Vector analysis 1 1/26/2021 Finite spherical square well potential: deuteron, with the use of Mathematica Masatsugu Sei Suzuki and Itsuko S. Suzuki Department of Physics, SUNY at Binghamton (Date: January 26, 2021) Here we discuss the bound states of deuteron in a three-dimensional (3D) spherical (attractive) square well potential with radius (a) and a potential depth ( 0 V ). This problem is of interest because it is mathematically straightforward and approximate a number of real physical situation. This problem is so familiar and has been used as exercises in standard textbooks of quantum mechanics. In spite of this fact, we present our results of detailed calculations using the Mathematica (ContourPlot) in the this note. It may be useful to understanding the role of angular momentum and potential depth in this model. The energy eigenvalues of the system are determined as a function of the potential depth and angular momentum l 0, 1, 2, 3, 4…. The nature of the wave functions, which depend on the depth of potential well and the orbital angular momentum, is also examined. To this end, we use the ContourPlot (Mathematica) for the boundary condition of the wave functions at the radius a. The boundary condition is given by (1) 1 1 (1) () ( ) () ( ) l l l l xj x iyh iy j x h iy with 2 2 0 x r y for l 0, 1, 2, 3,…. (L.I. Schiff). This condition for l = 0. reduces to a simple form cot y x x . Notations of this boundary condition will be discussed below. For the first time, we found such a form of the boundary condition, in a book, Problems and Solutions of L.I. Schiff Quantum mechanics 3 rd edition (Yoshioka, 1983) [in Japanese]. Typical result derived from this method, is shown below. These results are useful to the understanding of the phase shift analysis (of scattering) such as Levinson’s theorem.
Transcript
Vector analysis 1 1/26/2021
Finite spherical square well potential: deuteron, with the use of Mathematica
Masatsugu Sei Suzuki and Itsuko S. Suzuki
Department of Physics, SUNY at Binghamton
(Date: January 26, 2021)
Here we discuss the bound states of deuteron in a three-dimensional (3D) spherical
(attractive) square well potential with radius (a) and a potential depth ( 0V ). This problem
is of interest because it is mathematically straightforward and approximate a number of
real physical situation. This problem is so familiar and has been used as exercises in
standard textbooks of quantum mechanics. In spite of this fact, we present our results of
detailed calculations using the Mathematica (ContourPlot) in the this note. It may be
useful to understanding the role of angular momentum and potential depth in this model.
The energy eigenvalues of the system are determined as a function of the potential
depth and angular momentum l 0, 1, 2, 3, 4…. The nature of the wave functions, which
depend on the depth of potential well and the orbital angular momentum, is also
examined. To this end, we use the ContourPlot (Mathematica) for the boundary condition
of the wave functions at the radius a. The boundary condition is given by
(1)
1 1
(1)
( ) ( )
( ) ( )
l l
l l
xj x iyh iy
j x h iy
with 2 2
0x r y
for l 0, 1, 2, 3,…. (L.I. Schiff). This condition for l = 0. reduces to a simple form
coty x x .
Notations of this boundary condition will be discussed below. For the first time, we found
such a form of the boundary condition, in a book, Problems and Solutions of L.I. Schiff Quantum mechanics 3rd edition (Yoshioka, 1983) [in Japanese].
Typical result derived from this method, is shown below. These results are useful to the understanding of the phase shift analysis (of scattering) such as Levinson’s theorem.
Vector analysis 2 1/26/2021
r02 2 a2
2V0
y 22 a2
2E
2
2 23
2
22
25
2
23
27
2
2
E1
E2
E3
l 0
0 20 40 60 80 100 120
10
10
20
30
40
50
Fig.1 l = 0. Typical example. E : energy eigenvalue of the bound states which
is the closest to E = 0 among bound states, but E<0. 0V is a potential depth
and a is a radius. The ground state with energy 1E . The first excited state
with energy 2E . The second excited state with energy 3E .
1 2 3 0E E E .
((Levinson’s theorem))
Levinson's theorem is an important theorem in non-relativistic quantum scattering
theory. It relates the number of bound states of a potential to the difference in phase of a
scattered wave at zero and infinite energies. It was published by Norman Levinson in
Nucleus of deuterium (heavy hydrogen) that consists of one proton and one neutron (the notation: d or D). Deuterons are formed chiefly by ionizing deuterium (stripping the
single electron away from the atom) and are used as projectiles to produce nuclear reactions after accumulating high energies in particle accelerators. A deuteron also results
from the capture of a slow neutron by a proton, accompanied by the emission of a gamma photon.
