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High Energy Physics (3HEP)
aka “particle physics”
Dr. Paul D. Stevenson
2007
Outline of Course
1. Basic building blocks, particles and antiparticles, hole theory, feynman
diagrams
2. forces and particle exchange, Yukawa potential, units
3. leptons: their interactions and decays, lepton universality,
4. neutrino mass, neutrino oscillations
5. quarks, mesons and hadrons, baryon spectroscopy
6. baryon quantum numbers and selection rules
7. space-time symetries, color
8. QCD, jets
9. partons & deep inelastic scattering
10. lepton-quark symmetry, quark mixing
What we know about the universe
This course is about the following basic building blocks:
Electron, electron neutrino, up & down quarks
Muon, mu neutrino, charm & strange quarks
Tauon, tau neutrino, bottom & top quarks
And…
Photon, W & Z bosons, gluons
Higgs boson (maybe)
But not…
The other 96% of the universe
Results from WMAP
http://map.gsfc.nasa.gov/m_mm.html
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We don’t know much about Dark Energy, but will touch on
some Dark Matter candidates later on in the course
Antiparticles (see sec 1.2)
Assume that a particle in free space is described by a de Broglie
wavefunction:
(x,t)  N exp(i(p x  Et)/ )
Then using E=p2/2m, it is seen that the wavefunction obeys the
Schrödinger equation:

2

i
(x,t)  
 2 (x,t)
t
2m
But we now know (from relativity) that E2=p2c2+m2c4
So what is a suitable form for a relativistic wave equation?

Klein-Gordon Equation

2
 2 (x,t)
2 2 2
2 4


c

(x,t)

m
c (x,t)
2
t
But for any plane wave solution, there is an equivalent one with the

opposite sign of energy.
The negative energy solutions are a consequence of the quadratic
mass-energy relations and cannot be avoided
Dirac extended the above equation to apply to spin-1/2 objects
(e.g. electrons), and the results agreed impressively with
experiment (e.g by predicting the correct magnetic moment).
Dirac Hole Theory
The negative energy states remained, and Dirac picture
the vacuum as a “sea” of negative energy electron
states combining to give a total energy, momentum
and spin of zero.
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Vacuum state means negative
energy states filled with electrons
Positrons
When a “hole” is created in the negative energy sea, Dirac
postulated that this corresponds to a particle with positive
charge, but with mass equal to the electron, called the positron
or anti-electron.
γ
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• Note that removing an electron with energy E=-Ep<0,
momentum -p, and charge -e from the vacuum (which has E=0,
p=0, Q=0) leaves the state with a positive energy and
momentum and with a positive charge.
• This state cannot be distinguished in any measurement from the
situation in which an equivalent positive energy particle is added
to the system.
• Dirac postulated the existence of the positron based on this hole
theory in 1928
• Not everyone took this idea seriously:
– “Dirac has tried to identify holes with antielectrons … We do not
believe that this explanation can be seriously considered”
(Handbuch der Physik, 24, 246 (1933))
• The positron was discovered experimentally in 1933
Anderson, Physical Review 43, 491 (1933)
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Antiparticles in General
• Any particle with spin-1/2 obeys the Dirac equation, and the hole
theory applies. This means every spin-1/2 particle has an
antiparticle.
• Spin-0 particles do not obey the Dirac equation and do not feel
the Pauli Exclusion Princliple on which hole-theory depends,
and do not (necessarily) have antiparticles.
Need to extend the list of fundamental particles:
e , e  ,  e , e , u, u, d, d
 ,   ,   ,  , c, c , s, s
  ,   ,   ,  , t, t , b, b
 , Z 0 , W  , W  , g, H
Feynman Diagrams
Feynman introduced a way of drawing processes in elementary
particle physics which can be interpreted in an physical way, as
well as being used in calculations.
e.g, consider e- → e-+ γ
We draw it like this, with time going left to right:

