General Field Theory -Weak Interactions
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Binder of handwritten study notes by Phil Lucht, dated February 1978 to May 1979, with a cover listing General Field Theory, Weak Interactions, Math, Commins, and old and new field theory notes. The legible parts include an induction proof of the Feynman parameter identity and a worked Feynman integral from mass renormalization in scalar-boson QED, including UV and infrared limits. Other pages cover rotation and angular-variable integrals but are largely unreadable in the OCR.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
General Field Theory
Weak Interactions
Math
Commins
Old Field Theory Notes
New Field Theory Notes
Phil Lucht notes Feb 1978 -May 1979
MATH
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page 1
Atechnique for proving the Finkman Identity.
fe) 1.Hereiswhatwewanttoproveforanypositive integern: a
on3 =mn!\Ga (S,)"
ayaa °
where we have defined:
GNv 7 o~ Yegae TT(Ste)82-1) Sa=Zia
2.Iproposed aproof byinduction. Assume itistrue for n-l, then wmrikyt
verify that itistrue for n. We start inthis way:
“on mL ow ew at
od!VQ,(STEGen!TW(Yass)an)SBsitxty-) ° aseVST]ASay ONT
at -hn
x\Zat4QaKG
Sofar Ihave done nothing more or less than rename the last two integration
variables, Iwish to examine the dxdy integrations, holding all the other z's
(o) fixed. Iwanttodefinenewvariables toreplacexandylikeso:
kexty dud=$fdadk *eECs)=% yeECs)S=X %
OHO)= SHEM)SG49) =OCE+S) OC~)BCE)
a 8 2s 2S
was tes .aSTS Yad ayFQ)=Eyasyak $8, 2). hosas
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Baek the dtoff tothe left, and put twothetas into endpoints:
et ee am tan s*
=Gey!W(Qaa)\akS(t). Nas|=ardentlyinti] re) wrSe4S‘ et=r
so
page 2
Next, explicitly execute thedsintegration toget: cet
mast fe)zso.pe |Bawta(rssia @sy\/Em) L(derIn) i -t
‘ =nAl mt -n4l <.m4 i|Baas+4,-[Zaxat] =) n>An) m ‘
Wowtake thevariable tandrename it2, 60answer becomes:
° L <nay> aanahpa’ toJoot Sao,[Baal =[goaaraan] HarAo (Aran) 8 vay
2. Now what should Ido next? So far Ilike it because Ihave caused the
once-lowered (dz),_,integration toappear. Ihaveyettoprovethatthis
isequal tothe thing onthe left. But now lets use the fact that this thing
is supposed tobetrue for n-l. Then:
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ry 6° Py ~~ \ Jie,sensoeabet 2BAIae\= (x)!\@aen [Axe ‘aaeeeAneIn PABA np94% °
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<G@u-\ |\ 4.= . =| oe aeoAGMae Anan \G>4e)
Thus itisproved. Iftrue for nel, them mst betrue for n=2, and soon. But is
this proof valid for lown? Ishowed that (n-1) implies n. Must show true for
ne2 explicitly’:
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~\ =\[=\11i\Y1...3{al-Sf=KG: \=a A> &rvb boa ¢ab
Well the case ne? issame asany higher n,soIalready proved n=2 when Idid
the general induction proof. Now lets show for n=l:
eS AayHNa SQ) Cah =wl 4
°
And socompletes this induction proof ofthe finkman identity. There was no
need tosumover permutations orapything fancy like that.
Eyaluation ofaFeynman Integral #i.
(e) 1,Inthestudyofmassrénormalization inscalar-boson QED,yousoonericounter
this integral byconsidering the one photon loop onthe electron propagator:
T=Yka (A =Ig,%,hw) —KEX eK)=my Way
Hereyouseeaphoton propwithamassinserted, theelectron propatmyanda
factor put intoremove the logarithmic UVdivergence. For small k,itdoes not
appear tomethat you get anIRdivergence. But lets just treat this ananintegral
topractive on.
2.First, use Feynman parameters toget:
\ \ \
2T=2NSaag dayB(lesedees) -Qaheco)4¥(Gak we2p)a
so ‘. . . es -3 ~FER aT?
\ tog_
2 @=anaaa \vka° ° s ku . 3[itsaacpee=East} dened =e)N3]
Nowcomplete thesquare, then shift ktonewKs
4 ER.
=ataedyosSK
7 + xe rec os.(ays fated (rtp eaatert) KY
eee.
=A
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vane B=aeed(Mewe~ pS“Kh ~
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\ c+)
=NardaCatt(Sag-x-N) =TER)
Jon .
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IN4 Te4intEan)A[Gea’)aCe)=a;Sana= iiat
hast Ay=EGA) Elan a3:
day=E@C-I-N) Bh)20a, ND
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toget the UVbehavior: ‘
cs ie ws yt a=“V8Te a=New- TE=Ah
sagt$0 ae A) an =Te4h}Ean(ME)—OREO)\/*
. . 2 e 2 ©7 ORBAN) tf
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ee
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Thisisthecontrolling termforlarge|\..Thenextlowerterm iaisagonstant whichI-couldfindifIwanted.Ihavereplaced Toxpintdn(4) thep*scdlaingfactor.Youseethelogdivergence asr expected atthestart, andyouseethatnothing veryspecial
happens asyou gotothe mass shell,
Finally,letsholdJLfixedandgototheIRlimitwhereXO.Thenifindtht:°
VaanA=LOWS ah(eraety «oY*
=(ews ty2awe Latarws) +or)
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dy>[MEANAVee+(5)| =naiepeEE=H]
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a Thia\says thdtfoxtixea/N fn,
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7.Suppose wetake ourgeneral result first tothemass shell, holding |\.and\ finite.Notethetmassshellmeansm2=1,Thenwegets
cua=kite ~ we - a) \Rega =8,-\0@)
Os=-EXet}sope ="ae a mat)
Recompute these with care please:
aL
A= Ehee Chan’) ASV =Jae
» yt a Ss a=£LN=Aan) |=EAT Or)]
.
athhi-dire Ge)tte G4]
= ~& 1 a Va,= -\4ce=ah] Rte -itte a\h\+%
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:
Rot .
akJK=~adi~2e=arinXY=a
Ssjaxth)5 \Ba\=C@)LACE) suosigtoQa) Joy=24| eVdal—Cinta(id)
Soeven ifyou gotothe mass shell first, this integral isstill not IRsingular.
Somehow itdifferes from the true spinor-electron integral.
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Comming WeakInteractions |Y73, Reread Feb1978 -
o/ 1.Introduction. Briefhistory:-in 1932thebetadecayspectrum wasseento‘be continuous; this seemed toviolate energy conservation since reaction was
supposed toben-Ppe. Pauli suggested aneutrino, then Fermi made atheory
using thisideain1934. Fermi's lagrangian wasjustki,where J"issimple
vcurrent,for nucleon andj,isthesimple 4,current forelectron-neutrino.
Ofcourse this does not show V-A form, but itworked for beta decay.
There are many unstable particles. The ones which are stable against both
strong andEMdecay live 1071°seconds which isalong time; ie,awidth of
10ev.ifthere issomeEMdecay, likePI-2%{ thenexpect shorter life10716
sec,orwidth of1eV.Strong decay particles are107° secorwidth 100MeV.
In1958 Feynman and GellMann came upwith our present form ofthe lagrangian;
‘ recall bythewaythat itwas1953 when Gelllann Nishijima posulated thestrange
ness quantum number toexplain the low production rate ofsingle strange particles.
Thus the Feynman GellMann theory had strangness changing ourrents etc etc.
Infact, theweak current issum offour currents: electronic and muonio
.Surrents, anddS=0 anddS=1 hadronic currents. Each ofthese hadronic currents
has aVand Apiece, whereas the leptonic currents are each VA, Separate electron
@) 224-monconservation ignoted, endthehelicity story isnoted, Figure 1.4on‘Tyraecay makesthecamplete PalldCviolation ofweakinteractions veryclear.
Neutrinos are ofhelicity -Eonly; electrons are also, tothe extent you can
neglect their mass. Ie, left-handed neutrinos. Bythe way, semileptonic decay
means bothhadrons andleptons in-silvitheretate. Alsobytheway,the0%
paradox was that the Kdecays both into 2and 3pions which seems tobreak
parity conservation. Thus Lee and Yang. This was 1956. Feyn Gell was 1958.
Commins has huge table showing all process types. Blectron neutrino elastic
scattering has now been done Ithink. Muon decay isthe only .all-leptonic decay
process. Semi~leptanic includes beta decay, neutrino absorption (inverse beta decay),
pion decay, the pp process. Then you goto semi-leptonic stuff with strange
particles: kaon, sigma and lambda decays here. Also, the direct interaction of
the hadronic weak currents should effect npelastic reactions.
The dS=4Qrule applies toreactions which are strangeness-violeting and
which are semileptonic. The changes apply tothe hadron involved. This rule
isbuilt into the Cabibbo form,
_Theax\=#rule.Thisrulesaysthatwhenyouhaveareactionwitkahadron ©) —schanging/as{= 1,itsisospin mst change by+$unit. Again theCabibbo form
will incorporate this rule. .
Since K(long) goes into PI-PI, wehave GPviolation, hence Tvidolation.
This still amystery,,
Gh2: Muom Decay. Theamplitude Miswritten in(2.6). After squaring, you have
the produce oftwo traces Tland 12all ofwhich boils down to(2.26). Notice
theneutréno projection operator (2.22) isjust¢since nohelicity option, eo
whereas for‘lepton like electron you have mass and spin. Next thing you want to
doisintegrate over thé di-neutrino phase space weighted bysome momenta as
in(2.28) .Byfooling around youdeduce thetensor I,which iethiephase
space and you are left with (2.37). Finally you gotomuon rest frame with
the mon ‘polarized along the plus 2,décay rate is(2.40).
In(2,40) quantity€ isthe fraction ofmax ofelectron energy. You see that
_the degree 6ffront-back assynetry (confficient ofcosd) isenergy dependent as
plotted onnextpage.Sincethisformula ‘neglectes electron mass,youseethat
*pate is zero into electrons with positive helicity. ~ .
:
Next section shows how Lisnon Cinvariant, Then comes the piece about
trying toconstruct avery general amplitude, more general than V-A. You now
. have many constants Casin(2.449) which issame asconstants aandbof(2.45).
Théfinal decay rate ingeneral schewe is(2.53). The4parameters called 9,7,
Yana§aretheoretically givenfromtheconstants asshown. Thefunctions
g(x) and h(x), explicitly stated innext section, are the radiative :corrections.
What aré these Michel parameters? Clearly §controls thescaleofthe
fore-aft assymetry term,andis-1intheV-A-theory. Parameter ¢controls thee
*shape ofthe igh end ofthe electron energy spectrum inunpolarized decay (where
thecdsterm averages away andx=1roughly.) Then 4doesthe lowenergy end,
andfinally $idtheassymmetry shape parameter. Experiment shows that these
4parameters are all consistent with the V-A theory, see page 49.
Next subject isthe magnetic moment anomoly 3(g-2) ofthe muon. The first order
théoretical preidetion isjust{/2# “the same asfor thé electron, but the order
4”ana4?values aredifferent forelectron andmon.
The Garwin-Lederman method isthisf you stop mons end make them rotate
attheir Larmor frequency like a‘compass needle rotating. Using one ‘fixed detector
and noting the muon arrival time, the time distribution ofcounts ‘in the detector
tells you a)the gyromagnetic‘g, ie, the anomoly a,from the observed frequancy
ofthe decay distribution; b)the magnitude ofthe assymetry. Obviously ifthere
were nofront-back assymetry this would not work, This experiment has been used
tomeasure theassymetry shape factor §mentioned dove. AG-Lexpéiiemtm can
also beused togetBy, themagnetic moment ofthemon, Butthis type exper is
not good enough toget the anomoly because you need accurate muon mass beyond
ability tomeaure. Whathoppens. infactisthattheanonoly “ismeasured inanoth()
experiment very accurately, then thejust~mentioned fameasurement gives aacourate
massforthemuon, Theexperiment which getsap,isthefamous onethatuses
muons rotating inaring..
On a
: -2-
Finally, Gene notes that youozhmake long-lived’ muonium atoms ‘te. Aswith
hygrogen, there! issomehyperfine splitting Which youcancaloulaté‘in‘GED.Unlike fe) hydvogen, there isnbfhing unknown here(1ike préton shayie andsizeeffects)
Ofcourse you get the theoretical résiflt very accurately, and this isused as
ameasurement of%.Another measurement of4comes fromJosephson experiment.
Now{ismmeasuréd two ways and they agree to1ppm. Bythe way, Josephson
isthis: you put avoltage VonaJjunction, and you should see ‘anac signak
atfrequency W=26V/K . .
..
3‘Recall that the muon decay /N/* =64G" (peq)(p.q) .Byjust changing names
. around alittle yougetthevarious seattering andannihilation. processes, ie,
anything youcandowithtwoharrenta. . ‘
+.However, there -isaslight difference inthecioss sections. ForLy-process,
youfindthat theDCSis,onlyenergy. dependent, noangular. See(3.19). However,
forthe1% process there issoma angle dependence intheDOS. Reason igthat here
youaregoing through a.vector resonance andthis’picks outJ=0andJ=1,whereas
inthe$yprocess jouareexchanging thevectof andyougetthegontact effect
80 only J=0..
Obviously, with either ofthese elastic cross sections you.areintrouble’
©)_withunitarity because only1or2partial wavesinvolved andefergyiaupstairs.
Ofcourse this isjust first order perturbation theory, you might. hope that higher
orders would fix things up. But higher 4-point interactions just make things
worse. Evenwhenyouput.intheWboson, youstill violate ‘unitarity: insay
theseroth partial wave, although thetotal cross section goes to,a constant
instead ofblowing up. Theproblem with theWis that.the k"k” term in,the .
+ propagator-ruins your ability torenormalize. Sothe Wimproves unitarity a
little butdoes notdoitcompletely. Bytheway, TheWthedry givesBgwhich
differs alittle from’.75 oftheV-AFermi. theory. :
Chapter 4: Hadronio Weak Currents, ‘
Previous chapters were: introduction, mondecay, other purely: leptonic
weak processes. Forfirst time wenowlook into hadronic decays.
FirstsubjectisthefamousTyr_piondecay.Yourpiongoesintovacuum
_through ahadronic current: this current mst be A,type since axial andno
strangeness change. Whyaxial? Because only 4-vector around isq"which ispolar.
Thus matrix clement ofcurrent mustbepolar, butpion ispseudoscalar, so”current
operatormustbeaxialvector.Piondecayconstantf,isdefined.Youcan fe) termine this constant from thedecay rate ofpions through this mode (which,by
theway, isessentially 100% efcharged pion decays). *
.
Next, consider the,,decaymodel.Calculate insamewaywithsame
constant! Result differs from the[ys onlybymass factors, hence youknow
theratioasin(4.9);branchfraction isamere1074,Although electron has(#)more phase space, this mode issupresséd because youaretrying toforce electron
_into wrong helicity state (pion. hasnospin) ~
Nextcomes thedecays Ky,andKey.Compute inexact sameway,except
constant nowcalled fy.TheKy, modélxis nowonly 63%, andKpiswaydown
forséme reasoh asbefore, even more so.Sofrom rate into theK,,mode youcan
compute fy. Theratio offy/fq istand =.27,Cabibbo. later wewill learn
vthat you canallways associate sing with non-strahge and cos@ with strange.
changing orstrangeness violating reactions.
Nowmoveontoother decays. Consider those likeW->-w°E'Ve which is
called pion beta decay. Nowyouneed adifferent matrix element between two
pion, states. There arenow two4-vectérs available, both’ polar, socurrent
chosen will.be V,. For’each 4-Vector youwill have aform factor, usually
called £,andf_andusually defined with asin@ showing.
Digression: ifweak hadronic vector cirrent israisiig/lowering part of
isotopic cirrenttriplet whose third component istheI,piece of.theEMcurrent,
+(isovector t#iplet hypothesis), then what éan you say about pion beta decay?
-Firstofall,sinceEMcurrent isconserved, thesecond EMformfactor vahishes (>)
andaswewell know there isonly pion EMforma factor Called FiIniso-notation
“you canwrite EMcurrent mat}rix element asg*1, withPF).Tmstoget
thevector, fionstaange ‘element V, youfeplace with I,say. —
Beware: itisthe1;“compenents weighted bycos that aréassumed to
. bethe isospin partners ofthe EMcurrent. This isCVC, This, the constants
£,and£_mentioned above forpionbetadecay are: f,=Mer, addf=0.
Thus, itissimply the EMpion form factor that enterw inthe weak pion
*peta decay, the linkage being CVC.
Bytheway,pionbetadecay is1078 fraction ofcharged piondecays, Allch
pion decay modes areweak andinvolve nelithinos. Incontrast, allneutral pion
decays arebasically eléctromagnetic.” Thereason ofcourse isthat fora+fdecay
youhave’todumpthecharge onsomelighter particle, andthe’only candidates
are ieptons which impliés a’weak decay.
Imthejanemodeasabove youcanhavext. Yourcurrent isstill
. ¥,strangeness clanging, socosisreplaced with asign.” However, thedifference
nowisthet’ Idont think CVCapplies toayycurrent, onlytoaVycurrent, so
IT’wouldguessthatyoustillhavetwoformfactors todealwithsificeno~ Q“¥nownconserved current. GCofcourse doesnotapplytooeither A,orA),
-3-
Pofinish offpionbeta decay, Gene goes ontocompute therate. Since
theq"transferred offthepionissosmall,[email protected],ie, O nem pion formfactor normalization. hen your pionbeta decay rate
involves onlyG*oos*9 asshown in(4.63). Now, wemeasure Gfrommuon
decay, and wemeasure @from ratio ofkaon/pion decay constants, sothis
result isapure prediction. Agrees with experiment to10%exper error.
