10.3 Solutions 567
The ratio of cross sections is
σ−
σ^0=
|a 3 + 2 a 1 |^2
∣∣
∣√
2 (a 3 −a 1 )∣∣
∣
2Puta 3 =a 1 eiΦwithΦ=± 30 ◦σ 1
σ 2=
1
2
∣
∣eiφ+ 2∣
∣^2
∣
∣eiφ− 1∣
∣^2
=
1
2
(
eiφ+ 2)(
e−iφ+ 2)
(
eiφ− 1)(
e−iφ− 1)=
1
2
( 5 +4 cosφ)
( 2 −2 cosφ)= 15. 8
10.29 Deuteron has spin 1 and is in the s-state (l=0) mostly, with an admixture
ofl = 2. The total angular momentumJ =1 so thatJp = 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 theπ−–d system. The time for aπ−
to reach theK−orbit in the deuterium mesic atom is estimated as∼ 10 −^10 s.
Furthermore, direct capture ofπ−from 2p level is negligible compared to the
transition from 2p to 1s level. It is therefore concluded that allπ−’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
pπ−is the parity ofπ−then the parity of the initial state will be
π(initial)=pd.pπ−. 1 =pπ−
In the final state the neutrons obey Fermi – Dirac statistics and must be
in the anti symmetric state. The two neutrons can have total spinS=zero
(singlet, antisymmetric) orS=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
partitis(−1)L. Hence the overall symmetry of the wavefunctionψunder
the interchange of both space and spin will be (−1)L+S+^1.
The initial state ofπ−−d system is characterized byL=0 andS= 1
(∵the deuteron spin=1 and the pion spin is zero) andL=0. Hence the
total angular momentum of the initial state is 1. By conservation of angular
momentum, final state must also haveJ=1. The requirementJ=1 implies
L= 0 ,S=1orL= 1 ,S=1orL= 1 ,S=0or1,orL= 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 combinationsL = S= 1
alone givesL+Seven.
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 thatpπ−=−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.