Concepts

Particle Physics I
I.
II.
III.
IV.
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Particle Physics II
Introduction, history & overview (2)
Concepts (3):
Units (h=c=1)
Relativistic kinematics
Cross section, lifetime, decay width, …
Symmetries (quark model, …)
Quantum Electro Dynamics: QED (6-7)
Spin 0 electrodynamics (Klein-Gordon)
Spin ½ electrodynamics (Dirac)
Experimental highlights: “g-2”, ee, …
Quantum Chromo Dynamics: QCD (3-4)
Colour concept and partons
High q2 strong interaction
SB
Structure functions
Experimental highlights: s, ep, …
V.
VI.
VII.
VIII.
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Quantum Flavour Dynamics: QFD (4)
Low q2 weak interaction
High q2 weak interaction
Electro-weak interaction
Experimental highlights: LEP
Origin of mass? (2)
Symmetry breaking
Higgs particle: in ee and in pp
Origin of matter? (6)
K0-K0, oscillations
B0-B0 oscillations
JvdB
Neutrino oscillations
Fantasy land (2)
File on “paling”: z66/PowerPoint/ED_Master/Concepts.ppt
II. Concepts
Particle Physics 2002/2003
Part of the “Particle and Astroparticle Physics” Master’s Curriculum
1
2
Units (h=c=1)
r12
Coulomb:
q2
q1


q1 q 2 r 12
force-law: F 
3
4  0 r12
3
Units
units
2
2

Coulomb  T 
 0 
M L3
0 gives you the freedom to decouple unit of [q] from [Time], [Mass] and [Length] units

SI (so be it!):
Heaviside -Lorentz (simple field equation):
 0 1
Gauss (simple force equation):
 0
unit of charge:
Ampère:

j1
1
4
  
E 

  0
E  
 
E  4
qSI = 0 qHL = 04 qG
 force-law:
j2
  

0
j 1 j 2r 12  3  3 
F 
d r1 d r 2

3
4
r12
SI (so be it!):
Heaviside -Lorentz (simple field equation):
Gauss (simple force equation):
 

B   0 j
  1
B  j
c
  4 
B 
j
c
with
0 0 
1
c
2
4
Natural units: h=c=1
Fundamental quantities:
Units:
Length [L]
Time
[T]
Mass
[M]
h  1.0566  1034 Js
c  299792458 m/s
(1 Js=1 kg m2/s)
(definition of the length unit)
Simplify life by setting:
c1
h1
 [L]=[T]
 [M]=[L]=[T]
Remaining freedom: pick one unit Energy! [E]=eV, keV, MeV, GeV, TeV, …
Energy [E]
Mass
[M]
Momentum
Lenght [L]
Time
[T]





GeV
GeV/c2
GeV/c
GeV1 hc
GeV1 h
(via E=1/2 mv2)
(via p=mv)
(via 1=[hc]=Jm and [E]GeV)
(via 1=[h]=Js and [E]GeV)
e
(Gaussian units!)
potential energy of electron in Hydrogen

rest energy of an electron
+e
h
mec
e2
1
e2
2
 
mc 


h
c
137
( h / mc )
the fine-structure constant, , sets
the strength of the e.m. interaction
Maxwell equations in Gaussian units
5
 
  E  4

 
1 B
 E  
c t
 
B  0
In vacuum; d.w.z. vrijwel altijd:
 = 0 en j = 0
(In particle physics we do not need to worry
about e.m. behavior in matter!)

  1
 1 E
  B  4 j 
c
c t
Cross section: area m2  GeV2
conversion factor : h c  
2
typical
cross sections
1b
1 nb
1 pb
1 fb
Physics!
(this course)
=
=
=
=
1024
1033
1036
1039
10.031052 J 2 m 2
1.610
cm2
cm2
cm2
cm2
19
J/eV

2
 0.389  10
 31
2
2
2
GeV m  0.389 GeV mb; 1 barn  10
LEP: 10+31 cm2s1= 0.1 nb1s1  100 pb1year1
“B-factory”: 10+33 cm2s1= 100 nb1s1  10 fb1year1
LHC: 10+34 cm2s1= 0.1 pb1s1  100 fb1year1
Lifetime: time s  GeV1
1.05661034 Js
 25
conversion factor : h 
 6.6  10 GeV s
19
1.610 J/eV
 24
cm
2
typical
“luminosities”
Technology!
(other course)
6
Relativistic kinematics
Lorentz-transformations
Co-moving coordinate systems
S’
S
v
x
x’’
Lorentzcontractie: Lengteverkorting gezien vanuit bewegend stelsel
Events:
S: @ t=0: (0,0) & (0,L)
transform to S0: L0=L  L= L0/
Tijdsdilatatie: Tijdsverlenging gezien vanuit bewegend stelsel
Events:
S0: @ z0=0: (0,0) & (t0,0)
transform to S: t=t0
Voorbeeld: Muonen, 210-6 s, m106 MeV in kosmische straling
p=10 GeV  l   v  60 km i.p.v. 600 m
7
Notation:
8
4-vector
Invariant (“boosts”, rotations, …):
Introduce metric:
Contra-variant 4-vector
Co-variant 4-vector
Generally for 4-vectors a and b:
Lorentz-invariant  formulate in terms of scalars, 4-vectors,…
9
Lorentz-transformations: Rotations
Lorentz-transformaties:
These matrices must obey:
For infinitesimal transformations you find:
Because:
0
Rotation Z-axis:

