Transcript Document 7192543

Particle Physics -I
• Nucleon-nucleon interaction; meson
exchange; Exchange giving rise to the four
forces of nature; properties of pion
• Pion-nucleon interaction; Collision
kinematics and conservation laws;Phase
shift Analysis; More conservation laws; N*s
and s; Photon-nucleon interaction
1
Book: K R Krane, Introductory
Nuclear Physics, Wiley
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Nucleon-Nucleon force -Chapter 4
Intro/ the deuteron pp 80-81
Spin and Parity pp 83-84
Magnetic Dipole moment pp 84-85
Electric Quadrupole moment p 85
Properties of force pp 100-102
Exchange force pp 108-112
Isospin pp 388-389
Properties of pions pp 656-665
Pion-proton interaction pp 671-675
Scattering theory - partial wave analysis pp 408-411
Conservation of Baryon number p715
2
NUCLEONS
Nucleon is the generic name given to the
proton(p+) and the neutron(n0) - particles
with essentially the same mass but different
charges. We view the proton and the
neutron as different CHARGE STATES of
the nucleon.
3
Properties of Nucleons
Property Unit
Proton
Neutron
Charge
|e|
+|e|
0
Mass
MeV/c2 938.3
939.6
Spin
h/2
½
½

N
+2.793
-1.913
Radius fm(10-15m)
1
1
Stability
Stable >1032 y  ~ 11.1 min
Decay Mode
----- n p + e- +e
4
Di- Nucleons
Only ONE stable - the DEUTERON (D)
Charge
+|e|
Mass(MeV/c2) 1875.7 = 938.3 + 939.6 - 2.2
Binding Energy
2.2 MeV
Spin (h/2)
1=½ +½
( N)
0.857 ~ +2.793 -1.913
Radius fm(10-15m)
2 ~ 1 +1
5
Deuteron Cross- section
 ~  rD2 ~ 12.5 fm2 = 125 mb
Deformation from circle ~ 3mb
P
N
6
What is the deuteron state?
• From ~ p + n we infer that the proton
and neutron are in an orbital angular
momentum L=0 state most of the time
• This would imply a spherical nucleus. From
the fact there is a deformation this must
mean for a small fraction of the time the
proton and neutron are in a different orbital
angular momentum state
7
Unstable Di - nucleons
D* - Unbound excited state of Deuteron.
Eex ~ + 70 KeV. Spin=0.  ~ 10-22 sec
Di-proton - Unbound state of two protons.
Spin=0.  ~ 10-22 sec
Di-neutron - Unbound state of two neutrons.
Spin=0.  ~ 10-22 sec
8
Evidence for existence of di-proton (2-p)
Nn
En(MeV)
p+Dp+n+p
p + D  n + (2-p)
9
Possible combinations of two S=½ Nucleons in L=0 state
10
States of two nucleons with L=0
Only the deuteron - 1 CHARGE STATE exists with S=1, Sz = +1,0, -1
i.e. 1 CHARGE STATE, 3 possible spin
orientations
D*, 2-p and 2-n - 3 CHARGE STATES -exist
with S=0, Sz =0
i.e. 3 CHARGE STATES, 1 spin orientation
There is a certain symmetry here!
11
2-p
D*
2-n
S=0
Sz = 0
D
S=1
Sz = +1,0,-1
12
Introducing ISOSPIN
Formally, Isospin is a vector in a ‘Charge Space’ orthogonal
to normal space, whose orientation in ‘charge space’ defines
the charge on the particle.
As a vector it obeys the same rules as the Spin vector hence its name.Other than that it is totally unrelated to spin.
For the nucleon, the Isospin vector t = ½, tz = ½.
The relation between charge and tz is:
Q/e = tz + ½
For two nucleons Isospin T = t1 + t2, Tz = t1z + t2z.
There are two possibilities:
T=1, Tz = +1, 0, -1
T=0, Tz = 0.
and
Q/e = Tz + 1
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2-p
D*
2-n
S=0
T=1
Sz = 0
Tz = +1
Tz = 0
Tz = -1
D
Sz = +1,0,-1
S=1
T=0, Tz = 0
14
COMPONENTS OF NUCLEON-NUCLEON FORCE
•Spin dependent - D bound, D* unbound , only difference
orientation of spins
•Charge or Isospin Independent - 2-p, D*, 2-n have same
mass
•Range of force 1- 2 fm
Evidence: RD = 2fm
RA = 1.2A1/3 V= 4/3RA3 ~ A  V(Nucleon)
•Attractive - nuclei exist
Repulsive core
Evidence: RD= 2fm R(3He) = 2fm R(4He) = 1.7 fm
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D = 2  ( 2 Nucleons)
(3He) = 1.33  (3 Nucleons)
(4He) = 0.7  (4 nucleons)
D
3He
4He
Binding Energy per nucleon pair B/½A(A-1)
2.23
2.53
4.67
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The -particle (4He nucleus) is the smallest and most strongly
bound nucleus in the periodic table. The nucleons within it
clearly overlap. Why doesn’t it become even smaller?
The N-N interaction appears to have a Repulsive Core of
range ~ 0.7 fm.
The magnitude of Binding Energies (MeV) means that the
force must be a STRONG interaction (~ 100 times stronger
than the EM interaction).
Tensor Component
To explain the non-spherical deuteron the N-N force must
have a component which is non-central ie depends on the
orientation of spins relative to spatial position.
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Deuteron
r
r
18
Force between two dipole magnets
N
N
d1
d2
r
S
S
N
V(r ) = -1/r3{3(d1.r)(d2.r) - d1.d2}
d1
r
S
where r is the unit vector
N
d2
S
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Deuteron
r
r
The EM potential V(r ) = -1/r3{3(p.r)( .r) - p.  } leads to a
n
n
deformation which is of order b.
A strong interaction potential is needed:
V(r ) = VT(r) {3(sp.r)(s .r) - sp.s }
2sps VT(r)
to give a deformation of order mb.
-sps VT(r)
n
n
n
n
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Meson Exchange
What can give rise to the tensor potential
part of the N-N interaction?
The answer is thought to be the virtual
exchange of mesons between the nucleons.
What do we mean by this?
21
The 4 exchange forces
Force
Range Exchange particle
Strength
Gravitational

Graviton,g, 0
10-39
Electromagnetic

Photon, , 0
10-2
< 2fm Pion, , 140 MeV/c2
1
Strong
Weak
~ 0 W-boson,W,81 GeV/c2 10-5
22
The Electromagnetic Interaction
In the interaction between two charged particles (eg an electron
and a proton) photons are exchanged between the two particles.
However we do not observe any photons being exchanged. As
long as the exchange is described by the Uncertainty Principle
it can still occur even if it is unobservable.
Et ~ 
R~ c t =  c/E
If E = 0 (photon) the range R = .
Also as E2 =p2c2 + m02 c4 and m0 = 0 then the change in
momentum of the emitting particle will be p= E/c and F =
p/ t =  c/R2
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The relativistic relation between total energy, momentum and mass
helps us to visualise what may be happening in this virtual ( ie real but
non - observable) exchange of photons. We see that:
E =  (p2c2 + m02c4)
The negative solution implies that we can view the vacuum as a sea of
negative energies:
p
e

