An Intro duction to QED and QCD

Je Forshaw

Department of Physics and Astronomy

University of Manchester

Manchester M PL

Lectures presented at the Scho ol for Young High Energy Physicists

Rutherford Appleton Lab oratory September

Contents

Intro duction

Units and Conventions

Relativistic Wave Equations

Wavefunctions and Fields

The KleinGordon Equation

The Dirac Equation

Free Particle Solutions I Interpretation

Free Particle Solutions I I Spin

Normalisation and Gamma Matrices

Lorentz Covariance

Parity

Bilinear Covariants

Charge Conjugation

Neutrinos

Cross Sections and Decay Rates

The Number of Final States

Lorentz Invariant Phase Space LIPS

Cross Sections

Twobody Scattering

Decay Rates

Quantum Electro dynamics

Quantising the Dirac Field

Quantising the Electromagnetic Field

Feynman Rules of QED

Scattering

ElectronElectron Scattering

ElectronPositron Annihilation

e e e e

e e and e e

Intro duction to Renormalisation

Renormalisation of QED

Renormalisation of Quantum Chromo dynamics

A Pre Scho ol Problems

B Answers to the Exercises

Intro duction

The traditional aim of this course is to teach you how to calculate amplitudes cross

sections and decay rates particularly for quantum electro dynamics qed but in principle

also for quantum chromo dynamics qcd By the end of the course you should be able

to go from a Feynman diagram such as the one for e e in Figure a to a

number for the

We will restrict ourselves to calculations at tree level but at the end of the course

h amongst other things we will also lo ok qualitatively at higher order loop eects whic

are resp onsible for the running of the qcd coupling This running means that the qcd

coupling app ears weaker when measured at higher energy scales and is the reason why

we can sometimes do p erturbative qcd calculations As you might guess the sort of

diagrams which are imp ortant here have closed lo ops of particle lines in them in Fig

ure b is one example contributing to the running of the qcd coupling the curly

lines denote gluons

In order to do our calculations we will need a certain amount of technology In

particular we will need to describ e particles with spin esp ecially the spin leptons

and We will therefore sp end some time lo oking at the Dirac equation After

this well work out how to go from quantum mechanical probability amplitudes to cross

sections and decay rates Then with these to ols in hand we will lo ok at some examples of

tree level qed pro cesses Here you will get handson exp erience of calculating transition

amplitudes and getting from them to cross sections We then moveontoqcd This will

entail a brief intro duction to renormalisation in both qed and qcd We will intro duce

the idea of the running coupling and lo ok at asymptotic freedom in qcd

In reference you will nd a list of textb o oks which may b e useful

Units and Conventions

I will use natural units c h so mass energy inverse length and inverse time all

have the same dimensions

vector a a a a

scalar pro duct ab a b ab g a b

From the scalar pro duct you see that the metric is

if

g diag g g

if

For c g and g are numerically the same

e

e

a b

Figure Examples of Feynman diagrams contributing to a e e and b the

running of the strong coupling constant

From the ab ove you would think it natural to write the space comp onents of a vector as

i

a for i However for vectors I usually write the comp onents as a This ought

i

not to cause confusion since for vectors the metric is just the unit matrix Generally

sp eaking a little care must b e taken in getting the sign right for the comp onents

of a vector

Note that isacovector

x

x

i i

so r and r

My convention for the totally antisymmetric LeviCivita tensor is

if f g an even p ermutation of f g

if an odd p ermutation

otherwise

Note that and p q r s changes sign under a parity transformation

which is obvious b ecause it contains an odd number of spatial comp onents

Relativistic Wave Equations

The starting p oint for this course is the Schrodinger equation which can b e written quite

generally as

H i

t

where H is the Hamiltonian ie the energy op erator In this equation is the wave

function describing the single particle probability amplitude I shall usually reserve the

Greek symbol for spin fermions and for spin b osons So for pions and the like

I shall write

H i

t

In this course wewant to extend the nonrelativistic quantum mechanics familiar to

undergraduates into the relativistic domain For example in nonrelativistic quantum

mechanics you are used to writing

H T V

where T is the kinetic energy and V is the potential energy A particle of mass m and

momentum p has nonrelativistic kinetic energy

P

T

m

where capital P is the op erator corresp onding to momentum p For a slow moving

particle v c eg an electron in a Hydrogen atom this is adequate but for relativistic

systems v c the Hamiltonian ab ove breaks down For a free relativistic particle the

total energy E is given by the Einstein equation

E p m

Thus the square of the relativistic Hamiltonian H is simply given by promoting the

momentum to op erator status

H P m

So far so go o d but now the question arises of how to implement the Schrodinger equation

which is expressed in terms of H rather than H Naively the relativistic Schrodinger

equation lo oks like

p

t

P m ti

t

but this is dicult to interpret b ecause of the square ro ot There are two ways forward

Work with H By iterating the Schrodinger equation we have

t

H t

t

This is known as the KleinGordon KG equation In this case the wavefunction

describ es spinless b osons

Invent a new Hamiltonian H which is linear in momentum and whose square is

D

equal to H given ab ove H P m In this case we have

D

t

H t i

D

t

which is known as the Dirac equation with H being the Dirac Hamiltonian In

D

this case the wavefunction describ es spin fermions as we shall see

Wavefunctions and Fields

You may be wondering why I am talking ab out wavefunctions while in your eld theory

course Dave Dunbar is telling you ab out elds Some of you may even be wondering

what is the dierence between awavefunction and a eld

The bottom line is that single particle wavefunctions work just ne if you want

to describ e systems where the particle number is conserved Problems come when you

wanttoallow for relativistic eects In particular antiparticles and hence the p ossibility

that particles can annihilate or be pair pro duced In fact these new concepts can be

accommo dated within the wavefunction approach but inaway which is not really very

satisfactory Well take a lo ok at the problems encountered in trying to cling to the

wavefunction way of thinking shortly

Rather than patch things up its much more app ealing simply to ditch the usual

interpretation of as a wavefunction and to identify it as a eld This eld is then

sub jected to the usual laws of quantum mechanics This means elevating the eld and

its canonical momentum to the status of op erators which are then deemed to satisfy the

usual commutation relations This is what Dave has b een doing in his course In the

limit that particle number is conserved the theory is equivalent to the single particle

w avefunction approach

It is imp ortant to emphasise that a eld is very dierentfromawavefunction Think

rst of a classical eld I nd it easier to picture a eld by rst dividing space into

innitesimally small b oxes In each box is a ctitious particle and the amplitude of the

particles displacement from equilibrium is the value of the eld as the point where the

small b ox is If the eld satises some wave equation then you can think of the motions

of the individual particles as being inuenced by whats going on in the neighb ourho o d

eg consider a vibrating membrane the particles can b e thought of as little harmonic

oscillators all coupled together The eld is dened in the limit of vanishingly small

boxes and so describ es a system with an innite number of degrees of freedom

The Maxwell eld of electromagnetism is a famous classical eld We can go ahead

an demand that it also be consisten t with the laws of quantum mechanics This means

we must quantise the motion of the innite number of little harmonic oscillators To do

this we simply imp ose the commutation relation x p i where x is the displacement

i i i

th

of the i oscillator and p is its momentum Note that x is simply the value of the eld

i i

th

at the point where the i box is lo cated After doing this we have a theory which is

consistent with b oth relativity and quantum mechanics Photons emerge as the quanta of

the electromagnetic eld and so the theory naturally describ es systems with anynumber

of particles

Maxwells wave equations are not the only equations we can write down which are

consistent with relativity We can also write down the KleinGordon and Dirac equations

to o there are others but these are the most relevant ones for Quantis

ing the corresp onding elds leads to quanta whichhave spin and spin resp ectively

the Maxwell eld leads to spin quanta Letting the spin quanta carry electric

charge means that they can interact with the spin quanta and the formalism has no

problem dealing with varying numb ers of quanta Of course Im skipping the details so

as to giveyou an overview DaveDunbars course aims to provide a fairly comprehensive

intro duction to this whole area

A nal word of caution Notation can be a little confusing People often use the

same symb ol for b oth the wavefunction and the eld You should always b e aware of the

dierence and be able to sp ot from the context which is meant

The KleinGordon Equation

Lets now take a more detailed lo ok at the KG equation In p osition space we

write the momentum op erator as

p ir

so that the KG equation b ecomes

m x

where we have intro duced the box notation

t r

and x is the vector t x

The op erator is Lorentz invariant so the KleinGordon equation is relativistically

covariant that is transforms into an equation of the same form if is a scalar function

That is to say under a Lorentz transformation t x t x

t x t x t x

so is invariant In particular is then invariant under spatial rotations so it represents

a spinzero particle more on spin when we come to the Dirac equation there b eing no

preferred direction which could carry information on a spin orientation

The KleinGordon equation has plane wave solutions

iEtpx

xNe

p

where N is a normalisation constantandE p m Thus there are b oth p ositive

and negative energy solutions The negative energy solutions p ose a severe problem if you

try to interpret as a wavefunction as indeed we are trying to do The sp ectrum is no

longer b ounded from b elow and you can extract arbitrarily large amounts of energy from

the system by driving it into ever more negative energy states Any external p erturbation

capable of pushing a particle across the energy gap of m between the p ositive and

negative energy continuum of states can uncover this diculty Furthermore we cannot

just throw away these solutions as unphysical since we need them in order to dene

a complete set of states Note that if one interprets as a quantum eld there is no

problem ie the p ositive and negative energy mo des are just asso ciated with op erators

which create or destroy particles

A second problem with the wavefunction interpretation arises when trying to nd

a probability density Since is Lorentz invariant jj do esnt transform like a density

To search for a candidate we derive a continuity equation rather as you did for the

