Quasi-exact solution of sextic anharmonic oscillator using a quotient polynomial

Abstract

Among the one-dimensional, real and analytic polynomial potentials, the sextic anharmonic oscillator is the only one that can be quasi-exactly solved, if it is properly parametrized. In this work, we present a new method to quasi-exactly solve the sextic anharmonic oscillator and apply it to derive specific solutions. Our approach is based on the introduction of a quotient polynomial and can also be used to study the solvability of symmetrized (non-analytic) or complex PT-symmetric polynomial potentials, where it opens up new options.

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Science Front Publishers Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
(sciencefront.org) ISSN 2394-3688
17
Quasi-exact solution of sextic anharmonic oscillator
using a quotient polynomial
Spiros Konstantogiannis
4 Antigonis Street, Nikaia 18454, Athens, Greece
Corresponding author E-mail: spiroskonstantogiannis@gmail.com
(Received 14 October 2018, Accepted 13 December 2018, Published 15 December 2018)
Abstract
Among the one-dimensional, real and analytic polynomial potentials, the sextic anharmonic
oscillator is the only one that can be quasi-exactly solved, if it is properly parametrized. In this
work, we present a new method to quasi-exactly solve the sextic anharmonic oscillator and
apply it to derive specific solutions. Our approach is based on the introduction of a quotient
polynomial and can also be used to study the solvability of symmetrized (non-analytic) or
complex PT-symmetric polynomial potentials, where it opens up new options.
Keywords: quasi-exactly solvable potentials, sextic anharmonic oscillator, quotient
polynomial, energy-reflection symmetry
1. Introduction
The search for exact solutions to the Schrödinger equation has resulted in the discovery of a
new class of potentials, for which only a finite part of the energy spectrum, along with the
respective eigen functions, can be found in closed-form [1-4]. These potentials are called quasi-
exactly solvable (QES) and occupy an intermediate place, between the few which are exactly
solvable, such as the harmonic oscillator and the Coulomb potential, for which the entire energy
spectrum and all eigen functions are known, and the many which are non-solvable, for which none
eigen value and none eigen function can be exactly determined.
The sextic anharmonic oscillator is the only one-dimensional, real and analytic polynomial
potential that can be quasi-exactly solved if its parameters are properly chosen [2]. Depending on
the coupling constant, the system can be a single, double, or triple-well potential. Single and

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
18
double-well oscillators are among the simplest interacting quantum systems with numerous
applications in diverse areas of physics and chemistry, ranging from atomic and molecular physics
to particle physics, quantum field theory, and cosmology. For instance, as single-well potentials,
sextic anharmonic oscillators can be used to verify results derived from approximation methods,
give a better insight into the
6
ϕ
quantum field theory, and model nonlinear fields, while as double-
well potentials, they are useful to describe the motion of a particle in the presence of two centers of
force, Bose-Einstein condensates, quantum heterostructures, chemical processes in molecular
systems, quantum tunneling effects, and so on.
Herein, we adopt an approach that uses elements of both the Bethe ansatz method [5, 6] and the
recursion relation method [7] to study the sextic anharmonic oscillator by means of a quotient
polynomial.
The rest of the paper is organized as follows: in the next section, making an ansatz for the wave
function, we introduce the quotient polynomial, transform the Schrödinger equation into a
differential equation containing the quotient polynomial, and derive the sextic anharmonic oscillator
to be quasi-exactly solved. In sections 3 and 4, we quasi-exactly solve the system in the cases of
even and odd parity of the wave function, and give examples of specific solutions. In section 5, we
show that the system exhibits energy-reflection symmetry if the coupling constant vanishes and, in
section 6, we summarize and outline future developments.
2. Introducing the quotient polynomial
Choosing a length scale
l
and doing the transformations
x lx
→
,
2 2
2
E E ml
→h, and
(
)
(
)
2 2
2
V x V x ml
→h, the position
x
, the energy
E
, and the potential
(
)
V x
become
dimensionless, and then the energy eigenvalue equation – i.e. the time-independent Schrödinger
equation – for our system reads
(
)
(
)
(
)
(
)
0
x E V x x
ψ ψ
′′
+ − =
,
where
(
)
x
ψ
is an energy eigenfunction of the system.
Establishing our ansatz scheme, we seek definite-parity eigenfunctions of the form
(
)
(
)
(
)
(
)
4
exp
n n
x A p x g x
ψ
= (1)

