Quantum field theory

815 Pages·3.58 MB·English

by  Peskin, Schroeder.

Part I
Feynman Diagrams
and
Quantum Electrodynamics

Chapter (cid:0)
Invitation(cid:0) Pair Production
in e(cid:0)e(cid:0) Annihilation
The main purpose of Part I of this book is to develop the basic calculational
methodofquantum(cid:0)eldtheory(cid:1)theformalismofFeynmandiagrams(cid:2)Wewill
thenapplythisformalismtocomputationsinQuantumElectrodynamics(cid:1)the
quantum theory of electrons and photons(cid:2)
QuantumElectrodynamics(cid:3)QED(cid:4)isperhapsthebestfundamentalphys(cid:5)
ical theory we have(cid:2) The theory is formulated as a set of simple equations
(cid:3)Maxwell(cid:6)sequationsandtheDiracequation(cid:4)whoseformisessentiallydeter(cid:5)
mined by relativistic invariance(cid:2) The quantum(cid:5)mechanical solutions of these
equationsgivedetailedpredictionsofelectromagneticphenomenafrommacro(cid:5)
scopic distances down to regionsseveral hundred times smaller than the pro(cid:5)
ton(cid:2)
Feynmandiagramsprovideforthiseleganttheoryanequallyelegantpro(cid:5)
cedurefor calculation(cid:7)Imagine aprocessthat can be carriedout by electrons
and photons(cid:1) draw a diagram(cid:1) and then use the diagram to write the mathe(cid:5)
matical formofthe quantum(cid:5)mechanicalamplitude forthatprocesstooccur(cid:2)
In this (cid:0)rst part of the book we will develop both the theory of QED
and the method of Feynman diagrams from the basic principles of quantum
mechanics and relativity(cid:2) Eventually(cid:1) we will arrive at a point where we can
calculate observable quantities that are of great interest in the study of ele(cid:5)
mentary particles(cid:2) But to reach our goal of deriving this simple calculational
method(cid:1) we must (cid:0)rst(cid:1) unfortunately(cid:1) make a serious detour into formalism(cid:2)
The three chapters that follow this one are almost completely formal(cid:1) and
the readermight wonder(cid:1) in the courseofthis development(cid:1) where wearego(cid:5)
ing(cid:2) We would liketo partiallyanswerthat question in advance by discussing
the physicsof an especiallysimple QED process(cid:8)onesu(cid:9)cientlysimple that
many of its features follow directly from physical intuition(cid:2) Of course(cid:1) this
intuitive(cid:1) bottom(cid:5)up approach will contain many gaps(cid:2) In Chapter (cid:10) we will
returntothis processwiththefull poweroftheFeynmandiagramformalism(cid:2)
Working from the top down(cid:1) we will then see all of these di(cid:9)culties swept
away(cid:2)
(cid:0)
(cid:0)
(cid:1) Chapter(cid:0) Invitation(cid:1) Pair Production in e e(cid:0) Annihilation
(cid:0) (cid:0)
Figure (cid:2)(cid:3)(cid:2)(cid:3) Theannihilationreactione e(cid:0) (cid:0) (cid:0)(cid:0)(cid:2)showninthecenter(cid:3)
(cid:0)
of(cid:3)mass frame(cid:4)
The Simplest Situation
Since most particle physics experiments involve scattering(cid:1) the most com(cid:5)
monly calculated quantities in quantum (cid:0)eld theory are scattering cross sec(cid:5)