1
      
12
2
21
2
1
Finite rotation  around Z-axis yields:
1

2
3
0
1
2
3


S
S’
 x
x
e  Lim  1 
N
N  
Infinitesimal  macroscopic:
1 0

0 1
Lim 
N   0  

 0 0N

0


N
1
0
N
0
 1 0


0
0 1
  Lim  

N   0 0
0
  0 0


1 


0 0 0




 0 0 1
 Lim 1  
N 0 1 0
N 


0 0 0

1
0

0
0
0 0 0
0
1 0
0
0 1
0 0
0

0
0
  0
0
 
1 0
0
1

 0 cos

0 sin 

0
0


0
0
0
0
sin 
cos
0
0
0
0

0

0

0  
0


N
0
0
N
N
0
0


0
0
 exp 0


0
0

 
0
0

0

1
0

0
0

1
0 0
0 0


0 0
0 0
 

1 0
0 


0 1
 0 0N

0
1 0
2 0

0
0
0
 2
0
0
 2
0
0
0
0
0


0
0
0
0
0

0

1
0

0
0

0 
0
1 0
6 0

0
0
0
 3
0
0
3
0

0
0
0


1
0
N
10
Lorentz-transformations: Boosts
Boost along Z-axis:

0
3        
03
30
3
0
Finite boost  along Z-axis yields:
De relatie tussen x0=ct, x3=z en x’0=ct’, x’3=z’ is:
De relatie tussen  en de snelheid v/c is:
Oftewel:
cosh  
1
1  2
and
sinh  

1  2
Dus:
11
12
Lorentz group: Generators
Lorentz-group parameters:
Dus: 4 x 4 – 10 = 6 parameters:
3 rotaties: around X-, Y- & Z- axis
3 boosts: around X-, Y- & Z-axis
Generators for infinitesimal transformations:
  
J
Lim  1 J i   exp i
N 
N  
N
Angle  :
  
K
Lim  1 K i   exp i
N 
N  
N
Boost  :
Commutation-relations:
Representations:
spin j and mass m2
The velocity 4-vector
13
S
Standaard definitie van snelheid: x/t:
Gezien vanuit S: snelheid u= x/t
Gezien vanuit S’: snelheid u’= x’/t’
En dus
Dit is geen goede keus i.v.m. Lorentz invariantie
Betere definitie van snelheid (t= ):
Definieer nu de impuls als p=m=(E/c,px,py,pz), dan:
Waarom p0E/c?
Deeltjes met m=0:
u
x
14
Derivatives
Afgeleiden:
en
Een contra-variante vector transformeert als:
Een co-variante vector transformeert als:
Het invariante product:
voor de afgeleide
analoog voor
Covariante producten:
Transformatie naar zwaartepunts systeem
m1
m2
m3
m4
m5
Systeem van N deeltjes; transformeer naar het c.m.
Lorentz boost met
Dan inderdaad:
Lab-frame
15
16
Verval van deeltjes
mB
mA
mC
c.m. frame
Met
volgt
m  140 MeV
m  106 MeV
me  0.511 MeV
m  m 
2
Voorbeelden:
E
Ee
E 
2 m
2
 110 MeV
Ee53 MeV3 deeltjes
-verval: experimental result!
17
M/253 MeV
Electron energy spectrum


  e  e
Drempel energie
18
EA
B
Lab-frame
In het c.m. systeem is deze reactie juist mogelijk indien alle geproduceerde deeltjes in rust zijn
Dus:
C.M. energie
Voorbeeld: anti-proton ontdekking
p ontdekking : p  p  p  p  p  p
Voorbeeld: LEP,
e + e-
versneller met Ecm~90 GeV:
drempelenergie:
4m p 2m2p m2p
2m p
7 m p
19
Pion verval

0
12
Lab-frame

0


E2
E
Distribution?

c.m. frame
Openingshoek:
E
E1
m  140 MeV
Experimentele detectie van hoog energetische pionen moeilijk!
Neutrino bundel
20
(theory)

K
Generatie neutrino bundel via K of  verval
cm
-
c.m. frame
En dus:
Van cm naar lab

KLab-frame
lab
-
Voorbeeld; pK=p=200 GeV, m=139 MeV, mK=494 MeV:
GeV
GeV
Neutrino bundel
21
(experiment)
192 GeV
R
84 GeV
K    

     

50
100
150
200

K


CDHS
detector
E
22
Compton verstrooiing
P’
,E
k
Foton verstrooiing: licht op elektronen
   
Lab-frame

2

h
Js
kg
/s
m
Als h en c weer teruggestopt worden:


 kg m 
c
m/s
m/s
P

’,E’
k’
’
0.06
0.04

2h

2
 0.022  0.0485 sin 2
sin
2
2
me c
Compton’s
original data
0.02
0.00
0


23
Mandelstam variabelen (I)
Reactie A + B  C + D gekarakteriseerd door 2 variabelen
D
cm
A
C
C
B
c.m. frame
In het C.M. frame: bijvoorbeeld EAcm en
verstrooiingshoek cm
Lorentz-invariante variabelen:
Er geldt:
A
B
lab
D
Lab-frame
In het lab. frame: bijvoorbeeld EAlab en
verstrooiingshoek lab
24
Mandelstam variabelen (II)
Handige uitdrukkingen in termen van s, t & u:
Lab. frame:

p A  E A, p A   E A,  p 
en

p B  m B,0 
Energie van A:
C.M. frame:

p A  E A, p A   E A,  p 
C.M. energie:
E  E A  EB 
en
E A E B 2 


p B  E B , p B    E B ,  p 
s
Energie van A:
Want:
A
Voor A + A  A + A, in het c.m. frame geldt:
A

p  E , p 

p  E , p 
A

p  E , p 


p  E , p 
c.m. frame
A
25
Cross section,
lifetime,
decay width,
…
26
Levensduur
Bijna alle elementaire deeltjes vervallen! d.w.z.
Levensduur:
Vertakkingsverhouding
Eenheden
b.v.
mB
mC
mA
c.m. frame
verschillende
vervalskanalen, b.v.
27
Werkzame doorsnede
Ingrediënten om de telsnelheid A + B  C + D te bepalen:
1. De overgangswaarschijnlijkheid Wfi
De kans op een i  f overgang per tijdseenheid per volume eenheid
W wordt bepaald door de dynamica!
2. De experimentele omstandigheden (flux factor)
a)
b)
Bundel intensiteit
Dichtheid van het target
# per tijdseenheid per oppervlakte eenheid
# per volume eenheid
3. De fase-ruimte n
W geeft de kans voor exact gedefinieerde toestanden i en f. Omdat niet
elke toestand gemeten kan worden wordt over enkele (bijna identieke)
toestanden geïntegreerd.
Voor A + B  C + D:
[]=oppervlakte
v.b.: Verstrooiing aan een harde bol
Verstrooiing aan een massieve bol:
Berekening werkzame doorsnede