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
19
where
n
A
the normalization constant,
(
)
n
p x
an
n
-degree polynomial of definite parity, and
(
)
4
g x
a fourth-degree polynomial of even parity with negative leading coefficient, so that (1) is
square-integrable, which is necessary for a confining potential. Since
x
is dimensionless, the
dimensions of
(
)
n
p x
is carried by its coefficients. Thus, incorporating the leading coefficient of
(
)
n
p x
into the normalization constant
n
A
, we make
(
)
n
p x
both monic and dimensionless. As
exponent, the polynomial
(
)
4
g x
must be dimensionless too, and since
x
is dimensionless, the
coefficients of
(
)
4
g x
are also dimensionless. The constant term of
(
)
4
g x
is a multiplicative
constant that can also be incorporated into the normalization constant
n
A
. Finally, choosing the
length scale
l
appropriately, we can set the leading coefficient of
(
)
4
g x
to a desirable negative
value. Then, without loss of generality, we write
(
)
4
g x
as
( )
4 2
2
4
1
4 2
g
g x x x
= − + (2)
where
2
g
is a real parameter.
The second derivative of (1) with respect to the dimensionless
x
is
( ) ( ) ( ) ( ) ( ) ( )
(
)
( )
(
)
( )
( )
2
4 4 4 4
2 exp
n n n n
x A p x g x p x g x g x p x g x
ψ
′′ ′ ′ ′ ′′
′′ = + + +
Plugging the previous expression along with (1) into the energy eigenvalue equation, then
dividing by
(
)
(
)
4
exp 0
n
A g x
≠
and solving for the potential, we end up to
( ) ( ) ( ) ( )
( ) ( ) ( )
42
4 4
2
n n
n
p x g x p x
V x g x g x E
p x
′′ ′ ′
+′ ′′
= + + +
Since the potential
(
)
V x
and
( ) ( )
2
4 4
g x g x E
′ ′′
+ +
are both polynomials, the expression
( ) ( ) ( )
(
)
( )
4
2
n n n
p x g x p x p x
′′ ′ ′
+ is a polynomial too, and it is quadratic, as
(
)
(
)
(
)
( )
(
)
( )
4 4 4
deg 2 deg 2 deg deg deg
n n n n n n n n
p g p p p g p p g p p
′′ ′ ′ ′′ ′ ′ ′ ′
+ = + − = − =
(
)
(
)
( )
4
deg deg deg 3 1 2
n n
g p p n n
′ ′
= + − = + − − =
Also,
( )
4
g x
′
is of odd parity and
( )
n
p x
′
has different parity from
(
)
n
p x
, thus
( ) ( )
4n
g x p x
′ ′
has
the same parity as
(
)
n
p x
, and
( )
n
p x
′′
has also the same parity as
(
)
n
p x
, and then

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
20
( ) ( ) ( )
4
2
n n
p x g x p x
′′ ′ ′
+ has also the same parity as
(
)
n
p x
, and thus the polynomial
( ) ( ) ( )
(
)
( )
4
2
n n n
p x g x p x p x
′′ ′ ′
+ is of even parity. Therefore, we can write
( ) ( ) ( ) ( ) ( )
4 2
2 ;
n n n
p x g x p x q x n p x
′′ ′ ′
+ = − (3)
where
(
)
(
)
(
)
2
2 2 0
;
q x n q n x q n
= + (4)
We’ll refer to
(
)
2
;
q x n
as the quotient polynomial. The minus sign on the right-hand side of (3) is
put in for convenience.
Using (2) and (4), (3) takes the form
( )
(
)
( ) ( ) ( )
(
)
( )
3 2
2 2 0
2
n n n
p x x g x p x q n x q n p x
′′ ′
+ − + = − + (5)
Besides, equating the coefficients of the highest-order terms in
x
on both sides of (5) yields
(
)
2
2
q n n
=
(6)
Then, the quotient polynomial (4) and the differential equation (5) become, respectively,
(
)
(
)
2
2 0
; 2
q x n nx q n
= + (7)
( )
(
)
( ) ( )
(
)
( )
3 2
2 0
2 2
n n n
p x x g x p x nx q n p x
′′ ′
+ − + = − + (8)
In terms of the quotient polynomial, the expression of the potential takes the form
( ) ( ) ( ) ( )
2
2 4 4
;
V x q x n g x g x E
′ ′′
= − + + +
(9)
The potential (9) is expressed up to an additive constant and to determine it uniquely, we choose
its value at zero to be zero. Then, using the expressions of the quotient polynomial and of
(
)
4
g x
,
we obtain that the energy is given by
(
)
0 2
E q n g
= −
(10)
Substituting (10) into (9) and taking into account the expressions of the quotient polynomial and
(
)
4
g x
, we end up to the potential
(
)
(
)
(
)
6 4 2 2
2 2
2 3 2
V x x g x g n x
= − + − + (11)
with
0,1, ...
n
=
For every value of
n
, the potential (11) describes a symmetric sextic anharmonic
oscillator that, as we’ll show, is quasi-exactly solvable.

