Finite Volcano Potentials Admitting a Rational Eigenfunction

Abstract

Starting with a general rational wave function, we search for potentials admitting it as a bound energy eigenfunction. We thus derive singular and regular potentials asymptotically decaying as the inverse of x squared, with the latter being simple or multiple volcanoes having a finite number of bound eigenstates. We present specific examples and examine the transition from singular to volcano potentials.

Full text

1 22 March 2018
Finite Volcano Potentials Admitting a Rational Eigenfunction
Spiros Konstantogiannis
spiroskonstantogiannis@gmail.com
22 March 2018
Abstract
Starting with a general rational wave function, we search for potentials
admitting it as a bound energy eigenfunction. We thus derive singular and
regular potentials asymptotically decaying as the inverse of x squared,
with the latter being simple or multiple volcanoes having a finite number
of bound eigenstates. We present specific examples and examine the
transition from singular to volcano potentials.
Keywords: volcano potentials, bound states, rational wave functions, nodes, poles

22 March 2018
2
Contents
Finite Volcano Potentials Admitting a Rational Eigenfunction........................... 1
Contents ............................................................................................................ 2
1. Introduction................................................................................................... 3
2. Rational eigenfunctions.................................................................................. 3
3. The potential.................................................................................................. 4
4. Distributing the zeros of the wave function as typical nodes or simple poles in
the potential....................................................................................................... 9
5. Examples of volcano potentials.....................................................................10
The case m=4, n=1............................................................................................10
Transition from a singular to a regular potential, in a m=4, n=1 case.................17
A double finite volcano with four bound eigenstates, in a m=6, n=3 case..........19
A triple finite volcano with five bound eigenstates, in a m=6, n=4 case.............23
The case n=0.....................................................................................................29
The case m=4, n=2............................................................................................31
6. References....................................................................................................40

22 March 2018
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1. Introduction
Volcano potentials frequently appear in braneworld scenarios [1-4], while they’ve
also been used in non-relativistic quantum mechanics [5].
A typical volcano potential includes a well, usually at the origin, having finite bottom
and height on both sides, while at infinity, it can be unbounded from below (minus
infinity) or it can tend to a constant, finite value (including zero) [6].
Herein, we examine volcano potentials admitting a rational bound energy
eigenfunction. These volcano potentials, which include a simple or multiple well and
decay asymptotically as
2
1
x
, have a finite number of bound eigenstates, with the
highest of them having fixed energy that can be set to zero. Since they are finite at
infinity, we’ll refer to them as finite volcano potentials.
2. Rational eigenfunctions
We consider the function
( )
0
0
n
m
x
P
x
x A
x
Q
x
y
æ ö
ç ÷
è ø
=
æ ö
ç ÷
è ø
(1)
where
,
n m
P Q
are polynomials of degrees
,
n m
, respectively, with
2
m n
³ +
[7], and
0
x
is a positive length scale.
The polynomial
m
Q
has no zeros
1
, i.e. it has constant sign in
¡
, so that
(
)
x
y
has no
singularities. Then,
m
must be even
2
.
Notes
1. By zero, we mean real zero.
2. Every odd-degree polynomial has at least one zero, since it is continuous and its
values at minus and plus infinity have different signs.
A
is the normalization constant with dimensions of wave function, i.e.
[
]
1 2
A L
-
=
.
Since
0
x x
is dimensionless, all coefficients of
n
P
have the same dimensions and,
likewise, all coefficients of
m
Q
have also the same dimensions.

22 March 2018
4
Then, since the normalization constant has dimensions of wave function, the
dimensions of the coefficients of the two polynomials are the same and they are
eliminated.
Thus, without loss of generality, we assume that the coefficients of the two
polynomials are dimensionless.
Besides, if
n
p
and
m
q
are, respectively, the leading coefficients of
n
P
and
m
Q
, then
0 0
0 0
n n
n
m
m m
x x
P P
x x
p
q
x x
Q Q
x x
æ ö æ ö
ç ÷ ç ÷
è ø è ø
=
æ ö æ ö
ç ÷ ç ÷
è ø è ø
%
%
where the polynomials
n
P
%
and
m
Q
%
are monic in
0
x x
.
Incorporating the dimensionless factor
n m
p q
into the normalization constant, we
make both polynomials monic, and thus, without loss of generality, we assume that
both
n
P
and
m
Q
are monic.
Since
m
Q
has no zeros,
(
)
x
y
has no singularities.
Moreover, as we see from (1),
(
)
x
y
is infinitely many times differentiable on
¡
, i.e.
it is
(
)
C R
¥
.
Also, since
(
)
(
)
deg deg 2
m n
Q P
³ +
,
(
)
x
y
tends to zero faster than
2
1
x
, as
x
® ¥
,
and thus it is square integrable.
Therefore,
(
)
x
y
, as given by (1), is eligible to describe a bound energy eigenstate of
some potential.
We’ll see that, although the wave function
(
)
x
y
is
(
)
C R
¥
, the potential can have
singularities, particularly simple poles.
3. The potential
Assuming that the wave function (1) is an eigenfunction of energy
r
E
, of some
potential
(
)
V x
, it will satisfy the time-independent Schrödinger equation, i.e.
( ) ( )
( )
( )
0
2
2
0
r
m
x E V x x
y y
¢¢
+ - =
h

22 March 2018
5
We denote the mass by
0
m
, so that we don’t confuse it with the degree
m
of
m
Q
.
Solving the previous equation for the potential yields
( )
(
)
( )
2
0
2
r
x
V x E
m x
y
y
¢¢
= +
h
Since the potential depends on the ratio
(
)
(
)
x x
y y
¢¢
, the normalization constant
A
is eliminated and we can omit it in the calculation of the potential.
Introducing the dimensionless variable
0
x
x
x
=
%
(2)
we have
2 2
2 2 2
0
1
d d
dx x dx
=
%
and, in terms of
x
%
, the potential is written as
( )
(
)
( )
2
2
0 0
2
r
x
V x E
m x x
y
y
¢¢
= +
%
h
%%
(3)
where now the prime denotes differentiation with respect to
x
%
.
In terms of
x
%
, (1) is written as, omitting the normalization constant,
( )
(
)
( )
n
m
P x
x
Q x
y
=
%
%
%
(4)
Differentiating with respect to
x
%
, we obtain
( ) ( )
( ) ( ) ( )
( )
2
n n m
m m
P x P x Q x
xQ x Q x
y
¢ ¢
¢= -
% % %
%% %
Differentiating once again yields
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( )
2
2 2 3
2
n n m n m n m n m
m m m m
P x P x Q x P x Q x P x Q x P x Q x
xQ x Q x Q x Q x
y
¢¢ ¢ ¢ ¢ ¢ ¢¢ ¢
+
¢¢
= - - + =
% % % % % % % % %
%% % % %
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( )
2
2 3
2 2
n n m n m n m
m m m
P x P x Q x P x Q x P x Q x
Q x Q x Q x
¢¢ ¢ ¢ ¢¢ ¢
+
= - + =
% % % % % % %
% % %

22 March 2018
6
( ) ( ) ( ) ( ) ( ) ( ) ( )
(
)
( ) ( )
(
)
( )
2 2
3
1
2 2
m n m n m n m n m
m
Q x P x Q x P x Q x P x Q x P x Q x
Q x
¢¢ ¢ ¢ ¢¢ ¢
= - + + =
% % % % % % % % %
%
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
(
)
( )
(
)
( )
2 2
3
1
2 2
m n m m n m m m n
m
Q x P x Q x Q x P x Q x Q x Q x P x
Q x
¢¢ ¢ ¢ ¢ ¢¢
= - + - =
% % % % % % % % %
%
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )
22
3
22
m n m m n n
m m m
n m
Q x P x Q x Q x P x P x
Q x Q x Q x
P x Q x
æ ö
¢¢ ¢ ¢
-¢ ¢¢
ç ÷
= + - =
ç ÷
è ø
% % % % % %
% % %
% %
( ) ( ) ( ) ( ) ( )
(
)
( ) ( ) ( ) ( ) ( )
( )
2
2
22
m m n m n
m m m
n m
Q x Q x P x Q x P x
x
Q x Q x Q x
P x Q x
y
æ ö
¢¢ ¢ ¢
-
ç ÷
¢ ¢¢
= + -
ç ÷
ç ÷
è ø
% % % % %
%
% % %
% %
where in the last equality, we used (4).
Thus
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
2
2
2 2
m m m m n m n
m m n
x Q x Q x Q x Q x P x Q x P x
x Q x Q x P x
y
y
¢ ¢¢ ¢¢ ¢ ¢
¢¢ - -
= +
% % % % % % % %
% % % %
Substituting into (3) yields
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
2
2
2 2
0 0
2 2
2
m m m m n m n
r
m m n
Q x Q x Q x Q x P x Q x P x
V x E
m x Q x Q x P x
æ ö
¢ ¢¢ ¢¢ ¢ ¢
- -
ç ÷
= + +
ç ÷
è ø
% % % % % % %
h
%% % % (5)
In (5), the term
( ) ( ) ( )
(
)
( )
2 2
2
m m m m
Q x Q x Q x Q x
¢ ¢¢
-
% % % %
is regular and does not depend
on
(
)
n
P x
%
, while the term
( ) ( ) ( ) ( )
(
)
( ) ( )
( )
2
m n m n m n
Q x P x Q x P x Q x P x
¢¢ ¢ ¢
-
% % % % % %
has
singularities at the zeros of
(
)
n
P x
%
(if any).
We’ll consider only removable singularities – which are practically no singularities –
and simple poles in the potential (5).
If at the zeros of
(
)
n
P x
%
, the potential (5) has removable singularities, it can be written
as a regular function, which is actually
(
)
C R
¥
. In this case, all zeros of
(
)
n
P x
%
are
simple [8], i.e. of multiplicity 1.
Next, we’ll derive the expression of the potential at long distances.
To this end, we observe that
( )
(
)
( )
2
deg 2 1
m
Q x m
¢
= -
%

