3.3 Solutions 171
Fig. 3.7Deuteron wave
function and energy
whereμis the reduced mass=M/ 2 ,M,being neutron of proton mass. With
the assumption of spherical symmetry, the angular derivatives in the Laplacian
vanish and the radial equation is
1
r^2
d
dr
(
r^2
d
dr
)
ψ(r)+(M/^2 )[E−V(r)]ψ(r)=0(2)
With the change of variable
ψ(r)=
u(r)
r
(3)
Equation (2) becomes
d^2 u
dr^2
+
(
M
^2
)
[E−V(r)]u=0(4)
The total energy=−W, whereW=binding energy, is positive as the poten-
tial is positive
V 0 =−V, whereV 0 is positive
Equation (4) then becomes
d^2 u
dr^2
+
(
M
^2
)
(V 0 −W)u=0;r<R (5)
d^2 u
dr^2
−
MWu
^2
=0;r>R (6)
whereRis the range of nuclear forces, Fig. 3.7.
Calling
M(V 0 −W)
^2
=k^2 (7)
and
MW
^2
=γ^2 (8)
(5) and (6) become
d^2 u
dr^2
+k^2 u=0(9)
d^2 u
dr^2
−γ^2 u= 0 (10)