E=0
The photon interacts with the vacuum locally. Energy is
then transmitted out in all directions. When another charged
particle is encountered this is then transferred to the particle24
by a photon.
The Strong Interaction
Meson Exchange


E=0
Et ~ 
R~ c t =  c/E
For the strong interaction the long range part of the interaction is
produced by the exchange of the lightest meson the pion, .
M ~ 140 MeV/c2 ie E  140 MeV
R = 6.6 10-22 MeV sec  3 1023 fm sec-1/ 140 MeV
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ie R = 1.4 fm
Properties of Pions
Pions exist in three charge states + , - and 0
Therefore the isospin of the Pion T = 1 with orientations
in charge space Tz = +1, -1 and 0 respectively.
Q/e = Tz
M+ = M- = 139.57 MeV/c2 M0 = 134.97 MeV/c2.
The spin of each pion is S = 0.
Another important property of the Pion is its
PARITY
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PARITY
This is defined for wave-functions in a symmetric potential.
E.g. The N-N potential is symmetric about the mid-point of
the two nucleons: V(r) = V(-r). Then for the probability density of
the two nucleons:(r)2 = (-r)2
and:
(r) = P(-r) where P = 1
A wave-function has parity P = +1 if it is even under reflection ( i.e. a
swap) of both the particles co-ordinates and parity P = -1 if it is odd
under reflection.
In a symmetric potential wave-functions with Angular Momentum
L=0,2,4,… are Even Parity, those with L=1,3,5,… are Odd Parity.
All Strong and EM potentials are symmetric and each energy
state has a definite parity - either Even or Odd. Particles of finite
size with internal structure held together by a symmetric interaction
to be discussed later - e.g. Nucleons, Pions - have INTRINSIC
27
PARITIES. Parity is conserved in Strong or EM interactions between
Intrinsic Parity of the Consider the reaction AT REST between a - and a deuteron:
- + D  n + n
Spin (S)
0
1
Angular Momentumtum (L)
LD= 0
½ ½
Lnn?
Now Total angular Momentum J is conserved in all interactions.
Therefore for
Total Angular Momentum(J)
1
=
1
What is the value of Lnn?
As the neutrons are like particles with S= ½ they are
subject to Pauli Principle. i.e. the overall wave-function
must be anti-symmetric
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The overall wave-function of two neutrons consists of two
parts:
nn = (L) (S)
• a space wave-function produced by solution of
Schrodinger’s equation in a potential Vnn (r)
i.e. containing a Legendre polynomial dependent on
angular momentum state L.
Wavefunctions with L= 0,2, .. Are Symmetric Even Parity. Those with L=1,3,.. Are Anti-symmetric Odd Parity.
• a spin wave-function dependent on the spin of the two
neutrons, S. This too can be symmetric or anti-symmetric.
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The Spin Wave-function, (S)
Two neutrons can exist with their spins parallel:
This is a Symmetric state S = 1, Sz= +1,0 0r -1
Or anti-parallel:
This is an anti-symmetric state S=0, Sz =0
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For two neutrons J= 1 cannot occur if either:
•L=0 S=1 or L=2 S=1- Symmetric Space Wavefunction,
Symmetric spin Wavefunction ie Overall Symmetry
•L=1, S=0 - Anti-Symmetric Space Wavefunction, AntiSymmetric spin Wavefunction. ie Overall Symmetry
The ONLY possibility is L=1 S=1 - Anti-Symmetric Space
Wavefunction, Symmetric spin Wavefunction. ie Overall AntiSymmetry.
J=0
J can then have values J=2 L=1
S=1
L=1
S=1
or importantly J=1
L=1
S=1
J=1
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So now, in the reaction AT REST between a - and a deuteron:
- + D  n + n
Spin (S)
0
Angular Momentum (L)
LD = 0
Total Angular Momentum(J)
1
1
½ ½
Lnn?
=
1
We see that the value of Lnn has to be Lnn =1.
Thus in the reaction, an initial even parity WF (LD = 0 ) goes
to an odd parity Lnn=1 WF. But parity is conserved in the Strong
Interaction.We must thus consider Intrinsic Parities.
Then: Pinitial = PPnPpP(LD = 0 ) = Pfinal= PnPn P (Lnn
=1)
The intrinsic parities of the nucleons cancel each other. So:
PP(LD = 0 ) = P (Lnn =1) i.e. -1  +1 = -1
The INTRINSIC PARITY of the PION is NEGATIVE.
Incidentally the INTRINSIC PARITY of the NUCLEON is
32
POSITIVE - by convention.
Pion Exchange between
Nucleons
The negative intrinsic parity of the pion has important consequences
when it is exchanged between nucleons.
It gives rise to the TENSOR force.
When the pion is emitted by a nucleon, as the nucleon continues to have
positive parity, the pion must be emitted with L =1 - to cancel out its
intrinsic negative parity. In order for this to happen, to conserve total
angular momentum either:
(a) the emitting and receiving nucleons must reverse their initial spin
directions
(b) the emitting and receiving nucleons must recoil with L=1.
Depending on their initial spatial orientation (a) or (b) happens. 33
Pion
transfer
occurs by
nucleons
changing
spin
direction
Lnp=0
Lnp=0
0
L=1
Lnp=2
Lnp=0
Pion transfer
by angular
momentum
change
0
L=1
L=1
L=1
34
So in the loosely bound deuteron ONE PION exchange is the
dominant process leading to the tensor force.
In 3He, with nucleons closer together, TWO PION exchange
is more likely. Here no nett angular momentum need be
transferred between two nucleons - i.e. a central force
0
0
This has a range ~ 0.7 fm.
In 4He, with the nucleons overlapping, even heavier
mesons can be exchanged. The , and  mesons have
masses between 600 and 900 MeV/c2, negative parity
and S=1. When these are transferred (with L=1 to
conserve parity) nucleons are forced to L=2 state -i.e .
Must move apart, are effectively repelled.
35
Pion-Proton interaction
We can create beams of pions by bombarding
a heavy nucleus with a high energy proton
beam from an accelerator.
+
0
Proton beam
T= 3000 MeV
n
Tungsten (W) target
Beams of charged
pions are formed
using electric and
magnetic fields.
36
Liquid hydrogen target
+ beam
Area AH,
length
tH,density H
Pion detectors
Number of scattered pions = N(IN) –N(OUT) =
N(IN)* (NAV*H*tH *AH )*(T /AH)
By analogy with throwing point-like darts at a board T would just be the crosssectional area of a proton, but this analogy needs to be extended as
•there is an interaction between the pion and the nucleon, predominantly the
STRONG force with a certain strength and range, but there is also a small
contribution from the Coulomb force.
So we extend the definition of T to:
THE TOTAL CROSS-SECTION FOR THE STRONG INTERACTION OF A PION
WITH A PROTON
T is measured in barns (10-24 cm2) and, when divided by AH, is the probability 37
for a
pion to interact with a proton. It is energy dependent.
Pion-proton total cross-sections
195 600 900 1380
T  (MeV)
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Peaks in the p cross-section are interpreted as resonances in the p
system:
 +p  ( p)*   +p
•The lowest resonant state occurs in both the + p and the -p crosssections at a pion incident laboratory energy of T = 195 MeV.
•The width of this resonance is ~ 100 MeV. Using the Uncertainty
Principle  E t ~ we find that the lifetime of the resonance is
~ 10-23 sec.
•What is its mass?
This can be calculated by applying the laws of Conservation of
Energy and Momentum to the formation of the resonance.
T-, p-,m-
p
mp
(-p)*
T*,P*,m*,v*
39
C.of E.
C.of M.
T- + m-c2 + mpc2 = m*c2 + T*
p- = p* and hence p-2c2 = p*2c2
(1)
(2)
Relativistic kinematic relationships
Total Energy E = m0c2/ (1-v2/c2)
Momentum p = {m0/ (1-v2/c2)}v
Eliminating v, E2 = p2c2 + m02c4.
Also E= T + m0c2 and hence E2= T2 + 2m0c2T + m02 c4
Thus
p2c2 = T2 + 2m0c2T
In (2) we then have T-2 + 2m-c2T- = T*2 + 2m*c2T*
and substituting for T* from (1):
m*c2 =  {(m- + mp)2c4 + 2T-mpc2}
Substituting pion, proton masses and T- = 195 MeV we find:
m* = 1235 MeV/c2
40
The Total centre-of-mass Energy is E =m*c2 = 1236 MeV.
To see this impose a back velocity on the laboratory such that the
resonant state is brought to rest ( i.e. - v*). Then the pion and the
proton will have equal and opposite momenta.
pCM
pCM
m*c2
To find the spin state of the resonance we need scattering theory.
41
Scattering Theory -Spin-less particles,
Elastic Scattering,CM reference frame
r
Incident pion wave
Proton