tions(cid:2) We will now calculate the cross section for the simplest of all QED
processes(cid:7) the annihilation of an electron with its antiparticle(cid:1) a positron(cid:1) to
form a pairof heavierleptons (cid:3)such asmuons(cid:4)(cid:2) The existence of antiparticles
isactuallyapredictionofquantum(cid:0)eldtheory(cid:1)aswewilldiscussinChapters
(cid:11) and (cid:12)(cid:2) For the moment(cid:1) though(cid:1) we take their existence as given(cid:2)
Anexperimenttomeasurethisannihilationprobabilitywouldproceedby
(cid:0)ring abeam of electronsat abeam of positrons(cid:2)The measurablequantityis
(cid:0) (cid:0)
thecrosssectionforthereactione e(cid:0) (cid:0) (cid:0)(cid:0) asafunctionofthecenter(cid:5)of(cid:5)
(cid:0)
mass energy and the relative angle (cid:1) between the incoming electrons and the
outgoingmuons(cid:2)TheprocessisillustratedinFig(cid:2)(cid:13)(cid:2)(cid:13)(cid:2)Forsimplicity(cid:1)wework
in the center(cid:5)of(cid:5)mass (cid:3)CM(cid:4) frame where the momenta satisfy p(cid:1) (cid:14) p and
(cid:1)
k(cid:1) (cid:14) k(cid:2)WealsoassumethatthebeamenergyE ismuchgreaterthaneither
(cid:1)
the electron or the muon mass(cid:1) so that p (cid:14) p(cid:1) (cid:14) k (cid:14) k(cid:1) (cid:14)E Ecm(cid:2)(cid:11)(cid:2)
j j j j j j j j (cid:2)
(cid:3)We use boldface type to denote (cid:12)(cid:5)vectors and ordinary italic type to denote
(cid:15)(cid:5)vectors(cid:2)(cid:4)
Sinceboththeelectronandthemuonhavespin(cid:13)(cid:2)(cid:11)(cid:1)wemustspecifytheir
spinorientations(cid:2)Itisusefultotaketheaxisthatde(cid:0)nesthespinquantization
of each particle to be in the direction of its motion(cid:16) each particle can then
haveitsspinpolarizedparallelorantiparalleltothisaxis(cid:2)Inpractice(cid:1)electron
and positron beams are often unpolarized(cid:1) and muon detectors are normally
blind to the muon polarization(cid:2) Hence we should average the cross section
over electron and positron spin orientations(cid:1) and sum the cross section over
muon spin orientations(cid:2)
For any given set of spin orientations(cid:1) it is conventional to write the
di(cid:17)erential cross section for our process(cid:1) with the (cid:0)(cid:0) produced into a solid
angle d(cid:18)(cid:1) as
d(cid:3) (cid:13) (cid:1)
(cid:14) (cid:5) (cid:3)(cid:13)(cid:5)(cid:13)(cid:4)
(cid:1) (cid:1)
d(cid:18) (cid:19)(cid:15)(cid:4) Ecm (cid:3) M
(cid:0) (cid:0)
(cid:0) (cid:0)
(cid:0)
Chapter (cid:0) Invitation(cid:1) Pair Productionin e e(cid:0) Annihilation (cid:4)
(cid:1)
The factor Ec(cid:0)m provides the correct dimensions for a cross section(cid:1) since in
(cid:1) (cid:1)
ourunits(cid:3)energy(cid:4)(cid:0) (cid:3)length(cid:4) (cid:2) Thequantity is thereforedimensionless(cid:16)
(cid:4) M
it is the quantum(cid:5)mechanical amplitude for the process to occur (cid:3)analogous
to the scattering amplitude f in nonrelativistic quantum mechanics(cid:4)(cid:1) and
we must now address the question of how to compute it from fundamental
theory(cid:2)The other factorsin theexpressionarepurelya matter ofconvention(cid:2)