22 March 2018
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( ) ( )
(
)
( )
deg 2 2 1
m m
Q x Q x m m m
¢¢
= + - = -
% %
(
)
(
)
2
deg 2
m
Q x m
=
%
Since
m
Q
is monic, the leading term of the polynomial
( ) ( ) ( )
2
2m m m
Q x Q x Q x
¢ ¢¢
-
% % %
is
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 1 2 1 2 1 2 1
2 2 2
2 1 2 1
m m m m
m x m m x m m m x m m x
- - - -
- - = - - = +
% % % %
.
The leading term of the polynomial
(
)
2
m
Q x
%
is
2
m
x
%
.
Thus, at long distances,
( ) ( ) ( )
( )
(
)
( )
2 1
2
22
2 2 2
2
m
m m m
m
m
m m x
Q x Q x Q x
m m
Q x x x
-
¢ ¢¢ +
-
+
=
%
% % % :
% % %
In the same way, we have
( ) ( )
( )
}
*
deg 2 2
m n
Q x P x m n m n
¢¢
= + - = + -
% %
* If
0,1
n
=
, the polynomial
( ) ( )
m n
Q x P x
¢¢
% %
vanishes.
Also
( ) ( )
( )
}
**
deg 1 1 2
m n
Q x P x m n m n
¢ ¢
= - + - = + -
% %
** If
0
n
=
, the polynomial
( ) ( )
m n
Q x P x
¢ ¢
% %
vanishes
Thus
( ) ( ) ( ) ( )
(
)
2, 0
deg 2
0, 0
m n m n
m n n
Q x P x Q x P x n
+ - ¹
ì
¢¢ ¢ ¢
- = í
=
î
% % % %
Also
(
)
(
)
(
)
deg
m n
Q x P x m n
= +
% %

22 March 2018
8
Since the polynomials
n
P
and
m
Q
are monic, the leading term of the polynomial
( ) ( ) ( ) ( )
2
m n m n
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
is
(
)
(
)
2
1 2
m n
n n mn x
+ -
- -
%
, and vanishes if
0
n
=
, as it
should, since then the previous polynomial is zero.
The leading term of the polynomial
(
)
(
)
m n
Q x P x
% %
is
m n
x
+
%
, and thus, at long distances,
( ) ( ) ( ) ( )
( ) ( ) ( )
(
)
( )
2
2
1 2
2 1 2
m n
m n m n
m n
m n
n n mn x
Q x P x Q x P x n n mn
Q x P x x x
+ -
+
¢¢ ¢ ¢ - -
- - -
=
%
% % % % :
% % % %
We observe that both terms of the potential (5) decay as
2
1
x
%
and thus, at long
distances,
( )
(
)
2
2
2 2
0 0
1 2
2
r
m m n n mn
V x E
m x x
¥
+ + - -
= +
h
%
%
Since
( ) ( ) ( )( )
2
2 2 2
1 2 2 1
m m n n mn m mn n m n m n m n m n m n
+ + - - = - + + - = - + - = - - +
we end up to
( )
(
)
(
)
2
2 2
0 0
11
2
r
m n m n
V x E
m x x
¥
- - +
= +
h
%
%
(6)
We see that, at long distances, the potential (5) is symmetric and, since
(
)
(
)
1 0
m n m n
- - + >
*
, the term
(
)
(
)
2 2 2
0 0
1 2
m n m n m x x
- - +
%
h
is repulsive.
We also see that, at long distances, the potential depends only on the difference
m n
-
of the degrees of the two polynomials
(
)
m
Q x
%
and
(
)
n
P x
%
. It does not depend on the
coefficients of the two polynomials, thus it does not depend on the number of the
zeros of
(
)
n
P x
%
either.
* Actually, since
2
m n
³ +
,
(
)
(
)
1 6
m n m n
- - + ³
.
Therefore, all potentials (5) – regular or singular – with the same difference
m n
-
have the same symmetric asymptotic form
(
)
(
)
2 2 2
0 0
1 2
m n m n m x x
- - +
%
h
, plus a

22 March 2018
9
constant
r
E
, which can be set to zero choosing the infinity as reference point and
(
)
0
V
±¥ =
.
In any case, since
(
)
r
V E
±¥ =
, the energy
r
E
is the highest bound energy of the
potential (5).
Proof
Indeed, assuming that there exists a bound energy
r r
E E
¢
>
, then if
(
)
x
j
is the
respective eigenfunction
( ) ( )
( )
( )
0
2
2
0
r
m
x E V x x
j j
¢¢ ¢
+ - =
h
(7)
From (5), we see that if the polynomial
(
)
n
P x
%
has no zeros, the potential is
continuous everywhere, while if it has, then if
(
)
{
}
max
max | 0
i n i
x x P x
= =
, the
potential is continuous for
max
x x
>
.
In any case, the potential is continuous at long distances, and then
(
)
(
)
r
V x V E
±¥ =
;
, and thus from (7) we obtain
( ) ( ) ( )
0
2
2
0
r r
m
x E E x
j j
¥ ¥
¢¢ ¢
+ - =
h
Then, since
0
r r
E E
¢
- >
,
( ) ( )
0
2
2
exp r r
m
x i E E x
j
¥
æ ö
¢
± -
ç ÷
ç ÷
è ø
:h.
Thus, the probability density
( )
2
x
j
¥ is either a non-zero constant or it oscillates.
In either case, the probability density does not tend to zero, and thus the eigenstate
j
is not bound.
Therefore, the wave function (4) describes the highest bound eigenstate and the
energy
r
E
is the highest bound energy of the potential (5).
4. Distributing the zeros of the wave function as typical nodes or
simple poles in the potential
From (3), we see that at each simple zero of the wave function
(
)
x
y
%
which is not a
zero of its second derivative, the potential has a simple pole, or equivalently, at each

22 March 2018
10
simple zero of the polynomial
(
)
n
P x
%
which is not a zero of the polynomial
( ) ( ) ( ) ( )
2
m n m n
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
, the potential (5) has a simple pole.
If at a zero of
(
)
x
y
%
– equivalently of
(
)
n
P x
%
– the potential (5) has a simple pole, that
zero is considered as a special zero, not as a typical node [8].
If
(
)
x
y
%
– equivalently
(
)
n
P x
%
– has no zeros, then
(
)
x
y
%
is the ground-state wave
function [8, 9] of the potential (5), which is then regular and has only one bound state,
of energy
r
E
. The potential in this case is a simple or multiple finite volcano.
If
(
)
x
y
%
– equivalently
(
)
n
P x
%
– has one zero which is a simple pole of the potential
(5), then
(
)
x
y
%
is the ground-state wave function, while if the zero is not pole of the
potential (5),
(
)
x
y
%
is the first-excited-state wave function of the potential (5). In the
first case, the potential is singular, as it has a simple pole, and it has only one bound
state, of energy
r
E
, while in the second case, the potential is regular and has two
bound eigenstates, the ground state and the first-excited state, with the energy of the
second being
r
E
, and the potential in this case is a simple or multiple finite volcano.
In the general case, where
(
)
x
y
%
– equivalently
(
)
n
P x
%
– has
r
zeros, with
0,1,...,
r n
=
, some of them may be typical nodes in the wave function and the others
may be simple poles in the potential.
5. Examples of volcano potentials
The case m=4, n=1
We’ll examine the case where
4
m
=
and
1
n
=
. We remind that
2
m n
³ +
and even.
For simplicity, we’ll assume that
(
)
4
Q x
%
is symmetric, i.e.
(
)
4 2
4 2 0
Q x x q x q
= + +
% % %
(8)
Since
1
n
=
,
(
)
1 0
P x x p
= +
% %
(9)
The polynomial (8) must have no zeros, and since it is monic, it must be positive for
every
x
Î
%
¡
.
The non-normalized wave function (4) is then written as

