1 Introduction

The study of Elementary Particle Physics continues the great tradition of Physics to explorethe frontiers of human knowledge. In the case of Particle Physics we explore the realm ofvery small. Along the way we have discovered a basic understanding of the fundamentalforces of nature. However MANY puzzles remain. We will look at these puzzles and limitsto what we know as well as finding out what we we do know. I have very deliberately takenan experimental approach to this course. I really want to emphasize how much ParticlePhysics is an experimental science. Although this subject is an extremely active region ofTheoretical Physics research, Particle Physics is very much founded on fact. Many beautifuland elegant theories have died when confronted with experimental fact.

Particle Physics is a dynamic field.

1.1 Basic Concepts

We start by reviewing or introducing some basic concepts. With a good understanding ofjust some of these ideas it is possible to understand many points of Experimental ParticlePhysics.

A very brief history of Particle Physics

• 500 BC Greeks postulate the existence of atoms

• 1600 - 1900 Chemical elements are discovered. They’re identified as atoms.

• 1897 electron discovered

• 1900 Quantum Theory invented to explain Black Body Spectrum.

• 1905 Atomic Nucleus discovered by Rutherford.

• 1911 Planetary atom postulated with discrete atomic levels

• 1911-1928 Quantum Theory developed.

• 1928 Positron Postulated by Dirac.

• 1932 Neutron discovered.

• 1933 Neutrino Postulated.

• 1933 Positron discovered.

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• 1935 Pion’s existence postulated by Yukawa.

• 1937 Muon discovered.

• 1947 Pion discovered.

• 1947 Kaon discovered.

• 1953 Neutrino discovered.

• 1950 - 64 Strange Baryons discovered.

• 1957 Parity found to be violated.

• 1963 Eight-fold way predicts the existence of the Ω−.

• 1964 Ω− discovered.

• 1965 CP found to be violated in k0 decays.

• 1968 Deep inelastic e scattering shows substructure in Protons. “Bound” quarks dis-covered.

• 1961-1970 Electroweak Unification postulated. The W and Z predicted and charmquark predicted.

• 1974 Charm Quark discovered.

• 1970-75 Quantum Chromodynamics postulated as the strong force.

• 1977 Bottom or Beauty Quark discovered.

• 1983 W and Z discovered. “Standard Model” firmly established.

• 1997 Top Quark discovered.

• 1999 Neutrino Oscillations observed.

The discovery of neutrino oscillations is the first indication of physics beyond the standardmodel. Physicists continue to look for more evidence of physics beyond the standard modelsince such discoveries provide hints of what theory the Standard Model is a subset of.

Sizes

size Equivalent Photon Energy

ATOM 10−10 m = 105 fermi 0.00001 GeV = (10 KeV Xrays)Nucleus 10−14 m 10 fermi 0.1 GeVProton 10−15 m 1 fermi 1 GeVQuark ≤ 10−18 m 0.1 fermi 100 GeV

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Quarks and Leptons (e, µ, τ) are observed to be structure-less down to 10−18 = 100 GeV.

1.2 Reactions and Decays

We learn about particle properties by their interactions with other particles.

Example of a reaction

p + p → d + π+

Example of a decay

π0 → γ + γ

The basic conserved quantities are Energy, Momentum, Charge, Angular momentum. Theseare always conserved and are the same on either side of the arrow. Initial state quantities =Final state quantities.

Conserved integral quantities imply the existence of a Quantum Number. We shall learnabout other conserved quantities in this course.

1.3 Spherical Harmonics and Ortibal Angular Momentum

I shall take some time to review the concept of orbital angular momentum.

The Classical definition of Angular momentum about a point is:

~L = ~r × ~p (1)

Where ~r is the distance from the point to a particle of momentum ~p. This quantity isnon-zero if the distance of closest approach of the particle to the point is non-zero. If welook at an interaction between two particles it is clear we can define an angular momentumexactly as above. Now very interestingly the uncertainty principle can give us a non zeroangular momentum even if the two particles originate from the same point. ie.

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∆x∆p ≥ h

2(2)

implies a potential non-zero angular momentum for any two interacting particles since theuncertainty principle allows a non-zero distnace of closest approach. This type of angularmomentum is called the “Orbital Angular Momentum” of interacting particles and is alwayspresent is system of two or more particles.