Equation (cid:3)(cid:13)(cid:2)(cid:13)(cid:4) is actually a special case(cid:1) valid for CM scattering when the
(cid:0)nal state contains two massless particles(cid:1) of a more general formula (cid:3)whose
form cannot be deduced from dimensional analysis(cid:4) which we will derive in
Section (cid:15)(cid:2)(cid:10)(cid:2)
Now comes some bad news and some good news(cid:2)
The bad news is that even for this simplest of QED processes(cid:1) the exact
expression for is not known(cid:2) Actually this fact should come as no sur(cid:5)
M
prise(cid:1) since even in nonrelativistic quantum mechanics(cid:1) scattering problems
can rarely be solved exactly(cid:2) The best we can do is obtain a formal expres(cid:5)
sion for as a perturbation series in the strength of the electromagnetic
M
interaction(cid:1) and evaluate the (cid:0)rst few terms in this series(cid:2)
The good news is that Feynman has invented a beautiful way to orga(cid:5)
nize and visualize the perturbation series(cid:7) the method of Feynman diagrams(cid:2)
Roughlyspeaking(cid:1)thediagramsdisplaythe(cid:20)owofelectronsandphotonsdur(cid:5)
ingthescatteringprocess(cid:2)Forourparticularcalculation(cid:1)thelowest(cid:5)orderterm
in the perturbation series can be represented by a single diagram(cid:1) shown in
Fig(cid:2)(cid:13)(cid:2)(cid:11)(cid:2)Thediagramismadeup ofthree typesofcomponents(cid:7)externallines
(cid:3)representingthe fourincoming and outgoingparticles(cid:4)(cid:1) internallines (cid:3)repre(cid:5)
senting(cid:21)virtual(cid:22)particles(cid:1)in thiscaseonevirtualphoton(cid:4)(cid:1) andvertices(cid:2)It is
conventionaltousestraightlinesforfermionsandwavylinesforphotons(cid:2)The
arrows on the straight lines denote the direction of negative charge (cid:20)ow(cid:1) not
momentum(cid:2) We assigna(cid:15)(cid:5)momentum vectorto eachexternalline(cid:1) as shown(cid:2)
In this diagram(cid:1) the momentum q of the one internal line is determined by
momentum conservation at either of the vertices(cid:7) q (cid:14) p(cid:23)p(cid:1) (cid:14) k (cid:23)k(cid:1)(cid:2) We
must also associate a spin state (cid:3)either (cid:21)up(cid:22) or (cid:21)down(cid:22)(cid:4) with each external
fermion(cid:2)
Accordingtothe Feynman rules(cid:1) eachdiagramcanbe translateddirectly
intoa contributionto (cid:2)Therulesassigna shortalgebraicfactorto eachel(cid:5)
M
ement of a diagram(cid:1) and the product of these factors gives the value of the
corresponding term in the perturbation series(cid:2) Getting the resulting expres(cid:5)
sion for into a form that is usable(cid:1) however(cid:1) can still be nontrivial(cid:2) We
M
willdevelopmuchusefultechnologyfordoingsuchcalculationsin subsequent
chapters(cid:2) But we do not have that technologyyet(cid:1) so to get an answerto our
particularproblemwewillusesomeheuristicargumentsinsteadofthe actual
Feynman rules(cid:2)
Recall that in quantum(cid:5)mechanicalperturbation theory(cid:1)a transitionam(cid:5)