22 March 2018
11
( )
0
4 2
2 0
x p
x
x q x q
y
+
=
+ +
%
%
% %
(10)
Assuming that the potential vanishes at infinity, i.e.
(
)
0
V
±¥ =
, yields
0
r
E
=
.
The wave function has a zero, at
0
x p
= -
%
, which can be a typical node in the wave
function or a simple pole in the potential.
For
0
p
-
to be a typical node, it must be a zero of the polynomial
( ) ( ) ( ) ( )
4 1 4 1
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
.
Since
( )
1
1
P x
¢
=
%
and
( )
1
0
P x
¢¢
=
%
, the previous polynomial equals
( )
4
2
Q x
¢
-
%
, and thus
0
p
-
must be a zero of
( )
4
Q x
¢
%
, i.e.
( )
4 0
0
Q p
¢
- =
.
Using that
( )
(
)
3 2
4 2 2
4 2 2 2
Q x x q x x x q
¢= + = +
% % % % % , we obtain
(
)
2
0 0 2
2 0
p p q
+ =
and thus
0
0
p
=
or
2
2 0
2 0
q p
= - £
Every even-degree polynomial
(
)
2n
R x
%
with positive leading coefficient has a
global minimum, i.e.
(
)
2 0
n
R x R
³
%
, for every
x
Î
%
¡
.
This happens because
(
)
2n
R x
%
is continuous and finite for every finite
x
%
, and it
tends to plus infinity as
x
%
tends to minus or plus infinity.
Then, if
0
e
>
, the polynomial
(
)
2 0n
R x R
e
- +
%
, which differs from
(
)
2n
R x
%
only
in the constant term, is positive for every
x
Î
%
¡
.
Thus, if the constant term of an even-degree polynomial with positive
leading coefficient is big enough, the polynomial is everywhere positive, no
matter what its intermediate coefficients are.
Therefore,
(
)
4
Q x
%
can be positive even if
2
q
is negative.
For
0
0
p
=
, the wave function (10) becomes
( )
4 2
2 0
x
x
x q x q
y
=
+ +
%
%
% %

22 March 2018
12
For
2
2 0
2
q p
= -
, the wave function (10) becomes
( )
0
4 2 2
0 0
2
x p
x
x p x q
y
+
=
- +
%
%
% %
Both wave functions describe the first-excited state of a respective finite volcano
potential having only two bound eigenstates, the highest of which has zero energy.
Observe that if
0
0
p
=
and
2
0
q
=
, the two previous cases are merged.
In any other case, i.e. if
0
0
p
¹
and
2
2 0
2
q p
¹ -
, the wave function (10) describes the
ground state of a singular potential with only one bound state having zero energy.
Next, we’ll examine the case where
0
0
p
=
and
2
0
q
=
.
The wave function (10) then becomes
( )
4
0
x
x
x q
y
=
+
%
%
%
(11)
with
0
0
q
>
, so that the denominator does not vanish.
For
0
0
p
=
,
(
)
1
P x x
=
% %
and thus
( )
1
1
P x
¢
=
%
,
( )
1
0
P x
¢¢
=
%
.
For
2
0
q
=
,
(
)
4
4 0
Q x x q
= +
% %
and thus
( )
3
4
4
Q x x
¢=
% %
,
( )
2
4
12
Q x x
¢¢ =
% %
.
Plugging into (5), we obtain, for
0
r
E
=
,
( )
( ) ( )
( ) ( )
2
3 4 2
2 3
0
2
24
4
0 0 0
0
2 4 12 8
2
x x q x x
V x m x
x q x
x q
æ ö
- + -
ç ÷
= + =
ç ÷
+
+
è ø
% % % %
h
%
% %
%
( ) ( )
( ) ( ) ( )
2
3 4 2 6 6 2
2 3 2 2
00
2 2
2 2 4
4
4 4
0 0 0 0 0
0
0 0
4 6 16 6 64 4
x x q x x x q xx x
m x m x x q
x q x
x q x q
æ ö æ ö
- + - -
ç ÷ ç ÷
= - = - =
ç ÷ ç ÷
+
+
+ +
è ø è ø
% % % % % %
% %
h h
%
% %
% %
( ) ( )
6 2 6 2
2 2 2 2
0 0
2 2
2 4 2 4
4 4
0 0 0 0 0 0
0 0
10 6 5 34 2 2x q x x q xx x
m x x q m x x q
x q x q
æ ö æ ö
- -
ç ÷ ç ÷
= - = - =
ç ÷ ç ÷
+ +
+ +
è ø è ø
% % % %
% %
h h
% %
% %
(
)
( ) ( )
6 2 2 4
6 2
2 2
0 0 0
2 2
2 2
4 4
0 0 0 0
0 0
5 3 2 3 5
2 2
x q x x x q
x q x
m x m x
x q x q
- - + -
= =
+ +
% % % %
% %
h h
% %
That is
( )
( )
6 2
20
2
24
0 0 0
3 5
2
x q x
V x m x x q
-
=+
% %
h
%% (12)

22 March 2018
13
The potential (12) is symmetric as a result of the wave function (11) having definite
(odd) parity.
If an eigenfunction has definite parity, its second derivative has the same parity,
then the ratio
y y
¢¢
is an even-parity function, and thus the resulting potential is
also of even parity, i.e. symmetric.
At long distances, the potential (12) becomes
(
)
2 2 2
0 0
6
V x m x x
¥
=
% %
h
, in agreement
with (6) for
3
m n
- =
and
0
r
E
=
.
The potential (12) vanishes at
6 2
0
3 5 0
x q x
- =
% %
Thus
0
x
=
%
(double zero),
0
4
5
3
q
x= ±
%
The derivative of the potential (12) is – omitting the factor
2 2
0 0
2
m x
h
which does not
affect the sign of the derivative –
( )
( )
(
)
( ) ( )
6 2 3
5 9 5
05
0 0
0
2 3 2 4
4 4 4 0
0 0 0
3 5 4
18 10 12 20
2
2 9 5
x q x x
x q x x q x
V x x q x x q
x q x q x q
-
æ ö
- -
¢
= - = - - =
ç ÷
+
+ + +
è ø
% % %
% % % %
% % % %
% % %
( ) ( )( )
(
)
5 4 9 5
0 0 0
3
40
29 5 12 20x q x x q x q x
x q
= - + - + =
+% % % % %
%
( ) ( )
9 5 5 2 9 5
0 0 0 0
3
40
29 9 5 5 12 20x q x q x q x x q x
x q
= + - - - + =
+% % % % % %
%
( ) ( ) ( ) ( )
9 5 2 8 4 2
0 0 0 0
3 3
4 4
0 0
2 2
3 24 5 3 24 5x q x q x x x q x q
x q x q
= - + - = - + - =
+ +
% % % % % %
% %
( ) ( )
8 4 2
0 0
3
40
23 24 5
x
x q x q
x q
= - - +
+
%
% %
%
That is
( )
( ) ( )
8 4 2
0 0
3
40
23 24 5
x
V x x q x q
x q
¢= - - +
+
%
% % %
%
Setting
4
y x
=
%
, the trinomial
8 4 2
0 0
3 24 5
x q x q
- +
% %
is written as

22 March 2018
14
2 2
0 0
3 24 5
y q y q
- +
Its discriminant is
2 2 2
0 0 0
576 60 516 0
q q q
- = >
, thus it has two zeros, at
2
0 0 0 0
1,2 0 0 0
24 516 24 22.72
4 3.79 0.21
6 6
q q q q
y q q q
±±
= ± =; ; ,
0
7.79
q
Since both zeros are positive,
4 2 4
1,2 1,2 1,2 1,2 0
, 0.21
x y x y x y y q
= Þ = Þ = ±
% % %
;
,
0
7.79
q
The trinomial
2 2
0 0
3 24 5
y q y q
- +
is then positive for
1
y y
<
or
2
y y
>
, and negative
for
1 2
y y y
< <
.
Thus, the trinomial
8 4 2
0 0
3 24 5
x q x q
- +
% %
is positive for
4 2 4 4 4
1 1 1 1 1
x y x y x y y x y
< Þ < Þ < Þ - < <
% % % %
or
4 2 4 4
2 2 2 2
x y x y x y x y
> Þ > Þ > Þ < -
% % % %
or
4
2
x y
>
%
and it is negative for
4 4
2 1
y x y
- < < -
%
or
4 4
1 2
y x y
< <
%
Then, we have
For
4
2
x y
< -
%
,
(
)
0
V x
¢
>
%
and the potential (12) is strictly increasing.
For
4 4
2 1
y x y
- < < -
%
,
(
)
0
V x
¢
<
%
and the potential (12) is strictly decreasing.
For
41
0
y x
- < <
%
,
(
)
0
V x
¢
>
%
and the potential (12) is strictly increasing.
For
4
1
0
x y
< <
%
,
(
)
0
V x
¢
<
%
and the potential (12) is strictly decreasing.
For
4 4
1 2
y x y
< <
%
,
(
)
0
V x
¢
>
%
and the potential (12) is strictly increasing.
For
4
2
x y
>
%
,
(
)
0
V x
¢
<
%
and the potential (12) is strictly decreasing.
Finally, at
4
1,2
y
±
,
(
)
0
V x
¢
=
%
.
Therefore, at
4 4
2 2
,0,
y y
- the potential has local maxima, while at
4 4
1 1
,
y y
- it has
local minima.

