Gauge Theory Notes I & II

Gauge Theory Notes I,II Notes (1979) Ward Identity Slavnov Group [dA] detM f New Notes: (1979) Working Index toNewNotes Section . index, made 5.27.79 g “. . wee fe)1.Basicfactsheets. : 2.Origins oftheFpdeterminant (FPobject =detM ongauge surface) 3.Comparison oftheFPObject vsthe FPDeterminant. 4.Extraction ofgroup volume (Fadeev-Popov trick) 5.Finite Agauge transformation. 6.SEcond-order. term inA-transformation (but Inever used this). 7.Buler-Lagrange equations forYangMills Theory 8.Rules forSpace-Matrix Manipulation (also color-matrix -stuff). 9.Discussion ofkindsofgaguetransformations: global, local,ultrae-locdl (constrained). 10. Renorst parts ofYMT. 11. Old Color-Matrix notation. 12. 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VdaLAO)_ ,eo )¥ 6 8. / .*hutadativetncoerinantineasessestontnahessamtEI)git Idea: when wewrite A8(x) ="object", wemust understand that "object" isJdear, reallyafunctional ofthegaugefunctiong.Why?YouknowthatA®(x)involves 5derivatives ofthegauge function, evaluated atx.Thus, youneed aknowledge ofthe gauge function g(y) foryintheneighborhood ofx,notjust atx.That iswhyAS(x) isafunctional ofthefunction g.Inprinciple, you-might want toimagine that A®(x) depends on.thevalue g(y)takes atally,notjustynear x.Certainly ifyoutryto write down things indiscrete spacetime, you will have todosomething like this to get the idea ofderivatives! With this inmind, suppose youhave agauge condition £(A®(x))= a.Suppose wewrite thisinanotation whichshowsthe functionalness!* £(A8(x)) =F(e/a(x))whereanything appearing befére theslash indicates afunctional dependence. Nowsuppose youhave twospacetime points x,andx5. Thenyouhavetwogauge conditions. Instead ofshowing functional dependence, Iwill showexplicit dependente: 6 F(8,1853AQ)=&orF(&/Ata)ea F(B;sBp$Ay)=@ orFCE/Nos)=0 This means that ifyouknow thetwonumbers. A(x,) andA(x,), then youcould presumably solvethesetwogaugeequations forthesolution function g(Gz) Another. thing wecan doisdefine two variables: Fy=F(gys893 A(x) Fy=F(a,189A(x) Ofcourse forthesolution function @=(#185) theseguysareeachequaltoa.Butwe can define these functions for any function g.Then wecan doachange ofvariables like ™ aRah=YT:dgvag Wha, o}Y==:|=yipa,9/] . Of gL wos ai HereIemphasize thattheJacobian isafunctional ofg(ie,depends on.thefunction .gatallitspoints), ofA(depends onAatallx,),andalsoofF,depends onallie]valuesofitsx,argument.-Ofcourseifyouwant.youcanevaluatethefunctional S[F,A,g] atthefunction g=%. : What isusually done, I'think, istogoack tothegauge condition f(A8(x))-0 and insist thatthesolution function be@=(1:8) =(e,e). Probably youcan’solve this gaugeequationtoget:H=alt,g).te,thefunction isafunctional ofthefunctions f£andg-Orf, youcould solve theother way(aswedidabove) tosay%=g[f,A]. Whatdoesthissay?Youaregiventhefunction A(ieyyouknowA(x) andA(xg))« You arealso given the form ofthe gauge condition f. Then you can solve for the gaugefunction Zwhich solves 4t.Intheotherdirection, X=A{f,g] saysthatgiven sone functions.fandg,thefieldmustbe4tosatisfy thegaugecondition, Presumably thereisnotaunique1,buttheredsaclassofsolutions sothegaugecondition ig C3really «constraint, Usually, one fixes thegauge function g=eforeach x,andthig implies some fields 4=af,g-e]. Now,since“object” =A8(x)4safunctional ofg,wemeanthatitdepends onallg(y). Thus,itgpends(anourexample) ong(x;)andg(x)whichwehavecalledg,andgp. So,goingtoafinitenumberofspacetime pointsn,andstillonlyonecolor, we haver m. 7 ” ~E(aFGiaw)) =|2aLAw| T(agen) . dgey a a RNC / =bckWGK) se MK =O 4FCACH) /9g.0) Joo. -—3- Asyouletthenumber ofpoints gotoinfinity, theobject M(x,y) becomes afunctional =derivative inthesense ofmyVsymbol .Moreover, youusually endupevaluatingle]detM=J[f,A,e/] atg=wherethegaugeconditionissatisfied, ie,"onthegauge surface". Also, usually youchoose &=e so =0forparamters, . Mykeyerror before wasfailing torealize that £(A&(x)) ,andforthatmatter even A®(x) itself, isafunctional ofthefunction g,notjust afunction ofg atthepoint x.Iusedtosaythings likethis: "consider £(A8(x))=a. Foreachpoint xfindagthatworks, thus £(08D(x)easeIe,Iwasthinking thateachpoint xacted byitself and points were not mixed. Thus, Iwas unable tocome upwith aJabobiant! Infact, you have aseparate gauge condition ateach point x, and each ofthese (infinite number of)gauge conditions involves g(y) atall points y. Thus, each & isafunction ofall the values olf g(y) and you can define atransformation ofvariables from theg(xj) tothe£(x5) using theJacobian. Iamwriting all these words because Inever want tobeconfused bythis again. oOThisisthefourthorfifth/timeIhave"done"thisFPdeal,butthisisthefirsttimeIhave really understood how itreally words inthe spacetime sector. The color part iseasy. Infact, here isthe generalization tocolor: (Moo) =Qa eV A ~ ow oc ~yopid x THiat(Rea =TAs) THf aEday te\PNOT YINYY ay Agednee wes x ==[ave =Us) Une Y= dkHac) asa & fe)Map uxy)=IRA Gy) S4eeH Sohereisthechangeofvariables anditsinverse: . oe) TRO) =TAS) Lash i) “ -\ a(acdy- @)) =THA, SG@3) @) Toshow thevalidity ofthis (2), apply (1)inthevariable gtoboth sides then Aintegrate.over df(A8) onleft, which isdgontheright. Or,apply (1)invarialbe g' andintegrate overg'. Theprimes oneach side of(2)must match, butarearbitrary. In(2),replace thegwithg'inside J”’(forfreesince delta). Thenintegate both sides ondgtogets a ‘TUAgT= \usiS02) Thisistrueforanyg'.."Now, however, assume thatthegauge condition said £(A°(x)) =a. ‘Thenvecansetg’=identity onbothsides. Theleftsideisnolonger afunctional. ofg EYwutigainsfunction dependence ona(notfunctional, aiseconstant independent ofx). Thus we haver a Talal =Stay 8(F08)- 2) 7a teTHEN SuadsGOM-2) | . “Aaah A _— — —t aweff ro |JIKANal=akY\uaGas) fo} | IS\\Fe O)>|May)=(WAMLGN©)YaeCs) -5- folIfwelikewecanwriteMasaproductofderivatives asfollows: MaGis)=StSa0[gaa] (gay. VA) Taps) Inthefirst factor wehave gone ahead and-set g=esoyoudon't seeany A®objects there. But Ican't simplify the second factor without forgetting what Iamdoing! You cannot set gee first then dothe functional derivative! ‘The second factor may becomputed without knowing the gauge function. But first afew comments: our gauge condition is f(A) =a, but that doesnt tell you anything about the derivative appearing above asthe first factor. Itisnot zero! Ifyou ever get confused onthis, put back the previous form with g-e tacked onthe outside. First lets consider another question: does Mand thus JedetM really "know" 6aboutthevalueofa?Lookingattheabove,youwouldthinknotbecause df/dAknows nothing about a.Also, looking atthe integral representation for J, ifyou hold your fields Afixed andvaryaalittle, Idon't seehowJ”!canchange. Ofcourse ifyou consider f(A8(x))=a andyoukeep Afixed andvary a,(also keeping ffixed), then you arevarying thesolution @away from itsoriginal value e.ButIdon't think equation (3)cares what the hitvalue is. For the original a,maybe the integral in(3)“pit” at gee, and for anew "a" ithits atsome other g. The result should beindependent ofa! This may behelpful inderiving identities! So, lets return toour idea ofwriting the second term above ingeneral case. First letmedoafew changes: 1)put inLorntz uindex onAfield 2)change toa,b,c color indices; 3)change from gtoparameter ©. ‘Thus the above reads: “4 @=0Q|Maen=2\u[vEGe). (Why) > * VAcpn@ We) we® -6- ForYang-Mills theory weknow what A®looks likenear theidentity, sowecancompute .lo}thedesiredderivative: . Wl a” . aA A@®=N@- 5Diy(N@)Q@) +onde(8) a 620 at = (ME =-1°Dy3@4) Yee Gs) x Here Ihave used atheorem (seelater) that says: functional derivatives pass right through regular derivatives. (Remember these isaregular derivative inD). Now install this result inour general furmla: A Ww Nees) =-%ZNa0.[VO|Tas),Ses) oe VAR® <b ©)aside: 1couldif1wanted atthispointpartstheoperator Dontothe[],switch its color labels, and thus get anoverall minus. This would free upthe delta and getridoftheintegral. Butnotprofitable because youendupjust reversing the process togetwhatyoulaterwante(paareahgey-) Example: [thecovariede’ gouge| ” eo E(K®) =43.M@) 3 WRK) 29,.'Se2) Sac ~Ry oe PY >VAC &) Here again weuse fact that functional derivative reaches right through aregular derivative. : Ifyounowinsert this into theabowe, what happens? Convert d,tod,onthedelta, getting aminus sign. Butthen parts d,over totheright side getting acompensating sign. Then the’ freed delta sets z=x. Obvious result isthen": Mac SSDs LAO)Se-s)| sta ato =7o at©) cS =<8) ed De Ls twalL] 4 ld pow faRksd. ie}IfMisevaluatedonthegaugesurface(aswehaveclaimed)youcouldactuallymake useofthefactor that d,AY=a inside your expression forM. Recall how this detM gets into the generatin gfunctional. Itgets inthere through explicituseofequation(3)above,sodetMasitappearsthereisevaluatedonthe le) gauge surface. Even ifyou later "replace" the gauge delta with something else, your Jisstill evaluated onthe gauge surface. Example 2:|axialgauge. K (ha) =MAC) =a. ¥(ne)\:mySea)Sue YN) . YaLAS => MaGay =>tyDalLA@s) 8G-4) a. s *=~Mm\3—kay@arK@\Sey)3\se A wg |. =-+LyigGJaeOe)Soe3): fe) Since Ihave been talking pure Yang-Mills theory, these Gmatrices must bethenéjoint representation generators, related totheCy). Later Iwillhave toseewhat happens whén you have matter fields also. Thepoint here isthat inaxial gauge Mandthus detM which appears inthe FPquantization (and isongauge surface) isindependent ofAandacte asanoverall constant. Hence, noghosts! AReview ofDefinition ofFP_Object. (theinvariant, thing)and-Comarisin toEPDeterminant. fe)1.First,assummarized inthepreceding pages,starting withafunction £(A8)weare able todefine achange ofvariables and aJacobian for this change. Roughly, the idea wastosayf,=£(A&(x)) =afunctional ofg,soyoucompute df,product against dg(x;) product andyougetaJacobian: [Tac®oa)] =BLLFYages] S\= .R) . TW) =okWaQy tyX) ® pee 3 p wos MaGayliM)=VAIN) a aa Yeu (4) rs Notice that this Jacobian isafunctional ofASwhich issome point intheA-plene. If wearbitrarily saythat Ais apoint onthegauge surface, then A8ingeneral isapoint not onthe gauge surface, but isoff somewhere onthe orbit passing through A.This det~ Cee eA]isnottheFPdeterminant. Thatwillbedefinedlater.Inthe preceding notes wealso showed that theabove variable change implies asort ofintegral representation for J: “1 4HU=Star 8C8) -408) 2. Here is the definition of the FP determinant: AIA)=Va80-2) =ATR] ee"ATA) =ytsy] =4 < Here the symbol acan beany function ofspacetime. WEuse itabot tospecify, the chosen gauge surface f(A) =a.Itisalsoacolor vector. Sometimes itischosen zero, sometimes itistaken tobeaconstant. The main point isthat itisindependent ofA! From that fact, you see atonce byHaar invariance that the FPDet isaconstant over any given orbit, hence the rightmost equality above. 3.Clearly, based onthe above discussion, you can see that the FPdeterminant equals the Jacobian thing when the jacobian thing isevaluated atanAonthe gauge surface! tae -k aSTBY =oth =HT FA\= 4 oFThus,theFP"object" anywhereonthe orbit shown isequal tothe Jacobiandeterminantonlyattheintersection ta fea There isnosimple relation between theFPobject andtheJacobian evaluted atsome point notonthegauge surface! (reddotonlast page). That isbecause theJacobian dethasthat extra term initsdg’ integral representation which theFPthings does not have. Conclusion: The’FPobjectisconstant overanorbit.ItisequaltotheJacobian determinant only when this Jacobian thing isevaluated attheorbit point which lies onthegauge surface. TheFPobject hasnoconnection totheJacobian evaluated at some other point onthe orbit. io] 6.Ihaveestablished thatthéifearetwoobjetts andthattheyaredifferent. The first object isthe Jacobian determinant, the second isthe FPobject. Toreview, the FPobject isconstant onorbits, the Jac det thing isnot. These two objects agree at the intersection ofthe gauge surface, but not elsewhere. See other papers onthis. Here ismynew worry and concern: 1.Intherather correct-looking Lagrange miltiplier quantization procddure ofFraitkin and Tyutin, itsuddenly appears the that object which appears indide the path integral isinfact the Jacobian object and not the FPobject. Inparticular, see their equation (2.37). Notice that there isnogauge surface delta, rather there isthe usual "gauge fixing term" inthe Lagrangian. The significance isthis: you dointegrate over fields that are not onthe gauge surface here, although their contribution isexponentially damped asyou leave thegauge surface (that iswhat the gauge fixing term effectively does). Therefore, during theportions ofthat integration where youarenotonthe gauge’surface, theJacdetobject whichappears inthereisdefinitely notequalto the FPobject which FTdefine in(2.43). Itwasmyimpression that the PPobeject appeared andnottheJacdetobject, butapparently Iamwrong.| eo)yy Thismakesmegobackandexamine whatIthinkIknowabout"exponentiation of the delta". First Ilooked atmyown "version 3"ofthis. Inow think myarguemtn was invalid for the following reason: here iswhat Idid: . 3s ineWYss|eaudeyLe] Ee1 STA) -Yueela)SayGLY) 2) x bbywom Shepr nr ow=(Sa)Lary cisasta]. Seay Theidea wasthat G(a) beadecaying exponential sotheparth integral [da] G(a) be normalizeable, and then atthe end you knowck offthe dgintegration. The result looks good, butIthink Ihave screwed up: theFPobject asdefined abot really depends on "a, which Ihave gone back andleblled inred: ‘hus, youcan't move that daintegral totheright,sothederivation isjustplainwrong. fo)dd Next,IlookedbackatAbersandLeeandlearned thattheyonlydoexponentiation inthesense thatq-» 0.Sestheir page 87.They never claim anyvalidity except asd=0. OniyTcleinedvaliditybasedonmyomfalsederivation asabove. 