Unscattered pion wave
z
Scattered pion wave scatt
inc = eikz
Solution of: -2/2m(2 inc/z2) = E inc.
As E=pCM2/2m and pCM = k this reduces to
2  /z2 = -k2 .
42
scatt is a solution of: -2/2m(2scatt) = (E -V(r))scatt
scatt= f() eikr/r
•at large r where E>> V
•Solution independent of  as there is cylindrical
symmetry about the incident beam direction.
The total wavefunction after the p interaction
consists of an unscattered part (eikz) as well as scatt.
final = eikz + scatt
43
How are inc, scatt related to observables?
We know  * =  2 is the probability of a
particle’s existence at a point in space. If there are:
• n pions/unit area in the incident beam and
•the CM beam velocity is v
then the number of pions incident per second
per unit area is: nveikze-ikz = nv.
The number of pions elastically scattered at
angle  per unit area per second is :
nvf()f*()eikre-ikr/r2
44
Experimentally we measure the number of pions
scattered into unit solid angle (unit area /r2) at angle
 per second. This is: nv f() f*().
Then the differential cross-section for elastic
scattering del/d() =
number of pions scattered into unit solid angle at
angle  per second / number of pions incident per
unit area per second
i.e. del/d() = f() f*() .
el is the cross-section for elastic scattering.
el/unit area is the probability for scattering into 4
45
solid angle.
el = 4 d el/d () d  = 4 f()f*() d 
Now, we cannot specify the x- and y- positions of
the incident pions, momentum p, relative to a target
proton.
p angular
d
1013 fm
2 fm
momentum state
lp
=pd
46
The pions can interact, in principle, with the
protons in ALL possible angular momentum states
lp. This allows us to rectify the asymmetry in the
scattering - initial plane wave, final spherical
scattered and plane wave - by expressing the plane
waves in terms of ingoing and outgoing spherical
waves representing a pion-proton state of angular
momentum number l = lp.
inc = eikz = eikrcos =
1/kr 0 (2l +1) Pl(cos){eikr - e-i(kr-l)}/2i
where Pl(cos) is a
Legendre polynomial.
47
outgoing wave ingoing wave
When the p interaction occurs only outgoing
waves will be affected. Assuming only elastic
scattering occurs ( i.e. no inelastic scattering) then
these will be advanced (if p interaction attractive)
or retarded (if p interaction repulsive) compared to
the situation if there were no p interaction. i.e.
there will be a phase shift in the outgoing waves. By
convention this is 2(l). Thus the wavefunction
describing the total situation after scattering:
final =1/kr0(2l +1) Pl(cos){ei(kr+ 2(l)) - e-i(kr-l)}/2i
= eikz +f() eikr/r
(3)48
As: eikz = 1/kr 0 (2l +1) Pl(cos){eikr - e-i(kr-l)}/2i
we then find that, by rearrangement of (3):
f() = 1/2ik 0 (e2i(l) -1)(2l +1) Pl(cos)
=1/2ik 0 ei(l) (ei(l) - e-i(l) )(2l +1) Pl(cos)
= 1/2ik 0 ei(l) 2isin (l) (2l +1) Pl(cos)
= 1/k 0 ei(l) sin (l) (2l +1) Pl(cos)
Now el = 4 d el/d () d  = 4 f()f*() d .
As 4 Pl(cos) Pl’(cos) d  = 4/(2l+1) when l=l’’
= 0 otherwise
49
el = 4/k2 0 sin2(l) (2l +1)
Applying scattering theory to the
1235 Mev/c2 resonance
• Scattering theory applies to spinless
particles only and predicts:
el = 4/k2 0 sin2(l) (2l +1)
• We measure T = el + R
• When does T = el ???
• i.e.When is R = 0 ???
50
When a pion interacts with a proton many events other than elastic
scattering may occur :
- p  - p
Elastic
 0 n
Low Energy
High Energy
T- <300MeV
T- >300MeV
Low Energy
High Energy
 - 0 p
 - + - p
+ p  + p
Elastic
T+ <300MeV T+ >300MeV
 + 0 p
 + + - p
51
When pions interact with nucleons:
as many pions are produced as there is energy available. The number of
nucleons remains constant.
Analogy: Turning on an electric light.
As many photons are produced as there is energy available. The number
of electrons remains constant.
Conservation Laws
Baryon Number
B= +1 Nucleons
B=0 Pions
B is conserved in the All Interactions (Strong, EM, Weak, Grav.)
Epton Number
Le = +1 Electrons
Le = 0 Photons
Le is conserved in All Interactions
We will generalise these laws to include other particles later.
52
We can only apply our calculation of
el = 4/k2 0 sin2(l) (2l +1)
to + p scattering at low energy where T =
el .
How many l-values contribute?
lmax= pCM  d+ p
where d+ p is the range of the + p interaction (~1fm) and pCM
is the momentum of both the pion and the proton in the CM reference
frame. The Centre-of -Momentum ( Centre-of-Mass, CM) reference
frame is formed by imposing a back velocity on the laboratory such
that the CM momenta of the pion and the proton are equal and
opposite.
To work out lmax we need to calculate first pCM.
Pcm
(+ p)*
+
1 fm
Pcm
53
p
We know m* (T+ = 195 MeV) = 1236 MeV/c2.
E+ + Ep = m*c2
Then
i.e.
 (pCM2c2 + m+2c4) + (pCM2c2 + mp2c4) = m*c2
Rearranging:  (pCM2c2 + m+2c4) = m*c2 - (pCM2c2 + mp2c4)
