10.3 Solutions 561
(f)σ.E→−σ.Eunder both P and T-operations.
(g) Similarlyσ.B→−σ.Bunder both p-operation and under T-operation.10.9 The deuteron which is the nucleus of deuterium (heavy hydrogen) consists of
one proton and one neutron. Since the parity of neutron and proton are+1,
that of deuteron is also+1. Spin of deuteron is 1, andl=0 mostly, with
4% admixture ofl=2 so that the parity determined by (−1)l =+1for
l=0 or 2. The deuteron is in a state of total angular momentumJ =1.
ThusJp= 1 +. Using the spectroscopic notation2s+^1 LJ, deuteron’s state is
described by^3 S 1 and^3 D 1.10.10 (a) Forρ,ηandπ,theIGvalues are 1+, 0 +and 1−respectively. There-
fore G-parity is violated. Therefore, the decay ofρis forbidden by strong
interaction.
(b) Forω, ηandπ, the parities are−1. The overall parity ofηandπwill
be (−1)(−1)=+1, as they are emitted in the s-state of relative angular
momentum. Therefore, both strong and em interactions will be forbidden.
The strong interaction will be forbidden also because of non conservation
of isospin.
10.11 (a) In the decayρ^0 →π+π−, all the quantum numbers (Q/e, B, I, G,π)
are conserved as required by a strong interaction. Therefore its lifetime
(∼ 10 −^23 s) is characteristic of a strong interaction. On the other hand the
decayK^0 →π+π−, violates strangeness. It is therefore a weak decay,
characterized by relatively long lifetime of the order of 10−^10 s.
(b)Δ^0 →p+π−
→n+π^0
TheΔ^0 hasT= 3 /2 andT 3 =−^1 / 2. In both the decaysT 3 is conserved
(for the first one p+π−system hasT 3 =+^1 / 2 − 1 =−^1 / 2 , and for the
second onen+π^0 system hasT 3 =−^1 / 2 + 0 =−^1 / 2 ). Therefore, the
decay proceeds through strong interaction with a characteristic life time
of∼ 10 −^23 s.
In the case ofΛ,
Λ→p+π−
→n+π^0
ΛhasT 3 =0, and thereforeT 3 is violated. The decay being weak has the
characteristic life time of 10−^10 s.
10.12 (a) Forρ^0 (770 MeV),Jp = 1 − andIG = 1 +while for π(139 MeV),
Jp = 0 −andIG = 1 −. In the decayρ^0 →π+π−, all the quantum
numbers (Q/e, B, I, G,π) are conserved. Note that the decay involves
a large Q-value (491 MeV) so that the pions will come off with relative
angular momentum ofl=1, contributing (−1) to overall parity. Thus,
the overall parity is conserved (each pion has intrinsic parity−1).
(b) Decay is forbidden because of Bose symmetry.