5-4 finite spherical well revised 2-16-15bingweb.binghamton.edu/~suzuki/quantummechanicsii/...vector...
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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
1949.
https://en.wikipedia.org/wiki/Levinson%27s_theorem
((Deuteron))
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!
http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/deuteron.html
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 ,
leading to
Vector analysis 18 1/26/2021
(1)00 (1)
0
'( ) '( )( ) ( )
l l
l l
k iqj k a h iqa
j k a h iqa ,
where
( )'( )l
l
j zj z
z
,
(1)(1)( )
'( )ll
h zh z
z
.
Spherical Bessel function;
1/2( ) ( )2
l lj z J zz
SphericalBesselJ[l, z] (Mathematica)
Spherical Hankel function:
(1) (1)
1/2( ) ( )2
l lh z H zx
SphericalHankel1[l, z] (Mathematica)
________________________________________________________________________
Here we use the recursion formula for (1)( ) ( ), ( ),l l lf j h ,
1 1
2 1( ) ( ) ( )l l l
lf f f
,
1 1( ) ( 1) ( ) (2 1) '( )l l llf l f l f .
From these equations, we have
1
( 1)'( ) ( ) ( )l l l
lf f f
Leading to the relations,
1
1'( ) ( ) ( )l l l
lj z j z j z
z
(1) (1) (1)
1
1'( ) ( ) ( )l l l
lh z h z h z
z
Then, the boundary condition can be rewritten as
Vector analysis 19 1/26/2021
(1) (1)0 1 0 0 1
0
(1)
0
1 1[ ( ) ( ) [ ( ) ( )]
( ) ( )
l l l l
l l
l lk j k a j k a iq h iqa h iqa
k a iqa
j k a h iqa
or
(1)
0 1 0 1
(1)
0
( ) ( )
( ) ( )
l l
l l
k j k a iqh iqa
j k a h iqa
(l = 1, 2, 3, ).
Finally, we get
(1)
1 1
(1)
( ) ( )
( ) ( )
l l
l l
xj x iyh iy
j x h iy
2
2 2 2
0 02
2 ax y r V
ℏ
(L.I. Schiff, Quantum mechanics)
Note that 1
cos( )
xj x
x (APPENDIX-A)
In summary, we have
(a) l = 0
coty x x for l = 0
(1)
1 1
0 0
cos
( ) ( )cot
sin( ) ( )
xx
xj x iyh iyxx x
xj x h iy
x
.
(b) l = 1, 2, 3, 4, …
(1)
1 1
(1)
( ) ( )
( ) ( )
l l
l l
xj x iyh iy
j x h iy
, for l = 1, 2, 3,…
with 2 2 2
0x y r . We make a plot of 2 2 2
2
2( )y qa E a
ℏas a function of 2 2
0 02
2r V a
ℏ
with the use of ContourPlot of the Mathematica.
Vector analysis 20 1/26/2021
9. The condition for the bound state energy 0E
Using these boundary conditions, we can determine the values of 0V and l such that
the energy of the bound state E becomes zero. We note that the energy 0E is
equivalent to 0y . Note that l = 0 (s), 1 (p), 2 (d), 3 (f), and 4 (g).
For l = 0, we get the condition cos 0x , leading to
3 5, , ,...
2 2 2x
or
[0 1, ] (2 1)2
z n n
with n = 1, 2, 3,…..
and
2 (0)
2 20
2
2[ (0 1, )] [ (2 1)]
2
a Vz n n
ℏ (l = 0)
For l = 0, 1, 2, 3,…, we have 1( ) 0lj x . The zeros of 1( )lj x is defined by
( 1, )x z l n ,
or
2 (0)
20
2
2[ ( 1, , )]
a Vz l n
ℏ
where n = 1, 2, 3,,…, (0)
0V is the potential depth for E = 0, l (=1, 2, 3,…) is the orbital
angular momentum, and ( 1, , )l n is the zero of the spherical Bessel function 1( )lj x .
Table 1: Zeros of the spherical Bessel functions.