4)Inowunderétand thecorrect waytodothisstuff. Hereisthesequence ofevents: 6 A.Dotheextraction ofthegroupintegral, seesinglesheetinthesenotes. This step makes useonly oftheinvariance oftheFP"object" andofthe measure dAunder regular, unconstrained gauge transformations. You are left then with the FPobject and agauge surface delta function inside your path integral. B.Thenext step istonow replace theFPobject bytheFPdeterminant. After this replacement ismade, younext show that theproduct [dA] detu[A] is invariant under constrained gauge transformations. This fact then establishes that thegenerating functional isindependent ofparamter a.Then youcan goahead andeffect "exponentiation ofthegauge delta". This isjust how IeeZinn-Justin-4 doit,with theresult He. 11. Your det then becomes the ghosts and you are all set. C,Recall that Fradkin andTyutin obtain theidentical result bydoing a Lagrange multiplier trick. Just another way todoit. fe) ‘e] Seas.4GougValue eS ° sence~sanefa.)Sugsfon-af2 onSafinadhowothaeshask[eeoe »§yg(useson SLe9)-2] geAsK® Ra ia\-lan\ [suworcntan Cr.a SUPY=STL fem esaSaguagins Veall=TelAl [ameampereA Dove, weop oO =Yan\naoA S[MA)- 44 Nw \peaue ssanoapic, Jack. Conant: QQbose STAYachetauoeiinch wade iSAlees =QUALTRTASE)-a])ok Yeswyarvenusconloubving \ fe) Foto:Quyibanalpown.foA. TFrtenoooat :- -. :,:no ee eahp oeSN=WERYAee anUWaxp(18) imYedeonag Se EGR EVE SRV etekwotSabet yGE.Wedotnebogh TaAye=teeEVEN yas RIV" ,Mase=abe EVAT ByHakN | SD Go , nusNas(°*), =or). oO Goadudemer,JS20.WnSeEhRE)~ Qt oy of a Cee ~ ~\AG [email protected]+ay kGayy a Conctude , A ey wo oyRo=URKL©EG) fon|vYI ~K+@eR\- 2) fe] 7’ iaAws 72OBnwiae we regain Woego|” SeenOnde,TranalorneatinndlGanga.Eats 9©VoAdondnawegougefonelonwebunaQk (KosUSORL+FFwt]a@vy@v| WAM] =TIM, sar 4.Te Inday,apyreAdele ay) t -\[os=RITA] tegene TochadMieRattAmu,gedelowedondedoQeonmGuak . 2 USS =te ©oedormbt collebgk )we =Se810"), -recallSPU JHXEQW] --1@S)+ 45[@8)x3] Pikquanoatdegk Ap=An~ie\GAn] -tecLEGA { re! =mewgoneMIAdoarceadounte2,| — 5.8.14 Buler-Lagrange Equations ofMotion for Yang-Mills field theory. o 1.Theresultappears intheFradkin/Tyutin paper,hereisthederivation. SianLaqongeag»(225\ =otdasaAvanontginal, ICH) CAS) danundwuinr wat acs Seve. Behe=Hape. whe: ab2 Ind,hae.B\SySpv—pees.fe3Fre Coequhe: adw =SalSySp—3a8, Sy_ 2a|Sy m3(daNp)\mt°ir a j 4, ah 2~2% GQ =Ths-2% Nets[bar72 roe O wa Je.7OQ a abd \ i‘ See LBpA-- Fedv| 71 Qn: W 2-1 CORR SewkOd) bgt +ageybed od doea xVg CVA,Re=LitsaCTAMt#2 DallTee=o. So aah fe} DalAlh =© RutssfxSpacstadsng (Marengudater 6 ® be Sey) =Lay % A®Sees) =Whey =Tay anaAdjp&dpmclninuedas RUD Bog=MWe, Joronyopmnda “I| aBkMyodMyleeomy,boreogentne, hans SiniMaltle =Ula =Rabie See S Sey =Why, 6.Gps sume omnepualah undone re} WhaleyTe=WaeeeWagesonwallnoaleaBeeastkaged TTT)=Teh »AM=YayTAL,=JasAly=[Al=eos #8k FE=Laymesyhi) 9ESL=LyWhythy = Sl Wty) . HLMMs)=AloS609),Bane FO=Layogre) =Sahity Seeay\hs)=JaySeng)TRE)KG) =Rake, 9.Roschfarckionmistaansarctan—vuvoaadk:WbyOPE) fo] TAVB] -TSU =Teh -(het) =FART -15] soopus lo}1.Generally speaking, theproductoftwospacematricesinvolvesacomposition of operators. Here isanexample: Im=[at+A\ Ute) =a+8 QML =JaeTee Libey =Sa[pheKolSOPHO GO|rest ®)=\a2So-2)L922+AR](innsrea304) =LahKe](18+.60)8e-3)) * =LRN [R*KO] S@a) Oo-Lae |, Twolines back Ihave putina"composition operator", What this means isthat youtake theoperator ontheright andactonwhat istotheright ofit(here adelta), andthen youactonthat result with theoperator ontheleft .This isJust theusual thing youmean when youcompose operators. Eg,intheabove case theexplicit result is: =PEAK (556)+BC)B6es)) =TESteen)+MAYSH4)+B"RO)3G)+G09}TO) +KORAMES), 2.Sowhybother with aspecial notation? Consider this simpler example: fy=96)5°(pte), =Gd+(8,4) 56-3)hel=Be . Oo=(PHY) +BWI? Goes) Qua, SB.Fes) =S18. 3e-)\ F(PRY) T~5) You could always keep the distincition clear byputting incorrect parentheses, but oOtheproblemisthatthisisusuallynotdone.Forexample, inthecovariant gaugewhen you compute Mhere iswhat you really get: (} a” Macys) =>5° Da@) Bes) . #-4(FB) Fo-s). 3.Inspacematrix notation, the above equality iseasily stated becayse recall that itisoperator composition that goes with matrix multiplication. Thus: A x Meal -4LaeOh) =-4a)TS, 4.Therefore, wehave now aprecise enough notation totell these objects apart: ®—LeSsy =Gio)Bees) fo) \es\#[eelsl: * &Lac8)y =&once it Las) =()fed Infact, here isalittle identity involving these look=alike objects: lace) =fon) +esl Qua, acomdoomonaAuamanhs |(aMel= Telay +Bs] 0) YewAomagghrenbin ALO) [DEMel =TedlBe] +8.84 ‘e) Calon Mabie Vistation 5OTverdauble~ewnuad umdaloa dovadictacalammahix, Exons, IG=ee Cate ©dcombs Dalewil daletshQk undebin : ~ a Sas Wahy .-[WieSex| -Lots @Coward romarcanmitie wilhspreerahe Sams ISllea =tel€l oe)@r “tepe. ~daca,ok"Te" dotelbe soa aspot bat, o"pn ack, Grange TMNTR =ELIA @Fadoneport isan: Caso) =Whese). Oo 5.18.79 ADsicourse onGlobal, Locel,andUitra-local gaugetransformations, andtheir telationtothenotionof2GaugeSurface 6 1,Orbits: let Rectangle bethe space ofall fields. Each field Aisrepresented by a simgle point inthe rectangle. Inthis discourse wearenever going totalk about the value ofafield function A(x) atsome spacetime point,; the point inthe rectangle represents the function asawhole, ie, objects defined with Rectangle asadomain will befunctionals ofA.Weshould still regad each point Aasavector insome color space, however. Bysaying this, weimply there isa"gauge group" ;ie, there isaway totake any field vector Ainthe Rectangle and turn itinto some new field AT via some kind of gauge transformation which isrelated toagauge group hwich weassume has been chosen, Once there isagauge Broup and the possiblility ofgauge transformations; you tan see that the Rectangle can bevisualized asthe uniion ofabunch oforbits, where anorbit is the locus ofall fields you can get tofrom agiven field. Below isanillustration of the Rectangle and afew orbits. The space isdefomred until the orbits are verticdél lines (orwafers, ifyou want more dimensions). Furthermore, wemake agraphical simplification: ifyoustart atsome field A,ontheorbbt labelled by"c" andyou gotoanewfield vector A,'=U(g)A, viasometransformation labeled bygroup element g(oritsparameters), thenthedistanceyoumovealongtheorbitisthesameforall fo)orbits, for given g. Obviously this isnot really grue, you can't say this nomatter how you distort the Rectangles “surface metric” orwhatever. But Iwill doitanyway because ithelps illustrate some points below. Asanaside, itshould bementioned bet, like A,the symbol grepresents avector ofparamters each ofwhich isafunction of spacetime, but weignore the spacetime aspect here: everything below isreally a functional ofarguments, not afunction. Noneed ever toexpose the spacetime arguements. One final detail: Imay want tohave asort of“reference” field for each orbitg Lets take these tobethepoints along the bottom ofthe rectangle. Iwill call thereference point onorbit ¢bythenamea,.Anyfield onorbit ¢canbegotted likesorA,=U(g)a,. ovek b ork « w he Ale3 \ a a -2- a 2.Global Gauge Transformation. Under anykind ofgauge transformation (global, locel, C7)&mbotever), cachfieldpointin2givenorbitgetsmapped intosoneotherfieldpoint inthegame orbit. Thats thedefinition oforbit: nogauge transformation nomatter how fancy can.téke afield point overintosomedifferent orbit. . Now,by"global" gauge transformation wemean simply that thegauge transformation ineach orbit islabelled byaglobal, overall group label g. Te, the group transformation onthefields U(g) isgoing tobethesameineach orbit. Thepreceding drawing shows theeffect ofsuchaglobal gauge transformation ontwopoints called a,andA\'.The effect istheteachpoint is“raised” by,3cmonthegraph. (Presumably thegraph is spacewarlike soifyouraise through thetope youreappear onthebottom) : Suppose for some reason wewere interested inaparticular locus offield pointsthakxintmxxerimixeax .The effect ofaglobal gauge transformation onsuch a locus would besimply totranslated thelocus (the figure) vertically without distortion. Theglobel transformation isgivensimply ty: Ay=U(g)A,where A,igafield point inorbitc,andwhereA,‘isthenewpoint. Thisapplies toeachorbit,andthe point.is that gisthe same for all orbits. 3.Local gauge transformation. This isthe same thing asthe above except that now we oOallowthegaugeparameter gtobedifferent foreachorbit.Thus,wehave.A,'=U(g,)AlsGraphically.this means that ifyou doalocal gauge transformation onafigure, the figure gets vertically shifted and also distorted because the amouné ofvertical shift varies asyou gofrom one orbit tothe next.— Lo IK] Po Ae, 1 Fig2:(a)shows theeffect ofaglobal (b)Hereiseffect oflocal gauge gauge tronsformation onthe green figure; transformation onsame green figure. the result isthe purple figure, nodistortion. Orbits onthe right "upshift" more than onleft, sofigure distorts, But distance from top tobottom along same orbit of eachfigureissame,byourgraphconventi re)on, -3- kk,Ultra-localGaugeTransformation. NowimaginethatsomeonehandsyouafunctionH(A,) fe)ie,afunction ofthepoints onorbit c.Suppose youdoagauge transformation where, foreach orbit, youchoose g=H(A,). Then youaresaying: A =U(gsH(A,)) Ag Ihave called this ultralocal because: even within agiven orbit, the value ofgwhich tells you how much totransform isdifferent for each point inthe orbit. Ie, ifyou want toseewhere point A,goes (ofcourse itstill ends upinthesame orbit csomewhere), youhave tofirst goandcompute g=H(A,). Then youusethat gtogetA',. What does anultra-local transformation dotoafigure? The figure below gives abetter comparison otthe three kinds ofgauge transformations: 3 ~ ] i La | La : : \ || | ‘ |H Hl ! |, ] °rl| IE fo) a)global |b)lobal. e)ubtralocal Inthe global case, each "slice" ofthe figure israised bythe same amount. Inthe. local case, each slice israised byadifferant amound, but the distance between points on each slice ispreserved. ("Slice" refers toorbital portion ofthe figure). Inthe third case, the whol idea of“slice-being-raised" breaks down. Oryou could say” :each slice is raised and streched orshrunk. Vertical distagne isnolonger preserved beaause the effective gforthetwodifferent points onedchsliceisnotthesame,Imaéthe earlier graphical convention just soIcould illustrate these three cases asabove. ~he 5s,Notion of!aGaugeSurface.Considerafunctionalf[A]definedovertherectangle.Imagine CFtnstforeach pointA,£4ssomerealinimber, soyoucanimagine £[A]asasurface hovering above theRectangle ,possibliy intergecting it.‘Rmmxexamptex Actually, f[A] isreally avector-valued function incolor ‘space, soeach point onthis hovering surface isreally avector, butcannot visualize such a,thing soimagine just each f[A] asa.real number. LetssaythattheA-rectange ltesinthex,yplane andthatthefaxisisthe a-axis. Then points Aforwhich f-0areactually points intheRectangle. Forsome _ arbitrary functional f[A], youmayormaynotgetsome "figure ofintersection" with the A-plane. Porcarefully chosen functionals f,thehovering “surface fmight intersect theA-plane rectangle’in acurve which intersects eachorbit exactly once. Onemight wonder, forsomegiven f,isthisalikely thing tohappen? Consider £[A,] where A, issomepdint ontheorbit ¢.Assume thatf[A,] #0. Itdoesseemvaguely likely to methat asyoumove A,around ontheorbit, somewhere youwill getf-0. Infact, it 7 even seems likely that itwill happen more tlian once onthe orbit. But lets leave this for alater discussion. Fornowassume that f[A] =0describes acurve intheRectangle which docs infactintersect each orbit exactly once. Such asurface iscalled agauge surface. Itserves asasort ofbackbone through therectangle, sort ofathruway. Youcanget toanyplaceintheRectanglebydoingagaugetransformationoffthegaugesurface. \e] 6.Gauge surfaces under gauge transformations. Suppose wehave somegauge surface defined bythepoints Awhich satisfy £[A]=0.Butsuppose wenowdoagaugetransformation on thefields soweend upwith new, transformed fields A'. Suppose wenow insist that e[ar]=0.‘IntheA'-plane, thissurface isthesameasitwasbefore, butyoumight,wonder whatthissurface looks likeintheAcplane, First, consider global transformation. Then youareinterested inthesurface A,=U(g)* ‘Xwhere Xs thepointonorbit¢whichMesonthegaugesurface, ie,f[A,] =0.Te,youhavetomapbackwards toseewhatthesurface looks likeinA-apace. Asyou can see, itibjust the surface lowered byaconstant amount, because this is aglobal gauge transformation. Thus, wehaver . vo vk) =2 .es ws 7| : \e) : (ger) ~5- Inotherwords,wearedoing.agaugetransformation onagaugesurface.Whenyou. * oOset’f[A] =-0yougotonesurface, when yousetf[U(g)A] =f[A®]-0 youget_a different * surface. For aglobal transformation, the two surfaces are "parallel". But for either thelocal: or-ultralocal transformations, thenewsurface isacompletely newsurface, ~ bearing notesemblace tothe original surface. . . Exactly how 60you find the new surface, given the original one? Well, ifyou know’ theoriginal surface, then youknow allpoints Awhich lieonthesurface. Then to get’ the new surface for global case, you just examine the transformation: - . ae UE . . Thus, you just plug inand youve'got your new surface. The same goes for aaocal transformation where gbecomes g,+Youknow g,foreach orbit, plug énandyour done. +However, for the ultralocal case its not quite soeaéy. You have: : A=ufetaly? ¥ : Here, foreachAyouhavetosolve anon-linear equation ofsomesorttogetA. : 2.dacobiangoingfromfieldsAtofieldsA'.Recallthatéachfieldis:acolorvector, OP)20there isacolor matrix connecting A'toA,foragiven gauge transformation. (think ofaparticular spacetime point ifyoulike). .Butalso, since"A isagauge field (aYang mille field), there isthe"extra piece” inthetransformation law:Specicially,'* Ma “ye Ley z Q)AeUPON +ipG4)wal 6.