22 March 2018
15
Since the potential is symmetric, its values at the two symmetric minima are equal, as
they are at the two non-zero symmetric maxima too.
The potential (12) is then a symmetric double finite volcano that, at long distances,
decays as
2
1
x
%
. It has only two bound eigenstates, a ground state and a first-excited
state, with the energy of the second being zero.
In Figure 1, the potential (12) is plotted for
0
0.1
q= (red line),
0
1
q
=
(blue line), and
0
10
q
=
(green line), in units
2 2
0 0
2 1
m x
=
h
.
Figure 1
We see that, as
0
q
increases, the minimum value of the potential increases,
approaching zero from below.
The value of the potential at the two minima is
( )
(
)
(
)
( )
( )
(
)
( )
6 2 3
4 4
2 2
1 0 1 1 0 1
412 2
2 2
4
0 0 0 0
41 0
1 0
3 5 3 5
2 2
y q y y q y
V y m x m x y q
y q
± - ± -
± = = =
+
± +
h h
( )
(
)
( )
2 2
1 0 1
1 1 0 1
2 2
2 2
0 0 0 0
1 0 1 0
3 53 5
2 2
y q y
y y q y
m x m x
y q y q
--
= =
+ +
h h
Thus

22 March 2018
16
( )
(
)
( )
2
1 0 1
412
2
0 0 1 0
3 5
2
y q y
V y m x y q
-
± = +
h
(13)
Substituting
1 0
0.21
y q
;
, we obtain
( )
(
)
( )
2 2 2
0 0 0 0 0
402
2 2 2 2
0 0 0 0 0
0 0 0
0
0.63 5 0.21 4.37 0.46
2 2 2.75
0.21 1.46
1.21
q q q q q
V q m x m x q
m x q
q
-
± - -
h h h
; ; ;
The minimum value of the potential is then
2 2
0 0 0
2.75
m x q
-h, and it increases as
0
1
q
-, as
0
q
increases.
Since the ground-state energy must exceed the minimum value of the potential [10], it
will also approach zero from below, as
0
q
increases.
We expect that, as
0
q
® ¥
, the ground-state energy will approach zero as
0
1
q
- or
faster, otherwise the minimum value of the potential will exceed the ground-state
energy.
On the other hand, the first-excited-state energy does not depend on
0
q
, it is always
zero.
Then, as
0
q
increases, the difference between the two bound energies decreases, the
ground-state energy comes closer to the first-excited-state energy.
Similarly to (13), the value of the potential at the two local maxima
4
2
y
± is
( )
(
)
( )
2
2 0 2
422
2
0 0 2 0
3 5
2
y q y
V y m x y q
-
± = +
h
Substituting
2 0
7.79
y q
;
, we obtain
( )
2 2
0 0
402 2 2
0 0 0
0 0 0
18.37 2.79
2 1.33
7.79 77.26
q q
V q m x q
m x q
±
h h
; ;
Since
2 2
0 0 0
1.33 0
m x q
>
h,
2 2
0 0 0
1.33
m x q
h is the global maximum of the
potential, and thus it is the height of the two volcanoes.
The depth of each of the two volcanoes is then

22 March 2018
17
2 2 2
2 2 2
0 0 0 0 0 0 0 0 0
1.33 2.75 4.08
m x q m x q m x q
æ ö
- - =
ç ÷
ç ÷
è ø
h h h
and it goes to zero as
0
1
q
, when
0
q
® ¥
.
The width of each volcano is
4
0
7.79
q
(see Figure 1), thus it increases and goes to
infinity as
4
0
q
, when
0
q
® ¥
.
As
0
q
increases, the two volcanoes become shallower and wider and as
0
q
® ¥
, the
height of the two volcanoes vanishes, while their width becomes infinite, i.e. the
volcanoes decay.
Transition from a singular to a regular potential, in a m=4, n=1 case
We showed that if
0
0
p
¹
and
2
2 0
2
q p
¹ -
, the resulting potential has a simple pole at
0
p
.
For
0
0
p
¹
and
2
0
q
=
, the two previous conditions are satisfied and we have the case
of a singular potential, with a simple pole at
0
p
.
In this case, we have
(
)
1 0
P x x p
= +
% %
(
)
4
4 0
Q x x q
= +
% %
with
0
0
q
>
, so that
(
)
4
0
Q x
¹
%
.
For simplicity, we’ll further assume that
0
1
q
=
.
Then
( )
1
1
P x
¢
=
%
,
( )
1
0
P x
¢¢
=
%
and
( )
3
4
4
Q x x
¢=
% %
,
( )
2
4
12
Q x x
¢¢ =
% %
Plugging into (5), we obtain, for
0
r
E
=
,

22 March 2018
18
( )
( ) ( )
( ) ( )
( )
2
3 4 2
2 3
2
24
4
0 0 0
2 4 1 12 8
21
1
x x x x
V x m x x x p
x
æ ö
- + -
ç ÷
= + =
ç ÷
+ +
+
è ø
% % % %
h
%% %
%
( ) ( )
( )
( ) ( )
( )
2 6 6 2 3 2 6 2 3
2 2
2 2
4 4
4 4
0 0 0 0
0 0
32 12 12 8 20 12 8
2 2
1 1
1 1
x x x x x x x
m x m x
x x p x x p
x x
æ ö æ ö
- - - -
ç ÷ ç ÷
= + = - =
ç ÷ ç ÷
+ + + +
+ +
è ø è ø
% % % % % % %
h h
% % % %
% %
( ) ( )
( )
( ) ( )
( )
2 6 2 3 2 6 2 3
2 2
2 2
4 4
4 4
0 0 0 0
0 0
10 6 4 2 5 3 2
1 1
1 1
x x x x x x
m x m x
x x p x x p
x x
æ ö æ ö
- -
ç ÷ ç ÷
= - = - =
ç ÷ ç ÷
+ + + +
+ +
è ø è ø
% % % % % %
h h
% % % %
% %
(
)
(
)
(
)
( )
( )
( )
( )
6 2 3 4 7 6 3 2 7 32 2
00 0
2 2
2 2
4 4
0 0 0 0
0 0
5 3 2 1 5 5 3 3 2 2
2 2
1 1
x x x p x x
x p x x p x x x
m x m x
x x p x x p
- + - + + - - - -
= = =
+ + + +
% % % % %
% % % % % %
h h
% % % %
( )
( )
7 6 3 2
20 0
2
24
0 0 0
3 5 5 3
2
1
x p x x p x
m x x x p
+ - -
=+ +
% % % %
h
% %
That is, in units
2 2
0 0
2 1
m x
=
h
, the potential is
( )
( )
( )
7 6 3 2
0 0
2
40
3 5 5 3
1
x p x x p x
V x x x p
+ - -
=+ +
% % % %
%% % (14)
In Figure 2, the potential (14) is plotted for
0
1
p
= -
(red line),
0
0
p
=
(blue line), and
0
1
p
=
(green line).

22 March 2018
19
Figure 2
When the parameter
0
p
takes the critical value
0
0
p
=
, the two open edges of the
potential pole, one at plus/minus infinity and the other at minus/plus, are “glued”
together forming a volcano which is added to the existing volcano, resulting in a
symmetric double volcano, and at the same time, one more bound eigenstate is
created, with zero energy.
A similar formation of a volcano takes place at every value of
2
n m
£ -
, with
2
m
³
and even, when the coefficients of the polynomial
( ) ( ) ( ) ( )
2
m n m n
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
take such values that one of the zeros of the previous polynomial becomes equal to
one of the zeros of the polynomial
(
)
n
P x
%
, but the correspondence between the zeros
of
(
)
n
P x
%
and the finally formed volcanoes is not one-to-one, as we’ll show below.
If the polynomial
(
)
n
P x
%
has
n
zeros and the coefficients of the polynomial
( ) ( ) ( ) ( )
2
m n m n
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
can take such values that there exists an
(
)
2
m
-
-
degree polynomial
(
)
2m
R x
-
%
such that
( ) ( ) ( ) ( ) ( ) ( )
2
2
m n m n m n
Q x P x Q x P x R x P x
-
¢¢ ¢ ¢
- =
% % % % % %
(15)
with
(
)
0
m
Q x
¹
%
, then we can transform the
n
simple poles of the initial potential,
which has only one bound eigenstate, with zero energy, into a multiple finite volcano
with
n
bound eigenstates, with the highest of them, i.e. the
n
th-excited state, having
zero energy.
By means of (15), the potential (5) is written as, for
0
r
E
=
,
( ) ( ) ( ) ( )
( ) ( )
( )
2
22
2 2
0 0
2
2
m m m m
m m
Q x Q x Q x R x
V x
m x Q x Q x
-
æ ö
¢ ¢¢
-
ç ÷
= +
ç ÷
è ø
% % % %
h
%
% %
(16)
with
(
)
0
m
Q x
¹
%
.
A double finite volcano with four bound eigenstates, in a m=6, n=3 case
We’ll consider a case where
(
)
3
P x
%
is of odd parity and
(
)
6
Q x
%
is of even parity, i.e.
(
)
3
3
P x x x
= -
% % %