plitude can be computed(cid:1) to (cid:0)rst order(cid:1) as an expression of the form
(cid:0)nal state HI initial state (cid:6) (cid:3)(cid:13)(cid:5)(cid:11)(cid:4)
h j j i
(cid:0)
(cid:5) Chapter(cid:0) Invitation(cid:1) Pair Production in e e(cid:0) Annihilation
(cid:0)
Figure (cid:2)(cid:3)(cid:6)(cid:3) Feynman diagram for the lowest(cid:3)order term in the e e(cid:0)
(cid:0) (cid:0)
(cid:0) (cid:0)(cid:0) cross section(cid:4) At this order the only possible intermediate state is a
photon(cid:5)(cid:1)(cid:6)(cid:4)
where HI is the (cid:21)interaction(cid:22)part of the Hamiltonian(cid:2) In our casethe initial
(cid:0) (cid:0)
stateis e e(cid:0) andthe(cid:0)nalstateis (cid:0) (cid:0)(cid:0) (cid:2)ButourinteractionHamiltonian
j i h j
couples electrons to muons only through the electromagnetic (cid:0)eld (cid:3)that is(cid:1)
photons(cid:4)(cid:1)notdirectly(cid:2)Sothe(cid:0)rst(cid:5)orderresult(cid:3)(cid:13)(cid:2)(cid:11)(cid:4)vanishes(cid:1)andwemustgo
to the second(cid:5)order expression
(cid:0) (cid:0) (cid:0)
(cid:0) (cid:0)(cid:0) HI (cid:7) (cid:7) HI e e(cid:0) (cid:0)(cid:5) (cid:3)(cid:13)(cid:5)(cid:12)(cid:4)
M(cid:4)
This is a heuristic way of(cid:1)writing(cid:0) the(cid:0)co(cid:2)nt(cid:1)rib(cid:0)utio(cid:0)n to (cid:2) from the diagramin
(cid:0) (cid:0) (cid:0) (cid:0)
M (cid:0)
Fig(cid:2) (cid:13)(cid:2)(cid:11)(cid:2) The external electron lines correspond to the factor e e(cid:0) (cid:16) the ex(cid:5)
(cid:0) j i
ternal muon lines correspond to (cid:0) (cid:0)(cid:0) (cid:2) The vertices correspond to HI(cid:1) and
h j
the internal photon line corresponds to the operator (cid:7) (cid:7) (cid:2) We have added
j ih j
vector indices (cid:3)(cid:0)(cid:4) because the photon is a vector particle with four compo(cid:5)
nents(cid:2) There are four possible intermediate states(cid:1) one for each component(cid:1)
and according to the rulesof perturbationtheory we must sum overinterme(cid:5)
diate states(cid:2)Note that sincethe sum in (cid:3)(cid:13)(cid:2)(cid:12)(cid:4) takesthe form of a(cid:15)(cid:5)vector dot
product(cid:1) the amplitude will be a Lorentz(cid:5)invariant scalar as long as each
M
half of (cid:3)(cid:13)(cid:2)(cid:12)(cid:4) is a (cid:15)(cid:5)vector(cid:2)
(cid:0)
Let us try to guess the form of the vector (cid:7) HI e e(cid:0) (cid:0)(cid:2) Since HI cou(cid:5)
h j j i
ples electrons to photons with a strength e (cid:3)the electron charge(cid:4)(cid:1) the matrix
elementshouldbeproportionaltoe(cid:2)Nowconsideroneparticularsetofinitial
and (cid:0)nal spin orientations(cid:1) shown in Fig(cid:2) (cid:13)(cid:2)(cid:12)(cid:2) The electron and muon have
spins parallel to their directions of motion(cid:16) they are (cid:21)right(cid:5)handed(cid:22)(cid:2) The an(cid:5)
tiparticles(cid:1) similarly(cid:1) are (cid:21)left(cid:5)handed(cid:22)(cid:2) The electron and positron spins add
up to one unit of angular momentum in the (cid:23)z direction(cid:2) Since HI should
conserveangularmomentum(cid:1)thephotontowhichtheseparticlescouplemust