@V)U | Ifyou want toconvert some integration from dAtodA', you have todoaJacobian which schematically isJ=det(da'/dA). Ifyouaredealing withaglobal GT,thenA'(x)is just afunction ofA(x) andyouseethat J=ltrivially (ie,Uisaspecial matrxi since ~ geuge group islikeSU(n), eodetU=1; notice thet the"extra piece" does notcontain Aatellandthus does notcontribute tothis Jacobian intheglobal case.) This then leads tothe usual result that [dA'] =[dA] inthe sense ofpath integrals, for global transformations. What about locals? Again, for each orbit you have some g,, $0foralltheA'sonthat orbit, youtransform with g,. Nowtocompute the full J,you want toconsider all Apoints inthe Rectangle. However, you know that aGPcannot take some point A,offtheorbit ofA,. Thus, when youtalk dA'/dA, you knowyouaretalkingonlydA'c/dd,,.Youmightsay:whycan'tyoutalkabout oO varying dA, offtheorbit c?! Well, Ithink itisthis: wearegoing todoachange ofvariables from Ato A’, but weare going byfiat toonly dothis 6othat At isa 6+ function ofA'sonsame orbit. Infact that must bebecause that iswhat theabove gmoxpression says:.Wearetalking aboutdA(A),ie,thedifferential at,somepointinlo}theRectangle, alittlevolume(area)atsomepointA.ThispointAliesonsomeorbit, socallitA,.Thenewvolume ofinterest, dA",liesonthesemeorbit. Somehow itseem stomethat, even though g,does vary with thefields Aifyoumove horizontally in theA-plane (ie,ifyoumoveacross theorbits) g,.doesnot.vary ifyoumovevertically. Bytheway, Ialso imagine g,aschanging smoothly asyouchange toadjacent orbits. Thus,whenyoucomparedAanddA"inthefigurebelow,theginthe,transformation Sixx We) isreally aconstant. Thus theJacobian, which inprinciple might havebeen afunction ofposition inthe A-plane, isagain just unity for alocal GTaswell asfor aglobal one. Ifyou doanintegration using dA,yes thedAmight run over theentire Rectangle ,,andsodothedA". Butforeachparticular dA,dA"liesonthesameorbit. SoyounevertalkaboutaJacobian relating d(volumes) ondifferent orbits. Ie,you deal with figure (a)below, butnever with figure (b). . : < < & tax} 2 aa \o] _maAy. .haa yes,thisisit this4pjustnonsense, QIthink this isallright, butmaybe later I'11 change mymind. Nowfinallyconsidertheultralocal gaugetransformation case.Nowg=(A) andisafunetion ofyqur position onaagiven orbits, Now-all theg'sappearing «in~ formula (1)on-preceding ‘page arereally. g(A). Thus, when youcompute dA'/dA, you have toinclude the fact that gdepends onA.In-fact, since the "extra term" involves derivatives ofgjitwill involve derivatives ofA,and this means that inthe ultralocal case Atisa-functignal ofA,soyou have tocompute J=VA'/YA and.you nowhave tododeterminant inthe(x,y) spatime sense aswellascolor. Itisveryeasytosee how _,inthe ultralocal case, the ‘Jacobian Jmight not be1,and infact might bea functional ofA,yourlocation intheRectangles. - Lay : - Bytheway,this“ultrlocal "gaugetransformation ideaiw,Ithink, what Lee and Zinn Justin and Slavnov are talking ebout when they speak ofa“constrained” or"nonlinear" gaugetransformation./They haveafunctiong=H(A)inmind,so-on o sach-orbit g=H(A.) andyouhave exactly this ultralocal case. So,[dA] isnotinvariant for anultralocal gauge transformation, but. itisinvariant for aglobal orlocal ‘one! oSOK $22.94 Revs PanterdVirtte Expraed! lols(emareack gangs) D=4-& ---!@---- D=2 When pea ooon ve4 ~—o~ p=0 le) . ? _q-- ©Vabix Neledver O&ResROS <qepom-Ke COCOG +qetmrde -CF CReaume QkALE \=RS™onwoh Leaspeek=|. a)coda “aN aA a ”Y TRB=RRICO\ =RRicaGe =iGe-caMA =aGeREN” Que Fu=eS =(ARVAD£4.00VAAad, S ~ = Ye=Se=nN2A)~tgUA ona aSaw, cr2 fog AU =BOPRG =k Due, _ ;fay PelLPL =o ) init man RMPwae dye=keOADY'R) ashe WA ~+afear BR, beL@KT]= KLQMI VAI] =GALWAY bk8 =&ORR’) SQ|bse-sheteLQ] Oyaves Aja@sel= Loewe Le ~- feLee LAG) =BAO HLA L660) >5KCaeLE\LOMAE| iCrateLaci) ERCoa ~ARCave Qa &gh: ofa=+ble ore\-yGe)kyeaye =|har EALRCH [email protected])WC CompoGossWieserDayanekek AZBR___—- C~¢ ceage ©~B® ame=ERa.go +a é Cae=Ge|7% Tv a . wor Ey: ¥=aes Olaniainvaso. [De=(Ba—35GaA)anne (.D.-\y,-ig BeL®8S&Ny. CoS Cammde: |LABV= 3GCA*R) \o) tofTAREf=5(A)S nn =i QransBp=Exh-HS Le} ~Shk=1LOAN-4 3,6 he©=SG Fuddpideob: 1 -\\ ~\ |Ap=TAY sayWwe we UsogGSB)=og8) fe) 0; ; Preliminary Ghost Stuff: how tododetM without ever usi: lost fields. io}1,Hereisourstartingpoint: ae 5 n ma-Cge)lis-4 ¢Balla : at a={&£-9¢ (STEALS =[47 Here wehave dropped afinite constant and the d”operator. These things just make overall, infinite constants which play norole. These constants could beregulated to befinite byputting finite number ofpoints inspacetime orsomething. Orbysetting delta function equal toagaussian. 2.Next, wecompute detM inthis way: teAt hay] dan WRA(\-) -25ayy wre © ee -e Now, even though wehave noghost fields per se, wecan still iden tify the inverse ~ laplacian there asthe coordinate space ghost propagaotr. Iwill force this propagator fo)tohavesamenormalization asmy“feynman rules"ghostpropagator. Thus: 2 ae Dy=Sa@D oR, DaGa)=BaGr)Fes). . tov G = ble +giles [Ad a ¥ @) Cal>\rganCny)-\e9)DoarHALF S@rg0)Ass)da, ay wd) Ae==Cap[9aPhaoy]AG,(g) tobe. Feynman.RulesforGauge.TheorywithCorrectNormalizations. eo) Geepage\eSnasad.off) 1,The Bare Propagator. Let usassume the alpha-landau gauge here. Wethen wish to combine theusual F,, lagrangian with thegauge fixing term. Here ishowitgoes: B= PP- ROA AK | Bere) =3A0- Ap Sugrae saleSalada,frave, Quan: =-4pA, ~WARY AA=WAR) —deCB.AWIQVAY) Deparks dogk L=+eNyGd-WAD*3R(AnRIA) re)LotAp|Bap~Q-b)3alAy Rounasik colateeh oaaba a hoaSO(Tydaa h+AT . oe ale, cs—— -Ss Sesk One® =8”CTgw—Ck)38") Qos aDe ibaA Veo: FANE tate Fay Wl Qua, ES 2Sw a TSobre) -Bwesd|= edSSN >CYCED(Yort(s) -2- ne « ats es} LEBwGa] =<ITOR@MEDD =GyrGos) sou woe =iGw@ay =Cs /soy Whonw Se Nee;Ow yrGos)=3°SaSees) Tomake more sense out ofthis, gointo fourier transform. Todose, use standard transform conventions innotes, assume Disfunction ofx-y. Then: be TR OES) be DrG@ey = \ae © Dux Ce) “iyrGas)Sde. v0 Sn~ik & a be an2|(SRy+ODel)DGD=S75, be be, 0.>=@)=-2=([9~(2)hike| abhitytodauckea no Therefore, hereisthecorrectly normalized Feynman ruleforthebarepropagactr. + om k)- ap wy .neep(B) =SD =[ide(w)| sae =-id -Q-2)k| oe Lar aS Asacheck, setalpha=1 forFeynman gauge. Ignore color, andyouduplicate photon Feynman rule asinBjorken andDrell, including thesign. (see page 382volume 2) -~3- 2.TheFourGlueVertex.“Let'sconsideronlythisparticulerinteractionterm.. 0Ingeneral you can say this: : aiSanfsttete] ; Payee4. wholes¢€ Sakae atSI .gen WEGana TD T - 4Coesee yy.(Sumdlagquestost). Here Ihave included consideration oftheghosts. Butforthemoment wearenotinterested intaking derivatives wrttheghost source j,sosetthat factor always. ="unity. Then we are left with this: - . . . . a cy e, <-RacyYOBoGaWg) whyl=ovpi}SanRaGSe , a poke patld eo A ye o = -k aoehk(Ma GFOO ALAARAY Ame a ya =-4¢CaranCagayanSysSpee”AwAmAsAw Thismaynotbethemostefficientway,toexaminetheFeynmanrule,butletsexamine oO the 4-point Greens function like so: (a)Wibverby wt otSrunn Corer =GOSw DW)TTFGH)STFGSBIOLG) ‘ | NSpeby,AaALP.adeba.peee 6LeLON] Daw CDEGao] | *[Dm] (ypenaefavate) Inother words, wealready knowtheFeynman ruleforagluepropagator: putiniD . foreachline. Thus, whenyouconsider this lowest offergraph, whet ever is"left over" intheparenthesis must betheFeynman rule forthevertex. Soallwehave todoissee what goes inthere: Oy ‘12 wt 0°: +BYdeFEGCanaComaYeePayee|CDCA) bh ba bs ‘ a cs c %WED MOAMGBACH) FO)BUHBAGIA) i a ae » 4 »GE)bal)agieFOSG VE)] he Now the first thing todohere islet the “external” functional derivatives come in fe)anddotheirthing.YourJ,external derivatives actonaproductoffourfactors.Consider thed(x,) derivative. Itcanactinhways, ie,itcanhitinto anyofthe four factors. Then thed(xz) canhit’onanyofthethree remaining factors, andsoon. Inthis wey you generatoe a4! which cancels the one shown from the exponentiad. Nowconsider d(x,) hitting just oneofthefactors. Itcenhiteither Jinside thefactor. Ineffect, this makes afactor of2which cancels thetwo inthe -i/2. So, the net result isthis, after the external pincers have acted: «) tl-2e a?SeTELAT CraCapers YyeYryee| COGax oaaa Ncom(i)bo(FE)te +YG)FeOYIps,@)YayC 0)Iya3sI sens, CY s Sst abe :-2Cayee al\:RVayJpGy)DyvsGs,)| Now consider the action ofthe "internal" functional,derivatives. Again, each of : the derivatives shown comes inone atatime. The first one can hit one offour factors, 0thesecond hitsoneofthereminaing three, andsoon.Thus,therewillbe4!terms corresponding toallthese different hits. Ifyou know oneterm, you get allthe others byjust sunming permutations. Solets pick asouronetermtheonewhere thea,hits thebythings andsoon.Then weget: oy aye »(gtGmYan[DinCOILTEIL.(4CAS)CheeCanerYun3pm) Inthis term, a,gottied tobj,awastied tobyandsoon. Mnsome other term perhaps b,ties toayandbyties toay. Inthat therm youwould rename thesummation variabies a,andap;theeffect would betopermute theobject inparenthesis. Thus, here ismytentative concludion: . +(nbgt fae=4°C8)ZiCoanCare’yl z 4a, yaya ‘ys3yes Thereare24=4!termshere.Theusualthingpeopledoisclassifytheseterms ‘o}into three groups of8terms asfollows: 5- fe)CBW) standaGaBdomeyrghbydovig,asfacled pare. =iEgt)CageCava*\QsSpy apsSues tCaet) cy Sym Sys SpySits SpamGaye +Sonne] t rae) Garver (en) (ets)aedeas) =-igtConan { =FYCosaCaraLgyngps—SitsSe Thus the8terms inthis group make the above pair ofterms four times, cancelling the factor of1/4th. Soweend upwith this statement ofthe rule: ae|Qu=hg{Gaza)+(3,29)+cna§ :i ranGass)=CaneCaelVisYap—Spyssym]Sanne Notice that (12,34) =(12,43) because youchange thesign ofboth factors. Thus, itisjust thepairings youneed toremember, nosigns. Inthefirst term group, 12 are together, inthe second its 13, inthe third its 14. This rule agrees exactly with what Igot last time Idid this. And last time Icehcked itagainst various other sources soI'mhappy with it. -6-+ 2.5TheFourGlueVertex,MoreDetail. Oo Ithink itwould begood tochange the definition ofthe "\-vertex rule" sothat itisdefined tobethe thing which appears inhere: _ HHP (a)Bibosby rarSh Syn Coketaxy) =Go)!Svr abive. aby oe \\ (9)aaragay, = i a4a)\|*3a| ByYs, \asydgetgy LE! GasolléBarGass[ Fypgagin LESS Ifother terms were added, this would really bethedefinition oftheproper 4-point function. Here ofcourse these are allbare propagators andthis vertex will be the"coordinate space feynman rule". Ofcourse wealready know theanswer from preeding pages.Reulstisthis: weAusgactian _ — — - Bere OHaaagay o& & Sy Thries (SHH) =~ASSGI Gey)BGG) {Grats+of Here%,plays theroleofvariable called xbefore. Youcanseethatthisduplicates earlier result when inserted theré. Notice bhat this thing isreally afunction of thre three difference coordinates, asisalwasy thecase foratranslation invariant function. Nowletsfourier transform alloftheabove. Weget: , Ca eaey cabbababy _ate or_ Sovak (Poketein)=LeDeeCPO\LAUBYES] Tango Ctateita>) Hrethetwopi fourth delta overall hasalready beenremoved, sothese guysareonly functions ofthethreemomenta shown, thefourth being determined byconservation, The object here isfourier ofthe above like so: 4 . .San SHEECtetpoks Zu Mstrar) =\eanag ©ee V8.8.) =~igt5Quaa)+«ae [e) coeranclones SPRL RoutFeaeragay 1rs ' 3Vapi Costs) =“igco)wag | ~~ 3.TheThree GlueVertex. Asbefore, wehaveW[J]given by A,vee =-9CPQM)AN . Gnowwefocusonthetripleglueinteractiontermwhichis:Kedah is)LayFOGleave whl= © ge Webegan with~3,there were twocross terms, thenwetensor-doubled so-1.Byjust fiddling wecan rewrite this same object asfollowsL: AMA; cD) aN =Vayguag\s3©SpmOjSealSees Ct aa axAysGA yee) Apis8) *Now wesearch for the following object: @)eorby 3 53SFyvaryi%yrs) =CASwiWHS SW) SHEGad 2ales OCC POSagres HOSGalil EEE Gasegd aby ‘ : Conn ws—“e“=CDSagidyedyelisedysMesOL3a,Stor}|by be »: a oe 9:»DGD WO WE JeCoyBale) FaGq) : wes & 43»(4)Hay Katee Vo] Remember whatyouaredoing: computing G3)bytaking three external derviatives ontheW[J]given bove. Nosteps areleft outhere. Nowexactly asbefore wecan apply theexternal (say) derivatives tokill thefactorial andget: AAAS “(9 COSagagags ligCSpySey1ayrsih| abs oyam 3 : “ a |fas 0.JENBGOIMG)«[+SagheeyHCalsial -8- Nowwetaketheinternalderivatives.Theycangoinallpossibleways,hereis oo)one ofthe ways: ae yo agb3 Sy)dydyysbDawGuarli DavGui DavGaso| : wag Cs) .{igCogynBC4e)[SpeFewest s Asbefore, therearegoingtobenow6terms, oneofwhichyousee.Ifa,hadgone. ontoby»wecould justrename things etcetc. Sameworkings asbefore. Notice * that groups oflabels allmove together. Soourresult isthis coordinate-space result:c olespe, Maass eon Fandg ) PpapasChuguts)=vy‘ZzSoSpynsBCgpeyp) \SaSy] There arevarious waystowrite outthese 6terms, butletsinstead gofirst to momentum space. Wehave; @. Ase (8)Ore _: Ke . oO Me etAESgeFER3H) ¢ AAs, ®@ WakESMes =kgZC pyre BGSFee(&) heyoys —y® HERG BREE —— =ata=Sagage Me pie PEa WAQ ‘=-igae *aes rer eee ~7auaaay maaTa Got) ==9ST 36 =8099 0 =¥3FS gueGn whan Unk Aah L292, <a ee' m9 Tq Writingoutthetermsweget: =+4amCQSpe(PyeswRson, lo] ve |ewanphe _——_________. ie‘spre.