Schrodinger equation in the prescho ol problems Dening and J by

i

t t

J i r r

you obtain see problem a covariant conservation equation

J

where J is the vector J It is natural to interpret as a probability density and J

as a probability current However for a plane wave solution jN j E so is

not p ositive denite since weve already found E can be negative

Exercise

continuity equation Start with the KleinGordon equation multiplied Derive the

by and subtract the complex conjugate of the KG equation multiplied by

Thus may well be considered as the density of a conserved quantity such as

but we cannot use it for a probability density To Dirac this and the

existence of negative energy solutions seemed so overwhelming that he was led to intro

hoping duce another equation rst order in time derivatives but still Lorentz covariant

that the similarity to Schrodingers equation would allow a probability interpretation

Diracs original hop es were unfounded b ecause his new equation turned out to admit

negative energy solutions to o Even so he did nd the equation for spin particles

and predicted the existence of antiparticles

Before turning to discuss what Dirac did Lets put things in context Wehave found

that the KleinGordon equation a candidate for describing the quantum mechanics of

B

e

e

e

time

A

Figure Feynman interpretation of a pro cess in which a negative energy electron is absorb ed

Time increases moving upwards

spinless particles admits unacceptable negative energy states when is interpreted as

the single particle wavefunction We could solve all our problems here and now and

restore our faith in the KleinGordon equation by simply reinterpreting as a quantum

eld However we wont do that There is another way forward this is the way followed

in the textb o ok of Halzen Martin due primarily to Feynman Causality us to

ensure that p ositive energy states propagate forwards in time Feynman sp otted that

the negative energy states cause us problems only so long as we think of them as real

physical states propagating forwards in time If we these negative energy states

only to propagate backwards in time then we nd a theory which is consistent with the

requirements of causality and which has none of the aforementioned problems

According to Feynman we should interpret the emission absorption of a negative

energy particle with momentum p as the absorption emission of a p ositive energy

antiparticle with momentum p So in Figure for example an electronp ositron

pair is created at p oint A The p ositron propagates to point B where it is annihilated

by another electron Another way of describing this picture is to say that the incoming

electron interacts to pro duce an outgoing photon and a negative energy electron This

electron travels back in time where it scatters o the incoming photon to pro duce the

outgoing electron To someone observing in real time the negative energy state moving

backwards in time lo oks to all intents and purp oses like a p ositively charged electron

with p ositive energy moving forwards in time

The Dirac Equation

Dirac wanted an equation rst order in time derivatives and Lorentz covariant so it had

to b e rst order in spatial derivatives to o His starting p ointwas to assume a Hamiltonian

of the form

H P P P m

D

where P are the three comp onents of the momentum op erator P and and are some

i i

unknown quantities which as will b e seen below cannot simply b e commuting numbers

When the requirement that the H P m is imp osed this implies that and

i

D

as we shall discuss The rst step is to write the must be interpreted as matrices

momentum op erators explicitly in terms of their dierential op erators using equation

then the Dirac equation b ecomes using the Dirac Hamiltonian in equation

i i r m

t

which is the p osition space Dirac equation Remember that in eld theory the Dirac

equation is the equation of motion for the eld op erator describing spin fermions In

order for this equation to b e Lorentz covariant it will turn out that cannot b e a scalar

under Lorentz transformations In fact this will b e precisely how the equation turns out

to describ e spin particles We will return to this b elow

If is to describ e a free particle it is natural that it should satisfy the Klein

Gordon equation so that it has the correct energymomentum relation This requirement

imp oses relationships among the and To see these apply the op erator on each side

of equation twice ie iterate the equation

i j i j i i i

r r i mr m

t

with an implicit sum over i and j fromto The KleinGordon equation by comparison

is

i i

r r m

t

i

and commute then the KG will clearly b e satised if If we do not assume that the

i j j i ij

i i

for i j It is clear that the and cannot b e ordinary numb ers but it is p ossible

i

to give them a realisation as matrices In this case mustbeamulticomp onent spinor

on which these matrices act

Exercise

Prove that any matrices and satisfying equation are traceless with eigenvalues

Hence argue that they must be even dimensional

In two dimensions a natural set of matrices for the would b e the Pauli matrices

i

i

However there is no other indep endent matrix with the right prop erties for so

the smallest dimension for which the Dirac matrices can be realised is four One choice

is the Dirac representation





Note that eachentry ab ove denotes a twobytwo blo ck and that the denotes the

identity matrix

There is a theorem due to Pauli which states that all sets of matrices ob eying the

y y

relations in are equivalent Since the Hermitian conjugates and clearly ob ey

the relations you can by a change of basis if necessary assume that and are

Hermitian All the common choices of basis have this prop erty Furthermore we would

like and to b e Hermitian so that the Dirac Hamiltonian is Hermitian

i

Exercise

Derive the continuity equation J for the Dirac equation with

y y

J J

We will see in section that J transforms as it must as a fourvector

Free Particle Solutions I Interpretation

We lo ok for plane wave solutions of the form

p

iEtpx

e

p

where p and p are twocomp onent spinors which dep end on momentum p but are

indep endentofx Using the Dirac representation of the matrices and inserting the trial

solution into the Dirac equation gives the pair of simultaneous equations

m  p

E

 p m

There are two simple cases for which equation can readily b e solved namely

p m which might represent an electron in its rest frame

m p which might represent a massless neutrino

For case an electron in its rest frame the equations decouple and b ecome

simply

E m E m

So in this case we see that corresp onds to solutions with E m while corresp onds

to solutions with E m In light of our earlier discussions we no longer need to recoil

in horror at the app earance of these negative energy states

electron with p for which the The negative energy solutions p ersist for an

solutions to equation are readily seen to b e

 p  p

E m E m

p

Thus the general p ositive energy solutions with E j p m j are

iEtpx

e x

 p

E m

p

while the general negative energy solutions with E j p m j are

 p

iEtpx

E m

x e

for arbitrary constant and Clearly when p these solutions reduce to the p ositive

and negative energy solutions discussed previously As an aside it is interesting to see

how Dirac cop ed with the negative energy states

Dirac interpreted the negative energy solutions by p ostulating the existence of a

sea of negative energy states The vacuum or ground state has all the negative energy

states full An additional electron must now o ccupy a p ositive energy state since the

Pauli exclusion principle forbids it from falling into one of the lled negative energy

states On promoting one of these negative energy states to a p ositive energy one by

supplying energy an electronhole pair is created ie a p ositive energy electron and a

hole in the negative energy sea The hole is seen in nature as a p ositive energy p ositron

This was a radical new idea and brought pair creation and antiparticles into physics

Positrons were discovered in cosmic rays by Carl Anderson in

The problem with Diracs hole theory is that it do esnt work for b osons Such

particles have no exclusion principle to stop them falling into the negative energy states

releasing their energy

Recall that Feynman tells us to keep both typ es of free particle solution One is to

be used for particles and the other for the accompanying antiparticles Lets return to

our spinor solutions and intro duce some basis spinors Take the p ositive energy solution

of equation and dene

p

r

ipx ipx

e u pe E m

r

 p

r

E m

For the negative energy solution of equation change the sign of the energy E

E and the threemomentum p p to obtain

 p

p

r

ipx ipx

E m

E m e v pe

r

r

In these two solutions E is now and for the rest of the course always p ositive and given

by E p m The subscript r takes the values with

the For the simple case p we may interpret as the spinup state and as

spindown state Thus for p the comp onent wavefunction has a very simple

interpretation the rst two comp onents describ e with spinup and spindown

while the second two comp onents describ e p ositrons with spinup and spindown Thus

we understand on physical grounds why the wavefunction had to have four comp onents

The general case p is slightly more involved and is considered in the next section

The uspinor solutions will corresp ond to particles and the v spinor solutions to

antiparticles The role of the two s will b ecome clear in the following section where it

will be shown that the two choices of r are spin lab els Note that each spinor solution

dep ends on the threemomentum p so it is implicit that p E

Free Particle Solutions II Spin

Now its time to justify the statements we have b een making that the Dirac equation

describ es spin particles The Dirac Hamiltonian in momentum space is given in

equation as

H P m

D

and the orbital angular momentum op erator is

L R P

Normally you havetoworry ab out op erator ordering ambiguities when going from classical

ob jects to quantum mechanical ones For the comp onents of L there is no ambiguity

Evaluating the commutator of L with H

D

LH R P P

D

R P P

i P

we see that the orbital angular momentum is not conserved otherwise the commutator

would be zero Wed like to nd a total angular momentum J which is conserved by

adding an additional op erator S to L

J L S JH

D

To this end consider the three matrices



i



where the rst equivalence is merely a denition of and the last equality can read

ily be veried The have the correct commutation relations to represent angular

momentum since the Pauli matrices do and their commutators with and are

i

i j ij k k

Exercise

Verify the commutation relations in equation

From the relations in we nd that

H i P

D

Comparing this with the commutator of L with H in equation you readily see

D

that

i h

L H

D

and we can identify

S

as the additional quantitywhich when added to L in equation yields a conserved

total angular momentum J We interpret S as an angular momentum intrinsic to the

particle Now

 