So far we have done Meson to Meson +Pair, What about Meson to Meson +
Meson +Pair? Nowyouhave 3polar vectorm since 34~momenta, andyousuddenly
also have anaxial vector oftheformé.,. kp. Thus, ifyouaretalking
xsttree yourA,current has3terms, andyourV,ourrent has1term,
"so nowfour form factors toplay with, atthestart. Gene does notpursue this
decay verymet? since aSe1there is.no CVC,sowillbeamess. Geneomits things
that are amess. Fraction isof course way down onthis.
Nawwemove tobaryon stuff. Regular beta decay isBtoB+pair.
Nowbaryons havespincohavetodealwiththat. TheB/Bmatrix element must
always beoftheform U(p)..... u(p"); this isjust analternative tothe
M-function formalism. Byplaying with things, youseethat with your Dirac
%anayourtwoavailable momenta, youcanmake3polarvectors and3axial
0 asin(4.29), withf,andg,asthe6formfactors.Now comes CVC again. The EMcurrent would have the 3f-type form factors
sincecurrent isoftheV,type, ButEMcurrent conservation reulces thisto
theusual twoform factors, Aswritten in(4.39), thefirst form factor gives
charge plus usual moment, andsecond form factor gives themoment anomoly. If
you procedd towrite this EN matrix element iniso-notation, you get (4.45).
Whereas thecorresponding pionmatrix element forEMhadonlyanI,,piece,
thenucleon EMmatrix element hasan1;pieceandanIs0piece. ThisI=0
Piece arises from the hypercharge yofthe baryon via the GMNish formula, whereas
pion had noy.
Soexamilte (4.45). There are2form factors ineach term. Only thesecond
termwillbeassumed byCVCtobetheI,pieceofatriplet. Sowhathappens?
Well, for the vector piece ofthe current matrix element (which isdS=0
sincenuclear betadecay) youcanset‘fy=0bycurrent conservation, andthen
Py)=f, andBy=f) asin(4.52). Ie,thwtwo vector couplers getconnected
with EMform factors, the third isset tozero. You are ingreat shape, except
you dont know the axial form factors.
WhatareargumentsforCVC?TheWweakformfactorf,isexperimentally le] found tonormalize outtounity (cy=1),just like theEMform factor. I
think this isthe main motivation. .
Now, what are second-class currents? One presumes that ake, ofthe 6form
factors appearing innuclear beta decay, 4are completely caused bystrong
ifiterdcfions (a11butflandgl).Thus,youmightexpectthefourother r?)terms tobeG-parity invariant. This assumption combined with Tinvariance
(which forces allformfactors fobéteal) yields result 6=0. Theg,
term, ifitisexperimentally present, iscalled asecond class current,
Assuming nog,,thebeta decay current nowhasthetwoEMform factors
andthestillunknown g,)and&,axialformfactors. Bytheway,@,iscalled
"the induced pseudoscalar form factor” ofbeta decay. .
Finally, wehave Goldberger Trieman, Consider nucleon beta decay. We
havejustargued thatintheexial pieceyouhaveonlytheg,andtheg,terms.
Theone-pion exchange model isamodel forthe&,term, Thismodel ofcourse
involves the‘pion decay constant fy (ofweakinteractions, appearing inTe)
andthehadronic ion-nucleon coupler called g,here. Thus, themodel here
says precisely (4.74), where you put inalittle adjustment tokeep inmind
that youshould have used ¢(q°) asarunning PI-N constant. Sowenowhave
statement (4.76) withg,nowréplaced by(fy‘eye
Next,lookbackatregularTyedecayandobserve‘thattheA,current(which was involved tlere) isconserved to the extent that m= 0, Now since weare
dealing withthesamecurrerit operator A,(butbetween nucleon states) wewill(*)assume conservation ofthis current, PCAC, This ofcourse makes arelation
detween g,‘and (fey) +But@,=-C,=1.23 asmeasured, Youcanmeasure
each ofthese three constants separately and the result works!
*What does Goldberger-Trieman prove? Itisevidence for(PCAC)A (g.=0)A
*(OPEmodel). Another waytosayGTRisthatg,isrelated tog)-
*Obviously, after all this preparation onthe form ofthe nuclear beta
decay matrix’ element, weshould dobeta decay. Next chapter.
Chapter 5:Nuclear Beta Decay.
Thisisa‘long andcomplicated subject, butanexcellent exercise inapplying
what has been learned inteearlier chapters. Here are the basic facts: you are -
dealingwithahadronicweakcurrent.Inprinciplethatmeans,sincemoolear_,aimplies baryons, there could bethe6formfactors £1,f2,f3,g1,82,g3+ Bythe aN .way,theformfactors f2andg2arécalledthe"inducedtensorcoupling” we
and, Iguess, the "induced pseudotensor coupling" because these mitiply
theoP?ana#45 terms inthebaryon ‘weak current. Ofcourse theg,termis
alsocalled thesecond-class current termandissupposed tovanish byG-parity @)
argument. Thef3andg3arecalled induced scalar alldinduced pseudoscalar.
Theworkinduced means theyarecaused bypresence ofstrong interactions.
~4-
Inthe pure lepton world you have only fland gl non-induoed couplings. Any
+othercouplings are"induced". OfcourseinCVCtheoryyousetf,=0andyou e *identify flandf2withyourEMformfactors. Ata=0fliscalled Cyand
tpiscalied¢,forobvious reasons, .Inaplane-wave theory, the beta decay amplitude isgiven by(5.1). There
you see the usual leptonic V-A current hitting onto the 5~term hadronic current
(recall £30), But inthis decay you have totreat the nucleans ashaving
definite wave-functions, they art not plane waves. Plane waves are only useful
when something isnon-localized. Anucleus is*localized andhassome definite
wavefunction for neutron distribution in it, So'we can rewrite the amplitude
“putting inwavefunctions andreplacing themomenta (which appear inboth
imtineed tensor ters andtheinduced pseudoscalar term, ie,ing2,f2,g3 terms)
withposition gradients.’ fhesymbol jy(x) isshorthand fortheweakleptonic
current. :
Nowthekeysimplification inbeta detay isthie: the@values (energy released)
are onthe order of 1MeV for all nuclear beta decays, Since’nucleuses are mich
heavier thanthis, theydontrecoil much; Thus, youcanflakethemnon-rel and
use thelarge-small spinor reduction. Also, obviousiy themomentum transfer
+forthenucleus(theq”whichappearsinthevariousformfactors)isneatly0. fe) Thus,ellformfactors reduce toconstants like£,(0)=Cyadd8,(0)=Cys
Moreover, onee you write out the various terms inthe: reduced~Pauli
language asin(5.8) ,yourealize tateachgradient: appearing is,in
momentum space, 2 butthis issmali, Sointhe "allowed approximation” you
Justdropallgradient terms,But,theonlynon-gradient termsaretheCy‘andOy
*texms; the £2,g1,g2,g3 terms then all goaway (note that (Eiw) terms also go
away since this isq°); Therefore, onceyougointo the"allowed approx", you
canlearn nothing atallsbout these other. formfactors. Youcanlearn only
about Cy,Cy)andGy=theweakcoupling appropriate forbetadecay.
So, everything reduces to(5.14) which shows thaamplitude (called L):
|88toterms. “Basically, U=(Goos0/fi) (CAIP59(0)=©,A#7-5(0) ),where
41) isthenuelaer waffefunction value ofoperator “Z+( andigthus always some
numberlike1,00or{2),and4#)isthesamethingbutwithaPauli@operator
inthere, ie,this issort ofthenuclear spin. Forannucleus, yousumover
“all nucleons andyouhave little tfoperators whichalléwthepossibilityof "eachmucleon converting. Ie, you get acoherent sum ofamplitudes asthe total
amplitude, Theisospin stuff enters just asnotation, ie,wavefunctionXis 8anisotopic vector, doublet forneutron decay eg.” °
,Sothe big thing welearn isthat there are basically two terms, the CVand
the CAterms. These terms are called Fermi aiid Gamow-Teller asweshall see.
Sinceweassumedthatleptonwavefunctions areplanewavesovernucleus, eo Geneargues that this means L=0 for leptons. W1l thats not right. Really
the variation ofthe plane waves ofthe leptons issosmall over the nucleus
that you appréximate j(x) byj(0). Ie, the actual @ecay occurs inasmall
localized place; roughly speaking, the leptons originate atthe same spatial
point and therefore can have noorbitral L. They can have S=0 orS=l, and
hence leptons have J=0 orJel. Ifleptons carry off Jal, then muclear spin
must have changed byJ=1,0. What Imean isthie: J;c Jp@ 1.
Thisrequires alittle diression. Consider themtrix element 45(¢|%7.
ThePaulicurrent operator transforms asaspin-l object. Thestates iaddf
also have rotational transformation properties, ie,nuclear spin J,andJp.
This matrix element will vanish unless JC J,@1 .Ingeneral, ifyou
have atransition J,=1toJ,=1, this matrix element will have some value,
albeit unknown, Theother element will alse have avalue, ie,GUWi) where I
have omitted thet*inboth cases, Thus, this isa"mixed FandQTtransition".
Ifyouhave J=0goes toJ=0, that mst bepure Fermi, because then theKF)
. vanishes from rotation group theory. Or, ifyou have Je0 toJ=l, then pure
(GT,sincetheFvanishes fromsamegrouptheory. YeSo, depending onhow the nuclear spin changes, you can have F,GTormixed
transifitions, huxkeyxiaxtkxt In(5.14) these facts areobvious without
any talk about the lepton L, Its just that the L-0 disoussion explains why
there arenodJ=2 beta decays, eg+ an
Another detail: ‘electron wave function ,since itischarged, is"distorted"
bythe nuclear electrostatic field, soyou need F(Z,E) tocorrect for this. This
correction isafunction ofnuclear oharge Ze, aiid electron energy EB. 80with
this correction weriowhave Jamp/” as'in (5.17) with (5.18).
Fact: since leptons always have the(1+%%) factor, thelongitudinal
polarization ofany electron emitted inany weak interactiom, such asbeta decay,
is<v/c. HowdidInotkriow this’ Intheneutrino limit youof‘course get~l.
‘This fact has been tested experimentally invarious beta decays.
OK, now put inyour amplitude, square it, dothe traces eto and get the
rate shown in(5.30) which “applies only forJ=$toJef. Discuss this rate: _
You see the obvious phase space and Coulomb correction F(Z,E) already mentioned,
Theoveral ratenormalization shows 0”aswellas$=(5.31). Inthebrackets,
youseeal,thenallkinds ofangular terms. Thetheoretical coefficients ofe
these angular terms aregiven asa,4,B,D; theidea istomeasure these coceficiants
-5-
experimentally then compare. .
6 Butfirst, ifyouintegrate overeverything togetthetotaldecayrate,
none ofthese coefficients matters. Suppose you integrate over everything
except the electron energy, The (5.40) gives you the theoretical "electron
energy spectrun". Ie,youlook at1,000,000 beta decays andrecord theelectron
energy ineachcase,this,givesyou aspectrum, Experimentally, whatyoulearn
from this spectrum isthat ithas the right shape. This fact then tells you
thattheemitteed neutrino Vemusthavebeen,lighter than60eV,orelse
spectrum would have been wrong. Asimilar beta decay experiment which emits
apsitron only gives && 4.1Kev.
Somch for the electron energy spectrum, Now integrate this toget your
total decay rate, The relevwant dEintegratign iscalled the Fermi Integral
andiscalled f,,Thus, thelifetime isgiven by(5.43) andinvolves only
Ge:B,$. This implies ofcourse acertain" comparative half-life" which
isthehalf-1iRe uébhs@ insone sense ahlef life; this thingiscalled"rt" andft=(5.44). Thus,youcan,determine theproduct Gg§bymeasuring lifetimes.
Soifyou somehow mew§youcoulddeduceOp.So,golookatsomeJ=0toJ=0 transifionswhichareofcoursepureFermi,thenyouknow4t};butyou a)donotknowC,yet,nordoyouknowL*7, theWigner-Eckart reduced. 4#>.
But, ifyou gooffandmeasure thefront-back electron assymetry inanother
experiment, youknow Aof(5.33) andtherefore §of(5.47) atidtherefore$. Thus, you can measure @for various decays, result shown inFig. 5.2. Natice
thatallG'saresmaller than frommondecay, evidence forcospresence.
OnceyouknowGg,yougobacktoregular neutron decay where 47=3(since
nomessy wavefunction anymore) andyouconclude that C,=1.23, thefamous eresult.
. Bytheway, neutron lifetime is11minutes half-life.
_Nowlets goback andlook again atthose angular terms in(5.30). Note
that ingeneral J,,J,case youhave toaddinterms onbottom ofpage 110. Look
atthe terms, The aterm isascalar under papity, asisthe Dterm. But
the Aad Bterms are pseudoscalars. Thus, ifall these terms are preserit,
you have aparity violating interaction. Bg, parity invariance would tell you
that A=-A, Similarly, ifDisthere, you have Tviolation. Confirm this.
bylooking at(5.35). :
80, first angular thing isthe "electron-neutrino angular correlation" as
seen in(5.52). Inthetheory, a=asshown interms of¢.Thus, foreach
A decayaandG(orx)msthaveacertain relation, ie,theymstlieonthe
main straight line ofFig. 5.3. Experimentally youmeasure aamt bymeasuring
electron pandrecoil ionpanddeducing py.Asnoted above, youseparately
measure §from thefront-back electron assymetry inpolarized decays, the
coefficient A, Since points docome out onstriahgt line, you like.
:Nextsection dealswith5+(A,B,D terms). Hereofcourseyouneedto ()
use polarized nuclei inyour experiments Obviously, the fore-aft coeffient
Ainvolving only the electron iseasiest to get: Since this ispseudoscalar
term, thefactthatA40inCodecay started theparity revolution. (1957)
Ifyoumeasure AandB,you'get aseparate detenmination ofC,=-1.26.
Dalways comesoutnear0,indicating thatTandhencePCare@®inbeta
decay. Ofcourse small uxxaxxxinrxummams inconsistencies can always be
attributed to neglected terms in the allowed transition,
Now, when you elevate back up fro the Dirac-Pauli level to the Gamma matrix
level, you realize that, after dropping the various terms, your true amplitude
for the hadronic current isthe same asfor the leptonic, ’except the (1+3$)
isreplaced with(14¢,4y ).Butdontforget thatthisisonlyinthe"allowed
approximation", SoitisaV,A theory again. Ingeneral you gight have
expected P,S,? terms (note: weassumed vector—vector atthe start, way
. backwiththose formfactors). Theexperiments odmbined withgeneralforms
of course support but donot*prove the V,A weak hadronic current .
‘What the hell is"weak magnetism" ?This isaneffect which involves the
fpformfactor "induced tensor" “termwhich iscompletely neglected inthe (*)
allowed approx. Ifyou include this term and look atcertain beta decays
‘which tend toisolate it, itcauses alittle correction tothe electron energy
spectYud, ‘This correction involves £,(0), andtherefore (recall CVC)‘the
*mucleon magnetic moments. The prediction is(5.70). Measurements ofthis
shape correction agree well. Nore support for CVC. InEMcurrent, you
would call this fytensor term the“magnetic” form factor since itgives
the magnetic anomoly. Thus, this f2term iscalled the "weak magnetism" effect.
*Sdmich forf,, Another corrective term isthesecond-class gp.Bvidence
forthis term isinconclusive todate. Nocomments onthe83term.
{aes“Pa.Faw.Kanscheene,camedasacdion.MabeAnadditiontoMonta3,35, *tFayGT)ReaPe..
Ghapter 6:Nuon Capture, Basic reaction here isfa+p~yn+Vp. Ifyoudid
this atFNAL, you would get and compute across section. But, this chapter concerns
instead this process occuring asfollows: amuon iscaptured byanatom and
replaces a18electron, Since muon mass is,larger than eleetrop, the Bohr
radiusisverysmall;,ie, thiemuonspendstimeinthenucleus,sothecapturefo) rate issignificant. Mymain difficulty inunderstanding this isthat you get
arate, not, across section. Ie, somehow this rate islike across section
=6-
multiplied byaflux; the flux iecaused by.the presnce ofthe mon right there
intheatom,Iamsousedtoplanewavesandcrosssections, butIknowitis fo) possible to.fold imthe various particle wavefunctions (if they are not plane
waves) toget the total transition probability, which inthis case is,a capture
rate. Infact we had todo some ofthie wavefunction folding back innuclear
eta decay also. Notice that this folding gives you avery particular rate:
from som initial muon, 18wavefungtion and some nucleon wavefunction, inté some
particular final nuclear wavefunction; the final mon Iguess can excape in
aplane wave; itisthe only plane wave here among the 4particle states.
. So, forma (6.5) really gives you.a rate into that final nuclear state;
you need only, to integrate over neutrino phase space toget that rate.
Lets take time out to work on the matrix element for awhile. You see
in(6.6) that the amplitude isthe spatial integral ofthe imkmx overlap of
the four wavefunctions ateach point x;ateach point, however, the amplitude
isjust the leptonic current dotted into. the 4-term hadronic current (here we
seeonly f1,f2,g1,g3; ‘theg2second-class. term isdropped. )Thekinematics
is such that you are looking atthis 2-tor2 process atthreshold inthe initial
channel. This fact means that variables arefixed. Thetransfer t= q*offthe
nucleonisfixed.as shownin(6.13);the,neutrino energyisfixed,see(6.12). 6 You are inthe oms frame of course. Thus, all form factors are evaluated not
atq°=0butatsomespacelike fixed point. Notethat%=4isthetransfer,
;not the mon 4-mementum.
So,from your EMtheory andCVyoudeduce what f,andf,.are at.therelevant
points. -g)=C,to.999; theonly other constant isthet g,. Youdefine g.=ng,
swhere m=mon mass} thenuseGoldberger Trieman togetthis g..=7C,approxy.
