10.3 Solutions
567
The ratio of cross sections is
σ
−
σ 0 =
|a 3 + 2a 1 |
2
√
2 (a 3 − a 1 )
2
Put a 3 = a 1 e
iΦ with Φ = ±30
◦
σ 1
σ 2
=
1
2
e
iφ
+ 2
2
e iφ − 1
2 =
1
2
e
iφ
+ 2
e
−iφ
+ 2
e iφ − 1
e −iφ − 1
=
1
2
(5 + 4 cos φ)
(2 − 2 cos φ)
= 15.8
10.29 Deuteron has spin 1 and is in the s-state (l = 0) mostly, with an admixture
of l = 2. The total angular momentum J = 1 so that J
p
= 1
+ . Thus
deuteron’s state is described by
3 S 1 and
3 D 1 . Next we consider the parity
arising from the angular momentum of the π
− –d system. The time for a π
−
to reach the K − orbit in the deuterium mesic atom is estimated as ∼10
−10 s.
Furthermore, direct capture of π
− from 2p level is negligible compared to the
transition from 2p to 1s level. It is therefore concluded that all π
− ’s will be
captured from the s-state of the deuterium atom before they decay. Therefore,
the parity arising from the angular momentum is (−1)
l
= (−1)
0
= +1. If
p π − is the parity of π
− then the parity of the initial state will be
π(initial) = p d . p π − .1 = p π −
In the final state the neutrons obey Fermi – Dirac statistics and must be
in the anti symmetric state. The two neutrons can have total spin S = zero
(singlet, antisymmetric) or S = 1 (triplet, symmetric). Now the total wave
function which is the product of space and spin parts must be antisymmetric.
ψ = ϕ(space)χ(spin)
Now the symmetry of the spin function is (−1)
s+1 and that for the spatial
part it is (−1)
L . Hence the overall symmetry of the wavefunction ψ under
the interchange of both space and spin will be (−1)
L+S+1 .
The initial state of π
−
− d system is characterized by L = 0 and S = 1
(∵ the deuteron spin = 1 and the pion spin is zero) and L = 0. Hence the
total angular momentum of the initial state is 1. By conservation of angular
momentum, final state must also have J = 1. The requirement J = 1 implies
L = 0, S = 1 or L = 1, S = 1 or L = 1, S = 0 or 1, or L = 2, S = 1. The
possible final states are enumerated
3 S 1 (symmetrical);
3 P 1 (antisymmetrical);
3 D 1 (symmetrical);
1 P 1 (symmetrical). Of these combinations L = S = 1
alone gives L + S even.
Thus the only correct state is
3 P 1 with parity (−1)
l
= −1, which also
requires negative parity for the initial state as parity is conserved in strong
interactions. It follows that p π − = −1.
It may be pointed out that the assumed parity of +1 for neutron and
deuteron is immaterial because the parities of baryon particles get cancelled
in any reaction due to conservation of baryon number.
567
The ratio of cross sections is
σ
−
σ 0 =
|a 3 + 2a 1 |
2
√
2 (a 3 − a 1 )
2
Put a 3 = a 1 e
iΦ with Φ = ±30
◦
σ 1
σ 2
=
1
2
e
iφ
+ 2
2
e iφ − 1
2 =
1
2
e
iφ
+ 2
e
−iφ
+ 2
e iφ − 1
e −iφ − 1
=
1
2
(5 + 4 cos φ)
(2 − 2 cos φ)
= 15.8
10.29 Deuteron has spin 1 and is in the s-state (l = 0) mostly, with an admixture
of l = 2. The total angular momentum J = 1 so that J
p
= 1
+ . Thus
deuteron’s state is described by
3 S 1 and
3 D 1 . Next we consider the parity
arising from the angular momentum of the π
− –d system. The time for a π
−
to reach the K − orbit in the deuterium mesic atom is estimated as ∼10
−10 s.
Furthermore, direct capture of π
− from 2p level is negligible compared to the
transition from 2p to 1s level. It is therefore concluded that all π
− ’s will be
captured from the s-state of the deuterium atom before they decay. Therefore,
the parity arising from the angular momentum is (−1)
l
= (−1)
0
= +1. If
p π − is the parity of π
− then the parity of the initial state will be
π(initial) = p d . p π − .1 = p π −
In the final state the neutrons obey Fermi – Dirac statistics and must be
in the anti symmetric state. The two neutrons can have total spin S = zero
(singlet, antisymmetric) or S = 1 (triplet, symmetric). Now the total wave
function which is the product of space and spin parts must be antisymmetric.
ψ = ϕ(space)χ(spin)
Now the symmetry of the spin function is (−1)
s+1 and that for the spatial
part it is (−1)
L . Hence the overall symmetry of the wavefunction ψ under
the interchange of both space and spin will be (−1)
L+S+1 .
The initial state of π
−
− d system is characterized by L = 0 and S = 1
(∵ the deuteron spin = 1 and the pion spin is zero) and L = 0. Hence the
total angular momentum of the initial state is 1. By conservation of angular
momentum, final state must also have J = 1. The requirement J = 1 implies
L = 0, S = 1 or L = 1, S = 1 or L = 1, S = 0 or 1, or L = 2, S = 1. The
possible final states are enumerated
3 S 1 (symmetrical);
3 P 1 (antisymmetrical);
3 D 1 (symmetrical);
1 P 1 (symmetrical). Of these combinations L = S = 1
alone gives L + S even.
Thus the only correct state is
3 P 1 with parity (−1)
l
= −1, which also
requires negative parity for the initial state as parity is conserved in strong
interactions. It follows that p π − = −1.
It may be pointed out that the assumed parity of +1 for neutron and
deuteron is immaterial because the parities of baryon particles get cancelled
in any reaction due to conservation of baryon number.