S

 

and recalling that the eigenvalue of J for spin j is j j we conclude that S represents

spin and the solutions of the Dirac equation have spin as promised

Weworked in the Dirac representation of the matrices for convenience but the result

is necessarily indep endent of the representation

Now consider the uspinor solutions u p of equation Cho ose p p

r z

and write

p

E m

p

C B B C

E m

C B B C

p

C u u pB u u pB C

A A

E m

p

E m

It is easy to see that

S u u S u u

z z

So these two spinors represent spin up and spin down along the z axis resp ectively For

the v spinors with the same choice for p write

p

E m

p

C C B B

E m

C C B B

p

C C v v p B v v pB

A A

E m

p

E m

where now

S v v S v v

z z

This apparently p erverse choice of up and down for the v s is actually quite sensible when

one realises that a negative energy electron carrying spin backwards in time lo oks

just likeap ositive energy p ositron carrying spin forwards in time

Normalisation and Gamma Matrices

p

We have included a normalisation factor E m in our spinors With this factor

y y rs

u p u p v p v pE

r s r s

This corresp onds to the standard relativistic normalisation of E particles per unit vol

y

ume we shall justify this a bit later on It also means that u u transforms like the time

comp onent of a vector under Lorentz transformations as we will see in section

Exercise

Check the normalisation condition for the spinors in equation

There is a much more compact way of writing the Dirac equation which requires

that we get to grips with some more notation Dene the gamma matrices



In the Dirac representation







In terms of these the relations b etween the and in equation can b e written

compactly as

f g g

Combinations like a o ccur frequently and are conventionally written as

a a a

pronounced a slash Note that is not despite app earances a vector It just

denotes a set of four matrices However the notation is delib erately suggestive for when

h transform likevectors and combined with Dirac elds you can construct quantities whic

other Lorentz tensors see the next section

Lets close this section by observing that using the gamma matrices the Dirac equa

tion b ecomes

i m

or in momentum space

p m

The spinors u and v satisfy

p mu p

r

p mv p

r

Exercise

Derive the momentum space equations satised by u pand v p

r r

Lorentz Covariance

W e want the Dirac equation to preserve its form under Lorentz transformations

LTs Let represent a LT

x x x

A familiar example of a LT is a b o ost along the z axis for which



B C

B C

A



with as usual v in units of c and  L Ts can be thought of as

generalised rotations

The requirementis

i m x i m x

where This last equalityfollows b ecause

x

x x x x

and equation has been used in the last step

We know that vectors get their comp onents mixed up by LTs so we exp ect that

the comp onents of might get mixed up to o

x x S x S x

where S is a matrix acting on the spinor index of Note that the argument

of writing x so each comp onent of x is transformed into a x is just a fancy way

linear combination of comp onents of x

It is helpful to recall that for a vector eld the corresp onding transformation is

A x A x

where x x This makes sense physically if one thinks of space rotations of a vector

eld For example the wind arrows on a weather map of England are an example of a

vector eld at each point on the map there is asso ciated an arro w Consider the wind

direction at a particular p oint on the map say Abingdon If the map of England is rotated

then one would exp ect on physical grounds that the wind vector at Abingdon always p oint

in the same physical direction and have the same length In order to achieve this both

the vector itself must rotate and the p oint to which it is attached Abingdon must be

correctly identied after the rotation Thus the vector at the point x corresp onding to

Abingdon in the rotated frame is equal to the vector at the p oint x corresp onding to

Abingdon in the unrotated frame but rotated so as to keep the physical sense of the

vector the same in the rotated frame so that the wind always blows towards Oxford say

in the two frames Thus having correctly identied the same p oint in the two frames all

we needtodoisrotatethevector

A x A x

A similar thing also happ ens in the case of the comp onent spinor eld ab ove except that

wedonotyet knowhow the comp onents of the wavefunction themselves must transform

ie we do not know S

Wenow need to gure out what S is To determine S we rewrite the Dirac equation

in terms of the primed variables just a mathematical substitution

i m x

The key step is to realise that matrices satisfy the same anticommutation

relations as the s in equation ie

f g g

Exercise

Check relation

This is an imp ortant result since it means that the matrices constitute an accept

able representation of the gamma matrices Now any two equivalent representations of

the gamma matrices are related by a transformation

M M

This allows us to rewrite equation as

i mM x

So we see that if S M then it follows that

i m x

e its form in the primed frame and the Dirac equation do es indeed preserv

We still havent solved for S explicitly wehave just shown that there is a solution

To nd S we need to solve equation which maybe written as

S S

For an innitesimal LT it can be shown that

j i i j

g g

ij

where is an innitesimal parameter and the pair i j lab el the six typ es of transfor

ij

mation ie boosts and rotations

For example a b o ost along the z axis corresp onds to i j since in this case

 g  g



C B

C B

A



which is the usual matrix for an innitesimal Lorentz b o ost with and

The combinations i j and corresp onds to b o osts along the x and y axes

resp ectively The remaining combinations and corresp ond to innites

imal anticlo ckwise rotations through an angle ab out the x y and z axes resp ectively

ij

Its a nice exercise to check this out

We are at lib ertyto write

ij

S i s

ij

ij

and our task is to determine the set of matrices s To do this substitute the expression

ij

for S into equation and remember that S i s It is then not to o

ij

hard to show that the solution is

i

ij i j ij

s

ij

Here I have taken the opp ortunity to dene the matrix Thus S is given explicitly

in terms of gamma matrices for a general LT

Exercise

ij

Verify that equation relating and is satised by s dened through equa

tions and The result

A B C ffA B gCg ffA C gBg

mightprove useful

We are aiming to nd quantities which are Lorentz invariant or transform as vectors

or tensors under LTs To this end well nd it useful to intro duce the Pauli and Dirac

adjoints The Pauli adjoint of a spinor is dened by

y y

The Dirac adjointofa matrix A is dened by

A A

For Hermitian it is easy to show that

y

A A

Some prop erties of the Pauli and Dirac adjoints are

A B A B

AB B A

A A

With these denitions transforms as follows under LTs

S

Exercise

y

Verify that This says that

y

Using and verify that S S ie S S So S is not

unitary in general although it is unitary for rotations when i and j are spatial

indices This is because the rotations are in the unitary O subgroup of the

nonunitary Lorentz group Here you show the result for an innitesimal LT but it

is true for nite LTs

satises the equation Show that

i m

where the arrowover implies the derivative acts on

transforms as in equation Hence prove that

Note that result of the problem ab ove can be rewritten as S S and

equation for the similarity transformation of to takes the form

S S

Combining the transformation prop erties of and in equations and

we see that the bilinear is Lorentz invariant In section well consider the trans

formation prop erties of general bilinears

Let me close this section by recasting the spinor normalisation equations in

terms of Dirac inner pro ducts The conditions b ecome

rs

u pu p m

r s

u pv p v pu p

r s r s

rs

v pv p m

r s

Exercise

Verify the normalisation prop erties in the ab ove equations

Parity

In the next section we are going to construct quantities bilinear in and and classify

them according to their transformation prop erties under LTs We normally use LTs

which are in the connected Lorentz Group SO meaning they can be obtained by

a continuous deformation of the identity transformation Indeed in the last section we

considered LTs very close to the identity in equation However the full Lorentz

group consists not only of the SO transformations but also includes the discrete

op erations of parity space inversion P and time rev ersal T

B C B C

B C B C

B C B C

P T

A A

T

LTs satisfy g g see the prescho ol problems so taking determinants shows

t since the identity do es but the that det LTs in SO have determinan

P and T op erations have determinant

Lets now nd the action of parity on the Dirac wavefunction and determine the

wavefunction in the parityreversed system According to the discussion of the previ

P

ous section and using the result of equation we need to nd a matrix S satisfying

i i

S S S S

Its not hard to see that S S is an acceptable solution from which it follows that

the wavefunction is

P

t x t x

P

In fact you could multiply by a phase and still have an acceptable denition for the

parity transformation

In the nonrelativistic limit the wavefunction approaches an eigenstate of parity

Since

the uspinors and v spinors at rest have opp osite eigenvalues corresp onding to particle

antiparticle having opp osite intrinsic parities and

Bilinear Covariants

Now as promised we will construct and classify the bilinears You mightlike to p onder

why we are so interested in bilinears ie ob jects which carry no spinor indices and

which involve only spinor elds recall that the starting point of any eld theory is

writing down a Lagrangian density To b egin note that by forming pro ducts of the

gamma matrices it is p ossible to construct linearly indep endent matrices Any

constant matrix can then be decomp osed into a sum over these basis matrices In

equation we have dened

i

and now it is convenient to dene

i

where the last equality is valid in the Dirac representation This new matrix satises

y

f g

Now the set of matrices

f g

form a basis for gamma matrix pro ducts

Tosee that there are matrices there is unit matrix matrix matrices and

matrices and matrices see equation for the denition of Note

that is antisymmetric to avoid double counting with the unit matrix A little thought

is needed to convince yourself that any constant matrix can b e obtained from linear

combinations of these matrices

Using the transformations of and from equations and together with

the similarity transformation of in equation the fermion bilinears and their

transformation prop erties can be written as follows

S scalar

det P pseudoscalar

V vector

det A axial vector

T tensor

Exercise

Verify the transformation prop erties of the bilinears in equation

Observe that J is just the current we found earlier in equation

Charge Conjugation

There is one more discrete invariance of the Dirac equation in addition to parity It is

charge conjugation which takes you from particle to antiparticle and vice versa For

scalar elds the symmetry is just complex conjugation but in order for the charge conju

gate Dirac eld to remain a solution of the Dirac equation you have to mix its comp onents

as well

T

C

C

T T

and C is a matrix satisfying the condition Here

T

C C

In the Dirac representation

i

C i

i

I refer you to textb o oks such as for details

When Dirac wrote down his equation everyb o dy thought parity and charge conju

gation were exact symmetries of nature so invariance under these transformations was

essential Now weknow that neither of them nor the combination CP are resp ected b y

the standard electroweak mo del

Neutrinos

In the particle data book you will nd only upp er limits for the masses of the three

neutrinos and in the standard mo del they are massless Lets lo ok therefore at solutions

of the Dirac equation with m From equation we have in this case

E  p E  p

These equations can easily b e decoupled by taking the linear combinations and dening

in a suggestive way the two comp onent spinors and

L R

R L

which leads to

E  p E  p

R R L L

Since E jpj for massless particles these equations may be written

 p  p

L L R R

jpj jpj

 p

is known as the helicity op erator ie it is the spin op erator pro jected in the Now

jpj

direction of motion of the momentum of the particle we see that the corresp onds to

L

solutions with negative helicity while corresp onds to solutions with p ositive helicity

