10.3 Solutions
571
K
0
+ p → K
+
+ n allowed
K
0
+ p → Σ
+
+ π
0
K
0
+ n → K
−
+ p
forbidden because of strangeness non-conservation
K 0 + p → Σ
+
+ π
0
K 0 + n → K
−
+ p
allowed
K 0 + p → K
+
+ n forbidden
10.39
4 He is singlet and so T = 0, while K
− has T = 1 / 2 . Therefore, the initial
state is a pure I = 1 / 2 with I 3 = − 1 / 2 . In the final state
4
Λ He and
4
Λ H form a
doublet, with T = 1 / 2 . As pion has T = 1, the final state will be a mixture
of I = 3/2 and 1/2. Looking up the Clebsch – Gordon Coefficients 1 × 1/2
(given in Table 3.3 of Chap. 3), we find the
final state ∼
1
√
3
π
0 H Λ
−
2
3
π
− He Λ
σ ( Λ
4 He)
σ ( Λ
4 H)
=
2
3
2
1
√
3
2 =
2
1
10.40 As both K
− and π
0 have zero spin, l is conserved. Thus the orbital angular
momentum l must be the same for the initial and final states. Since this is a
strong interaction, conservation of parity requires that
(−1)
l P k = (−1)
l P π = (−1)
l (−1)
Therefore, P k = −1
10.41 The deuteron spin J d = 1 and the capture of pion is assumed to occur from
the s-state. Particles in the final state must have J = 1. As the Q- value is
small (0.5 MeV), the final nnπ
0 must be an s-state. It follows that the two
neutrons must be in a triplet spin state (symmetric) which is forbidden by
Pauli’s principle.
10.42 Referring to the two nucleons by 1 and 2, + for proton and – for neutron,
and using the notation |I, I 3 for the isotopic state and the Clebsch–Gordon
Coefficients for 1 / 2 × 1 / 2 we can write the p–p state:
| p |p ≡ |1+ |2+ = |1, 1
The p–n state
| p |n ≡ |1+ |2− =
1
√
2
[|1, 0 + |0, 0]