3.3 Solutions
171
Fig. 3.7 Deuteron wave
function and energy
where μ is 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 , where W = binding energy, is positive as the potential is positive
V 0 = −V , where V 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 −
M W u
2 = 0; r > R
(6)
where R is the range of nuclear forces, Fig. 3.7.
Calling
M(V 0 − W )
2
= k
2
(7)
and
M W
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)