Squaring:
pCM2c2 + m+2c4 = m*2c4 + pCM2c2 + mp2c4 - 2m*c2 (pCM2c2
+ mp2c4)
Rearranging: 2m*c2 (pCM2c2 + mp2c4) = m*2c4 + mp2c4 - m+2c4
Squaring again:
4m*2c4 (pCM2c2 + mp2c4) = (m*2c4 + mp2c4 - m+2c4)2
And: pCM2c2 =(m*2c4 + mp2c4 - m+2c4)2 - mp2c4
4m*2c4
Substituting the mass values we find: pCM = 240 MeV/c 54
Now
lmax= pCM  d+ p
Thus lmax = 240 MeV/c  1fm/6.6 10-22 MeV sec
=240 MeV1 fm/ (3 1023 fm sec-16.6 10-22MeV sec)
~ 1.2
Therefore lmax = 1
And el = 401 sin2(l) (2l +1).
k2
55
How does sin2(l) vary with pion energy T ?
sin2(0)
At T=0, k=0 and for el to be finite this means that
sin2(0)  0 and hence (0)  0.
As T increases l= pCM  d+ p = 0
then d+ p = 0 at all energies.
I.e. l=0 waves ‘see’ the same amount of the + p
potential at all energies in ahead-on collision.
+
p
56
Therefore the phase-shift (0), which is caused by the potential
seen, is a constant at all energies except T=0.
90
(0)
T
This is non-resonant behaviour. I.e. the l=0
phase shift cannot cause the peak in the crosssection - sin2(0) is a constant at all T.
57
sin2(1)
Again, at T=0, k=0 and for el to be finite this means
that sin2(1)  0 and hence (1)  0.
As T increases l= pCM  d+ p = 1
I.e. as T and hence pCM increases d+ p must
decrease.
low T
d+ p
high T
I.e. as T increases the l=1 wave ‘sees’ more and more
58
of the potential and hence (1) increases.
180
(1)
90
0
T
sin2(1)
(1) = 90
1
0
195 MeV
T
59
This is resonant behaviour. Thus the l=1 phase shift
behaviour causes the resonance at T = 195 MeV.
Thus el = 4 sin2(l)  3 + small l=0 bit
k2
At the peak of the resonance (l) = 90 and
k = 240 MeV/c / 
= 240 MeV/ (3 1023 fm s-1 6.6 10-22 MeV s)
= 1.2 fm-1
I.e. 1/k2 = 6.94 mb
[as 1fm2 = 10 mb]
Thus T = el ~ 255 mb at the peak whereas it is
measured to be only 200 mb.
Why do we get an over prediction?
60
In the formulation of our scattering theory we
assume the pion and proton to be spin-less - true
for the pion but the proton has spin-1/2.
Thus the interaction with l=1 can be in either of
two total angular momentum states j = 3/2 or 1/2.
Thus our formula becomes:
el = 4/2j=1/2,3/2 sin2(j) (2j +1)
k2
Is the resonance in the j=1/2 or the j=3/2 state?
Assume (j=3/2) is small and (j=1/2)= 90at T
= 195 MeV then:
el = (4/2k2)2 ~ 85 mb
61
Clearly this is far too much of an under prediction!
However if we assume (j=1/2) is small and (j=3/2)= 90at
T = 195 MeV then at the peak:
el = (4/2k2)4 ~ 170 mb + small j=1/2 and l=0 bits.
This is then in good agreement with the measured crosssection of 200 mb.
The resonance in the +p cross-section at T = 195 MeV is in
the j=3/2 state.
What is its isospin?
As it appears in the +p total cross-section and tz
(+) = +1, tz(p) =+1/2 then tz ( resonance) = +3/2 and hence t 
3/2. It also occurs in the -p, +n and -n cross-sections so tz =
+3/2, +1/2, -1/2, -3/2 (i.e. only 4 states) and hence t = 3/262.
It is called the (3/2,3/2) - or in even shorter shorthand
- the (3,3) resonance and as it occurs in four charge
states is an example of a - resonance - the (1236).
There are many others at higher excitation energies
( see particle data sheet handed out).
In the -p cross-section the next higher peak at ECM ~
1.5 GeV) does NOT occur in the +p cross-section . It
only occurs in only two charge states - in the -p and
+n cross-sections i.e. it has isospin t=1/2 and tz = 1/2, +1/2. This , and similar resonant states, are called
N*s.The peak at 1.5 Gev is in fact due to a
combination of 3 states the N*(1440), N*(1520) and
N*(1535) - these can only be sorted out by doing 63
many different -p scattering experiments.
Now we think the peaks in the -p and +p crosssections are due to excited resonant states of nucleons
- rather than compound -p states. Evidence for this is
available in p total cross-section measurements.Both
N* and  resonances are produced -in fact to a good
approximation:
T(p)  {T(-p ) + T(+p ) }
( Incidentally this is an example of Tz but not T being
conserved in the EM interaction).
As s are pure energy the states can only be nucleon
excited states.
64
Photon-Proton Total Crosssections
E (GeV)
65
Particle Physics -II
• Particles and Anti-particles. Kaons and
Hyperons. Associated Production and
Strangeness. Properties of hyperons and
kaons. Hyperon excited states. K-p and K+p
cross-sections. Isospin, Strangeness and
Charge. Hypercharge. Families of particles.
• Quarks …...
66
Particles, anti-particles and the vacuum perceived as a
‘sea of negative energies’ are not discussed in Krane.
Kaons, hyperons and strangeness - Krane pp 686-689
The rest of particle physics is covered in Chapter 18.
67
Particles and Anti-particles
• We are familiar from the undergraduate
laboratory with two phenomena:
• Pair production:
e-
e+
• Positron annihilation
e+