Geometrie
Berekening werkzame doorsnede:
(vgl. Rutherford verstrooiing)
Totale werkzame doorsnede,
oppervlak zoals de bundel die ziet:
b


R
28
29
Fermi’s ‘golden rule’ (klassiek)
Storingsrekening:
Schrödinger vergelijking
‘ongestoorde’
toestanden
Veronderstel:
selecteer toestand af(t)
Dan volgt voor da(t)/dt:
kies alleen n=i
En voor a(t) (laagste orde):
De overgangsamplitude:
met fi
Voor een tijdsonafhankelijke storing:

i  E f  E i t 
e
30
Fermi’s golden rule
Geeft |Tfi|2 de kans voor if overgang?
Neen!
De overgangswaarschijnlijkheid per tijdseenheid wordt dan:
i.h.a.: - welgedefinieerde begintoestand (bepaal jij!)
- scala van eindtoestanden (bepaalt de natuur!)
Fermi’s ‘golden rule’:
31
Relativistisch A + B  C + D
 N e
i
Voor relativistische vlakke golven:
Volgt voor de overgangswaarschijnlijkheid:
Per tijd- en volume-eenheid wordt dit:
Behoud van 4-impuls
En de werkzame doorsnede:
A
A

pA x  E At 
i
N e
A
pA x
N-particle phase-space
Klein-Gordon:
32
(volgende college)
Dus voor het geheel in een doos LxLxL=V:
take this
simplest
Faseruimte voor 1 deeltje
Faseruimte voor 2Ec en 2ED deeltjes:
Wfi met gekozen
normalisatie:
gebruik periode randvoorwaarden:
 pi=2 ni /L
The flux factor
De volumes V kunnen voor de bundel en target
verschillend zijn. Ze vallen uiteindelijk weg.
33
v
In het lab stelsel
(B in rust en A heeft snelheid v)
bundel
target
De cross sectie is onafhankelijk van volume & tijd!
bundel
Voor d:
target
(N=1/V)
Wfi  V4
2  V+2
1/Flux  V+2
Voor d:
(N=1/2EV)
Wfi  V4
2  V+2
1/Flux  V+2
Note: even the factors 2E come out right i.e. the same!
Lorentz invariant form
34
35
Flux factor voor A + B …
c.m. stelsel:
P1=(E1,+p)
P2=(E2,p)
note :
lab stelsel:
P1=(E, p)
P2=(m2,0)

 p
v
E
Generic Expressions:  and 
verstroooiing
A
A + B  C + D werkzame doorsnede
A
B
A
36
D
B
C
B
verval
A  C + D vervalsbreedte
C
D
A
Faseruimte 2 deeltjes
Bepaling van de faseruimte met impulsbehoud
Doe de integraal over de 3-momenta van p4 m.b.v. de -functie
Merk op dat E4 niet onafhankelijk is:
Herschrijf in bol-coördinaten, d3p3=|p3|2ddp3
Dus:
In cm stelsel geldt E3dE3=E4dE4=pdp
Met E’=E3+E4 nog niet gelijk aan EE1+E2
p 1

d
2
2
E 4 
Note: 6 vrijheidsgraden
waarvan er 4 vastgelegd
worden door
impulsbehoud: 2 over
37
38
Voorbeeld: verstrooiing A+B  C+D
In het c.m. frame
Fase ruimte
(2dLips)
Flux
Werkzame doorsnede
Identieke deeltjes:
Identieke deeltjes in ‘final state’ kunnen
niet onderscheiden worden
A+B  1+1
A+B  1+1+2+2+2
1
S=1/2!=1/2
S=1/(2!3!)=1/12
2
Statistische factor:
2
1
Voorbeeld: verval A  C+D
39
In het c.m. frame
(gelijk lab frame)
Fase ruimte
Flux
Voorbeeld: 0
Note S=1/2
u
u




Schatting:
geeft:
10-17 s, terwijl PDG geeft: =8.7 10-17 s
40
Toy-model: ABC theory
Feynman regels
41
Berekening van de amplitude M:
Hiervoor is de dynamica van wisselwerking nodig. In het vervolg zullen we de
amplitudes berekenen voor elektromagnetische, sterke en zwakke wisselwerkingen.
Om een idee te krijgen eerst de amplitude voor een hypothetisch model, 3 toy-model:
Feynman regels ABC theorie:
p2
p1
A
ig
B
p3
C
B p1
p2
A
q
ig C ig
p3
B
p4
A
Levensduur
42
(3 toy-model)
B
The matrix element
A
is simply [g]=[GeV]
g
C
De vervalsbreedte wordt
met Fermi’s regel:
8
En de levensduur
8
van deeltje A is dus
De impuls van B (of C) kan bepaald worden, in c.m. systeem:
8
Verstrooiing A+A  B+B (3 toy-model)
A
p2
(A)
p4
B
Tweede diagram!
A
C q
A
p1
(A):
(B):
(A+B):
(B)
p2
C
q
p3
B
A
p1
B
p4
p3
B
43
44
Verstrooiing A+A  B+B (3 toy-model)
B

A
In het c.m. stelsel
Veronderstel mA=mB=m en mC=0
Twee identieke deeltjes in eindtoestand
(dus een extra factor ½!=½)
Werkzame doorsnede:
B
A
c.m. frame
Verstrooiing A+B  A+B (3 toy-model)
A
p2
(A)
C
B
p1
(A):
(B):
(A+B):
p4
B
q
p3
Two diagrams
contribute!
A
p2
p1
A
B
(B)
q
C
B
p4
p3
A
45
Verstrooiing A+B  A+B (3 toy-model)
B