Number
of zero; 0 ( )j x 1( )j x 2 ( )j x 3( )j x 4 ( )j x
n
1 3.14159 4.49341 5.76346 6.98793 8.18256
2 6.28319 7.72525 9.09501 10.4171 11.7049
3 9.42478 10.9041 12.3229 13.6980 15.0397
4 12.5664 14.0662 15.5146 16.9236 18.3013
5 15.7080 17.2208 18.6890 20.1218 21.5254
___________________________________________________________________
Vector analysis 21 1/26/2021
Table 2: Zeros of the spherical Bessel functions.
Number
of zero 0 ( )j x 1( )j x 2 ( )j x 3( )j x 4 ( )j x
n
1 1.4303 1.83457 2.22433 2.60459
2 2 2.4590 2.89503 3.31587 3.72579
3 3 3.4709 3.9225 4.36022 4.78727
4 4 4.4774 4.9385 5.38696 5.82547
5 5 5.4815 5.9489 6.40497 6.85175
((Mathematica))
Zeroes of the Spherical Bessel functions
Vector analysis 22 1/26/2021
Here we make a plot of 2 (0)
0
2
2 a Vℏ
where E = 0, as a function of l and n.
Vector analysis 23 1/26/2021
l 0 l 1 l 2 l 3 l 4 l 5
2 a2 V00
2
0
100
200
300
400
(a)
Vector analysis 24 1/26/2021
l 0 l 1 l 2 l 3 l 4 l 5
2 a2 V00
2
0
50
100
(b)
Fig.11(a) (b) Plot of 2 (0)
2002
2 a Vr
ℏ where E = 0, as a function of l and n: l = 0, 1, 2,
3,…. n = 1, 2, 3, 4, ….. Note that the bound state (negative energy
eigenvalue) exists only when 2
0 ( 1)r l l from the discussion of the
effective potential.
(a) l = 0: 1
cos( ) 0
zj z
z cos 0z
n (0 1, )z n 2[ (0 1, )]z n
1 / 2 (= 2.46740
2 3 / 2 (= 22.2066
3 5 / 2 (= 61.6850
4 7 / 2 (= 120.903
5 9 / 2 (= 199.859
(b) l = 1: 0
sin( ) 0
zj z
z ; sin 0z
n (1 1, )z n 2[ (1 1, )]z n
1 (= 9.86960
2 2 (= 39.4784
3 3 (= 88.8264
4 4 (= 157.914
5 5 (= 246.740
(c) l = 2: 1( ) 0 (2 1, )j z z z n
Vector analysis 25 1/26/2021
n (2 1, )z n 2[ (2 1, )]z n
1 4.49341(= 20.19073
2 7.72525(= 59.6795
3 10.9041(= 118.8994
4 14.0662(= 197.8580
5 17.2208(= 296.5560
(d) l = 3: 2 ( ) 0 (3 1, )j z z z n ;
n (3 1, )z n 2[ (3 1, )]z n
1 5.76346(= 33.2175
2 9.09501(= 82.7192
3 12.3229(= 151.854
4 15.5146(= 240.703
5 18.6890(= 349.279
(e) l = 4: 3( ) 0 (4 1, )j z z z n ;
n (4 1, )z n 2[ (4 1, )]z n
1 6.9879 (= 48.83075
2 10.4171(= 108.5160
3 13.6980(= 187.6352
4 16.9236(= 286.4082
5 20.1218 (= 404.8868
9. Solution with l = 0 (s-wave)
We now discuss the solution of the Schrödinger equation with l = 0 (s wave). We have the wave function,
0 0( ) sin( ) sin[( ) ] sin( )r
u r A k r A k a A xa
for r a .
( ) exp( ) exp[ ( ) ] exp[ ]r
u r C qr C qa C xa
for r a
0k a , qa
2 2 2 2 2 2
0 0 02
2( ) ( )k a qa V a r
ℏ
The boundary condition:
Vector analysis 26 1/26/2021
0sin( ) exp( )A k a C qa ,
sin( ) exp( )A C ,
0 0cos( ) ( )exp( )Ak k a C q qa , cos( ) exp( )A C ,
cot .
