ofp,since that istheonly vector there is,andthat canbeputin terms ofgamma-u. Hence, theabove definition ofZ,-/ (which Ihavebeencalling Zs"!inothertheories). 2,isthusascalar andcandepend onlyonmandcutoff. Sofarthen both Z3andZ,have been defined. Thescaling rules@as : ° s- yk 2 Se- R-JER => (°0=2%0 T=27 QA *- var 4°S=BS ort 2 \f=2Z,2,7 (sex -—3- Lakconewits, sooumaodhaatly + vot \ on RC¥2 ° .>a Oyweheody Trane=e% 7. + of ™_ ~ y aFGM =Wp Banh =y : aA 4 A\ aGrowers: ~ ak s xamy =hae w= e-UBAa =Be So,asusual, thedefinition of2andthefield rescalings lead toavertex rule which inturn tells howthecharge rescales. Ward says 24-2 butIwont useityet. Perturbation theory: the lowest graph for the vertex isobvious, but you have to putinaphoton IRcutoff. Same goes forZ1=Z2, andyouhave todeal with thesoft- photon problem etc. Idon't want toget involved inthat now, sonow make no statement. Obviously, the vertex behawes asUVlog ofcutoff since D=-0. 3.Kernel: This isdefined toomit single horizontal photon andelectron pair. Then itwill prove useful ineliminating overlap. Rescaling rule is: 62 \ K=2, K @orm)4Se:f-cnergy, Dyson. Again, nomasscounterterm so: 5 ~~= Tw) =e te)%S'S -\ 7=2%{PWoelyss BYep SY =Two We =Ee) = ECworWOR EpFy“Wwwe °ay 2 are Binhdoe|. Que, “Wey=EDGp)~Se@) \ 2=WH= SLEG-EO) =SOE (). oo —\ =\ OAs:|T=2YEO=B-\ > Y= (-FEO -h- Comments: first, because there isnomass counterterm, the overall subtraction does not (e)appear.Thesecondary subtraction thenyieldsAEfortheself-energy renormed, andthisisfinite apar tfrom possible IR.Ie,theBlogA cancels here. Also, notice that E(o) islogdivergent, certainly doesnotvaanish. Thus, 2Aand (0)#©forsure. 5.Vertex Dyson. Very easy, same asallprevious theories: wel +eC °y,=+5Rest (Vn aN ®SSK ° oy%=Whe SESSK Yonvagh: 2 Ved PSyEsST \Ewy= Yhasth —SREKED 2wemete! fe)&.WardIdentity andElectron Self-energy Avoidance. According the BD, and me, the unrenormalized Ward reads: © = =on os ©. =phpeCat)=OSG)OS) qmee ( Woo o! a 3 ewwodHPCae) =S@M- Se) Jrsuovwsd., ace Tee / \ >SGV= gplaCeifew) tolaxSeow : Here istheidea: using the vertex Dyson, self energy Dyson forphoton, andthis connection ofStothevertex viaWard, youcaniteratively build upyour finite renormalized theory inthesamewayalmoast asIdidfor#.Inductive proof, etc,asshown inBDalso. There isanelectron self-energy Dyson also, but you can ignore itbecause ofthis formula, Jim's Self Energy Renormalization Question 1.Hecomputes lowest order electron self-energy inQED. Hehas quadratic divergent term. How should herenozmalize it? AsIlook atmynotes, Iseems that the renormalize self-energy isgiven by: a]to AQ) =LO) L+Gm) ZO) vn ZA(w’)=O—[ovaasmama. cans] us au yYz Zar =&- aXe) rad SY,e nlouask aden,waeweirdogudrudn se ey °Se ©ge LaRae YLese RF-Yeoh = Theproblem Jimisrunning intoisofcourse this: whenyoucompute &(p“=m") inorder togetDX(p*), youhitanIRsingularity. What should hedo? a)put inasoft photon regulator. b)change his renormalization pointg. 2.How doyou dothis second choice? Itseems tomethat you change one ofthe two renorm conditions fortheelectron self-energy: youkeepthepoleatn°,but youredefined 2,sothatatp’=u”youhave“unitreside" sotospeak. Then . the renormalized self energy isfinite. But what effects result from thus changing the renorm point inthis way!!!??? voF ~S- As usual we start with the basic definitions:a °°S S28 A =_\ s=B8 Uy ‘Pum—2H) .ws¢a owe 52Q)= ZGOL +E-HZR) * =_4 Wy om BH ZBA~ Lopes Gm) Fey | Qanaw wonte bobohonepodckamse “AyGoe)=Zion)=o.7 S(fep)=a Quen ° \22S== 3 By=\-PCw) Gy L2G] S= \ w= ZY] 6.... Ze rarBry Uetoabersalseatbe EE)=WAP) 2085] Quon: . .+@r =Lee. +tA BA =a\Qrson 4be =oyTu%YYS ofaeSantSS——~- KE) =Ew) KE)ceasd- Ovndag wquark: mwywodaueqegh WornB=Jed.Then: 3 Or> re nyVere oe) "tay=kh =|BH=- €aXe) =Yon a =b- ToseasOKaJp, odesmedlege \s O w= Se ae owe a Qe Kel=ree -BB—-o—@ a) K The claim here isthat the overlapping singularities cancel. OnBIAN: Qa- +os anraebe pod = Here you see how itworks in4th order: the third graph has both divergent subintegrations, and these cancel with the first two graphs sothe sum has only overall divergences, which are removed when you make delta~X. Ofcourse all this really doesnt matter inQED since you have the Ward identity telling you that Sisfinite ifthe vertex is: Yo RR ow ontCe TAC) = OScey-Se) we -< jewFar =Say -$a) Jpsscnons5 ~ CPLY+HGR| =prt+2Q~2¢) ~ fo) 3 CRAY NwQe)=2Q)\-2q') = BR m=FOGn) \quar‘Rate’=LEY—Lwy| s3 ~j\ Qomadebrin Blom Jaumgossthle UaDealowner,lottdeQue ‘ohaa ne ug,Seynn 2 }Re RYWY=2G) God MF) =fhtayi(ep)ug)=Reyeke Awe) st) Youn,Oiapasieywpe prow’ssgkaamt.Pikgeobotwihqas0 SoapeSaisie.” Qe a)2 ~, wg’)Beg) =2m. gayKune Ce) The spinology isalittle vague here, but itseems clear that you can compute the renormalized self-energy completely ifyou know the renormalized vertex. That isall Iwant tosay for now. ve -¥- Suge TamatoxhvonDrobo fe) Te AO pe (\woie) “De oy|(Guu) vamo. D —(anemal) ve: , 3 Coon) Se “SC dase) Se Ss (o1e~)) S ° rs) 5 Cpamsins\hoy!)eh Is (samen,obeys) i h (quasiwch Ca Ka ive . °.48) (sons) za) mca) co Camaxce) TR) Ge. EKG) >Weyetiqy ate Bey (rey) EAs)