}aaa aes ||Pano Cnty) =+gC Syl aa}+este S| Notice that, aswas notthe case with the4~glue vertex, allthe terms inthe3-glue have acommon color factor which can beextracted asshown. Eg, the cyclic terms all have same ©factor. -10- 4..General. Method to"READ OFF" therules. Now that Ihave’ worked afew examples, it @) 3clearthetyoudon'thavetoredoallthisworkeverytine youwanttoreadafeynnan rule from alagrangian. a)propagators: First, examine thelagrangian tofind theoperator inL=+406. HereIamcalling theoperator 0asintheexample. Thecoordinate-space propagator, D, isprecisely the "inverse" ofthis operator, ie,itsGreens function. Itisusually easiest toconvert 0tomomentum space using therule d=-ik,. Then getits inverse inmomentum space which isnotvery hard. Using this method, Icould immediately getthepropagator just byaread-ff, butitisdone indetail inpreceding pages. b)vertices: First youwrite down your interaction Lagrengian vertex ofinterest. You thenmultiply by+i(caused because itis+iLint 8indetail). Thenreplace each derivative with -ip where the prefers tothe field acted upon. Atonce you arrive atthemomentum-space feynman rule. Togetthecoordinate space rule it's notquite soeasy, plug Idon't care somuch about that soforget it.Iforget tosaythat after youmultiply byplus i,andreplace derivatives by~ip, youthenhave tosun ‘over allpermutations ofthefield labels. This, yourecall, arises from thed/aJ's going _ inin'the last stepinallpossible ways. Example: 3-glue Crone 3.\aoy f=-4C pa12AyAyeApis 4 Arras :s .=GG 2a ~i®a | >Rate GiIGa) =Ccays\Qa Youcanseethatitdoeswork. Ofcourse youusually procede towrite outthepermutation suminsoneniceway,Notice thatyoumustalwasy startoffbywriting yourLy,in suchawaythateachfield getsitsomsetoflabels. Te,youinsert g,,'8 asneeded, etc. Thederivatives canjust sitthere acting ontheir chosen field. Recall also that "Rule" refers tothelowest order term intheproper vertex with noadditional factors! (as was noted above). Example: J-glue AAAgy Byoeagwy La-kefo)=89CeByesAryeApAwAunAp Ana aaaaAke =GNGCES) BCtO) 4 Say Spy ue “beGhosts. Asanapplication oftheabovemethod of"reading offfeynman rules" lets fo)dotheghostsinthecovarsantgauge.Thegeneralideaisthis:to me SEYCe)MEWGd)CoO) acl] =\atae © and incovariant alpha-landau gauge wehave: a a ays at, et H (W-Cys 2P-Col Ts-4 FBAY %, o ~ , “ aaWe -[Osa- 5CuXX]. QR) Asusual inspacematrix notation, theayhere acts oneverything toitsright, notjust the asitting there. Letsnowdoaparts oneach termanddroptheoutside factor. These factors have nosignificance, since they just make anoverall constant which cannot affect any rules, and will becancelled inthe divied byWstage anyway. Thus wehave: . a ot a a. ys ai\=\adae.a(=RGTer+3Coe@QHALcd fo) foe0dt OK = hs +QV + we(HQYA / >&e&sy 23Ca+gCoe GG)<G Now wecan read off the rules. Onedetail isthat the propagator here does not have a %factor. This compensates forthefacto that thed/dJ's nowcanonly goinoneway instead oftwo ways topick off the propagator. Sowefind: “ ~ . \ Ca=Bad 2 Ga=-Bak¥ 2 Da=-Bae T, or BjGere) =Dalit) =-2 BAe BARE) :rs a=eyCaeB =BabeCaP") Sp ‘ Ward Identity Slavnov WardIdentitySection |finishedJune14,1979 , Comment:ThissectionwasdoneasIwasreadingtheSlavnovandLeeZinnJustin lo) sequence. TheWard Identities areneeded todorenormeliztion ofYWT. Allthis stuff isinthe covariant gauge because Slamov's work was. Gontesnts: Page Inverse ofGauge Propagator prlim Notations fortheGhost Propagator, relation tow+,ete prelim Analysis ofWard Identity 1 a.general Ward was derived byme, see LZJ~4 notes b.specialized here toalpha-landau gauge ¢.thebasic nonrenoym ofpropgator facts stated forfirst time 2d.replacement ofWw? byghost propagator . 3 e.the graphical notation for Ward's rs e1) details onthe first-level Ward ha f.the second-level Ward (first time) 5 £1)Review ofWard Identities (levels 0,1,2,3 end (n-1) cases) Saybyc acs £2) Review ofthe M-identities (level 1,2 and n-1) 5c psa £3)HigherDerivatives ofWardIdentities, Examples 5d,e,f .£i,)Fourier Transform onallIdentities *5g.h g.Thew7identity (original derivation, graphical) 6 h,CombinelowestWardwithlowestMidentity forprop.result. 7 fo)i.Restate the Second and Third Level Wards & i1) Summary ofMidentities offirst two levels 8b i2) First Definition ofSlavnov Object 8c J.Fourier Transform ofthe4,4, G54 object 9 *x*herebegins thederivation oftheSlavnoy Identity xx Byfiddling around, Iget toaresult onbottom ofpage 12which isclose to the desired identity ,but isnot quite it. There isanextra term asshown on page 13. k.Review ofsequence togetSlavnov (prior toredefinition ofSO) 13d 1. Definition of Gamma -slash in terms of unknown F. m,Final result inbox. There exists some object Fwhich works. 17 n,The b-glue Ward Identity. 18,19 The Inverse Gauge Propagstor. le]Thisisveryeasy,Ijustwanttohaveitonaseparate sheet.Wehave: L DwOd= TF]AGhe+2weange Da&=-¥Feher? DuDr=Ya=A&Oy+acey =ABWCB®W~ ben)“acta ~ARRE +haGc-@s Ne)arA ‘eek 6Ss svLWe= AWB abel ~\ sv L re ee’od=Se>a) Os weee Ss & a CRY) =PF fe) i=Pe=o cere gage \llnate, The Dressed Ghost Propagator and its Relation fo M. 6 1.First, thesefactsareself-evident: AS WHA) =(agdctae© * . Sm8.+Saray CoyMAG gig)AG)Ala[Cahe +eies] So=SQ)-EPS SoeBW4G) a tee v ant, pt 2S mre> LATCacho>=4DaGy)=[SasTeSe) 8EO)Ieee, This object isofcourse thedressed ghost propagator. Asdefined here,thesources 7 of the other fields has been left on. 2.Onthe other hand, wecan gobackwards and "do" the ghost-field integrations to getL: 5 + 4 ASQ) _—RYSrde,Wo(g)Wide(yin)da©) 6why] =\agsna)© e s whuaeVowaiaDaingamerranUnbeyral: \4anVax Fg ~=* \ Yadan otSarLdomc+jergte|=Gam)© 3 Now wecan take the second derivative with respect tojtoget another expression for the ghost propagator: - -\ ~\ 4SQ) DKS =|whi!VagSkACQ) WE(4,258)© Co =whetetic, $G)- wbyl oT = -\ j aeS =4GIT(‘itesO) ; =sO, AvetaWandAust, . ©ogFnsoaen,aunt reste GIL)nearenvariehgenet dondaUkee eS>: -. hk): xSOculth—lian tO$whl=o | (eke EUROS Scere SOBNDwel=e| [reasasised gongs, pantsongOAs SxWang. 2.Suppose youtakenoderivatives, thensetJ=0.,The-entire second termvanishes since itislinear inJ.The first term gives you theVEV ofagauge field allbyitself, and this must vanish bysymmetry Ithink. Only scalar fields canhave VEV.Thus, youget 0=0 which isfine. _ oa . : 3.Nowtakeasinglegaugederivative: Fisk,repackWecoatoe-Donel — B=9Kpeat.7G) e)|22h oeyam TO I -i 9 ix wily =RY|:al ., | ByOF(Hs) TH Cons+Be)a+SasIEG)DA(F49)MibeCanny+3)ty). AptsPPWAG &yor LLwW=Cin)dx o. = = swt nan -a m yay __|a>laWEA ol v =HDae(45,24)Mtoe.C05M65)woBY ) A +5 dy”gebothSuue.B @® 7 , RNS= Hy YDa) Mela) ©=-2S..8G-9) wl. 7 -2- Danawnsgd: O4ao [atsubi_] _-8Sq) wm«7[saesal aaal on on Sodsduegsbywo Ll=Gace =iDa&s) SGpseu co ~) 2a2navVa(a4)=Ss3Ges) Qewate: . _4aha” [Dew] =SaTory) ®For daekone: . y 4ShGea) ” LiDacesy) =\g&C€ [ADA@)|Coe\* dpe(x3) fe) Borsy=Jde,wmCons + ARNEL) EDAGI= Sa | es |> 4bak,[EDA (Ol=Sa @]WaskieDaaneegoog?Warsentre udernehion woaddheer Wis\=)TAN) ovpd\f-k ALLOSTA 4+VORGx .eis We)BXGos)weil) wynagmolaowVeta.Cetha ©Bors 4 .alo Lieut, [iDalia]=&|gR]=Ba7 -3- ‘Thus, ourfirst Ward Identity makes astatement about apiece ofthegauge propagator. Itwaysthatthelongitudinal partofthe(unrenormalized butdressed)isthesame fo) asthelong part oftheware propagator :"thelong part isnotrenormalized atall". This isnot asobvious asyou would think. Inatheory with only glouons and noghosts you get this result quickly from the self-energy Dyson. Here you would have toinclude the ghosts insuch ananalysis. Even inthe pure gauge thing you have to assume that the self-energy ispurely transverse. I'm note sure why that should be obvious here. But the result stands: itcomes right from the generalized Ward so ithas tobecorrect. Bythe way, Idon't think the gauge self-energy istransverse! Bg, inaxial gauge Iknow itisnot because there isanPIq equation, etc. 2.Ghost.Se insertion. Letusnowrewrite thegeneral Wardreplacing theobject M” with theghost propagator according toseperat sheet. Just insert wir? ondusethatniceresult. Youget: Taye an wetee[222=2\yttoBion)fvieciei| BO] 3)SLWIKY .Ihavenotbotheredtoshowtheghostssourceswhichmustbethereintherightmostfactor fe)inorder for you dodifferentiate. Once this second derivative iscomputed, the jand j* have been settozero. Thus, inthis entire equation jej*=0 soyoucandonofurther little jderivatives. Now lets apply (-i)d/dJ toboth sides again, then setJ=0: ia] 2, Op rot v iS PWLy-5»jetau |esDass,9)fuIVEYa xyes iSLOWS ¥ SY.Bes) Except forthelittle transversailty restriction ofrenormalization dealabove, Ithink nowthere isnopurpose tokeeping thepieces ofDtogether, soletsgobacktothe start und writ ethings out indetail: >). ww 2h oa3°ee|=3Sag3»is3 RP Leyes dy 3).@)Vit@) rs4seo re) +iaySoCs)Cav\232 .A :BIFG) Bj) BG) =y- 3.Simple Graph Notations.: le) solidline=ghost(barindicates thec*field) . ‘ squiggle line ~gauge particle (maybe anyother particle) ‘ ball =WJ] : attachment =(-i) d/ad standard label order": Lorentz, color, space i Inthis notation, the original Ward reads asfollows: : . *pee oe Ay °o 4ey 9) # a -& = - : ae -Sash] [gate vc ood . : “5WardIdentity * IfnoSSB, obvious howsatisfied ifyouletJ-0. Nowtake onederivative toget: -. codvide an oO. us 4) Vt 7 =4) XSwt 3 a Fe =VokEol 3-99 +ON .vd a vide+ by on : aw \*+{-aS sh vfe BY ae be IfweletJ=O anddivide through byW(first), weget arelation between some objects: vie poe =| @Tex |£x —— se-+t ai~*~ Ul be By[-O- oy[Sora] +Ct[poe Funak-Sawd Work ww ta More Detail onthe Level-1 Ward Identity. Asnoted onthe last page, the identity fe)reads:»aay aelyaa‘iwe)=-+9 KayKe -—Qe[aDCxxe)\ goo DGye Saba*Career Gope(2 2) »: : ‘al i .4,Ke,=eBamaDenDoyenGoons)=DeeD*GsmBaer+CagoierGeerCeveere) First-level Ward Identity, Coordinate Space Representation The Fourier onthis isstraightford and gives: wh ; m2) ae, GEGniBaanLOW]Dugas) =LEShaTike] DQ)+CaperAeGoa(iyeyr) rd3 — le) » SaazWo"DMR)+C JkoeCha,,-) FSee K'Dale)=+eDawe ie)+Carbscr\\MeGop»Qs,he 72Baa,EDyalk) =+4Bans we First-level Ward Identity, Momentum Space Representation. When welater apply another ktothis thing and use the first-level Midentity, welearns something about the longitudinal part ofthe gauge propagator. Namely, we learnthat:Tm mea k1KDy) ==@(%<) \ Rk SuKUW=HRSu(K)- Cul® 6x) +aSy =+a(XK)-ved\AE,GnGR y Sek ® cy First~level Ward Identity, Momentum Space fe)IfIdotthisthingintoKY,IgettheFirst-level M-identity, butthisaboveisa more powerful result! -5- ksTripleGaugederivative.Nowletsapplyasecondd/aJtoouronce-diff'dWard o)and see what happens: cn yte o °a sh i) es a aanYe, =\ILC ++ +ob . ueog ay(EMD5% .® ye *y moOR *yey ~*~) cmwoy'x _y°cS£ vit +~gs «Ca +3% +Ca Aco” woe da be Now set J=0 and write result compactly as: : on * ve poxYew ‘6) dev 1° ¢ -49=3-7 +Cy +9(Vdaes(dew) my 3 va ew a be Rooms: a an 4%1a” ;x 13” £ at ae a|e mM 3om Pik+Coe preehe pPaks Saks HatsBX res areaSoukwork, +Lest, Note how the 2,3 symmetry isautomatic. Aswas always claimed, the Ward relates various kinds ofgreens functions. Lets duplicate here the previous one: any aam myfm @) a | =eae mhoe =73Se +CapDake aux pads, " ynak- Lowel word. be, You can see how these things are going tolook. There isatendency ofthe Ward to turngulonsintoghosts, andtomixinthen#lpointfunction withthenpointers. OD ether nessy .Tsuspect you have touse the Dysons toget these into standard forn?? -5a- ReView ofthelowest levelWardIdentities ingood notation, 6(0)_Level Zero mWaxy, im ‘ . ~My &xJamLeal aK, aK Be ak ban, Mee%e (1) Level One axe xy ay, hypase, me =eyby={5uy°+CaderROak. bx, Mem ax, ojuan BSMy,Pedr, ad e*Saskes3{5SeayA+Cagbaes&{aux by 8% (2) Level Two bay amay 95%; aywyJly as poate aadn hs =3ey +Cates Pasexe, ade odPaeeke £3203t{ ax, BK, hs pats Bs9s% "+\\an(iseon]-¥Ney par+Cone, pak aut v.PaceyKy ~5b- Theidea isthis: each time index the subbcript onthe Jintegration labels. Also, push re)downformergaugelinestomakeroomforthenewones.Onemoreforfun: (8) Level Three AX, aX a a3cr deaay pean-Ly" mos =J-1) PsasksHECR) nan, 5 Pyar Case, KL akebie Bese +$2293} ~tree xKy anayXy \ptt ae Dsas" patsXs +\an\pVag(a~F%0)cae +Casbsey praekeasxs eeMSCS 0Notehowthesymmetrization comesoutatezchlevel: youneedsixtermstobecompletelysymmetric ingluons 2,3,4, but the terms incurlies are already 3-1, symmetric soonly need to add the two terms shown. Ithink &can see the level nresult: (N)Level (r-) aK Ba a xia 3seC Print ~2 : 1 3 Ow = 1b dG, Pay beee meeae 9 Fm feesciteate bane~Metere &Preah +Jreauk aJeerst ee4front ee avemahbowut,ak +Sna,\BSauHm)\ ~5e- Review of the Midentities. ie](1)LevelOneM ax, an i de .ey}.Me°+Caer& =ideSQ-%)Daa.1)ati iene (2) Level Two M te ane pie Hs42% cs Aapats :_ Ss wyisoyer+Corre =ASRBOa-x,)dae am wy Ate These are obtained bysimply injecting. General one isthis: (n) Level..(n=1)_- fo) =Paar adn Paankn,a. te A ¢ . ‘eee‘4ou<prot,Caterson{~1FAZGere)SaoeKinaOke ba Ps ~5a - .Higher Derivatives_of WardIdentities making useoftheMidentities. Comment: Now isthe time tointroduce ultracompact notation sothings can bewritten infinite amount oftime. I'l] bet Iwill find something like this inthe literature. a AL Oy=dey port . phe Ho og aks am eaNonprone ahxm3) 3. Fam =4-2%y aC 5 FO PSs oats Py ake ‘ by Sth QO *Haymn=iS%SQur%e) SeeGain *CER) (te=id8mm)Saar:‘) Ha =o 5.Ryaxdrangye erybwapart Lyingbeeamgr. Inthis compact notation wecan write the level n-1 identities like so: Level _(n-1) Ward Identity 3)Gan=UrRarerFallaon Level _(n-1) MIdentity OnFaw =Hm Thecompactionfactorisratherimpressive, Ithink.NowIcanstartapplying Ooderivatives onthe first Identity and make use ofthe second like so: -Se- ‘ ro Grom =LiPagePartPagto+Paw]Fron AAGam =a[Porter tics +PelFo Hae 328.9, Game =aa[Pretec ++Pan]Fen +UFPe]95Han DeP323 Gem =949sDa[Pag ee.