22 March 2018
20
(
)
6 4 2
6 4 2 0
Q x x q x q x q
= + + +
% % % %
with
0
0
q
>
, otherwise
(
)
6
Q x
%
has zeros.
Then,
( )
3
P x
¢¢
%
is of odd parity, thus
( ) ( )
6 3
Q x P x
¢¢
% %
is also of odd parity, while
( )
3
P x
¢
%
is
of even parity and
( )
6
Q x
¢
%
is of odd parity, thus
( ) ( )
6 3
Q x P x
¢ ¢
% %
is again of odd parity,
and finally,
( ) ( ) ( ) ( )
6 3 6 3
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
is of odd parity.
Then, since
(
)
3
P x
%
is of odd parity,
(
)
4
R x
%
in (15) must be of even parity, and thus it
has the form
(
)
4 2
4 4 2 0
R x r x r x r
= + +
% % %
Plugging into (15), we obtain
(
)
(
)
(
)
(
)
(
)
6 4 2 5 3 2 4 2 3
4 2 0 4 2 4 2 0
6 2 6 4 2 3 1x q x q x q x x q x q x x r x r x r x x
+ + + - + + - = + + - Þ
% % % % % % % % % % % %
(
)
7 5 3 7 5 5 3 3
420 4422
6 6 6 6 2 18 6 12 4 6 2
x q x q x q x x x q x q x q x q x
Þ + + + - - + - + - =
% % % % % % % % % %
7 5 5 3 3
4 4 2 2 0 0
r x r x r x r x r x r x
= - + - + - Þ
%%%%%%
7 5 3 7 5 5 3 3
4 2 0 4 4 2 2
6 6 6 6 36 12 24 8 12 4
x q x q x q x x x q x q x q x q x
Þ + + + - + - + - + =
% % % % % % % % % %
7 5 5 3 3
4 4 2 2 0 0
r x r x r x r x r x r x
= - + - + - Þ
%%%%%%
(
)
(
)
7 5 3 5 3 7 5 3
4 2 0 4 2 4 2 4 0 2 0
30 18 6 6 12 8 4
x q x q x q x x q x q x r x r r x r r x r x
Þ - - - + + + + = + - + - - Þ
% % % % % % % % % % %
(
)
(
)
(
)
(
)
(
)
7 5 3 7 5 3
4 4 2 0 2 4 2 4 0 2 0
30 6 2 3 2 4 3 2 3 2
x q x q q x q q x r x r r x r r x r x
Þ - + - + - + + = + - + - -
% % % % % % % %
Comparing the same-degree terms in both sides of the last equation, we obtain
4
30
r
= -
(17)
(
)
2 4 4
6 2 3
r r q
- = -
(18)
(
)
0 2 4 2
2 4 3
r r q q
- = -
(19)
(
)
0 0 2
2 3 2
r q q
- = +
(20)
Substituting (17) into (18) yields
( )
2
2 4 4 4 2 4
30 6 2 3 12 18 18 18 1
18
r
r q q q r q
+ = - = - Þ = - - Þ = - -
Substituting
4
q
into (19) yields

22 March 2018
21
2 2 2
0 2 2 2 2
2 4
2 4 1 3 2 4 3 8 6
18 9 9
r r r
r r q q q
æ ö
æ ö æ ö
- = - - - = - - - = - - - Þ
ç ÷ ç ÷
ç ÷
è ø è ø
è ø
0
2 2 2
2 0 2 0 2
4 5 5
4
6 8 8
9 9 3 54 6
r
r r r
q r r r q
Þ = - - - + = - + - Þ = - + -
Substituting
2
q
into (20) yields
0 0 0
2 2 2
0 0 0 0
2
5 5 104 8 16
2 3 2 2 3 6
3 54 6 3 27 3 3 27 3
r r r
r r r
r q q q
æ ö
æ ö æ ö
- = + - + - = - + - = - + - Þ
ç ÷ç ÷ ç ÷
è ø è ø
è ø
0 0 0
2 2 2
0 0 0
210 10 5
16 16 8
6
3 27 3 3 27 3 9 81 18
r r r
r r r
q r q
Þ = - + - + = - - Þ = - -
If the intermediate coefficients
2
q
and
4
q
of
(
)
6
Q x
%
are non-negative, and the
constant term
0
q
is positive,
(
)
6
Q x
%
has no zeros.
For
4
q
to be non-negative,
2 2 2
1 0 1 18
18 18
r r r
- - ³ Þ £ - Þ £ -
For
2
q
to be non-negative,
0 0
2 2 2
0
5 5 5
4 4
0 8
3 54 6 6 3 54 9
r r
r r r
r- + - ³ Þ £ - + Þ £ - +
For
0
q
to be positive,
0 0
2 2 2
0
5 5 10
8 8
0 16
9 81 18 18 9 81 9
r r
r r r
r- - > Þ < - Þ < -
We observe that if
0
r
is enough negative and
2
18
r
£ -
, all three inequalities are
satisfied.
In the limiting case, where
2
18
r
= -
and
(
)
0
5 18
8 8 10 18
9
r-
= - + = - - = -
,
the third inequality holds, since
(
)
10 18
18 16 18 16 20 36
9
-
- < - Þ - < + =
(it holds)
Then

22 March 2018
22
(
)
0
5 18
8 18 8 10
1 3
9 81 18 9 9
q--
= - - = + + =
and
2 4
0
q q
= =
Thus
(
)
6
6
3
Q x x
= +
% %
and
(
)
4 2
4
30 18 18
R x x x
= - - -
% % %
Substituting into (16) yields
( )
( ) ( )
( ) ( )
2
5 6 4
2 4 2
2
26
6
0 0
2 6 3 30
30 18 18
23
3
x x x x x
V x m x x
x
æ ö
- + - - -
ç ÷
= + =
ç ÷
+
+
è ø
% % % % %
h
%%
%
( ) ( )
( ) ( )
2
5 6 4
2 4 2
2
26
6
0 0
6 3 15
15 9 9
3
3
x x x x x
m x x
x
æ ö
- + + +
ç ÷
= - =
ç ÷
+
+
è ø
% % % % %
h
%
%
(
)
(
)
( )
10 10 4 4 2 6
2
2
26
0 0
36 15 45 15 9 9 3
3
x x x x x x
m x x
- - - + + +
= =
+
% % % % % %
h
%
( )
2 10 10 4 10 4 8 2 6
2
26
0 0
36 15 45 15 45 9 27 9 27
3
x x x x x x x x
m x x
- - - - - - - -
= =
+
% % % % % % % %
h
%
( )
2 10 4 8 2 6
2
26
0 0
6 90 9 27 9 27
3
x x x x x
m x x
- - - - -
= =
+
% % % % %
h
%
( )
2 10 8 6 4 2
2
26
0 0
6 9 9 90 27 27
3
x x x x x
m x x
- - - - -
=+
% % % % %
h
%
That is
( )
( )
2 10 8 6 4 2
2
26
0 0
3 2 3 3 30 9 9
3
x x x x x
V x m x x
- - - - -
=+
% % % % %
h
%% (21)
At long distances, the potential (21) takes the form
( )
2 10 2
2 12 2 2
0 0 0 0
3 2 6 1
x
V x
m x x m x x
¥
= =
%
h h
%
% %

22 March 2018
23
in agreement with the general expression (6) for
3
m n
- =
and
0
r
E
=
.
In Figure 3, the potential (21) is plotted in units
2 2
0 0
3 1
m x
=
h
. It is a symmetric
double finite volcano potential with four bound eigenstates, the highest of which, i.e.
the third-excited state, has zero energy and is described by the wave function
(
)
(
)
(
)
2 6
1 3
x A x x x
y
= - +
% % % %
, where
A
is the normalization constant.
We see that, although the polynomial
(
)
3
3
P x x x
= -
% % %
has three zeros, the potential is
still a double and not a triple volcano.
Figure 3
A triple finite volcano with five bound eigenstates, in a m=6, n=4 case
We’ll consider a case where
(
)
(
)
(
)
2 2 4 2
4
1 2 3 2
P x x x x x
= - - = - +
% % % % %
(
)
6 4 2
6 4 2 0
Q x x q x q x q
= + + +
% % % %
with
0
0
q
>
, otherwise
(
)
6
Q x
%
has zeros.

22 March 2018
24
Both polynomials are of even parity, thus
( ) ( ) ( ) ( )
6 4 6 4
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
is now of
even parity, and since
(
)
4
P x
%
is also of even parity,
(
)
4
R x
%
must be of even parity too,
and thus it has the form
(
)
4 2
4 4 2 0
R x r x r x r
= + +
% % %
If
(
)
m
Q x
%
is of even parity and
(
)
n
P x
%
has definite parity, i.e. it is of even or odd
parity, then
( )
n
P x
¢¢
%
has the same parity as
(
)
n
P x
%
, and thus
( ) ( )
m n
Q x P x
¢¢
% %
has
also the same parity as
(
)
n
P x
%
, while
( )
n
P x
¢
%
has different parity from that of
(
)
n
P x
%
, and since
( )
m
Q x
¢
%
is of odd parity,
( ) ( )
m n
Q x P x
¢ ¢
% %
has the same parity as
(
)
n
P x
%
, and thus the polynomial
( ) ( ) ( ) ( )
2
m n m n
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
has the same
parity as
(
)
n
P x
%
, and then from (15) we derive that
(
)
2m
R x
-
%
must be of even
parity.
That is, if
(
)
m
Q x
%
is of even parity and
(
)
n
P x
%
has definite parity,
(
)
2m
R x
-
%
is of
even parity.
Plugging into (15), we obtain
(
)
(
)
(
)
(
)
6 4 2 2 5 3 3
4 2 0 4 2
12 6 2 6 4 2 4 6
x q x q x q x x q x q x x x
+ + + - - + + - =
% % % % % % % % %
(
)
(
)
4 2 4 2
420
3 2r x r x r x x
= + + - + Þ
% % % %
8 6 6 4 4 2 2
4 4 2 2 0 0
12 6 12 6 12 6 12 6
x x q x q x q x q x q x q
Þ - + - + - + - -
% % % % % % %
(
)
8 6 6 4 4 2 8 6 4 6 4 2
4 4 2 2 4 4 4 2 2 2
2 24 36 16 24 8 12 3 2 3 2
x x q x q x q x q x r x r x r x r x r x r x
- - + - + - = - + + - + +
% % % % % % % % % % % %
4 2 8 6 6 4 4 2 2
0 0 0 4 4 2 2 0 0
3 2 12 6 12 6 12 6 12 6
r x r x r x x q x q x q x q x q x q
+ - + Þ - + - + - + - -
% % % % % % % % %
(
)
(
)
8 6 6 4 4 2 8 6 4
4 4 2 2 4 2 4 4 2 0
48 72 32 48 16 24 3 2 3
x x q x q x q x q x r x r r x r r r x
- + - + - + = + - + - + +
% % % % % % % % %
(
)
2
2 0 0
2 3 2r r x r
+ - + Þ
%
8 6 6 4 4 2 2
4 4 2 2 0 0
36 66 20 42 4 18 12 6
x x q x q x q x q x q x q
Þ - + - + - + + - =
% % % % % % %
(
)
(
)
(
)
8 6 4 2
4 2 4 4 2 0 2 0 0
3 2 3 2 3 2r x r r x r r r x r r x r
= + - + - + + - + Þ
% % % %
(
)
(
)
(
)
8 6 4 2
4 4 2 2 0 0
36 2 33 10 2 21 2 6 3 2 6
x q x q q x q q x q
Þ - + - + - + + - =
% % % %
(
)
(
)
(
)
8 6 4 2
4 2 4 4 2 0 2 0 0
3 2 3 2 3 2
r x r r x r r r x r r x r
= + - + - + + - +
% % % %
Comparing the same-degree terms in both sides of the last equation, we obtain
4
36
r
= -
(22)