Vector analysis 3 1/26/2021
Fig.2 Deuteron consisting of one proton and one neutron in a nucleus.
The reduced mass of deuteron (neutron and proton) is
2
p n P
p n
M M M
M M
.
1. Schrödinger equation for the finite spherical square well
r
V HrL
a
-V0
E
Fig.3 Spherical (attractive) square well potential ( )V r as a function of r. The
potential depth 0V and radius a. 0 0V . 0E E (bound state).
The Hamiltonian for the deuteron in a finite spherical square well potential is given by
Vector analysis 4 1/26/2021
)(2
)1(
2 2
22
rVr
llpH r
ℏ
,
with the radial linear momentum,
rrri
pr
1ℏ
.
V(r) is the potential energy for the spherical square well,
0 ( )( )
0 ( )
V r aV r
r a
,
where r is the distance between proton and neutron, and is the reduced mass,
1
2
n P
n P
M MM
M M
(of proton or neutron).
Then the Schrodinger equation can be written as
)()()()(2
)1()]([
1
2 2
2
2
22
rErrVrr
llrr
rr
ℏℏ
,
where E is negative and numerically equal to the binding energy.
0E ,
and
)]([1
)()1
(1
)(2
222
rrrr
rrrri
rrri
rpr
ℏℏℏ
.
(i) For r<a,
)()()(2
)1()]([
1
202
2
2
22
rErVrr
llrr
rr
ℏℏ
,
or
0)()(2
)()1(
)]([1
0222
2
rVErr
llrr
rr
ℏ.
Vector analysis 5 1/26/2021
We put
)()( rurr ,
or
0)()(2
)()1(
)( 0222
2
ruVErur
llru
dr
d
ℏ
(for r<a).
(ii) r>a,
0)(2
)()1(
)(222
2
rEurur
llru
dr
d
ℏ
(for r>a).
The effective potential is defined as
2
0 2
2
2
( 1)
( )2
( )( 1)
2
eff
l lV
r ar
V
r al l
r
ℏ
ℏ
The normalized effective potential effVɶ can be rewritten as
2
2
220
2
2 2
0
2
2( 1)( )
( 1)( )
eff
eff
a VV
a V al l
r
ar l l
r
ɶ
ℏ
ℏ for r a
and
2
2
2
2
2
2
2
( 1)( )
eff
eff
a VV
a V
al l
r
ɶ
ℏ
ℏ for r a
Vector analysis 6 1/26/2021
When / 1r a ,
2
2
/ 1 02
2| ( 1)
eff
r a
a Vr l l
ℏ
So that the bound state (negative energy eigenvalue) exists only when
2
0 ( 1)r l l .
We make a plot of
2
2
2 eff
eff
a VV
ɶ
ℏ as a function of /r a for each l, where 0r is changed
as a parameter.
r a
2 a2 Veff2
l 0
r0 2
3
4
50.5 1.0 1.5
30
20
10
0
10
Fig.4 l = 0. Effective potential
2
2
2 eff
eff
a VV
ɶ
ℏ. 0 0r for the bound state.
r a
2 a2 Veff2
l 1
r0 2
3
4
5
0.5 1.0 1.5
30
20
10
10
20
30
Vector analysis 7 1/26/2021
Fig.5 l = 1. Effective potential
2
2
2 eff
eff
a VV
ɶ
ℏ. 0 ( 1) 2 1.4142r l l for
the bound state. 0r is changed as a parameter; 0 2 5r . 0 0.5r .
r a
2 a2 Veff2
l 2
r0 2
3
4
5
6
0.5 1.0 1.5
30
20
10
10
20
30
Fig.6 l = 2. Effective potential
2
2
2 eff
eff
a VV
ɶ
ℏ. 0 ( 1) 6 2.4495r l l for
the bound state. 0r is changed as a parameter; 0 2 5r . 0 0.5r .
r a
2 a2 Veff2
l 3
r0 3
4
5
6
0.5 1.0 1.5
30
20
10
10
20
30
Fig.7 l = 3. Effective potential
2
2
2 eff
eff
a VV
ɶ
ℏ. 0 ( 1) 2 3 3.4641r l l for
the bound state. 0r is changed as a parameter; 0 3 6r . 0 0.5r .
