10.3 Solutions
561
(f) σ.E → −σ.E under both P and T-operations.
(g) Similarly σ.B → −σ.B under 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, and l = 0 mostly, with
4% admixture of l = 2 so that the parity determined by (−1)
l
= +1 for
l = 0 or 2. The deuteron is in a state of total angular momentum J = 1.
Thus J
p
= 1
+ . Using the spectroscopic notation
2s+1 L J , deuteron’s state is
described by
3 S 1 and
3 D 1 .
10.10 (a) For ρ, η and π, the I
G values are 1
+
, 0
+ and 1
− respectively. Therefore G-parity is violated. Therefore, the decay of ρ is forbidden by strong
interaction.
(b) For ω, η and π, the parities are −1. The overall parity of η and π will
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 decay ρ
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
decay K
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
The Δ
0 has T = 3/2 and T 3 = − 1 / 2 . In both the decays T 3 is conserved
(for the first one p + π
− system has T 3 = + 1 / 2 − 1 = − 1 / 2 , and for the
second one n + π
0 system has T 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 of Λ,
Λ → p + π
−
→ n + π
0
Λ has T 3 = 0, and therefore T 3 is violated. The decay being weak has the
characteristic life time of 10
−10 s.
10.12 (a) For ρ
0 (770 MeV), J
p
= 1
− and I
G
= 1
+ while for π (139 MeV),
J
p
= 0
− and I
G
= 1
− . In the decay ρ
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 of l = 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.
561
(f) σ.E → −σ.E under both P and T-operations.
(g) Similarly σ.B → −σ.B under 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, and l = 0 mostly, with
4% admixture of l = 2 so that the parity determined by (−1)
l
= +1 for
l = 0 or 2. The deuteron is in a state of total angular momentum J = 1.
Thus J
p
= 1
+ . Using the spectroscopic notation
2s+1 L J , deuteron’s state is
described by
3 S 1 and
3 D 1 .
10.10 (a) For ρ, η and π, the I
G values are 1
+
, 0
+ and 1
− respectively. Therefore G-parity is violated. Therefore, the decay of ρ is forbidden by strong
interaction.
(b) For ω, η and π, the parities are −1. The overall parity of η and π will
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 decay ρ
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
decay K
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
The Δ
0 has T = 3/2 and T 3 = − 1 / 2 . In both the decays T 3 is conserved
(for the first one p + π
− system has T 3 = + 1 / 2 − 1 = − 1 / 2 , and for the
second one n + π
0 system has T 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 of Λ,
Λ → p + π
−
→ n + π
0
Λ has T 3 = 0, and therefore T 3 is violated. The decay being weak has the
characteristic life time of 10
−10 s.
10.12 (a) For ρ
0 (770 MeV), J
p
= 1
− and I
G
= 1
+ while for π (139 MeV),
J
p
= 0
− and I
G
= 1
− . In the decay ρ
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 of l = 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.