A
In het c.m. stelsel
Veronderstel mA=mB=m en mC=0
Werkzame doorsnede:
A
B
c.m. frame
46
47
Symmetries
•
•
•
•
(…, quark model)
Symmetry  conserved quantity & classification of states
Mathematical background: groups & representations
Examples
Quark model: Baryons & Mesons
Symmetry 
48
conserved quantity
classification of states
Imagine Hamiltonian invariant under
some kind of (coordinate) transformation
Physics invariant  1=U+U  U+=U1
 x  x
x  x’  Rx
2
   x    x 
2
  x U U  x

(x)  ’(x’)  U(x)
For matrix elements Hamiltonian H:
(H=H’)
Via de Schrödinger equation:
conserved quantity U<U>
 x H  x
2
   x  H   x 
2
2

   x  U HU   x 
 H  U  HU  U 1HU  UH  HU  U , H 0
dU
d

  x  U   x     x  HU  UH   x   0
dt
dt
For an infinitesimal transformation, one can write:
U  1  i O

1
U  U  1i O   1i O 

1
  i  O    i O  O  O 
Operator O hermitean and commutes with H  use eigenvalues to classify states!
2
Continuous space-time symmetries
49
Space: physics invariant under xx’x+ (if possible always study infinitesimal transformations)
 ( x)   ( x)   ( x   )   ( x)  

 ( x)

  ( x)  ii  ( x)
x
x
i
 p x commutes with Hamiltonia n  momentum conserved
x
Space: physics invariant under rr’(xy,y+x,z)


 
 ( r )
 ( r )
  







 ( r )   ( r )   ( x  y , y  x , z )   ( r )  y
 x
  ( r )  i  ix iy  ( r )
x
y
 y x 
 ix


 iy
 l z commutes with Hamiltonia n  angular momentum conserved
y
x
Time: physics invariant under tt’t+
 (t )   (t )   (t   )   (t )  

 (t )

  (t )  ii  (t )
t
t
i
 H commutes with Hamiltonia n  energy conserved
t
inversion
reflection
50
y
z
x
y
z
 x  x 
x
    
r
  y   y 
Coordinates:
 x  x
    

r
Coordinates:   y     y    r
 z  z
   

2
Operators: P  P , with : P  1  P  P
hence P has eigenvalues 1

 z
 
2
Operators: S  S , with : S  1  S  S



 





 
 

pˆ  pˆ   Spˆ S  pˆ x, pˆ y , pˆ z
S
1
 0

0
0
1
0
  1
00
 
1  0
0
1
hence S has eigenvalues 1
Vectors: rˆ  rˆ   Srˆ S   xˆ , yˆ ,  zˆ 

Vectors: rˆ  rˆ   P r̂ P    rˆ


 

pˆ  pˆ   Ppˆ P   pˆ




Pseudo-vectors: lˆ  lˆ  Plˆ P    lˆ


 

sˆ  sˆ  Psˆ P   sˆ
1
 z
 
0
1
0
  1
0  0
 
1  0
0
0
0
0
1

1 0   P R z  as   

51
Zeeman and Stark effects
energy levels
|JM>
Note:
• vectors: E-field, r, p, …
• axial-vectors: B-field, rp, …
2J+1 times degenerate
rotations to change in between states |JM>
(symmetry group: SO(3))
|JM>
E0
Stark effect
2 times degenerate
reflection in plane changes |J+M> into |JM>
(symmetry group: SO(1), reflection)
B0
|JM>
Zeeman effect
no degeneration
can not change between states |JM>
(symmetry group: SO(1), parity)
52
Is physics
mirror P
invariant?
C.S. Wu:
60Co
60Co
5
4
+
e
e
53
B
asymmetrie in e
hoekverdeling?
5

60Ni*
+1
e
e
Experiment!
60Ni*
4
+1
e
e
Sketch and photograph of apparatus used to study beta decay in polarized cobalt-60 nuclei. The
specimen, a cerium magnesium nitrate crystal containing a thin surface layer of radioactive cobalt60, was supported in a cerium magnesium nitrate housing within an evacuated glass vessel (lower
half of photograph). An anthracene crystal about 2 cm above the cobalt-60 source served as a
scintillation counter for beta-ray detection. Lucite rod (upper half of photograph) transmitted
flashes from the counter to a photomultiplier (not shown). Magnet on either side of the specimen
was used to cool it to approximately 0.003 K by adiabatic demagnetization. Inductance coil is part
of a magnetic thermometer for determining specimen temperature.
L. M. Lederman: +  + + 

+
+
Exp
OK
+

Intrinsic
spin
P
+

CP
C

54
Experiment:
-
+ beam stopped in target
+  + +  decays
study +  e+ + e +  decays
shows + polarisation
infer  polarisation since S=0
Experiment shows that parity
transformed configuration does
not exist!

