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web site Part I Feynman Diagrams
Quantum Electrodynamics Chapter 1 Invitation: Pair Production
in e+e; Annihilation The main purpose of Part I of this book is to develop the basic calculational
method of quantum eld theory, the formalism of Feynman diagrams. We will
then apply this formalism to computations in Quantum Electrodynamics, the
quantum theory of electrons and photons.
Quantum Electrodynamics (QED) is perhaps the best fundamental physical theory we have. The theory is formulated as a set of simple equations
(Maxwell's equations and the Dirac equation) whose form is essentially determined by relativistic invariance. The quantum-mechanical solutions of these
equations give detailed predictions of electromagnetic phenomena from macroscopic distances down to regions several hundred times smaller than the proton.
Feynman diagrams provide for this elegant theory an equally elegant procedure for calculation: Imagine a process that can be carried out by electrons
and photons, draw a diagram, and then use the diagram to write the mathematical form of the quantum-mechanical amplitude for that process to occur.
In this 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. Eventually, we will arrive at a point where we can
calculate observable quantities that are of great interest in the study of elementary particles. But to reach our goal of deriving this simple calculational
method, we must rst, unfortunately, make a serious detour into formalism.
The three chapters that follow this one are almost completely formal, and
the reader might wonder, in the course of this development, where we are going. We would like to partially answer that question in advance by discussing
the physics of an especially simple QED process|one su ciently simple that
many of its features follow directly from physical intuition. Of course, this
intuitive, bottom-up approach will contain many gaps. In Chapter 5 we will
return to this process with the full power of the Feynman diagram formalism.
Working from the top down, we will then see all of these di culties swept
away. 3 4 Chapter 1 Invitation: Pair Production in e+ e; Annihilation Figure 1.1. The annihilation reaction e+ e; !
of-mass frame. The Simplest Situation + ; , shown in the center- Since most particle physics experiments involve scattering, the most commonly calculated quantities in quantum eld theory are scattering cross sections. We will now calculate the cross section for the simplest of all QED
processes: the annihilation of an electron with its antiparticle, a positron, to
form a pair of heavier leptons (such as muons). The existence of antiparticles
is actually a prediction of quantum eld theory, as we will discuss in Chapters
2 and 3. For the moment, though, we take their existence as given.
An experiment to measure this annihilation probability would proceed by
ring a beam of electrons at a beam of positrons. The measurable quantity is
the cross section for the reaction e+ e; ! + ; as a function of the center-ofmass energy and the relative angle between the incoming electrons and the
outgoing muons. The process is illustrated in Fig. 1.1. For simplicity, we work
in the center-of-mass (CM) frame where the momenta satisfy p0 = ;p and
k0 = ;k. We also assume that the beam energy E is much greater than either
the electron or the muon mass, so that jpj = jp0 j = jkj = jk0 j = E Ecm=2.
(We use boldface type to denote 3-vectors and ordinary italic type to denote
Since both the electron and the muon have spin 1=2, we must specify their
spin orientations. It is useful to take the axis that de nes the spin quantization
of each particle to be in the direction of its motion each particle can then
have its spin polarized parallel or antiparallel to this axis. In practice, electron
and positron beams are often unpolarized, and muon detectors are normally
blind to the muon polarization. Hence we should average the cross section
over electron and positron spin orientations, and sum the cross section over
muon spin orientations.
For any given set of spin orientations, it is conventional to write the
dierential cross section for our process, with the ; produced into a solid
angle d, as
64 E 2
cm Chapter 1 Invitation: Pair Production in e+ e; Annihilation 5 ;2 provides the correct dimensions for a cross section, since in
The factor Ecm
our units (energy);2 (length)2 . The quantity M is therefore dimensionless
it is the quantum-mechanical amplitude for the process to occur (analogous
to the scattering amplitude f in nonrelativistic quantum mechanics), and
we must now address the question of how to compute it from fundamental
theory. The other factors in the expression are purely a matter of convention.
Equation (1.1) is actually a special case, valid for CM scattering when the
nal state contains two massless particles, of a more general formula (whose
form cannot be deduced from dimensional analysis) which we will derive in
Now comes some bad news and some good news.
