CP Violation
in
Neutral K-Meson Decay
[Secs 16.8 Dunlap]
CP Violation
James Cronin (1931- )
Val Fitch (1923- )
In 1980 Cronin and Fitch received the Nobel Prize in Physics for
and experiment they completed in 1964: Cronin and Fitch were
physicists from Princeton University, working at Brookhaven
National Laboratory. What did they discover?
The detection of CP violation in the decay of neutral K mesons (K0).
Quark constituents of K0 and K0
From the quark model we know that the , K0 an K0 are :
K ( 4 9 8)  d s
0
and
K (498)  sd
0
which have strangeness S=+1 and S=-1 respectively. However,
because strangeness is not conserved in weak interactions,
these states can be converted into each other
What is special about K0 – K0 system?
This interchange is in marked contrast to most other
particle-antiparticle systems, for which such transitions
are forbidden due to some major conservation law.


For example  and  have opposite charge, while
n and n have opposite Baryon no.
K
0
and K
0
K0 and K0 are not the free-space
eingenstates of the K0 system
K0 and K0 are the eigenstates of the (K0 system) when “molded”
by the STRONG force – under production in a target
H Strong K
0
 ES K
0
H Strong K
0
 ES K
0
However when the K0 or K0 are out in “free space” the WEAK
force is also at work
PRODUCTION
DETECTED
by STRONG
Interaction
by STRONG
Interaction
Out in Free Space
What are “free space” eigenstates?
Thus in free space the K0 system will be expected to be
eigenstates of the WEAK interaction. That is they should
have natural eigenstates with good CP eigenvalues:
0
0
( Hˆ Strong  Hˆ W eak ) K
 E SW K
This is because we expect
Hˆ
.Cˆ Pˆ  Cˆ Pˆ . Hˆ
W eak
W eak
0
But K0 and K0 are not eigenstates of CP !
CP K
CP K
0
0
 C ( P ds  C sd  sd  K
0
 C ( P sd  C ds  ds  K
0
The CP operator just converts K0 ↔ K0
What are “free space” eigenstates?
No, The “Free Space” eigenstates are:
KS 
1
KL 
1
 K0  K0 

2 
 K0  K0 

2 
CP=+1
CP= -1
Can you see why?
What is the meaning of the “S” and
the “L”. It stands for Short-lived and
Long-lived. Thus we can conclude
that when the weak interaction is
operational the there are two natural
eigen-states- KS and KL
Why is the decay time different for
KS and KL ?
KS   
CP=+1
(0.89 × 10-11s)
KL    
CP= -1
(5.2 × 10-8 s)
It is a little difficult to prove (although you can find the
proof in many books) that these are the CP values of
these final states. Lets assume it to be the case.
It is the difference in energy available (output phase
space) for the products that causes the difference in
lifetimes.
These states also differ very slightly different in mass
(3.5x10-12MeV/c2)
How we detect the K0
K  p
0
K
0

1
2
K
0

1
2


 K S  K L 
 K S  K L 
CP violation
Cronin and Fitch in 1964 discovered CP violating
decays have been observed at the 0.3% level
KL   
indicating that CP is not perfectly conserved in the weak
interaction. CPT invarience is expected though under
relativistic quantum electrodynamics.
This implies a breaking of T conservation too, which has
implications with regard to why we live in a universe that
is largely matter and not anti-matter!