Nowyouhave agood idea ofallthe constnet involved, assuming nog,term.
The mon wavefunction isgiven by(6.18), has only upper component because
solow energy. For the neutrino, pisnot negligeable soyou have the usual
(6.21). Note that this @isamatrix inthe lepton Pauli space. Meanwhile
WAIT; maybe the reason the muon has only anupper component isthat experimentally
youareforced toaverage over-monhelicities. 50% neverappears anywhere.
In.contrast,, the nucleon before ang after spin.does appear, asdoes the final
neutrino spin; ie,youcould propare-your nucleus inpolarized state, before
: yon shoot inyour mons, . . -_
. When thedust settles, theamplitude you get isshown in(6.28). ,You, have
thatfolding ofthewavefunctions andaspinormatrix,element ofanperator given 8.in(6.29), Theconstants GyandG,ang.6, areallknown asdiscussed above.
Youmust remember that %operates inthe,nucleon 2x2space, whereas gin the lepton,
OK, sothe amplitude isnow clear, Toget rate, use upthe eldta function
and exhibit neutrino direction integral asin(6.35). Bythe way, notice in
(6.33) for the general nucleus that Gene has gone back tothe isonotation, so
younowhavethreespacestoworryabout.Thenotation of(6.36)ismtquite(*)
clear; the nuclear wave functions f(x) now appear asstates, soaspatial
integratio is implicit; you even see the xinside the mon wavefunction, Now
theToperator does itsusual lowering function, Soyougetanuclear {jyand
4 typematrix elenent, asinbetadecay. Asnoted, these nould beexactly
the beta decay objects ifmon were aplane wave; the exiting electron in
beta decay isaplane wave, but here the absorbed mon ésnot.
Now comes atrick; you want to sum over available. final nuclear states.
Ifyoufudge alittle bypulling bracketed factor outofZig,thenthissum
goes directly onto matrix elements squared asshown in(6.40). But then you
can use unitarity tomxx absorb this sum; your squared matrix elements then
contain double-sums over all nucleons inmuoleus, each has adouble spatial
integral, sonon-local. However, ifyou execute the neutrino direction integral
before doing the.double coordinate space integrals, you reduce the latter to
adouble-radial integral asin(6.45). The final rate result isin(6.45),
and the double nuclear sum isstill inthere ofcourse. Ifyou can.ignore
.the non-diagonal terms inthis sum, you simplify to(6.46). For light small nucleimyourmon-nuclear overlapissmallsoget(6.47).Simplifythissomemoreto_*)approximate endresult (6.48), which shows that Z4rate increase!
Comments: for larger Z,your 15wavefunction.is heavier atthe origin, .
asshown in(6.20); this gives 2°.The4thZarrives fromthediagonal sun
of(6.47). Thecrude result (6,48) tells youthat forappatom (monic
+hydrogen gas) the capture rate is1000 times less than the mon decay rate.
But for large Zatoms, even Z=10, itseems that the mon capture rate will
exceed the mon decay rate.
Notice one bad aspect ofthis process; inessence there-is only one
final particle, the neutrino, soyou are forced todo the complete Shase-space;
thus, the only data you can measure isthe total rate; it is unlike beta decay
- where you get assymetries, angule correlations etc. Not avery rich experiment,
What does.experiment say? First, you candothe gaseous pt atoms. But
the two fermions can have spin S=0 or1,and these two states have different
matrix elements, Gen goes back and undoes some ofthe rate, showing Wonce again.
Well, the conclusions are not tery exciting. You can dothis muonic hydrogen,
orcapture inEe?orinnet, Section closes with "radiative muon capture" and
«somehard-to-do experiment withmmmnueleus recoil aggular distribution. Doese
: not prove ordisprove agything; iscompatible with the theory.
~T-
Chapter7:theGabibboTrick.Multipletsreviewed.Enpkiricalselectionrules fe) arenoted: fora@Sa0decays, /aI/=1ie,bothV,andAbehave asisovectors.
foxdS=1dedays, aS=4Q ;other tule? 10(@8u2-o7- morddecays. fordS=1decays,/dI/= +iey.bothyandA,behaveasisospinors, ThesearerulesIshowld keepinmind.Examples: 2-7WKYisblocked byd8=a9
rule; KS"WEY alsoblocked bysame. Notice thatthe/aI/=% rule contains
the dS=dQ rule due toGiN forma, However, there issome evidence of/dl/=b
rule violation. .
‘Theexample given forvalid /@l/<} rulé isthedecay KZ4¥F~ Since final
state mst have L-J=0, isospin state can be only, I=0 or2.But.if current operator
isaInd object, only the I=0 amplitude isnon-vahishing. Therefore, using
trivial clebsches, you can deduce that K-short should gointo charged pair
twice asoften asneutral pair, ie, ratio of branching fractions should be 2.
Nytables show this fraction isinfact 69/31 =2.2.
Consider nowKf»WF.Again, asabove,yourequire. 1-0bytheI=}
+rule; butI=) forfinal state, sorule says this isblocked! However rate
shwsupas.30%oftheKy rate. Isthie really aweak decay? It's rate is.«,
intheweak time scale ball park. Iamused toconsidering this asaZweig .,”,
A)_—forbidden hadronic decay.YaybeZneigprohibition isexactanditonly.leakethrough weakly; then cylinder correction should bezero.
Anyway, intheKta-ef youmustconclude thatitssleaking through in
. theI2channél, Doingisospin comparisons, ‘youconclude thatA,/A,=$f,a
measure ofthedegree of/dl/=$ rule violation.
Next Gene goes into SU(2) endSU(3) discussion. Hewrites down theoctet
3x3matrices inusual way. Ihave reviewed this but have notset myown
conventions yet. :
Now, howdoyouassociate thecurrents V,,V,,4,;4) .with octet currents?
Well, inCVCyouhavealready saidthatVV=j,4iJ, andJ,=d3+Te.you
have anisovector, sofar. Now, recall that they key SU(3) assumption isthe
identification ofFgwithhypereharge y.=B+S. Consider theweekdecayktsFy.
(strange version 6fpionbetadecay*) Thisinvolves only,thecurrent V,(seep84).
Clarly this current isacting like V-spin raise/lower, .soyou conclude that
vy=dgtiy +Bytheway,J,=neutral ourrent soislikeU-spindirection.
Perhaps J,,,=mixofj;andjg} theCVConly says: that V,currents arepartners
ofthe isovector portion ofthe EMcurrent; as usual, thing GMI. formla, -
eo) WhatabouttheU-spinraise/lower operators which’would‘relatetocurrentsJgand7which nooneevermentions? These would beinvolved in8decay say+ofK%9 Wb) ;thecatch isthat this would violate charge conjugation; sothere
seems tobe noway to "access" those members of the current octed. That makes me
uncomfortable. Atany rate, Inow understand how the various identifications
aremade,Similar tothedssocociation ofthecurrent withaniso-spthor in °“beta decay] ie,recall presence ofTin there. Seems solid.
>The&.andA)axialvector currents arecomectea withtheirownaxial
octet, ofcurrents. Gene calls these g,+ig, =ay etc. Earlier Iasked: does
axial d5=0 current look like avector im isospin; answer isyes.
Now comes the Cabibbo stuff. Goback first to the leptons. Just asyou
didformandpinhadronic beta decay current, you canput ©andYeinadoublet
andigagine &group "weak isospin" SU,. Since Weisdefinitely lefthanded,
youmight pititonly with e,component of¢,Inthis- viewpoint, theregular
Vaiid Aportions of the weak leptonic current look like weak-isospin raising
operators, sothecharges connected withthesecurrents aisiaAlepandYepsare called F,andF,? where the5reminds youofaxial. Youcanofcourse
imagine diagnoal currents and charges (these would beneutral currents) which
havecharge L,and1?+Clearly youhaveyourself analgebra SU(2) xSU(2)
where one SU(2} isfor the vector charges and the other for the qxial charges.
f Tyis idthe chiral group. That the charges respect chiral algekra isnotBquestionable,youshowitdirectly,Geneisnotsayingyetwhatyoushoulddowiththisgrouprelevant totheweakleptanic currents, Hementions this(*)because henext sfas: presumably the hadronic weak currents (ie, the related
charges) form SU(3) xSU(3), where again, oneSU(3) isforthevector octet
andtheother fortheaxial octet ofourrents; these arethe.j, andtheg,currents.
Obviously this extension of CVC tothe "octet hypothesis ofthe weak
hadronic currents" isnot exact norprovebble, sotheSUSxSU¥ things isnot
as good as the SU2xSU2 chiral algebra. .
The Cabibbo business now arrives. Youknow experimentally that @Sq1 ~
decays are suppressed relative to dS=6, arid there are no dS=2. The Cabibbo
Hypotehsis isreally the octet hypotheses; obviously ifweak hadronic Rie
currents transform asU,I,V spin raise andlower operators, youcangetonly”
: dS=0anddS=1. Suppose therewerenodS=l., ThenyouwouldsaythatTusSytiig
: period, and by "universality" you would give this current exactly the same
strength astheleptonic: current Sep=51,10p+31ep+Ie,weak-isospin ;
..vaise and lower operators, same G. But then you imagine
that somehow the presence ofstrong interactions causes arotation ofthis
J(nad) vector imthe 8-dimensional space; the length ofthe vector does not
changehowever. Youchoosejusttherightrotation sothatyoupickupa (a)component along thej,+ij, axisinthisspace. Youchoose theroatation angle
sothat d5<0 and.dS=1 parts ofthe current have correct experimental stregnth.
-8-
Moreover, youassume that theaxial current isrotated exactly asmuch, as
.oe) thevectorhadronic current. "Thisleadstotheconciseresult(7.79)whichimthereal thing. Obviously tons ofpredictions willfoloow.
"What about charm. Gene's book is1973. Iwould guess offhand: put
weakcurrents intoa15repofSU(4). DothesameCabibbo rotation toget
(7.19), then rotate again tobring insome d@s1 weak currents. Ie,charm-changing
weak currents, These will then beinvolved indecays ofDpesons. Not clear,
‘butseems youwill need atleast twoCabibbo angles. Maybe more. Youmight
usetheBox fy/t decay constants ratio togettang, where $issecond Cabbibbo
angle. . *
The Cabibbo business isfinally very clear andstraightforward.
*Chapter 8:Baryon Semileptonic Decays. This chapter opens witha review of‘the
Adler-Heisberger relation. Lets see ifIcan't penetrate this and see how it
really goes. You start with acommitator ofaxial +and -charges put between
two protons. Put insame intermediate states and ofcourse you get matrix
elements ofthea8O<thcomponent oftheweak, axial, hadronic current. For
neutron intermediate state, this isprecsiely the axial part ofbeta decay and
)*yourelatetoC,,TheRHSofthis{ese-charges comrelgivesaconstant. ‘Thus,
Ithink you get the left hand side ofthe AWrelation.
Now the problem iswhat todoabout higher intermediate States? Instead
ofwriting thecharge %asaspatial integral ofA,thistimeyoucen
introduce a@/dt atthe expense ofenergy denominator asin(8.12), then you
canreplace d/dtwith®yAS .Finally, thislastisreplaced with’pionfield
via the Goldberger Trieman. . .
Pause toreedll howGTRworks. Youhake aOPEmodel toexplain g,interms
offyandYyyw-ThomyouusePOACtorelate g,toB32sincesecond-cldbs g,=0.Butg=C,.Thus,youhavesimplerelation betweentheseguys:+,SueCA.Now,ifyoulikeyoucanuse(8.10) version which saysthatJyN* «!pionfield
with constant asshown. From all this, you get for_your “higher intermediate
state" sum the form (8.15). But since pion field imthere, the only possible
contributing intermediate states are PI-N states, Somehow (Gene skips this)
you can then relate this term ofthe Adler Weisberger relation tointegral
*over PI-N cross section, whichcertainly sounds teasonable. Thenetresult is
(8.16). Amazingly, when you numerically integrate the PI-N’ cross section difference,
e youdogettheright answer forG,. ‘
What does this verify? First, weused aportion ofthe SU3xSU3 algebra, but =~
only tie non-strange portion. This whofe thing does not involve strangeness atall.
Second, weused Goldberger Trieman, but wealreagy know that works. Weonly
uses PCAC inasmuch asitgives the GPrelation, Really, the main thing this
testsisthealgebra ofcharges, onlyinthenon-strange sector. Ie,weare e
testing theconcept that weak hadronic currents have isospin structure. By
thewey, strong interaction currents must allbeisoscalars! Just because
weak interactions violate isospin does notrule out arole for isospin. The
statement that’ the43-0 currents areisovectors simply bursts with predictive
:
power..
Now wego tothe baryon decays. Obviously the simplest weak baryone
decays are those emitting aQY¥ pair. There are 10such decays. The mtrix
elements areasshown in(8.19) foraSq0 and(8.20) fordS=1decays. You
preselect your baryons asband b*‘. The currents inthere must have various
terms involving BB since you have todestroy one baryon and create another.
Question: howdoyoucombine thebaryon fields inthie bilinear wayso
that what youhave isanoctet? (Recall that Cabivbo Says currents aresupposed
totransform asoctet.) Obviously therearetwoways,the8,anathe8.inthe
product of8x8. Question: what are the entries inthe 3x3 bilinear BB matrices
thatyouconnect withthe8,and8,?Typical entries aregivenim(8.26).
Nowasubtle point. Consider your dS=0 ourrents, both vector andaxial.Thevectorarelikej,+ij andaxielarelikeg,+ig,. Thus,the45-0,wectore
weak hadronic current looks like isospin-raising operator, and weknow that such
anoperator can never change total isospin I. Recall that apure Fermi beta
ecay involves only theCyterm, ie,only thedS=0 vector term, andforsuch
decays dI=0. Smae idea.
Incontrast, your axial dS<0 current looks like anaxial-isospin raiser,
andthus cannot change axial-ieospin. Butaxial-isospin isnotthesame as
regular isospin, Thus, the dS=0 axial current can change total isospin.
Sowhat? Thepoint comes now: when youmake your 8,and8,entries of
BEterms, youfind that allyour 8,entries contain atleast oneterm which
changes total isospin ofthe baryon. ‘Thus, for the vector current you mst
ruleoutall8,typeBB”terns. a -
But, according toCabibbo, once you have removed the Symmetric combination
from thedS=0 vector current, youshould also remove itfrom thedS-1 current
because this thing isjust aninduction from the aS=0 via the Cabibbo rosation.
Ontheother hand,adirectdS=1argument wouldsay:well,theseourrents are
notI,s$yle currents, soyoucould notaprioiri saym#I-0.°
_Soyimagine putting intheBBlinear combinations andletting thefields nit@
onto thestates tomake u(b) spinors. Youthenget(8.30) and(8.31). The
~9-
symbols aandsinthere arejust thecoefficients like 2/{ appearing in(8.26).
,o)YouseethatthedS=0anddS#1matrixelementseachhaveonlyonevectortermwhichistoway,the8,term. Notealsothatthesetermseachhave‘overall
coefficient: 1,asside from thecosandsin, This 1is£,(0) ofregular nuclear .
Reta decay. Recall that this was the experimental fact that suggested CVC, ic,
thatf,=D, ie,thatthebetadecay Gissameasmuondecay @aside fromcosine.
However, for axial currents, the isospin argument given above fails and
youarestuck with both 8,and8,type terms, theusual FandDterms, Again,
looking etreguler ndecay, you seethat F+D=C,. .
So, after alotofyak, Inow see what (8.30) and (8.31) are saying. All
10allowed baryon simple (semileptonic) decays can befitted with just two
constants, DandFf .
Table 8.1infactliwts theeffective CyandC,forallthedecays. Bythe
way, notice that allthis business ofcombining BBinto8,and8,implicitly
means weare assuming exact SU(3) invariance. Inessence weare applying SU3
Wigner Eckhart theorem tothe matpix elements here. However, because there are
two octets inthe 8x8, you cannot predict everything from SU3, ‘ever’ exact SU#3.
Notice that some times Cy=0 for some processes.' This issimply because
[e) thisprocess’ isSU(3)-absent fromthe8,term.Sinceaxialhasmoretermsthanvector, ieaxial has8,+8.,ikisofcourse morelikely thataneffective
cyorfyshould vanish than fortheeytovanish. *
Bythe way, itshould also benoted that weare treating akl “these baryon
decays jnthe same approximation aswe treated the beta decay, ie, weassume
small momentum transfer, evaluate allform factors atq2=0, drop allsmall
terms such as'thef,weak-magnetism. Ie,inthis limit thedecay theory is
just af),g, theory.
Now, lets doafit. Choose Fand Dasshown in(8.58),’ use the usual
Cabibbo angle. This then enables ustocompute all these decays. Ie, wecan
get all the decay rates. Most total rates fit well.
+But are there other things you can measure experimentally totest this
Theory ofBaryon Decays? Sure, everything you measured inbeta decay you can
@lsomeasure here, Bg,the@9correlation, whichisafunction osfheangle
©between Fe,andy -(thiswasthecoefficient ainbetadecay) issomedefinite
function ofyour fland gl. Bymeasureing correlation, you can check tosee if
f1=0 inthose decays where itissupposed tobe. You cam-also look atthe
fo)leptonandrecoilbaryonenergyspectra; thefore-aftleptonassymetry; the recoil polarization; (these latter two inpolarized initial baryon decays).
