10.3 Solutions 571
K^0 +p→K++n allowedK^0 +p→Σ++π^0
K^0 +n→K−+p}
forbidden because of strangeness non-conservationK^0 +p→Σ++π^0
K^0 +n→K−+p}
allowedK^0 +p→K++n forbidden10.39^4 He is singlet and soT=0, whileK−hasT=^1 / 2. Therefore, the initial
state is a pureI=^1 / 2 withI 3 =−^1 / 2. In the final state^4 ΛHe and^4 ΛHforma
doublet, withT=^1 / 2. As pion hasT=1, the final state will be a mixture
ofI= 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 HΛ〉
−
√
2
3
∣
∣π−HeΛ〉σ(Λ^4 He)
σ(Λ^4 H)=
(√
2
/
3
) 2
(
1
/√
3
) 2 =
2
1
10.40 As bothK−andπ^0 have zero spin,lis conserved. Thus the orbital angular
momentumlmust be the same for the initial and final states. Since this is a
strong interaction, conservation of parity requires that
(−1)lPk=(− 1 )lPπ=(−1)l(−1)Therefore,Pk=− 110.41 The deuteron spinJd=1 and the capture of pion is assumed to occur from
the s-state. Particles in the final state must haveJ =1. As the Q- value is
small (0.5 MeV), the finalnnπ^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 −〉=