Exp
OK
Restore the symmetry using
particle  anti-particle
operation: charged conjugation
December 27, 1956: Fall of Parity Conservation
55
Symmetries have long played a crucial role in physics. Since 1925, physicists had assumed that our world is indistinguishable from its mirror image - a
notion known as parity conservation - and prevailing scientific theory reflected that assumption. Until a series of pivotal experiments at the National
Bureau of Standards in 1956 (now the National Institute of Standards and Technology), parity conservation enjoyed exalted status among the most
fundamental laws of physics, including conservation of energy, momentum and electric charge. But as with relativity, Nature once again demonstrated
that it is not always obliged to follow the rules of "common sense".
Parity conservation implies that Nature is symmetrical and makes no distinction between right- and left-handed rotations, or between opposite sides of a
subatomic particle. For example, two similar radioactive particles spinning in opposite directions about a vertical axis should emit their decay products
with the same intensity upwards and downwards. Yet although there were many experiments that established parity conservation in strong interactions,
the assumption had never been experimentally verified for weak interactions. Indeed, when the weak force was first postulated to explain disintegration
of elementary particles, it seemed inconceivable that parity would not hold there as well.
All that changed in the 1950s, when high-energy physicists began observing phenomena that could not be explained by existing theories, most notably
the decays of K mesons emitted in the collision of a high-energy proton with an atomic nucleus. The K meson appeared in two distinct versions,
decaying into either two or three pi mesons, (which necessarily had opposite parity), although in all other characteristics they seemed identical. In June
of 1956, theoretical physicists Chen Ning Yang and Tsung Dao Lee submitted a short paper to the Physical Review raising the question of whether parity
is conserved in weak interactions, and suggesting several experiments to decide the issue.
Lee and Yang's paper did not immediately spark more than passing curiosity among physicists when it appeared in October 1956. Freeman Dyson later
admitted that while he thought the paper was interesting, "I had not the imagination to say, 'By golly, if this is true, it opens up a whole new branch of
physics!' And I think other physicists, with very few exceptions, at that time were as unimaginative as I." Richard Feynman pronounced the notion of
parity violation "unlikely, but possible, and a very exciting possibility," but later made a $50 bet with a friend that parity would not be violated.
One of the simplest proposed experiments involved measuring the directional intensity of beta radiation from cobalt-60 nuclei oriented with a strong
magnetic field so that their spins aligned in the same direction. Parity conservation demands that the emitted beta rays be equally distributed between
the two poles. If more beta particles emerged from one pole than the other, it would be possible to distinguish the mirror image nuclei from their
counterparts, which would be tantamount to parity violation.
Between Christmas of 1956 and New Year's, NBS scientists set about performing beta decay experiments. The team was led by Columbia Professor C. S.
Wu. Professor Wu had been born in China in 1912, had received her PhD from the University of California in 1940, and had worked on the Manhattan
Project during World War II. In 1975 she would serve as the first woman president of the APS.
When the results were in, the NBS team arrived at a startling conclusion: the emission of beta particles is greater in the direction opposite to that of the
nuclear spin. Thus, since the beta emission distribution is not identical to the mirror image of the spinning cobalt-60 nucleus, parity was unequivocally
shown not to be conserved. Leon Lederman, who at the time worked with Columbia University's cyclotron, performed an independent test of parity with
that equipment, involving the decay of pi and mu mesons, and also obtained distinct evidence for parity violation.
In short, Nature is a semi-ambidextrous southpaw. And Feynman lost his bet. The result shattered a fundamental concept of nuclear physics that had
been universally accepted for 30 years, thus clearing the way for a reconsideration of physical theories and leading to new, far-reaching discoveries most notably a better understanding of the characteristics of elementary particles, and a more unified theory of the fundamental forces.
Further Reading: S. Weinberg, Reviews of Modern Physics, 52, 515 (1980); A. Salam, p. 525; S.L. Glashow, p. 539.
56
Is physics
CP
invariant?
57
The K0-K0 system: strong interactions : K0-K0
weak interactions : KS-KL
Kaons consist of s-quarks and u/d-quarks:
(more details next week)
 K    us 
 0   
 K   ds 
 K 0   u s 
   

 K    ds 
Production: strong interaction: ss quark pairs produced
E > 0.9 GeV:
  p  K0 
E > 6.0 GeV:
 p  K0 n n
E > 1.5 GeV:
  p  K0K p
Decay: weak interaction: s-quark  u-quark decay (via s  d quark mixing)
S = 0.89350.0008  1010 s
L = 5.170.04  108 s
The
K0-K0
58
system: CP eigenstates
Meson:
For mesons: P qq    1 qq 
L
C qq  1
L S
q
orbital motion: L
intrinsic spin: S
(KS)
(KL)
 Y LM  , 
S 0
  
S 1
  
2
2
P
C
0
0
0



qq
K
0
 K0
 K0
K
0
 K0
 K0
arbitrary
q’
|qq’> state
CP  0 0    0 0
CP        
CP  0 0 0    0 0 0
CP    0

    0
 
  0

   
L  even
L  odd
CP
K
0
0
K K

 KS
2
CP
K
0
0
K K

 KL
2
Start with a
K0
59
system: oscillations
intensity
The
K0-K0
beam
CP
CP

K K L
K
K
0


K 
 S
2
2

t=0
 I K ,t 0 
 I K 0,t 0 


N(t)
t=t
    1 
   0 
50%
K0
t
 p  K 


75%
25%
0
0
K0
100%
0
K 0
K
K L
2
 AK ,t t  AK ,t t 2 
1
S
L

 
2
2
2  AK S ,t t  AK L,t t  


 t  i m  1  t  i m  1  2 
 e  S 2  S   e  L 2  L  
1

 
N t  et /   et
2
4 t  i m  1  S  t  i m  1  L  
 e  S 2  e  L 2  




Schrödinger
 e S t  e Lt  2 costme S   L t / 2 
1
KL

   t  t
  S   L t / 2 



S
L
e
 2 costme
4 e

 A K 0,t 0

 
 A K 0,t 0

  0

 I  K ,t t  
 

  0

 I  K ,t t  
 
 

2

KL
t
m  3.5  10
6
eV while m K S  mK L  498 MeV
CP-violation in the
K0-K0
60
system
So for a while the concept of mirror symmetry appeared to be restored if we assume
that for reflections also particles  anti-particles
Angle (KL-beam, )
KL
However:
 

KL  
Br K L    e e   Br K L    e e 
(N+N)/(N++N)
K0
s  u  e e
s  u  e e
10%
5%
0%
KL
-5%
0
t
K K
KL 
2
0
0
KS 
KL 
1  K 0 1  K 0
2  2 2
1  K 0 1  K 0
2  2 2
61
Physics is
CPT
invariant!
Symmetries in (particle) physics

Space-time







Permutations identical particles



Fermions  anti-symmetric under interchange
Bosons  symmetric under interchange
Internal symmetries


Space translations  conservation of momentum
Time translations  conservation of energy
Rotations  conservation of angular momentum
Boosts
Space reflections  parity conservation (violated in weak interactions!)
Time reversal
SU(2), SU(3)
Gauge symmetries