e68
We can understand these phenomena in terms of the vacuum viewed as a
sea of negative energies.For the first, shown below, a photon is incident
on the vacuum, raising an e-, mass energy mec2, and producing
simultaneously a hole in the vacuum- an absence of negative energy -(mec2) - a positron e+.
In the second an electron fills the hole and the energy released is emitted
as two back-to-back photons ( annihilation photons) - to conserve
momentum.
e-
E=0
e+
69
We arbitrarily define the electron as a particle and the positron as an
anti-particle.With higher energy photons we can in principle produce
more massive particles and anti-particles than the electron and positron.
Anti-particles for the particles we have encountered thus far are shown
below. The anti-particle has the same mass as the particle but has
opposite scalar attributes e.g. charge (and hence for strongly-interacting
particles - TZ), Baryon Number and Epton Number. Vector attributes spin, isospin - are the same for both.
• Particles
Electron eProton p
Neutron n
Positive pion +
Neutral pion 0
• Antiparticles
Positron e+ (Le = -1)
Anti-proton p (B =-1)
Anti-neutron n (B=-1)
Negative pion Neutral pion 0
Photon 
70
Kaons and Hyperons
• At higher pion incident energies new particles
(Hyperons) such as the 0 and the +, - and 0
were discovered in the 1950’s in reactions:
- + p  - + K+
 0 + K0
+ + p  + + K+
• They are always produced in association with a
KAON - a heavy meson with mass ~ 500 MeV/c2.
• They have mass > mp and decay into protons and
pions.
e.g. 0  - + p
71
72
A side remark!
Knowing the kaon, pion and proton masses and the pion momentum
and being able to measure to the exit angles of the 0 + K0 we can
calculate - by applying conservation of energy and momentum to
the collision - the mass and momentum of the 0.
Knowing the momentum of the 0 and by measuring its flight
length before decay we can work out how long it lived.
Repeating this for many events we can work out the mean life of the
0 - 2.6310-10 sec.
An odd quantum mechanics phenomenon!
We can do the same sort of calculation for the K0 - we find for this
particular event and similar ones a time of ~10-10 secs. From other
events we find apparently much longer times ~ 10-8sec - as though
the K0 has two lifetimes. You can read about this (NOT
COMPULSORY!!) in Krane pp. 692-695. K+ and K- mesons have
73
-8
only one mean life - ~1.210 secs.
At even higher pion energies heavier hyperons were
discovered - the - and 0:
+ + p  0 + K+ + K+
The s are often called the ‘Cascade particles’ because of their
decay mode:
e.g. 0 0 + 0
p + -
Even if we use extremely high pion energies the Hyperons are
ALWAYS produced in the STRONG interaction with protons
with either one Kaon ( the 0 and  hyperons) or 2 Kaons ( the
- and 0 hyperons).
This phenomenon is called ASSOCIATED PRODUCTION.
The hyperons and the kaons decay via the WEAK interaction.
74
STRANGENESS
• To describe both the phenomenon of
ASSOCIATED PRODUCTION and the Weak
Decays of hyperons and kaons we need a new
attribute STRANGENESS and an associated
Quantum Number S:
• S= +1 K+, K0, 0 and +,-,0
• S= -1 K-, K0,0 and +,-,0
• S= 0 s and Nucleons
and a Conservation Law:
• S is conserved in the Strong Interaction but NOT
in the Weak interaction ( where in fact S= 1). 75
Other properties of hyperons and
kaons
Mass(MeV/c2)
B
T
TZ
Spin
0
1115
+1
0
0
1/2
0
1115
-1
0
0
1/2
+,0,-
~1190
+1
1
+1,0,-1
1/2
+,0,-
~1190
-1
1
-1,0,+1
1/2
0,-
~1320
+1
1/2
+1/2,-1/2
1/2
0,-
~1320
-1
1/2
-1/2,+1/2
1/2
K+,K0
~495
0
1/2
+1/2,-1/2
0
K-,K0
~495
0
1/2
-1/2,+1/2
76
0
 and  Excited States
As the K- has the same strangeness as the hyperons it should be possible
to observe any resonant states in the K- p total cross-section plotted
versus energy:
K- + p  0* or 0*  decay products.
As the kaon mass is ~ 495 MeV/c2 and the proton mass is ~938 MeV/c2
giving a total mass at rest of 1433 MeV/c2 the ground state 0 or 0
( masses ~ 1118 and 1192 MeV/c2 respectively) cannot be seen but
excited states should be visible.
77
Kp
Total Cross-section
1525 1670 1750
ECM
78
As we see there are resonant states around ECM= 1525, 1670 and 1750
MeV.
These are the 0(1520), 0(1670), 0(1660), 0(1670), 0(1750) plus
some other states which can be sorted out by also doing differential
cross-section measurements versus scattering angle.
The rising cross-section as zero momentum is approached is a reflection
of two ‘sub-threshold’ resonances - the 0(1385) {which has a spin of
3/2} and the 0(1405) which have widths of 36 and 50 MeV respectively.
The K+p Total Cross-section
K+ plus a proton has S= +1 and hence cannot give rise to S= -1 hyperon
resonant states. It cannot produce S= +1 anti-hyperon resonant states due
to non-conservation of Baryon Number. B = 1 initially (BK= 0, Bp= +1).
B = -1 for any produced anti-hyperon. We would expect a flat crosssection versus plab at low momenta due to K+p elastic scattering. At
higher momenta the cross-section should increase as there is enough
79
energy for more mesons to be produced.
+
Kp
Total Cross-sections
K+ + p  K+ + p + 1,2 or 3 s
80
Excited States of the  particles
• As these have Strangeness Number S= -2
and there is no S= -2 meson, they cannot be
seen in any meson-proton total cross-section
measurement. They can be produced in
collisions though:
- + p  -* + K0 + K+
0 + 0 + 0
p + and observed in - for example - a bubble
chamber.-* (1530), 0*(1530) have spin3/281
Isospin, Strangeness and Charge
For pions and nucleons and their anti-particles the charge on a particle
can be expressed as:
Q/e= TZ + B/2
This formula is too simple for hyperons and kaons . A more general
formula is needed:
Q/e= TZ + B/2 +S/2
This is often written as:
Q/e= TZ + Y/2
where
Y= B+S
is called the HYPERCHARGE.
82
Families of Particles
Historically the proton was discovered, followed by its charge
neutral ,spin-1/2 partner the neutron - different charge (ISOSPIN)
states of the NUCLEON.
Subsequently by adding energy to the nucleon via pion
bombardment more heavier spin-1/2 particles were found -the
hyperons - differing in Strangeness or HYPERCHARGE.
We can plot these on a graph with TZ and Y as axes:
+1
Y 0
-1
p
+
n
0,
0
TZ
+1 +1/2
0
-1/2 -1
83
An Analogy
In Atomic Physics a similar historical progression led to the
discovery of HYPERFINE STRUCTURE in the Spectral Lines
of Atoms.
•For example in the hydrogen atom characteristic lines were
observed (the Balmer, Lyman series etc) which were explained
as transitions between discrete electron energy levels in a
Coulomb Potential - the e-p interaction.
•Later, fine structure of these lines was observed - explained
as due to an extra interaction between the electron magnetic
moment e and the orbital magnetic field of the electron He e.He, splitting the energy level.
•Then hyperfine structure was discovered - explained as due
to a further interaction between (e +He) and the proton
84
magnetic moment p - (e + He). p .
e.He
(e + He). p
Coulomb
Basic energy level
Fine Structure
Hyperfine Structure
Can the spin-1/2 baryon octet be considered similarly?
E.M.
Strong
Super-strong
Basic spin-1/2 baryon
N
,