R

In other words describ es a lefthanded neutrino while describ es a righthanded

L R

neutrino and each typ e of neutrino is describ ed by a twocomp onent spinor

The twocomp onent spinors describing neutrinos transform very simply under LTs

i

  i

e

L L

i

  i

e

R R

where  n corresp onds to space rotations through an angle ab out the unit n

axis and  v corresp onds to Lorentz b o osts along the unit vector v with a sp eed

v tanh Note that these transformations are consistent with the fact that it is not

p ossible to b o ost past a massless particle ie its helicity cannot be reversed

However under parity transformations they b ecome transformed into each other

L R

h involves only without such as the standard mo del manifestly So a theory whic

L R

violates parity

Although massless neutrinos can be describ ed very simply using two comp onent

spinors as ab ove they may also be incorp orated into the fourcomp onent formalism as

follows From equation we have in momentum space

jpj p

For such a solution

p Sp

jpj jpj

using the spin op erator S with dened in equation But Spjpj

is the pro jection of spin onto the direction of motion ie the helicity and is equal to

Thus pro jects out the neutrino with helicity right handed and

pro jects out the neutrino with helicity left handed

R L

dene the fourcomp onent spinors and

R L

To date only left handed neutrinos have been observed and only left handed neu

trinos app ear in the standard mo del Since

any theory involving only left handed neutrinos necessarily violates parity as we saw

b efore in the twocomp onent formalism

Finally note that in the Dirac representation which we have b een using

and the relation between the twocomp onent and fourcomp onent formalisms is via the

change of variables in equation However there exists a representation in whichthis

change of variables is done automatically and the massless Dirac equation falls apart

to the two twocomp onent equations discussed ab ove In this chiral representation in

and hence

L

R

We have identied and as the twocomp onent spinors discussed previously These

R L

results are also applicable to the electron in the approximation that its mass is neglected

by the simple transcription e e In fact in the standard mo del the

R R L L

electrons start out massless so these results will b e of use to Tim Morris in his course

The standard mo del and the minimal sup ersymmetric standard mo del contains only left

handed massless neutrinos and neutrino mass terms are forbidden by gauge symmetry

at least given the limited number of elds present in the standard mo del If extra elds

eg right handed neutrinos are added then neutrino masses b ecome p ossible If neutrino

oscillations are conrmed as the solution to the solar neutrino problem or are discovered

in lab oratory exp eriments then such a mo dication would b ecome a necessity

a b

Figure Scattering a and decay b pro cesses

Cross Sections and Decay Rates

Dave Dunbar has already discussed how to compute scattering amplitudes in quantum

eld theory see Section ofhisnotes ie how to compute the matrix element

T

iM P P hf Tj U TTj i Ti

fi f i

where U TT is the op erator which determines how the initial state i makes the

transition to the nal state f ie

Z

T

U TT T exp i H dt

I

T

As Dave has illustrated its quite a lengthy pro cedure to derive particular scattering

amplitudes from rst principles ie expanding in a p erturbation series using Wicks

theorem etc Fortunately there is a quicker way which involves using Feynman rules

Later on well actually calculate some scattering amplitudes for qed but only after Dave

at the Feynman rules has illustrated how to get

For now let us assume that weve done the work and have computed M Our task

fi

in this section is to convert this into a scattering cross section relevant if there is more

than particle in the initial state or a decay rate relevant if there is just particle in

the initial state see Figure

The probabilityfor the transition to o ccur is the square of the matrix element ie

Probability iM P P

fi f i

Attempting to take the squared mo dulus of the amplitude pro duces a meaningless square

This is a technical problem b ecause our amplitude is expressed of a delta function

between plane wave states These states are states of denite momentum and so extend

throughout all of spacetime In a real exp erimentthe incoming and outgoing states are

lo calised eg they mightleave tracks in a detector Todealwiththisproperlyyou can

construct normalised wavepacket states which do b ecome well separated in the far past

and the far future We wont deal with wavepackets instead well put our system in a

boxofvolume V L We also imagine that the interaction is restricted to act only over

The nal answers come out indep endentofV and T repro ducing the a time of order T

ones we would get if we worked with lo calised wavepackets We are in good company

here Nob el Laureate Steven Weinberg says in his recent book when discussing cross

sections and decay rates as far as I know no interesting op en problems in physics

hinge on getting the ne points right regarding these matters

In innite spacetime with plane wave states the transition amplitude from i to f is

given by However in our b ox of nite size L and for our nite time T the amplitude

is given by equation but with the Dirac delta functions replaced by well b ehaved

functions

I E E TI P P L P P

i f i f i f

where for example

E E T

f i

I E E T sin

f i

E E

i

f

This function has the prop erty that as T

I E E T E E

f i f i

and also

I E E T T E E

f i f i

with analogous results for I P P L Thus in our spacetime box we have the ap

f i

proximate result

VT P P P P

f i f i

Now there is a further subtle p ointwhich needs to b e addressed If wehavechosen

to normalise the elds so as to corresp ond to E particles p er unit volume as we did for

the spinor elds earlier then we need to get rid of this factor by dividing the amplitude

squared by EV per particle

y

To see this note that the Dirac probability density when integrated over the

volume of the b oxgives

Z

y

u u EV

box

and wehave used equation

The transition rate ie the probabilityper unit time is thus

N

Y Y

jM j VT P P

fi f i

T E V E V

f i

in

f

The Number of Final States

For a single particle nal state the number of available states dn in some momentum

range k to k dk is in the box normalisation

d k

dn V

This result is proved by recalling that the allowed momenta in the b oxhave comp onents

which can only take on discrete values suchask n L where n is an integer Thus

x x x

dn dn dn dn and the result follows

x y z

For atwo particle nal state we have

dn dn dn

where

d k d k

dn V dn V

where dn is the numb er of nal states in some momentum range k to k dk for particle

and k to k dk for particle There is an obvious generalisation to an N particle

nal state

N

Y

d k V

f

dn

f

Lorentz Invariant Phase Space LIPS

The transition rate for transitions into a particular element of nal state phase space is

thus given by using equations and

N N

Y Y Y

d k V

f

dW jM j P P V

fi f i

E V E V

f i

in

f f

This can b e rewritten as

Y

LI P S dW jM j V

fi

E V

i

in

where the lips is

N

Y

d k

f

LI P S P P

f i

E

f

f

Observe that everything in the transition rate is Lorentz invariant save for the initial

energy factor and the factors of V using d kE d k k m k which is

manifestly Lorentz invariant where E k m For a one particle initial state

the factor of V cancels and we can breath a sigh of relief after all wewould not exp ect

physical quantities to dep end on the size of our articial b ox For a two initial particle

scattering situation the factors of V will also cancel in the physical cross section as we

will show in the next section

Exercise

Show that the expression for twob o dy phase space in the centre of mass frame is given

by

d k d k

P k k s m m d

E E s

where s P is the centre of mass energy squared d is the solid angle element for the

angle of one of the outgoing particles with resp ect to some xed direction and

a b ca b c ab bc ca

Cross Sections

The total cross section for a static target and a b eam of incoming particles is dened as

the total transition rate for a single target particle and a unit b eam ux The dierential

cross section is similarly related to the dierential transition rate We have calculated

the dierential transition rate with a choice of normalisation corresp onding to a single

target particle in the b ox and a b eam corresp onding also to one particle in the b ox

A b eam consisting of one particle p er volume V with a velo city v has a ux N given by

v

N

V

particles p er unit area p er unit time Thus the dierential cross section is related to the

dierential transition rate in equation by

V dW

dW d

N v

where as promised the factors of V cancel in the cross section

Now let us generalise to the case where in the frame where you make the measure

ments The b eam has a velo city v but the target particles are also moving with a

velo city v In a colliding b eam exp eriment for example v and v will p oint in opp osite

directions in the lab oratory In this case the denition of the cross section is retained as

ab ove but now the b eam ux of particles N is eectively increased by the fact that the

target particles are moving towards it The eective ux in the lab oratory in this case is

given by

jv v j

N

V

which is just the total number of particles per unit area which run past each other per

unit time I denote the velo cities with arrows to remind you that they are vector velo cities

which must be added using the vector law of velo city addition not the relativistic law

In the general case then the dierential cross section is given by

dW

d jM j LI P S

fi

N jv v j E E

where we have used equation for the transition rate and the box volume V has

again cancelled The amplitudesquared and phase space factors are manifestly Lorentz

invariant What ab out the initial velo city and energy factors Observe that

E E v v E p E p

In a frame where p and p are collinear

m m jE p E p j p p

and the last expression is manifestly Lorentz invariant Hence we can dene a Lorentz

invariant dierential cross section The total cross section is obtained byintegrating over

the nal state phase space

Z

X

jM j LI P S

fi

jv v j E E

nal states

Because the result is indep endent of the dimensions of the b ox you can think of making the b oxas

large as you like ie big enough to t your exp eriment inside This means that there is no reason to

worry ab out the b ox

p p

c a

p p

d b

Figure scattering

A slightword of caution is needed in deciding on the limits of integration to get the total

cross section If there are identical particles in the nal state then the phase space should

be integrated so as not to double count

Twob o dy Scattering

An imp ortant sp ecial case is scattering see Figure

ap bp cp dp

a b c d

Exercise

Show that in the centre of mass frame the dierential cross section is

d s m m

c d

jM j

fi

d s s m m

a

b

Invariant scattering amplitudes are frequently expressed in terms of the

Mandelstam variables dened by

s p p p p

a b c d

t p p p p

a c b d

u p p p p

a d b c

In fact there are only two indep endent Lorentz invariant combinations of the available

momenta in this case so there must be some relation b etween s t and u

Exercise

Show that

s t u m m m m

a b c d

Exercise

Show that for two body scattering of particles of equal mass m

s m t u

Decay Rates

With one particle in the initial state the total transition rate is

Z

X

LI P S jM j W

fi

E

nal states

Only the factor E is not manifestly Lorentz invariant In the rest frame of a particle

of mass m we have

Z

X

jM j LI P S

fi

m

nal states

This is the denition of the decay rate In an arbitrary frame we nd W mE

which has the exp ected Lorentz dilation factor In the master formula equation

this is what the pro duct of E factors for the initial particles do es

i

Actually although the result is correct our derivation is not quite right To get

the answer we needed to consider an initial state at sometime in the distant past but

this state is unstable so it would have decayed b efore the interaction I refer to Peskin