The bad news is that even for this simplest of QED processes, the exact
expression for M is not known. Actually this fact should come as no surprise, since even in nonrelativistic quantum mechanics, scattering problems
can rarely be solved exactly. The best we can do is obtain a formal expression for M as a perturbation series in the strength of the electromagnetic
interaction, and evaluate the rst few terms in this series.
The good news is that Feynman has invented a beautiful way to organize and visualize the perturbation series: the method of Feynman diagrams.
Roughly speaking, the diagrams display the ow of electrons and photons during the scattering process. For our particular calculation, the lowest-order term
in the perturbation series can be represented by a single diagram, shown in
Fig. 1.2. The diagram is made up of three types of components: external lines
(representing the four incoming and outgoing particles), internal lines (representing \virtual" particles, in this case one virtual photon), and vertices. It is
conventional to use straight lines for fermions and wavy lines for photons. The
arrows on the straight lines denote the direction of negative charge ow, not
momentum. We assign a 4-momentum vector to each external line, as shown.
In this diagram, the momentum q of the one internal line is determined by
momentum conservation at either of the vertices: q = p + p0 = k + k0 . We
must also associate a spin state (either \up" or \down") with each external
According to the Feynman rules, each diagram can be translated directly
into a contribution to M. The rules assign a short algebraic factor to each element of a diagram, and the product of these factors gives the value of the
corresponding term in the perturbation series. Getting the resulting expression for M into a form that is usable, however, can still be nontrivial. We
will develop much useful technology for doing such calculations in subsequent
chapters. But we do not have that technology yet, so to get an answer to our
particular problem we will use some heuristic arguments instead of the actual
Recall that in quantum-mechanical perturbation theory, a transition amplitude can be computed, to rst order, as an expression of the form
h nal statej HI jinitial statei
(1:2) 6 Chapter 1 Invitation: Pair Production in e+ e; Annihilation Figure 1.2. Feynman diagram for the lowest-order term in the e+ e; ! + ; cross section. At this order the only possible intermediate state is a photon ( ). where HI is the \interaction" part of the Hamiltonian. In our case the initial
state is je+ e;i and the nal state is h + ; j. But our interaction Hamiltonian
couples electrons to muons only through the electromagnetic eld (that is,
photons), not directly. So the rst-order result (1.2) vanishes, and we must go
to the second-order expression
M + ; HI HI e+ e; :
This is a heuristic way of writing the contribution to M from the diagram in
Fig. 1.2. The external electron lines correspond to the factor je+ e;i the external muon lines correspond to h + ; j. The vertices correspond to HI , and
the internal photon line corresponds to the operator j ih j. We have added
vector indices ( ) because the photon is a vector particle with four components. There are four possible intermediate states, one for each component,
and according to the rules of perturbation theory we must sum over intermediate states. Note that since the sum in (1.3) takes the form of a 4-vector dot
product, the amplitude M will be a Lorentz-invariant scalar as long as each
half of (1.3) is a 4-vector.
Let us try to guess the form of the vector h j HI je+e; i . Since HI couples electrons to photons with a strength e (the electron charge), the matrix
element should be proportional to e. Now consider one particular set of initial
and nal spin orientations, shown in Fig. 1.3. The electron and muon have
spins parallel to their directions of motion they are \right-handed". The antiparticles, similarly, are \left-handed". The electron and positron spins add
up to one unit of angular momentum in the +z direction. Since HI should
conserve angular momentum, the photon to which these particles couple must
have the correct polarization vector to give it this same angular momentum: Chapter 1 Invitation: Pair Production in e+ e; Annihilation 7 Figure 1.3. One possible set of spin orientations. The electron and the negative muon are right-handed, while the positron and the positive muon are
left-handed. = (0 1 i 0). Thus we have
H e+e; / e (0 1 i 0):
The muon matrix element should, similarly, have a polarization corresponding to one unit of angular momentum along the direction of the ;
momentum k. To obtain the correct vector, rotate (1.4) through an angle
in the xz -plane:
H + ; / e (0 cos i ;sin ):
To compute the amplitude M, we complex-conjugate this vector and dot it
into (1.4). Thus we nd, for this set of spin orientations,
M(RL ! RL) = ;e2 (1 + cos ) :
Of course we cannot determine the overall factor by this method, but in (1.6)
it happens to be correct, thanks to the conventions adopted in (1.1). Note
that the amplitude vanishes for = 180, just as one would expect: A state
whose angular momentum is in the +z direction has no overlap with a state
whose angular momentum is in the ;z direction.