+PruFem+[+PastPer] Idsthew These things are trivial toshow, using compact identities onlast page. Asyou move down you see that, ineffect, aterm istrasported from theFlisting tothe Hlisting via the Midentity. Here all terms except one are transferred: O120-- DeenGrom =ada. BuyPanFree ie) +DsReetage Papen)IadqBoryHeer And finally, here isthefull result with ellterms transported over: DidaDeGam=Usasebate+Paw|OydqdeWon Example 1: Write the preceding result incase n=2: Ye Su =Ha=—hd BGM) Baae This agrees with thesame result found earlier, seepage 7redbox(longitudinal prop deal). Example 2: Try case n=3 . ° —~ 2:93 Grs= l+talds tes =© . =5- Example 3:try the case n=4 e)323294Gray=UsPasPy|3:94Haas Ican see that this will lead tothe famous Ward relating the proper 4-poitn tolower objects. Return later tothis. Example _4: Take only two derivatives ona3-point function toget: KN Gra =2RaFag+es =92Fan This result appears ontop ofpage 9andisthe first step tnthelong journey of deriving the Slavnov Identity. Comment: Youcanseethat, unless thefull complement ofderivatives isapplied, you aregoing tohave some Fterms, which istosay, youaregoing toinvolve ghosts. -5g- Fourier Transformation ofthe Above Results. re) Ithinkitwillbesimplertonotexposetheoverallmomentumconservation delta.Then wewill use this direct fourier transformation: Yn atk ~ARXy, : FOdak) =Conyaa. deySS Fae, ks) With this fourier convention, the rule for momentum space rep ofaderivative isclearly: , aaSeyalli \ Nowletsexamine thetransforms ofourbasic objects. Theobject Fyy ,isalittle tricky because twolines have thesame space label x». Consider thetwoterms inF asshown onpage 5d. Expand them like this: ssaaah ~Ake]( Fata. Xe)=lesVaneaVd OF(ey,tee,beg=bes) v ‘| keg”7 oS t \ FG%eXaKsKe)=|awe ir(legeeMeaks,en 0Gy . ‘Thesecond term really isan)n+1( point function andhasthah many arguements. Ijust has anextra integration. SoImight define fourier transformed Flike this: aky ‘ akom +al aks, Bowfrp LEManche oe” Vkaye= wR +Nabe ay k.on $peaaky K as th bake Patek, park, Allthesereubitsareobtainedjustbegetting vel oxa thecoefficient ofthen-foldintegration. wesHankin, ~ 2 .: Hawn =CaBete) Gain *ChaSane) ine “(He=Gon"Stars) CidSe) Cha =0) =5h= : Nowitiscompletely triviel toexpress allourprevious identities interns fe)offourierobjects. Justputtwiddles overthingsendreplacederivetives accordingtothe rule, that isall there istoit. Example 1:twoderivatives onpropagator: L mS CSRRGr=Be eed O94 +=NE[CoedSeek)ED(|=CodSkate)LiaBane] mye 3PRR DyeCR)--«| HereThavejustelaborated theidentity. Example 2:single derivative onpropagator. ~ikiGe=Fe ‘This iswhat ihave called the First-Level Ward identity. However, wehave information )avoutk,6,, fromanentirely different source: tensor forms. Write: aa 2 ERS eSReed) LDiyaGed) =2 Ge =LANBethe KBase|keDyerCed| ~Want Ap aLob BeDADAD Were tok RR DG) a=4 " wos =Dnt) =-2Se ‘ Me .wt on Le La OnDEROE~2@e)3nik,Gu=CosBleate)Saas2Ge) ~ wy ak NWwe=eGE) |2 We | 0 wikiGp=eGa) GelSleeved) Fare||eee -6- 5.TheWtidentity.WehavebeenineffectreplacingMbytheghostpropagator.But fo) wemustkeepinmindthatytisalsotheinverse ofMandthisimplies something about the ghost propagator. Namely, weknow this: —— ®La -_!CFR) FDL) Matas ds)=FOBa VY as Was (4,39) a WA wos=>Gam) VAG) WPL(oEMeL Gasdd)WE)=SEEBeeWE]eee: 71DLGy=IsuoWeTeTywy + erste a&) »-WSS \=< 3Cie) ®\Ba%y ~4CaCodySeMey BQwGeW. Aer 32 wyto At ye * San SIFSLAs Poy gx s* 3. a Pow PNY pany“33859.arog ~3HeyO ey, 4 Ofcourse youcandifferentiate this togetafamily ofidentities. These identities canthen becombined with therawWard identities togetsimpler results, Ibet. -7- 6. Combine Lowest Ward with M-identity. Teke the lowest Ward and differentiate it: 3.G) a “a os xa‘ --4 es on =~2Doo +CauPeake aK, pater ee See =fFBr) Baw mYovoy Pen . =|Oyan =~haBCom) Saar PAaaK Asbefore, wecanfourier transform inobvious way.Ruleisthat.ay“ik, forincoming momentum, soweget: Fra ,=-iad to)KyRan$ a,PadeOW) Inmomentum space wecanexpress thegluon propagator like so: mak » kee . é-ibay=-iSonJNO(Byartate)‘Bley(Sele) Prac,i x Ie, there isonly one possible color tensor, and two possible Lorentz tensors, sothere are two scalar functions possible inthe propagator. The above identity then tells us that: .~iBaa, BES =~Ldaeat = Bax BNE, ant Rkwerueo Ook“AWS =1and RBM)= <, Thus, ourWard identity tells usthat, although weknow nothing about AG), theobject Bdoes not get shifted atell from its bare value! -8- 7.Restate the Midentity and ascend one level. Itisthis: ~ a,-bMe°+CarerYou =MTSGD, O x on bay Pao Vrouber,agphyadervatua toag aK at asks, a|pons Mepte. a ~Seyi +Carre960 =iTKTHex)SaOC am Bx Pek, Now gotoJ=0 with thése and divide byWtoget: + psa Me >aaaXs -5Seay +CabrerYay =o! a ba Ate he Second—levelM-identity. |a This identity makes adirect relation between the two greens functions shown. 8.Triple Gauge Identity. Nowrewrite thetriple gauge identity with 2,3switched: an ahpng axasks oea a0 -19 an=}-*Wy eeCaves os zo ag 3 pad OMe aaa, hy be Paeake a, ux, jAS pao proaake+g dex ~Ccaghst ars, bayMONS (o)Applyal)tobothsides.ThefirstcurlybracketwillthenvanishbytheMidentityabove leaving the following result: -ab - Summary ofM-identitys ofthefirsttwolevels: ©) 1.First, heretheyareincoordinate space: op 1 es Ca Be Aes|=Moo[5DaeGr3+CanesdeeLE(rand) =EIRSKK)ag | first level | .avaray beaycabsay \ ~¥DesUSLoyGerais) |©Cages BoyLEpopsQerenare)] =O| second level 2.Fourier transformation requires some care because space indices are pinned atsame value. For example, mR Gen) Serv=\She pat [rnttn. hs aexby HERG) wasLG Gummy] =ay\sitse6.paCPater) lo) Cons(WY ~ERGO) Ae’ _—s yehZ -\en.e 5jeselWasGipye(tutsx: Wwe, . oe ee .oY .aeabaeR DiC) +CannesGraSatSSCity?) = Bae.4first level Midentity ay soon) 1 aug may PROBS) desJe oe rh SIT Outi) +Camis,LaasLees" @ ; ; ; abies,Thislastthingisgarbage. Ihavetoswitch2,3 .Cla psCrakee,eg) inthe second level M-identity above first, then fourier transform: =Be - iB\o283 -ay,A,ACabs j 7sTeg, Gop? Gn%e,xa)+CapyesJonSFpaps(XX,%)=O| second level Midentit; aama . Ds AaaC3by wRGoneGrae)+CardyesLiCerend]) \eeGpany(RRs) =© x any dasa Rerkerky eo. Rauide an, | Baa: Scans ds chaste !patReGoesCiut-) +CarbsesLely“|cetFjasCeakesp) =© 8 _second levelMuidentity _. _ Intheintegration ontheright here, kyandkyareheldfixed, theother twomomenta vary. Thus, wecan only extract one ghost and one gauge propagator thus: a y AAAsOEBS VOBAW VayVTGk) dksFEaacabs 1 atpi She| - =CarreyLY]7DPC)Dp09\SG.day»(RoR) Aslong asthere issome longitudinal component onthegluon propagator, itsinverse exists. Thus wecancancel offtheD's toget this: xAe Bay are ade, norQarego3 ,ahVODVNGb)=RMCaneStySOTA Cees#) Divide:Owamnghdogh +a ‘aan by: OOPCae)=LR—Cares\Ha(ie(hye,Bo Kid (awl apa tyRe,Re SC Thisistheobjectdefined in aaa, 0Slavnov equation (27),orithad =Youn, (tor) better be! -9- ax am nauk © Oyoe ae Mey mesene me aoe oH)Or peste =On =-3easHX +CayaSexy 4ea 133Cy, Pes, However Iamnowstuck. Ihave noobvious waytoeliminate the2-glue/2-ghost greens function ontheright; some kind ofDyson equation isneeded here with thesimplification that twooftheparticles have thesamex,. Theelimination ofthis last termis accomplished byLeeandZinn Justing bytheir mysterious andunexplained equation (4.10). Since Idon't seehowthisgoes, Ihavetostill justaccept their result (4.15). Hopefully allthese things will bemore obvious intheproper-vertex formulation. However, Icanapply onemore derivative totheabove togets i aae _ -aBearyPonyDeagyRe =© . ‘ Baty tke 4 AOMeMs 3Preps” Cnet, mesespatsps) =O. re)Analogoustothepropagatorresult.Notsurewhatthisdoesforyou,keepforlater.Retry: lets gofourier transform the above result and see what the effect isofthe X= XeTgett phePSs pry ans - @) a TRBayGul) =“tRRSTa”(hiban) ‘ heAR,HKWACabs 1 -tky QR haRa,aj +Games EERO)aeEA a) While were here, lets write outtheMidentity ink-space which isrelevant here: LANA .|deYZaacabs t a0a x Oks deaMea,bes,- HEHGUCir)+Comey, EER\8.SoreCeutaks) Istill don't seehowtogetridofthelast J-particle object inthere. Lets goahead andreduce toproper objects with the3-point functions andseewhere this leads: -10- &ade.mona, mad wad myads naeeedGlas)=GenQe)Gyareey GpaasCer)PanesCala=) fe)=50 Now this object appears dotted into some k's, and wealready know from alower Ward identity that : aay vant w . » weSn(a)=EWDane)=kSag)RP=AleSa(e). « Rr MoyDe and. ~dard.. kke.393, WAAR UMS=495[riadealiaSadel(=)etDuaneGe)Pa(aka): But ofcourse. weknow that the propagator intoto isdiagonal incolor, soweget: 2 A AeoF (ke aym85 @ws=-2*Dass) CEYGE) PSCaste) ro)SimilarlywecanexaminethefirsttermontheRHSof(*).LetG(k)bethescalar ghost propagator (=iD), which weknowiscolor diagonal also. Then weget: ° en OS Qores @)gas?=[risa|GOW)GCs)By,Clete) Ls Status: although Ihave something that looks like the left side ofSlavnov's identity, the right side doesnt look right. Here Ishow two ghobs propagators, whereas Slavnov only has one. Plus Istill have that extra term with 2-glue/2-ghost. Here then isthe closest Iamable toget: : _— weyyRoM)@aenas~iaEDowd PRS]] GE]PasCho) | . Des ~ayarag =¢ans \feGCa)Gu)FALCaley) . ke aor VeAMCabs 1 fe)4CoajtgeyLWWasRC) a -l- But never say die. Suppose Icome inang apply atransverse projection operator: aa vGed=(aw7) JL Boa) =Sa. sr pv _9psw ve an TYLCpt)Y Goth=CayBEVganSe)(ana) ~Bo C=C") Lot v Lvs tWeta=ekee=evPaosPY=). oa ~ : T r FdaGl= Fue)LAGos eKO\= ARG) ~B- Soletsapplyatransverse projecti-n operator fromtheleftinkgs + BayaCd)DaadaCha=Dana(ks) é ©eyw=Caps sede =0 3 . Sothe second term dies and itisthe third term that issignificant. Wenow have: » ML ar ke Aaz03~ioDyasic)GS yddsCryke-) CRS keae dk2MASabs I =CaresERTLSasaS|VayGvepe.”(huke,Ry-) Nowtheleftsidehereis,apart fromconstant, precisely theLHSofSlavmnov's identity. Next, lets usetheM-identity second level onthesecond piece oftheRES: a\ AAAg RUSaC ka Gees Cenk) o.VQdks ares 1 Deis=CassaEETSBEEME”Catan). . -12-. Now goback topage 6cwhere Ihave defined theSlavnoy object: .0Drsacaba ' Le}MyonCala)=_¥_Cagbresate,SovagyChabe,bao) keADCs) ass Wasae aaa a9:BOSE Geky-) =akeWide Cee). Gas), Thesquiggle over theGhere means that oneghost andonegauge propagator iD,iD have been removed. Lets now put them back intoget: 5 aay Kaye COREE YG ~DW)DaaChadBae= CaabsesSahGna Chal) &UD*S) = Quawagi ; Casey SBSpeyts-ibn) =FEEDEYDCR)DpamC60)BeaneCebe) (o) _. — This atlast allows metoreplace the4-point function Gwith Slavnov's 3-point thing with theextra lorentz index. IfInowgoinsert this into myWard identity onpage 11, thek_"2 will hitthegauge propagator like sor whe vateLRa”Dparelee) =~iad} Jonompoya Therefore, myWard identity now reads:: EOor k oon |~2x™DyasCe)Gye) Fyxcdsuke) | Cry xr acu: ’ zy 3=4DCR)isXCD)Byeses)Vian(eke)| : + ke iae . aay k= eek ; . ky=P : -Bo- ItlooksverygoodexceptIseemtohaveanextraghostpropagator inthere.But, fejlets now goback and use the level-one Midentity like so: %v¥GDSan,+2 weyGaGREEgutoy=Fa ¥byD'S) Baa,+Carrer Qs) Ces US hey)=VASuey. Put htis into the Ward toget: » Waray Waye) ad asaus= aGE)De)Xa i dog andy mallgud Cab, 1 ke 295~&Carorer(a)NsCopaCae)|GIDH)Vpn IfIcould showthatthislasttermvanishes bycolor ossomething, thenmyresult would be: te Vay eeess mot ynetehDyatsChe)GY)Fyne(eka)=(te)ED(K)BoreChe.) fo)" Andthisisprecisely Slavnov's result! !!! Slavnov Debug Sequence. re)1.Letsquicklyrevicewthestepswhichleadtotheidentityofinteresthere.Istarted with the Second-Level Ward Identity shown onpage 5,copied over onpage 8.Itook aderivative onthisthing andthereby obtained result ontopofpage9;Imadeuseof the Skcon-Level Midentity tokill off two ofthe four terms. This identity isshown onpage incorrd space rep. Next, Iconverted the intermediate result —-top page 9-- toKespace. Then Itook the two 3-point functions and expanded them toproper vertices. OntheLHSthis meant thegeneration of3gauge propagators. Butthetwok"'s chopped two ofthem down and left the third. Onthe RHS, the single gauge prop got chopped, and there were two left over ghost propagators. Thus Idnded upwith result onbottom of page 10, Next Isimply applied ak,transverse projector toboth sides. Thefirst term onthe RHS was thereby killed off, result shown onpage 11. Atthis point, Ihad the correct LHS ofSlavnov's identity, but Idid not have thecorrect RHS. Iwasstuck with a2-glue/2-gauge piece ontheright which Slavnov did not seem to have. Next, Iplayed with the Second-Level M-identity totry and create Slavnov's fancy ghost vertex with the two lorentiz indices. This identity relates the 3tothe 4. Iexpanded the 3-vertex down toaproper vertex onpage 8cand then finally Icame 0upwithadefinition fortheSlavnov object, duplicated onpage12.Thus,Iwasfinelly aboe toget rid ofby4-point function infavor ofthis Slavnov object, hence result onbottom pag page 12. Finally Iused the first-level Midentity toreplace the extra ghost propagator, and Ifinally obtained the Slavnov Identity plus anextra term. Sothere igmysummary. le) ~1h- Ganma-slesh. Lets define anew Slavnov object like so: aay 4 etZ_tetacabs \ ( X 2 — ~ be) DeCit)©Ba Cases|BOO (Rk8S) m LDR) Cet ” ™ Fyay, aay xv AyanaKare Cee) =Bre Codie) +Blanes) FoGayta=) h2y, Ihave copied myearlier definition ofgamma-from page 12top, and have defined gamma-slash interms ofascalar function F,asyet undetermined. Since the extra piece ingemma-slash istransverse, wemaintain the desired considiotn that: AA2a, Ay Waras,Boy Olan) =iRSGuadeCaer), (14.3) Letusnoweliminat gamma infavor ofganma-slash inourfinal Slavnov Identity result, making use of(14.2) above. For ourRHS wethen get: a aaa 2 ee) s,6ws- @GE)PH)Fae xanesNe a Baarll)F 2keenC)Ee ° +k4) ee (ts hs dey8, ayes Be 1arabe, | i. dk - =<CEYD(K) Yea(Carrer \Bs,CG) Thefirst term istheonewewanttogettheSlavnov identity; thelast twotwoterms wewanttoarrangetocancel.Iwilltrynowtomakethemcancel.Factoroutalphaand