have the correct polarization vector to give it this same angular momentum(cid:7)
(cid:0)
Chapter (cid:0) Invitation(cid:1) Pair Productionin e e(cid:0) Annihilation (cid:7)
Figure (cid:2)(cid:3)(cid:0)(cid:3) Onepossiblesetofspinorientations(cid:4)Theelectronandtheneg(cid:3)
ative muon are right(cid:3)handed(cid:2) while the positron and the positive muon are
left(cid:3)handed(cid:4)
(cid:0)
(cid:8) (cid:14)(cid:3)(cid:24)(cid:6)(cid:13)(cid:6)i(cid:6)(cid:24)(cid:4)(cid:2) Thus we have
(cid:0) (cid:0)
(cid:7) HI e e(cid:0) e(cid:3)(cid:24)(cid:6)(cid:13)(cid:6)i(cid:6)(cid:24)(cid:4)(cid:5) (cid:3)(cid:13)(cid:5)(cid:15)(cid:4)
(cid:5)
The muon matrix e(cid:1)lem(cid:0) ent(cid:0) shou(cid:2)ld(cid:1) similarly(cid:1) have a polarization corre(cid:5)
(cid:0) (cid:0)
sponding to one unit of angular momentum along the direction of the (cid:0)(cid:0)
momentum k(cid:2) To obtain the correct vector(cid:1) rotate (cid:3)(cid:13)(cid:2)(cid:15)(cid:4) through an angle (cid:1)
in the xz(cid:5)plane(cid:7)
(cid:0) (cid:0)
(cid:7) HI (cid:0) (cid:0)(cid:0) e(cid:3)(cid:24)(cid:6)cos(cid:1)(cid:6)i(cid:6) sin(cid:1)(cid:4)(cid:5) (cid:3)(cid:13)(cid:5)(cid:10)(cid:4)
(cid:5) (cid:1)
To compute the am(cid:1)pl(cid:0)itude(cid:0) (cid:1) w(cid:2)e complex(cid:5)conjugate this vector and dot it
(cid:0) (cid:0) M
into (cid:3)(cid:13)(cid:2)(cid:15)(cid:4)(cid:2) Thus we (cid:0)nd(cid:1) for this set of spin orientations(cid:1)
(cid:1)
(cid:3)RL RL(cid:4)(cid:14) e (cid:3)(cid:13)(cid:23)cos(cid:1)(cid:4) (cid:5) (cid:3)(cid:13)(cid:5)(cid:19)(cid:4)
M (cid:0) (cid:1)
Ofcoursewe cannotdeterminethe overallfactorbythis method(cid:1) but in(cid:3)(cid:13)(cid:2)(cid:19)(cid:4)
it happens to be correct(cid:1) thanks to the conventions adopted in (cid:3)(cid:13)(cid:2)(cid:13)(cid:4)(cid:2) Note
that the amplitude vanishes for (cid:1) (cid:14) (cid:13)(cid:25)(cid:24)(cid:2)(cid:1) just as one would expect(cid:7) A state
whose angular momentum is in the (cid:23)z direction has no overlap with a state
whose angular momentum is in the z direction(cid:2)
(cid:1)
Next considerthe case in which the electronand positronareboth right(cid:5)
handed(cid:2)Nowtheirtotal spinangularmomentumiszero(cid:1)andtheargumentis
moresubtle(cid:2)Wemightexpecttoobtainalongitudinallypolarizedphotonwith
a Clebsch(cid:5)Gordancoe(cid:9)cient of (cid:13)(cid:2)p(cid:11)(cid:1) just aswhen weadd angular momenta
in three dimensions(cid:1) (cid:14) (cid:3)(cid:13)(cid:2)p(cid:11)(cid:4) j (cid:14)(cid:13)(cid:6)m(cid:14)(cid:24) (cid:23) j (cid:14)(cid:24)(cid:6)m(cid:14)(cid:24) (cid:2) But we
j(cid:6)(cid:7)i j i j i
are really adding angular momenta in the four(cid:5)dimensional Lorentz group(cid:1)
(cid:3) (cid:4)
so we must take into account not only spin (cid:3)the transformation properties of
statesunderrotations(cid:4)(cid:1)butalsothetransformationpropertiesofstatesunder
boosts(cid:2)Itturnsout(cid:1)asweshalldiscussinChapter(cid:12)(cid:1)thattheClebsch(cid:5)Gordan
(cid:0)
