10.3 Solutions 565
10.20 The nucleon hasT = 1 /2 and pionT =1. TheΔ+hasT = 3 /2 and
T 3 =+ 1 /2. Using the C.G.C. for 1× 1 /2 (Table of Chap. 3), we have
∣
∣
∣
∣
3
2
,
1
2
〉
=
√
2
3
| 1 , 0 〉
∣
∣
∣
∣
1
2
,
1
2
〉
+
√
1
3
| 1 , 1 〉
∣
∣
∣
∣
1
2
,−
1
2
〉
∣
∣
∣
∣Nπ;3
2
,
1
2
〉
=
√
2
3
∣
∣Pπ^0 〉+√
1
3
∣
∣nπ+〉∴
Γ(Δ+→π^0 +p)
Γ(Δ+→π++n)=
(√
2 / 3
) 2 /(√
1 / 3
) 2
= 2
Note that for the charge statesΔ++, the strong decay is through only one
channel (Δ++→p+π+), so also forΔ−,vizΔ−→n+π−.10.21 X+→n+π+
→p+π^0
X+can have eitherT = 3 /2or1/2.ForT = 3 /2 the predicted ratio
Γ(X+ →nπ+)/Γ(X+ → pπ^0 )= 1 /2 (as in Problem 10.20) which is
in disagreement with the experimental ratio of 36/18 or 2.
If we assume the valueT= 1 /2forX+the state
∣
∣^1
2 ,
1
2〉
must be orthogonal
to the state∣
∣^3
2 ,
1
2〉
. Therefore
X+
∣
∣
∣
∣Nπ,1
2
,
1
2
〉
=−
√
1
3
∣
∣pπ^0 〉+√
2
3
∣
∣nπ+〉upto an overall phase factor.
The branching ratio
∴Γ(X+ →nπ+)/Γ(X+ → pπ^0 ) = 2 /1, which is in agreement with
experimental ratio of 36/18 or 2/1.10.22τ = Γ = Γcc = 158 (MeV)^197 .3MeV× 3 ×− 10 fm (^8) (m/s) =^197.^3 ×^10
− (^15) (MeV−m)
474 (MeV)× 108 (m/s) =^4 ×^10
− (^24) s
a lifetime which is characteristic of a strong interaction. Therefore, theρ-
meson decays via strong interaction.
10.23 We first write down the isospin for a pair of pions and then combine the
resultant with the third pion. (a) Each pion hasT=1, so that theπ+,π−
combination hasI= 2 , 1 ,0. Whenπ^0 is combined, possible values areI=
3 , 2 , 1 ,0(b)Twoπ^0 ’s giveI=2. When the thirdπ^0 is added totalI= 3
or 1.
10.24 From the conservation laws for strong interactions the quantum numbers for
XareB=+ 1 , Q/e=− 1 , S=−2. It is Xi hyperon (Ξ−) and decays
weakly with lifetimeτ∼ 10 −^13 s. Its quark structure is dss; the quantum
numbers forYareB =+ 1 , Q/e= 0 , S=−1. It is a sigma hyperon
(Σ^0 ) which decays electromagnetically with lifetime of the order of 10−^20 s.
Its quark structure is uds.