9 Symmetry Violations and Quark Flavour Physics
543
|K L =
1
1 + |ε − δ| 2
(|K 2 + (ε − δ)|K 1 ) .
(9.74)
with a CPT violating parameter δ.
9.5.2 Isospin Decomposition
In K S,L → 2π decays, the angular momentum of the pions vanishes. The spatial
part of the wave function is therefore symmetric, and since pions are bosons, the
isospin wave function must be symmetric too. The two symmetric combinations of
two I = 1 states have I = 0 and I = 2, and the four transition amplitudes that exist
are
|K S ,
|K S ,
|K L ,
|K L .
(9.75)
These can be reduced to three complex numbers by normalizing to the amplitude
|K S
ε 0 =
|K L
|K S
,
(9.76)
ε 2 =
1
√
2
|K L
|K S
,
(9.77)
ω =
|K S
|K S
.
(9.78)
The experimentally observable quantities are
η +− =
+ π − |T |K L
+ π − |T |K S
,
(9.79)
η 00 =
0 π 0 |T |K L
0 π 0 |T |K S
,
(9.80)
δ L =
Γ (K L → π − l + ν) − Γ (K L → π + l − ¯
ν)
Γ (K L → π − l + ν) + Γ (K L → π + l − ¯
ν)
.
(9.81)
Relating the isospin states to the physical 2π states
=
1
√
3
− π
+
| −
1
√
3
0 π
0
| +
1
√
3
+ π
−
| ,
(9.82)
=
1
√
6
− π
+
| +
2
3
0 π
0
| +
1
√
6
+ π
−
| .
(9.83)
543
|K L =
1
1 + |ε − δ| 2
(|K 2 + (ε − δ)|K 1 ) .
(9.74)
with a CPT violating parameter δ.
9.5.2 Isospin Decomposition
In K S,L → 2π decays, the angular momentum of the pions vanishes. The spatial
part of the wave function is therefore symmetric, and since pions are bosons, the
isospin wave function must be symmetric too. The two symmetric combinations of
two I = 1 states have I = 0 and I = 2, and the four transition amplitudes that exist
are
|K S ,
|K S ,
|K L ,
|K L .
(9.75)
These can be reduced to three complex numbers by normalizing to the amplitude
|K S
ε 0 =
|K L
|K S
,
(9.76)
ε 2 =
1
√
2
|K L
|K S
,
(9.77)
ω =
|K S
|K S
.
(9.78)
The experimentally observable quantities are
η +− =
+ π − |T |K L
+ π − |T |K S
,
(9.79)
η 00 =
0 π 0 |T |K L
0 π 0 |T |K S
,
(9.80)
δ L =
Γ (K L → π − l + ν) − Γ (K L → π + l − ¯
ν)
Γ (K L → π − l + ν) + Γ (K L → π + l − ¯
ν)
.
(9.81)
Relating the isospin states to the physical 2π states
=
1
√
3
− π
+
| −
1
√
3
0 π
0
| +
1
√
3
+ π
−
| ,
(9.82)
=
1
√
6
− π
+
| +
2
3
0 π
0
| +
1
√
6
+ π
−
| .
(9.83)