Vector analysis 8 1/26/2021
r a
2 a2 Veff2
l 4
r0 4
5
6
0.5 1.0 1.5
10
10
20
30
Fig.8 l = 4. Effective potential
2
2
2 eff
eff
a VV
ɶ
ℏ. 0 ( 1) 2 5 4.4721r l l for
the bound state. 0r is changed as a parameter; 0 4 6r . 0 0.5r .
3. Solution with l = 0 case
We consider the case when l = 0, for which there is no centrifugal barrier.
)()()(2
)(2
0022
2
rukruVErudr
d
ℏ
(r<a)
and
)()(2
)(2
22
2
ruqrEurudr
d
ℏ
(r>a)
where
2
2
0
2
0
kVE
ℏ ,
2
22q
Eℏ
.
with 0E (bound state). This leads to the condition,
2 2 2 2
0 0 02
2( ) ( )k a qa V a r
ℏ.
The solution of u(r) is obtained as
)sin()( 0rkAru (r<a)
)exp()( qrCru (r>a)
Vector analysis 9 1/26/2021
The continuity of u(r) and u'(r) at r = a, leads to
qaCeakA )sin( 0
qaeqCakAk )()cos( 00
From these two equations, we have
)cot( 00 akakqa (2)
We assume that
yqa , xak 0
Then we have
)cot(xxy , (3)
with
2
0
2
02
22 2raVyx
ℏ
(4)
where 0r is the radius of the sphere. Figure shows a plot of Eqs.(3) and (4) in the x-y
plane.
((Note))
For 0l ,
1
0
( )cot
( )
xj xx x
j x
, (1)
1
(1)
0
( )
( )
iyh iyy
h iy
leading to the relation
coty x x
Vector analysis 10 1/26/2021
Fig.9 l = 0. A plot of )cot(xxy and 2
2 2 2002
2 V ax y r
ℏ. akx 0 .
qay . The intersection of two curves leads to the solution (graphically
solved). 2
22q
Eℏ
. The curve [ )cot(xxy ] crosses the y=0 line at
( 1, ) (2 1)2
x z n n
; x = /2, 3/2, 5/2.
There is no bound state for
222 0
0 2
22.4674
2
V ar
ℏ
.
There is a single bound state for
2 2
2
0
3
2 2r
.
Vector analysis 11 1/26/2021
There are two bound states for
2 2
2
0
3 5
2 2r
.
4. The binding energy and potential depth for deuteron
A. Goswami Quantum Mechanics second edition (Wavelend, 2003).
The binding energy of deuteron can be evaluated as follows.
M(H1): mass of proton 938.27208816 MeV
M(n): mass of neutron 939.56542052 MeV
M(d): mass of deuteron 1875.612928 MeV
M(e); mass of electron 0.510998950 MeV
Binding energy = M(H1) +M(n)-M(d) 2.22458 MeV.
M(H1) +M(e) = 938.783087 MeV
M(n)- [M(H1) +M(e)] = 0.7823335 MeV
The stability of the deuteron is an important part of the story of the universe. In
the Big Bang model it is presumed that in early stages there were equal numbers
of neutrons and protons since the available energies were much higher than the
0.782335 MeV required to convert a proton and electron to a neutron. When the
temperature dropped to the point where neutrons could no longer be produced
from protons, the decay of free neutrons began to diminish their population.
Those which combined with protons to form deuterons were protected from
further decay. This is fortunate for us because if all the neutrons had decayed,
there would be no universe as we know it, and we would not be here!
It is known that the deuteron bound state has an energy
2.226E MeV. (we use this value hereafter).
The range of nuclear potential is approximately equal to the Compton wavelength of the pion, which, to a first approximation, meditates the interaction between nucleons, that
is
1.46 fmam c
ℏ
≃
Vector analysis 12 1/26/2021
where 2135.0 MeV/m c . Using this value of a, and p n
p n
M M
M M
, we have
22
0 247.89
2 2V
a
ℏ MeV
which is much larger than 0E . We make a plot of 2 2
0 02
2r V a
ℏ
as a function of
2
2 Ey qa a
ℏ. This can be obtained using the ContourPlot of the Mathematica;
2 2 2 2
0 0cot( )y r y r y .
r02
y qa
l 0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
0
2
4
6
8
Fig.10 The range-depth relation for the square-well potential. The plot of
2 2
0 02
2r V a
ℏ
as a function of 2
2 Ey qa a
ℏ. This figure is the same
as that shown by Goswami in his book (Quantum Mechanics, 2nd edition,
Wavelend, 2003), except for the scale along the y axis ( 2
0r ).