Normalization:
2 2 2
2 20
2 2
0
sin ( ) exp( 2 )1 4 4
a
a
A k r C qrr dr r dr
r r
,
or
2 22 2 20
02
0 0
2
sin ( )4 4 sin ( )
2 sin(2 )4 [ ]
4
a aA k r
r dr A k r drr
A a
2
2 2
2
22
exp( 2 )4 4 exp( 2 )
42
a a
C qrr dr C qr dr
r
eC
or
22 22 sin(2 )
1 4 44 2
eA a C a
,
sin( ) exp( )A C ,
22 sin cos 2 ( sin )A a
,
2
sin
2 sin cos 2 ( sin )
eC a
.
10. The energy eigenvalue E vs potential depth with l = 0.
Vector analysis 27 1/26/2021
We now use the ContourPlot to determine the relation of 2y vs 2
0r , from the equation
2 2 2 2
0 0cot coty x x r y r y with 2 2
0x r y
which depends only on 2y and 2
0r . Figure 12 shows the scaled energy eigenvalue
22 2
2
2( )
ay qa E
ℏ, as a function of scaled potential depth
22 0
0 2
2 a Vr
ℏ
for l = 0.
r02 2 a2
2V0
y 22 a2
2E l 0
E1 E2 E3 E4
2
2 3
2
2 5
2
2 7
2
2
0 50 100 150 200
20
0
20
40
60
80
100
Fig.12 l = 0. 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
.
0E at 0 ( 1 0 1, );r z l n where 1 0( ) 0j r .
0 01 02 03 04
3 5 7, , ,
2 2 2 2r
,…
As is shown From Fig.10, there is no bound state for
22
0
2
2
2
V a ℏ
.
Vector analysis 28 1/26/2021
There is a single bound state (ground state) for
2 22
0
2
2 3
2 2
V a ℏ
. There are
two bound states (ground state and the first excited state) for
2
2
2
0
2
2
72
2
5
ℏ
aV.
There are three bound states (ground state, the first excited state, and the second excited
state) for
2 22
0
2
27 9
2 2
V a ℏ
.
Here we define the minimum energy eigenvalue E among the allowed states for fixed
0V . The value of E is the closest to E = 0 (but E<0). We make a plot of the minimum
eigenvalue E as a function of potential depth 0V .
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.13 The energy eigenvalue ( E ) of the bound states which is the closest to E =
0 among bound states, but E<0. The minimum energy (2
2
2 ay E
ℏ
) as a
function of the potential depth (2
2 00 2
2 a Vr
ℏ
).
0 01 02 03 04
3 5 7, , ,
2 2 2 2r
,…
11. Wave function for l = 0.
(a) The wave function ( )u r of the ground state for l = 0.
Vector analysis 29 1/26/2021
Below we show the plot of the wave function ( )u r as a function of /r a for the
ground state for l = 0 (s wave). The two nucleons have a substantial probability of being
separated by a distance that is greater than the range of the potential well. The shape of
the wave function is, correspondingly, not very sensitive to the detailed nature of the
potential. The wave function for s = 1.0 ( 02
r s
)is similar to the figure obtained by
Bethe and Morrison. For the ground state with l = 0,
( ) exp( ) exp[ ( ) ]r
u r C qr C qaa
for r a
The quantity 1/ ( )qa can be taken as a measure of the size of the deuteron (the effective
radius of the deuteron). It is shown that the effective radius is considerably larger than the
range of nuclear force.
11
qa .
Thus, most of the area under ( )u r occurs for / 1r a .
r0 2 s
s 1.0
1.41.82.23.0
r a
4 a u rGround state
0.5 1.0 1.5 2.0 2.5 3.0
0.2
0.4
0.6
0.8
1.0
1.2
Fig.14 l = 0. Wave function ( )u r for the ground state (s = 1 -3) as a function of
/r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
. s is changed as a parameter. 1 3s .
0.2s .
Vector analysis 30 1/26/2021
r0 2 ss 3.05.0
r a
4 a u rGround state
0.5 1.0 1.5 2.0 2.5 3.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
Fig.15 l = 0. Wave function ( )u r for the ground state (s = 3 -5) as a function of
/r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter; s = 3.0 –
5.0, s 0.2.
r0 2 s
s 5.07.0
r a
4 a u rGround state
0.5 1.0 1.5 2.0 2.5 3.0
0.5
1.0
1.5
Fig.16 l = 0. Wave function ( )u r for the ground state (s = 5 - 7) as a function of
/r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter; s = 5.0 –
7.0, s 0.2.