ngD(ky)-Definetheparenthetical groupigg tobeHELA(ey).Probably thisisproportional “\ todag butdoes notmattedr fornow. Then wehaver Ps ad” meas Moor AHAnAS aya 2 FeReFeaGosdPI—haBhs(8)Yana” Hs) =OF 9sx aN295|MyayaDe,A492, CO” Bslid PRP RD HM eR, =07 ~- Now expand the underlined vector into its tensor components: Da,ayaeay, 9203s Oaras, ni 15.1 Ooysee | mash Bh (5) Thetransverse projector will kill theBterm here andwell get: na Bs, A003 Aq ne) _2 Pray(is)$chy(2FANBayA=o2 Clearly this will besatisfied ifwechoose Flike so: tBards 7ayes O49Aa, 15.LEO ew A oe So,witholut going another step, Inowseethat itispossible todefine gamma-slash such that theSlavnov identity works out! Iwill continue andattempt togetan explicit expression forFandthen for gamma~slaxh. First, solve (15.1) forAas follows: Da ONGeaa) WAZ OT ByO° ysGerleYe)=ANBC Ceo =A Ta elRI]is 3 veantes ae (Chey Ske Playa CeadkeXpre=A eoAP) “ispCes)eaOp weAE 2Ayar03, aBsor KayyBnethaye e A =CkIaPsa(ts)Re”Gate) ARESCar's) Qua Yye + ards aay sr asOr ereleyFo=SeAGe)Pande) fa’Pip(lo)kaORK, fe) eneKies ~(Axes) -16m Wecannowcombine the55andtheP’andthegammatogetanother result forA: 5 oa osTs b= (ee) <)Mxazay lace:xT Aw™ aT)RL eaA \ From the first-level M-identity weknow H, W'SGe)=BML -D's thus wemay conclude that Fis given by: . Dye7pARS, lack,ayaayd PA2LRT Ea]EDERTVG BEYORE Iwould prefer toseetheSlavnov object combined with some propagators like so: me, hae ~\ <\-\ aNd ae,OReSFESOSDAGoKOHDHHBote)VN“SF .)arcs ' ba,nayGodresViGspa(elkeks,) eral nea’ ReDyeGe=kk Gros,DaindWominFalomnte ho =ake7\Sa-3Weed] ayCasbsca oh oven by SS ndLS!we|Na,GopeeCeekh»\COME) Cn - -i7- Solets define ashorthand for this nasty J-point function integra}: .aaa, agb,ca 1 ArCyba 'QingGok-)sQ deSG.oe(a~)~ Cary COUCeY + ary astog,clyeFRR._ 4 mA ALR =ga T[SPD] DUM). ReekeQjae(bh-). ara = »4 uy AAAS, fy=siaTLR-Z VG) ROOTS Nowtake equation (14.2) andmultiply bythevarious propagator factors toget: ed 4 ya2% aaadsLBD&)D Os)Dpara(le) Hyd =Qs (bo)¥ ‘ a8 aay24 ai e) *ahVK)D>(ke)Darr) Pyac(es) TR+Feyf End: OK, IHAVE HAD ENOUGH. The fact isthat the object you add, which Ihave called E,israther complicated. Itcauses the definition ofgamma-slash tobecomplicated. This complication isreflected inLee ZJ's unexplained equation (4.10) which Inow realize isinfact adefinition ofthe object Icall gamma-slaxh (they call gemma with the two subscripts). The expression for gamma-slash isnot really significant for the implementation ofthe Slavnov identity. Just knowing that the definition exists issufficient. Here then is the final result: Aaras,T DapMAMD N%, > .ay(er vee)=(2)iD)Bode”Cele.) danshe)(Se)AE) (ten)=Re } Bey VRvhed3, 7 By,HAAS Aas Phe RaBade Cyan) =BonLChae) . Slavnov Identity!!! omk%acawayrmrany dali. CeeeesSaw.) The keglue Ward identity. fe)Thisisalreadystatedonpage5f.Wemayconverttofouriernotation toget: 8 x a XsGO)RRRaeGray =COR: kyHreay +(209) +(294) Here wehave chosen toputonthe"full complement” ofderivatives inorder tonot involve ghosts. Wewill find that the4-glue connects tothe3-glue and2-glue. Ourfirst task istodecompose theobject G13, into itsIplcomponents. Recall that "connected" inthe sense ofgenerating functions means ‘no concomitant vacuum bubbles". Graphs that aredisconnected intheS-mattix sense azestill present in Gyp9,, +Thus, graphically andasanequation wehave: Oy \~m—2 \-oeta= 234=ho-@—y +Ged+GOH +oes +\ 223-0 Ws oe+(re3) +(24) 2 pee) nen e oeGuy=¢GaGa*G+ent+GyGadasGwNase ~vyw ~ wy a 2+GyGatCasGat|VseegTay+Gs)+os) Notation: thetwiddle asusual means k-space object. Since momdeltas arestill inside objects, repeated indices include momentum integrations with PIfactors, aswell as Lorentz and color index sums. Our next task istoapply the four k's tothe above expression. Aswas shown on5h, when ak, hits aG,,, yougetcertain factors plus atwopifourth momentum delta. In the second two terms above, these deltas remove all k-integrations, but inthe frst bracket you will have some deltas left over. Taus wehave: ShineCady=TKR. Lik,Glee CERMR. CrGay=CORReetbsCre-ilesGrae a° =SWPP E+HOR Bn] :a3 eh4wee LeCEZ\OS DCeatle)Oye Bod +e ees Met Bas _-\s= a ny (t,t g EM) BREE, Cuban =CTW SaneSosyGH)Fert) Sates) This isthefirst term inthefirst bracket ontheLHS. Nowbefore doing therest of theLHS, lets examine thefirst term ontheRHSofourWard identity: You —---——. —ra CB BatyCvS(karly) G.GiaSacer) nSad ~ 4 . BE CoFa,Gye=Hy=CatTester) iaBaae)—/pageSy Sowagk Lu ti=Cit)SaarSagar(ZRYSChetter) SGeyHler) Thus, the first term onthe right balances the first term onthe left. Sowesee that the purpose ofall the terms onthe right ofthis Ward identity istobalance the disconnected terms onthe left. Inother words, wejust showed this: aw sé as ~Gi)RyReRSRy GeGe=Gi RyRy Hieay These terms having cancelled off, wemaynowstate theresidual Ward Identity as: Hes ay Toe R22*kb NaasheGeely+G8)+QO=0 BYORDkeke Inthe groupings there aretwointegrations butthree delta groups, onebeoing the overall. Pulling all this stuff weget: =ee BeHatyteYoauayay hs 5 porseayRykaBaRy[nenCalesr)+;SaneCha=)eDaguse)|psprets(Rekas) ; (ea)4coh=0 This result appears inLZJ-1. Itrelates the k-proper, the 3-proper, and the 2-glue. Asalready noted, noghosts. Much simpler toderive than the Slavnov!! Group GroupSoowmaryRogge | (oebone aaaneywoKonnyaiget”searsearth Gy=vanmihion We Case = *aan eo=TAL+teed roeae dae =TS. +ceak yr- 3 ooieoehaG.)O.=ce. SGb=CaGo|| 4,w&. “(Goe) =FS e 0,_ s 4°(Gl@Gd) =iF?." ao, weo (646,63) =D'da Dsawd4(GGeG) =EF” o” Ae& eeetED Aareeter f4 Ste7 a =imadecible (Qua(" r @® on.* : Aare aFeanFO) Fo)pOms: zZ.y®ao.» (mnt)- :. yl_.aSLindhupvolluce. |ww _ ew aela=tah(G.0G8) OX)(Eee iCae OT” eo“° 6ek. -4°(GeGs)=ConyConyJ7CryC RAE? (GAC) =Cray©ee Kay,yorMack=~ekCove 1 1b r(G,G,) forIie ,. fe)1.Asuumethatyouhavesomegenerators ofsomedimension N(notnecessarily the adjoint rep, nospeciel rep) andthey satisfy alieAlgebra like so: [65105] =455.5 Assune thattheconstants cj, erereal andformatotally antisymmetric tensor. Iwill postpone till later thequestion ofhowyouarrange todothis. Probably foranyLieGroup youcanshuffle thegenerators until this istrue. Assume itstrue. Because thefinite transformations U=exp(-i0.G) aresupposed unitary, the generator matrices inany rep must beHermitian. 2.Recall that alie Algebra isinvariant under asimilarity onthe generators. Ie, after youdoasimilarity, thestructure constants arestill thesame astheabove. (Not used) 3.Nowdefine gy,=trace(G,G5). Obviously gyissymmetric. Andbecause thegeneratorsareHermitian, gj)happens tobereal.Recallinproving thisthattr(ABC) =tr(cTB7A7), 1,Supposethetensorgyisnotdiagonal.Sinceitisareal,symmetricmatrix,itcan re)bediagonalized byareal,orthogonal transformation. Thus wewrite: x x&5oRGG)Ry=+fRnGeRyOne Wire 2+ a 22GeKG. Gan age#ESG “ue a2 root svoh:LG,Ele RuRyGe,Gy)=RieRyteeGe . > = 4=ilcaeRaReRealGe Rate Ye 2 = Cu Thus, wehave come upwith new generators (twiddled ones) and new structure constants. Notice that the new c's are still real and totally antisymmetric. Inour new Lie Algebra, the trace oftwo generators isatleast adiagonal real matrix. Sodrop twiddles fron now on.—_— Le)5.Wenowwanttoshowthatthisdiagonalmatrixisproportional totheidentity. Consider: aos —1— + .3. dge-iTeLG:L6,,G0) =Cydiy 4 ‘eoRoy gusPSaa Aye=FBCyei iVaG: cil Be, de=-CTrLGiG,Ge- KORO =-CTr[=OyeGesGye =—dk Q der eyk Faquim k#\\onoeQh\ ieae otkuk.CieOy mage=Qk=FLGe egr=F. =G-R)t=0- aES Que,anne SUICieraidsQn, gazF&, Tosummarize, ifweassumethat: (e) a)LieAlgebra canbecastintoformwhere theconstants 4jyFerealandantisymmetric b)Foranyi,jwithi/jthereexistssomeksuchthat¢4¢£0. Then wecan produce new structure constants and new generators such that 1)the new structure constants are still real and antisymmetric. 2)the new generators are still Hermitian (sum ofHerms with real coeff =Herm!) 3)theobject tr(a,G;) =Fdj,whereFissomeconstant. 6,.Now,letsfocus onaparticular representation. Forthatrepresentation, youcould ifyou liked rescale the generators tomake Fbewhatever you like, However, this would require scaling the structure constants. This isOKand does not affect their properties £6reality and antisym and Jacobk. However, suppose you dothis and you get Ftobe what you want for some rep. Then the structure constants are gixed and you are stuck with constants Ffor other reps which can nolonger beadjusted! 7.Fortheadjoint representation, thematrix wecalled g,, isprobably themetric tensor oftheRie Algebra (Killing form). Weare claiming that wecanfind abasis in whichitisproportional totheidentity matrix. Obviously, groupmstbesmmi-simple © sor tnis to works re. ’[Leos eaeee Raeresaad,nding wapSolu. cial 2 a an g nr HeLECY=PSH. amdteLRLOGY =nPhen, soe —3- Conclusion: Lie Algebra can always bearranged sothat the following istrie: ie) Cc,al-4CapeGy«Wherecarerealandantisymmetric |trace(GyGp)=F8, wee. ea ¢ trace(G,[G.6,]) «iP cay Perea YG,Heme Here¢isarepresentation label andIimagine thegenerators to.bealwaysHermitian sothegroup transformations areunitary. Inthe special case that the generator matrices ofsomerephappentobepureimaginary (andhenceantisymmetric tobeHermitian), the last result above simplifies tothis: pe 6 fo trace(G,G,6,)=i4F Cabe Thisisthecasefortheadjoint representation wherein onehas: (Gayo ste. ce. add one. C3 1.tngenerat, suppose youconsider thistracet ¢eg. Re g . from(EPG, CF)=Dive [BasincedYEonYoun Atthe moment, itseems tomethat Disarepresentation-dependent, totally symmetric tensor. You can show its total symmetry byplaying with the cyclicness oftrace and nothing more. With the commutator inthere, you could reduce the thing and get a tensor that was rep~independent, namely, the structure constants. But here you seem toget adifferent symmetric tensor for each representation. Maybe the tensor Disthe same for all representations, but for now Iwill assume itisdifferent. Ihave afeeling that inSU(2) theDtensor iszero forallreps; forSU(3) I know itisnotzeroforthe3rep;itisthe45thing. 2.Given the above result and the other one with the commutator, you can conclude that: dd of ¢tron||GGeGeh =LAFCave++Dans <ots eats Riad 3.Another misc. result: define the Casimir tobethe sym ofthe squares ofthegens inany particular rep. Then let N=number ofgenerators ingroup. You have: nre ¢ ca too|ZAG) |=hoe[easy]=ee = e za ue =. [Pe walO] a CeAY f aG) gy 4 . Fe =u, Ca TPeaay Poa B=4iQn)- GH) Favre, GaToLbySen.Cn aVER, G=Ts4by Som.Ova, 2N)~~ fo)T=NF Use=daindep N= gprs. ATrace Theorem for the Rotation Group Lie Algebra. 01,First, recall thefactthattheClebsch-Gordon coefficients bringthedirect product representation into block form. This also applies therefore tothe direct-product. generators. Thus wesay: Wey aie Lat QM -CR EC Ser Yaugly Sy7 ok, LOY mma =SaCaneWearNee|Creat, ye by .oon Camom=SyemmlT hy=dosing, ewShawe. Now consider anarbitrary product ofdirect product generators put into atrace. Eg, Wey Won ~' <Iwoe PY] =ea[LES SHC] =e[EB] vr.v ves fe)=ALBoe Gow SepGwe] =51,WoeLEE]TK TY ow Ifwenow label the space ofthe operators onthe trace, weget this result: os Gen)» a tec (UR) =Bate || 2.Asthe second part ofthe theorem, wecan write out the direct product trace as follows: . WL |= am AM)20 QB)de WeLUGateddQeeteHGH HD)=ASR) ATO) od +RGseR)- ah 4OF Ifyoutrace only three objects, only twoofthe8terms survive because theJyare traceless inany rep because any group rep isassumed tobeunimodular. For four or more objects, you will pick uplots ofcross terms inanobvious way. Sokeep to3: Gey . Qo lo) face PUK. =Byer)beaseWI | . oy ,+H) wa) WK] -2- 3.“ssatestofthetheorem, consider thefollowing: omevan! (TTI) =Fee seFleF3Gn@ny. GraaegardeoWarShimaneangguikQa ;Ba) ers) eaeP™+yeyiCae=So.iPeaene Bayh sGayPe=is Tosee ifthis isreally true, dothe j-sum using notes inSeries section ofMath Binder: AGP) =Zyywaqyny =21Ug+3yy] CABSDV= GA) AbeARe) odde‘nueqa Fam=ery)Myer«GenJ re) . fo) add 1:generalization oftheorem toSU(3) and beyond. (eo)1.Foranygroup,thereisalwaysgoingtobeaunitaryClebshmatrixwhichbringsyour direct product toblock form, The reps are now labelled conventionally bytheir dimensionality. Generators are always traceless and Hermitian. The "rows" within a repwill require more than oneleable, Ithink itstwo forSU(3), eg. Part Iofthe theorem should gothrough with noproblems whatsoever. Ofcourse the sum isonly over those reps inthe direct product ofyour starting reps. The second part ofthe theorem isextremely general, applies for any operators (traceless) indirect product ofany two spaces. Thus here ismytheorem: (ren) 28 .ho. Tot [GG] #SW LOO. ,Denon Crow) ey oY TMM LEGG] =mWLose] +KLE 64] (2.Application 1A!Consider theconmutator-trace relation andseewhatyouget: 5.onay \e)BOP oP ak Renom nsx\\: Aggy m8)dogkR=Ram aFeo Congk) 7 . * & 3 2 Ast set dbgh Fo=3h +3F ¥e Fa 3 aBIGP =\oe (FP= Qa,wb|S \QeY neywen) so(FPoF)yevntitosd z a . 