ee-
Positrons
• In Feynman diagrams, an antiparticle is drawn with an arrow
pointing to the left, i.e. back in time, e.g.
e- + e+ → γ is drawn as
e-
There are 8 basic process
where electrons and positrons
interact with photons.
e+
The arrow direction convention reflects the conservation
of charge. What about energy and momentum?
Energy and Momentum Conservation
• Consider our first process (e- → e-+ γ) in the rest frame of the
electron. We write their four-momenta to describe the
conservation of both energy and momentum:
• e-(E0,0) → e-(Ek,-k)+ γ(ck,k)
Where we have already made sure momentum is conserved by
construction. In free space (e.g. not in a an electric field from
which can do work on the electron):
E0=mc2, Ek=(k2c2+m2c4)1/2 so, for finite k, ΔE≡Ek+kc-E0 satisfies
kc < ΔE < 2kc
And energy is not conserved.
Real and virtual processes
These 8 basic vertices all fail to conserve energy, and are called
virtual processes. They cannot happen in isolation, but must be
combined with other processes so that the violated energy is
recovered within a time
τΔE∼ħ
a nd the initial and final states in particular must have equal
energy and momentum. You can think of the initial and final
states as being able to be studied in the distant past and future
respectively
Example: elastic electron scattering
e
-
e
e-
In principle, both these “timeorderings” should be included, but in
practice, usually one example timeordering is shown and the others
are implied.
-
ee
e-
e-
-
e-
Higher-order Diagrams
e-
e
e-
e-
-
Each vertex in the 8 diagrams
contributes a probablility of
e2
1


40 c 137
1

Hence the more vertices, the smaller
the contribution to the scattering
amplitude
Particle Exchange
e-
e
e-
-
eInteractions take place by exchange of other particles. In this
case, electrons are interacting by exchanging photons.
Photons are massless, but in general, the exchanged particle
can have mass.
Particle Exchange
suppose two particles of mass MA and MB
interact by exchanging a particle X of mass
MX.
The reaction at the upper vertex, including the
four-momentum in the rest frame of particle

A is
A(MAc2,0) → A(EA,p) + X(EX,-p)

and the energy difference between initial and
final state is
ΔE = EX+EA-MAc2 →2pc as p → infinity
→ MXc2 as p → 0
This is always at least MXc2
A
A
X

B
B


Range of Force
The energy violation can persist only for a time
τ∼ħ/ΔE
during which time the force-carrying particle can travel a maximum
distance
R ∼ c τ = ħ/MXc
This maximum distance over which the particle can propagate
gives the range of the interaction.
A photon is massless, so electromagnetic interactions are infiniteranged. The weak force is mediated by W and Z bosons which
have MW = 80.3 GeV/c2 and MZ = 91.2 GeV/c2 so
RW = ħ/MWc ~ 2 x 10-3 fm
Point interaction approximation
In the case of the weak force, the range is so short - often smaller
than the de Broglie wavelength of the particles involved in an
interaction (e.g. a nucleon undergoing beta decay) so that the
interaction can be approximated to a zero-range point
interaction corresponding to the MX → infinity limit
A
A
A
A
X


B

B




B


B
Yukawa Potential
In the limit that MA becomes large, we can regard B as being
scattered by a static potential of which A is the source. In
general this potential will be spin-dependent, however we can
consider the simplest case of neglecting spin (so applying to the
exchange of spinless bosons) obeying the Klein-Gordon
equation

2
 2 (x,t)
2 4
2 2 2


c


(x,t)

M
X c  (x,t)
2
t
Which for static (in time) solutions becomes

  (x,t) 
2

M X 2c 2
2
 (x,t)
Yukawa Potential
In this equation, φ plays the role of a static potential. Note that if
MX=0 the equation is Laplace’s equation for the electrostatic
potential, in which case the potential is
e2 1
V (r)  e (r)  
40 r
In the case of finite MX, the solution turns out to be

g 2 er / R
V (r)  
4 r
where R = ħ/MXc as expected and a coupling strength g, analogous
to the electric charge has been included.
 potential, introduced by Hideki Yukawa in 1935
This is the Yukawa
Scattering Amplitude
To obtain quantitative information from the Yukawa picture, we
need to calculate scattering amplitudes. These are the
probabilities of scattering a particle from an initial momentum qi
to a final momentum qf by a potential V(x). In first-order
perturbation theory, this is given by
M(q) 
3
d
 x V(x) exp(iq x / )
where q=qf-qi is the momentum transfer
This gives…

g 2 2
M(q) 
| q |2  M X2 c 2
This is the Born approximation and corresponds to the Feynman diagrams with one particle being
exchanged
Zero-range Approximation
If the range R = ħ/MXc is indeed small compared with the de
Broglie wavelength of any particle involved and we make the
zero-range approximation, then we have |q|2<<MX2c2 and the
scattering amplitude reduces to a constant M(q)=-G where
G
1  g 
4 X
 