Vector analysis 31 1/26/2021
r0 2 s
s 7.09.0
r a
4 a u r
Ground state
0.5 1.0 1.5 2.0 2.5 3.0
0.5
1.0
1.5
Fig.17 l = 0. Wave function ( )u r for the ground state (s = 7 - 9) as a function of
/r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter; s = 7.0 –
9.0, s 0.2.
(b) The wave function ( )u r of the first excited state for l = 0.
The wave function ( )u r of the first excited state for 0l is shown below, as a
function of /r a , where 0r is changed as a parameter. The wave unction ( )u r with
0
3
2r ≃ , is similar to the figure obtained by Bethe and Morrison. Since the energy
eigenvalue of the first excited state is still negative (bound state), but is very close to
zero. So that, the wave function does not decay over a distance much longer than the
radius a.
r0 2 s
s 3.03.43.84.25.0
r a
4 a u rFirst excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.0
0.5
0.5
1.0
Vector analysis 32 1/26/2021
Fig.18 l = 0. Wave function ( )u r for the first excited state (s = 3 -5) as a function
of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter; s = 3.0 –
5.0, s 0.2.
r0 2 ss 5.07.0
r a
4 a u rFirst excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.5
1.0
0.5
0.5
1.0
1.5
Fig.19 l = 0. Wave function ( )u r for the first excited state (s = 5 - 7) as a function
of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter; s = 5.0 –
7.0, s 0.2.
r0 2 ss 7.09.0
r a
4 a u rFirst excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.5
1.0
0.5
0.5
1.0
1.5
Vector analysis 33 1/26/2021
Fig.20 l = 0. Wave function ( )u r for the first excited state (s = 7 - 9) as a
function of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter;
s = 7.0 – 9.0, s 0.2.
(c) The wave function ( )u r of the second excited state for l = 0.
r0 2 s
s 5.07.0r a
4 a u r Second excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.0
0.5
0.5
1.0
Fig.21 l = 0. Wave function ( )u r for the second excited state (s = 5 - 7) as a
function of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter;
s = 5.0 – 7.0, s 0.2.
r0 2 ss 7.09.0
r a
4 a u r Second excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.5
1.0
0.5
0.5
1.0
1.5
Fig.22 l = 0. Wave function ( )u r for the second excited state (s = 7 - 9) as a
function of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter;
s = 7.0 – 9.0, s 0.2.
Vector analysis 34 1/26/2021
r0 2 ss 7.09.0
r a
4 a u r Third excited state
0.5 1.0 1.5 2.0 2.5 3.0
1.0
0.5
0.5
1.0
Fig.23 l = 0. Wave function ( )u r for the third excited state (s = 7 - 9) as a
function of /r a . 2
2 00 2
2 a Vr
ℏ
. 02
r s
and s is changed as a parameter;
s = 7.0 – 9.0, s 0.2.
12. ContourPlot of y vs x for l
x k0a
y qa
l 1
r0 11 r0 12 r0 13 r0 14
0 2 4 6 8 10 12 14
0
2
4
6
8
10
12
14
Vector analysis 35 1/26/2021
Fig.24 ContourPlot for the boundary condition. y vs x. l = 1. 0E at
0 (1 1 0, )r z n , at which 0 0( ) 0j r . 0 11 12 13 14, , , r
x k0a
y qa
l 2
r0 21 r0 22 r0 23 r0 24
0 2 4 6 8 10 12 14
0
2
4
6
8
10
12
14
Fig.25 ContourPlot for the boundary condition. y vs x. l = 2. 0E at
0 (2 1 1, )r z n , at which 1 0( ) 0j r . 0 21 22 23 24, , , r .
Vector analysis 36 1/26/2021
x k0a
y qa
l 3
r0 31 r0 32 r0 33 r0
0 2 4 6 8 10 12 14
0
2
4
6
8
10
12
14
Fig.26 ContourPlot for the boundary condition. y vs x. l = 3. 0E at
0 (3 1 2, )r z n , at which 2 0( ) 0j r . 0 31 32 33 34, , , r .