These peripheral experiments oftenyield information onthec,/Cypation asf.
|, Apparantly nothing drastically wrong has yet been found,
Now afew more topics are inthis chapter. Why does SU3 invariance work
sowellhereifmazses aresoseverly broken? Somepeople wetouttofind [*)
out, leading to the Ademollo~Gatto Theorem, This theorem isdezived later in
this book butsays this: theSU3-breaking term inHaniltonien must beFg=¥
_=hypercharge (octet dominance), This term has relative coefficient .1if
superstrong Hhasweight 1.0. ‘Theorem claims this: theFgcorrection has
noeffect onour analysis tofirst order inparameter .1, 0expect corrections
tobeamere 1%,
_Another subject concerns generalized GTrelations. For regular beta decay
werelated ¥y,Jpn, C{. Starting withanyother baryon decay youcanmake
asimilar theorem. Eg, inastrangeness violating defay, the meson-pole term
which gives g,willbeaKsoyouwillinvolve fy,Eng MEgy=Cyfor
this decay. Youcangetsome formulas like (8.35) which relate ratio ofC,
for two different processes tothe strong coupling constant ratio.
How suddenly Gene writes down the two strong coupling legrandian terms
asin(8.3%). The two relative couplings are here called fand d.By
justexamining theterms, youcangetthestrong-SU3 prediction forstrong
coupling constant ratios, Interestingly when you are alld done, you find thataand£ofthehadronic lagrangian arethesaneacDandPofthevaryon weak @
decay legrangian. Not too surprising wince itisall su3 8x8 business.
‘The final punch line then isthis: you fit all your baryon decays via
certain value ofFand D, These values agree with hadronic process measurements
of fand a,
.
Sothis chapter has been: generalized beta decay. Roughly, the generalization
isSU2toSU} forbehavior ofweak currents (flavor groups), with, Cabibbo
relative weights. .
Ghapter 9:Nonleptonic baryon decags. Consider nowthepossibility ofheavy
stable baryons weak decaying into lighter ones with noleptonic emission. Since
initial baryons are"steble", youknow youaretalking weak orEMdecays. Also
dont care about decays with photons. So, any purely nonleptonic weak decay of
@baryon into abaryon must be atmost dS=1 since weak. But the only mass that
will fit inthere isapion mass. Soall decays inthis chapter are hyperon
into baryon +pion, (All are dSql.
This isthe first time inthis book wehave looked atweak decays with noleptons atall.GobacktoadecaylikeAw@(4,thereyouwereabletoQe
~lo-
essentially "factorize" the final state, and group the leptons off bythemselves
@_—_—s24Lertonic ourrent, andgroupthetwobaryonstogether inahadronic current.Inthereaction N*M QW with no‘leptons what areyou going todo?‘ Gene
suggests youwrite ew] J,J,\N>. Ie,nowboth currents arehadronic andgot
inthe hadronic space, soIguess the point isyou cannot factorize this thing
into,sayyGe\%|AYo|Talwy. thedifference mustbethatthepandqfinthefinal
state interact strongly, unlike the interaction oftHe pand &¥ofaleptonic
decay, This must bewhat precludes this factorization, Related toperturbation
theory question. . .
+Sothink about oneofthese decays, sayRew .In-partial wave terms,
initial state has J=$, final state has S=}, soyou ‘must have L=0 or1.Ie, for
all these pion decays ofbaryons into baryons, there isonly ans-wave and a
powave amplitude, -
. Now, independently, you may goontoshow that the amplitude has amost
general form (9.3) with amplitudes AandB,Not immediately obvious that Ais
the s-wave and Bisthe p-wave, Where are the 4other terms you usually start
With? Answer: because only three momenta heré, you can always write pion p
interms ofthe fermion p's, and these then kill extra temms asillustrated with
ro) example onpage199. .
Sogreat, for each decay there are just two complex amplitudes tofind.
Now lets gothrough the Pauli reduction. Your initial and final baryon are
polarized, say,along directions 4;andawhichyouarefreetoshoose.
The ‘amplitudes Aand Bare converted tosand p. Then from (935) "you see very
* clearly why sand pare the s-wave and p-wave amplitudes.
Wearenot even going totrytopredict total rdtes; there isnoway to
‘connect sand ptothe constant G. So, weonly care bout the relative rate
‘which isgiven by(9.7). 2g,forTinvariance, youexpecth=0.Youshould
beable tomeasure the various coefficients toget ratio s/p type information.
*Notice aproblem, however. After the ‘and-€ are created “at apoint",
they are certainly going todosome PI-N scattering, why not, for awhile they
are within afermi, This effect is called "final-state interaction" and Thave
never leamned much about it. But Iamnot atall surprised toLearn that wewill
getinvolved withonlythesandpwaverphaseshifts, sincethosearetheonly
‘waves there are. . :
Inthedecay ofunpolarized A.,youexpect theproton tohave noaverage
@ _—sttensverse polarization because theinitial state hasnopolarization, Thisis
atrivial electrostatic symmetry argument.‘ Bymeasuring the longitudinal
“polarization, you can deduce 4=.65. Somhow you can measure the 3parameters
iP,¥%whichisineffect3ofthe4realnumbers present insandpsThe4th
number isthe total rate.
. Now, there mist besome waytorelate these various decays. Since allaredS=1,youknowthattheweakcurrent oreffective hamiltonian hereshould e
transform asInj, Ifitdid, youwould "see" thedl=} rule being repected in
these interactions, So for the moment lets just assume this isso. Then you
install thespinor behavior ofthehamiltonian byadding a“spurion" tothe
initial state; trivial.
.
Once the spurion isinthere, isospin isnow conserved, soyou use standerd
isospin techniques to get ratios of things. Inthe end, these isospin ratios
_give you 6conditions onthe 2x7=14amplitudes involved inthe 7decays. In
particular there are twobranshing fractions that should come outbeing 2,
butwhich areexperimentally only close (to10%.)
Soisthereanything elseyoucansqybeyond thisashruleimplicafions?
Ie, ifyou assume acertain SU(2) flavor behavior ofthe effective hamiltonian,
yougetthésedl=}-rule predictions. WhynotmakesomeSU(3)~behavior assumption
forthehamiltonian? Andbytheway, since theeffective Aissupposed tobe
Jody) smbyouwould expect toseesomedl=3/2 anddl}; butyoudont seothe
41=3/2, Whyisthis? Well wealready assumed intheabove paragraph that no
aI=3/2 whenweputU~spurion, . e
. Perhaps youaresupposed totake acertain combination oftheJ,andJ,
currents for Hwhich combination goes asanoctet member? Obviously you want
tochooge anoctet member that hasIs}. Byindisting that #have properties
@Q+0, dSal xmbxix, youconclude that only menbers 6and7arepossible. Later
itwill beshown that can beonly one ortother; both would yield CPviolation,
So tryHasa7-menber or6-menber ofanoctet. Writeoutallthe
SU3scalar combinations ofB,B, M,and h=spurion looking like a7-matrix.
UseCPtoreduce number ofamplitudes to3instead of9(among thes-waves).
,Then examine expanded H to get ratios onpage 209. Most are duplicates otthe
QI=} rule spyrion method above, but clearly you should get something more. The
oneextra relation forthes-waves (forthep-waves ifyouchose member~7) is
(9.47) found byLee-Sugawara. .
Goback: wehave 7processes ofinterest. ThedI~$ rule gave 3relations
among the7s-wave amplitudes, Lee-Suzawara isa4th relation, Arethese relations
trueornot? Figure 9.1plote for§ofthe7processes (amv andSA”
arenotthere because these arethetwowith nocharged particles, experiment
nearlyimpossible). Sothefigureshows5arrows,eachonerepresenting the[o}sand pamplitudes, assumed real. One arrow isrepeated twice because ofsome
experimental ambiguity. All relations geem tohold within experimental error.
-l-
Notice thatthesimple (Lrules arenotopthispicture because, they inyolve
thenon-charged decays. Thispicture isatestoftherule9.21from,dI=}, and 6theLee-Sugawara 9.47rule, forbothsandp,infact. Thesimple {2rules
are born out by compaming brafiching fractions and they look OK.
Chapter 10:Kaonsemileptonic decays. Westart offwiththeoldstory ofKandK,.
Fromtheviewpoint ofstrong interactions, andem,thestates K,andK,are,distinct
states; ie,strong andemdonotatallmixsuch states (eg, youcould notcouple
K,toR,because this would bedS=2). Butofcourse theweak interection canmix
. these twostates via anintermediate state ofS=0, say PI~PI. Note thet this
mixing mst besecond order inG,soisvery small.
.Now when you speak ofastate having adefinite. lifetime, what.do you mean?
,lifetime means lifetime with respect todecay. Since kaons decay only byweak
interactions, only these interactions areinvolved inthequestion., A"state"
,Should bediagonal inallquantum nuybers whichcommute withtherelevant hamiltonian.
Thus, youwant neutral kaon states that arediagonal inCP;you dont care about
diagonal inSbecause (8,H,)40 +Infact, CPisallyouhave inthewayofquantuam
numbers conserved inweak decays. This iswhyyou make theKLatidKScombintions.
Sowhichoneislongandwhichisshort? ThestatewithCP=+l candeéay 6into PI-PI, sothis will beK,with lifetime 100picosec. Skmxmiharxstate This
K,state virtually 100 into charged andneutralwe’, with 107? rate towa¥.
Obviously anydecay canaddbremstraullung andbedown order «
TheCPe-1 state isblocked intoww andasaresult lives 600times longer,
.dey60,000 psec=60nanosec. Theobvious decays hereareKiw(ey).which together give 66%ofthedecays. Theremaining 34%goes intoW¢ which isCP=-1.
There aresoemdecays intoPI-PI indicaténg small CPviolation. Alsodecay into
YN.Notice thattheK,allousmorevision intovarious decayproducts because
. experiment hasmore time, not soswamped . . . :
So,onceyoudecide that kKeidKyarerelevant, thenext topic iswhat is
meantbyregeneration, Verysimples Youproduce sayK,inWfscattering. Notebythewaythat youcannot produce Kebecause there arenoSat] baryons. Soyou
makeyourself anice Kebeam. The“wavefunction" forthisbeamisKetKyas acoherent amplitude sum, Butofcourse your Kydecay quickly leaving relatively
moreK,downthebean, After awhile youhaveonlyK,.Butnowrunthisbeam
ofK,into some crud andofcourse theKecomponent canreact with|protons whereastheKocannot. So,ineffect,theordremoves moreoftheKecompanent éofthebeam, thus unbalancing themtrture equal mixture ofKe-+Ke®KL. Soyou
ineffect causesomeKs4.4reappear, theyare"regenerates".
Clearly the KI-KS mass difference can be roughly explained by the CP
selection of intermediate states, and result should be quadratic inG.Thus,
expect masadifference tobesaneorderofmagnitude asanyweakdecay rate. GyInfact, aM=.44%5~-(S'asZ .OFcourse thisisveryemall compared tothe
absolute kaen mass which isroughly 10° sec, ie,typical hadronic size.
What can wededuce about decays ofthe Gorm K--WRP? The first fact
. isthat ,Sor all such decays, there are only 3amplitudes involved, called
f,g, andA*onpage 219. Ofthese, gviolates thedS=aQ rule, Isthere any¢
present? Comment: when you write the hadronic current matrix element, you
want tobeusing isospin anddSetc, so-you putinK,,notK.,asin(10.19).
Nowconsider K7-9WEV .This decay isreally your only source ofdecay electrons.
Youcancompute therate ofe*toe”production asafunction ofdistance down
the beam, asin(10.23), Then doexperiment toget exper measure ofg/f. You
find .04, soseems tobe4%violation ofthedS=dQ rule. Buterrors arelarge.
So,presumably g=0andweareleft with amplitudes fandA*, These are
then trivially related bythedI~} rule, ie,byxisospin, soA*=£/{%. Thus,
you ean equate some rates ‘and mekem afew predictions. Itworks.
Wehave nowboiled things down tothestudy onjust oneanplitude, namely
.At=GP\K'D. Wenowwritethisintermsofthetwoformfactors f,and
f_.IfthiswereaV,currentwewouldreplace f,ial(heENformatactor}eyusing CVC, and wewould also say things about vaiues atq°=0. Ie, wecould
. useforV,current allthepower ofCVC. Now, canweapply CVCtoV,current?
This isnotaxial, sowearenottalking PCAC. Inperfect SU(3), since Vv,and
V1aveinsaneactet according toCabibbo Hypothesis, snswer is’yes. theconcept ofperfect SU(3) justification wasthat order N Ademollo-Gatto theorem,
- Soletsanalyze Kvrowev. Westartwiththetwoformfactors f*andf
butbecause eissolight, thef_termgoesaway(notet2#/fr).Theamplitudetheninvolves onlycosinet4¢(a°). Now,smallbytfinite aisinvolved because
there islarge Qvalue, butset£,(0) =1byCVGapplied toV,asdiscussed above,
andinclude @small empirical q°linear terminf,(q°). Compute total rateand
.compare to exper to get still another measure ofCabibbo angle. Oruse old Cangle
andconclude that theory andexper agree forKY9WEY process. Once ggain,
basically only one constant eneters here and you set itto 1by CVC ,this allowing
total rate computation. By looking only atthe pion-energy spectrum, you can
evaluate thatsmall linear termin£,(a°). Thenlookatelectron energy spectrum
atfixed pion energy, This gives confirmation ofveotor type theory and conflicts
violently witheither SorTtheory asseeninfigure -p231. (*)+Whatabout theK+ PHYaocay? Wold, ithasthatextra f_or§ term
80obviously what youwant todoismeasure €,This isdone bycomputing the :
-l-
muonpolarization intermsof¥andthenmeasuringit.Resultsareshown le) imtable10.25Roughly ¢=£7/f*=-1,dimensionless, althoggh youhaveto,
account forq°dependence somehow. Then, using this{informl (10.79) for
them/@ ratio inK*decays .(orK,decays, since ratio) youcompute .53for
K*and.50forK,(keepinmindthatsuchratios always involve massdifferences
andsometimes radiative corrections). Mydata book says 32/48 forK*which is.7,
and27/39 forK,which isagain «7, Strangly results areboth thesame butare
bothoff, Noreason given yetforthisdiscrépancy! (rate onwita?)
Sowhere are wé? Gene discussed the simple Ky, decays back inchapter 4
interms ofthesimple constant fg. Inthis chapter hehas sofandone theKy
decays which involve onepion. Thetrick here wasaversion ofCVCtomy f,=1.
Thelast semileptonic decay iatheKyyinvolving twopions. 4snoted back in
chapter 4,there will be 4form factors here, and both Vand Acurrents operate.
Ithink theuseful fact, here isthis: your finals state ofT#eN has200MeV
ofkinetic energy, but myintuition says that momenta will beequipartitioned,
sopions may only typically, have 10MeV ofenergy, ie, they are relatively
soft. Ingeneral, all momenta pare relavively small for both pions? Anyway,
Iamtryingtojustifytwofacts:(1)thethrowingawayofthea,termin(10.98) fe)since bilinear inpion momenta; (2) the near constancy ofthe other form factors
. intheir arguments. . :
Now, wehave three form factors with noknowledge ofthem yet. Byteking .
the two different soft-pion limits afd using the soft-piop thearem, you conclude
thata,=0anda)=a5,atleastwhenallareevaluated atthatsoftpionpoint.
«Fromtheotherlimitweconclude thata)+ay=48Jo+»floodgrief, wheredid
these old form factors come from? ,How did they get into soft pion theorem? Well
the SPT has fq sitting initintrinsically bepaiise the SFT contains Goldberger
: Trieman/PCAG stuff inits derivation. Onthe other hand, the SPT removes apion
from final state, andthis brings inty. Soresult nosurprise.
Soofthefour caonstants a,through a4,youseta,qa,a0 ineffect, and
yousetaj=ay= et,/te»numbers whichyouknow,andthenyoucancompute the
total rate! compare with exper: itworks, agreat victory for the Soft Pion
Theorem! ' .
So lets take alook atthis Soft Pion Theorem, Stated onpage 234. You
pullyour softpionoutofthefinal state andyougot18UFeStILES,oflliy -
Howdoes itwork? Obvieusly bypulling thefinal state pion, yougenerate apion
@_—e148whichcanbeputintoaconmtator Zorfree,see(10.85). Thenyou+replace thepionfield with Se'-3%, ie,thedivergence of:theaxial
current; thisisjustGoldberger trieman andyouseeSgappear. Butinthe
Limit ofqo (softpion) youcaneffectively replace Yai withg,integrated
overspace asin(10.88), which yields thecharge F?,ThusQED, Onlyproblem
isthat Idont think the oosine should ‘te inthere. Later when he applies the
rule, Gene does not show the cosine. :
‘what about Ademollo-Gatto which Iskipped. This is another one ofthose
things which start with acurrent algebra statement with charges F,.Veyy simiar
+ to Adler Weisberger: you sandwich inbetween pion states, not proton states, put
inintermediate state set. When the intermediate stete@ isaK,you get
term involving t,andfofcourse, Ifthere were noother terms atall, you
could conclude that2,(0) =1 (thisstepinvolves going toinfinite momentum
frame). Now consider those extra terms The only available states are single
. particle states out ofthe }xx 0”octet, and many particle states (recall in
aWitisthe PI-P states that are important.) Look at(10.44). For lerge
: energy difference, itisclear that the “other terms" will all beorder
where X=«1isthe SU3 breaking Hamiltonian scale. Thus wehave proved
that £,(0) =1+order(%). Harlier Iconjectured thisbyarguing thatCVC
should apply totheAycurrents, based onSU(#3). Here weseethat even if
there isSU(3) breaking, youstill havef £,=1 tolf» :
. This isthe general theorem Ithink which says couplings obey SU(3)
muchbetter thanmassdifferences. Ie,massdifferences areorder =10% e
whereas couplings are SU(3) invariant tooraer =1%. Good argument toknow.
Chew used italot, but never used phrase Ademollo-Gatto.
Final chapter comments are onneutral currents. Processes ofinterest
-hereareKPyay orGE. Clearly these decays violate strangeness andmst
therefore beweak, and you can see that the current involved must beneutral.