In Quantum Mechanics the quantization of angular momentum in integral units of h arisesnaturally from the mathematics of wave functions in spherical coordinates.

The Schrodinger Equation in a potential free region, a long way from the interaction pointreduces to:

∇2Ψ = 0 (3)

which can be solved by the method of speration of variables. In Spherical coords we look forsolutions of the form:

Ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) (4)

In Spherical coords ∇2Ψ = 0 becomes:

1

r2

∂

∂r

(r2∂ψ

∂r

)+

1

r2 sin(θ)

∂

∂θ

(sin(θ)

∂Ψ

∂θ

)+

1

r2 sin2(θ)

∂2Ψ

∂φ2= 0 (5)

Since r,θ and φ are totally independent variables equation 6 can only solved if:

1

R(r)

d

dr

[r2dR(r)

dr

]= a (6)

1

Θ(θ) sin(θ)

d

dθ

[sin(θ)

Θ(θ)

dθ

]= b (7)

and1

Φ(φ)

d2Φ

dφ2= c (8)

where a, b and c are each speration constants that satisfy.

a+ b+ c = 0 (9)

Now define c = m2l and equation 8 becomes:

d2Φ

dφ2+m2

l Φ = 0 (10)

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Which has the solution:

Φml =1√2πeimlφ (11)

Where ml = 0, ±1,±2,.. The equation for Θ(θ) becomes legendre’s equation:

1

sin(θ)

d

dθ

[sin(θ)

Θ(θ)

dθ

]+

[l(l + 1)− m2

l

sin2(θ)

]Θ(θ) = 0 (12)

Where l =0,2,1,3... and ml=0, ±1,±2,..,±l.The solution Φlml(θ) can be expressed as a

polynomial of degree l in sin(θ) or cos(θ). Together, and normalized, Φmland Θlml

givethe sperical harmonics Ylml

(θ, φ).

Each distinct value of l correponds to a unit of orbital angular momentum h. Each value ofml corresponds to the spin projection of the mth substate. In addition these functions havethe symmetry property:

Ylmm(θ, φ) = (−1)lYlml(−θ, φ+ π) (13)

This change in coordinates corresponds to interchanging two particles in a two body system.

1.4 Bosons and Fermions

Consider a wavefunction made of 2 or more identical particles Ψ(n1n2).

Then| Ψ |2=| Ψ(n1n2) |2=| Ψ(n2n1) |2 (14)

ie we cannot distinguish the two situations. That implies

Ψ(n1n2) = ±Ψ(n2n1) (15)

ie. The wave functions can be either symmetric or anti-symmetric with the interchange oftwo particles.

The following rule holds always and divides all particles into two classes.

Bosons: Integral Spin 0h, 1h, 2h, ...Have Ψ(n1n2) = Ψ(n2n1) ie symmetric upon interchange.

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Fermions: 12

integral spin 12h, 3

2h, 5

2h, ...

Have Ψ(n1n2) = −Ψ(n2n1) ie anti-symmetric upon interchange.

We split the wavefunction into at least two parts:

Ψ = α(space)β(spin) (16)

Theα(space) (17)

gives the orbital motion with respect to each other. Because of the features of sphericalharmonics this gives a factor of (−1)l upon interchange of identical particles where “l” isthe orbital angular momentum of the two particles.

The β(spin) function requires either spins aligned (symmetric upon interchange) or anti-aligned (antisymmetric). These are the only two possibilities given

| Ψ(n1n2) |2=| Ψ(n2n1) |2 (18)

.

So for Bosons:

either Ψ(n1n2) = α( l even) β(↑↑)symmetric symmetric

or Ψ(n1n2) = α( l even) β(↑↓)anti-symmetric anti-symmetric

The result is always symmetric.

And for Fermions:

either Ψ(n1n2) = α( l odd) β(↑↑)anti-symmetric symmetric

or Ψ(n1n2) = α( l even) β(↑↓)symmetric anti-symmetric

The result is always anti-symmetric.

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1.5 Side Note.

The requirement that a fermion wavefunction be antisymmetric automatically gives the Pauliexclusion principle. “No two fermions can occupy the same state”.

Occupying the same state implies l = 0 (no motion relative to each other) and spins aligned(the same state remember).

SoΨ(n1n2) = α(symmetric)β(symmetic) (19)

= Ψ(n2n1) (20)

Which is not allowed.