Vector analysis 37 1/26/2021
x k0a
y qa
l 4
r0 41 r0 42 r0 43
0 2 4 6 8 10 12 14
0
2
4
6
8
10
12
14
Fig.27 ContourPlot for the boundary condition. y vs x. 4l . 0E at
0 (4 1 3, )r z n , at which 3 0( ) 0j r . 0 41 42 43, , r .
13. The energy eigenvalue E vs potential depth with l = 1, 2, 3, and 4.
We now use the ContourPlot of
(1)
1 1
(1)
( ) ( )
( ) ( )
l l
l l
xj x iyh iy
j x h iy
with 2 2
0x r y
We get the scaled energy eigenvalue 2
2 2
2
2( )
ay qa E
ℏ as a function of scaled
potential depth 2
2 00 2
2 a Vr
ℏ
for l = 1, 2, 3, and 4.
Vector analysis 38 1/26/2021
r02 2 a2
2V0
y 2 qa 2 2 a2
2E
l 0 l 1 l 2 l 3
0 20 40 60 80 100
0
20
40
60
80
100
Fig.28 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
for l = 1, 2, 3, and 4.
0E at 0 ( 1, )r z l n , at which 1 0( ) 0lj r
Vector analysis 39 1/26/2021
r02 2 a2
2V0
y2 qa 2 2 a2
2E
l 0 l 1 l 2
l 3
0 10 20 30 40
0
5
10
15
20
Fig.29 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
for l = 0, 1, 2, 3, and 4.
0E at 0 [ ( 1, )]r z l n , at which 1 0( ) 0lj r
Vector analysis 40 1/26/2021
r02 2 a2
2V0
y22 a2
2E
l 1
0 20 40 60 80 100 120
0
20
40
60
80
100
Fig.30(a) 1l . 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
. 2 2 2 2
0 11 12 13, , r
(vertical blue lines).
Vector analysis 41 1/26/2021
r02 2 a2
2V0
y22 a2
2E
l 2
0 20 40 60 80 100 120
0
20
40
60
80
100
Fig.30(b) l = 2. 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
. 2 2 2 2
0 21 22 23, , r
(vertical blue lines).
Vector analysis 42 1/26/2021
r02 2 a2
2V0
y22 a2
2E
l 3
0 20 40 60 80 100 120
0
20
40
60
80
100
Fig.30 (c) l = 3. 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
. 2 2 2 2
0 31 32 33, , r
(vertical blue lines).
Vector analysis 43 1/26/2021
r02 2 a2
2V0
y2 2 a2
2E l 4
0 20 40 60 80 100 120
0
10
20
30
40
50
60
70
Fig.30(d) l = 4. 2
2 2
2
2( )
ay qa E
ℏ vs
22 0
0 2
2 a Vr
ℏ
. 2 2 2
0 41 42, r (vertical blue
lines).
As is shown From Figs.30, there is no bound state for
2
20
2
2[ ( 1,1)]
V az l
ℏ.
There is a single bound state (ground state) for
2
2 20
2
2[ ( 1,1)] [ ( 1, 2)]
V az l z l
ℏ.
There are two bound states (ground state and the first excited state) for
2
2 20
2
2[ ( 1, 2)] [ ( 1,3)]
V az l z l
ℏ.
There are three bound states (ground state, the first excited state, and the second excited
state) for
Vector analysis 44 1/26/2021
22 20
2
2[ ( 1,3)] [ ( 1, 4)]
V az l z l
ℏ.
Here we define the minimum energy eigenvalue E among the allowed states for fixed
0V . The value of E is the closest to E = 0 (but E<0). We make a plot of the minimum
eigenvalue E as a function of potential depth 0V .
14. The effect of the angular momentum
The minimum value of V0a2 for the p-wave binding (l = 1) is larger than that for the s-
wave binding (l = 0), and so on.