Schro eder for a nice discussion of this subtlety

Quantum Electro dynamics

Quantising the Dirac Field

Dirac Field Theory is dened to be the theory whose eld equations corresp ond to the

x as b eing dynamically Dirac equation We regard the two Dirac elds x and

indep endent elds and p ostulate the Dirac Lagrangian density

L xi m x

Its easy to show that the EulerLagrange equation

L L

x

leads to the Dirac equation

The canonical momentum is

L

y

x i x

x

and the Hamiltonian density is

y

H L i

t

Rather than interpret as a wavefunction and therebyhavetokeep in mind notions

of negative energy states moving backwards in time we shall follow Dave Dunbar and

regard as a quantum eld We need to quantise this eld Now naively we would try

to imp ose the usual equal time commutation relations ie

xt y t i x y

i j ij

xt y t

i j

xt y t

i j

where i and j lab el the spinor comp onents of and This is a recip e for disaster In

particular there is no ground state ie excitations of the vacuum can have negative

energies The only way to cure the problem is to imp ose anticommutation relations

f xt y tg i x y

i j ij

y tg f xt

i j

f xt y tg

i j

There is a very nice discussion in Peskin Schro eder on this Chapter In particular

they show how anticommutation relations really are the only solution

As in Daves course the Heisenb erg equations of motion for the eld op erators have

solution

Z

X

d k

ik x ik x y

kv ke xt b ku ke d

E

Z

X

d k

y ik x ik x

b k xt u ke d kv ke

E

and the anticommutation relations imply that

n o

y

b kb k E k k

n o

y

d kd k E k k

fb kb k g

n o

y

y

kb b k

fd kd k g

o n

y

y

k kd d

The total Hamiltonian is

Z

H N d x H

The prex N denotes normal ordering This is the way we remove the ambiguity asso

ciated with the order of op erators Normal ordering means we are to put all creation

op erators to the left of all the annihilation op erators If this means moving an anticom

muting fermion op erator through another such op erator then we need to remember to

pick up a minus sign

After a bit of algebra we can get

Z

X

d k

y y

E b kb kd kd k H

E

y

If we had tried to imp ose commutation relations the dd term would have entered

with a minus sign in front which would signal that something has gone wrong In

y

particular it would mean that d creates particles of negative energy This is not supp osed

to happ en in the quantised eld theory Note that we could not x the problem by simply

y

relab elling d d since that would not be consistent with the commutations relations

imp osed on and

So in order to quantise the Dirac eld we are necessarily led to the intro duction of

anticommutation relations Remarkably we nd that we have automatically taken into

account the Pauli inclusion principle For example

n o

y

y

b kb k

implies that its not p ossible to create two quanta in the same state ie

y y

b kb kji

This intimate connection between spin and statistics is a direct consequence of desiring

our theory to b e consistent with the laws of relativityand quantum mechanics

The charge op erator is

Z Z

y

Q N d x j x d x

which in terms of the creation and annihilation op erators is

Z

X

d k

y y

E b kb k d kd k Q

E

y y

which shows that b creates fermions while d creates the asso ciated antifermions of

opp osite charge

Using the techniques discussed in Daves eld theory course we can go ahead and

compute the propagator for a Dirac particle

S x y hj x y ji x y

F

hj y xji x y

Skipping details see section ofDave Dunbars course this is

Z

ip m p d

ipxy

e S x y

F

p m i

Quantising the Electromagnetic Field

Dave showed us how to derive the Maxwell equations from the Lagrangian density

L F F j A

where the eld strength tensor is

F A A

See the prescho ol problems for an intro duction to this way of formulating classical

electro dynamics He also highlighted the gauge invariance ie that Maxwells equations

dont change under the transformation

A x A x x

where x is some scalar eld This gauge invariance allows us to imp ose the Lorentz

gauge condition ie without loss of generalitywe can x

A

Note that even after xing the Lorentz gauge we can p erform another gauge transfor

mation on A ie A x A x x where xmust satisfy the wave equation

Immediately we try to quantise the electromagnetic eld we hit a problem To see

this note that the canonically conjugate eld to A is

L

F

A

and from this it follows that This means we have no p ossibility to imp ose a

nonzero commutation relation between and A which we would need if we are to

quantise the eld

Fortunately all is not lost Let us incorp orate the gauge condition into the La

grangian density ie write

L F F j A A

The new term xes the gauge and is a dimensionless Lagrange multiplier as such it

can takeonanyvalue wecho ose Atrstsight it do esnt lo ok likeweve done anything

useful since F g A and so This is certainly true classically

however we need to b e a bit more careful with the quantum theory How are we supp osed

to interpret the Lorentz gauge condition If we assume it means that the op erator A

ever this is to o restrictive We need only ensure vanishes then we cant quantise How

that the gauge condition holds for matrix elements of A and now we can imp ose

nonzero commutation relations The quantisation condition then leads to

A xt A y t ig x y

with all other commutators vanishing The Heisenberg op erator corresp onding to the

photon eld is putting

Z

h i

X

d k

ik x y ik x

A x k a k e k a k e

E

where are a set of four linearly indep endent basis vectors For example

If we might cho ose and

k is along the z axis then and are p olarisation vectors for transverse p olarisations

whilst is referred to as the timelike p olarisation vector and is referred to as the

longitudinal p olarisation vector

The commutation relation implies that

h i

y

a k a k g E k k

y

terpret a k as an op erator which creates quanta At a glance this lo oks ne ie we in

of the electromagnetic eld photons with p olarisation and momentum k However

for we have a problem since the sign on the RHS of is opp osite to that of

the other p olarisations This shows up in the fact that these timelike photons make a

negative contribution to the energy

Z

d k

y y y y

H E a a a a a a a a

E

Fortunately although we might not realize it yet we have already solved the problem

The Lorentz gauge condition implies that for all physical observables the contributions

from the timelike and longitudinal photons always cancels More explicitlyby demanding

that

h j A j i

it follows that

y y

h ja a a a j i

This is nice b ecause it is in accord with our knowledge that free photons are transversely

p olarised

This is often referred to as the Feynman gauge

Here is a nice pro of that the classical electromagnetic eld is p olarised transversely The

classical eld can b e expanded thus

Z

i h

d k

ik x ik x

A x a k e a k e

E

and I have absorb ed the sum over p olarisation vectors into the Fourier co ecients a

The Lorentz gauge condition says that k  a ie a k k  ak and hence that

the time comp onent of a equals the longitudinal comp onent So we are already down

to indep endent comp onents The next step comes on realizing that we have still the

residual freedom to shift A A for any that satises the wave equation Since

satises the wave equation we can p erform a mo de expansion just like we did for the

A eld Thus we are free to p erform the replacement a a k For k k k

and a a a a a the choice a k transforms a a a and we have

nished

Having convinced ourselves that we can go ahead and quantise the Maxwell eld we

can now pro ceed to lo ok for the photon propagator Again Im going to skip the details

and refer back to Dave Dunbars course

iD x y hjTA xA y ji

F

Z

ik xy

d k e

ig

k i

This is the Feynman propagator Generalising away from is a bit more

tricky it gives

Z

ik xy

e k k d k

g iD x y i

F

k i k

Feynman Rules of QED

We are now ready to let our fermions and photons interact with each other The inter

action is describ ed by the Lagrangian

L e A

int

Suchaninteraction maybeintro duced by the concept of minimal substitution familiar

from classical electro dynamics

p p eA

E E e

In four vector notation

p p eA

Applying this classical concept of minimal substitution to the Dirac equation gives

iD m

where we have intro duced the covariant derivative

D ieA

The qed Lagrangian describing electrons photons and their interactions is then given

by

iD m L F F A

where A isthe gauge xing term for the Feynman gauge

The qed Lagrangian is invariant under a symmetry called gauge symmetry which

consists of the simultaneous gauge transformations of the photon eld

A A

and a phase transformation on the electron eld

ie

e

I refer to Dave Dunbars course for more details on gauge symmetries

For us the imp ortant thing is that we have got our hands on the qed Lagrangian

densit y From it we can gure out the Feynman rules Wehave already written down the

fermion and photon propagators and can more or less write down the other Feynman rules

without further ado I refer back to Dave Dunbars course Section for the detailed

discussion of how to get Feynman rules from the Lagrangian density The Feynman

rules for qed are summarised in Table

The rule for the vertex can b e obtained directly from L The external line factors

int

are easily derived by considering simple matrix elements in the op erator formalism They

For every draw write

ig

Internal photon line

q i

ip m

Internal fermion line

p

p m i

Vertex ie

u p Outgoing electron

s

Incoming electron u p

s

Outgoing p ositron v p

s

Incoming p ositron v p

s

Outgoing photon

Incoming photon

Attach a directed momentum to every internal line

Conserve momentum at every vertex

Table Feynman rules for qed are Lorentz indices and are spinor indices

p  p 

a c

e e

q

p  p 

b d

Figure Lowest order Feynman diagram for electronmuon scattering

are what is left b ehind from the expansions of elds in terms of annihilation and creation

op erators after the op erators have all been anticommuted until they annihilate the

vacuum

The spinor indices in the Feynman rules are such that matrix multiplication is p er

formed in the opp osite order to that dening the ow of fermion number The arrowon

the fermion line itself denotes the fermion number ow not the direction of the momen

tum asso ciated with the line I will try always to indicate the momentum ow separately

as in Table This will b ecome clear in the examples which follow We hav e already

met the Dirac spinors u and v I will say more ab out the photon p olarisation vector

when we need to use it

ElectronMuon Scattering

To lowest order in the electromagnetic coupling just one diagram contributes to this

pro cess It is shown in Figure The amplitude obtained by applying the Feynman

rules to this diagram is

ig

ie iM ie up up up up

fi c a d b

q

Note that for clarity I have dropp ed the spin lab el on the spinors Ill restore it when