22 March 2018
25
(
)
2 4 4
3 2 33 10
r r q
- = -
(23)
(
)
4 2 0 4 2
2 3 2 21 2
r r r q q
- + = -
(24)
(
)
2 0 2 0
2 3 6 3 2
r r q q
- = +
(25)
0 0
2 6
r q
= -
(26)
Substituting (22) into (23) yields
(
)
2 4 2 4 4 2
108 2 33 10 108 66 20 20 42
r q r q q r
+ = - Þ + = - Þ = - -
Thus
2
4
21
10 20
r
q= - -
(27)
Substituting (22) and (27) into (24) yields
2 2 2
2 0 2 2 2
21 2121 21*21 21*21
72 3 2 21 2 2 2 4
10 20 10 20 5 10
r r r
r r q q q
æ ö
æ ö æ ö
- - + = - - - = - - - = - - - Þ
ç ÷ ç ÷
ç ÷
è ø è ø
è ø
2 2
2 2 0 0
21 9
441 81
4 72 3
5 10 5 10
r r
q r r r
Þ = - - + + - = - + -
Thus
0
2
2
9
81
20 40 4
r
r
q
= - + -
(28)
Substituting (28) into (25) yields
0 0
2 2
2 0 0 0
3
9 2781 243
2 3 6 3 2 6 2
20 40 4 20 40 4
r r
r r
r r q q
æ ö
æ ö æ ö
- = - + - + = - + - + =
ç ÷
ç ÷ ç ÷
è ø è ø
è ø
02 0
93* 27
3* 243 12
10 20 2
rr q
= - + - + Þ
0 0
2 2
0 2 0
9 3
3* 27 41
3* 243 729
12 2 3
10 20 2 10 20 2
r r
r r
q r rÞ = - + + - = - +
Thus
0
2
0
3
41
729
120 240 24
r
r
q= - +
(29)
Substituting (29) into (26) yields

22 March 2018
26
0 0 0
2 2 2 2
0 0
3 3 11
41 41 41 41729 729 729 729
2 6 11
120 240 24 20 40 4 40 20 4 10 5
r r r
r r r r
r r
æ ö
= - - + = - + - Þ = + Þ = + Þ
ç ÷
è ø
2
0
41
729
11
5 10
r
rÞ = - +
Thus
2
0
41
729
55 110
r
r= - +
(30)
Substituting (30) into (29) yields
2
2 2 2
0
41
729
3
41 41 123
729 729 2187
55 110
120 240 24 120 240 1320 2640
r
r r r
q
æ ö
- +
ç ÷
è ø
= - + = - - + =
2 2 2 2
123 41 328 41
729 2187 5832 729
120 120*11 240*11 240 1320 2640 165 330
r r r r
= - + - = - = -
That is
2
0
41
729
165 330
r
q= -
(31)
Substituting (30) into (28) yields
2
2 2 2 2 2
2
41729
9 9 41 9 4181 81 729 729 81
55 110
20 40 4 20 40 220 440 20*11 20 40 40*11
r
r r r r r
q- +
= - + - = - + + - = - + - =
2 2
58 29
162 81
220 440 110 220
r r
= - + = - +
That is
2
2
29
81
110 220
r
q= - +
(32)
Since
0
q
must be positive, from (31) we obtain
2 2 2
41 41
729 729 240570
0 35.56
165 330 330 165 6765
r r r- > Þ < Þ < ;
To simplify things, we consider the case where
2
r
vanishes.
Then, (31), (32), and (27) give, respectively,
0
729
0
165
q
= >

22 March 2018
27
2
81
110
q= -
4
21
10
q
= -
Thus, the polynomial
(
)
6
Q x
%
is
( )
6 4 2
6
21 81 729
10 110 165
Q x x x x= - - +
% % % %
The polynomial
(
)
6
Q x
%
is plotted in Figure 4 and, as we see, it is everywhere positive,
i.e. it has no zeros.
Figure 4
For
2
0
r
=
, (30) gives
0
729
55
r= -
Then, since
4
36
r
= -
(eq. (22)), the polynomial
(
)
4
R x
%
is
( )
4
4
729
36
55
R x x= - -
% %

22 March 2018
28
Substituting into (16), we obtain, in units
2 2
0 0
2
m x
h
,
( )
2
5 3 6 4 2 4 2
2
6 4 2
42 81 21 81 729 126 81
2 6 30
5 55 10 110 165 5 55
21 81 729
10 110 165
x x x x x x x x
V x
x x x
æ ö æ öæ ö
- - - - - + - -
ç ÷ ç ÷ç ÷
è ø è øè ø
= +
æ ö
- - +
ç ÷
è ø
% % % % % % % %
%
% % %
4
6 4 2
729
36 55
21 81 729
10 110 165
x
x x x
- -
+
- - +
%
% % %
The previous potential is plotted in Figure 5. It is a symmetric triple finite volcano
potential consisting of two deep, symmetric-around-zero volcanoes and one shallow
volcano which is also symmetric around zero.
The potential has five bound eigenstates, the highest of which, i.e. the fourth-excited
state, is described by the wave function
( )
(
)
(
)
2 2
6 4 2
1 2
21 81 729
10 110 165
A x x
xx x x
y
- -
=
- - +
% %
%
% % %
, where
A
is the normalization constant,
and it has zero energy.
Figure 5

22 March 2018
29
The case n=0
Since
(
)
0
1
P x
=
%
, the wave function (4) is, including the normalization constant
A
,
( ) ( )
m
A
x
Q x
y
=
%
%
(33)
with
2
m
³
and even.
Then, the potential (5) becomes
( ) ( ) ( ) ( )
( )
2
2
2 2
0 0
2
2
m m m
r
m
Q x Q x Q x
V x E
m x Q x
¢ ¢¢
-
= +
% % %
h
%%
Since
(
)
m
Q x
%
has no zeros, the potential has no singularities. Actually it is
(
)
C R
¥
.
The wave function (33) has no zeros, thus it describes the ground state of the previous
regular potential [8, 9], which has only one bound state, of energy
r
E
.
Using that
0
n
=
, from (6) we derive that, at long distances, the previous potential
takes the form
( )
(
)
2
2 2
0 0
11
2
r
m m
V x E
m x x
¥
+
= +
h
%
%
(34)
Assuming that the potential vanishes at infinity,
0
r
E
=
, and then it is written as
( ) ( ) ( ) ( )
( )
2
2
2 2
0 0
2
2
m m m
m
Q x Q x Q x
V x m x Q x
¢ ¢¢
-
=
% % %
h
%%
(35)
As an example, we’ll examine the case where
(
)
2
0
m
m
Q x x q
= +
% %
, with
1, 2,...
m
=
, and
0
0
q
>
, so that
(
)
0
m
Q x
¹
%
. In this case, the wave function has definite (even) parity.
We have
( )
2 1
2m
m
Q x mx
-
¢=
% %
,
( ) ( )
2 2
2 2 1 m
m
Q x m m x
-
¢¢ = -
% %
Plugging into (35) yields