2
20012
2 V a
ℏ (l = 0)
2
20112
2 V a
ℏ (l = 1),
2
20212
2 V a
ℏ (l = 2),
2
20312
2 V a
ℏ (l = 3),
2
20412
2 V a
ℏ (l = 4),
where
012
, 11 , 21 4.49341 1.430297 ,
31 5.76346 1.83457 , 41 6.9879 2.22432
Physically, the meaning of this is very clear. In the case of 1l , there exists a centrifugal
barrier and, therefore, a particle requires stronger attraction for binding. In fact, it can be
shown that the strength of the spherical potential well, 2
0V a , required to bind a particle of
arbitrary l increases monotonically with l. This system does not show any degeneracy in
the l quantum number. (Das).
15. The use of Heisenberg’s principle of uncertainty.
David Bohm, Quantum Theory (Dover, 1079).
The fact that no bound states are possible unless
Vector analysis 45 1/26/2021
2
20
2
2( )
2
a V
ℏ,
is easily understood in terms of the Heisenberg’s principle of uncertainty. To have a
bound state, a particle must be localized roughly within the radius of the well a. To have
a wave function large only in a region of the size of the well, there must also be a range
of momenta /p a ℏ and, therefore, the energy is
2 21 1( )
2 2p
a
ℏ.
Before a particle can be trapped within the well, the potential energy given up when the
particle enters the well must be greater than the kinetic energy that the particle obtains
merely because it is localized within the radius a. Thus, no bound states at all are possible
unless
2
0
1( ) 0
2E V
a
ℏ, or
2
0
2
21
a V
ℏ (bound state)
If 0V is barely great enough to provide the kinetic energy necessary to localize the
particle within the well, then the binding energy E will be very small.
If 0V is increased, the binding energy becomes greater, and eventually 0V becomes so
great that it can supply the kinetic energy necessary to make the wave function oscillate
once within the well. At this point, a new bound state becomes possible. If 0V is made
greater still, eventually a third oscillation becomes possible, then a fourth, etc. Thus, the
number of bound states depends on how much deeper the well is than the minimum
amount needed to contain the particle within the well. We now apply the Heisenberg’s
principle of uncertainty to the case of 1, 2,3,4,l …. Suppose that the linear momentum
is approximated as
0paℏ
,
where 0 is on the order of unity. Then energy E is evaluated as
2
2
0 02[ ( 1)]
2E l l V
a
ℏ≃ .
Vector analysis 46 1/26/2021
The bound state with 0E , appears, only if
2
2 200 02
2( 1) ( 1)
a Vr l l l l
ℏ.
which is consistent with the condition that the effective potential at r a is negative.
2
0 ( 1)r l l .
We make a plot of 2 2 2
0 02 /r a V ℏ as s function of l (l = 0, 1, 2, 3, and 4).
l
r02 2 a2 V0
2
1 2 3 4
10
20
30
40
50
Fig.31 2
2 00 2
2 a Vr
ℏ
vs the orbital angular momentum l. 2
01 2.4674 ..
2
11 9.8696 . 2
21 20.1907 . 2
31 33.2175 . 2
41 48.8307 .
As shown in Fig.31, the smallest value of 2 2 2
0 02 /r a V ℏ for which exists a bound state
(ground state) with l = 1, is greater than the corresponding value of 2 2 2
0 02 /r a V ℏ for
0l . This is due to the additional repulsive centrifugal potential energy. A particle
possessing angular momentum requires a stronger attractive potential to bind it than a
particle with no angular momentum number. Indeed, it turns out that the minimum square
well potential strength 2 2 2
0 02 /r a V ℏ required to bind a particle of orbital angular
momentum quantum number l increases monotonically with increasing l.
16. Summary
Although the spherical square well is not a realistic model for the internuclear
potential it does give us some useful information on the nuclear force. We note that the
proton and neutron have spin 1/2. In fact, the deuteron has an intrinsic spin (S = 1), but
not 0;
Vector analysis 47 1/26/2021
1/2 1/2 1 0D D D D .
This indicates that the nuclear force is spin dependent. The deuteron has a magnetic
moment and an electric quadrupole moment (Q = 0.0027 x 10-24 cm2). The existence of
the quadrupole moment tells us that the system is not strictly expressed by a spherically
symmetric force. However, the departure of spherical symmetry turns out not to be large.
