568 10 Particle Physics – II
10.30 In the initial stateT(K)=^1 / 2 andT(p)=^1 / 2 , so thatI =0or1.Inthe
first two reactionsI=2, 1 or 0; so that these reactions can proceed through
I=1 or 0. In the third reactionT(Λ)=0 andT(π^0 )=1sothatI=1 only.
Since the resonance does not go through, the conclusion is that the resonance
hasT=0.
10.31 K−+p→Σ++π− (1)
K−+p→Σ^0 +π^0 (2)
K−+p→Σ−+π+ (3)
K−+n→Σ−+π^0 (4)
K−+n→Σ^0 +π− (5)
The initialK−+psystem of two particles ofT=^1 / 2 hasI 3 =0 and
consists equally ofI=0 andI=1 state. The finalΣ+πstate will be|ψ〉=1
√
2
(a 0 |φ( 0 , 0 )〉+a 1 |φ( 1 , 0 )〉)ForI 1 =I 2 =1, referring to the C.G. Coefficients (Table 3.3) we can write|φ( 0 , 0 )〉=1
√
3
(∣
∣Σ+π−〉
−
∣
∣Σ^0 π^0〉
+
∣
∣Σ−π+〉)
|φ( 1 , 0 )〉=1
√
2
(∣∣
Σ−π+〉
−
∣
∣Σ+π−〉)|ψ〉=1
√
2
[(
a 0
√
3−
a 1
√
2)∣
∣Σ+π−〉−√a^0
3∣
∣Σ^0 π^0 〉+(
a 0
√
3+
a 1
√
2)∣
∣Σ−π+〉]
∴
(
Σ+π−)
:
(
Σ^0 π^0)
:
(
Σ−π+)
=
1
2
(
a 0
√
3−
a 1
√
2) 2
:
a^20
6:
1
2
(
a 0
√
3+
a 1
√
2) 2
The reactions ofK−with n go throughI=1 only. SinceI 3 =−1 and the
finalΣπstate isa 1 |ψ( 1 ,− 1 )〉=√a^12[∣∣
Σ−π^0〉
−
∣
∣Σ^0 π−〉](
Σ−π^0)
:
(
Σ^0 π−)
=
a 12
2:
a 12
2
If K−is incident with equal frequency on p and n, thenΣ−+Σ+=
1
2
(
a 0
√
3−
a 1
√
2) 2
+
1
2
(
a 0
√
3+
a 1
√
2) 2
+
a 12
2=
a 02
3+a^21Σ^0 =
a 02
6+
a 12
2
It follows thatΣ−+Σ+= 2 Σ^010.32 B+→ω^0 +π+
Jp 1 − 0 −
I01