On the other hand Bosons can happily sit right on top of each other because Ψ(n1n2) issymmetric. So cool bosons enough and they all end up with the same wavefunction.

1.6 Example of an application to Particle Physics

The decay of the neutral ρ0 meson which has spin 1. The π0 has spin 0. Look at the decay:

ρ0 → π0π0 (21)

Since the π0 has zero spin the decay ρ0 → π0π0 needs the π0 to have orbital angularmomentum l = 1 to conserve angular momentum.

However this implies the final state wavefunction has the property Ψ(π01π

02) = (−1)1(symmeticspin) =

−Ψ(π02π

01). This is not allowed!

The decay ρ0 → π0π0 has never been observed but ρ0 → π+π− is the main way theρ0 decays. This happens because the π+ and π− are not identical and so do not have thisrestriction.

This is typical of one sort of problem you’ll be given in this course. Use selection rules todecide if a process is allowed.

1.7 Particles and Antiparticles

A consequence of relativistic quantum mechanics is the existence of antiparticles for everyparticle.

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Antiparticles have the same mass but opposite charge and magnetic moment. All Bosonsand fermions have anti-particles.

A consequence of the fact that fermions have half-integral spin is that they are only createdor destroyed in pairs. Any process that created or destroyed a single Fermion would violateconservation of angular momentum since orbital angular momentum produces allows onlyintegral values.

There is no such problem for Bosons which can be created with the conservation of angularmomentum. So for example the reaction:

π+p→ π+π+n (22)

can procede.

1.8 Quarks and Leptons

Quarks

Quarks are the constituents of Hadrons. A hadron is any particle that feels the strong force.Hadrons are either Baryons = (QQQ) (3 quarks) proton, neutron or a meson = QQ (2quarks) π, k.

Quarks are observed to be structureless and featureless on a scale of less than 10−18m orabove 200 GeV.

They carry fractional electric charge. Either +23| e | or -1

3| e |. The occur in several

varieties or flavours.

u(up), d(down), s(strange), c(charm), b(bottom), t(top). Each quark has an internal quan-tum number that is conserved in strong and elecromagnetic interactions. They are groupedas doublets of generations.

u 23

7 MeV c 23

1.5 GeV t 23

174 GeVd -1

34 MeV s -1

3150 MeV b -1

34.5 GeV

Since quarks have half integral spin, Baryons with 3 quarks have half integral spin and areFermions. Basons have 2 quarks and so have integral spin and are bosons.

Examples

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Baryons Mesonsuud = p (proton) ud = π+ (pion)udd = n (neutron) ds = k0 (kaon)

uds = Λ (lambda hyperon) cc = J/Ψ (meson)

All quark quantum numbers are conserved in strong and electromagnetic reactions.

Example

π+ p → k+ Σ+

ud uud us uus

S 0 0 +1 -1

Leptons

The leptons have also been observed to be structureless and point-like down to 10−18 or upto 100 GeV. They’re

e µ τ

νe νµ ντ

They have integral charge 0 or ±1 and do not feel the strong force. Indeed the ν onlyinteract via the weak force. Each lepton has an internal quantum number that is strictlyconserved in all interactions.

Le, Lµ, Lτ = +1 for e−, µ−, τ−

= -1 for e+, µ+, τ+

= -1 for νe, νµ, ντ

= +1 for νe, νµ, ντ

Examples:Pair Production

γ → e+ e−

Le 0 -1 +1

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Pion Decay

π+ → µ+ νµ

Le 0 -1 +1

Muon Decay

µ+ → e+ νe νµ

Lµ -1 0 0 -1Le 0 -1 +1 0

While the decay:

µ+ → e+ γLe 0 -1 0Lµ -1 0 0

has a branching ratio below 10−13.

In 1999 SuperKamiokande observed that νµ neutrinos from cosmic ray interactions in theatmosphere oscillate into another (unknown) form.

νmu → νX

At this time (May 2000) we do not yet know what neutrino type νX is although the currentprejudice is that they are ντ . Ray Volkas and Robert Foote from the theoretical particlephysics group have postulated that νX are new massive neutrinos via a well formulatedtheory.

This process of neutrino oscillation of course violates lepton number conservation. In additionis requires that neutrinos have non-zero mass. Both these conclusion violate the StandardModel of particle physics and will be used to develop the superset of the standard model.