22 March 2018
30
( )
(
)
(
)
( )
( )
2
2 1 2 2 2
20
2
22
0 0 0
2 2 2 2 1
2
m m m
m
mx x q m m x
V x m x x q
- -
- + -
= =
+
% % %
h
%%
(
)
(
)
( )
2 4 2 4 2 2 2
20
2
22
0 0 0
8 2 2 1 2 2 1
2
m m m
m
m x m m x m m q x
m x x q
- - -
- - - -
= =
+
% % %
h
%
(
)
(
)
(
)
( )
2 4 2 2 2
20
2
22
0 0 0
8 2 2 1 2 2 1
2
m m
m
m m m x m m q x
m x x q
- -
- - - -
= =
+
% %
h
%
(
)
(
)
( )
(
)
(
)
(
)
( )
2 4 2 2 2 2 2 2
2 2
0 0
2 2
2 2
2 2
0 0 0 0
0 0
4 2 2 2 1 2 2 1 2 1
2 2
m m m m
m m
m m x m m q x m m x m q x
m x m x
x q x q
- - -
+ - - + - -
= = =
+ +
% % % %
h h
% %
(
)
(
)
(
)
( )
2 2 2
20
2
22
0 0 0
2 1 2 1
m m
m
m m x m q x
m x x q
-
+ - -
=+
% %
h
%
That is
( )
(
)
(
)
(
)
( )
2 2 2
20
2
22
0 0 0
2 1 2 1
m m
m
m m x m q x
V x m x x q
-
+ - -
=+
% %
h
%% (36)
The potential (36) is symmetric, as a result of the wave function having definite
parity.
At long distances, (36) becomes
( )
(
)
(
)
(
)
4 2 2
2 2
2 4 2 2 2 2
0 0 0 0 0 0
2 1 2 2 1 2 2 1
1
2 2
m
m
m m x m m m m
V x
m x x m x x m x x
-
¥
+ + +
= = =
%
h
h h
%
%%%
in agreement with (34) for
0
r
E
=
and
2
m m
®
.
The potential (36) vanishes at
( ) ( )
}
0
2 1 01
2 1
2 2 2
0 0 0
2 1 2 1
2 1 2 1 0 2 1 2 1
mq
mm
m m
m m
m x m q x q x q
m m
->
+
- -
æ ö
+ - - = Þ = Þ = Þ
ç ÷
+ +
è ø
% % %
1
2
0
2 1
2 1
m
m
x q
m
-
æ ö
Þ = ±
ç ÷
+
è ø
%
and at 0 if
1
m
>
.
Since
2 2
0
m
x-
³
%
and the denominator of (36) is positive, we have
(
)
0
V x
>
%
for
1
2
0
2 1
2 1
m
m
x q
m
-
æ ö
>
ç ÷
+
è ø
%

22 March 2018
31
(
)
0
V x
<
%
for
1
2
0
2 1
2 1
m
m
x q
m
-
æ ö
<
ç ÷
+
è ø
%
and
0
x
¹
%
(if
1
m
>
)
Thus, if
1
m
>
, the potential (36) is a symmetric double finite volcano, while if
1
m
=
,
it is a symmetric simple finite volcano, since then it does not vanish at 0.
In both cases,
1
m
=
and
1
m
>
, the potential (36) has only one bound state, its ground
state, with zero energy, which is described by the wave function
(
)
(
)
2
0
m
x A x q
y
= +
% %
.
In Figure 6, the potential (36) is plotted in units
2 2
0 0
1
m x
=
h
, for
0
1
q
=
and
1
m
=
(red line),
2
m
=
(blue line), and
3
m
=
(green line).
Figure 6
The case m=4, n=2
For simplicity, we’ll assume that both
(
)
2
P x
%
and
(
)
4
Q x
%
are symmetric.
Then, since both are also monic, they have the form
(
)
2
2 0
P x x p
= +
% %
(
)
4 2
4 2 0
Q x x q x q
= + +
% % %

22 March 2018
32
with
(
)
4
0
Q x
>
%
.
As explained, for every value of
2
q
, there exists a
0
q
such that the global minimum of
(
)
4
Q x
%
is positive.
Also,
0
0
p
¹
, so that the zeros of
(
)
2
P x
%
(if any) are simple zeros.
The wave function (4) is, including the normalization constant
A
,
( )
20
4 2
2 0
x p
x A
x q x q
y
+
=
+ +
%
%
% %
(37)
The wave function (37) has definite parity (even), thus
(
)
x
y
¢¢
%
has also definite parity
(even), and then the potential (3) is of even parity (symmetric).
Assuming, as usual, that the potential vanishes at infinity,
0
r
E
=
, i.e. the energy of
the eigenstate described by the wave function (37) is zero.
I. If
0
0
p
>
,
(
)
2
0
P x
>
%
and the wave function (37) has no zeros, and thus it describes
the ground state of the potential (5) (with
0
r
E
=
), which is regular.
In this case, the potential is a symmetric double finite volcano.
II. If
0
0
p
<
,
(
)
2
P x
%
has two zeros, at
0
p
± -
.
Then, the polynomial
( ) ( ) ( ) ( )
4 2 4 2
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
is
(
)
(
)
4 2 3 4 2
2 0 2 2 0
2 2 4 2 2 14 6 2
x q x q x q x x x q x q
+ + - + = - - +
% % % % % % %
That is
( ) ( ) ( ) ( )
(
)
4 2
4 2 4 2 2 0
2 2 7 3
Q x P x Q x P x x q x q
¢¢ ¢ ¢
- = - + -
% % % % % %
The previous polynomial is symmetric, thus its zeros (if any) will come in opposite
pairs.
Then, since
(
)
2
P x
%
has two opposite zeros, either none or both of them will be zeros of
( ) ( ) ( ) ( )
4 2 4 2
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
.
IIa. Substituting the two zeros of
(
)
2
P x
%
into the expression of
( ) ( ) ( ) ( )
4 2 4 2
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
, we obtain

22 March 2018
33
(
)
(
)
4 2 2
0 2 0 0 0 2 0 0
7 3 7 3
p q p q p q p q
± - + ± - - = - -
The discriminant of the previous trinomial is
2
2 0
9 28
q q
+
.
But
0
0
q
>
, otherwise
(
)
4
Q x
%
has zeros.
Thus, the discriminant is positive and the trinomial has two zeros, the product of
which is
0
7 0
q
- <
, thus one zero is positive and the other is negative.
The negative zero is
(
)
2
0 2 2 0
3 9 28 14
p q q q= - + .
Therefore, for every pair of the parameters
0 2
,
q q
, with
0
0
q
>
, there exists a
0
0
p
<
such that the polynomial
( ) ( ) ( ) ( )
4 2 4 2
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
vanishes at
0
p
± -
, and the
resulting potential is regular.
In this case, the potential is again a symmetric double finite volcano, but now it has
three bound eigenstates, with the highest of them, i.e. the second-excited state, being
described by the wave function (37), which takes the form
( )
2 2
2 2 0
4 2
2 0
3 9 28
x q q q
x A x q x q
y
+ - +
=+ +
%
%
% %
(38)
where we’ve incorporated into the normalization constant
A
the factor
1 14
appearing in the expression of
0
p
.
The wave function (38) has two zeros, thus it is the second-excited-state wave
function of the resulting potential, with energy
0
r
E
=
.
Next, we’ll calculate the potential for the case where
2
0
q
=
.
In this case,
0
0
7
q
p= -
, and then
( )
2
0
2
7
q
P x x= -
% %
,
( )
2
2
P x x
¢
=
% %
,
( )
2
2
P x
¢¢
=
%
Also
(
)
4
4 0
Q x x q
= +
% %
,
( )
3
4
4
Q x x
¢=
% %
,
( )
2
4
12
Q x x
¢¢ =
% %
Then, we have

22 March 2018
34
( ) ( ) ( )
( )
(
)
( ) ( )
6 2 4
2
6 2
0
4 4 4 0
2 2
24 4
40 0
32 12
220 12
x x x q
Q x Q x Q x
x q x
Q x x q x q
¢ ¢¢ - +
--
= =
+ +
% % %
% % %
% %
%% %
and
( ) ( ) ( ) ( )
( ) ( )
(
)
( ) ( )
4 4 4
0
4 2 4 2 0
4 2 4 2 4 2
0 0
0 0
2 16
214 2
7 7
x q x
Q x P x Q x P x x q
Q x P x q q
x q x x q x
¢¢ ¢ ¢ + -
-- +
===
æ ö æ ö
+ - + -
ç ÷ ç ÷
è ø è ø
% %
% % % % %
% % % % % %
( ) ( ) ( )
2 2 2
0 0 0
40
40
4 2 4 2
0 0
0 0
14 14
14
7 7 7
7
7 7
q q q
qx x x
x
x q
q q
x q x x q x
æ öæ ö æ ö
æ ö - - + +
ç ÷ç ÷ ç ÷
- -
ç ÷
è ø è øè ø è ø
= = = -
æ ö æ ö +
+ - + -
ç ÷ ç ÷
è ø è ø
% % %
%
%
% % % %
Substituting into (5), with
0
r
E
=
, we obtain the potential
( )
( ) ( ) ( ) ( )
2 2
0 0
6 2 6 22 2
0 0
2 2
2 2
4 4
4 4
0 0 0 0
0 0
0 0
14 7
7 7
20 12 10 6
2
q q
x x
x q x x q x
V x m x m x
x q x q
x q x q
æ ö æ ö
æ ö æ ö
+ +
ç ÷ ç ÷
ç ÷ ç ÷
- -
ç ÷ ç ÷
è ø è ø
= - = - =
ç ÷ ç ÷
+ +
+ +
ç ÷ ç ÷
ç ÷ ç ÷
è ø è ø
% %
% % % %
h h
%% %
% %
( ) ( )
(
)
(
)
( )
6 2 2 4
2
6 2
2 2 0 0 0
0
02 2
2 2
4
4 4
0 0 0 0
0
0 0
10 6 7 7
7 7
10 6 x q x x q x q
x q
x q x
m x m x
x q
x q x q
æ ö - - + +
+
-
ç ÷
= - = =
ç ÷
+
+ +
è ø
% % % %
%
% %
h h
%
% %
( ) ( )
6 2 6 2 4 6 4 2
2 2
0 0 0 0 0 0 0 0 0
2 2
2 2
4 4
0 0 0 0
0 0
10 6 7 7 7 7 3 7 13 7
x q x x q x q x q q x q x q x q q
m x m x
x q x q
- - - - - - - -
= =
+ +
% % % % % % % %
h h
% %
That is
( )
( )
6 4 2
2
0 0 0 0
2
24
0 0 0
3 7 13 7
x q x q x q q
V x m x x q
- - -
=+
% % %
h
%% (39)
At long distances, the potential (39) takes the form
(
)
2 2 2
0 0
3
V x m x x
¥
=
% %
h
, in
agreement with (6), for
2
m n
- =
and
0
r
E
=
.
In units
2 2
0 0
1
m x
=
h
, the potential (39) becomes
( )
( )
6 4 2
0 0 0 0
2
40
3 7 13 7
x q x q x q q
V x x q
- - -
=+
% % %
%%