Next consider the case in which the electron and positron are both righthanded. Now their total spin angular momentum is zero, and the argument is
more subtle. We might expect to obtain
p a longitudinally polarized photon with
a Clebsch-Gordan coe cient of p1= ;2, just as when we add angular momenta
in three dimensions, j"#i = (1= 2) jj = 1 m = 0i + jj = 0 m = 0i . But we
are really adding angular momenta in the four-dimensional Lorentz group,
so we must take into account not only spin (the transformation properties of
states under rotations), but also the transformation properties of states under
boosts. It turns out, as we shall discuss in Chapter 3, that the Clebsch-Gordan
coe cient that couples a 4-vector to the state je;R e+R i of massless fermions is
zero. (For the record, the state is a superposition of scalar and antisymmetric
tensor pieces.) Thus the amplitude M(RR ! RL) is zero, as are the eleven 8 Chapter 1 Invitation: Pair Production in e+ e; Annihilation other amplitudes in which either the initial or nal state has zero total angular
The remaining nonzero amplitudes can be found in the same way that we
found the rst one. They are
M(RL ! LR) = ;e2 (1 ; cos )
M(LR ! RL) = ;e2 (1 ; cos )
M(LR ! LR) = ;e (1 + cos ):
Inserting these expressions into (1.1), averaging over the four initial-state spin
orientations, and summing over the four nal-state spin orientations, we nd d = 2 ;1 + cos2
4Ecm (1:8) where = e2 =4 ' 1=137. Integrating over the angular variables and gives the total cross section,
total = 34E 2 :
Results (1.8) and (1.9) agree with experiments to about 10% almost all of
the discrepancy is accounted for by the next term in the perturbation series,
corresponding to the diagrams shown in Fig. 1.4. The qualitative features
of these expressions|the angular dependence and the sharp decrease with
energy|are obvious in the actual data. (The properties of these results are
discussed in detail in Section 5.1.) Embellishments and Questions We obtained the angular distribution predicted by Quantum Electrodynamics
for the reaction e+ e; ! + ; by applying angular momentum arguments,
with little appeal to the underlying formalism. However, we used the simplifying features of the high-energy limit and the center-of-mass frame in a very
strong way. The analysis we have presented will break down when we relax
any of our simplifying assumptions. So how does one perform general QED
calculations? To answer that question we must return to the Feynman rules.
As mentioned above, the Feynman rules tell us to draw the diagram(s) for
the process we are considering, and to associate a short algebraic factor with
each piece of each diagram. Figure 1.5 shows the diagram for our reaction,
with the various assignments indicated.
For the internal photon line we write ;ig =q2 , where g is the usual
Minkowski metric tensor and q is the 4-momentum of the virtual photon. This
factor corresponds to the operator j i h j in our heuristic expression (1.3).
For each vertex we write ;ie , corresponding to HI in (1.3). The objects
are a set of four 4 4 constant matrices. They do the \addition of angular Chapter 1 Invitation: Pair Production in e+ e; Annihilation 9 Figure 1.4. Feynman diagrams that contribute to the 3 term in the
e+ e; ! + ; cross section. Figure 1.5. Diagram of Fig. 1.2, with expressions corresponding to each
vertex, internal line, and external line. momentum" for us, coupling a state of two spin-1=2 particles to a vector
The external lines carry expressions for four-component column-spinors
u, v, or row-spinors u, v. These are essentially the momentum-space wavefunctions of the initial and nal particles, and correspond to je+e; i and h + ; j
in (1.3). The indices s, s0 , r, and r0 denote the spin state, either up or down. 10 Chapter 1 Invitation: Pair Production in e+ e; Annihilation We can now write down an expression for M, reading everything straight
o the diagram: ;
M = vs0 (p0 ) ;ie us (p) ;igq2 ur (k) ;ie vr0 (k0 )
= ieq2 v s (p0 ) us (p) ur (k) vr (k0 ) : (1:10) It is instructive to compare this in detail with Eq. (1.3).