p
n
+
0,0
0
-
85
What are the constituents of the basic spin-1/2
baryon?
We can study the proton. It is thought to be
composed of 3 point-like objects - the quarks.
The evidence for this comes from experiments analogous to
Rutherford back-scattering of -particles - which give us our
picture of the atom having a central small, heavy, nucleus (
radius a few fm) surrounded by a light, electron cloud( radius
many Angstroms).
To penetrate the nucleon very high energy electrons - or
neutrinos or anti-neutrinos - have to be used and backscattering
corresponds to a few degrees deflection from the incident
direction.
86
Deep-inelastic electron scattering
87
Main conclusions from deep-inelastic
electron-, neutrino - and anti-neutrino-proton
scattering measurements
•The nucleon contains three point-like objects quarks
•They have spin -1/2
•Two have fractional electric charges consistent
with +2/3e, the up-quark qu and one with
-1/3e, the down quark qd
88
To explain the properties of the hyperons there is a
need also for a third quark - the strange quark qs
The quark model
T
Tz
S
B
Y
Q/e
Spin
qu (u) 1/2
+1/2
0
1/3
1/3
+2/3
1/2
qd (d) 1/2
-1/2
0
1/3
1/3
-1/3
1/2
qs (s) 0
0
-1
1/3
-2/3
-1/3
1/2
There are also corresponding anti-quarks u, d, s with
corresponding attributes:
u
1/2
-1/2
0
-1/3
-1/3
-2/3
1/2
d
1/2
+1/2
0
-1/3
-1/3
+1/3
1/2
s
0
0
+1
-1/3
+2/3
+1/3
1/2
89
The nucleons are composed of the non-strange quarks:
p
(u, u, d)
in an L=0 state, T=1/2
n
(d, d, u)
The spin-1/2 hyperons include one or two strange quarks:
+ (u, u, s)
0
(u, d, s)
-
(d, d, s)
0 (u, d, s)
-
in an L=0 state, T=1
in an L=0 state, T=0
(d, s, s)
in an L=0 state, T=1/2
0
(u, s, s)
90
An Example
A proton consists of : u
u
+2/3
d
Charge
+2/3
-1/3 =1
Spin
1/2 + 1/2 + 1/2 = 1/2
B
1/3 + 1/3 + 1/3 = 1
T
1/2 + 1/2 + 1/2 = 1/2
TZ
+1/2
+1/2
-1/2 = +1/2
91
Families of Particles
The family of spin-1/2 baryons is thus:
+1
Y 0
p (uud)
+(uus)
n(ddu)
0,(uds)
0(uss)
-1
-(dds)
-(dss)
TZ
+1
+1/2
0
-1/2
-1
There is also a family of spin-0 mesons composed of quarkanti-quark pairs - whose spins must be opposed to give spin-0.
K0( d s)
+1
Y
0
-( d u)
K+( u s)
0,,’( u u, d d, s s)
K-( s u)
-1
-1
-1/2
+(u d)
K0( s d)
0
+1/2
92
+1
T
The  is a T=0 meson with a mass of ~550 MeV/c2
The ’ is a T=0 meson with a mass of ~960 MeV/c2
The 0 is formed from pairs of uu and dd quarks - its
wave-function is a linear combination of these quark
wave-functions
The  and ’ mesons are linear combinations of uu,
dd and ss wave-functions.
93
How are the quarks oriented inside
the baryons and what are their
masses?
Magnetic moments of protons and neutrons
The proton and neutron magnetic moments are anomalous in the
sense that if they were point particles like electrons you would
expect that: p = e/2mpc = 1 Nuclear Magneton (NM) and n = 0
In fact p = 2.8 NM, n = -1.9 NM
i.e. n /p = -2/3.
They both MUST contain some charged substructure. How does
the quark model do?
•Assume the u,d quarks ARE point-like and have same mass m.
Then u= +2/3 e/2mc and d= -1/3 e/2mc
•Two of the quarks must be parallel (spin 1) and one anti-parallel
94
(spin 1/2) to give the nucleons spin-1/2.
•Let the two quarks forming the spin-1 bit have net magnetic
moment a; the remaining anti-parallel quark b.
Then rules for vector addition say that:  = 2/3a -1/3 b.
•Then if a proton is ( u d ) + u as we might expect if Pauli
principle holds for spin-1/2 quarks:
p= {2/3(2/3 -1/3) -1/3(2/3)} e/2mc = 0!!
Similarly if the neutron is ( u d) + d then:
n= {2/3(2/3 -1/3) -1/3(-1/3)} e/2mc = 1/3 e/2mc
Both are obviously wrong.
•However if we take the proton to be ( u u) + d then
p= {2/3(2/3 +2/3) -1/3(-1/3)} e/2mc = e/2mc
and if the neutron is ( d d) + u then
n= {2/3(-1/3 -1/3) -1/3(2/3)} e/2mc = -2/3 e/2mc
Thus
n /p = -2/3
•This is an excellent result, implies quarks ARE point-like and
have charges +2/3e and -1/3e but seems to require them to 95
disobey Pauli principle! This may be very important.
Families of Particles
The family of spin-1/2 baryons is now:
+1
Y 0
p ( u u d)
+( u u s)
n(d d u)
0( u d s)
 ( u d s)
0( u s s)
-1
-( d d s)
-( d s s)
TZ
+1
+1/2
0
-1/2
-1
96
The spin-3/2 family of baryons
If we look at the first excited states of particles we find these too are
in agreement with the nucleon model that predicts magnetic
moments i.e Pauli principle appears to be violated for these too.
Also the - - hyperon is predicted.
Y 1 - ( d d d)
0( d d u)
-*( d d s)
0
+( d u u)
0*( d u s)
-*( d s s)
-1
++( u u u)
+*( u u s)
0*( u s s)
- ( s s s)
-2
-3/2
-1
-1/2
0
+1/2
+1
+3/2
97
TZ
Quark Masses
• From p = e/2mc = 2.79 e/2mpc we find that the
mass of both the up quark and down quark (assumed the
same) is m =336 MeV/c2
• We can get an estimate of the strange quark mass from
 = -0.61 e/2mpc. If we take the - hyperon to be
( u d)+ s
then the spin-0 (u d) combination does not
contribute to  and
 = (-1/3) e/2msc
From this we find ms=510 MeV/c2.
• This agrees with the mean estimate of the qs - qu,d mass
difference we get by comparing the mass increases e.g
M(-*) -M(-) = ms - m in the spin-3/2 decuplet:
ms -m = 160 MeV/c2
98
The space wave-function (L)
(L=0)
(L=1)
r
P= +1, Symmetric
r
P=-1, Antisymmetric
(L=2)
r
P= +1, Symmetric
99
Table of masses of s=3/2
100
Reconciliation of Pauli Principle
and Quarks
If we want to both
• retain the dictum that all spin-1/2 identical particles
obey Pauli Principle and
• retain the static quark model
then clearly the quarks must have some extra attribute
that nucleons, mesons, hyperons don’t have. If we
look at the point-like particle we are most familiar
with, the electron, it has the attribute of charge and
interacts with the exchange of mass-less photons.101
Color Charge and Gluons
For quarks which have to have:
• an extra attribute AND
• some mechanism for interacting with other quarks
it seems natural to assign the extra attribute as one
which could be associated with the interaction COLOR charge - and introduce a new mass-less
exchange particle to allow the interactions - the
GLUON.
To allow three otherwise identical u-quarks to exist
with parallel spins in the ++, COLOR charge
clearly needs to have THREE different forms. 102
When these three COLOR charges are added together
then the resulting COLOR charge must be zero - as
protons, neutrons etc do not have any COLOR charge.
This is how the word COLOR arose - by analogy with
red + blue+ green mixed together making white
white . The
quarks are deemed to have a r, b or g COLOR charge.
Also anti -quarks must have anti-COLOR r, b or g
charges so that the mesons don’t have any COLOR
charge.
103
The GLUONS like the photons have spin-1 but differ by also
having COLOR charge. There are three COLOR charges and
three anti-COLOR charges. As the GLUONS are exchange
particles like mesons they are given COLOR- anti-COLOR
combinations. There are eight of these:
rb rg b g b r g b
gr
(r r + g g -2bb)/6
(r r - g g)/ 2
The exchange of colored gluons gives rise to the COLOR force.
104
The COLOR force
105
Suppose the COLOR lines of force have been pulled together until
they form a tube AND the interaction energy is then so high the
quark masses are small by comparison. If this system is now
considered to be rotating we have a crude model for a meson with
spin. k2 is the energy per unit length of the force tube. The ends
of the tube rotate at velocity c, the half-length of the tube is .
Then the mass of the system is given by:

Mc2 = 2 0  k2dr/ (1-v2/c2)1/2.
This is true as k2dr is the REST MASS energy of dr so that the
relativistic total energy is k2dr/ (1-v2/c2)1/2.
At a distance r from the centre v = c r/ .

Thus Mc2 = 2 0  k2dr/ (1-r2/2)1/2 = k2
106
The angular momentum of the infinitesimal mass at distance r is
(k2dr/[c2 (1-r2/2)1/2] ) v r = (k2dr/[c2 (1-r2/2)1/2 ]) (cr/)r
= (k2r2/c (1-r2/2)1/2 )dr
Thus the total angular momentum of the tube, in units of , is:

J = 2/ 0  (k2r2/c (1-r2/2)1/2 )dr
= k22/2c
= (Mc2)2/ (2k2c)
107
108
We see that :
dJ/d(Mc2)2 = 0.9 Gev-2 = (2k2c)-1
and knowing  = 6.6 10-22 MeV sec
c = 3 1023 fm sec-1
we find
k2 = 1Gev fm-1
As the spin-1/2 baryons have masses ~ 1GeV and radii
about 1 fm this is not a bad result! At a distance of 1 fm it
means the energy holding quarks in the baryons is 1 GeV ~100 times nuclear binding energies. Put another way the
force holding quarks inside a hadron is 1015 Gev/m or in
traditional UK units -about 10 tons per pointlike quark!
109
Quarks on springs
We see now that the potential between quarks is of the form:
V=k2r
i.e. as the distance between the quarks increases so does the
magnitude of the potential binding the quarks in (e.g.) the proton.
A classical example of such a potential is a spring:
110
As two nucleons approach each other one can envisage gluon
transfer between quarks being attempted. However at distances
of 1 fm the energy involved ~ 1 GeV is sufficient to allow pair
production to occur:
Thus mesons are emitted rather than gluons! Three quarks
111
always remain confined in the nucleon.
Evidence for the COLOR Force
In addition to the spin-0 mesons:
K0( d s)
+1
Y
-( du)
0
K+( u s)
0,,’( u u, d d, s s)
K-( s u)
-1
-1
-1/2
+(u d)
K0( s d)
0
+1/2
+1
there are the spin-1 mesons:
K*0(d s)
+1
Y
0
- ( u d)
K*+(u s)
0,,( u u, d d, s s)
K*-(s u)
-1
+( d u)
K*0(s d)
112
-1
-1/2
0
+1/2
+1
TZ
TZ
These have masses of: K* s ~ 890 MeV/c2
 s ~770 MeV/c2