I need to In constructing this amplitude we have followed the fermion lines backwards

with resp ect to fermion ow when working out the order of matrix multiplication which

makes sense if you think of an unbarred spinor as a column vector and a barred spinor

as arow vector and remember that the amplitude carries no spinor indices

The cross section involves the squared mo dulus of the amplitude which is

e

jM j L L

fi

e

q

where the subscripts e and refer to the electron and muon resp ectively and

up up up up L

c a a c

e

with a similar expression for L

Exercise

Verify the expression for jM j

fi

Usually wehave an unp olarised b eam and target and do not measure the p olarisation

of the outgoing particles Thus we calculate the squared amplitudes for each p ossible

spin combination then average over initial spin states and sum over nal spin states

Note that we square and then sum since the dierent spin congurations are in principle

distinguishable In contrast if several Feynman diagrams contribute to the same pro cess

you have to sum the amplitudes rst We will see examples of this b elow

The spin sums are made easy by the following results I temp orarily restore spin

lab els on spinors

X

u p p m u p

r r

r

X

v p v p p m

r r

r

Dont forget that by m I really mean m times the unit matrix

Exercise

Derive the spin sum relations in equation

Using the spin sums we nd that

h ih i

X

e

jM j p m p m p m p m

fi e jk e li ab bc cd da

ij

a c b d

kl

q

spins

e

tr p m p m tr p m p m

e e

a c b d

q

Where in the rst expression I chose to make explicit the spinor indices in order that

you can see how the trace which app ears in the second expression emerges Since all

of traces of calculations of cross sections or decay rates in qed require the evaluation

pro ducts of gamma matrices you will generally nd a table of trace theorems in any

quantum eld theory textb o ok All these theorems can b e derived from the fundamen

tal anticommutation relations of the gamma matrices in equation together with

the invariance of the trace under acyclicchange of its arguments For now it suces to

use

trab ab

trabcd abcd acbd adbc

n



tr for n odd

Exercise

Derive the trace results in equation

Using these results and expressing the answer in terms of the Mandelstam variables

of equation we nd

X

e

m s um jM j m s u m

fi

e e

t

spins

This can now b e used in the cross section formula to give in the high energy

limit s juj m m

e

d s u e

d s t

for the dierential cross section in the centre of mass frame

Exercise

Derive the result for the electronmuon scattering cross section in equation

p  p  p 

a c a

p  e e e e

c

e e e e

p 

d

p  p  p 

b d b

Figure Lowest order Feynman diagrams for electronelectron scattering

Other calculations of cross sections or decay rates will follow the same steps wehave

used ab ove You draw the diagrams write down the amplitude square it and evaluate

the traces if you are using spin sumaverages There are one or two more wrinkles to

be aware of which we will meet below

ElectronElectron Scattering

Since the two scattered particles are now identical you cant just replace m by m in

e

the calculation we did ab ove If you lo ok at the diagram of Figure with the

replaced by electrons you will see that the outgoing legs can be lab elled in two ways

Hence we get the two diagrams of Figure

The two diagrams give the amplitudes

ie

iM up up up up

c a d b

t

ie

up up up up iM

d a c b

u

Notice the additional minus sign in the second amplitude which comes from the anti

commuting nature of fermion elds You should accept as part of the Feynman rules for

qed that when diagrams dier by an interchange of two fermion lines a relative minus

sign must be included you dont need to get the absolute sign of an amplitude right

just its sign relativetothe other amplitudes This is imp ortant b ecause

jM j jM M j

fi

so the interference term will have the wrong sign if you dont include the extra sign

dierence between the two diagrams

Squaring the amplitude and doing the traces yields in the limit of negligible fermion

masses

X

s u s t s

jM j e

fi

t u tu

spins

ElectronPositron Annihilation

e e e e

For this pro cess the two diagrams are shown in Figure with the one on the right

known as the annihilation diagram They are just what you get from the diagrams for

electronelectron scattering in Figure if you twist round the fermion lines The fact

that the diagrams are related this way implies a relation between the amplitudes The

is interchange of incoming particlesantiparticles with outgoing antiparticlesparticles

p  p 

a c

e e

e e

p   p

c a

p   p

e e

d b

e e

p  p 

b d

Figure Lowest order Feynman diagrams for electronp ositron scattering in qed

called crossing For our particular example the squared amplitude for e e e e is

related to that for e e e e by p erforming the interchange s u Hence squaring

the amplitude and doing the traces yields again neglecting fermion mass terms

X

u t u s u

jM j e

fi

t s ts

spins

e e and e e hadrons

If electrons and p ositrons collide and pro duce muonantimuon or antiquark pairs

then the annihilation diagram is the only one which contributes At suciently high

energies that the quark masses can b e neglected this immediately gives the lowest order

qed prediction for the ratio of the annihilation cross section into hadrons to that into

X

e e hadrons

Q R

f

e e

f

where the sum is over quark avours f and Q is the quarks charge in units of e The

f

comes from the existence of three colours for each avour of quark Historically this

was imp ortant you could lo ok for a step in the value of R as your e e colliders cm

energy rose through a threshold for pro ducing a new quark a vour If you didnt know

ab out colour the height of the step would seem to o large Incidentally another place

the number of colours enters is in the decay of a to two photons There is a factor of

in the amplitude from summing over colours without which the predicted decay rate

would b e one ninth of its real size

At the energies used at lep you have to remember the diagram with a Z replacing

the photon

Exercise

is equal to s neglecting the lepton Show that the cross section for e e

masses

Compton Scattering

The diagrams which need to be evaluated to compute the Compton cross section for

e e are shown in Figure For unp olarised initial andor nal states the cross

section calculation involves terms of the form

X

p p

 

e e e e

Figure Feynman diagrams for Compton scattering

where represents the p olarisation of the photon of momentum p Since the photon is

massless the sum is over the two transverse p olarisation states and must vanish when

contracted with p or p Moreover since the photon is coupled to the electromagnetic

current J of equation any term in the p olarisation sum prop ortional

to p or p do es not contribute to the cross section This is because the current is

conserved J so in momentum space p J The upshot is that in calculations

you can make the replacement

X

p p g

Intro duction to Renormalisation

Renormalisation of QED

Lets start by considering how the electric charge is dened and measured This will

bring up the question of what happ ens when you try to compute lo op corrections In

fact the expansion in the numb er of lo ops is an expansion in Plancks constanth asyou

can show if you put back the factors ofh

The electric charge e can be dened as the coupling between an onshell electron

and an onshell photon that is as the vertex on the left hand side of Figure with

p m where m is the electron mass and q The Lagrangian parameter e p

can never b e measured in an exp eriment since quantum uctuations are always present

Exp eriment tells us that

e

We call e the renormalised coupling constant of qed We can calculate e in terms of e

in p erturbation theory To one lo op the relevant diagrams are shown on the right hand

side of Figure and the result takes the form

h i

M

e e e a ln b

m

where a and b are constants obtained from the calculation The e term is divergent

so we have intro duced a cuto M to regulate it This is called an ultraviolet divergence

since it arises from the propagation of high momentum mo des in the lo ops The cuto

amounts to selecting only those mo des where each comp onentofmomentum is less than

M in magnitude Despite the divergence in it still relates the measurable quantity

e to the coupling e we intro duced in our theory This implies that e itself must be

For a sensible theoryinany relation b etween physical quantities the ultraviolet divergent

divergences must cancel leaving a relation which is indep endent of the metho d used to

regulate divergences This seems a very sensible demand of our theory Essentially we

exp ect that qed breaks down at very high energies eg when gravitational eects start

to b ecome imp ortant ie b efore M However we hop e that what is going on at

such ultrashort Planck length distances do es not mo dify physics as we know it to day

eg Quantum do es not destroy Law Demanding this of our theory

is equivalentto saying that we want our theory to be renormalisable

As an example consider the amplitude for electronelectron scattering which we

considered at tree level in section Some of the contributing diagrams are shown in

Figure where the crossed diagrams are understo o d weshowed the crossed tree level

diagram explicitly in Figure Ultraviolet divergences are again encountered when

the diagrams are evaluated and the result is of the form

h i

M

e c ln iM c e d

fi

m

where c c and d are constants determined by the calculation In order to evalu

ate M numerically however we must express it in terms of the known parameter e

fi

Combining and yields

h i

M

iM c e e c a c ln d b c

fi

m

q



p  p 



Figure Diagrams for vertex renormalisation in qed up to one lo op

Figure Some diagrams for electronelectron scattering in qed up to one lo op

where the ellipsis denotes terms of order e and ab ove Since jM j is measurable

fi

consistency renormalisability requires

c a c

This result is indeed b orne out by the actual calculations and the relation b etween M

fi

and e contains no divergences

iM c e e d b c O e

fi

To understand how this cancellation of divergences happ ened we can study the con

vergence prop erties of lo op diagrams although we shall not evaluate them Consider

the third diagram on the right hand side in Figure and the middle diagram in Fig

ure These both contain a lo op with one photon propagator behaving like k at

large momentum k and two electron propagators each behaving like k To evaluate

the diagram we have to integrate over all momenta leading to an integral

Z

d k

I

k

large k

which diverges logarithmically leading to the ln M terms in and Notice

however that the divergent terms in these two diagrams must be the same since the

divergence is by its nature indep endent of the nite external momenta the factor of two

in equation arises b ecause there is a divergence asso ciated with the coupling of each

electron in the scattering pro cess In this way we can understand that at least some of

the divergences are common in both and What ab out diagrams such as the

third b oxlike one in Figure Nowwehavetwo photon and t wo electron propagators

leading to

Z

d k

I

k

large k

Figure Primitive divergences of qed

Figure Diagram containing a primitivedivergence

This time the integral is convergent

Detailed study like this reveals that for qed ultraviolet divergences always cancel

in relations b etween physically measurable quantities We discussed ab ove the denition

of the physical electric charge e A similar argument applies for the electron mass the

