564 10 Particle Physics – II
Therefore,σ 1 =C|M 3 |^2
whereC=constant
For reaction (2) we have the mixture ofI= 1 /2 andI= 3 /2 states.
Refering to the table for 3/ 2 × 1 /2 C.G. coefficients we can write|ψi〉=∣
∣ψf〉=√
1
3
∣
∣
∣
∣φ(
3
2
,−
1
2
)〉
−
√
2
3
∣
∣
∣
∣φ(
1
2
,−
1
2
)〉
Therefore, σ 2 =C〈
ψf|H 1 +H 3 |ψi〉 2
=C
∣
∣
∣
∣
1
3
M 3 +
2
3
M 1
∣
∣
∣
∣
2For the reaction (3) we have|ψi〉=√
1
3
∣
∣
∣
∣φ(
3
2
,−
1
2
)〉
−
√
2
3
∣
∣
∣
∣φ(
1
2
,−
1
2
)〉
∣
∣ψf〉
=
√
2
3
∣
∣
∣
∣φ(
3
2
,−
1
2
)〉
+
√
1
3
∣
∣
∣
∣φ(
1
2
,−
1
2
)〉
Therefore,σ 3 =C∣
∣
∣
√
2
9 M^3 −√
2
9 M^1∣
∣
∣
2The ratio of cross-sections areσ 1 :σ 2 :σ 3 =|M 3 |^2 :1
9
|M 3 + 2 M 1 |^2 :
2
9
|M 3 −M 1 |^2
Ifa 1 / 2 <<a 3 / 2 ,M 1 <<M 3 ,σ 1 :σ 2 :σ 3 =9:1:2
And ifa 1 / 2 >>a 3 / 2 ,M 1 >>M 3 ,σ 1 :σ 2 :σ 3 =0:2:110.19 From the knownπ−Nscattering cross-sections in terms of the amplitudes
a 3 / 2 anda 1 / 2 obtained in Problem 10.18 we can construct a diagram for the
amplitudes in the complex plane, Fig. 10.3
Fig. 10.3Diagram of
amplitudes
√
σ+=|a 3 |
√
σ−=1
3
|a 3 + 2 a 1 |
√
2 σ^0 =2
3
|a 3 −a 1 |where for brevity we have writtena 1 fora 1 / 2 anda 3 fora 3 / 2
From the triangle the required inequality follows from the fact that the sum
of two sides is equal to or greater than the third side.