~ 782 MeV/c2 ~ 1020 MeV/c2
These spin-0 and spin-1 mesons exhaust the possibilities for u,d,s
and u,d,s combinations.
The  in particular is an ss meson - which we might have
expected given a static model s-quark mass of 510 MeV/c2.
Its lifetime is very short - from its energy width 4.4 MeV - using
Et ~  - it is ~ 1.510-22 secs. It decays into K+K- predominantly
rather than +-0 .
The only known explanation for this is in terms of gluons!
113
Zweig’s Rule
In a strong decay completely ‘disconnected’ qq decays qq gluons  q’q’ - are strongly suppressed relative to
direct qq production - qq  qq q’q’.
114
The Discovery of the J/
It was quite a surprise then when in 1975 a new spin-0 meson was discovered
simultaneously in both e+-e- collisions at high energies (shown below) and in the
reaction:
p + Be J/ + anything
 e+-ewith an energy of 3.1 GeV/c2.
115
As all spin-1 mesons predicted by the quark model
had been found the only way the J/  could be
understood was if another heavier quark existed
called qC - the charm quark. Associated with this is
another quantum number C=+1.
The J/  is a
cc state ie C=0. It is said to have ‘hidden charm’. It
must have J=1 as e+-e- annihilate to produce a
virtual photon (spin 1) which forms the J/  in a
3S state. By analogy with the  -meson - an ss
1
state- we infer the mass of the c-quark to be 1550
MeV/c2. Its other properties are:
T
qC (c) 0
Tz
S
B
Y
C
Q/e
0
0
1/3
1/3
+1
+2/3
Spin
1/2
116
Other states involving the charm quark
Subsequently many other meson states made from cu (the D0), uc
(D0), cd (D+ ),dc (D- ) - all with masses around1870 MeV/c2, and cs
(DS+) ,sc (DS-) pairs - with masses around 1970 MeV/c2.
There are also charmed baryons :
C+ (udc), C++(uuc) , C+(udc), C0(ddc), C+(usc), C0(dsc),
C0(ssc).
Also other ‘hidden charm’ cc states like the 3S1 J/ except in
configurations 1So , 3P0,1,2 were found - the  and  mesons
respectively. As these are made of very heavy quarks they are
likely to be moving very slowly and hence can be compared with
states formed by another particle-anti-particle pair - the e+ e- pairwhich form a set of very loosely bound states called positronium.
The new ‘hidden charm’ set of meson states is called
charmonium.
117
Lifetime ( energy width) of the J/
The -meson (ss) has sufficient mass to decay directly into a pair
of K+(us) and K- (su) mesons and hence, as we have seen this is
not suppressed by Zweig’s rule, has a broad width - when
produced in e- - e+ collisions - of 4.4 MeV.
The J/ (cc ) however has insufficient mass to decay directly into
D+(cd) and D- (dc) mesons. Also - see examples below- other
possible decay modes are inhibited by Zweig’s rule. Hence it has
a very narrow width - 88 keV.
118
Charge of Charmed Particles
The much-modified expression for charge is now further
generalised to:
Q/e = TZ+ (B + S + C)/2
119
Charmonium
109
difference
in
energies!
Positronium
120
Potential deduced from charmonium states
121
QUARKS
• Evidence for point-like objects in nucleons from deep inelastic
lepton-nucleon scattering
• 3 flavor (u,d,s) static quark model of baryons-3q & mesons- qq
• qq orientation in mesons - anti-parallel (spin-0), parallel (spin-1)
• Magnetic moments of baryons  q orientation in baryons
• Parallel orientation of like flavor quarks in L=0 state  Apparent
violation of Pauli Principle
• Magnetic moments and J=3/2 baryon masses  u,d,s quark
masses in static model
• Reconciliation of Pauli Principle and Quarks  Need for
COLOR charge  COLOR force through exchange of
massless gluons
• Constant Energy / unit distance of COLOR force from simple
interaction model  quarks bound by spring-like force
• Need for fourth flavor quark qC  charmonium  evidence
122 for
shape of COLOR force potential.
As a confirmation that the interaction between quarks is due to the
special property of COLOR charge (that ALL quarks have
regardless of flavor) AND the exchange of COLORED gluons,
at even higher e+- e- collision energies another meson was found
the  (Upsilon) with a mass of 9460 MeV/c2. This could not be
explained as a combination of u,d,s,c quarks and anti-quarks.
Another quark attribute was needed beauty and a quantum
number b. For qb, b=-1and hence, with an even further revised
expression for charge:
Q/e = TZ+ (B + S + C +b)/2
the charge of the b-quark is -1/3 e - as it is a singlet with Tz = 0.
Its mass is I/2 the upsilon (bb) mass- 4730 MeV/c2. Otherwise:
T
qb (b) 0
Tz
S
B
Y
C
b
0
0
1/3
1/3
0
-1
Q/e
-1/3
Spin
123
1/2
Another set of states was then discovered - bottomonium.
It can be fitted with exactly the same potential as for charmonium!
124
Leptons
Just as more and more massive hadrons - strongly interacting
particles - were discovered as higher and higher energy
accelerators became available, a heavier charged lepton was
discovered in 1975 - the -lepton or (tauon) with a mass of 1784
MeV/c2.
There are thus three known charged leptons:
Electron e-
Lifetime
(sec)
Mass
(MeV/c2)
Stable
0.511
Muon
-
2.210-6
105.66
Tauon
-
3.410-13
1784
They of course all have antiparticles. Both the s and the s
125
decay via the WEAK INTERACTION.
The muon decays: -  e- +  + e
The e-anti-neutrino, e ,conserves epton number in the decay;
the -neutrino,  conserves mepton number:
Le
0  1 + 0 + ( -1)
L
1 0+ 1 + 0
The masses of  and e are near zero.
The tauon can decay in many ways (three shown here):
-  e- +  + e
 - +  + 
 - + 
The tau-neutrino, , conserves tepton number in the decays
- L = 1
(-,
), L = -1
(+,
)
126
Families of Quarks and Leptons
‘Normal’ matter thus consists of the up and down quarks, the
electron and e. As more energy is available the strange and
charm quarks can be created, together with the muon and :
ud
sc
e- e
- 
At even higher energies we find the b-quark, the tauon and 
Is there a further quark - the top quark (t-quark) with the extra
attribute truth?
bt
-

127
Such a quark has been found in extremely high energy protonantiproton colliding beam experiments (Tp =Tp = 900 GeV) . In
these experiments quarks in the proton annihilate with antiquarks
in the antiproton:
128
The tt pair then decay leading to lighter quarks which initiate ‘jets‘
of hadrons by a complicated process called ‘quark fragmentation’.
Tracking back the ‘jets’ to their origin the top quark mass can be
determined as:
Mt ~ 175 GeV/c2
The other properties of the top quark are:
qt (t)
T
Tz
S
B
Y
C
b
t Q/e
0
0
0
1/3
1/3
0
0
+1 +2/3
Spin
1/2
129
Quark counting!
Production of hadrons or muons in e+-e- collisions can be
viewed as Rutherford Scattering.
e
e
q
e

e+
q
e-
e+
q
q
e-
e-
-
-

+
130
As there are different quark species:
(e+-e-  hadrons) = i (e+-e-  qiqi)
The ratio of the cross-sections for hadron production and
muon production at a particular energy will take out any
kinematic factors AND the cross-sections will be
proportional to the square of the charges of the scatterer
(quarks or muons) - Rutherford scattering formula.
Therefore we have:
R = (e+-e-  hadrons) = i Qqi2
(e+-e-  + -)
Q2
i.e. R counts the number of quarks produced at a
given energy!
131
We see that if there are only 3 flavors of quark -u, d and s and COLOR is not a real attribute then:
R= (2/3)2 +(1/3)2 + (1/3)2 = 2
12
3
If COLOR is a real attribute the we have :
ur,ub, ug,dr,db,dg,sr,sb,sg quarks and:
,
R= 3{(2/3)2 +(1/3)2 + (1/3)2 } =2
What do we get experimentally?
132
133
When we have five flavors of quarks we get:
R= 3{(2/3)2 +(1/3)2 + (1/3)2 + (2/3)2 +(1/3)2 } = 11/3
in excellent agreement with the data.
Clearly with the t quark:
R= 15/3 = 5
but we do not have data at sufficiently high energies yet.
We do though have specific evidence of the reality of both
quarks of different flavors and color charge - evidence as
strong as for the existence of nuclei.
134
The Weak Interaction
Examples of this we have met already are neutron decay
and muon decay :
np + e- + e
-  e- +  + e
These are both very slow in particle physics terms - mean
lives 887 s and 2.10-6 s respectively.
Another example is neutrino scattering:
e.g. e + e-  e + eThe cross-sections for such reactions are incredibly small
indicating near-zero range for the interaction.
These two facts indicate that the reaction must be mediated
by the production of a very heavy particle .
135
These have been discovered in PP collisions at very high
energy and are the INTERMEDIATE VECTOR BOSONS:
W MW ~ 80 GeV/c2
Z0
MZ ~ 91 GeV/c2
The Z0 mediates such reactions as:
 + e-  + e-
e-
eZ0


136
When both leptons are of the same generation
(e.g. eptons) both the Z0 and the W- can
mediate:
e + e- e+ e-
e-
eZ0
e
e-
eW-
e
e
e
137
In weak decays of particles only the W+ and W- are
involved:
138
139
Strong decays occur by the excited state energy being
converted into quark-antiquark pairs.
140
Strong interaction reactions: annihilation is followed by pair
production:
141
Electromagnetic Decay Example
142
And finally: Decay by Annihilation!
143
144
145
146
147