Lagrangian bare mass parameter m is divergent but we can dene a nite physical mass

m

In fact you nd that all ultraviolet divergences in qed stem from graphs of the typ e

shown in Figure and are known as the primitive divergences Any divergent graph

will b e found on insp ection to contain a divergent subgraph of one of these basic typ es

For example Figure shows a graph where the divergence comes from the primitive

divergent subgraph inside the dashed box Furthermore the primitive divergences are

always of a typ e that would be generated by a term in the initial Lagrangian with a

divergent co ecient Hence by rescaling the elds masses and couplings in the original

Lagrangian we can make all physical quantities nite and indep endent of the exact

details of the adjustmentsuch as how we regulate the divergent integrals This is what

we mean by renormalisability Put slightly dierently a renormalisable theory is one

which needs as input only the values of those observable ie renormalised parameters

which have bare unobservable counterparts sitting in the original Lagrangian density

This means that a renormalisable theory has real predictive power

This should b e made clearer by an example Consider calculating the vertex correc

tion in qed to one lo op

q

h i

up A B q Cq up

p p 

The calculation shows that A is divergent However we can absorb this by adding a

cancelling divergent co ecient to the A term in the qed Lagrangian The B

and C terms are nite and unambiguous This is just as well since an innite part of B

for example would need to be cancelled by an innite co ecient of atermofthe form

F

which is not available in

In fact the B term gives the qed correction to the magnetic dip ole moment g of

the electron or muon see page of the textb o ok by Itzykson and Zub er These

are predicted to b e at tree level You can do the onelo op calculation it was rst done

bySchwinger b etween September and Novemb er with a few pages of algebra to

nd

g

This gives g which is already impressive compared to the exp erimental

values

g

electron

g

muon

Higher order calculations show that the electron and muon magnetic moments dier at

two lo ops and ab ove Kinoshita and collab orators have devoted their careers to these

calculations and are currently at the four lo op level Theory and exp eriment agree for

the electron up to the th decimal place

The C term gives the splitting between the s and p levels of the hydrogen

atom known as the Lamb shift Bethes calculation of the Lamb shift done during a

train ride to Schenectady in June was an early triumph for quantum eld theory

Here to o the current agreement between theory and exp eriment is impressive

In discussing the vertex correction in qed we said that the divergent part of the

A term could be absorb ed by adding a cancelling divergent co ecient to the A term

in the qed Lagrangian When a theory is renormalisable al l divergences can be

removed in this way Thus for qed if the original Lagrangian is ignoring the gauge

xing term

F F i eA m L

then redene everything by

Z A Z A

R

R

Z

e Z e e m Z m

e m

Z Z

where the subscript R stands for renormalised In terms of the renormalised elds

Z F F iZ Z e A Z Z m L

R R R R m R

R R R

R

Writing each Z as Z Z reexpress the Lagrangian one more time as

F F e A m Z terms L i

R R R R R

R R R R

Now it lo oks like the old Lagrangian but written in terms of the renormalised elds with

the addition of the Z counterterms Nowwhenyou calculate the counterterms giveyou

new vertices to include in your diagrams The divergences contained in the counterterms

cancel the innities pro duced by the lo op integrations leaving a nite answer

The old A and are called the bare elds and e and m are the bare coupling and mass

Note that to maintain the original form of L you want Z Z so that the and

eA terms combine into a covariant derivative term This relation do es hold and is a

consequence of the electromagnetic gauge symmetry it is known as the Ward identity

Let me stress again that renormalisation is not ab out sweeping innities under the

carp et It is ab out saying that we dont need to understand physics at the Planck scale

in order to interpret lep data

Renormalisation of Quantum Chromo dynamics

Qcd is a theory of interactions between spin quarks and spin gluons It is a

nonAb elian gauge theory based on the group SU with Lagrangian

X

gauge xing and

a a

L G G iD m

f f

f

ghost terms

f

Here a is a colour lab el taking values from to for SU and f runs over the quark

avours The covariant derivative and eld strength tensor are given by

a a

D ig A T

a a a abc b c

G A A gf A A

abc a

a set of eight in where the f are the structure constants of SU and the T are

dep endent Hermitian traceless matrix generators in the fundamental or dening

representation see the pre scho ol problems and the quantum eld theory course

As in qed gauge xing terms are needed to dene the propagator and ensure that

only physical degrees of freedom propagate The gauge xing pro cedure is more compli

cated in the nonAb elian case and necessitates for certain gauge choices the app earance

of FaddeevPop ov ghosts to cancel the contributions from unphysical p olarisation states

in gluon propagators However the ghosts rst app ear in lo op diagrams which we will

not compute in this course

There are no Higgs b osons in pure qcd The only relic of them is in the masses for

the fermions which are generated via the Higgs mechanism but in the electroweak sector

of the standard mo del

A fundamental dierence b etween qcd and qed is the app earance in the nonAb elian

case of interaction terms vertices containing gluons alone These arise from the nonvan

ishing commutator term in the eld strength of the nonAb elian theory in equation

The photon is electrically neutral but the gluons carry the colour charge of qcd sp ecif

ically they transform in the adjoint representation Since the force carriers couple to

the corresp onding charge there are no multi photon vertices in qed but there are multi

gluon couplings in qcd This dierence is crucial it is what underlies the decreasing

strength of the strong coupling with increasing energy scale

In qcd hadrons are made from quarks Colour interactions bind the quarks pro duc

ing states with no net colour three quarks combine to make baryons and quarkantiquark

pairs give mesons It is generally b elieved that the binding energy of a quark in a

is innite This prop erty called connement means that there is no such thing as a free

quark Because of asymptotic freedom how ever if you hit a quark with a high energy

pro jectile it might b ehave in manyways as a free particle For example in deep inelastic

Figure Schematic depiction of deep An incident lepton radiates a

photon which kno cks a quark out of a The struck quark is detected indirectly only

after hadronisation into observable particles

Figure Additional diagrams for vertex renormalisation in qcd up to one lo op The dashed

line denotes a ghost For some gauge choices and some regularisation metho ds not all of these

are required

scattering or dis a photon strikes a quark in a proton say imparting a large momentum

to it Some strong interaction corrections to this part of the pro cess can be calculated

p erturbatively As the quark heads o out of the proton however the low energy strong

interactions cut in again and hadronise the quark into the particles you actually detect

This is illustrated schematically in Figure

Wenowtry to rep eat the pro cedure we used for renormalising the coupling in qed

but this time in qcdwhich is also a renormalisable theory If we dene the renormalised

couplingg as the strength of the quarkgluon coupling then in addition to the diagrams

of Figure with the photons replaced by gluons there are more diagrams at one lo op

shown in Figure Lo oking at the second of these new diagrams it is ultraviolet

divergentcontaining a ln M term and is also infrared divergent since there is no mass

tum mo des In qed all the lo op diagrams contain at least to regulate the low momen

one electron propagator and the electron mass provides an infrared cuto you still have

to worry when the electron is onshell but this is not our concern here In the second

diagram of Figure there is no quark in the lo op Now we can cho ose to dene the

The nite value of q provides renormalised coupling oshell ie at some nonzero q

the infrared regulator and the diagram has a term prop ortional to lnM q

an onshell vertex Thus in qcd we cant dene a physical coupling constant from

This is not really a serious restriction since its up to us how we dene the coupling

We could equally well have dened the qed coupling oshell its just that the value

of is easy to extract from low energy exp eriments which are close to the onshell

limit Now the renormalised coupling dep ends on how we dene it and therefore on at

least one momentum scale in almost all practical cases only one momentum scale The

renormalised strong coupling is thus written

gq

When physical quantities are expressed in terms of gq the co ecients of the p ertur

bation series are nite

You can dene counterterms for qcd in the same way as was demonstrated for

qed Now the gauge coupling g enters in many terms where it has the p otential to get

renormalised in dierent ways This would be a disaster In fact the gauge symmetry

imp oses a set of relations b etw een the renormalisation constants known as the Slavnov