Bxperimentally ,Gene says less than 1079, Butmy1976 data book says these
events doocour atrate107%,Ie,di~mon events inK,decays isgoodevidence
that wenowhave neutral currents. Maybe this nowagrees with theKy» WY.rate.
Some experimenters mst have made amistake in thés business.
Ghapter 11: Keg and Ker3 decays. FactL: two pions can only beinstate I=<0
or2for the KW2 decays. Exper seems toindicate a5% I=2 amplitude, which
isinviolation oftheai=}rule. Hg,theratio ofKshort intoWatAW’should
be.2.0 but seems to be .2.2
. How should you analyze things? For Ky, you cannot doaourrent=current
thing with Gbecause this is nonleptonic. Similar to the hyperon decays where
youhavetajustmakeupsomeparatmerts andseehowfaryouget.Obviously efor the Kyq you have L=O soonly s-wave amplitude. But asnoted you can have
I+0,2, sothere aretwoamplitudes called 4)anaao,
-13-
Nowobviouslyyouaregoingtowanttomaecomparisons hetweehallthe (o)Kqy decays, half ofwhich involve the neutral kaons. Thus, you immendiately
. have, the additional, complication ofthe k*=K¥ system. You have to.understand
that "system" before yon can make phenomonologicai analysis.
Yuch of this chapter is;about that. system. We start off with amass mtrix
which would bediagonalized bythestates K,andRjwere there-noH, .The
weak interaction, asitviolated strangenesa conservation, connects these two
.States caugeing offdiagonal entries inthemass matrix intheK,,KR,basis,
see(11.29). Soobviously, youre-diagonalize andcall your newstates K,and
Kg.Howaretheseexpressed interms ofK,"4K, 7,Tygeneral form(11.37)
and CP? then leads you tothe most general form (11,46), showing the one
parameter @.Clearly, if€$0 you have some CP-violation.
From (11.50) you.can sebt that. CKs\Kur#O when EO. Inother words,
these two states can couple inpresence ofCPviolating interaction; ifthere
were no OPviolation, each would be aGP eigenstate. Actually, Ithink they
are CPeigenstates anyway, CP-violation: justallows themtocouple.
-Ifthere were nocoupling,.no CPviolation, then you.know that KyandKy
wouldhavetohavesamemassto2partsin109,AssumingCPT,theyhavetolwe (oe)exactly the same mass, Thus, the fact that there even.is aK-long/K-short mass
difference. isevidence ofCP-violation. .
Gene throws inalittle analysis of regeneration, though not used: later.
Ie, what happens when you run aneutral kaon beam down through some material?
Invacuum youknow that mass eigenstates willbe K,andK,. But, insone
material thenew,eigenstates will bedifferent, namely K,'andK,". Inthe
. end, these new eigenstates are given.as in(11.67) and (11.68) .Phere you
see the regeneration parameter r, This arises because, thescattering .onprotons
*
ornuclei ofK,differs from xfor..reasons that areclear. Ie,'you rediagonali,e
your mass.matrix once again and get these new eignstates. .
Besides the regeneration complication which you cannot avoid inanexper,
therearealsostrong interaction s-wave placeshifts intheI=O'and I=2
channels. “Iheffect, these phasé shifts’ change thephase ofthe2=pidn final
statebyei8E,soturnsouttobaverysimplecorréctioh %6‘actount ‘for.I
dontknow thetheory ofthis, however. Sd,whenyouinoludes tHeoé phase shifts
atidtheusual isospin Clebéch's, your final possible pion states areasin(11.70)
and(11.71).Combinethiswith’yourearlierconstiuctioh oftheKyandKy le]states and yéu end upwith main results (11.77) through (11.80). Asnoted,
.
everything isnowdescribed byAp,4yanaE.
Recall that A,ismeasure ofviolation ofthedIw} rule, and@ measures the
CP violation. -
What canbeexperimentally measured ?Consider K,/K, ratios. Obviously
such ratios are ‘small because inexact OPconservation, K,cannot gotoPI-PI.
Tworatios are.called hae andne. See(11.81). IfdI=}isOK,thenno
A, andthen these ratios should both equal €.
Experiments say\9'~| =2x1077, although somerecent controversy in1976
particle data book, ‘The mass difference ismeasured bythe very clever gap~
+regenerator method toaccuracy of1%. Notice from from (11.21), bythe way,
that intoorder tobeable tocompute dM,youneed tocompute M,,, that
-off diagonal strength. That inturn requires you to compute quadratic terms
like that shown in.(11.21) (first term vanishes since dS=2), Since, well,
theresult will beorder G?andprobably propertional to€,.but youcannot
really compute itbecause wnkrown intermediate states, ie, a.long calo isneeded
totry to compute dM. No trivial relation between dMand€ .
Thephese ofM*" iscalled ye andismeasured bysimply looking
atthePI-PI production rate versus time inavacuum regenerator. Youget45°
with error of4°, Youcanalso trytoget.\Nee| andphase by ;also
people can measure Re({) to10%.
Theresults whowclearly thatthereis.CPviolation, butnotmuchelse(*)
can besaid, Obviously there isalso some Tviolation going on. What isthe
- cause of this CPviolation? Isitjust that the weak interaction violates CP?
Orisitmaybe that strong orem are violating GPand itshows uphere? Or
is there some superweak interaction which isdoing the CPviolation? Strong
and EM. are obsérved torespect CPP and Ptohigh degree, soonly possibility
issome small C/T violation there? Noone really knows. InWolfensteins simple
“superweak” thregry, you just assume there issome extra dS=2 interaction which
letsK.~yKgwhich thengoestoPI-PI, Thislittle theorylet predicts *~=A”
which looks asthough itmay betrue. But dis} aso predicts this.
So,tosummarize theMw story, there arethree parameters inthe
phenomological “theory” called4),4,and©.Whatyoucanmeasure areA"PS”
nN”, q”, Reb, QM. There isno"test" ofanythreory here, youjustdefine
some parameters and measure them. You, observe tix both from mess difference and
_factthat MLS WWthatthere isCPviolation. SoCPviolation isreally the
main point ofinterest intheKy, Story.
,
WhataboutKyg?Obviously evenlesswillbedoable, ‘Thereareseveral)
such decays ofkaons, obviously try tomake some predictions using isospin etc.
-u-
Sincecondarethree-body decays with(néar) equalmasses, youcdnplotevents
inaverysimpleDalitaplot,seefigure11.8.Asshownbg(11.163), the 6 density ofpoints atsome place inthe’diagram (sayT,,7, orr,®) tells
you the square of the amplitudé. Notice that there isno Dalitz diagram for
a2-body decay since omskinematics fixes T,and7,bymasses. Ie,forKye
decay, Tl=T2= kndwn number. _
Sowhat canyoudowith amplitudes A,(7,9) where i=process label?
First there issymmetry 9to-@ because two pions are usually the same.
Second isthi: pions havé atmost the entire Q=75MeV, usually mich less,
typically 25KeV which means pions are slow. Ie, not much phase spac eavailable
for these decays, soyou might say this: the amplitude atsome point inthe
Dalitz diagram isthe amplitude atthe celter r=0 plus asmall correotion.
Hence (11.166) which incorporates this statement with the 9symmetry. The
correction turns out tobe afunction ofyercos®, the size ofthe correction is
measured byparameter g” called “slope parameter".
* Isospin and centrifugal barriér also have somthing to say here. Infact,
considering pion plus dipion, you can correlate GP, Iaid L‘of final state; the
netresult isshownintable11.4‘which alsoshawsthedI-$rule.Result
isthatKYandK,caifbothgohealthily intoIeltripionstate,whereas loaeK, cant goinany I-spin final state, Ofcourse there will besome bleedthru
because dix} isviolated, CPisviolated, and barrier isviolated.
*Asusual, the dis} rule byitself makes various ratio predictions, see
Table 11.5 page 283. Obvious small violations ofthe rule, asusual.
* Finally, soft-pion theorem has some stuff tosay here, because ifallows
*
youtoextract pions from final state torelate Ky toKg decays. And
pions are definitely soft. Results seem towork OK.
Tosunmarize, theKyyamplitudes areparametrized byAatcenter
of diagram, plus aslope parameter. There isno theory to compute either
parameter (foreachdecay). Relations between decays intheKyg class arise
from isospin, dI=$ rule, barrier effect, 2~pion symmetry, softness ofpions.
Obviously apainful subject. Youhave 4theory which works nicely forsemi
leptonic and pure leptonic sectors; but itsays little about nonleptonic
dewayse
Chapter 12: Parity Violation imNuclear Forces. The point here isvery simple:
Thereshouldbeaweakinteractionoftheformkne|Sods+3,5ls92.>ieon ©_topofthehugeparityconserving stronginteractiom weshouldexpectaparity
violating weak interaction. Recall that EMHamiltonian also conserves parity.
How are you going tofind this effect? Certainly not looking atthe nbcross
section, Clearly, what you doissay that the effective nuclear potential
nowhas @small, parity violeting term. This then admixes states ofopposite
parity, whichwereformerly unmimed. Thus,thisadmixing causesphoto-decays [*)which youwould normally think notpessible.
_Soonetrick istolookforsomestrictly forbidden transition under
strong +BM, and see ifinfact itleaks through alittle. Bythe way, the
wavefunction admixing effect isabout F=107’, soyour leakthru transition
Willbedownbyhugefactor of10¢ fromaregular allowed transition. Iel,
this requires good experimntal technique tosee. Asexample, Gene uses a
certain forbidden decay Go» C'*4 4where the0! isprepared inacertain
excited state. Ifyouseealphas ofthischaracteristic energy, thenyouare
‘seeing the forbidden transition. This has been seen and measured, and result
seems about what you would expect.
"
Besides looking forforbidéen events, youcanlookatallowed transitions
which mayhave mixtures offorbidden events inthem, Ie,itisprobably hard
tofind things thet are reall forbidden, eoyou have toaccept amixture. But
thatisenough.
There aretwoClassic experinénts. Inone,youlookatasimple photon
emitted inanuclear transition, Ie, agamma ray, Ifthe ‘states involved
aregetting parity mixedbypresence ofparity violating terminhamiltonian, @thenyouexpect toseecircular polarization ofemitted photan (ie,emitted
fromunpolarized nucleus, andfinalnuclear polarization isnotmeasured). Just
draw picutre and you can see thet there ehoutd be nonet longitudinal polarization
ofaphton in‘reaction W9W°X. Circular pol=longitudinal polarization.
So set upyour polarized iron target and look for this gamma ray polarization,
‘Typical netcircular polarizations are107! effect, soyouneedmanymany
events togotstatistics onthis, Often your process isstarted bythermal
neutron hitting some nucleus, getting "captured" into ‘an excited state which
them dumps “agamma ray. .
The other obvious experiment isto look at fore-aft gamma asymmetry when
‘initial nucleaus ispolarized (captures apolarized neutron). ‘This effect is
sometimes relatively large, 1074, andhasbeen ‘seem. Chapter ends with quick
discussion ofthe "theory" involved, not’ very exciting atpresent.
Bythe way, what about parity violation inatomic transitions? Now you
areasking aboutCEP\WwiEG7 .Since lepton factors off,youseethatthis
ieamoutral current effect. Soparity violation inatomic transitions has to
dowithneutral weakcurrents, andisreally adifferent subject altogether (nog
-inour theory sofar); ofcourse experiments are. very similar. Ithink this is
what Commins isnow up towith Steve Chu. :
-15-
Ghapter13:meutrino scattering. Obviously youdontwanttolookforweak 8 scattering inaprocess that isdominated byother forces, like ep. That is
whythereal thing todoisYp4e%. (Standard charged-current deal), The
Kinematics here isthe same.as depp inelastic epscattering, more orless. Ie,
‘ interesting variables areq”andJ.
First consider elastic Vp? Fe. Obviously youputinyourusual
BtoBweak hadronic current with flthrough g3form factors. Now however
youwill needthelarge q°behavior ofthese form factors. Analogously tothe
form usually taken and/or measured for the BMformfactors (vector dominance etc)
youmighttakeformsasshonin(13.19). Youmightassume that£32220 (though
not obvious that CVC still works at"high energy"). Ifprecision experiments
could bedone, you could test theory: Atthe time ofwriting, only apoor
statistics CERN 1967 experiment had been done with neutrinos.
What about inelastic stuff, like nucleon throws off apion. Adler has some
current algebra relation about composition of products indeep inelastic. Inthe
deep inelastic, ingeneral you donot imow the tensor that getts mltiplied
*bythe(weak orEM)lepton trace; hence theusual structure functions Wyand
W,forBM;andalsoW,forneturinos, whicharisesfrompresence ofaxial © sicce not present inEi, This isthe sane old hadronic stroy: when you have too
many hadrons, you cannot make upform factors, soyou set upsome pheonom.
parameters, Aand Binhyperon decay, structure functions here. All structure
functions are expected toshow Bjorken scaling.
Chapter 14: Neutrino Astrophysics. You can compute solar neutrino flux; Davis
etalmeasured itin cave inSouth Dakota, number was not inagreement with
solar theory. (the famous cleaning fluid experiment). Gene goes ontodothe
neutrinos out ofhot stars, and cosmological topics. Iamnot reading this
chapter now.
Excellent book! However, most recent references are about 1971; itisnow 1978.
Much has happened inbetween, Neutral currents found, sothe 1958 Feynman-Gellman
theory now obselete; must use something with weak currents, like Weinberg-Salam.
Ofcourse any new theory must duplicate all the steble-particle decay results
described inthis book! Charm also found, sothe whole theory has tobe
e generalized tounclude bothstrangeness andcharm changing currents. Then
there isQCD for strong interactions, quarks looking like leptons, and soon
andsoforth. Areview oflast 7years would behelpful atthis point, rather
than detailed reading ofspecial papers.
4 Vouume: 4,Number 7 PHYSLCAL REVIEW LETTERS Arne1,1960Fyouustn
‘- —
i Watanabe.” Itshouldbenoted,however,. that “Thepossibility thatweakintergetions aremediated onth
;somevariantsoftheirtheorymaybefound byabosonhasbeencongideredbymanyauthors:H.calcur (oe)which-cannot bedisproved byonenegativeex- Yukawa,Proe.Phys.=Math, Soc.Japan17,48(1938;|PUh periment alone. Therefore, itisdesirable that J+Schwinger, Ann.Phys.2,407(1957);T.D.Lee Helatk4 othoftheseexperiments beperformed. It Rae eee taeeeasdfi . 4toberatherdifficulttoInventa.theoryof|_evnmanandM-Gell-Mann, Phys.Ree etsealan | seems if Y¥.Tantkawa, Progr.Theoret.Phys.(Kyoto)3,338{|sealar,
: theTantkawa typewhichdoesnotgivearesonance (1545);Y,TantkawaandS,Watanabe, Phys.Rev.113,eleona }
inanyoftheprocésses v+p,v+n,e+p,and 1244(1959). Wew |1 +n, where»andedenotecittierparticles or $5,L,Glashow, Phys.Rey.(to’bepublished). astricti. antiparticles. LeeandYang(reference2)showedthattheproducfF-f,2050| ‘Weconcludethatitisprobablyworthwhile to_#08erosasectionofsuchabosonintheneutrino tateq search forpossible anomalies inelectron and ‘nucleus collision isabout10-%om!foranIncident bei: Saree erecta oucrtheenergy rangewhich "outsin0 energymuchgreater than2Bev.Thisis myays m ey morepractical thantheresonance scattering method {poseth4 inay becovered with thepresent accelerators ‘asadirect testoftheintermediate boson hypothesis. beta de
before looking into weak processes athigher Irshould benoted thattheabsence of1~e+isnotproper! I.energies. conclusiveevidenceagainsttheintermediate-boson 4h Ishould liketothank C.R.Schumacher for hypothesis. SeeG.Feinberg, Phys. Rev. 110,1482 rie checking someofthecalculations. (ag56). fateii Ty,Tantkawa andS.Watanabe, Phys.Rev.113,
ay a 1944(1958).
aye. “ThetheoryofTanikawaandWatanabeleadstothe Hi ‘Supported inpartbythejoint program oftheOffice V-A theory onlyafter spinors arerearranged bythe
iy ofNaval Research andtheU.8.Atomic Energy Com- Fiera transformation. Inthiatheory, therefore, it
{re mission, 15noteasy (ifnotimpoesible) tofindanatural ex-
ti 1K,P.Feynman andM.Gell-Mann, Phys. Rev. 109, planation fortheconservation ofthevector current
iy 193(1956); E.C,G,Sudarshan andR.-E, Marshak, partoftheweak interstion, iftisInfactconserved4 Phys.Rev.109,1860(1958). (R.P.Feynman andM,Gell-Mann, reference 1).If q
he 2M,Schwartz, Phys.Rev.Letters4,360(1960); thepresenceofthewealmagnetism[ff,Gell-Mann, Ye ‘T.D,LeeandC.N.Yangy Phys. Rev.Letters 4, Phys. Rev.111,362(1958)] 1sestablished exper!~ ‘Thepi
367-(1960).- Similar considerations haverecently been mentally, itmayagain behardtoexplain by‘Fanikawa's} tumm4s radebyY.Yamaguchi (fobepublished), andN,Cabbibo theory.Theexperimental resultisnotconclusive yet.|thedis|
H feiausuallyaseumedthatneutrinos emittedinthe _slightlytotakeaccountofanadditional decaymodeof|garmni 11-4decay andBdecay areofthesame kind. Thereis theBparticle.
i jopositive proot fordhie, however. Obviously, itis ""R,Holetadter, F,Bumiller, andM,R,Yearian, Fl@l
i desirablethatthisbeexaminedexperimentally. Reva.ModernPhys.30,482(1958). iiti
ji AXIALVECTORCURRENT CONSERVATION INWEAKINTERACTIONS*| Yoichiro Nambu ji EnricoFermiInstitute forNuclear StudiesandDepariment ofPhystes 1aii University ofChicago, Chicago, Ilinois where
i - (Received February 23,1960) + low m|
q for 14
Inanalogy totheconserved vector current in- momenta. Suchanattempt hassome appeal in thatF]
Jkr téractioninthebetadecaysuggested byFeynman viewoftheapparently modestrenormalization thenw 4 ‘andGell-Mann, some speculations havebeen effect ontheaxial vector betadecay cons weald
Aye current." Oneeanformally construct anaxial_point,’namely, thepossible forbidding ofne+y,} imme
Me vector nucleon current, which satisfies acon- hasnowlostitsrelevance. piondWd tinuity equation, (xe,1sConsents) ‘Theexpression (1),unfortunately, cgmbe pion-atN A easilyruledoutexperimentally, as-#aspointedi}ph = Av pl-ivgy 2Mgt,/a',api, (1)outbyGoldberger andTreimgnr*sinceitintro- iy # # i ducesalargeadmixture ofseudoscalar interac} wy,4 wherepandofseethesandfinalnucleon|tion. wid lie 2LMjoo CACurnct)\\afock fla.- i 380 Byaad ih p88
AOO) tonGTR &Dede YO
160°|Vorume4,NUMveR 7 PHYSICAL REVIEW LETTERS ‘APM1,1960 eS
_ ie
aOntheotherhand,Bq.(1)arousestheoreti- secascompared withtheobserved value2.56 iBjy;|calcuriosity astotheoriginofthesecondterm x10"sec. Hee“Witreallyexists; according toourconventional Goldberger andTreiman’® havearrived atthe (Sea.