The ground state of the deuteron is a mixture of 96 % ( 0l ) and 4% ( 2l ) states. This
mixing is due to a spin-orbit coupling in the nucleon-nucleon interaction which is
neglected in our simple model. Further discussion will be given elsewhere.
REFERENCES
H.A, Bethe and P. Morrison, Elementary Nuclear Theory, 2nd edition (Dover, 2006).
L.D. Landau and E.M. Lifshitz, Quantum Mechanics, 3rd edition (Pergamon, 1977).
D. Bohm, Quantum Theory (Dover, 1979).
L.I. Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955).
E. Merzbacher, Quantum mechanics, 1st edition (John Wiley & Sons, 1961).
J.S. Townsend, A Modern Approach to Quantum Mechanics, 2nd edition (University
Science Books, 2012).
A. Goswami, Quantum Mechanics, 2nd edition (Waveland Press, 2003).
N. Zettili, Quantum Mechanics, Concepts and Applications, 2nd edition (Wiley, 2009).
A. Das, Lectures on Quantum Mechanics, 2nd edition (World Scientific, 2012). P.223 -
230.
F.S. Levin, An Introduction to Quantum Theory (Cambridge University Press, 2002).
R.L. Liboff, Introductory Quantum Mechanics, 4th edition (Addison-Wesley, 2003).
________________________________________________________________________
APPENDIX-1
Spherical Hankel function of the first kind
(1)
1 ( )ixe
h xx
. (which will be derived below)
xiexh ix 1
)()1(
0 .
2
)1(
1 )(x
ixexh
ix.
3
2)1(
2
33)(
x
ixxiexh
ix .
Vector analysis 48 1/26/2021
4
23)1(
3
15156)(
x
ixixxexh
ix .
Spherical Bessel function
1
cos( )
xj x
x (which will be derived below)
x
xxj
sin)(0 ,
21
cossin)(
x
xxxxj
,
3
2
2
cos3sin)3()(
x
xxxxxj
,
4
22
3
cos)15(sin)25(3)(
x
xxxxxxj
.
Recursion formula:
1 1 0
1( ) ( ) ( )j x j x j x
x .
Rayleigh formula:
x
x
dx
d
xxxj lll
l
sin)
1()1()( .
x
e
dx
d
xxixh
ixlll
l )1
()1()()1( .
Mathematica
( )lj z : SphericalBesselJ[l,z]
(1) ( )l
h z : SphericalHankelH1[l,z]
((Note)) Derivation of the expression of 1( )j x and (1)
1 ( )h x
Using
Vector analysis 49 1/26/2021
1/2( ) ( )2
l lj x J xx
, 1/2
2( ) cosJ x x
x ,
we get
1 1/2
2 cos( ) ( ) cos
2 2
xj x J x x
x x x x
.
We also get
1 0 1
1 cos( ) ( ) ( )
xj x j x j x
x x ,
(1) (1) (1)
1 0 1
1( ) ( ) ( )
ixeh x h x h x
x x .
Plot of 1 0 1( ), ( ), ( )j x j x j x as a function of x
j 1 x j0 x j1 xx
2 3 2
1 2 3 4 5 6
0.5
1.0
Fig.A-1 Plots of 1 0 1( ), ( ), ( ),j x j x j x as a function of x.
Plot of 1 0
(1) (1)( ), ( )ih ix h ix
as a function of x
Vector analysis 50 1/26/2021
ih 1
1ix
h01ix
x1 2 3 4 5
0.4
0.2
0.2
0.4
Fig.A2: Plots of 1 0
(1) (1)( ), ( )ih ix h ix
as a function of x.
APPENDIX-2 Mathematica
((Mathematica-1))
l = 1
Vector analysis 51 1/26/2021
Vector analysis 52 1/26/2021
x k0a
y qa
l 4
r0 4,1 r0 4,2 r0 4,3
0 2 4 6 8 10 12 14
0
2
4
6
8
10
12
14
Vector analysis 53 1/26/2021
((Mathematica-2))
Vector analysis 54 1/26/2021
Vector analysis 55 1/26/2021