3
2 
( c)
c M X c  (M X c 2 ) 2
in units of inverse energy squared. This approximation is very
useful in the theory of beta-decay, in which the appropriate
coupling constant which has been measured is the Fermi
coupling
 constant
2
GF
5
-2

1.166
10
GeV
( c) 3
Units
In particle physics, a somewhat curious system of units is used
which are called “natural” units and set the units of velocity to be
the speed of light, c, and the standard unit of action to be ħ. In
other words ħ=c=1.
The units of energy used are electron volts (eV) and its multiples
(keV, GeV etc).
By choosing these “natural units,” all occurrences of ħ and c are
omitted from equations so that
E2=p2c2+m2c4
becomes
E2=p2+m2
and the Fermi coupling constant is now written
GF=1.166 x 10-5 GeV-2
Units
All quantities now have the dimension of some power of energy
since they can be expressed as some combination of ħ, c and
energy. For example, mass, length and time can be expressed
as
M=E/c2, L=ħc/E, T=ħ/E
Because of this, note that if the SI dimension of a quantity are
MpLqTr then in natural units they are En=Ep-q-r.
To convert expressions back to “normal” units, factors of ħ and c
are inserted by dimensional arguments and then
ħ = 6.528 x 10-22 MeV s
Example
The cross section for Thomson Scattering (photon scattering from
free electrons when the photon energy is much less than me) is
8  2

3 me2
To turn this into practical units, we write

8  2

3 m e2
a
cb
and demand that the cross section has units of length-squared.
We find a=2, b=-2 so

8  2 ( c) 2
29 2


6.65
10
m
2 2
3 (me c )
Leptons
The fundamental fermions (particles with spin-1/2) which do not
feel the strong force are called leptons. They are

e
e






electron: Charge -e, mass 511 keV, stable
electron neutrino: charge 0, mass ?§, stable#
muon: charge -e, mass 105 MeV, lifetime 2x10-6s
muon neutrino: charge 0, mass ?§, stable#
tau: charge -e, mass 1.78 GeV, lifetime 3x10-13s
tau neutrino: charge 0, mass ?#, stable#
§ neutrino masses non-zero but small. See later.
# neutrinos do not decay, but oscillate. see later, too.
Interactions
The charged leptons interact via the electromagnetic interaction
(like all charged particles) and the weak interaction (like all
particles)
The uncharged leptons (i.e. the neutrinos) interact only via the
weak interaction, mediated by W and Z bosons
In all interactions, it is observed that the following numbers are
conserved:
Le = N(e-) - N(e+) + N(ve) - N( ) e
Lµ = N(µ-) - N(µ+) + N(vµ)-N( ) 
L = N(-) - N(+) + N(v) - N( ) 



Lepton processes
Since neutrinos do not feel the electromagnetic interaction, in
electromagnetic processes, the Le conservation rule reduces to
the conservation of N(e-)-N(e+). This implies that in
electromagnetic processes, electrons and positrons can only be
created or annihilated in pairs.
In weak interactions, other possibilities exist. For example, the
following beta-decay process is allowed:
n → p + e- + e
because the total Le number on the left (0) is equal to the number
on the right (1-1=0)
Charge is also conserved, which is required in all physical

processes
Disallowed processes
Various processes are allowed by energy conservation, but not by
some of the empirical conservations laws, e.g.
   e   n
  e   e   e 
  e   e  
   p    n

Neutrino interactions
Neutrinos are very hard to detect, because they interact only via
the weak interaction. This means that the scattering amplitude
(determining the probability of reactions to occur) is small
because of the large mass of the W and Z particles. Its
existence was postulated in 1930 by Fermi to appear in ßdecay.
If ß-decay had a two-body final state
(Z,N) → (Z+1,N-1) + ethen the energy of the electron would be uniquely determined by
Ee  M  M(Z,N)
 are
M(Z
1,N 1)
However, experimentally,
electrons
observed
to have a range of
energies. If a neutrino is also present the energy of the electron
can lie in the range