Finally, a consequence of fermion number conservation is that Baryon number is observedto be conserved.

Each Baryon is assigned +1 Baryon Number:Each anti-Baryon is assigned -1 Baryon Number.

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1.9 Cross Sections and Decay Rates

The strength of an interaction can be infered by the decay rate of a particle decay or thesize of the cross section for the particle reaction.

Decay Rates

P (t) = e−tτ (23)

where P(t) = Probability of decay per unit time and τ = mean life or lifetime.Faster decay rate ⇒ stronger interactions.

If a decay rate is very fast it is easier to determine the “width” of the particle. This is theuncertainty in the mass of the particle. These are related through the uncertainty principle.

τΓ = h (24)

(cf. ∆ t∆ E ≥ h2) Where Γ is the width of the particle.

For example, the π which decays via π → µ+νµ has a meanlife, τ of 2.6× 10−8 secs. Thiscorresponds to a width, Γ of:

Γ =h

τ=

6.22× 10−22MeV − secs

2.6× 10−8secs= 2.5× 10−8eV (25)

Slow decay ⇒ weak interaction.A different example is the decay of the π0. This decays as π0 → γγ. For this reaction:

τ = 0.87× 10−16secs⇒ Γ =h

τ=

6.58× 10−22MeV − sec

0.87× 10−16secs= 7.6eV (26)

This is considerabally faster than the decay of the charged pion as it occurs via the electro-magentic interaction.Finally consider the decay: ρ0 → π+π−. This decay has a width Γ = 150 MeV.

τ =h

Γ=

6.582× 10−22MeV − secs

150MeV= 4.4× 10−24secs (27)

Very fast decay ⇒ strong interaction.

Cross SectionA cross section gives a measure of the probability of an interaction. For the reaction:

a+ b→ c+ d

where a is the projectile, b is the target, c and d are the final state reaction products, thecross section is defined as:

σ =Y ield(c+ d)

NaNb

(28)

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where Na = number of particles a fired at the target.Nb = number of target particles b per unit area.Yield(c+d) = number of times the reaction products (c+d) are produced.

The dimensions of σ are area. An interpretation of σ is the effective area blocked by eachb for the process: a + b → c + d.

Then the yield of the reaction is given by:

Y ield = σ ×Nb ×Na (29)

The units used for σ are barns,

barn b = 10−24cm2

millibarn mb = 10−3 b = 10−27cm2

microbarn mb = 10−6 b = 10−30cm2

nanobarn mb = 10−9 b = 10−33cm2

Some typical cross sections:

Strong Interaction: π+ p→ anything, σ(1 GeV) = 20 mb

Electromagnetic Interaction: γ p→ anything, σ(1 GeV) = 0.130 mb

Weak Interaction: νµ p→ anything, σ(1 GeV) = 10−11 mb

1.10 Special Relativity

In typical Particle Physics situations the kinetic energy of a particle is often at least as largeas its mass. Consequently the low energy Newtonian Approximations to the equations ofmotion do not work.

In this course we will typically work with Relativistic formulae. The basis of these are thefollowing. Let:

E = Total energy of the particle

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m0 = Rest mass of the particlep = momentumc = speed of lightγ = Relativistic Gamma factorv = velocity of the particleβ = v

c

Then the basic equations of Special Relativistic kinematics are:

E2 = p2c2 +(m0c

2)2

(30)

E = γm0c2 (31)

γ =1√

1− β2=

1√1− v2

c2

(32)

Take (30), rewrite it as:

E2 −(m0c

2)2

= p2cc

⇒ (E −m0c2)(E +m0c

2) = p2cc

Let T= kinetic energy of the particle then

T = E −m0c2

ie. Kinetic Energy = Total Energy - Rest Energy. So:

p2cc = T (T + 2m0c2)

⇒ p =

√T (T + 2m0c2)

c(33)

**This is the relation between kinetic enrgy and momentum.

Turning it around gives:T 2 + 2m0c

2T − p2c2 = 0

Solving for T gives:

T =−2m0c

2 ±√

4(m0c2)2 + 4p2c2

2

Clearly only the +ve gives +ve T so

T =

√((m0c2)

2 + p2c2)−m0c

2 (34)

13

Next T = E - m0c2 by definition so T = γm0c

2 −m0c2 ⇒

T = m0c2(γ − 1) (35)

γ = 1 +T

m0c2(36)

γ =

√√√√ p2c2

(m0c2)2+ 1

− 1 (37)

Then finally we can derive:

β =pc

m0c2 + T=pc

E(38)

Work with E,T,p, m0, γ, β with these expressions.

