10.3 Solutions 581
The initial state is a mixture ofI=0 andI=1 states. The final state can
exist only inI=1.|Λ,π^0 >=1
√
2
[a 1 | 1 , 0 >+a 0 | 0 , 0 >]λ(Ξ−→Λ+π−)
λ(Ξ^0 →Λ+π^0 )=
1
(1/
√
2)^2
= 2
10.72 Let us introduce a fictitious particle called spurion ofI= 1 / 2 ,I 3 =− 1 /2,
and neutrally charged and add it on the left hand side so that the weak decay
is converted into a strong interaction in whichIis conserved. The reactions
withΣ+can proceed inI= 1 /2 and 3/2 channels while that withΣ−through
pureI= 3 /2 channel. We can then write down the amplitudes for the initial
state by referring to the C.G. coefficients for 1× 1 /2:
Σ++s→√
1
3
a 3 +√
2
3
a 1Σ−+s→a 3wherea 1 anda 3 are theI= 1 /2 andI = 3 /2 contributions. Similarly, for
the final state the amplitudes arenπ+→√
1
3
a 3 +√
2
3
a 1nπ−→a 3pπ^0 →√
2
3
a 3 −√
1
3
a 1Therefore,a+=〈
Σ+
∣
∣nπ+〉=^1
3 a2
3 +2
3 a2
1
a−=〈
Σ−
∣
∣nπ−〉=a^2
3
a 0 =〈
Σ+
∣
∣pπ^0 〉=√
2
3 a2
3 −√
2
3 a2
1
givinga++√
2 a 0 =a−10.73 Converting the decay into a reaction
S+Λ→p+π−
→n+π^0wheresis the fictitious particle of isospinT=^1 / 2 andT 3 =−^1 / 2. Because
ΛhasT=0 andT 3 =0, the initial state will be a pureI =^1 / 2 state with
I 3 =− 1 /2. In the final state the nucleon hasT=^1 / 2 and pion hasT=1,
so thatI= 3 /2or^1 / 2 .IandI 3 conservation require that the final state must
be characterized byI=^1 / 2 andI 3 =−^1 / 2. Looking up the table for C.G.
Coefficients for 1× 1 /2 (given in Table 3.3) we can write down