. fieldtheory, wewouldhavetointerpret thede- same'relation Eq.(4)(inthelimitoftheirself-" By.
“95077nominator q?asimplying amassless, pseudo- ‘energyintegral J=«)fromanentirely different ‘athmyscalar,andchargedquantumbridgingthenu- approach. Inouropinion,thisisnotacoinci- 2 18|cleoandleptoncurrents. dence,aswillbeexplginedelgewhere, On+Wewould liketosuggest thatthere may notbe Wearetempted toéxterd thisapproximate BY
.that.wemayhaveanapproximate conservation alsothevectorcurrent) tothestras 4‘=‘whichbecomesrigorousinthelimitq™>>m.7, conservingbetadecays.Wetake,forexample, iiizm,belngthepionmass.Specifically, wepro-“theANaxialvectorintheform R= {posethattheaxialyector partofthenucleon Ot+m) iebetadecayvertexhasthefollowing formand r4p, Pe NS 6) OHSproperties: wen Pg Gm hs=
Bete: afAlpp) andattributethesecondtermtothepseudoscalar. ifee #Kmeson.*Thedegreeofaccuracyoftherela- A 2My-4tion,(5)willbepoorerthanintheprevious case i =eylivgyFle) a Ra), inviewoftheA-Nmass difference (which de- 6) VS LO) am" a Bstroys vectcr conservation) andthelargeK- iesmesonmass.Atanyrate,weobtainananalogGR: HAM FO), /ay=F(0); ofBq.(4): fi
: +My) B4'=G, a a| F(Q)-FA@) forgt>>m_2. @) Og+My)Bg"=Cbg (6) ie=
which relates theAbetadecay-axial vector et ‘Thepionis.then theanalog ofthemassless quan- aes*,theANK Gx,andtheK, oCaaa vs{tummentioned above.‘Thisisconsistent with teenythmecouplingGxandtheKy Aiei & ion rel ::
i B ieeee Faas ectedoren WiththeobservedKlifetime2.1%10°*sec CHG, Aamely,F,andF,shouldhave ingeneraltheandatentativevaluedy/4x=46,2/42, weget ne
. Lege. 7) Pres FF (-m2) ayy ¢ “net
iS Thisisnotinconsistent withtheobserved beta. ie
©p.m)dm? decayofAwhichseems‘anorderofmagnitude a =(qtemafwas lessthanpredictedfromauniversalcoupling aied TSne" 7 scheme gy’=g4'=ay." 2a
WecanStillgofurther,thoughtheargument a= { (i=1,2), (8) becomes morearbitrary. Letusassume thata Rsfundamental weakcoupling(WNW: riset ie wherem3m,unless therearenewparticles of"damental weakcoupling (NNNA) gives riseto i foe SmsI aneffective V-Ainteraction (oratleast partof alene lowmass.ThustheF'swillbeslowlyvarying it)oftheform ahs forIq*i<cm3. Theconditions inEq.(2)imply Pla thatF,/F,=1 forallg,Ii'm,=0andF,/F,=1, eneerAeYepAL ts ha thenwerestoreexactcurrentconservation,? and BRONNSwpOPIAT yie: wealso-expect F\(0) =g4/gy=1. ‘ v a)1 Here ©, =iy,, which isapproximately con- heA iY PP: y nsatiadopt Ba.(2),thesecondtermofTA servedbyitself,and7,4standsforE49.(2)or tt4= .‘loatelyciesaiter)eotants,fareth (5).WeseeeasilythatEq.(8)containsinfor- MigBiondecay(Pgeudovector) constantgp,andtherationabouttheA~N+zdecaymatrixelement: is pion-nucleon (pseudoscalar) coupling Gy: fips
vA ag =-m_#)= . (21NE (a - 9) ome 2Mg,=2MgyF(-m2)v8Gg..(4)(2Mals264tahynae(9) ,iai 7Y)Withgq=1.25gy=1.75x10-" ergom’,G,2/40 Combinedwiththeassumption ofAT=}selec- ktfy:<13.5, thisgivesar-udecaylifeof2.7x10"* tionrule,thisgivesalifetime of2.5%107 sec PU .
381 the
alk
= .
: |Votume4,Numere7 PHYSICAL REVIEW LETTERS Apri.1,i9@You iki
- a Micefor8"»gyascompared withtheobserved value fieldisnotrigorous, possibly becauseofasmi tayt2.810" sec, baremassoftheorder.of thepiohmass. = FARIt‘spossibletoapplythiskindofconsidera- Theabove-mentioned modelofelementary pe iC)tiontootherhyperons. Moreover, iftheFeyn- particleswillbestudiedinaseparatepaper. at iman~Gell-Mann couplingschemesuchas = by Hie(nné)isformallyextendedto(Krev),etc.as ThisworkwapsupportedpysheU.S.Atomic ing Aulihasbeentriedbysomepeople,alltheobserved: EnergyCommisajop. fore 44 decay,processes maybecovered. Herewe $3:G.Taylor,Phys.Rev,110,1216(1958): eren| wouldliketopointoutthatifallbaryonsshould j1..C-Polkinghorne, Nuovoelmento8,179and761bpub,ig satistyEqs.(4)and(6),theratiosg4/G,and Oegolbergerant6.B,TrPhys.Rrays £4'/Gx mustbeapproximately common constants. 410,1478(asa) “eSB: Treiman,Phys.Rev, at.otfinalremarkconcernsthetheoreticalbasis, aaInternational Con|NUC (|!|g__fortheassumptionsmadehere.Ifthebaryons ferenceonHigh-EnergyPhysics,Kiev,1989(unMA: TAT!,yrage278derivedfromsomefundamental fieldywhich ubiiches) StucFiar‘gh’vossessesaninvarianceunderatransformation 5M.L.Goldbergerand’s,B,Trelman,Phys,Rev.|TllintH{|Stigofthetypey-explia-7y,)y," thentherewillbea.«110,1178(1958);M.L.Goldbérger,Neva,nadeborHIT!42conservation ofthepseudovector change-current. PAYS.2,7971950. Jan)IAIigd Afiniteobserved masscanbecompatioic with withtheiFacerareaculateslatmebonighA tyi theconservation iftheparticleiscoupledwitha ytaxialvectorunaccounted for7 resi BeitHI bosonaswas-noted inEq.(1).. "AgainEq.(5)andthesubsequent conclusions are The! sa)if—7ThissituationmaybeunderstoodbymakinganessentiallythesamesetreeC.H,Albright,Phys.toieiyi SnenGeytothetheoryofsupercondctivity origi-av.114,1648195)sodBySabla,Piys,hesteeebati, natedbyBardeen,Coopér,andSchrieffer,* and1650(1959),whicharebasedontheGoldberger Thebay refinedbyBogoliubov.!® Theregaugeinvariance, TFeimanmethod.FortheA-dueaysavebelow,oeteft theenergygap,andthecollectiveexcitationsare,Tenaglla,Nuovocimento14,499(1959). La'F.Giirsey(privatecommunication) hasrecently F iAtf logicallyrelatedtoeachotheraswasshownbyobtainedsimilarreaults.onthexdecaybaseg'onthisWeis -theauthor." InthepresenteaseweRaveonlytoy, invariance, Wenseonteshecifythe'interactionAEN replacethemby'y,invariance, baryonmass, ofthe#field,which'may.be ofthenonlinear Heisen- peandthemesons. Infact,themathematical me- bergtype,ofduetoanintermaiate boventard om ry: thodused insuperconductivity may betalten over fromame Ang"tostudytheself-energy problemofelementary 9,RationUN.Cooper,andFRSobran) |R1 eae particles. itisinteresting thatpseudoscalar Phys:Rev.106,162(1957). EYE.meson snomateny epee ane, Comaht meme fon ghtP Shirkov, ANewMethod intheTheory ofSuporeon. Weyboundstatesofbaryonpairs.‘Thenonzero ductivity(AcademyofSelencesofUSSR,Mascon ots ‘alimeson-massesandbaryonmasssplittingwould1958) ofa eyfi“ndizatethatthey,invarianceoftheparebarjor_ "ty.nambe,Phys.Rev.11,648seo, late cae
-the| alWaataenguin aatouly Queonidtonpin gougeuvoromer iSlaroken,)te] HH\ ~ SoyAyPr- tion cea OO
of| BSii ERRATUM hoch«oteb a4ene taeroilasclastprckayofene BR;iHELICITYOFNEGATIVEMUONSFROMPIONmeasurements haveyieldedanatuendeaovatconsan) DECAY. W.A.Love,8,Marder, I.Nadethaft, fortheasymmetry parameter a,as.defined in‘|toa|IandR.T.Siegel(Phys.Rev.Letters2,107thigLetter,of@=(0.64+0.58)%. Ttthusappearsave] geeft(1959). thatinpentanetheB’?isdepolarized’ byoneofmat SeHi thevarious.interactions (quadrupole coupling, conigal| JpthisLetterwepresented preliminary results, multipleelectroncapturewnaloss,etc.)which-fconaflHofanéxperimentdesignedtomeasuretheforward-mightcausespinreorientation. Therefore,ajfoteal|| backward asymmetry of#raysemittedfromBY,definiteconclasion aboutthehelicityoftheneg-“ifSan]gen [ishfasbeenproducedbyabsorption ofpolar~ ativemuoncannotbedrawnfromcorcennieto“fslidMaul! izedmuonsincarbon(inC,H,,).Continuing date. ofles:Shed 3a2
Sault
it
OldFieldTheoryNotes
Btn f088SshDharDodeanOlokn Ande ke eae a -©Raga.shoankeasond _-sinumehiea _&,OZ_woe
=ORipatiecedee a
—--® Somdorsd Tuwetio 8 ee
-—-—-f Tpananenass chonere: -ee
Iii Rely
Gontents ofthis section:
re)1,Variouspageinvolvingdiscretesymmetries likeP,C,T,G: a)how the field operators transofmr
b)howthecreation operators transform 2c)whattodoabout squares ofoperators likeP”,7”etc.
d) time reversal
e)G-parity
f)parity andeffect onlagrangian and current; pairs like BBandFF .
g)charge conjugation and pairs
2.Reviews offield theory method
a)the two methods offield theory
b) the U-matrix and the S-matrix
c)another review ofthe two methods, including Wick
d)eescattering asanexample
3.Feynman rules for QED.
1,,‘The Reduction Formulas
5.Propagator functions
6,Pield theory Norms (like C,etc.)
7.More commutators and Ppopazators8.Thephoton spinsumbisiness9.The transverse photon propagator
10, The old pion nucleon page.
11, Some detail on the Dirac current.
12.Dimensions infieldtheory. fo)13. Unitarity infield theory.
14. Poincare versus Lorentz groups.
15. Crossing infield theory.
16, Noether Theorem for internal and external symmetries (various verions)
17, Feynman graph combinatorics, loops, internal lines etc .
18,Feynman rule forvertex inspinor QEDandscalar QED.
PeProofbyinduction ofFeynmanIdentity.
Field Operator Transformations
S 1.Parity: eat6's+4%)
OXG4)&=to(-%,t)PAGHF =~Al-x,t)
2.Charge Conjugation: i"Bamc’ =go _BHO =CVMH= ivtth= CV
GA@C =-A®)
3.Time Reversal:
S4caad =29%-t)
SVGNS =THe -t)=i8887K-t) SAGNS' =-AG-b)
—heTCP: 2 Cees)8)(83S==a*-*) .
(©63) Yq)(63 =4b¥FV(-x)
(PG3) A®(ees)! =-ACX)
5.Poincare:
UWA) 9OSAA =O(Ax+a) VAsa)WG)UICA)=SCA) (Anta) Us) AG) Ua) =Cry AYAx+e)
Additional information:
.
-\
- CHC =-H WaoHiVVeM-THTesHaye “oy yon Savsay=Nv a=Av% : a pe=ieBy S=Peleg
mcahaGlaTaHi? ;
ateTete -f Ny= Sy+40%
=-¥
Pam G=+aGk Recen=wGRs) GB vaevt. +doy Raceag =—ows
Savy =aw gba S= AQ 4
GawHC =ao cata T= lps)
Be UK 4
Ublesd K=—b(-¢-8)
Wdips)«=—dtpas)
ES
ro)o_See=6SgeNCseseat ee
_@SxS"=ve=2BQ ee A ———j,9 NON = GO)2NEA CRA
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—— ye Ln gy-220. Ge\BS230f GB_SH1O= 2wit=ae Fe
Sap iat ftaesGames [Ra aen A =a\ Jospsaye soda] ze
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i
aa eo * ; °oy @d==ceegh=R=-1%*
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a a.
G=*l Geet - .
@R=eRh)y .RS= -BK Vv - -
2 Che-ac | Re=-EK v
~ 8ts-ce RG =+OK ;
‘Sone ofthese equa ionsagree, withLeeandWick, somedonot. Eg,(2.18) through _-tose (2.21)-0fLee-Wick-agree-with-the-abover -I-think-we vanBetrid‘Ufthediscagreement by_redefining theToperator sothat.newT=GR,orsomething like.that.OK,but. - Tdont want todothat now. What tokeep the above relations strictly for the BD
‘conventions+
.Time Reversal.
= 1.Weknow theeffect oftheoperator Son‘Yand bandd:
=\ >|
sys =TY Sb =bEP HS) de.
2.Now consider a"process" and the defined "time reversed process" +
' es re! We~<e8'| GIp.S> ewe@QO< hs
, '
~P3 ~33 mia es MW~BS)%\-253> ~eOF
Notice that electrons are not turned into positrons. It's just that
¢the final electron reverses its momentum and becomes the intial electron.
‘The energy does not change sign.
3.Now dothe usual trick toobtain:
Ss ro} Slesr =sBENT ID=BEeD0> =|-2-D
-\ 1. aa m= 88) S%3 13>
Well, notice that script T,as Ihave it,does not change anINstate toanOUTstate. Itjustchanges themonentum andspinofaplane wave
state and the state remains inthe same picture itstarted in.
4.Now assume QEDsowrite HasJ,AU where Jisthedirac current. What can
be said?
1 nn Vays molMM~LES)LIS \PPPBA'S
By ae ~Ss|TAs? Aw
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oO FieldTheorySession April24through264\QQ5”
Motivation: Iwas trying to_read p139ofFeynman onparton calculations
‘and was unalbe torecall clearly the trace business. Aiso, Ihave been ~
wanting tosee howfield thery looks inmyall—norm notation. Finally, I
have been interested inthe various pseudoféeldtheoretic cdlodlations ~
,ofthe massive vector mesons, All these things justified this long session.
Applied Field Theory. Ireviewed thebare bones offield theory, skipped over “'postderivations anddetails, otherwise I-Would ‘néver havefinished. 1found~that ‘there are basically two methods ofdoing calculations:- Rist Hephog: inthiswethow (1SZ)youuse’thereduction-formikas-to -~|oxpreas esEnatriz elenent interasofagiantvacuum expectation valueof -«>4-9tineordered product oflotsoffivtdss Prior-to~perturbation-theoryy - -~ithere isafield for each external particle, and these are all “full” fields,
7 —{One then goos-to aperturbation-theory- by-transforming-these full- --~
|Heisenberg Piture fields toINficldé viatheoperator U(t,+').. This~~~ |-caunes-all the-external -particle fielde-to become“in“fields (free. fields). -:andyou are left with U(~inf,+inf) insidé theT-product with these éxternalpofielas: ‘This:operator-U-is-then replaced-with-the. usual-exponentiated .—.|interaction Hamiltonian offree fields, and each ocourrence ofHisnormal
— {ordered to-prevenit the possibility -of-tadpole type graphieto.... _._ -
You are now ready todo“contractions”. Choose an"order" and insert
1the corresponding number..of H's-and d4x integrals. Then pair offthe. |_
fieldsinallpossible waysusingWickstheorem. Inthecaseofeeelastic @),seattering insecondorder, éachcontraction isaproduct of5propagators. |After doing all contractions, you put the huge mess back into your LSZ;Teduction whereupon alltheexternal_propagators are"amputated". Then _
you canconvert thewhole thing tok-space andgetoutthe feynman rules.