coe(cid:9)cient that couples a (cid:15)(cid:5)vector to the state e(cid:0)ReR of massless fermions is
j i
zero(cid:2)(cid:3)Forthe record(cid:1)the stateis asuperpositionofscalarandantisymmetric
tensor pieces(cid:2)(cid:4) Thus the amplitude (cid:3)RR RL(cid:4) is zero(cid:1) as are the eleven
M (cid:0)
(cid:0)
(cid:8) Chapter(cid:0) Invitation(cid:1) Pair Production in e e(cid:0) Annihilation
otheramplitudesinwhicheithertheinitialor(cid:0)nalstatehaszerototalangular
momentum(cid:2)
Theremainingnonzeroamplitudescanbe foundinthesamewaythatwe
found the (cid:0)rst one(cid:2) They are
(cid:1)
(cid:3)RL LR(cid:4)(cid:14) e (cid:3)(cid:13) cos(cid:1)(cid:4)(cid:6)
M (cid:0) (cid:1) (cid:1)
(cid:1)
(cid:3)LR RL(cid:4)(cid:14) e (cid:3)(cid:13) cos(cid:1)(cid:4)(cid:6) (cid:3)(cid:13)(cid:5)(cid:26)(cid:4)
M (cid:0) (cid:1) (cid:1)
(cid:1)
(cid:3)LR LR(cid:4)(cid:14) e (cid:3)(cid:13)(cid:23)cos(cid:1)(cid:4)(cid:5)
M (cid:0) (cid:1)
Insertingtheseexpressionsinto(cid:3)(cid:13)(cid:2)(cid:13)(cid:4)(cid:1)averagingoverthefourinitial(cid:5)statespin
orientations(cid:1) and summing overthe four (cid:0)nal(cid:5)state spin orientations(cid:1) we (cid:0)nd
(cid:1)
d(cid:3) (cid:9) (cid:1)
(cid:14) (cid:13)(cid:23)cos (cid:1) (cid:6) (cid:3)(cid:13)(cid:5)(cid:25)(cid:4)
(cid:1)
d(cid:18) (cid:15)Ecm
(cid:1) (cid:3) (cid:4)
where (cid:9) (cid:14) e (cid:2)(cid:15)(cid:4) (cid:13)(cid:2)(cid:13)(cid:12)(cid:26)(cid:2) Integrating over the angular variables (cid:1) and (cid:10)
(cid:8)
gives the total cross section(cid:1)
(cid:1)
(cid:15)(cid:4)(cid:9)
(cid:3)total (cid:14) (cid:1) (cid:5) (cid:3)(cid:13)(cid:5)(cid:27)(cid:4)
(cid:12)Ecm
Results (cid:3)(cid:13)(cid:2)(cid:25)(cid:4) and (cid:3)(cid:13)(cid:2)(cid:27)(cid:4) agree with experiments to about (cid:13)(cid:24)(cid:28)(cid:16) almost all of
the discrepancy is accounted for by the next term in the perturbation series(cid:1)
corresponding to the diagrams shown in Fig(cid:2) (cid:13)(cid:2)(cid:15)(cid:2) The qualitative features
of these expressions(cid:8)the angular dependence and the sharp decrease with
energy(cid:8)are obvious in the actual data(cid:2) (cid:3)The properties of these results are
discussed in detail in Section (cid:10)(cid:2)(cid:13)(cid:2)(cid:4)
Embellishments and Questions
WeobtainedtheangulardistributionpredictedbyQuantumElectrodynamics
(cid:0) (cid:0)
for the reaction e e(cid:0) (cid:0) (cid:0)(cid:0) by applying angular momentum arguments(cid:1)
(cid:0)
with little appeal to the underlying formalism(cid:2) However(cid:1) we used the simpli(cid:5)
fying featuresof the high(cid:5)energylimit and the center(cid:5)of(cid:5)massframe in a very
strong way(cid:2) The analysis we have presented will break down when we relax
any of our simplifying assumptions(cid:2) So how does one perform general QED
calculations(cid:29) To answer that question we must return to the Feynman rules(cid:2)
Asmentionedabove(cid:1)theFeynmanrulestellustodrawthediagram(cid:3)s(cid:4)for