me  E e  (M  m e )
Neutrino detection
Though postulated in 1930, neutrinos were not detected until 1956.
They can be detected in the following processes:
 e  n  e  p
e  p  e   n
These have a cross section of around 10-47m2 which means that a
neutrino would typically have to travel through many light years
of matter before interaction. If the neutrino flux is large enough,
 sufficient events can be seen in much smaller (!)
however,
detectors.
see article from G. L. Trigg, Landmark Experiments in 20th Century
Physics
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Heavier Leptons
electrons are the lightest charged particles, and as such can not
decay (since charge must be conserved).
Muons behave very much like electrons, except that their mass is
much larger (105.7 MeV/c2 compared to 0.511 MeV/c2).
In particular, they satisfy the Dirac equation for point-like spin-1/2
particles, and their magnetic moment is
µ=(e/mµ)S
indicating that there is no substructure, and the muon is an
elementary particle.
Lepton Decay
Muons decay with a lifetime of 2.2x10-6s via
  e   e  
   e      e
Taus decay to many different final states. This is because there mass is
sufficiently high that many different allowed combinations of particles
have a lower rest-mass energy than the tau.
The different decays are characterised by their branching ratio, which
gives the fraction of decays to a given final state compared to all
decays. For taus we have
   e     e
(17.8%)
        
(17.4%)
      hadrons
(~ 64%)
Lepton Universality
electrons are light and stable, and are stopped by a modest thickness of lead.
muons are about 200 times heavier and are very penetrating. taus are much
heavier still and has a lifetime many orders of magnitude below the muon.
Nevertheless, all experimental data is consistent with the assumption that the
interactions of the electron with its neutrino, the muon with its neutrino and the
tau with its neutrino are identical, provided the mass difference is taken into
account. This is know as Lepton Universality.
e.g. the decay rate for a weak process is predicted to be proportional to the Q-value
multiplied by the same GF2 independent of kind of lepton: for example,
5


(  e  e    )
Q
6

1.37
10




Q 
(


e




)
e

  
In excellent agreement with experiment



Neutrino Mass
The masses of particles is not predicted by the standard model.
Neutrinos certainly have a very small mass (e.g. as measured
by the electron energy distribution in beta decay, and had
usually been considered to be zero. If the neutrino masses are
not zero, neutrino mixing can occur.
This happens because the weak neutrino states (e) need not
be the same as the mass neutrino states (123) but linear
combinations of them. For simplicity, let’s consider the mixing of
two states in the following way:
 e  cos sin  1 
  
 

   sin  cos  2 
Oscillations
Consider an electron neutrino created at time t=0 with momentum
p as
 e ,p  1,p cos    2p sin 
After a time t it will be in a state
exp(iE1t / ) 1,p cos   exp(iE 2 t / )  2p sin 
corresponding to a probability of having changed weak eigenstate
of
P( e    )  sin 2 (2 )sin 2 ([E1  E 2 ]t /2)


Detection
When cosmic ray protons collide with atoms in the atmosphere,
they create many pions which decay to neutrinos via
      ,
      
One would expect then to have twice as many muon neutrinos as
electron neutrinos, but the observed ratio was about 1.3:1
QuickTime™ and a
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are needed to see this picture.

Neutrinos
do oscillate.
Quarks (section 2.2)
The six quark types, or flavours, know to exist are given below:
Name
Symbol
Mass (GeV) Q/e
Lifetime(s)
Down
d
≈0.3
-1/3
Up
u
≈0.3
2/3
Strange
s
≈0.5
-1.3
10-8-10-10
Charmed
c
≈1.5
2/3
10-12-10-13
Bottom
b
≈4.5
-1/3
10-12-10-13
Top
t
≈180
2/3
~10-25
Quarks
Unlike the leptons, quarks are never seen as free particles but only
in the following combinations:
baryons: qqq
antibaryons: q q q
mesons: qq
Collectively, these resulting particles are known as hadrons.

No other combination are seen, though it is thought that other

possibilities could exist (e.g. pentaquark states)
Quarks and Interactions
Quarks (and hence hadrons) feel the weak interaction (because all
known particles do), the electromagnetic interaction (because
they are charged) and also the strong interaction, mediated by
gluons.
Despite the fact that they are never seen in isolation, the evidence
for their existence is compelling because of:
• hadron spectroscopy
• deep-inelastic scattering
• Jets
Spectroscopy



