566
10 Particle Physics – II
10.25 (a) From Problem 10.18, we have the result
a 3/2
2 ∝ 190 mb
1
√
3
a 3/2 + 2a 1/2
2
∝ 70 mb
Dividing one by the other, and solving we find a 1/2 = 0.0256 a 3/2 . Thus
the amplitude a 1/2 is negligible. The resonance is therefore characterized
by I = 3/2.
(b) The fact that π
+ p scattering cross-section has a fairly high value at
p = 230 MeV/c implies that a 3
2
amplitude is dominant.
10.26 From the analysis of π-N scattering (Problem 10. 18) the ratio
σ (π
+
+ p → Δ)
σ (π − + p → Δ)
=
a 3
2
2
1
√
3
a 3
2
+ 2a 1
2
2
If we put a 1/2 = 0, we get the ratio as 3.
10.27 The analysis is identical with that for π p reactions (problem 10.18). The
Σ’s are isospin triplet (T = 1) with the third components T 3 = +1, 0, −1
for Σ
+
, Σ
0
, Σ
− similar to π
+
, π
0
, π
− . Further, the K
+ and K
0 form
a doublet (T = 1/2) with T 3 = +1/2 and −1/2, analogous to p and n.
Therefore the result on the ratio of cross-section will be identical with that
for pion – nucleon reactions as in Problem 10.18
σ (π
+
+ p → Σ
+
+ K
+ ) : σ (π
−
+ p → Σ
−
+ K
+ ) :
σ (π
−
+ p → Σ
0
+ K
0 )
=
a 3/2
2 :
1
9
a 3/2 + 2a 1/2
2 :
2
9
a 3/2 − a 1/2
2
If a 1/2 << a 3/2 , then the ratio becomes 9:1:2.
10.28 Pions have T = 1 and nucleons T = 1 / 2 , so that the resultant isospin both in
the initial state and the final state can be I = 1 / 2 or 3/2. Looking up the table
for Clebsch – Gordon Coefficients of 1 × 1/2 (Table 3.3) we can write
π
− p
=
1
3
3
2
, −
1
2
−
2
3
1
2
, −
1
2
π
0 n
=
2
3
3
2
, −
1
2
+
1
3
1
2
, −
1
2
Therefore,
π
− p |H | π
− p
=
1
3
a 3 +
2
3
a 1
π
0 n
H
π
− p
=
√
2
3
a 3 −
√
2
3
a 1