the process we are considering(cid:1)and to associate a short algebraic factor with
each piece of each diagram(cid:2) Figure (cid:13)(cid:2)(cid:10) shows the diagram for our reaction(cid:1)
with the various assignments indicated(cid:2)
(cid:1)
For the internal photon line we write ig(cid:0)(cid:1)(cid:2)q (cid:1) where g(cid:0)(cid:1) is the usual
(cid:1)
Minkowskimetrictensorandqisthe(cid:15)(cid:5)momentumofthevirtualphoton(cid:2)This
factor corresponds to the operator (cid:7) (cid:7) in our heuristic expression (cid:3)(cid:13)(cid:2)(cid:12)(cid:4)(cid:2)
(cid:0)j ih j
Foreachvertexwewrite ie(cid:7) (cid:1)correspondingtoHI in(cid:3)(cid:13)(cid:2)(cid:12)(cid:4)(cid:2)Theobjects
(cid:0) (cid:1)
(cid:7) area setoffour(cid:15) (cid:15)constantmatrices(cid:2)Theydothe (cid:21)additionofangular
(cid:9)
(cid:0)
Chapter (cid:0) Invitation(cid:1) Pair Productionin e e(cid:0) Annihilation (cid:9)
(cid:2)
Figure (cid:2)(cid:3)(cid:1)(cid:3) Feynman diagrams that contribute to the (cid:2) term in the
(cid:0) (cid:0)
e e(cid:0) (cid:0) (cid:0)(cid:0) cross section(cid:4)
(cid:0)
Figure (cid:2)(cid:3)(cid:4)(cid:3) Diagram of Fig(cid:4) (cid:0)(cid:4)(cid:7)(cid:2) with expressions corresponding to each
vertex(cid:2) internal line(cid:2) andexternal line(cid:4)
momentum(cid:22) for us(cid:1) coupling a state of two spin(cid:5)(cid:13)(cid:2)(cid:11) particles to a vector
particle(cid:2)
The external lines carry expressions for four(cid:5)component column(cid:5)spinors
u(cid:1)v(cid:1)orrow(cid:5)spinorsu(cid:1)v(cid:2)Theseareessentiallythemomentum(cid:5)spacewavefunc(cid:5)
(cid:0) (cid:0)
tions of the initial and (cid:0)nal particles(cid:1) and correspond to e e(cid:0) and (cid:0) (cid:0)(cid:0)
j i h j
in (cid:3)(cid:13)(cid:2)(cid:12)(cid:4)(cid:2) The indices s(cid:1) s(cid:1)(cid:1) r(cid:1) and r(cid:1) denote the spin state(cid:1) either up or down(cid:2)
(cid:0)
(cid:2)(cid:10) Chapter (cid:0) Invitation(cid:1) Pair Production ine e(cid:0) Annihilation
Wecannowwritedownanexpressionfor (cid:1)readingeverythingstraight
M
o(cid:17) the diagram(cid:7)
s(cid:0) (cid:0) s ig(cid:0)(cid:1) r (cid:1) r(cid:0)
(cid:14)v (cid:3)p(cid:1)(cid:4) ie(cid:7) u (cid:3)p(cid:4) (cid:1) (cid:1) u (cid:3)k(cid:4) ie(cid:7) v (cid:3)k(cid:1)(cid:4)
M (cid:1) q (cid:1)
(cid:5) (cid:6) (cid:3)(cid:13)(cid:5)(cid:13)(cid:24)(cid:4)
ie(cid:1) s(cid:3)(cid:0) (cid:0)(cid:4)s r r(cid:0) (cid:3) (cid:4)
(cid:14) (cid:1) v (cid:3)p(cid:1)(cid:4)(cid:7) u (cid:3)p(cid:4) u (cid:3)k(cid:4)(cid:7)(cid:0)v (cid:3)k(cid:1)(cid:4) (cid:5)
q
(cid:3) (cid:4)(cid:3) (cid:4)
It is instructive to compare this in detail with Eq(cid:2) (cid:3)(cid:13)(cid:2)(cid:12)(cid:4)(cid:2)
To derive the cross section (cid:3)(cid:13)(cid:2)(cid:25)(cid:4) from (cid:3)(cid:13)(cid:2)(cid:13)(cid:24)(cid:4)(cid:1) we could return to the an(cid:5)
gular momentum arguments used above(cid:1) supplemented with some concrete
knowledge about (cid:7) matrices and Dirac spinors(cid:2) We will do the calculation