Combining quarks in the allowed forms of baryons or mesons
correctly reproduces all known hadrons, e.g.
p = uud
n = udd
 = uds
+ =
ud
K+ =
us
B =
etc. bu
Additive properties like charge are seen to agree straightforwardly
with the quark model. A large part of the mass is associated
with the binding energy, and is not easy to calculate.
Young man, if I could remember the names of all these particles, I would have been a botanist -- Enrico
Fermi
Hadron quantum numbers
Various quantum numbers are associated with hadrons. Some are related
to symmetry operations, and some are intrinsic to the quarks. The
quark quantum numbers are the electric charge Q, and baryon number
B, which are conserved in all known interactions, and the strangeness
S, charm C, beauty B’ and truth T, which are conserved in strong and
electromagnetic, but not in weak, interactions
quark
Q/e
B
S
C
B’
T
d
-1/3
1/3
0
0
0
0
u
2/3
1/3
0
0
0
0
s
-1/3
1/3
-1
0
0
0
c
2/3
1/3
0
1
0
0
b
-1/3
1/3
0
0
-1
0
t
2/3
1/3
0
0
0
1
Isospin
For partly historical reasons, the internal quantum number associated with
the up and down quarks works a little differently.
Heisenberg thought that the similarity in mass between the neutron and
the proton meant that there was some underlying symmetry which was
only broken by the charge on the proton vs the lack of charge on the
neutron. He called this isospin symmetry, because it behaves like an
angular momentum, and thought of the nucleon as an isospin-1/2
system, with the proton being the +1/2 projection and the neutron the 1/2 projection on the third axis.
Similarly for other families of particles with similar masses and other
properties:
Isospin
Particle
B
Q
I
I3
(1116)
1
0
0
0
p(938)
1
1
1/2
1/2
n(940)
1
0
1/2
-1/2
K+(494)
0
1
1/2
1/2
K0(498)
0
0
1/2
-1/2
+(140)
0
1
1
1
0(135)
0
0
1
0
-(140)
0
-1
1
-1
quark isospin
u & d quarks are an isodoublet, both with I=1/2, but d has I3=Iz=1/2, u has I3=Iz=+1/2
the equivalent antiquarks are also an isodoublet with I=1/2, but with
opposite third-components.
Another quantum number is often used, which is derived from the
known ones, called hypercharge:
Y=B+S+C+B’+T
and then for a hadron, I3=Q-Y/2
Example: How new particles can be identified by the
quantum numbers and conservation laws.
When a K- is fired at a proton target, they can interact (with a cross
section appropriate to the strong interaction to give an - and a
new particle +, which decays with weak interaction lifetimes to
+ + + n, or + 0 + p.
Applying conservation of B,S,C,B’,T to the K-+p  - + + reaction,
and using the known values for K- (B=0, S=-1, C=B’=T=0), p (
B=1, S=0, C=B’=T=0) and - (B=0, S=C=B’=T=0) tells us that for
the +, B=1, S=-1, C=B’=T=0. From the relationship between
hypercharge and isospin, we also infer I3=1. It must therefore
be part of an isotriplet, and partners of similar mass but different
isospin should exist, as indeed they do.
We can infer the quark structure from the quantum numbers as uus
(n.b components of vectors are always additive)
Space-Time Symmetries
M&S Chapter 4
As well as the quantum numbers associated directly with intrinsic
quark properties, particles (including the leptons) have quantum
numbers associated with spatial properties. These are Spin, J,
Parity, P, and Charge Conjugation, C.
SPIN
We use the word spin to characterise the intrinsic angular
momentum of a particle whether or not it is composite.
If the particle is elementary, the total spin J is equal to the intrinsic
spin S, e.g. 1/2 for any of the fundamental fermions (quarks &
leptons)
Meson Angular Momentum
Mesons are made a quark and an antiquark which have intrinsic spin
of their own, and can have orbital angular momentum about their
centre of mass
If we first consider the case that there is no orbital angular
momentum (this is the case for the lowest masss mesons) then
we just have J=S and S is made by coupling together two single
spins of 1/21.
When coupling an angular momentum I1 to another I2 the allowed
values of the result I are
I = |I1-I2|, |I1-I2|+1,…,I1+I2-1, I1+I2
Coupling 1/2 with 1/2 gives only 0 or 1 as possibilities for S and
hence for J when L=0.
1
we are not including units of c or hbar. Angular momentum has units of hbar
Meson Angular Momentum
If the quark and antiquark in the meson have a relative angular
momentum (which is quantized in integer units of hbar) L=1, 2,
3, … then this L couples with the coupled intrinsic spin S
according to the normal rules of AM coupling, I.e. the total spin J
can lie anywhere (in integer steps) between |L-S| and L+S.
We label the mesons in spectroscopic notation:
2S+1L
J
where L is written as the usual identifying letter: S=0, P=1, D=2,
F=3, G=4 etc.. and S and J are given as numbers. So a pion,
which has L=S=0 has an angular momentum configuration 1S0 .
For a given quark-antiquark pair, there are many allowed
angular momentum configurations, which are identified as
different particles. For example there is a state made of u (like
a +) but with J=1, which is labelled as +. It can be thought of
as an excited state of +, butdsince it has different quantum
numbers, it is also thought of as a particle in its own right.