- ! "What égn besaid about nopns for this first method? First ofall, the“square root factor" or“external wayefunction” owwhatever youwant tocall|dtmustbethesameasthefactor occurring in(§;a)field/creator commutation|Felation, This"external wavefunotion"is détermined ineffect bythésizes ~~
|ofsingle particle states andbythesize ofthe.spinors forDiracions.|Remember that fields arethesame"size™ foreveryone. This, ingeneral
|"external wavefunction factor" ¥C,. Nevertheless, itigetill Cythat|Sours inthedefinition ofT, 7° “= aNowinmost norm systems C4=extersial wavefunction factor, butGas,|aeWeialisanotable exception. (seeKinewheetdaggers). ----—
‘ When all issaid and done, wehave amatrix clement Mpeyn given
~
|exattly by“the‘foynnan ruleswith-no external leg-factors+-the connection --- |between Mpand the all-norm Tp; ist
tTe=+469) Gel (“Gas)~Mean .| - vs cx
HInanyofthestandard (daggar) normsystems thisreduces to:
i :oO- Th=HCG)Wee _ |For Gasiorowies the result isinstead,
| -—-— | wee eee ee
| Tks +¢ Tlam:|ket ffA@atMine, a
‘——
re
4ae -e2-
oO ForGasiorowics itisasifhehadanextrafeynmanrulewhichreads:"Fora ~GS0hi ‘exterrial”lifie iiGlude afactor of (~)2r -
ForKallen theresult is:
~--) Ske=CY)Mhe, —Geate
-—~ ——~,,~,¥e should noteinpassing thatTo,isnotalways C¢independent. ‘hisisbecause thephasespace“knows” aboutCyinGertin novesyste, ——
- . like Bjorken Drell_and Gasiorowids.Inthisfirst method (1SZ) eachtimeyou“pull” aparticle outorthe~~~ state and into afield, you get afactor ofi. This issimply the ithat ~
occurs inthe inversion formula “(1279)page 27resulting fromatine——— derivative onafree field, The renormalization factor ofZIalways set
~~" "-““gqGal tootis,atidIuséVedormulized (physical) masses-amt couptings-in— ~— any. caloulation.
Second Method. This technique ismuch more efficient for doing acalculation.- -eeToate“wayINcreaton-and annihiletion-operators-for yourexternal -
particles, and you replace Sdirectly with the.exponentiated interaction
- ~-Hamtltonkany-Then“atonceyowdothe-contractionsyEach timeyou-contract-—— afield with acreation operator, you pick upone ofthose “external
wavefungtion-factore,-The-nuiibet:of-spacetime integrationsequalsthe- —— order of the calculation. It is this method that allows afast derivation
ad -of-the feynman-rules—f: theory. This isthe method used-by-Sakurai, —
fe) ‘peepages130etc.The factorial factor, Ineither perturbation method, there isalways an—~-—____—_n inthedenominator_for_an_nth order_caloulation, butsomehow thisnever__appears inthe result. The reason isthat there are n!"contractions" which
——--- —~belong tothesametopologically distinot_feynman graph, so.the factorial __
isalways cancelled.
Wicks Theorem, This isoneofthose combinatoric things that isbest proven
.—nlybyexamples, because thegeneral proofistoohardtounderstnad for more than one day. First, you prove Wicks theorem for two fields. Todo ~~___._this,youchoose aparticular timesequence soyoucanremove theT-product.Thenyouobserve thatthetimeordered thing differs fromthenormal ordered
__ _____._thing only bysome field commutators, and these are always o-numbers, 80 .
a the TorderandWorder canonlydiffer byac-itiiber. “Thenyoutakea-VEV~ toshow that that c-number isjust the VEVoftheTOP. ~“Next, YoumoveonWothreefields. Agaid; ‘picks definite tiveorders =~The game istoget "from" this particular order, "to" the normal order._-—"~"~ “This Tuvelves shifting“ all-plus fields to~the-R'LEFT where-they-can-do ~no harm. Each such "shift" causes aterm with apropagator for the two
— ---— interchanged"fielde; and-alt-the-other-fietds~just sit -theres—Eventuallyy—-- all fields that donot appear inpropagators end upnormal ordered. Ishould
_- ~~~“ do-a better proof-some-day-on-thisy—— ——
Note that any VEVofaTOPofanodd number offields must vanish, due—--—-- +directly to-Wieke—theorem—————— = =--——- —___-- —-.
a Oo
*
a
e
-3-
BisBNdlea quoerrcnser+ ie)Braces. Whenever there isaDirac particle inafeynman amplitude, tracesHIToccurwhentheamplitude iesquared, Theseariséfromtheprojectionoperators and the Inversion theorems. tt; -
te SS [+%$ z =m) (1+8H
~ ef - ~~ letye- [x@X sO] =AQXa X=KY
Rather than first write te amplitudes ,then squere them, then convert
|totraees, ste, "itis eaéiesttousethe("trave dtaprans'} wsillustrated énthe back ofpage 12, there for electron electron scattering insecond-“order QED;Theideaisto“form allthe-"unitarity- like"digrams-with =—~-appropriate statistics minus signs. Then each "line" may beclosed onto
|ttself-ané stands fora trace. Zachtime-you-hit-an external particley—
insert aprojector. Each time you hit avertex, insert the feynman rule
- +vertex: factor-(egy ~ie{*-)+—The-trace-cleses-when-you-return-to-starte --
‘Sometimes there isonly one trace, sometimes two orthore. Don't forget~ +‘the-propagators apropo. henyou:do--all~this, -you-have -at-once _/M/2and .|youarereadytogoto&crosssectionorwhatever. a - = | Compton provides another-example -ofthe.technique.—(See. page-15). -Again|you use vertex factors inthe various traces, except here ithappens|thatpolarization 4-vectors-are. tied_to eachvertex.. Also,in.this.example.
|therearerealfermionpropagators andthesetwomustbeinsidethetraces, (@!: Ingeneral, any quantity from the "top". unitarity graph_should_be.
|complex conjugated, eg, propagators with ie's,polarization vectors,
|coupling. constants. Usually..we donot worry_about such things. LoL
| In actually exeouting the traces, there are very important tricks to|beobserved. SeeBDonthis,Ihavenotgonethrough allthesethings here.
|
— |SpinSuge:ZonDinac_clentnonas Spinsummingistrivial, Aspinsummerély_ {clears the.projector ofits spin portion. Ifyou are averaging over initial. .spins, donotforget toputinaPoreachinitial fermion. Forfinal. fermions, summing isdonebysimply ignoring thespinpartoftheprojector.
. Porphotons sndmassive vector mesons thespinsumisnotsotrivial.Forexternal photons, thereabeOnlytwotransvérse polarisation states ~~although you could combine them téget circulars. Itisbest todophoton
polarizationsumsasthéVeryénd-of-a calculation (asis’not~thecase=~ for electrons). Basically you have todothe -sum explicitly. Ihave done
anexample on&separate sheet (ie;—the Compton-case)s "Note that-the three-1aafourdotproducts6fapolarization vectorwithitsmomentumvanish :-Cthe‘three-dot vaniswing isnotrelativistio- statement «+.-we-adjust -our-|frame byarranging forthistobetrue, thisistheradiation gauge) -+ For massive: vector mesons; there-are three polarizations toworry--about~
.|The four-dot condition still holds. The parallel polarisation vector gets
- !stretched; seedack-of page20:Againy-you~have-to.do-spin sumsexplicitly.
i|par, Howdoyoucomputefieldcomiiitateredtiméqiat‘tiie?You"~~~|Girfteyourlagrangian andcomputeyourcanonicalmomenta.Thenimpose oO |theequdltinecémPélé. Makemomentum éxpansionsandinversions; andther |Compute the creator comrels bylooking atequal tine. Then, using these
comrels, the théqual time fiéld Conimitators folaow, See pags 20. ~~~ ~
1
- \a.:
H -
i
2-
“
~
ad-4- . O ‘The REO decay. Now weare getting into the realm ofphenomonological= “field theory calculations, asin-wéakinteractions, For-the tho ~ _interaction with two pions, wepostulate aldgrangian,
. 4— SevigderdBsh4sshown onpage 24, this contains two terms involving the neutral rho.-——~- —-Thenyusing~the “"second method" we-oan-get-the-feynman-rutes-and—in- ==particular the decay matrix element, Inadaggar standard norm wefind:
~~ ~ Ty Here, pisthe4-momentum ofeither of__[Tal =aype tne tinal'pions, Thefactor of2arises“because therearetwocontributing terms~ Using this amplitude, we inthe lagrangian.°GanqaLckIy compatetie— ~ns width ofthe RHO )into these two pions:
eeeg WE(|—Sm - an Tgete =spegtte oe _
-From_theLagrangian youcanshowthatgisdimensionless. Here,Ihave — xumumixarerxtiaxt trivially averaged over initial polariaations ofthe RHO,
a+ —.—+Theresult isindependent ofpolarisation anyway. 9_ Usingthefirstmethod, weoanshowthat: —— ee
Scale wee a - OCF. ©AONFOSE ND=_AMLA(S)- AigREGAL _
Except for normalization, this isthe result quoted inGas page 455 top.
--=~ —“We-eould havearrived atthisformfrouLorentz—invariance-without— ——
ever using alagrangian and perturbation theory. Thus, wecan justify
— - that phenom, lagrengiar-in this-ways——————— ———-———- —- ~~
———-———Rhe-production. Itried doing-the-i#ttie-perturbation-calc-thet-Gas—shows——~ but Idid not get the right thing exactly. Atthat point Istarted this
summary "~~ -- oe
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oS
1.Adjointing operators.
+ae ABle> =\q(AS)=BA Aine v+
Whenyouadjointaproductoftwo <elSe=<aloperators, the order reverses and =thereare'nominussignsevenif Salat=<4these are anticonmuting dirac
fields. Also, this reversal ofthe operator, the adjoint operation
order has nothing whatsoever todo has nothing todowith transposing
with matrices. When you adjoint an and complex conjugating anything.
2.Adjointing matrix representaions ofoperators..
Square matrices have the above property which isone reason that matrices
can beused torepresent operators inthe first place. With matrices, the
conventional adjoint operation applies: twiddle and *.
3.Playing with the dirac current operator.
P Inwither the first orsecond quantized theories, the dirac current jyis& hermitian. Theonlypossible pointofconfusion isthatinthesecond
quantized theory you donot throw inaminus sign because atnopoint
are the fields anticommuted. See 1above.
+at oentAg yee A=LAG =CDs MV
it. + A“fe LTH =de
Notice that the adjoint ofanumber isjust its *. Asshown inlecture 22
of230A, the hermiticity ofjisunaffected bynormal ordering.
hsWhen does that anticonmutation minus sign come into play? when you
charge conjugate the'direc current!
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4.22.76
Unitarity inField Theory - -
Ce 1.Because youaredealing withahermitian lagrangian, unitarity inafield
theory isautonatically repected: ‘So 18Lérénts invariance and cotiger'ved
quantum numbers; Wealways make’suchabigdealabouthowhardit“isto
satiefy unitarity, but here in-field-theory-you dont-even give-it a-thoughty
--‘Formally, unitarity follows from$=el,Butyoucanshowunitarity. onder
ee byorder in,say,QED.Herearesomegraphs:
.
.- - ae eee
-OreTy=YadTeTey - :
dort =0 ~Lordsr &..
- dmFA=ade (ed)x(1) 7]ondana
2.Infact, unitarity istrivial inQD. ItHolds for each graph. #11 you ao
isputtheintermediate lines-on massshell€utkosky-like. -—~
— 3.Ofcourse theothersymmetries follow frominvariances of.theLagrangian.6 Suchas:lorents invariance, isospin invariance, andsoon._
:
4.Well, what ever happened topion-nucleom field theory? Igather that the _
large dimensionless coupling ruins perturbation theory, but what about
theaxiomatics? Andthenthere isthéproblem offundamental entitities
which appear inthe lagrangian. . :
Comments onthePoincare groupversus Lorentz groupproblem,
.‘Theproblemhereisthis:whyistheRHOsometimesavector(4,3)oftheLG,bat fo}other times itisa(0,1) ofthe same group, and other times amass=m, spinal
rep ofthe Poincare group. How are these ideas all related?
Idont have afull answer yet. Here are some ideas though.
1,IkmowthatIcanconstruct spinor states oftheform\p™*) whichtransform as
the vector rep ofIG. Insuch acase, neither ofthese indices isahelicity. T
canthenmakeavector wavefunction bychanging basis, sothat wehave|p*) and
@Xp) asourmomentum-space wavefunction. This transforms inobvious way
u(t)Ite) =2%ite)
orsomething like that. This thing isjust afunction. Obviously, you also know how
the state transforms:
uJ u(L)/*)=LiAp’)
Next, you should beable tofind out how the creation operator transforms. Then finally
you find out how £he field operator goes.
‘The conclusion isroughly this: the field operator (x) offield theory
transforms just like the wavefunction (spinor) @(p), except perhaps under translations.
e Sincethesetwothings areroughly related bysimplefourier transform, thatisjust what you expect. So, both wavefunctions and fields should belong torepresentations
ofthe Lorent group rather than the Poincare group. Atleast wecan see the Lorentz
group inboth pand xspace.
2.Onthe other hand, ifyou take any massive particle and put itatrest, itmst
have some spin sdue tothe Poicare group. Ie, itmust belong tosome rep ofthe Poincare.
The RHO has s=1 for sure. The spinor state Iwould normally make for the RHO would
transform with (0,s)= (0,1), not (3,4). This latter hasthe wrong number ofstates (4).
Nevertheless, intheories ofthe RHO, the three states are often embedded into a
vector field. Ithink this iswhy you need guage invariance, tocompensate for this
erreonsous, embedding ofthree objects into a4-plet.
InS-matrix theory, Idont think there isany reason atoll tothink ofthe
RHO as a4-vector.
lo]
NewFieldTheory Notes
Sadow$nNous Jd.Momoles) ee
OoSoran)oa;EstiGovaney
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fo oN
Grossing infieldtheory. ° duly7,1977
ra)1.Considerthemainreductionformila(16.81)ofBjorkenDrell:thefieldsinsidethe Toperator have adefinite order. Now lets goatonce toperturbation theory
where this T-function isexpanded asin(17.22). The INfields in(17.22) are
ordered inthe same way asthe full fields in(16.81).
Now, the interaction hamiltonian musk contain aneven number offermion
fields (as, eg, inQED) toconserve fermion superselection. Thus, you can move
afield through thesymbolic exponential in(17.22) topline. Thefact that Hy
isnormal ordered makes nodifferen to,this argument. Thepoint isthat ithas
even number of fermion fields.
2.Now wehave to say something about how weshall write states. Notice that:
aati =Wy
= +
SB sett =Auey = Sola
This isimportant. Wehave defined astate /1,2) andnever deal with astate /2,1),
atleast not here. When you conjugate the state, the ordering ofthe labels inthe
Ce)finalstateisthereverseoftheorderoftheannihilation operators! Oftenpeopledonot order things inthis wy, but itseems right tome.
3.Therefore, itwould bemyinclination toreverse the label order for the final
state in(16.81). Then this equation combined with the other equation would give:
poe7CytanSeas)
me ~ — <— x =GY SeeTSagQed(Ot*)4sew) SeGy) BrSey
rR
¢SAT(ASMHEY-BYR)QO)+QGm))A) BAAS Pm pes —viK
=o) (Qala) Malye) =MACgn) QabG)-- PsdOn)S ‘Yt
Now you can see what crossing will do. Without changing any orders, you can cross
particle y,,=P, from thefinal state toaposition totheleft ofa,=x;inthe
initiel state. The total 4-momentum gets negated because fgoes to£8.Any
6 spinindex.. well,lookatself-reduction page.Au(p,s)infbecomes u(~p,s)which isv(p,s) orsomething, soessentially spin stays put. This crossing is
inaccord with Taylor's crossing rule, except they have nophase "over the top"
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C=Ws =aR Py =SyOetoe) +2
a=-98
Yofoy=3h=fee LErsvoo =©hh
Te oth Ferd, =Cer woryh - ;aa Wakexhotennoi=dn[ear xe )¥seroea)aiisXg™t=0 fe)=i
So,ecmamench cansat
P=Wsy-d =TELiga -egxgnst
= ye[eva -eal~(pew)
Soy
S vByeWa, eT wT) =WwW”
3,100
Brrovss) heoryea. ove~
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SeaG9-M =a(GO=-BH) 2CA®)O(KAR) =HO]
B® dab+oye ~ACAyaX)O)
Que, FaLaeyal
oo
owdofindAWolk.CavatsmutMeedeasiperrnetay”
LagIpeya tH,soy=LO)
$@)=Lar= wefe*e(eal
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DrmslaLeMaynatEL), Medat,OlearmswaleeseodeUnvariouk asin’ BikGeady NPUL.Wkadook
a"adord=Vek3[eeola*leeGo)
goes Seba EBAY
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Se,0Raspongian Like,LOR=EQAve-Aeussatsuvarah| aw back toMe:
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Prortoasln yawcheeseAaudetudkAueALaNSf
TPH [teev]h-PL=coanned AWelablan ourrod
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D=VAS=Decheye=sanke
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maWAwWae&wedex THhidend Sawn
EBtraclewel Dave LaHQ :
Va &varies- “Vowaracequoncinie aetehad7
Dust: fraYeonmuckd” Jupwman qragh[reamaJ2Spteedl, QuininQue: G-y4+r=L
PratsstedwithTratere, OfUfcoached.thawrmushbe okSackW-DuleuelMine.4THT),Denyohoe OF&SurqpegeweLerpe,SackoddahWl Laefrenan
Conlon FonSnoags,r=.
QnomornOe:Ae“yteyets¢,a=alt+s.