Taylor identities which generalise the Ward identities of qed

We have just seen that the renormalised coupling in qcd gq dep ends on the

momentum at which it is dened We say it dep ends on the renormalisation scale and

commonly refer to g as the running coupling constant We would clearly liketoknow

just how g dep ends on q sowe calculate the diagrams in Figures and to get the

rst terms in a p erturbation theory expansion

h i

M

gg g a ln b

where a and b are constants and g is the bare coupling from the Lagrangian

I have switched to using in place of q and have written g as a function of for

convenience From this equation it follows that

g

g a g

The discovery byPolitzer and by Gross and Wilczek in that a led to the

p ossibility of using p erturbation theory for strong interaction pro cesses since it implies

that the strong interactions get weaker at high momentum scales g is a stable

solution of the dierential equation Keeping just the g term we can solve

to nd

g

s

ln

where is a constant of integration and a Thus decreases logarith

s

mically with the scale at which it is renormalised as shown in Figure If for some

pro cess the natural renormalisation scale is large there is a chance that p erturbation

theory will b e applicable The value of is

n

f

where n is the number of quark avours The crucial discovery was the app earance

f

of the coming from the selfinteractions of the gluons via the extra diagrams of

Figure Quarks and other nongauge particles always contribute negatively to

NonAb elian gauge theories are the only ones we know where you can have asymptotic

freedom providing you dont have to o much matter eg providing that the num ber

of avours is less than or equal to for qcd

s

Figure Running of the strong coupling constant with renormalisation scale

What is the signicance of the integration constant The original qcd La

grangian contained only a dimensionless bare coupling g the quark masses dont

matter here since the phenomenon o ccurs for a pure glue theory but now we have a

dimensionful parameter The real answer is that the radiative corrections in all eld

theories except nite ones break the scale invariance of the original Lagrangian In

qed there was an implicit choice of scale in the onshell denition of e Lacking such a

canonical choice for qcd you have to say measure at M or nd the scale

s Z

where so that a scale is necessarily involved The phenomenon was called

s

dimensional transmutation by Coleman isgiven by

Z

g

dg

exp

g

and is indep endent The explicit dep endence is cancelled by the implicit dep en

dence of the coupling

Weve seen that the coupling dep ends on the scale at which it is renormalised

Moreover there are many ways of dening the renormalised coupling at a given scale

dep ending on just how you have regulated the innities in your calculations and which

momentum scales you set equal to The value of g thus dep ends on the renormal

isation scheme you pick and with it In practice the most popular scheme to day

ms in which integrals are evaluated in is called mo died minimal subtraction

n

dimensions and divergences show up as p oles of the form for p ositive integer n In

the particle data book you will nd values quoted for around MeV it also de

MS

pends on the number of quark avours Dont buy avalue of unless you know which

renormalisation scheme was used to dene it

In Figure you see that the coupling blows up at This is an artifact of

using p erturbation theory We cant trust our calculations if In practice you

s

can p erhaps use scales for down to ab out GeV but not much lower and GeV is

probably safer

Exercise

of g in terms of g in to two lo ops gives Extending the expansion

h h i i

M M M

a ln g g g a ln g b b ln c

with a similar equation for g in terms of g Renormalisability implies that g can

be expanded in terms of g

X

n

g g X

n

n

where the X are nite co ecients Show that this implies that a is determined once

n

n

the one lo op co ecient a is known In fact a determines all the terms ln called

s

the leading logarithms from a one lo op calculation you can sum up all the leading

logarithms

For qed there is no p ositivecontribution to the b eta function so the electromagnetic

coupling has a logarithmic increase with renormalisation scale However the eect is small

even going up to lep energies go es from to ab out The so called Landau

p ole where blows up is safely hidden at an enormous energy scale

Exercise

Any pro cess sensitive to strong interactions is in principle able to measure at any

s

scale However in practise there is some optimal choice Discuss this statement

Acknowledgements

This lecture course has a long history I have made a few changes to the course that

A

Steve King gave last year Id liketo thank Steve for supplying me with the L T X les

E

Im also very grateful to Jonathan Flynn Steves supplier

Many thanks to Steve Lloyd for inviting me to come and deliver the course and

for doing such a go o d job as director Thanks also to Ann Rob erts for organizing the

scho ol Id also liketo say how much I enjoyed myself I thank the students tutors and

other lecturers for that

References

I J R Aitchison and A J G Hey Gauge theories in particle physics nd edition Adam

Hilger

JDBjorken and S D Drell Relativistic Quantum Mechanics McGrawHill

F Halzen and A D Martin Quarks and Leptons Wiley

C Itzykson and JB Zub er Quantum Field Theory McGrawHill

F Mandl and GShaw Quantum Field Theory Wiley

M E Peskin and D V Shro eder An Intro duction to Quantum Field Theory Addison

Wesley

L H Ryder Quantum Field Theory Cambridge University Press

JSchwinger Physical Review

H A Bethe Physical Review

A Pre Scho ol Problems

The main aim of this course will be to teach the techniques required for p erforming

simple calculations of amplitudes cross sections and decay rates particularly in Quan

tum Electro dynamics but also in Quantum Chromo dynamics Some asp ects of quantum

mechanics sp ecial relativity and electro dynamics will b e assumed during the lectures at

the scho ol The following problems should b e helpful in consolidating your knowledge in

these areas The solutions can be found in many standard textb o oks

Probability Density and Current Density

Starting from the Schrodinger equation for the wave function xt show that the

probability density satises the continuity equation

rJ

t

where

h i

h

J r r

im

What is the interpretation of J

Rotations and the Pauli Matrices

Show that a dimensional rotation can be represented by a orthogonal matrix R

with determinant Start with x R x and imp ose x x xx Such rotations form

the sp ecial orthogonal group SO

For an innitesimal rotation write R l iA where l is the identity matrix and

A is a matrix with innitesimal entries Show that A is antisymmetric the i is there to

make A hermitian

Parameterise A as

ia ia

X

C B

a L A ia ia

A

i i

i

ia ia

where the a are innitesimal and verify that the L satisfy the angular momentum

i i

commutation relations

L L i L

i j ij k k

Note that the Einstein summation convention is used here In general I will switch

t notational conventions without warning You should b e able to around b etween dieren

tell from the context what is meant notation should be your slave not your master

The Pauli matrices are

i

i

i

Verify that satisfy the same algebra as L If the twocomp onent spinor

i i

y

transforms into l ia under an innitesimal rotation checkthat is invariant

under rotations

Raising and Lowering Op erators

From the angular momentum commutation relations

L L i L

i j ij k k

show that the op erators

L L iL

satisfy

L L L

L L L

and show that

L L

From the last commutator it follows that there are simultaneous L L where L L

eigenstates of L and L Let be such an eigenvector of L and L with eigenvalues

lm

l l and m resp ectively Show that each of L either vanishes or is an eigenstate

lm

of L with eigenvalue l l and of L with eigenvalue m

Four Vectors

A Lorentz transformation on the co ordinates x ct x can be represented by a

matrix as follows

x x

Foraboostalong the xaxis to velo city v sho w that

C B

C B

C A B

A

where vc and as usual

By imp osing the condition

g x x g x x A

where

B C

B C

g B C

A

that show

T

g g or g g

This is the analogue of the orthogonality relation for rotations Check that it works for

the given by equation A ab ove

Now intro duce

x g x

and show by reconsidering equation A using x x or otherwise that

x x

Vectors A and B that transform like x and x are sometimes called contravariant

and covariant resp ectively A simpler pair of names is vector and covector A particularly

imp ortantcovector is obtained by letting x act on a scalar

x

Show that do es transform like x and not x

Electromagnetism

The four Maxwell equations are

rE rB

E B

r B J rE

t t

Which physical laws are represented by each of these equations Show that

rJ

t

and explain the signicance of this equation Verify that it can be written in manifestly

covariant form

J

where J c J

Intro duce scalar and vector p otentials and A by dening B rA and E

r A t and recall the gauge invariance of electro dynamics whichsays that E and

B are unchanged when

A A r and

t

for any scalar function Using this gauge freedom we can set

rA

c t

Assuming that and A can be combined into a four vector A c A this can be

written as A which is known as the Lorentz gauge condition Dening

show that with this condition Maxwells equations are equivalentto

A J

The tensor F is dened by

F A A

How many indep endent comp onents do es F have Rewrite F in terms of E and B

Show that

E

B F F

c

F F EB

c

where

if is an even permutation of

if is an o dd p ermutation of

otherwise

This gives the relativistic invariants which can b e constructed from E and B

Group Theory in Particular SU N 

y

Unitary matrices U satisfy U U l Verify that they form a group by showing that

W UV is unitary if U and V are In general you should also show that there is an

identit y element and that every U has an inverse but these are both obvious U N is

the group of complex unitary N N matrices and SU N is the subgroup of matrices

with determinant

Let U be a U N matrix close to the identity Write

U l iG

where G has innitesimal entries Show that G is hermitian If in addition U has

determinantso U SU N show that G is traceless

N traceless hermitian matrix can be written as a linear combination of a Any N

chosen basis set So for any G we can cho ose innitesimal numbers such that

i

N

X

G T

i i

i

where the T are our basis Explain why the summation runs from to N

i

Show that T T isantihermitian and traceless and hence can b e written

i j

T T if T A

i j ij k k

for some constants f The commutation relations b etween the dierent T dene the Lie

ij k i

a of SU N The T are called the generators and the f are called the structure algebr

i ij k

constants

Find a set of indep endent matrices which are generators for SU and a set

of indep endent generators for SU

Verify the Jacobi identity

h i h i h i

T T T T T T T T T

i j k j k i k i j

and hence show that

f f f f f f

jkl il m kil jlm ij l klm

Dene a new set of N N matrices

i

T if

jk ij k adj

and show that they ob ey the same commutation relations as the T in equation A The

i

i

T dene the adjoint representation The W s of the weak interactions and the gluons

adj

of the strong interactions belong to the adjoint representations of SU lefthanded

L

weak SU and SU the strong interaction colour algebra resp ectively

The generators and hence the algebra were found by lo oking at group elements

near the identity Other group elements can be recovered by combining lots of these

innitesimal rotations

N i T

i i

U lim l i T N e

i i

N

where the are nite This construction generates what mathematicians call a simply

i

connected group There is a theorem stating that every Lie algebra comes from exactly

one simply connected group SU N and its algebra give us one example

However we have seen that b oth SU and the rotation group SO have the

same angular momentum algebra What is going on It must be that SO is not

simply connected In fact there is a mapping called a covering from SU to SO

which preserves the group prop erty that is if U SU is mapp ed to f U SO

then f UV f U f V In the SU SO case two elements of SU are

mapp ed on to every elementofSO Whenever a group G has the same Lie algebra as

a simply connected group S there must b e such acovering S G

The double covering of SO by SU underlies the behaviour of spin and

other halfo ddinteger spin particles under rotations they really transform under SU

and rotating them by only gets you half way around SU so you pick up a minus

sign A second rotation gets you back to where you started