2 We | FE=6F* ba Qe need, pen? a FeGoutiwiny A(RRYS FAO tEE = Fee, : x .Whi prayer Tha 8.Qaba Qaitfladk SLdLadscgk What this says isthat wehave a“building-up" principle here: one you scale your overallalbegra(andthusfixF*),alltheotherF'sfortheotherrppresentations ©) can bedetermined byapplying the above iterative formula, I'11 just bethis istrue. add 2: application #2 fe)3.NowconsidertheD-tensorsandseewhatthereistosee: V9aD ey vousangeGost ALELGG =Dae =TSaul Qrrown Dron,yd: By ~ es 4_Vi=aVasmeDae Freres Dav Me. 4 + Gavad, Dae=o. Wodo302 g 2 at e") or De= SDar3 Dee wa (D'= DEC] @) a oa=30-39 --d =o. C7)Tssarees withwhatTknows fortheregular repvhereG'saregtiven bystructureconstants, Iknowthat tr(G,,G_) =totally antisym, proportional toC,,4 with no Dpart. This result generalized quickly toanySU(n). The adjoint reporregular repisalways then’-1 repwhich appears innant=(n°-1)+1. Youwillalways findthat D,fecdoint rep)20 anygroup 4.More application: there isnodoubt abuilding upprinciple here also. For example, look at this case: * 6 oS 2ES 3@3~ 603 = Dat Dac =3D+3aD & 2 2 \ 3sDac=7Dac /Cons.sopDaceEdan I'LL bet myhat that you cen "build up" your D's inthis way for all the UIR's. This means that allDtensors areproportional tothe onefor thefundamental representation! Thus we can remove the tensors and claim this: ma ciwen v_Deaesees }a . fo)ioe=‘dee |D=ke&seyGRRMuwaleeriwJocksnap| s,A*eS i fee V=wO+emd Yolucug-I add3: still more [o) i=rep 5.ForSU(2), weknow byplaying with thePauli matrices that D’ =0.Thus, asyoubuild uptothehigher reps, youkeep getting zero. There isnosymmetric tensor toworry about inSU(2), andthere isonly oneinSU(3) oranyhigher group (Ithink). 6.Therefore, wearrive atsome nice results: @) : ehAe”(Ge16,62)| oy eS) | |WOCG016,68)=WPdnc | ™ Zk * ”i 2 W666 =£Poted ay | re) wy «FELT4° ° 3% x 3 * mE g 3 ° ForSU(2) ofcourse alltheD)arezero,andFY=(1/3)3(J+1)(2j+1) asfound onearlier sheet. Ihave checked the lowest few toverify that the building up principle doesinfactbuild upthePYcorrectly!!! . Arelation between theadjoint representation andanarbitrary representation. 1.Recall from SU(2) theory that a"vector operator" canbedefind intwoequivalent ways, oneisglobal and oneinfinitesimal: QO ry =\ Gs,(WM =beeWe > RURS ROW Gorivatigns boorbevhay opny re .aa eo par CN) * TEN sie Ne USVI’) =TU AN This, then, isthe basic true fact about an"adjoint-vector operator" ofany kind. 2.Inparticular, thegenerators themselves (inanyrepresentation youchoose) comprise an“adjoint-vector operator". Thus wecan say? Fr.)Wri dssigusdrvectn ote, coe ° io}FWaspaboef: 0)TGINe |=iCaceNe | eye ie CR) oe |eyWVU =Uw Avg awOVE =VAG Sons4 NeomthabbeyGoheedaegd—Verve =VETER east ay \@) Ry =POUa[US=fete lavev’| This isavery powerful statement because: 1)itrelatestheadjointreptoan.arbitraryrep co)2)itisaglobal statement, notjust infinitesimal. . (ade \ ovYates: GRD eo=TETee[G16] =Carve Va a oi3.Iwaswondering whatyoucouldlearnfromthesecond-order comparison ofthis ."last equation. After tedious calculation, here iswhat Igot: ¢ .(BO 2.9.81 iVreweh=i(LCase Aace¥CreeAdee crYS =Qabyed War —Nobche CaeCodess, Upon examination, you find that both tensors handQareantisym under exchange ofc,d: thus, you learn nothing atallhere! This second-order result isautomatitally true foranytensors cys, anddj, aslongastheyhavetheright symmetry. Itisamising towatch howthis equation elaborately arranges tobetrue without yielding upany information! Atfirst Ithought itwas ghing togive some relation between the d-tensor andthec-tensor, butnodeal. Probably these tensors areindependent objects! And,nodoubthigherordertermswillbezero(ie,equationwillwork)againby oe)symmetry: youwill inthird order have asymmetric rank-3 tensor 0,0,0, times somethi which has each term antisym insome pair soyou will get zero. Of course the left-hand side is not zero. 4,Iadd this merely asanaside: The Jacobi Identity says this: Hevea*LabePeas,tePadx*ed=o =heped *Mocyad *¥easna 5, Idon't knowofanywaytoexpress thetensor hy,.qintermsofotherobjects.I usedtothink youcould write itintheterms ofg,,€,q etc, butnowIdont think that iscorrect. Only forSU(2) doyougetthat simple relation. Note that theobject isnot adirect-product representation ofthe Casimir min the adjoint rep. That's because youhavetoremember thatG°%.g®x1+1xG” andyoudontgetG°xG inthisway,. Once Itried toget anexpression for hinthis way, but Igot nothing. . :ves fe) BL,yorSwenonduroUnduegreg Loqsaw, Wyma iCarCS| : Cot), oy|ConyCemy=gasFS. = [Sa by de= to= Troe. aryCony=Y + 0G PsdisteaBisSUG)NoosedonQueaswee eeeee OeGlinTEE RIOT.Rtsedgisth angenVorCoil, | a _ iegSnt*=g OMe, east dhtedBeakTs8aadoneetTo23 ———— . ---8 .. ts Sear Gy ewWHET coEe OND EGEASHy2 ATL wee Oey Jp806)ga SES WaseiGaaai ince eqn] 08) chm. vody/0ee)) Properties of the Fundamental Representation, Se O31. consider atieAlgebra forwhichN=numberofgenerators, andny=dimension of one ofthe fundametnal representations. Consider only those algebras where: ag. = Ithink this includes all the SU(m) albegras. Maybe more. Now, examine the fundamental representation which considts ofnpxne matrices. Assume these areHermitian and traceless. [Yes, that means group rep isspecial and unitary, soIamtalking about general SU(m) case]. Consider this: {GG =WacGe +Qub where wearetalking only about the fund rep. Easy toshow that theanticom isalso Hermitian, thusithasonlyn°realparameters, not2n*. Thismeans youcanwrite it aslinear combination ofthen°basic matrices, n°-1ofwhich aretheganerators which aretraceless, theother istheidentity matrix. Sofarweknow nothing about Hay. andQ,y+ However, take trace togets ms 3 O aha-aw 3(arGE}& Moat,SayecrtoyGaDundohak a x ny ge Ddaa=HaePSa=FOHava tf gene 5=WassOUT) dar Dw4Wave=\Feoboe eT J S:aatA)4a+LeS G5Qe Fe abe Te my ‘be Whoo =podogAWE)=>Solna) = auGyHO, OFRosesxTGs daGe$38 4& GaGeeEA50Drawt= Lda+2Bsa rf us = . FaSo@\, Deo,Fet3UCase} =(EENBo= ths Qk Gusto, so [oth =288. U7 oa 2. a.ndSWonGanda , LG,VnGs=[LGa,64], bef+TLGe6e),Gf [gpnvedeam “WtosJakhcosoaGrandvarSudagfate ~ o Gals, BedeGe)=Gyitare EGe,GhteiCrceEG,Oh -¥daeANCadGr=KoreGPaG+44)ANCoeGaFdaGe++2) A Tarbaa,ahFTa ArceactSAP=Cadede®Sat+CavedetSat re)=dkeeCard=Caveeed+Cacedebd = ViceCdce=eteCave*ArseCace >iceCalexdeseCate©AudeCece=o => CaveAde +CaceAue*Cadeduce = a— — This, then, is a/Jacobi-like identity which applies toany SU(m) group. aed Tas,bh+advo=0 Ie I'm not sure ofthe symmetry shown with the ?added. Anyway, this identity appears as(17.61)inGas.Nowtheideaisthatyoucancreateanew“adjoint-vector_" operator: ie) — ¢ve ee 7youl) DxontdacGe >[G,Di)= ica GmNaa a Nqewaan |Orsaloneoctal: -3- Thus inany representation there are two vector operators you can construct; one aisthegenerators, theotheristhiscombination ofthegenerators. Thus;whenyou doWigner Eckart you get two coefficients instead ofone when you consider the matrix element ofaphysics adjoint-vector operator. These are the famous dand f coupling, anditseems that youwill have them foranySU(m) group. Idon't want toget off onWigern eck right now. Probably this isthe same asthe adjoint appearing twice when you take adjxadj « le) lo) [dA] -detM , Fe S279 Géntents ofthissectiononInvariance oftheProduct_faA] detitfa] .(gupSeidoly) . -1 C1,Basic setups. foshowthat dJ=Tr(D (dg/dA) )anddirkettl]= tr(aMM™")and how you want these two sum gozero toshow desired invatiance ofproduct result. ‘Third pageonearlier paradoxes. Basically, detMdiffers fromtheinvariant FPobject. Also, detM isnotinvariant byitself even under free guage transformations. 2.ProofofTheoremintheCovariantGauge. | A.expression ford[1n(detM)] obtained { B.expression for dJobtained C.verification that they add tozero (Jacobi required here, 0factored out) D.astart atshowing the application which appears inFrakin and Tyutin 3.Proof ofTheorem for General Gauge A.expression for dJobtained, all-matrix notation B.same expression obtained ford[in(detM)], except minus sign sothey addtozero. Ji,Summary andRecap ofGeneral Gauge Theorem Proof. Byequation: 1. General notation oe} (1.1)showswhatTmeans(complete Transpose) Y 1.2) definition of $ 1.3) definition ofR 1.4) general. expression forM(1.5) relation DtoD 2. Calculation of dJ 2.1) the condition of constraint 2.2) differentiation of(2.1) with respect fofield A 2.3) definition ofscalar operator F 2.4) solve (2.2) togetexpression forSor(dg/dA) imposed byconstraint 2.5) dJ final result 2.6) general formula for dJ, used toget (2.5) 3.Calculation ofd[1n(detM)] 3.1) expresseion for dM bydirect variation 3.2) crucial Jacobi/[Dg] Identity 3.3) result for dM with identity inserted 3.4)generalformulaforaLin(aett) 3.5) explicit final result for d[In(detM)], cancels with (2.5) above (3.6) final conclusion ofinvariance. a 2 2. .. 2+Field Jacobian forinfinitesimal, constrained gauge transformation. Here aretwoCC generalfacts,validinanygaugeandforanygaugetransformation: ‘7 -t p*Ajne ~Ane —yDisOy ” || =~4CoreSwCAI .NAc Trosfar youAiendvate cy aLis\Yep[Ave|=Se8eltl —7CoSptel +glLag, VAue = This 4sofcourse through first order inthe parameters, weare doing aninfinitesimal transformation here. TheJacobian for going from dA® todAasintegration variables insome functional integral isgiven bythe full determinant ofthis object above. Ie, you want totie together the space indieces, Lorentiz indices, and color indiees. Consider forthemoment thattheobject ontheleftabove isamatrix ysin W= A-Q, @awed trQaAABa(l-@) ALGO) ~he awk=© =e -@ =e «*«\-d@, Therefore, thedeviation ofourdesired Jacobian from unity --computed tofirst order inthegroup parameters —isgiven byminus thetrace ofthequantity incurly brackets above. When you trace, you tieatoc which kills the first term because structure constants areantisymmetric. Thus wehave this general result: Yas\= Y-taay Qualifications: vyo. 1)infinitesimal gaugetransonly| YeA-h4}(Dl oe 2)constrained or:unconstrained i y Mt LV 3)valid inanygauge — 4)obviously J=1 ifnoconstraint. Basic Setup toExamine possible invariante ofcombination [dA] deté[A] CB1.singworkalready done,hereiswhatIknows v 5Ue)-Tha ve1$4SA “ ©Saw) =te[Sted]. s(antag)dak Now use these facts toget the first order expected shift inthe product ofthe two Jacobian objects: [aNskALAS] =(AALS: [ask+5cesta =(AALS. skMCA) CL&8G4Laan))) v ~ =a)akals \\~arlo alll+1{sm Sinceeachofthesetracesisfirstorderintheparameters, thefirst-order condition ofor invariance is this: ne ee 7 [St+Wun)=0| ~gRCOTR)+R(T) = jn . peut _ a]4{Gxartye] -*walle =o <>VAALARMLAL =Gruaniad vaJinstondan. This applies toconstrained ornonconstrained, infinistemisl, any gauge you want, sofar not specified. a Itisveryobviousthatifyouhaveanunconstrained transformation, Cene-dA-Jabocian change vanishes becuase @=constants, independent ofA.Thus, the second term in that trace vanishes. Iam not at all convinced that the first term even should vanish .Do I expect ittovanish for anunconstrained gauge transformation, ornot? Remember, this isnot the FPobject, itisthe Jacobial object. Apriori, Ihave noidea how this object changes asyou gauge-transform off the gauge surface. Have Iever before claimed that detM[A] should byitself beinvariant asyou move. off the gauge surface inan-unconstrained transformation??? The only place Imay have isinthe business of"extracting the group volume". But look atthat: asinlee, you extract the group volume while minatinting "the gauge surface delta function. Ie,youalways have afatd(f-a) sitting there. Thus, +inthe extraction operation, you have the FPobject, not the detM sitting inthere. Ie,you never putindetM[A] until after you have performed theextraction!!! Thus, during the extraction process, you never dealt with invariance ofthe object detM !!! So, i£isprobably not invariant ! This then answers part ofanold question: Iwas never able toshow that trace[ dmwy vanishes foranunconstrained transformation. Thatalways bothered me. CFNowIseethatitisnotzero,andisnotexprected tobezero!ThatiswhyIwasnever able toprove it!!! ~ le} (An)eateatation ofvariationintheJacobianobjectforconstrainedgaugetransformation, CyGoveriant gaugeonly,A-fields only. 1.Here iswhat wewant tocompute, and here isalso what weknow to start with: SSW) hoseLN88 IMs)=-2gq]a) ~ 3 = oR, ‘MaGer)=~5.|DELaw]Se) 3WMMaGs=SalYate]se Ve Qa, _:NMaGos) =2a|ERMElSe VAC@) + VAG) ro) =+Sn\-*CaveSeSQe)fosy| Maco) L ° = RG =Caedy[ease 3 Ava, BR@=-[Daesuc) Quer, cs vSMaGay)=SeeCareOvSeYTES|(saw roo) =Eu ave[po «eae v %)7Ge[Phoaels: Ses] .+(ar[Daweae)])Bong QO cy » SMecos)sosQeBv|@a®ea©)Seal Thus, wegetthe following formula fordetM's variation: ‘o) FLO T]=Lacey[BMaGosy ILMG0} ay a=\arey,[-sCaveY{Damaelse-s |Litos9| oO> ~\eay ehCave) [DUBE] BG) BWMbe(4 ® 3 y .=ShaySees)}\yM,9)|‘eyCae)Ldaores} ~ v Cal [GriGayDserescs)] =pLeryer’)| Sas 2 ©a Sab) One:as) 4CanTR|Meaao9)(Daaey369) nn wenTRLS@SY=NanaySoy)$04), | 2. —ee This result isvery close toFrad/Tyutin thing. This useofsymbol "TR" ishelpful because “ft!1etsyowavoty unpleasant barnotation toindicate derivative ofMt,Thisreally isthesimplestiesayit:ie,thisiswhatanyothernotationwouldreallymean. ;TSRC}kal 6 SseLieve re)hl Sede ob vos -3- Iamhopeful that hoped-for result may betrue for all 0,insome sense. Therefore I .ammotivated to“parts off" assuggested byFT tbfree upthe parameters sothey haveoOnotderivatives actingonthem.ThusIget: 8 aor-+Caray Fess(a“tiacs)| Dic)ae) ubeQoreeMevseide4gortsey +Wksw Ss: ad ®&Oe Baden] =—CarYrsoadDi[se(s‘ntos)| Calculation ofthe variation ofthe Field Jacobian for constrained gauge transformation. Covariant gauge only. A-fields only. oO1.Myearlierpageonthissubjectisstillcorrect.ThereIshowedthisgeneralresult valid for any gauge: =-i\ Ss. Yeacy_ byDanas, [Deodeos)|(SWee) easadeah90/94: S“MasQosy S604)dy=Ina) =\ZWitlteos)Sey)ay+MG) Vee =Via& “2 sy 4)YS) Vha(x) xe) s VAS@)*SAVE) Cae3)”[BorasThe-35) ie)FokJew CaveSdySuly)iBonyQeSeay4TOL5)[yyFoes} 9)=cohol U)\S02)8g)+JayeresSens)BMoh =CreSoePPeel*IwSoe Sve) =Car3Lsa-Vere} ~Cre3S?[sed 8@-2)) =Cae Laydos| ° Ye Vhe SousghCaeBye)+UtaSal=|et fo) Vo vay ~Gm +& ‘ ~).. Layee 2) =7CeLes os ae a Pe} The astFe]=WG]+QeBest Watsoa.agateacta: 82)yk. ‘ (te\sel-Cw“iwai)(Gls&,) We)- WIL] «CS) Dee) PoeLh == = oteagers Ooo videod -\Vha) No@) Bo. =\e‘Mad(4.2)esaoloathe’[aoie-s]| Geohranh, orks muss Veco. VeMeas.) Wha)VAL@ VAR® -\ve[ewer Gay]TacOv@Sa-x) \o] 3 Forthemoment,letsforgetabouttheh-term,Pretendh=constant, asinLeeZinn. fo}Then wehave only the second term which is: Sus) =~CoteOn)[307Wey(|VAG) Doe, AY=42Ga\nasOF(Va®ro)(3WHC) Oo4a 4 : ba > oa _ SY=&Caa\i (tee oo |=~£CaeYdr@(OH)(x3))(QfaweG) fe) Oo Proof of product invariance for the Lee case, covariant a CF1.Sofarwehavecomputed thevariation intheseparate pieces,nowweneedmerely" add them together. Here iswhat Iget: SLdw(aak De ®vant Deal = 8ay vIEXCOIN\N xCaseandewDaw[Sey(CaPam(9)Q) Wo- wyConeVaBxvyS ay ey Sade ydday ®) S@)eH oyWe(4.x) Idohave aparticular @inmind, but itturns out you can show that these two things sumtozeroforarbitrary parameters 0.