Baryon Spin
In the case of baryons, the three spins of 1/2 can couple to either 3/2 or
1/2 which then couples to the orbital angular momentum, which can be
any integer value.
q1
L3
q3
L=L12+L3
L12
q2
what are the allowed baryon states for L=0, L=1?
Parity
The parity transformation is xixi’=-xi, i.e. the position vector of
every particle is reflected in the origin. If we define a parity
operator P as
P(x,t) = Pa(-x,t)
where Pa is a phase factor. Since acting with the parity operator
twice must give us back the original state, we have
P2(x,t) = (x,t) =Pa2(x,t)
and so Pa=1
Intrinsic Parity
Consider an eigenfunction of momentum
 p (x,t)  exp[i(p  x  Et)]
then

Pˆ  p (x,t)  Pa p (x,t)  Pa p (x,t)
so when a particle is at rest (p=0), it is an exact eigenstate of the
parity operator with eigenvalue Pa. For this reason, Pa is called
 intrinsic parity of particle a, or more commonly just the parity
the
of particle a.
If a Hamiltonian is invariant under the parity transformation then
[H,P]=0 and in a reaction governed by the Hamiltonian, the initial
and final parities must be the same, and parity is conserved.
This is true in the strong and electromagnetic interactions, but
not the weak interaction.
Angular Momentum & Parity
orbital angular momentum wavefunction are written in spatial
coordinates as spherical harmonics in the angular part. These
feature in the usual separable solutions to central-field potential
in quantum mechanics:
(r,,) = R(r)Ylm(,)
The first few spherical harmonics are
Y00 = 1/sqrt(4)
Y10 = sqrt(3/4) cos
Y11 = -sqrt(3/8)sin exp(i)
Spherical Harmonics
Using x=rsincos, y=rsinsin, z=rcos
the parity transformation in polar coordinates become
r  r’ = r
  ’ = -
  ’ = +
from which it can be shown that
Ylm(,)  Ylm(-, +) =(-1)l Ylm(,)
and (r,,) is an eigenstate of parity with eigenvalue Pa(-1)l (where
Pa is the intrinsic parity ignoring angular momentum)
Fermion Parity
From Dirac’s equation one can obtain the following for any fundamental
fermion and its antiparticle
Pf Pf  1
Parity is not conserved in the weak interaction. In the strong
interaction quarks and antiquarks are always produced in pairs.
Likewise the leptons and their anitleptons. One can indeed
confirm the above relation experimentally but not determine the

parity of the individual fermions. By convention the particles
(quarks, electron etc.) are assigned P=+1 and the antiparticles
(antiquarks, positron etc.) are then P=-1.
The parity of the photon can be derived from electromagnetic
theory. It is P=-1
Charge Conjugation
Charge conjugation is an operation which replaces all particles by
their antiparticles in the same state (so that positions, momenta
etc are unchanged).
Like Parity, Charge Conjugation is a good symmetry of the strong
and electromagnetic interaction, but not the weak interaction.
The Charge Conjugation quantum number of a particle is
conserved in these interactions.
It can take on the values ±1. Its conservation can be used to
understand why some reaction will work but not others, and help
identify particles.
0 decay
for a quark antiquark pair with given values of S, J, and L, the quark
model predicts
Cˆ ff;J,L,S  (1)L S ff;J,L,S
It is observed that the dominant decay mode of the 0 is
0 . Then
Cˆ  0  C  0

0
Cˆ   C C   
So the final state has C=1, so too must the initial state, in
agreement with the quark model prediction (since a 0, in the
lightest multiplet
of mesons, has L=S=0).