To derive the cross section (1.8) from (1.10), we could return to the angular momentum arguments used above, supplemented with some concrete
knowledge about matrices and Dirac spinors. We will do the calculation
in this manner in Section 5.2. There are, however, a number of useful tricks
that can be employed to manipulate expressions like (1.10), especially when
one wants to compute only the unpolarized cross section. Using this \Feynman trace technology" (so-called because one must evaluate traces of products of -matrices), it isn't even necessary to have explicit expressions for
the -matrices and Dirac spinors. The calculation becomes almost completely
mindless, and the answer (1.8) is obtained after less than a page of algebra.
But since the Feynman rules and trace technology are so powerful, we can
also relax some of our simplifying assumptions. To conclude this section, let
us discuss several ways in which our calculation could have been more di cult.
The easiest restriction to relax is that the muons be massless. If the beam
energy is not much greater than the mass of the muon, all of our predictions should depend on the ratio m =Ecm. (Since the electron is 200 times
lighter than the muon, it can be considered massless whenever the beam energy is large enough to create muons.) Using Feynman trace technology, it is
extremely easy to restore the muon mass to our calculation. The amount of
algebra is increased by about fty percent, and the relation (1.1) between the
amplitude and the cross section must be modi ed slightly, but the answer is
worth the eort. We do this calculation in detail in Section 5.1.
Working in a dierent reference frame is also easy the only modi cation
is in the relation (1.1) between the amplitude and the cross section. Or one
can simply perform a Lorentz transformation on the CM result, boosting it
to a dierent frame.
When the spin states of the initial and/or nal particles are known and
we still wish to retain the muon mass, the calculation becomes somewhat
cumbersome but no more di cult in principle. The trace technology can be
generalized to this case, but it is often easier to evaluate expression (1.10)
directly, using the explicit values of the spinors u and v.
Next one could compute cross sections for dierent processes. The process
e+ e; ! e+ e;, known as Bhabha scattering, is more di cult because there is
a second allowed diagram (see Fig. 1.6). The amplitudes for the two diagrams
must rst be added, then squared.
Other processes contain photons in the initial and/or nal states. The Chapter 1 Invitation: Pair Production in e+ e; Annihilation 11 Figure 1.6. The two lowest-order diagrams for Bhabha scattering, e+ e; ! e+ e; . Figure 1.7. The two lowest-order diagrams for Compton scattering.
paradigm example is Compton scattering, for which the two lowest-order diagrams are shown in Fig. 1.7. The Feynman rules for external photon lines
and for internal electron lines are no more complicated than those we have
already seen. We discuss Compton scattering in detail in Section 5.5.
Finally we could compute higher-order terms in the perturbation series.
Thanks to Feynman, the diagrams are at least easy to draw we have seen
those that contribute to the next term in the e+e; ! + ; cross section in
Fig. 1.4. Remarkably, the algorithm that assigns algebraic factors to pieces
of the diagrams holds for all higher-order contributions, and allows one to
evaluate such diagrams in a straightforward, if tedious, way. The computation
of the full set of nine diagrams is a serious chore, at the level of a research
In this book, starting in Chapter 6, we will analyze much of the physics
that arises from higher-order Feynman diagrams such as those in Fig. 1.4.
We will see that the last four of these diagrams, which involve an additional
photon in the nal state, are necessary because no detector is sensitive enough
to notice the presence of extremely low-energy photons. Thus a nal state
containing such a photon cannot be distinguished from our desired nal state
of just a muon pair. 12 Chapter 1 Invitation: Pair Production in e+ e; Annihilation The other ve diagrams in Fig. 1.4 involve intermediate states of several
virtual particles rather than just a single virtual photon. In each of these diagrams there will be one virtual particle whose momentum is not determined
by conservation of momentum at the vertices. Since perturbation theory requires us to sum over all possible intermediate states, we must integrate over
all possible values of this momentum. At this step, however, a new di culty
appears: The loop-momentum integrals in the rst three diagrams, when performed naively, turn out to be in nite. We will provide a x for this problem,
so that we get nite results, by the end of Part I. But the question of the
physical origin of these divergences cannot be dismissed so lightly that will
be the main subject ...
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