[Thefactofconstraint has,ofcourse, already been used inthe derivation ofthe dJexpression. Constraint was not used toget the other one]. Inother words, itwill bepossible toshow this: aA xt CaehayDie[Ses(BhWeaCyd)] ah * ago0+CareSayDs@Scey)\|9Wika(0)=Oo: @) Ifequation (3)canbeshown, thenforsure (1)+(2)=0 foranyfunction 0since, inthe xintegration, @multiples anidentical zero function!!! First, examing theA-independent terms in(3), ie,theterms inD'swhich just involve derivative d.Youseethat there areessentially three terms. They are: ” balCasey itSeeai]Qe) ~~+CarYay[Seg] FEWAG) A ~ +CateSay(BRTeesy\84). 7 The first andthird terms exactly cancel, leaving only the second term which is: Zl+CotagSeg[ETEGs.) @) ro)Now consider the other remianing tems, the ones involving A's. From (3)weget atonce: s+ : OS \, -2- ~VWCaeCace +Care « *oni6laneCaee+CateCove|MOJagSens)L3EWELG0} Luckily these C's are inthe right form tocombine via Jacobi thus: QuveCede +CadeCree =thaed +Radve =-head =—Conc Core » ~>+4CrayCopyKooaySexy [3sas|a * ec a ~=¥Cava CoeACSaySoe)LAFMAG] Gs) Itwill now beshown that the sum ofQe's (4) +(5) equals zero and that therefore equation (3)istrue. Adding (4)+(5) wegets as Aya CashdaySG)[(hwec.s)~4GochLyeG|(6) ie) However, wecan compute the box object asfollows. Recall from Lee-paperl how the transpose business worked: So ren aYeWes)=|;4GSass)fomGs)=BacBry) Lit yeat >tsQoxJee) =Sn8e3) The reason this istrue isbest seen inspacematrix notation; also, recall how LEEZ used this alot. Note that There means transpose inboth space and color, sowehave: e\os ¢+ney) Wes(4)=See804) i) Lyn pa(5%Daw)MeeGs)=SacS(x-4) @ \e}Thus,wehaverulesforapplyingderivatives toeitherthefirstorsecondarguement ofthe1Greens fimetion. ; 3. Inparticular, write out(7)inmore detail:: \e]/a~\5.TyAPG)NE :2IEMEG] +CopeAeBeWares] =BehSos)an Y * : ~ a” ®=|TeMay =-388B64)+yCocoNe®19‘fee(s39\ (a) Ifwenowstuff thisresult (9)backintoequation (6)wefinally getournullresult: (6)=)Casaates) \sabeSeesy\, =nyCoad Se) =©leyoaler \ ve fo)Thuswehavefinallyshowthattheproduct[dA]detM[A] isinvariant infirstorder asyoumoveawayffomthegauge surface, provided youconstfain thewayLeeZinnsdo Q!!! Thisthenyields their nicederivation ofthealpha-Lendau gauge. a 8° / 7 1 Modification foruse inTuyin thing. + ‘e]1.MyresultsarefindforproofoftheLeeresult(covariantgauge),buttheydon'tquiteapplytotheTyutin thingbecause theirDisayN?transpose, soThavetomae@ chenge inmy’computation ofthevariation inJ.Theother piece stays thesame. 2.Soafter some fiddling here iswhat Ifind: Na7™ “1 = SayMabe) Wes)=Sashats)Wve(x) = on&=&“(wy .aE =\esBUX)WMealy,.x) =hak) \o4VousWhals.»)+\a4@)Vitaly) =WheeVA@ VAL) VAL) Ws .oo re) (cirrmudrJesse)» Cae Orta)Boa) TO OT Tyan ames|=FYOalsy =+CQaae 8.00)MaeCos)1VAC , \ :| \3Vha@y aq 1+ \ A) Maa (2 Sn) MayRe) MAO) | Ifyoucompare, youcanseethat this isvery close totheprevious Lee~mode result. You can almost connect them byreplacing Mwith its complete transpose everywhere. But, asyoucanseeinthefirst term above, there isalso aparts change. which Ihadknow waytoknow about without doing this calculation. Ifyoutake this result anddoan x-parts with thefirst term (without justification since nox-integral here), andif youreplace Mbyitscomplete transpose, youduplicate theprevious result exactly. v Teyworhal@\= —k-(2)Au@) . * : . RoseySQascelm inbg= k\9,Mexos) Ithen gets. : > ~\ s\=. ea we), &®We4g weeefae(FOO) MLCor)=eBE?ates(53) Isuspect that ifyou played with this you would get the same dJ asinthe mther problem, and they would then cancel inthe same way. But here there isstill alot ofwork to do. You have toremove all derivatives from 9's. Soyou have toparts out the fist term. That isnoreal problem. But the second term above does not contain 9explicitly, soyou have togo"find it"using sométhing like this: ” . - - . a” ®t SanRoo) Lee aeGa)) =ety). Idon't quite see how its going towork, but the thing isthere infull detail in Tyutin paper. Iwill accept their result. Mygoal was really toshow the thing for Lee only, where itisbasic towhat they are trying todo. The general gauge case, Lee mode. lo)1.SincewenowhaveanewandmoregeneralobjectM,wehavetowriteit,thencompute dM/dA togetanewmatrix equation forobject dg/da. Thisisneeded onthewayto computing the delta-J variation. Solets dothat for awhile: ™.The dJ-variation. First, here iswhat weknow: Wel=-+\Vh)Ip| mel TES OR => MaGasy =-2\skVERO PSay Se°5SIWRC)[m°>| > ; _, ;Wren) 2-Ae(wo \[be(e)84-3) 4 VAG) 30 NA@ wie ° -s\*(HQ)|WEGSes] 4ayWWAaw FAL i tt ne)93,CaBuoFae) ft= Ca PYRO 3-2)ten Itmight bebetter todoparts right atthe start indefinition ofMasfollows: <i “WeGos)=+abe()vo|VAs) VWabs) |(gojua)9A.OG)TSG)VAelay 3LA YAMS) VAIVALR) fe) +4CareSofOy-2) > = ®=CureVEOSGy- aDay ZYVEOabewae(-2)+5Da4)AimyVALE) Lo _— Sfe) teccvee vb Vita) =Care 3-2) VEM piDaw YY VALS Ye 3 YAS) Ve) —OY YE.)= Vara) +b BaDeeg) Sefeo.SDEOPKSTRG)TRO This isnice notation because now everything acts onthe gauge surface function which iswayover ontheright. Just forfun, here issome fancy notation that Imight consider using ifitisconveneient: .. x hasGes)=+40 VRe3 VAAL)ae a=4\H[Bio Mes], PEG | | YA) ie) YH 4 _ok wor) =RG») ‘ ant is x = Mud) =aye DoGt)VaGx) ov a ry =(Me) =2DalLR] xY-=~|WwWi- 2pix = +Sz Notice how this object Risdefined (it agrees exactly with Fradkins). Both the color and space indices are reversed from the way Iusually use this thing. Soroughly you .could sayR=(df/dA)" incomplete sense. I'mnotyetconvinced this notation helps. Nowlets goontocompute thedg/dA object: oo OO = °JayWaaAG)=Ie=caashadk > YMG) Sue) +S Wah,s) DEy) =oO.v4VAR) 8)oe °YALE) ) a) SaleWeVEO. “Sige) refie)——Yo™6){cass»peda)7aYAM(2) =~CareWeasSGy-2)Ran) -28 outs)Busy QLaxyNy1S)9)SR)de(2%) = ig v ©(tess)~~CmtanTRE (eoYes > = vKR~ -~ 1d yyDoel) 2 5‘rN(aewyon) (Rae Nowletsdefine anobject likeRfordg/daderivative. Dothis: Si(ay)=_VSsty) VAC(2) teYoMGs)DoW=LaySn)MEG) oh(x4) - cb(@0 Noa(4,x *Sateaie) ° rare vat ‘ y » tsehta~~Cuseaiey f~ oS v = ‘ v q so —Y4- BbOR. yNo. »vi. \ RLT vSire]~-tent)Ig)-4{Orpstioge, 4 y ba hyeTe mos x 2UZl=eating -41Geokod evar Vijee: SN S-3(\ogDutZe)=F=omc s("3 Ae) me Gran ¥. Sv.=ey, +.(Sd=-toaLVL bt)+(Fgh) fo)This,then,isoursolution fordg/DA.NoticethatFissomekindofscalaroperator. Iguess Icould represent itasacolor/space matrix, butI'll leame itasisbecause TIhope eventually itwaill cancel against asimilar object. For covariant gauge this second term vanishes. Let's pause tocheck that the first term agrees with what Igotbefore’ inthecovariant case(aconsistency check). It:certainly looks thesame: Onval se: Lw= 9Xo >VYu®.SapyS-5)\-Raa(5%)VAL) Y ‘s) =Rad ==Ban[”BE]. :.; 67)1Gt=~1aLat Jossoninkgouag ~*~ . ‘ °[she sidCtBapat) a oe y Ps O>SuGay= ha(Led (rian)Jay ~\ =CoaSokexval>? Bory] Was) “\ =CecaSakCoedTG) 3%an,Cay) “ ~CeedOG xWeGs) =CredO6@iwotha|: =—ChaOve)i?Wacy] =YEG) 4 VRC®) OryJomWashed, -\ fe) Soon) =ChaCoesdyeWeay~\ =GaoeWWE) 7 So,inthecovariant casethisdoes duplicate ourformer result, Iamglad. tosey. Now wehave ageneral forma for dJasfollows: v we-a (ax DAG) Vals)*Vy )DAY) v ~2(by GrySas) v ne”=~3Yee DeaGrySee(yx) . v v v. v=-$+UBUSAL -4([BITS fo}= 3 Ff vt=SPORTS) akACADauto. toe 4 o8 wr. - v > roy. VDaGs)= Dee) Mey) =-LoSq-*) Poxa-%yak eo SY=Vas)=~Dalyx)) ~ =~BuG) 86-5 Waeat (igitgi) a =~ fe)So,usingourcompactcolor/space matrixexpression forSappearing onpagehywehave acomplete result for dJinany gauge. : ns we 1+(IpACSIESV GND) | (ID) taaeiey 6) “3¥(WIE) (¥)) ‘o) 3.Now compute the variation inthe FPdeterminant object./ Acomment isinorder here. fe)AgainwewillneedtheobjectdM(x,y)/dA(z). InthedJbusinessgustdiscussed, this entity had tobeintegrated against afunction ofvariable y,and for that reason we wrote the object out asay-z operator acting onanxzmatrix (see page 2above). The reason was that then the xzmatrix just sitgs there, impervious tothe y-integrations. Now however, tocompute dMyeare going totake the same object and integrate itaginast afunction ofz,mamely dA(z). This suggests tomethat Itry towrite this thing asay-z operator times anxymatrix. Here iswhat Igets ~> VilaGoa) yRay 71CmSdGprgDGVkVEOv< VA@) VAs)Nena y on, Raalye) ~ SihGe)=VeeSKE|Wises) ST LSA) \> .-4 os \:caDe@s3\ Iwill parts thisD"overright awaytoget: a SMawsy=+1\aeeeVeeTrae)\. $ VR 5 =(VEN tags) Here Isuddenly realize that Ihave that same invariant operator FIhad before. Towrite this last line Idon't need the thing atthe top ofthis page (yet). Solets goahead and compute the variation inIndetM: = -,\ Spresay =—Te( Fawmry trlod = But this does not seem tohelp very much, soI'11 goback and use top-of-page result: so a. ¢ > ” le) BMGs) = CasedeS50[Pec®)3@-3}]Rac(yr) => = v v-4Dae(CF)Rao] Lets deal first with the second term which Iwill now express as: y.=v Lat)=-.h(Eign)ews YinskYer, 7 eR x PronQecation do3Ldea(ast))| ca ay +—T. SBgayl= TF|aati) -7%Le iw Oissonce. SeCAD=ACTAI=ACAT). Der: va) aT, Baas -4{iz(tg)Ls4 (a) This exactly cancels thesecond term inEq. (&), aswehadhoped!!! Thus, weneed only consider the first term above. Notice that this same cancellation occurred inLee appendix. NowIcanforget about thefunctional differential operator F.Noneed to worry about exactly what itdoes, since terms’ cancelled. So: . a' A” . fwd ==CuLbsse)LR) =Jeree 2 4 . a ateSTs 5CuetafLoon4)Omef 0 Togetthe. above "first term" Ihadtoparts theDover onto the ©(which Ithink is where itwas intHe first pléce). This result here issimilar tobut different from thefirsttermin(1).Iknowwesomehow willneedtheJacobis. -9- fe)Now,letsmomentarilyconsiderjustthed”containedintheD”inourresult:8 n < font==Cae}BaolRC)(ast Notice that the£color sumdied, andweputinanewdelta d,4bybhand. Thereson for this issoIcan use mytheorem derived elsewhere ,namely: ” 8aese] =[Dalle] ~le\(da}== Nenad mydeme AR hewn 48, Iwill now show that term (1B) infact vanishes. Watch: =\A” clown(\8)=aComeafteaLDS)[Rae|Hd.)ia bee Pe = aD =Titlesy/ fo) =Coch? teSei} =Cathie =obyote Thus weareleft with term (1A) plus theterm intheresult onpast page that wehave not yet included. Adding these two terms here, weare left with: a Ay pat~5Cop's{TOSMagLSI &x es w >~4CaebkfyCaltign Lyi y Notice that this first term isvery close toour page 6equation (1)result, Eq. (1), but not quite. For example, inthe above thing, acolor index onRdoes not tie tothe C, whereas itdoes onpage 6. Solets expand the first term sowethan have three terms: =~Str} [CatSa(a'eo9|-4(Cure Kowd|WbLi 0=-% eh(Ze84|-4(CasotsCanCan)|08d|RR]La] ao a oe”comeWeLomasJacdsr cones -og CoteCale+CatzCte=-CraComm [ssyung" DhanSeoSyrenendedIecomsas \cayewy+4aaCinus|(gs\ x be ~Che\WHS,~4Cunlstl|ter a a 4=CrLy)eb=-[Ox]LesCa Qheewsox 5 o Ostainje +ah{WeedteSaRVR Sept] When(C113wei,6) Youseesinatlythatthiecancelsthefirsttereof(1),sowearenowdone!!IIIffr11tt! Inother words, **(2)+(3)+(1) =0andwehave shown the product invariance that we set out toshow, inany gauge, Lee mode. 6 fos Asummary. ofthegeneral gaugeinvariance theorem, ®Thenotationisextremelycomplicated, soIwilltrytopickoutthecritical \o; equations. First, here are the basic definitions and relations: 2)Twain Athomepeas T: FaGos)=FoaCax) [e230 Qa) b)cewsravds cboik ve\ =Vem . aN Ww -9Lela Sram)=(Shs=SaGesy=Sk(yx) Qeeroaraar| Wei lS) | ® OFSse eect dk al»VEO. Reka)=ReesTAIeTAGS) SSCnENTe ~ RY- vant Dreeeont ‘7 (els-w2)]se¥ IgAia = as a)Dnaob,ekWM —afehiyy) | AT a Wr-4Sel] | [Bl--(3]SO = 211 Us) ww) csnn oO,Ml-si2itg OS (yy ST ®rinthefield~jacobien variation,weneedfirsttoknowhowtoimplementthe condition of.constraint. Here ishow itwent: . [Mlig: =ta ok ietlwW ey |Gd=z A a a € Vypoiele dsgi LSMw) +toil CURT =Fig] =o - ~ ~ Qa 5 ~ twal-elt ey-Ral=o one |Fe~pt{CALElige} eo) = — Qa)=-2(\%5Soy)Vee)a5) Sasongy Dodou SeCopa)weeo |i)-{@lg)- wag wt]- ca ~ * Qa) Wal~1tFlSl +Atciey Aw onyr o|t=+ewhlAig-exrcity) es =47(OLPLbssLedcltet} eh) Inthis last wehave used the general formula for avariation dJ: O|oesaeits?) --26GAS) |ay @nowweturntothesecondpart,thevariation indetM.FirstIfoundthis: Wael=-(oF ie)-Chale) TTdiamangendadion gieusdhOsanced Lan: {salve =[DaWe)-eAldd] -SagCamAreal exe >Corc{SatlDkes\, =CaeLated-Cacla|Dal 0 -4(CatCen+CapeCet)Wases) >Cale] --Caleta) -WICatel « This isthe line that really makes itgo. Itisacombination ofmyone theorem which pulls apart d,9- ,andtheJacobi identity. Substituting this back inthe above you gett (anid=4LR)(Feel)+p{CacCoaWaysWACteatLed Bikwoth DRoamTssOk Laat--leatwdl -p(w{(Fig)-createll owcomaSramaadCRYaidaPSweSkeme: iGhtysl. -Wii¢led- tateiy) ay: ; 7 Inowseehowtowrite allthese things inmatrix notation. First, bydirect computation olofthevariation inMweget: tah-3WF) +eLeiiwelig’ Gay Then weinsert into these these identities, inorder: a re PondIie -WLC Lel -L&E ILE | Ss i sails (32) acs~sw) Ineertion ofthesethenyields, « sw] =-tedl CU) | ; = “= G3)ie) -pte){Figl-terteiteh Theformaforvariation ofthedetMthingis: |S[ge(stwy) =Te(Laelbat\)(Qa) Inserting the above wefind: SLnuse) =-Te(teat) vy,= .’ = (as) ~atk FLR’ -falc vi5©(CO)LIS)ASL ow1) Thefirsttracevanishesbycolorythesecondisequalandoppositeinsigntothe C@Jvariation. Thus wecan add toget: BT+&LinQa),=0 Ta]Laskm(ayh=Laatsat] remy