From electromagnetism, it can be deduced that C=-1. One does
not see
0 
because of C-conservation in the electromagnetic interaction.
Indeed, searches for this decay mode are made to look for
potential C-parity breaking.
Colo(u)r
The quark model is successful, but why are there no states of the
form qq? Why is it always 3 quarks, 3 antiquarks or a quarkantiquark pair?
Also: according to quantum mechanics, the wavefunctions for
composite particles made of identical spin-1/2 quarks must be
antisymmetric under interchange of any two of the quarks. But
the observed properties of some particles clearly contradict this.
e.g. the Ω- is observed to have strangeness S=-3 and spin J=3/2,
L=0, indicating that it is made up of three identical strange
quarks with their spins all pointing in the same direction, and in
the same physical state. This wavefunciton is symmetric, and
violates the Pauli principle.
Colo(u)r
Another degree of freedom is at work. It is a property that quarks
have but leptons do not, called color. It is in some sense
analogous to electric charge, in that it has an associated color
charge which is what the strong interaction acts on, like the
electromagnetic interaction acts on charged things.
There are three kinds of color state for quarks, called red, green,
and blue (which do not have anything to do with real colours).
Antiquarks can be antired, antigreen or antiblue.
This solves the antisymmetry problem. In the Ω- the three quarks,
labelled 1, 2 and 3 exist in a linear combination of different
colour combinations which is antisymmetric with respect to the
interchange of any two labels.
C 
1
[r1g2b3  g1r2b3  b1r2 g3  b1g2 r3  g1b2 r3  r1b2 g3 ]
6
Color confinement
It was hypothesised that hadrons could only exist in colorless
states (either color-anticolor or rgb), which “explained” why only
(anti-)baryons and mesons are seen.
Quarks are colored and have non-zero color charge and are hence
confined to hadrons.
There is direct experimental evidence for color charge, coming from
electron-positron annihilation (M&S 7.2)
Consider reactions of the type e+ + e-  hadrons at rather high
energies (15-40 GeV) in a colliding beam experiment
Jets
The primary mechanism of the reaction is that the electron and
positron annihilate to a photon which then creates a (colourless)
quark-antiquark pair.
Because the quarks are generated with so much momentum, and
they are created moving away from each other, they soon
fragment into two jets of hadrons. They cannot keep travelling
as isolated quarks because of color confinement.
The total summed momentum for each jet lets one reconstruct the
trajectory of each quark. This can be compared with the branch
of
e+ + e-  + + -
Jet Distributions
From QED the differential cross-section for this process is wellknown
d
 2




2
(e  e     ) 
(1
cos
)
2
dcos
8E
to generate a quark-antiquark pair in this way (via the
electromagnetic interaction), the cross section is just reduced
slightly by the lower charges on the quarks compared to muons,
but multiplied by a factor NC which is the number of different
colours available for a quark of a given flavour. So the
differential cross section for generation of a quark anti-quark of
flavour a will be
d
ea2 2


2
(e  e  qa  qa ) 
(1
cos
)
2
dcos
8E
colour counting
If you perform this experiment, and measure these cross sections
as a function of energy (and integrate over angle) you should
see the ratio
 (e e  hadrons)
R
 (e e    - )
Increase as 
more quark channels open up. For example, when u,
d, s, c, b quarks are available, the ratio should be 11NC/9
The Strong Interaction & QCD
M&S Chapter 7
QCD stands for Quantum ChromoDynamics and is the theory
which governs the strong interaction.
The strong interaction acts by the exchange of gluons, which
interact with objects with color charge, so in particular with
quarks. Gluons are spin-1 bosons, which also have color
charge, so gluons interact with themselves. This gives rise to
curiosities such as glueball states which are made entirely of
gluons. no such state has been conclusively observed, though
there are some candidates.
In some circumstances the rather complicated QCD interactions
can be approximated by a static potential between quarks (esp.
if the quarks are heavy)
quark potential
At short distances, the interaction between a heavy quark and its
antiquark is dominated by single gluon exchange. This gives
rise to a coulomb-like potential
V(r)=-(4/3) (S/r)
for r<0.1 fm
because of color confinement, the strength of the interaction
actually increases quite rapidly as you increase the distance
between the quark and antiquark. This means a lot of energy is
stored in the field which can turn into quark-antiquark pairs to
stop the colored quarks from getting too far apart. Emprically
the potential looks like:
V(r)r
asymptotic freedom
The strong coupling constant S is not actually constant, but varies
as a function of momentum transfer
At high values of momentum
transfer (short distances) the
coupling constant gets weaker.
This means that when quarks
are close to each other, the
become more “free” from the
strong force, but feel it very
strongly when they try to
separate.
Strong interaction summary
The strong interaction acts between object with color. These are
the quarks and gluons. The force is mediated by the exchange
of gluons.
Conserved quantities in the strong interaction: Baryon number (B),
quark flavour numbers: (S,C,T,B), parity (P), charge conjugation
(C)
Also conserved (because all interactions conserve them): charge,
angular momentum, energy.
Colored objects are never observed in isolation, but only combined
into colorless particles
Weak Interaction with Hadrons
• There are two basic weak interaction vertices for interactions
between W± and leptons:
l
l
l
l