Global  conservation of charges
Local  interactions
62
63
Groups & representations
Group (in mathematics) set transformations G obeying:
Finite:
o
{1,a} with a2=1
o
Permutations of N elements: SN
Infinite:
o
Translations in 3 dimensions T(3)
o
Rotations in 3 dimensions SO(3)
o
Boosts in 3D space-time
Closure:  a,b in G cab in G
Identity:  1 in G with ( a) 1a=a1=a
Inverse:  a in G exists a1 with aa1=a1a=1
Associative:  a,b,c in G: (ab)c=a(bc)
Commutative:  a,b in G: ab=ba (Abelian)
1.
2.
3.
4.
5.
Representation: mapping of elements of G onto matrices obeying Mc=MaMb once c=ab
Ma
a
b
c=ab
Mb
Mc=MaMb
Example: group S3 three representations:
d
G
Example: group {1,a} two representations:
1 (+1)
and 1 (+1)
a (+1)
and a (1)
matrices
If all the matrices can not be broken down into
blocks of smaller dimensional matrices the
representation is called: irreducible
" (1)" " (12)" " (13)" " (23)" " (123)" " (321)"
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
   
1 0
1
0
0 1
0 1
  12
 3
 2

3
2
1
2
   12
  3
 2
3
2
1
2
   12
  3
 2
3
2
1

2

   12
  3
  2
3
2
1

2



64
Rotation group SO(3)
Always the same questions:
Specific for SO(3):
1.
J1, J2 & J3 for rot. x-, y- & z-ax
[Ji,Jj]=iijkJk and [Ji,J2]=0
spin j (J2) and projection m (J3)
multiplicity 2j+1
Find generators infinitesimal transformations and
their commutation relations (i.e. Lie algebra)
Find quantum numbers with which to label states
Find irreducible representationsmultiplet structure
2.
3.
Ansatz with jm :
J 3 jm m jm
2
J jm  j  j 1 jm
introduce :
J   J1  i J 2
and :
[ J 3 , J ]   J 
|j+j>
What can you conclude?
|jm+p>
J 3 J  jm  J   J 31 jm   m1 J  jm

J  jm  jm1
Find C  :

2
2
2
C   jm J  J  jm  jm J  J  jm  jm J  J 3  J 3 jm  j ( j  1)  m  m
J+
|jm+1>
2
|jm>

2
2
2
C   jm J  J  jm  jm J  J  jm  jm J  J 3  J 3 jm  j ( j  1)  m  m
2
 C 
j ( j 1)m( m1)  0
Note: phase convention required!

mmax  j
mmin  j
2j must be an integer!
|jm1>
J
|jj>
|jmq>
65
SO(3) representations & SU(2) symmetry
j  0:
J 3   0
J    0
1
j :
2
1  1 0 
J3  

2  0 1
SU(2)SO(3)
j  1:
0
0
0
0

J  
0
00
1
0

J  
1

0
1  0 1
J1  

2  1 0

 1
J3   0

0
J    0

0

1
0
J  0

0
0
0

0
1  0 i 
J2  

2  i 0 

2

0 
2

J  


0
0
0
0
2
0
0
0

1 1

2 2

1 1

2 2
 11
0   10

0
2 0

1 1
Addition of angular momenta (j1j2): j1j2= j1j2j1j2+1j1j2+2……j1+j21j1+j2
Example:
1
2
1
 2  0 1
two sets of wavefunctions : jm( j1 j 2 )
10  J  11 
2
j 1 states: 11  1 1  1 1
22
j 0 state: 00  10
22

00 
11
22
 11
2 2
2
1 1

2 2
 11
22
2

1 1

2 2

11
22
and
 11
2 2
2
j1m1 j 2m2  j1m1  j 2m2
11  J  10  1  1  1  1
2 2
2 2
2
 11
22
2
(phase convention used!)
Clebsch-Gordon coefficients
66
67
Constructing the SO(3) representations
General principle:
• Start with infinitesimal transformations: x-, y- and z-axis rotations:
X :
RX
1
 0

0
0
cos 
sin 

sin  

cos  
0
1
 0

0
0
1

 1
    0
 
1  0
0
0 0
0
1
0
0
0
0   0
 
1  0


0
   1   0


0 
0
0
0
0
1

1  1 

0
0
J1
• Find the matrices corresponding to these infinitesimal rotations
j  0:
j
1
:
2
J 3   0
J3 
J    0
1  1 0 


2  0 1

j  1:
 1
J3   0

0
0
0
0

0

1
0
J    0

J  
0 1
0

0
1  0 1
J1  

2  1 0
0
J  0

0
2
0
0
00

J  
0 0
1

0
1  0 i 
J2  

2  i 0 

2

0 
0

J  


0
2
0
0
0

1 1

2 2

1 1

2 2
 11
0   10

0
2 0

1 1
68
Constructing the SO(3) representations
• Construct the macroscopic rotations via exponentiation:
69
Isospin
Heisenberg (1932):
1. proton & neutron states
of one particle: the nucleon
2. physics invariant under
p  n transformation i.e.
an internal SU(2) symmetry
All this: strong interaction only!
q  1
proton :
s 1
2
m 938.28 MeV
p
1 1

2 2
q 0
neutron :
s 1
2
m 939.57 MeV
n
1 1

2 2
Name: “isospin” analogous to the normal (Euclidean) spin
Isospin multiplets:
I=1/2
I=1
I=3/2
nucleon (mp=938.28 MeV, mn =939.57 MeV):
pions (m0=135 MeV, m =140 MeV):
-baryons: (m1232 MeV):    
p  12  12
   11
3 3

2 2
 
n
 0  10
3 1

2 2
0 
1 1

2 2
   11
3 1

2 2
 
3 3

2 2
70
Isopin examples
Two nucleon system:
Only stable 2-nucleon system: deuterium 2H
What is the isospin of deuterium?
if I  1 : pp 
if I  0 : pn 
1 1