1.11 Relativistic 4-vectors

Define a four vector p = (~pc, E) where ~p is usual momentum vector and E is the total energyof the particle. Then:

p2 =| ~p |2 c2 − E2 = −(m0c2)2 (39)

is imvariant under Lorentz transformations. This leads to some very useful relations.

Example of 4-momentum use

We have two particles A and B that interact. How much energy is available for particlecreation?

p2 = (~pA + ~pB)2 − (EA + EB)2 = −m2Ac

4 −m2Bc

4 + 2~pA.~pBc2 − 2EAEB (40)

The center of momentum system (CMS) is defined as the reference frame where the totalmomentum is 0. ie (~pA + ~pB) = 0. If the total energy in this system is E∗,

⇒ p2 = (~pAc+ ~pBc)2 − (EA + EB)2 = −E∗2 (41)

Lets look at two special cases. (i) If the target particle mB is at rest, (~pB = 0) so

E∗2 = −p2 = (mAc2)2 + (mBc

2)2 + 2mBc2EA

the energy required to create a new particle is m∗c2 = E∗

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The initial energy EA required to create it is

EA =(m∗c2)2 − (mAc

2)2 − (mBc2)2

2mBc2(42)

EA ≈(m∗c2)2

2mBc2for EA ≥ mAmB (43)

ie the energy available for particle creation grows ∝√EA in fixed target reactions.

This is because a lotof the incident beam’s momentum is used to give the created particlesmomentum too.

(ii) If the two particles have equal and opposite momentum (~pA = −~pB). This is the situa-tion in many colliding beam experiments.

ThenE∗2 = −p2 = −(~pA + ~pB)2 + (EA + EB)2

⇒ E∗2 = (EA + EB)2

So E∗ = 2EA

So all the incident Energy is available for particle creation.

1.12 Units in Particle Physics

The fundamental units in Physics are length, mass and time. In MKS these are expressedin terms of meters, kilograms and seconds.

In particle physics, the study of the small, these units are inconvenient. Here we work withparticles with energies in the regions of MeV and GeV.

Here we work with particles with energies in the regions of MeV and GeV.

We express masses in terms of their Einstein equivalent rest energies m0c2.

For example m0 = (proton) = 938 MeV/c2.m0 = (electron) = 0.511 MeV/c2.

We express momentum in units of MeV/c. Typical lengths are 10−15 m = 1 fermi, 1 barn= 100 fermi2.

h = 1.054×10−34 Joule-secs

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= 6.582×10−22 MeV-secsc = 2.99×108 m sec−1

hc = 197.3 MeV-fermi

Natural Units are defined to have h = c = 1. All other quantities are then expressed in unitsof energy.

Conversion Factor h=c=1 Units Actual Dimensions1kg = 5.26×1026 GeV GeV GeV c−2

1 m = 5.07×1015 GeV−1 GeV−1 hcGeV

1 fermi = 5.07 GeV−1 GeV−1 hcGeV

1 sec = 1.52×1024 GeV−1 GeV−1 hGeV

Other useful conversion factors (1 GeV)−2 = 0.389 mb

1.13 Interactions and Fields in Particle Physics

This is a very brief, hand-waving introduction to this. See Girish Joshi’s course for the realstory.

Classically interaction between particles occur through the effects of a potential or field.

In Q.M. interactions occur through the exchange of quanta (bosons). Since the quantumcarries energy and momentum, conservation laws require that it can only happen over a timelimited by the Uncertainty Principle. That is

∆E∆t ≤ h

2

Such transitions are called VIRTUAL

Lets illustrate this concept through Electrostatics. Classically electrostatic force:

| F |= Q1Q2

r2

The force between charges comes from the exchange of virtual photons of momentum q, thechange of momentum as it absorbs or emits a photon.

Then, the uncertainty principle links, the position (r), with the momentum of the photon(p), p r ≈ h .

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Then for a photon the time of each transfer is t = rc

⇒ ptc ≈ h

so p =h

tc

| dpdt|= h

t2c

This is the rate of momentum transfer for each transfer which is also the force, (F) on theparticle. For a photon, t = r

c⇒

| dpdt|=| F |= hc

r2

The number of photons emitted per unit time ∝ Q1Q2 and the uncertainty principle givesthe r dependence.

