Download CP pheno 1/5 - CP violation: from quarks to leptons

Italian Physical Society
International School of Physics
“Enrico Fermi”
CP Violation
Topics to be covered
Lecture 1
1. Introduction
• Why study CP violation?
• Grand view - Sakharov's ideas
2. Symmetries
• Mechanics
• Electrodynamics
• Quantum mechanics
time reversal operator
anti-unitary operator
Lecture ２
•
•
•
•
•
Edm
Krammer’s degeneracy
Particle physics
How do you measure pion spin?
How do you show that pion is a
pseudoscalar?
• G-parity
• C,P,T in particle physics
Lecture 3
3. K meson system
•
•
•
•
τ-θ Puzzl
Weak interaction
CPT
Mixing
４．CP violation
• Asymmetry in partial widths
• Role of final state interaction
• Hyperon decays
• Watson's theorem
Lecture 4
•Regeneration
•Explain Eq. 7.8 and Fig 7,.1
•Discovery of CP violation Cronin-Fitch experiment
•Direct
•indirect
• In this section, we refrain from deriving the expression
for ε and ε' etc.
Lecture 5
CP violation in B decays
Towards building the B factory
EPR paradox
Experiments at the B factory
without much about the KM
formalism
Its goes back the basic question asked by man.
Why do we exist?
radius 104 cm
100Km
109
1013
104
We are here
1 billion
years
Supernova and heavy
elements
1026
1m
100
Stars and milky way
30
Temp(℃) 10
1010
1015
Light elements
1034
Nucleosythesis of
Helium
protons and neutrons
43
Time(sec) 10
5 billion years
13.7 billion years
270
1028 cm
Generation of light elements
Black-body radiation
The universe is filled with ３°Ｋ
（－２７０℃）photons
Expanding universe
Measuring the speed of expansion
universe
Theory predicts universe and anti-universe
anti-universe
1. You can see Armstrong’s foot step.
The moon is not made out of
anti-matter
2.No primary anti-particle in cosmic rays
BESS experiment KEK/IPNS
10-1
He/He limit (95% C.L.)
Smoot et al. (1975)
10-2
Evenson (1972)
Smoot et al. (1975)
10-3
Antihelium/helium flux ratio
Evenson (1972)
Aizu et al. (1961)
Badhwar et al. (1978)
Golden et al. (1997)
10-4
Buffington et al. (1981)
10-5
Ormes et al. (1997) BESS-95
T. Saeki et al. (1998) BESS-93~95
10-6
10-7
J. Alcaraz et al. (1999) AMS01
M. Sasaki (2000)
BESS-93~98
BESS-1993~2000
Preliminary
BESS-Polar (2003, 20 days)
PAMELA (2002~, 3 years)
10-8
AMS02 (2004~, 3 years)
10-9
10-1
1
10
102
Rigidity (GV)
KEK/IPNS Yoshida
103
3.You see galaxy collisions but
collisions of galaxy and anti-galaxy have been found
Really absent？
Alpha Matter Spectrometer
Space Shuttle NASA program
SEMINAR – EXPERIMENTS ON ANTIMATTER
SEARCHES IN SPACE
Battison
We have to understand how CP is violated in
the fundamental theory
t  1016 sec
r  1029 cm
T  270C
200 / cm 3
1080 p , n
0
p ,n
Why there is no anti-universe
三田一郎 名古屋大学大学院 理学研究科
1. Anti-baryon has to disappear. So, there has to
be baryon number violation
2. Only anti-baryon has to disappear. So, there
has to be CP violation
3. Created asymmetry will be washed out if the
universe is in equilibrium. So, baryon has to be
created out of equilibrium.
t  1016 sec
r  1029 cm
T  270C
X ® qe
X ® qe
N = 10
88
N = 10
88
These annihilations
Produce light
+ 10
80
200 / cm 3
1080 p , n
0
p ,n
N - N = 1088 + 1080 - 1088
N +N
1088 + 1080 + 1088
= 10- 8
strings
The origin of quark masses
The origin of CP violation
So, let us go back to symmetries
Mirror image
x  x
Time reversal
t  t
You can’t tell which is the
original
The symmetry is imbedded in
your brain
Newton’s Equation
2
d r
F m 2
dt
Show that this equation is invariant under parity
Show that this equation is invariant under time reversal
Maxwell Equation
  E  4
1 E 4
 B 

J
c t
c
B  0
1 B
 E 
0
c t
Parity
  E  4
1 E
4
 B 

J
c t
c
B  0
 E 
1 B
0
c t
F  ma
F  q( E  v  B)
  
E  E
BB

Charge conjugation

  E  4
1 E 4

J
c t
c
B  0
 B 
 E 
1 B
0
c t
F  ma
F  q( E  v  B)
E  E
B  B
  
J  J
Time Reversal
  E  4
1 E 4
 B 

J
c t
c
B  0
1 B
 E 
0
c t

d
d

dt
dt
EE
F  ma
B  B

F  q( E  v  B)
J  J
Our cells are controlled by
electrodynamics.
Why don’t we get younger?
Symmetries and quantum mechanics
Ω
i
[H,Ω]=0
 i
ψ and　Ω ψ are both solutions to the Schorodinger Eq.