2 2
1 1

2 2
, pn 
1 1

2 2
1 1

2 2
 11
2 2
1 1

2 2
1 1

2 2
 11
2 2
2
1 1

2 2
1 1

2 2
, nn 
1 1

2 2
1 1

2 2
 this corresponds to deuterium: I=0:
2
Energy levels in “mirror” nuclei:
nuclei with same total number of nucleons
but differences in number of protons/neutrons
What can you say about the isospin of these levels?
Branching ratio’s:
Iz=1
Iz=0
Iz=+1
4+
4+
4+
 I  1
2+
2+
2+
 I  1
0+
0+
0+
 I  1
18O
1+
18Ne
 I  0
67%
18F
What can you say about the relative occurrence of: p0 and

  p : M 
0
3 1 1 1

 10 
2 2 2 2
2
3


  n : M 
33%
n+ decays?
3 1 1 1

 11 
2 2 2 2
1
3
hence
0

  p
  2


n


71
Charge conjugation
particle  anti-particle
Cp  p
Define operation “C” which converts
a particle into its anti-particle:
particle  anti-particle
particle doublet: 

Cn  n
C e  e
1
C
1 1 C
C C C2=1
Not often particle = anti-particle. Exceptions:
photon:
particle/anti-particle
bound states
C = 

n
anti-particle doublet:
Group with two elements because C2=1
 C=C1=C†
C hermitean
eigenvalues +1 and 1
p
 n 
p
With this choice
Particles and anti-particles transform
alike under rotations in iso-space
e.g. rotation of  around x-axis
  p    0
  n   1
1  p   n 
 p  n 

     

0   n    p 
  n  p 
 n    0
  p   1
1 n    p 
 n  p 

     

0   p   n 
  p  n 
72
Isospin at the quark level:
u-quark & d-quark (flavors)
Messy situation in the sixties (compare pre-Mendeleev chaos atomic elements):
Many particles: p, n, , , , , , , , , , , , ……
e
p
too often large
angle scatterings
 proton has sub-structure
Progress:
1. Theory: ordering in terms of quark sub-structure: u-, d- and s-quarks (flavors)
2. Experiment:
a. Discovery of sub-structure in the nucleon (compare Rutherford atom)
b. Discovery of a fourth quark flavor in 1974: charm
Copy isospin to u- and d-quarks: quark doublet :
Baryons : qqq bound states
1 1 1 1 1 3
isospin :     
 doublets or quadruplets
2 2 2 2 2 2
example :  - baryon :
-

  ddd


  -  udd  dud  ddu 


3
 0 uud udu  duu 
 

3



 uuu


 u , q  23 

1
d
,
q



3
anti - quark doublet :
 d , q  13 

2

u
,
q



3
Mesons : qq bound states
1 1
isospin :   0  1  singlets or triplets
2 2
example :  - meson :
  -   du

  0  uu  dd

2
   ud
 






“Eightfold way”:
73
u-quark, d-quark & s-quark (flavors)
Gelmann & Zweig (1963):
1.
hadrons built with three
constituents: u-, d-, & s-quarks:
1.
2.
2.
Mesons: qq
Baryons: qqq
physics invariant under uds
transformation i.e. an internal
SU(3) symmetry
 u 
d

 u:

 d :
 s:
u quark, q   3 
2
d quark,
s quark,
1
q  3
1
q  3



Note: do not carry this beyond uds since the mass
differences between the quarks become way too large
All this: strong interaction only!
What does this buy you?
1.
2.
Order in the zoo of particles; compare p and n as two states of the nucleon
Expression of properties (masses, magnetic moments, …) in terms of a few parameters
But realize: SU(3) flavor is badly broken (mumdms) and applies only to strong interaction!
This opposed to exact color SU(3) symmetry of strong interactions we will encounter later!
74
SU(3) symmetry
J
For SU(2):
Multiplets |jm>, 2j+1 states; one traceless generator: J3
Step operators within multiplet: J  J1  iJ2
Build all representations from 2D (j=1/2) representation
J1
1
2
 1  12 
0
 1
1

0
J 2
1
2
 2  12 
0
 i
i 

0
J 3
1
2
j=1/2
J
 3  12 
0
1
j=1/21/2
j=0

1
0
j=1
For SU(3):
Three SU(2) sub-groups; two traceless generator: 3 and 8
Three step operator sets; 1  i2 4  i5 6  i7
Build all representations from 3D (“3 and 3”) representations
0
1   1

0
1 0
i 0 
0 1
0 i 
0 0
0
0
0
0
0
0
0
0 2   i


0
0
d
0
0
0 4   0


0
1
s
u
s
3
0
0
0 5   0


0
i
d
0
0
0  6   0


0
0
qq
u
1
0
1 7   0


0
0
0
0
i
0
0 i 

1 0 0 
1
0 1 0 
1 0   8 
3


0 0
 0 0 2 
0
0
combinatio ns : 3  3  1  8
qqq combinatio ns :
3
1
3   0

0
3  3  3  1  8  8  10
Mesons
qq
75

combinatio ns : 3  3  1  8

=
So what does this all mean:
•
Mesons in multiplets with 1 or 8 similar particles:
•
Same intrinsic spin, total spin, parity, …
•
Different quark decomposition
•
Compare: energy levels H all refer to H; same for particles in same multiplet
•
Symmetry not exact  slight mass differences, etc.
How do you find the quark wave functions?
1. Start with arbitrary one (normalized)
2. Apply step operators until you exhaust the multiplet!
Examples:
Octet:

K K
S  0 : ,'    0  

K K
0
*
*0
K K
S  1 : ,
 0 
*
K K
 ds
0
*0
Singlet:
uu  dd  ss
3
dd  uu
2
 us
 du
ud
dd  uu  2 ss
6
 su
sd
76
Meson masses
If SU(3) would be exact all particle masses within a multiplet identical
SU(3) symmetry broken by:
1. u-, d- and s-quark mass differences (singlet+octetmixed “nonet”)
In addition binding energy has contribution from quark spin-spin interaction