Neither the field nor the virtual particles are directly observable.

1.14 The Yukawa Theory

This theory was developed by Yukawa in 1935 to explain the short range nature of the strongforce. Suppose that instead of a photon, the particle exchange involves a particle of massm. The relativistic relation between the total energy E, the momentum p and mass m. Therelativistic relation between the total energy E, the momentum p and the mass m is:

E2 = p2c2 +m2c4

The differential equation describing the wave amplitude Ψ for such a free particle is obtainedby putting in the Q.M. operators.

Eop = ih∂

∂t;Pop = −ih∇

which give the Klein-Gordon equation:

∇2Ψ− m2c2

h2 Ψ− 1

c2∂2Ψ

∂t2= 0 (44)

We are interested in time independent, spherically symmetric solutions for a central forceU(r). The θ and φ components are given by the spherical harmonics as given earlier.

∇2U(r) =1

r2

∂

∂r

(r2∂U

∂r

)=m2c2

h2 U(r) (45)

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The solution of this equation is:

U(r) =g

4πre−

rR (46)

where

R =h

mc(47)

Here g is a constant of integration and is identified as the strength of the interaction. R =h

mc= range of the interaction. Historically the range of the strong interaction was measured

to be ≈ 1 fermi which implies m ≈ 100 MeV. In 1947 the pion with a mass of ≈ 140 MeVwas discovered. It was identified, too simplistically, as the strong force quantum.

1.15 The Boson Propagator

Let’s consider a particle scatered through an angle θ or equivalently the momentum transfer~q.

The potential U(~r) in coordinate space will have an associated amplitude f(~q) which is justthe Fourier transform of U(~r).

f(q) = g0

∫Ureiq·rdV (48)

where g0 is the intrinsic coupling strength of the particle to the potential.The volume integralis over all space.

q · r = qr cos θ

dV = r2dφ sin θdθdr

θ and φ are the polar and azimthal angles. We shall use units h = c = 1.

f(q) = g0

∫ ∞

0

∫ π

0

∫ 2π

0U(r) (cos(qr cos θ) + i sin(qr cos θ)) r2dφ sin θdθdr (49)

= πg0

∫ ∞

0

∫ π

0U(r) (cos(qr cos θ) + i sin(qr cos θ)) r2 sin θdθdr (50)

Let:

y = cos θ ⇒ dy

dθ= − sin θ ⇒ dθ = − dy

sin θ

⇒ f(q) = 2πg0

∫ ∞

0

∫ 1

−1−U(r) (cos(qry) + i sin(qry)) dyr2dr (51)

⇒ f(q) = 2πg0

∫ ∞

0−U(r)r2

[1

qrcos(qry)

i

qrsin(qry)

]−1

1

dr (52)

= 2πg0

∫ ∞

0

−U(r)r2

qr[sin(−qr)− i cos(−qr)− sin(qr) + i cos(qr)] dr (53)

= 2πg0

∫ ∞

0−U(r)r2

qr×−2 sin(qr)dr (54)

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= 4πg0

∫ ∞

0

U(r)r2

qr× sin(qr)dr (55)

Recall:U(r) =

g

4πre−

rR

and R = m since h = c = 1. So

f(q) = g0g∫ ∞

0

e−mr(eiqr − e−iqr)

2iqdr (56)

f(q) =g0g

2iq

∫ ∞

0(er(iq−m) − e−r(iq+m))dr (57)

=g0g

2iq

[1

iq −mer(iq−m) +

1

iq +me−r(iq+m)

]∞0

(58)

=g0g

2iq

[0− iq +m+ iq −m

(iq −m)(iq +m)

](59)

f(q) =g0g

q2 +m2(60)

Where g g0 = product of two vertex factors describing the coupling of the boson to scatteredand scattering particles.

dσ

dq2∝| f(q2) | (61)

1.16 Electromagnetic Interactions

The coupling constant for E-M interactions is:

α =e2

4πhc=

1

137.036

This is the so-called “fine structure constant”.

Some simple Processes

(a) Photoelectric effect.