i
|  ; t  H |  ; t 
t
 1
i
  |  ; t  H  1 |  ; t 
t

i
 |  ; t  H  |  ; t 
t
P:
C:
T:
So, you can’t tell the difference!
r  r
e  e
t  t
ψ ψ = ψ Ω†Ω ψ  Ω†Ω=1
X x x x
XP x   xP x
P x  x
P x = eid - x
so we take e i d = + 1
P2 =1, P=P-1, P=P†
P 1 XP   X
XP x   PX x   xP x

|  ; t  H |  ; t 
t
 1
Pi
P P |  ; t  PHP 1P |  ; t 
t

i
P |  ; t  HP |  ; t 
t
i
Show that:
(1) If [P,H]=0
xHx = - xH- x
H( x ) = H( - x)
¶
(2)
ih P |y ,t = HP |y ,t
x H y  H (x ) (x  y )
¶t
¶
x ih P |y ,t = ò dy x H y y P |y ,t
¶t
¶
then ih y (- x ,t ) = H(x)y (- x ,t )
¶t
(3)
y
±
=
1 é
y ±Py ù
ë
û is also a solition
2
Show that
(1)
- x y
±
= x P† y
±
= x Py
±
= ± x y
±
(2) y ± (- x ) = ± y ± ( x )
&
&
(3) y ± ( x ) is also solution to the Schordinger
equation.
p
2
2
e
e2


H 

A  p  p A 
A  e
2

2m 2mc
2mc
If charge of a particle is flipped, and the external
all fields are flipped, the motion is invariant.
e  e
A  A
  
CHC
1
H
 x, p  i
T  x, p T
1
   x, p   Ti T
1
i
Time reversal and quantum mechanics
|b>=O|b>
<a|O†O|b>=<a|b>
|a>=O|a>
|<a|b>|=|<a|O†O|b>|=|<a|b>|
<a|O†O|b>=<a|b>*
O=UK K is a complex conjugate operator
K α|a>+β|b>  =α *K|a>+β*K|b>
K † acts always to the left
[α <a|+β<b|]K † =α* <a|K † + β* <b|K †
2
K =1
-1
K =K
Ｔ＝Ｋ
Show that x,p =i is consistent with time reversal.
Transformation of Schrodinger equation under T

|ψ(t)>=H|ψ(t)>
t
 -1
Ti
T T|ψ(t)>=THT -1T|ψ(t)>

t
i
|ψ(-t)>=H|ψ(-t)>

t
-i
T|ψ(t)>=HT|ψ(t)>
t
T|ψ(t)>=|ψ(-t)>
i
Transformation of the wave function under T
ψ(-t) =ò dx' x' x' ψ(-t)
T ψ(t) =T ò dx' x' x' ψ(t) =ò dx' x' x' ψ(t) *
ψ(x,-t)=ψ(x,t)*
T ψ(t) =T ò dx' x' x' ψ(t) =ò dx' x' x' ψ(t) *
T f (t) =T ò dy' y' y' f (t) =ò dy' y' y' f (t) *
f (t) T †T ψ(t) =ò dx 'dy ' y ' f (t ) y ' x ' x' ψ(t) *
f (t) T †T ψ(t) = ψ(t) f (t)
1. If T=K, from the definition of
operator P, show that T-1PT=-P.
2. From the definition of J=rxP, show
that T-1JT=-J
 Vx '   cos sin  0  Vx 
 V '     sin  cos 0  V 
 y  
 y 
V '   0

0
0 
 z  
 Vz 
 Vx '   cos 0  sin   Vx 
V '    0

1
0 
 y  
 Vy 
 V '   sin  0 cos  V 
 z  
 z 
0
 Vx '   1
 V '    0 cos
 y  
 V '   0  sin 
 z  
0  Vx 

sin  
V
y
 

cos 
 Vz 
around the z axis
around the y axis
around the x axis
 0 i 0 
0 0 0 
 0 0 i
J z   i 0 0  J x   0 0 i  J y   0 0 0 
0 0 1
0 i 0 
 i 0 0 






TJT 1  J if T=K
Rotation operator
e
iJ z
 0 i 0 
0 0 0 
 0 0 i
J z   i 0 0  J x   0 0 i  J y   0 0 0 
0 0 0
0 i 0 
 i 0 0 






But in quantum mechanics
first question we ask:
what operator can we diagonalize
together with H.
[H , J 2]  0
[H , J ]  0
So we often use:
[Ji , J j ]  i ijk Jk
1 0 0 
0 1 0
 0 i
1 
1 



Jz   0 0 0  Jx 
1 0 1  Jy 
i 0
2 
2 
 0 0 1 



0 1 0
0 i
0
i 
0 
Then we have to change T=K
T  exp  i  J y  K