((Mathematica))
Vector analysis 13 1/26/2021
The radius a:
13
15
1.46168 10 cm
1.46168 10 m
=1.46168 fm
a
The reduced mass :
25
28
8.36887 10 g
=8.36887 10 kg
The potential penetration depth 0V :
0 47.894V MeV.
We use the cgs unit for Mathematica.
Vector analysis 14 1/26/2021
5. Solution of u(r) for r<a with l
We solve the differential equation for r<a
0)()(2
)()1(
)( 0222
2
ruVErur
llru
dr
d
ℏ
(for r<a).
or
2
2
02 2
( 1)( ) [ ] ( ) 0
d l lu r k u r
dr r
,
where the wave number 0k is defined as
Vector analysis 15 1/26/2021
22
0 02
E V k
ℏ
.
Now we introduce a dimensionless variable ,
0k r .
Then we have
0)(]1)1(
[ ,22
2
lkull
d
d.
The solution of this differential equation is obtained as
)()()( 21, lllk nAjAu .
where the spherical Bessel function and spherical Neumann function are defined by
)(2
)(2
1
l
l Jj ,
)(2
)(2
1
l
l Nn .
Note that )(ln becomes infinity in the limit of 0 . So we choose the first term
, 0( ) ( )k l l lR r A j k r .
where )()( ,, rurrR lklk and lA is constant.
((Mathematica))
Clear "Global` " ;
eq1 y'' x 1L L 1
x2y x 0;
DSolve eq1, y x , x
y x x BesselJ1
21 2 L , x C 1
x BesselY1
21 2 L , x C 2
Vector analysis 16 1/26/2021
((Note))
We note that
)(2
)(
)()( 2
1
2
1
l
l
lj
J
Ju
.
Thus we have
)()()(
)(
lju
r
rurR .
6. Solution of u(r) for r>a with finite l
We solve the differential equation for r>a
0)(2
)()1(
)(222
2
rEurur
llru
dr
d
ℏ
, (for r>a).
where
2
2
2E q
ℏ
,
where q is the wavenumber.
2
2
2 2
( 1)( ) [( ) ] ( ) 0
d l lu r iq u r
dr r
.
Now we introduce a dimensionless variable ,
iqr .
Then we get
0)(])1(
1[)( ,2,2
2
lklk u
llu
d
d.
The solution of this differential equation is obtained as
, 1 2
(1) (2)
1 2
( ) ' ( ) ' ( )
( ) ( )
k l l l
l l
R r B j iqr B n iqr
B h iqr B h iqr
Vector analysis 17 1/26/2021
where )()1(
xhl and )()2(
xhl are the spherical Hankel function of the first and second kind.
)()()(2
)()1(
2
1
)1(xinxjxH
xxh ll
ll
,
)()()(2
)()2(
2
1
)2(xinxjxH
xxh ll
ll
.
The asymptotic forms of (1) ( )lh x and (2) ( )lh x are given by
x
eixh
lxi )2/()1(
)(
ℓ
,
x
eixh
lxi )2/()2(
)(
ℓ
,
in the limit of large x. Then we get
( /2) ( /2)
(1)( )
i iqr l qr ile eh iqr i
iqr qr
ℓ
,
( /2) ( /2)
(2)( )
i iqr l qr ile eh iqr i
iqr qr
ℓ
,
which means that (2) ( )h iqrℓ
becomes diverging for large r, while (1) ( )h iqrℓ
becomes zero
for large r. So, we choose (1) ( )h iqrℓ
as the solution of )(, rR lk for r>a,
(1)
, ( ) ( )k l l lR r B h iqr .
where B1 is constant.
10. Boundary conditions with finite l
Using the boundary condition at r = a, we determine the energy eigenvalues. We note
that the wave function and its derivative should be continuous at r = a.
(1)
0( ) ( )l l l lA j k a B h iqa ,
(1)
0 0'( ) | ( ) '( ) |l l r a l l r aA k j k r B iq h iqr ,