Quran: imaCainsWideoaaduackin, Lagann DashUrb
pushLolde|re\rour:de@rj-Y+ 4
al:Aanetalemd,ogptinaenkadocomeched qeaghe.
ods ii offtogetafiniteintegral.
re) 1,Consider anintegral ofthefollowing formwherethereissomeparameter inside:a
Ie)=ANSe .pxra
Itisobvious from inspectiong that this thing islogarithmically divergence. Here
are three ways ofputting inacutoff:.
a)just stop the integral atafinite value:
(e .dx n Ata TG) BGA) =BE=AeG|y =a(S)0
b)doasubtraction: es
T@)> Te)-X(s) =de|Xa KER8
axa eth =aa(S)-2- (5)=dn(A).
. Bysubtracting theintegrand atsome point (the subtraction point) yougetanew
fo] integral whichisconvergent, although hereIevaluated thethingbyshortcut.
c)addamultiplying factor: ofa derNGeSdb(coe ) SM. (=) =Ge x » SER)=2G,A)=\xaGr aly SenKAQW
Moo. Gb aK ,A). Yan xem en xan)An* :
2.Comment: forfinite LAMBDA each ofthese methods gives aslightly different
result. Butforlarge LAMBDA allresults arethesame, basically 1n(I/a), which
shows upthelogarithmic divergence. Youcanuseanymethod youwant:
d)changethenumber ofdimensions ofspacetime, Idon'tknowhowtodothisnow.
‘TheMinimal Substitution andFeynman rules forvarious kinds ofQED.
1,First,calibratethingsbydoingregularQEDalaBD.Hereisthelagrangina +a)and the minimal substitution:
La+Fgm) Ap>AHR—Apsig3potted YS
“Tha, te= VK foniw(155).
Now get the feynman rule for aQED vertex byb soneidering asimple S-matrix element
(even though kinematically blocked):
,€N
y.‘we Dome=Qk S-alp) =i<phkl FA pS
Lo NN WBley wc)
=TG)Piey| wey€”* eLAS
LL,this istheFeynman Rule fortheQEDvertex.
2,Nowredo this foracharged scalar field. First, notice that althought there is
a44inthelegrangian foraneutral scalar field asin(12.2), there isnone in
thecharged case because youadded twoKlein Gordon fields, see (12.54). Also,
youmustkeepthedagger alsoond, togettherightresult:
a ~Pa an . o= Hg O*R)~ whFY >Brod,
v .Tan +A ave + a=Geld HeORVied =ied” [@da- F@e)]
e
—apne. a yop=2Gelied(@a)4FOND{—) &\¥*
fo} Sc) Cit)
. woh at . A”=iGiele* Ghtip) =[rie rte .
LyFeynman Ruleforscalar QEDvertex.
—~-DeakQusypmmadio Adincdomie sloademset.
Ss SpgLOcak ote” aSoOS)
Ws ete PA)TESeGW =eo.
.a gancoset, seamed - .
es Qe a
—@Noadij PY=Rahg=wksywere )eOe).
(RB soma,TP2oman ILEN, adsTT=8/3 Tank,
6YeaemDonfaseweshalloma
AN S804 aSoepee
Oy 2PyeM22le)=WGN)TG) Sal iat
Sa adsfodatas dlQain1ladVpconeocasatits Dersrboah
©NeaadoanSlsensebeg
[email protected] Pe OPPakCe a1East otaley ondasypendeaantee
6DDame: Yo? =iatae
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1Raaamose +(Colegun SaskJAN) ;ae {ee Oe ee |Radin@ouabe J —.. ee
~LQys.Ostasavers suannaakssano“VPCauiSsagaumahc). ouchfagafar|—onaareonuencsannoa WY
Weawea-y eh ae ee_ WP=Ga eT Wea a We .
8 OP2Cer) 53,Craeeweke)
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ns aN sR (Wer gatee Pesy)
>RA. See we
~ |S&B =Se =ee —
a
Howtodefine theLorentz andTranslation Generators. _
6 1.First,letsgetstraight thenotionofusingacoordinate transformation Rto
define atransformation inthe Q.M. Hilbert Space. You define Rinterms ofRin
this way: ot<ARley=CRED
Youcould, ifyouwanted, make analternative definition byusing Rinstead ofR7.
BUT, ifRinaddition tobeing atransformation isalso representing some group,
thenyoumustuseR™sothattheoperators Ralsorepresent thegroup, ie,you
want:RyRo=Rytobereflected byRR,=Ryfortheoperators. Notice, bythe
way, that the above setup isjust the usual "regular representation":
Ralm= g(a
22.Sohere ishowyoudoangular momentum. Define fil” like so:
iow show 4cas gy=CM axel
Here both Mand@Mwillsatisfy thesameLieAlgebra. (Ifyouweretouseopposite
signthatwouldnotbetrue).Gotoinfinitesimal tolearnthat: (nie:M=Meo).ne, 8<A ey=TW"Lysx2QL
vga sy . 7 Bae=ALSp SSI IEbE)=LGEI-W3*) GO=HMQO)
3.Since thetranslation group isabelian andtyto=toty, wedont have toworry about.
thedistinc£ion made above. This means that the sign ofoperator Pisjust mm
aconvention, soIwill doitthe way Ihave always done itand inaway thet
duplicates classical definition ofthe relation between angular and linear momentum.
Soz i rNwig RASIN Vey=Lavalay ok TINY =Sa) .
SRL PALEY =iee®e@ =Pde
“Men: WeCPLR eg, BY=RRP =Ls
4. Summ of results:
2 n SAPSY =PRG whew PAshy
fe) SOE =MIA wre AY=a(IEX’9")
GQawrabupition YoPndudssew. eo)
OoVsowsty a=(RSOIR oean c=ceefpguunbin space
S& spe agad:
we,= Wwe
EVAL= eaedo205 Ted=LooaieSoIC .
©Awraranetein witspineGomalrna De905
HGS=LRA =RAR) .
> . y a 5)5) z Nae:TRAE =KS,IRAE) =RPERRN(R KO=CDG(fer),
WeearwithMioe
SreLQ BD=RyRelea waAAHKEARY= CARL KyGlad.
OvcBad =<BR ond KEAADE RG ALES .
So,RyyissomeFDRofsomegroup, ie,oftheLorentz group (need notbeirreducible, eg,
might bethe4x4Dirac spinor rep), andRijactsonlyinthespinspace. Notethatthe
useofRj;andnotRyiscorrect intheconsistency sensementioned onearlier page.
Ie, now wehave three different representations ofthe Lorentz group acting indifferent.
spaces.
3.Nowwecandefine M°Ywhen spin ispresent:
—iot™ iow iow” <0)SMVay =LEM CE Tatex|badd
ConerDredenanstakionbosey we—a ~-is L= ev =oe am
sieln’@ we —e he”=Sowrowel So oeASR=ABI+108
AWww
Se\W olllas=HM Cadile
ion ~io,=SOMay TSP" Ysdead
Ohua walowe:
Oweeay =Gere aimtiay =feanay
éei\SoVes=jee"iydele =>delwiley=(weldels
Cordhanae :
SalWelt) =FRY,Beene)
ceded =UNI
Heil WAR =dre Mer +10171 &d9
=BUHRIARD Fd WYded)
6={Fed HUM }Grraceved
=(HM Goalde.dd
Qyue:
Sele =Wels a
waa LMS =AU Ly
fo]
WoswQunchuon Nenad:
6 '
BO =C= o@)
TeWYuaeyak
BOO=jo eleacy=~ie<xl tt=~10HL8
BM agt
BR) =rigGeilPQ =C1a)(*) Re)=+addi®)
CGdeggomarous usmbsyoSleeaspuvies. QunsWY.
WW@=-icWLO
ro) :
-
io]
_. Introduction ofFields. Global, comrels, variations, and2i”,
01.Wehavealreadyintroducedtheobject.Risasaspinmatrix,Rxasatransformation
onx,and asanoperator intheQMHilbert Space. Wenowdefine a"field with spin”
as follows:
u s =\
RE =KRROR =Ry (Rs)
Youhavetodoitjustthisway.If,eg,youweretoswitch theRandR-!which
surround the field, you would find agroup multiplication inconsistency. Notice
that ifyou take the VEV ofthis field, thus getting the classical field, you
would find aclassical field that fransforms consistently with out previous
wavefunction transformations.
2.For linear and angular momentum, then, wehave shown that:
cot” —iow” ~ieW™, ie <eCgee -le kh&l€ yer)
” te A|sige LieQeBe)© =&(xxa)
3.Now put insmall paramters toget the commutation relations:
wey] = . = Wr. seLW”,Resl=SHO =-lLeLiely&@
hap]P",BO)=SR =-ignP*GH)
v qe’ AW WD) 5 >(THe,HP]=Tg,BOD=fewwg+WETRie [8@, PA]=Baw ="bo
&Examine &O(8) doQooun
. wr to6 By=XIN"|g
: °) »ys—=—*Vva Son (Taree, w=4(perry +22.)RE
Comment: for afree field wecan make this expansion:
‘
Qa By ; a 6 ea oefoapean Ar) +<ppcnled ae,4
where you sum over all momenta and spin states n.Iamallowing here for achange
ofspin basis. Then wehave:
=Sy omot R&S=ridLR&)AG+gh"alGerd]
*, wor) > CABH MOMS =<olROlpH =QE)
Thus, byusing @single particle state onone side wecan get aconnection between
the regular wavefunctions and the fields; wecould have used this toshow the
transformation properties ofthe fields ifweliked.
,
Dilatakim. Gammatot
@©Wine 0wouelmetinn Seat Komakrrmnakin oefallows:
[oeCiad) a= 2e(ds)
Lenaval, RKaarsrmalhbogasrat
ADO =1CA4 BSH) =Bantad B=i(aetd,)
@Funitidamepriaim YunMoaDaFranoraadd:
iad had da 0ef ee =Sf&&
@®Godwn boe-dedmaln dogit
+eLD,GH]=BQ =-iBde
>(ae, DL=-Baw
6
©Wao: [BW =[Lean,iGevxH\] =0
LB,&=Licastan ,iat]=-(ia,a4]=atge-0
1B,BL=~em,%|=0
lo]
Diototin ondGrae Func.
AdGasesDestsadeaon emrevamancr FyyerDaaggeeae
LATAWN: KESNQ =ATTHYLor gay... QEeats
Sx deee Maw
. ey, derdytetde[scotavoor]> Cwm) =QQ &Om,eeAX)
©WowLekAfounsr:
Pra e=Yay«deoS SOen)
en
Vino ait XfAK so dike Malu. Oe
on Lala) dyes doe=GeO88"Vad ©‘ OQ) Otel.xh)
P Agenda LtA =ae erldsPe/A+Qald)-
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weraonhueiyQaoySrnareodswverauct,
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BPE br=GPRD EEGD Gp) TOCrates pow)bo,24 \ L v z\ aa
eA COMME CTS) U3
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& 4-Ba -Pe Meybs) =OD POCasey Pan)
@Suy toook dammed (Senay aMeer)
we|CPO, xd]=Ba Owe Sows a2alone \
©. Sage BILE rs Pe) = Ba—Ye
0dawLIroy=Y-Za
Sonmony JoPY,"ant Dz
6©Licgeo: (=Map=MCieemnat)
wvToe) =a|teeth int5gt =han?|
[em] =3{gyP- e*|
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aistt? ~ient” ie io @ €we =ek aew)
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e ke © =©d&(Cek)
@Goose :
Lae,me)=Uy RO Ly=Reay4Lady
Tao, MY=Baie Bnax
\%©O, Pl=Bea Dei(dsotan)
Odeo - e HEP=ieta x3) Iw" =BS
SMG =a(eroe3"] 84+28)
onEber Thomann ohThomann Paaniaait Baath
oeCit tySoda 2eSFGete
Gat) =AigiagengeedieSETSHES|SET
~Woguatuet andapOudkvuligake Sgh
0=eSMeena VigdeSSE RGge)
eae NeSeeSe
TG =GE)Se SSE FR
TDaatiges ioRG) inMallya polafe
Tiss. Fumckiows amdFou. Tromalown..
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Ge_iPew=Kooy)=LEDGrsi)i 2 Ou
Qe gy
FGreKa)=SaudyeRew).-BoGocyr)Peng,ts)laa
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BIKES SIM)
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Peak:doinormnakChenin, WateratkLogharr)dogel
tytyeiGaysPeg =—26%G2)
oR aL a 2 Ldydy, |S SY aien --\sv . \ws(Saal Saal esany\ (Baal
aa ann cae
9 les Youn
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Qo Ges Sp Ayres Syoe BIW)WOH)++8404]YsSSTAH)FA--39s
: SY =2» Qeed:dono BPSCoy) TR
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2? sayy
ro)TruncatedGreensFunctions.
1.RBcall BDdiscussion ofINfields, and OUT fields. These are supposed tobefree,
non-interacting asymptotic states. Wheo you examine anS-matrix element, you dosoin
terms ofthese free, asymptotic states, id,Sp=(£,0UT/i,IN) =(f,IN/S/i,IN). Ithink
the idea isthat the external lines ofadiagram representing anS-matrix element are
supposed torepresent these free states. Such "states" donot self-interact, andthus
donot“build up"thefactor Z,duetotheir self-interaction. Theexternal lines are
supposed tobepropagating asfree particles without even self-interactions.
Sohowwould yourelate J,totheunrenormalized field ?Well, the
propagator youwould getbyfourier transforming (,.,,,) would bejust 1/(p-n?),
whereastheunrenormed (but"interacting" )fieldswouldyield(9,f,)=2/(p?-n").
Thus, you can imagine how these two kinds offields might berelated byasquare root
ofZ,,assuggested inthe"asymptotic condition" (16.20) ofBR.
Dont beconfused. Here weare discussing the relation between afree-field (which
makes just the bare propagator) andaninteracting field. Inthepast Ihave often
compared twokinds ofinteracting fields: theunrenormalized which Icallg., andthe
renormalized whichIcall$.Thesetwointeractingfieldsaresimplyrelatedby fe)amiltiplicative factor.
2,Iclaim that the ISZ formula really says this: compute your momentum-space
Greens function, then knowk off the contributions which are due toself-energy
insertions onexternal lines. The result isthe "truncated oramputated" Greens function
and isprecisely the S-matrix element you seek. You knock off oramputated the
external Greens Propagators because you are supposed tobecomputigg S-matrix element
between asymptotic states. Inother words, anS-matrix element isamilti-pole residue
ofaGreens function. Then the object you get issimlar tothe 1PI function (which also
has its external self-energy insertions knocked out) but the amputated guy isnot IpI.
‘e)
Qseinask blesonoorSiderdogit:
vy wv rae(uty ge)=Ue)ew)... Qar’) aCOsbayseQn)
ERK Banos ae
=@)Nya dy,CS “ee oyes 4+COs) COeat)Shy Ye)
- =ZalhQeny---- feo)|B
The+choices depend whether particles isingo oroutgo anddepends onbabelling
convention, butthesignhasnoeffectonthesignofBox”operator. Sohefeyousee
the"amputation" going oninthex-space ibtelf. Thus, wearrive atprecisely
BD's LSZformila (16.81) except Idont have any2's. Thereason issimply that
Iamusing renormalized fields, whereas they areusing unrenormalized fields. Icould
gettheirresultifImaketheusualreplacementthat$=(23)?$,.Then 0O youshould imagine BD's fields as9's.
4.Bytheway, according tothewayIdidit,Ijust pulled offexternal propagator
insertions. Thus, the first term inmyS-matrxi element here called iT will bethe
1PIproper vertex. Te,thisistheconnected-part oftheT-matrix.- . -
5.‘TheseZ'ssitting outside theformula inBDaresometimes called“external waye-
function renormalization", which terminology relates toparagraph 1above. For me,
these Z's arejust the Z'sthat getyou toarenormalized Greens function, soIdont
have tocallthen anything. .
le 2
3.Implement theabove conments inscalar field theory, say#*orsomething. Then
a)letG(pjs-++-p,) betherenormalized andthusfiniten-pointGreensfunction.It contains the overall delta ofcourse. Now, remember that LSZ gives you amatrix
element ofS-1rather than 8,soyouarereally getting theT-matrix. Lets saythat.
S-1-iT which isastandard form. Then here iswhat wecansay: . .
Gy @ @) ‘Duby Pe)=Sigeage:MtCea)wodong 6Tq: i) . ‘ :: anCry an soigt<T
Here Iamusing all-script functions inmomentum space because these arethécomplete
ones which include the(2PT)* (...) overall. Nowlets expose allthose delta functions
and the result ismuch simpler, and wegotonon-script functions:
aay OG Cy ()GSConte,teal) =FRYGG SRY EVGates Bary,OD)
Here Iputthelast momentum arguement inparentheses because really p,,should beregarded
asafunction ofthe other n-1 momenta, since weare now conserving 4-momenta. So
invert togett,
fe)ce :@= ow=8ITCeyeyees Gad=LEI TE) GCate tes),
.,
4eS) Qytewey, LFFB8y=HM.) &)
@):
24a= |Hy)
e- Lo Qk Sw= 0G) wr B= LL.
: poe . =S
wot Qua, D= Get) amd:
&) woom. 1 + sozt RoKTGuta ts)=P) Gee) Gee) &Cerys few)
Ofcourse theimplication isthatyouaretaking alimit hereaspj”goonshell. This
ishowyou‘aretakingamiti-poleresiduetogettheS-matrixelement.Ifwewerenow oOotoputthings back into x-space, wewould get p*=-d2soeach factor makes a(-1)
which combines with each facotrs's 1/itomake aplus i.Thus:
QaQaeda Voge sO, ©acimRadiomanrmratizeMeowwebaad“Oued(cinecle). :
~\arleamgw+QTE:.” Spa@ %8@)°Me ;oy, e* [aveg BIH)
@AvDiormole + . _:
—Tk<ascain? SGI? . Nae:
:-Tin g
chu LAVAS =LADGUMMY =LawMGat
=LAWCAMDAN : : le):-EWG.” .
oa\iisy=LAM WYP : a :
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SoweconSay -
=Saredy[2QE)MEAG)+4OTOPT) \aa\© . —_7
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Yakm
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