in this manner in Section (cid:10)(cid:2)(cid:11)(cid:2) There are(cid:1) however(cid:1) a number of useful tricks
that can be employed to manipulate expressions like (cid:3)(cid:13)(cid:2)(cid:13)(cid:24)(cid:4)(cid:1) especially when
one wants to compute only the unpolarized cross section(cid:2) Using this (cid:21)Feyn(cid:5)
man trace technology(cid:22) (cid:3)so(cid:5)called because one must evaluate traces of prod(cid:5)
ucts of (cid:7)(cid:5)matrices(cid:4)(cid:1) it isn(cid:6)t even necessary to have explicit expressions for
the(cid:7)(cid:5)matricesandDiracspinors(cid:2)Thecalculationbecomesalmostcompletely
mindless(cid:1) and the answer (cid:3)(cid:13)(cid:2)(cid:25)(cid:4) is obtained after less than a page of algebra(cid:2)
But since the Feynman rules and trace technology are so powerful(cid:1) we can
also relax some of our simplifying assumptions(cid:2) To conclude this section(cid:1) let
usdiscussseveralwaysinwhichourcalculationcouldhavebeenmoredi(cid:9)cult(cid:2)
Theeasiestrestrictiontorelaxisthatthemuonsbemassless(cid:2)Ifthebeam
energy is not much greater than the mass of the muon(cid:1) all of our predic(cid:5)
tions should depend on the ratio m(cid:0)(cid:2)Ecm(cid:2) (cid:3)Since the electron is (cid:11)(cid:24)(cid:24) times
lighter than the muon(cid:1) it can be considered massless whenever the beam en(cid:5)
ergy is large enough to create muons(cid:2)(cid:4) Using Feynman trace technology(cid:1) it is
extremely easy to restore the muon mass to our calculation(cid:2) The amount of
algebraisincreasedbyabout(cid:0)ftypercent(cid:1)and the relation(cid:3)(cid:13)(cid:2)(cid:13)(cid:4)betweenthe
amplitude and the cross section must be modi(cid:0)ed slightly(cid:1) but the answer is
worth the e(cid:17)ort(cid:2) We do this calculation in detail in Section (cid:10)(cid:2)(cid:13)(cid:2)
Working in a di(cid:17)erent reference frame is also easy(cid:16) the only modi(cid:0)cation
is in the relation (cid:3)(cid:13)(cid:2)(cid:13)(cid:4) between the amplitude and the cross section(cid:2) Or one
can simply perform a Lorentz transformation on the CM result(cid:1) boosting it
to a di(cid:17)erent frame(cid:2)
When the spin states of the initial and(cid:30)or (cid:0)nal particles are known and
we still wish to retain the muon mass(cid:1) the calculation becomes somewhat
cumbersome but no more di(cid:9)cult in principle(cid:2) The trace technology can be
generalized to this case(cid:1) but it is often easier to evaluate expression (cid:3)(cid:13)(cid:2)(cid:13)(cid:24)(cid:4)
directly(cid:1) using the explicit values of the spinors u and v(cid:2)
Nextonecouldcomputecrosssectionsfordi(cid:17)erentprocesses(cid:2)Theprocess
(cid:0) (cid:0)
e e(cid:0) e e(cid:0)(cid:1) known as Bhabha scattering(cid:1) is more di(cid:9)cult because there is
(cid:0)
asecondalloweddiagram(cid:3)seeFig(cid:2)(cid:13)(cid:2)(cid:19)(cid:4)(cid:2)Theamplitudesforthetwodiagrams
must (cid:0)rst be added(cid:1) then squared(cid:2)
Other processes contain photons in the initial and(cid:30)or (cid:0)nal states(cid:2) The