240
3 Quantum Mechanics – II
binding energy W of the deuteron, which is much smaller than the well depth
(∼25 MeV). In the deuteron problem the outside function Ce
−γ r , where
γ =
MW / 2 , is matched with the inside function A sin kr . Here we
match the functions sin(kr + δ 0 ) and Ce
−γ r at r = R, both in magnitude
and first derivative.
This gives us k cot(kR +δ 0 ) = −γ . Further, R ≈ 0. This is also reasonable
since for the square well the main features of the deuteron problem remain
unaltered by narrowing the well width and deepening the well. It follows that
sin
2
δ 0 = k
2
/k
2
+ γ
2 )
But the s-wave cross-section is given by
σ = 4π sin
2
δ 0 /k
2
= 4π/(k
2
+ γ
2 )
Substituting k
2
= ME /
2 and γ
2
= MW /
2
σ =
4π
2
M
1
W + E
(1)
where M is proton or neutron mass, W is the deuteron binding energy
(2.225 MeV), and E is the lab kinetic energy.
Formula (1) agrees well with experiment at relatively higher energies (say
5–10 MeV) but fails badly at very low energies. For E W , for example,
(1) predicts σ = 2 barns which is far from the experimental value of 20
barns. Wigner pointed out that in n− p scattering the spins of the colliding
nucleons could be either parallel or antiparallel. Formula (1) holds for the
parallel case because the analogy is made with the deuteron problem which
has parallel spins. Now for random orientations of spins:
σ =
3
4
σ t +
1
4
σ s
(2)
where σ t and σ s are the cross-sections for the triplet and singlet scattering,
the factors
3
4
and
1
4
being the statistical weights. In (1), W is the binding
energy of the n− p system for the triplet state. Corresponding to the singlet
state the quality W s is introduced, although it is a virtual state.
Combining (1) and (2)
σ =
3π
2
M(E + W )
+
π
2
M(E + W s )
(3)
W s takes a value of 70 keV if agreement is to reach with the experiments.
Agreement at higher energies is preserved because for E W or W s , (3)
reduces to (1).
3.3.9 Scattering (Born Approximation)
3.113 (a) F(q) ≈
∞
0
ρ(r) sin(qr /)4πr
2
qr /
dr
ρ(r ) = A exp(−r/a)
F(q) ≈ 4π A
∞
0
r exp(−r/a)[sin(qr /)/(q/)]dr
3 Quantum Mechanics – II
binding energy W of the deuteron, which is much smaller than the well depth
(∼25 MeV). In the deuteron problem the outside function Ce
−γ r , where
γ =
MW / 2 , is matched with the inside function A sin kr . Here we
match the functions sin(kr + δ 0 ) and Ce
−γ r at r = R, both in magnitude
and first derivative.
This gives us k cot(kR +δ 0 ) = −γ . Further, R ≈ 0. This is also reasonable
since for the square well the main features of the deuteron problem remain
unaltered by narrowing the well width and deepening the well. It follows that
sin
2
δ 0 = k
2
/k
2
+ γ
2 )
But the s-wave cross-section is given by
σ = 4π sin
2
δ 0 /k
2
= 4π/(k
2
+ γ
2 )
Substituting k
2
= ME /
2 and γ
2
= MW /
2
σ =
4π
2
M
1
W + E
(1)
where M is proton or neutron mass, W is the deuteron binding energy
(2.225 MeV), and E is the lab kinetic energy.
Formula (1) agrees well with experiment at relatively higher energies (say
5–10 MeV) but fails badly at very low energies. For E W , for example,
(1) predicts σ = 2 barns which is far from the experimental value of 20
barns. Wigner pointed out that in n− p scattering the spins of the colliding
nucleons could be either parallel or antiparallel. Formula (1) holds for the
parallel case because the analogy is made with the deuteron problem which
has parallel spins. Now for random orientations of spins:
σ =
3
4
σ t +
1
4
σ s
(2)
where σ t and σ s are the cross-sections for the triplet and singlet scattering,
the factors
3
4
and
1
4
being the statistical weights. In (1), W is the binding
energy of the n− p system for the triplet state. Corresponding to the singlet
state the quality W s is introduced, although it is a virtual state.
Combining (1) and (2)
σ =
3π
2
M(E + W )
+
π
2
M(E + W s )
(3)
W s takes a value of 70 keV if agreement is to reach with the experiments.
Agreement at higher energies is preserved because for E W or W s , (3)
reduces to (1).
3.3.9 Scattering (Born Approximation)
3.113 (a) F(q) ≈
∞
0
ρ(r) sin(qr /)4πr
2
qr /
dr
ρ(r ) = A exp(−r/a)
F(q) ≈ 4π A
∞
0
r exp(−r/a)[sin(qr /)/(q/)]dr