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The photon couples to the electron with amplitude√α ⇒ σ(photo-electron) ∝ |

f |2 ∝ |√α |2 = α

(b) Rutherford Scattering.

q = momentum transfer (from the intermediate virtual photon)each vertex gives

√α so

f(q) ∝√α√α

q2 +m2γ

=α

q2as mγ = 0 (62)

sodσ

dq2∝| f(q) |2= α2

q4(63)

This is a second order process

(c) Pair Production near a Nucleus

The nuclear charge is Ze :

f ∝√α

3Z ⇒ σ ∝ α3Z2

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The field theory used to make exact calculations is called Quantum Electrodynamics orQED.

(d) Self Energy Process

The sum of all these terms is divergent. However QED is renormalizable which means thatall these divergent integrals are dumped into me and the charge e, and replaced by theirexperimentally determined values.

The result is that QED is incredibally precise and has been verified to the limit of experi-mental precision. (Currently 13 significant figures!)

1.17 Weak Interactions

Weak interactions take place between all quark and leptons. However it is normally swampedby E/M and strong interactions unless these are forbidden by conservation laws.

Therefore weak interactions are only observable in neutrino interactions (these have no elec-tric or strong charges).

Or quarks with a flavour change, (∆s = 1, ∆c = 1, forbidden in strong and electromagneticdecays). eg.

(a) n → p + e− + νe neutron β decay(b) νµ + p → n + µ+ anti- neutrino absorption(c) Σ → n + π− τ = 10−10 ss -1 0

Compare this last decay with the E/M decay of Σ0

Σ0 → Λ + γ τ = 10−19 ss -1 -1

This decay conserves strangeness but violates isospin ⇒ E/M decay. Relative strength ofweak / strong action:

√10−10

10−19≈ 10−5

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The weak interactions are mediated by massive bosons, the W± and Z0. For example,neutron β decay.

For example, neutron β-decay

Another example, neutrino absorption

Then early in 1973 neutral currents showing the existance of the Z0 were observed in

νµ + e− → νµ + e−

Since then many W± ’s and Z0 have been observed in pp colliders at CERN and FERMILABand at electron-positron colliders at LEP at CERN and SLAC (Stanford USA).

If we assign a single number, g, to the couplings of the W± and Z to quarks and leptons,

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the amplitude is:

f(q2) =g2

(q2 +M2WZ)

(64)

For q2 M2WZ the amplitude is independent of q and so looks pointlike (the amplitude has

no dependence on q2). In 1935 Fermi proposed just such a theory to explain β-decay. Thestrength of the Fermi theory is given by the quantity, G.

G =g2

M2W

≈ 10−5GeV −2 (65)

G is determined from the rates of beta-decays of atomic nuclei and from the decay rate ofthe muon

µ+ → e+νeνµ

The electroweak theory of Glashow, Salam and Weinberg proposed that

g =

√α

2 sin θw

(66)

for the charge changing weak interaction and

g =

√α

2 sin θw cos θw

(67)

for the neutral current weak interaction. Where θw is the Weinberg angle.

1.18 The Strong Force

First we find the magnitude of the strong interaction. Look at the reaction:

k− + p→ Σ0∗(1385) → Λ + π0 Γ = 36MeV ⇒ τ = 10−23s

Compare with E/M decay

Σ0(1192) → Λ + γ τ = 10−19s

⇒ αs

α≈√

10−19

10−23≈ 100

αs =g2

s

4π≈ 1 (68)

This gives the strength of the coupling of the strong charges, gs of the quarks. The mediatingboson is called a gluon.

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A gluon is massless and electrically neutral. The strong charges on the quark are calledquark colors.

This is just the name for an extra degree of freedom of quarks. Whereas in QED there arejust two kinds of electric charge +1 and -1, in Quantum Chromodynamics, (QCD) there aresix.

red blue green on quarksred blue greem on anti-quarks

These labels are in analogy with the 3 primary colours of light.

Also unlike the QED photon, gluons also carry color.

Because αs ≈ 1 multiple gluon exchange is highly probable and we can’t use perturbationtheory ie.

1 ≈ αs ≈ α2s ≈ α3

s = ...etc

At very energies αs is reduced enough to apply perturbation theory. ⇒ the existance ofjets of quarks at high energy.

The other different property of QCD, colored mediating bosons, has interesting properties.Each gluon interacts with every other gluon so:

This leads to color flux tubes and the color potential has the form:

Vs = −4

3

αs

r+ kr (69)

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