2
h
S 1: m1 m 2  A
 
s 1s 2 
4m1m 2
Ansatz : M ( meson )  m1  m 2  A

m1m 2 
3h 2
S 0: m1 m 2  A

4m1m 2
Fit this  m =m =310 MeV and m =483 MeV
u
d
Meson nonet
S=0
s
fit
mass
exp
mass
Meson nonet
S=1
fit
mass
exp
mass
140
138
 (3)
780
776
K (2x2)
484
496
K (2x2)
896
892
 (1)
559
549
 (1)
780
783
’ (1)
---
958
 (1)
1032
1020

(3)
Baryons
qqq combinatio ns :
77
3  3  3  1  8  8  10


= 


Same recipe as for mesons; bit more complicated
Singlet:
udd  dud  ddu
3
uds  dus  usd  sdu  dsu  sud
6
uuu
ddd
sdd  dsd  dds
3
Decuplet:
udd  dud
2
Octets:
need a convention!
“A” asymmetric in 12
“S” symmetric in 12
dsd  sdd
2
“A”
uds  dus  usd  sdu  dsu  sud
6
dss  sds  ssd
3
dsu  sdu  usd  sud  suu
2
usu  suu
2
uss  sus
2
suu  usu  uus
3
uss  sus  ssu
3
“S”
udu  duu
2
2(ud  du ) s  (us  su )d  ( sd  ds )u
12
dss  sds
2
duu  udu  uud
3
sss
The  baryon discovery
At the time of the proposal by Gelmann and Zweig were not all multiplets complete!
Nice example: baryons in decuplet had one missing member; characteristics were predicted!








0



0

1232 MeV

1385 MeV

0

1533 MeV
?
0p
1680 MeV
000
0
e+e
Bubble chamber experiment:
s
sss s
s
K+p  +K++K0
(strong interaction: s-quarks conserved)
 0
78
79
Adding the intrinsic spin
Mesons:
1 1
  0 1
2 2
00 
11  
Baryons:
10 
 
2
 
2
S=0
11  
S=1
1 1 1 1 1 3
    
2 2 2 2 2 2
Symmetric
103/2
8A1/2A+8S1/2S
  
3 1
 
3
2 2
3 3
  
2 2
" A" :
Anti-symmetric
13/2
8A1/2S+8S1/2A
" S" :
 
1 1
 
2
2 2
  2
1 1
 
6
2 2
  
3 1
 
3
2 2
 
1 1
 
2
2 2
3 3
  
2 2
S=1/2
  2
1 1
 
6
2 2
S=1/2
Remark: of course this only refers to the intrinsic spin (S) of a hadron.
In addition any hadron (meson and baryon) will have orbital spin (L) due to the quark motions.
S=3/2
80
Restoring anti-symmetry: color
little problem:

s
s
s
++
u
u
u
intrinsic spin :
quarks :
sss
3 3

2 2
   symmetric
 symmetric
   symmetric
intrinsic spin :
3 3

2 2
quarks :
 symmetric
uuu
particle with half-integer spin obey Fermi-Dirac statistics  must be anti-symmetric
brute force solution: invent hidden degree of freedom: color
physics:
1.
2.
3.
4.
q

q
q
q
invariant under color (RGB) transitions; exact SU(3) symmetry
23
compare the color charge and the electromagnetic charge
instead of the photon now eight force carriers: gluons: RB, RG, RB, RG, BG, BG, RR, BB, GG
hypothesis: all hadrons in color singlet state:
baryons: multiply with: (RGBRBGGRBBGR+BRG+GBR)/6
mesons: multiply with: (RR+BB+GG)/3
(symmetric in color)
(anti-symmetric in color)
81
Color in experiments
R
 e
quarks
 
 
e  qq
 
e e  
udsc
uds
udsc no color
decay width
color
no color
Z decay probabilities
Z e  e 
81 MeV
81 MeV
Z 
162 MeV
162 MeV
Z uu
276 MeV
93 MeV
Z  dd
360 MeV
120 MeV
Example: , p, and n wave functions
Conventions:

   sss  SSS
p 
82
quark with spin up: qQ
quark with spin down: qq
1  udu  duu    udu  duu  2 uud    2  





2
2
2
6
6


11
 udu 3 3    2  duu 3 3     2  2 uud    2 
26

11
 udu  422  duu  24 2  uud  2 2 4 
26

11
 4UdU  2 uDU  2UDu  2 DuU  4 dUU  2 DUu  2UuD  2 uUD  4UUd
26

1
 2UdU  uDU UDu  DuU  2 dUU  DUu UuD  uUD  2UUd
18

1
 2UdU  uDU UDu
18
n 
1
18
 2 DuD  dUD DUd
 2 dUU  DuU  DUu
 2 uDD UdD UDd
Wave functions full (123) symmetry
Complete wave function: multiply with:
 2UUd UuD  uUD

 2 DDu  DdU  dDU

(RGBRBG GRB BGR +BRG +GBR)/6
83
Baryon magnetic moments
Magnetic moment  of a particle is obtained using:
  jj
constituents
  i  3i jj
i
For a S=1/2 point (Dirac) particle with mass m and electric charge q one finds:
q

2m
With the quark-flavour wave functions one can calculate magnetic moments of mesons
and baryons with as only unknown parameters the u-, d- and s-quark masses!
Proton:
 p  p

3
 
i 1 i 3 i
p
2UdU  uDU UDu
 2 dUU  DuU  DUu
18
 2UUd UuD  uUD 3
2UdU  uDU UDu
 
i 1 i 3 i
4 u   d
3
 4( 2  u   d )  (  d )  (  d ) 
18
3
Neutron (simply ud):
 n  n
3

i 1 i 3i
n 
3
4( 2 d  u )(  u )(  u )  4 d  u
18
3
n
2

(exp : 0.68)
3
p
 2 dUU  DuU  DUu
 2UUd UuD  uUD
18
Assume now:
 mumdm  u=e/3m and d= e/6m
4 u   d
e

3
2m
4 d   e
e
n 

3
3m
p 
84
85
86
87