22 March 2018
35
For
0
1 7
q= and
0
4 7
q= the previous potential is written as, respectively,
( )
( ) ( )
6 4 2 6 4 2 2 6 2 4 2
2 2
24
4
42
13 1 13 1
3 3
7 *3 7 7*13 7
7 7 7 7
1
17 1
7 1
7
7
x x x x x x x x x
V x x
x
x
- - - - - - - - -
= = = =
æ ö +
+
+
ç ÷
è ø
% % % % % % % % %
%%
%
%
( )
6 4 2
2
4
147 49 91 7
7 1
x x x
x
- - -
=+
% % %
%
That is
( )
( )
6 4 2
2
4
147 49 91 7
7 1
x x x
V x x
- - -
=+
% % %
%%
And
( )
( ) ( )
6 4 2 6 4 2 2 6 2 4 2
2 2
24
4
42
4 4 52 8
3 2 13 2 3 2
7 *3 7 *2 7*52 7*8
7 7 7 7
1
47 4
7 4
7
7
x x x x x x x x x
V x x
x
x
- - - - - - - - -
= = = =
æ ö +
+
+
ç ÷
è ø
% % % % % % % % %
%%
%
%
( )
6 4 2
2
4
147 98 364 56
7 4
x x x
x
- - -
=+
% % %
%
That is
( )
( )
6 4 2
2
4
147 98 364 56
7 4
x x x
V x x
- - -
=+
% % %
%%
Both potentials are plotted in Figure 7 (red and blue line, respectively). Each of them
has three bound eigenstates, with the highest of them, i.e. the second-excited state,
being of zero energy.
We see that, as
0
q
increases, the bottom of the symmetric double volcano approaches
zero. Then, since the ground-state energy must exceed the minimum value of the
potential [10], both the ground-state and the first-excited-state energies approach zero,
as
0
q
increases. In other words, as
0
q
increases, the two lower bound energies of each
volcano potential get closer to the third, which is fixed to zero.

22 March 2018
36
Figure 7
IIb. If
(
)
2
0 2 2 0
3 9 28 14
p q q q¹ - + , the polynomial
( ) ( ) ( ) ( )
4 2 4 2
2
Q x P x Q x P x
¢¢ ¢ ¢
-
% % % %
does not vanish at the two simple zeros of
(
)
2
P x
%
and then the potential is singular, it
has two simple poles, at
0
p
± -
.
In this case, the two zeros of the wave function (37) are not typical nodes, they are
special zeros, and the wave function is the ground-state wave function of the potential
(5) (with
0
r
E
=
), which is a symmetric singular potential with only one bound state,
its ground state, having zero energy.
As the coefficient
0
p
reaches the critical value
(
)
2
2 2 0
3 9 28 14
q q q- + , the two
open edges at each of the two simple poles of the potential are “glued” together
forming a volcano and, at the same time, two more bound eigenstates are created, with
the highest of them, i.e. the second-excited state, having zero energy.
For the case where
2
0
q
=
, we saw that the critical value of
0
p
is
0
7
q-.
We’ll examine what happens if
(
)
0 0
1 7
p q
e
= - + , where
1
e
> -
, so that
0
0
p
<
.
For
0
e
=
, we obtain the critical value of
0
p
.

22 March 2018
37
The polynomials
(
)
2
P x
%
and
(
)
4
Q x
%
are, respectively,
( ) ( )
2
0
2
1
7
q
P x x
e
= - +
% %
,
(
)
4
4 0
Q x x q
= +
% %
The regular term of the potential (5), i.e. the term
( ) ( ) ( )
(
)
( )
2 2
4 4 4 4
2
Q x Q x Q x Q x
¢ ¢¢
-
% % % %
,
is the same as before, i.e.
( ) ( ) ( )
( )
( )
2
6 2
4 4 4 0
2
24
40
220 12
Q x Q x Q x
x q x
Q x x q
¢ ¢¢
--
=+
% % %
% %
%%
while the singular term
( ) ( ) ( ) ( )
(
)
( ) ( )
4 2 4 2 4 2
2
Q x P x Q x P x Q x P x
¢¢ ¢ ¢
-
% % % % % %
is
( ) ( ) ( ) ( )
( ) ( )
( )
( )
40
4 2 4 2
4 2 4 2 0
0
14
27
1
7
q
x
Q x P x Q x P x
Q x P x
q
x q x
e
æ ö
-
ç ÷
¢¢ ¢ ¢
-è ø
= -
æ ö
+ - +
ç ÷
è ø
%
% % % %
% % % %
Substituting the two previous terms into (5), we obtain, assuming again that the
potential vanishes at infinity, i.e.
0
r
E
=
,
( )
( ) ( )
( )
40
6 22 0
2
24
0 0 4 2 0
00
14
20 12 7
217
q
x
x q x
V x m x q
x q x q x
e
æ ö
æ ö
ç ÷
-
ç ÷
-
ç ÷
è ø
= -
ç ÷
æ ö
+
ç ÷
+ - +
ç ÷
ç ÷
è ø
è ø
%
% %
h
%%% %
To simplify the calculation, we further assume that
0
7
q
=
.
Then, in units
2 2
0 0
1
m x
=
h
, the previous potential is written as
( )
( ) ( )
( )
( )
( ) ( ) ( )
( )
( )
( )
4 4
6 2 6 2
2 2
4 2 4 2
4 4
14 1 7 1
1 20 12*7 10 6*7
27 1 7 1
7 7
x x
x x x x
V x x x x x
x x
e e
æ ö
- -
- -
ç ÷
= - = - =
ç ÷
+ - + + - +
+ +
è ø
% %
% % % %
%% % % %
% %
(
)
(
)
(
)
(
)
(
)
( )
( )
( )
6 2 2 4 4
2
4 2
10 42 1 7 1 7
7 1
x x x x x
x x
e
e
- - + - - +
=+ - +
% % % % %
% %
But

22 March 2018
38
(
)
(
)
(
)
(
)
(
)
(
)
(
)
6 2 2 4 4 8 6 4 2
10 42 1 7 1 7 10 10 1 42 42 1
x x x x x x x x x
e e e
- - + - - + = - + - + + -
% % % % % % % % %
(
)
(
)
(
)
8 4 8 6 4 2 8 4
7 6 7 10 10 1 42 42 1 7 42 49
x x x x x x x x
e e
- + - = - + - + + - - + =
% % % % % % % %
(
)
(
)
8 6 4 2
3 10 1 84 42 1 49
x x x x
e e
= - + - + + +
% % % %
Thus, we end up to the potential
( )
(
)
(
)
( )
( )
( )
8 6 4 2
2
4 2
3 10 1 84 42 1 49
7 1
x x x x
V x x x
e e
e
- + - + + +
=+ - +
% % % %
%% %
(40)
For
0
e
¹
, the potential (40) has two simple poles, at 1
e
± +
(with
1
e
> -
).
For
0
e
=
, a volcano is formed at each pole and the potential then becomes a double
volcano, with three bound eigenstates, the highest of which, i.e. the second-excited
state, has zero energy.
In Figures 8-10, the potential (40) is plotted for
0.01
e
= -
,
0
e
=
, and
0.01
e
=
,
respectively.
Figure 8

22 March 2018
39
Figure 9
Figure 10

22 March 2018
40
6. References
[1] Martin Gremm, Phys.Lett. B478 (2000) 434-438.
[2] W. T. Cruz et al 2011 EPL 96 31001.
[3] Lisa Randall, Raman Sundrum, Phys.Rev.Lett.83:4690-4693, 1999.
[4] Michael Martin Nieto, Phys.Lett. B486 (2000) 414-417.
[5] Sayan kar and Rajesh R. Parwani 2007 EPL 80 30004.
[6] Ratna Koley, Sayan Kar, Phys.Lett.A363:369-373, 2007.
[7] Zafar Ahmed, Eur. J. Phys. 37 (2016) 045404.
[8] Spiros Konstantogiannis, On the Ground State of Potentials with, at Most, Finite
Discontinuities and Simple Poles,
https://www.researchgate.net/publication/323749210_On_the_Ground_State_of_Pote
ntials_with_at_Most_Finite_Discontinuities_and_Simple_Poles.
[9] M. Moriconi, Am. J. Phys. 75, 284-285 (2007).
[10] L. D. Landau and E. M. Lifshitz, Quantum Mechanics (Pergamon Press, Second
Edition, 1965).