W
W
• Which give rise to 8 basic reactions when
pulled aroundin all possible time-orderings
Lepton-Quark Symmetry
The basis of weak interactions with hadrons depends on how the
W± bosons are absorbed or emitted by quarks.
For simplicity, we restrict discussion to the first two families of
quarks:
and
c 
u 
 
d 
 
s 
The two main ideas in the weak interaction of quarks are leptonquark
 symmetry
 and quark mixing. They describe to a good
accuracy the observed amplitudes of weak interaction events
with W± bosons, and can deduce the interaction of quarks with
the Z0 (see section 9.1.1 in Martin & Shaw for that).
Lepton-Quark Symmetry
Lepton-Quark symmetry asserts that the first two generations of
quarks, as on the previous slide, and the first two generations of
  
 e  interactions.
leptons:
&
have identical 
weak
  
  
e
 
 
i.e. one obtains the basic vertices for weak interaction with quarks
by the replacements:


e  u
e  d
  c
  s
leaving the coupling constants unchanged
Neutron decay
gw
Wd
d
u
e-
e
gud
u
d
u
and gud = gw = gcs
This idea works well in describing events such as neutron decay,
      
and pion decay:
However it does not allow observed decays such as

(why?)

K     
The Cabibbo Hypothesis
To explain events such as K- decay, the idea of quark mixing was
introduced. According to this idea, the d and s quarks take part
in the weak interaction via the linear combinations
d’ = d cosC + s sinC
And
s’ = -d sinC + s cosC
where C is the Cabibbo mixing angle.
This means that lepton-quark symmetry is assumed to apply to the
doublets
and
u 
 
d'


c 
 
s'
Cabibbo angle
Now the previously allowed ‘udW’ vertex is suppressed by a factor
cosC while the previously forbidden usW vertex is now allowed
with a coupling gWsinC and the couplings are now:
gud = gcs = gwcosC
gus = -gcd = gwsinC
This set of couplings successfully accounts for the interaction of
quarks with the W± - the so-called charged-current interactions
(since the W± is the only charged force-carrying particle).
The success is demonstrated partly by the fact that the only free
parameter - the Cabibbo angle - in this description can be
measured in different ways, and the values agree with each
other
Measuring the Cabibbo Angle
From the relative decay rates of
determine the Cabibbo angle:
K   to
 
  one
   can

2
(K      ) gus
2
 2  tan



  C
(     ) gud
Of course, one must also take into account the difference in the d
and s quark
 masses since this will have an effect on the rates.
When this is done, the value is
gus
 tan  C  0.226  0.002
gud
giving C=12.7±0.1 degrees

Measuring the Cabibbo Angle
Comparing neutron and muon decay depends on the ratio
g2ud/g2w = cosC, which gives a value in agreement with that from K- / pidecay.
Values of the Cabibbo angle should agree when obtained from the gcs
coupling and the most accurate value for C coming from gcs is C =
12±1 degrees, which is compatible with the value from measuring gud.
Because of the value of the mixing angle, the couplings which were not
allowed in the simple unmixed scheme are still heavily suppressed
compared to the the allowed couplings. The decay rates for cabibbosuppressed decays, which depend on the square of the couplings are
suppressed by tan2 C ≈ 1/20 compared to cabibbo-allowed decays.
An example of this is that decays of particles with charmed quarks
almost always have an s quark in the final state.
Third Generation
The idea of quark mixing can be extended to the third generation
with a mixing matrix
d' Vud
  
s'  Vcd
  
b' Vtd
Vus Vub d 
 
Vcs Vcb s 
 
Vts Vtb b 
and apply lepton-quark
symmetry to the doublets

u 
, 
d'
c 
& 

s'
t 
 
b'
To a reasonable
approximation
Vtb=1 and the other elements in the



third row and column are approximately zero