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
21
Before finishing this section, it is worth mentioning that the vanishing of the linear term
(
)
1
q n
in the quotient polynomial results in the potential (11) being dependent only on
n
and not on the
energy. Otherwise,
(
)
1
q n
would appear in (11) and its dependence on the energy would result in
the potential being also dependent on the energy, which is generally unwanted, because if the
potential changes with the energy, then only one eigenstate of it could be found.
3. The even-parity case
We’ll examine first the even-parity case for the polynomial
(
)
n
p x
.
Since the polynomial
(
)
n
p x
is of
n
degree,
n
must be even. Also, since the exponential factor
(
)
(
)
4
exp
g x
is of even parity, the ansatz eigenfunction (1) is also of even parity in this case.
Setting, for convenience,
2
n n
→
, with
0,1, ...
n
=
, the oscillator (11) takes the form
(
)
(
)
(
)
6 4 2 2
2 2
2 3 4
V x x g x g n x
= − + − + (12)
while the differential equation (8) is written as
( )
(
)
( ) ( )
(
)
( )
3 2
2 2 2 0 2
2 4 2
n n n
p x x g x p x nx q n p x
′′ ′
+ − + = − + (13)
where
(
)
2 2 2
2 2 2 0
...
n n
n n
p x x p x p
−
−
= + + +
The product
(
)
( )
3
2 2
2n
x g x p x
′
− + is of even parity, as are all other polynomials in (13), and thus
we have only even-degree terms in
x
. Then, equating the coefficients of the
2
k
-degree terms in
x
on both sides of (13) yields the following three-term recursion relation
(
)
(
)
(
)
(
)
(
)
2 2 2 0 2 2 2
2 2 2 1 4 2 4 1
k k k
k k p kg q n p n k p
+ −
+ + = − + − + − (14)
For
0
k
=
, we drop the coefficient
2 2
k
p
−
, whose index is negative, and (14) gives
(
)
0
2 0
2
2
q n
p p
= − (15)
For
1
k n
= +
, dropping the coefficients
2
k
p
and
2 2
k
p
+
, whose indices exceed the degree of
(
)
2n
p x
, we see that (14) holds identically, which is expected, since for
1
k n
= +
, from (14) we
calculate the leading coefficient of the quotient polynomial, i.e. the relation (6) for
2
n n
→
. Then,
we’ll use (14) for
0,1, ...,
k n
=
.

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
22
Using the
1
n
+
different values of
k
, we obtain a system of
1
n
+
equations with the
1
n
+
unknowns
( )
0 2*1
2* 1
, , ..., n
p p p
−
, and
(
)
0
2
q n
.
The relation (15) is written as
(
)
(
)
2*1 1 0 0
2
p f q n p
=
with
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − , a polynomial of degree 1 in
(
)
0
2
q n
. Using the previous
expression of
2
p
, we obtain from (14), for
1
k
=
,
( )
(
)
( )
(
)
(
)
4 2 0 1 0 0
4*3 4 2 2 4
p g q n f q n n p
= − + +
Now, setting
( )
(
)
( )
(
)
(
)
( )
(
)
2 0 1 0 2 0
4 2 2 4 4 * 3 2
g q n f q n n f q n
− + + = ,
the previous expression of
4
p
is written as
(
)
(
)
2*2 2 0 0
2
p f q n p
=,
with
(
)
(
)
2 0
2
f q n
a polynomial of degree 2 in
(
)
0
2
q n
.
We assume that
( )
(
)
(
)
1 0 0
2* 1 2
k
k
p f q n p
−
−= and
(
)
(
)
2* 0 0
2
k k
p f q n p
=,
with
(
)
(
)
1 0
2
k
f q n
− and
(
)
(
)
0
2
k
f q n
polynomials of degree
1
k
−
and
k
in
(
)
0
2
q n
, respectively.
Then, substituting into (14), we obtain
( )( ) ( )
(
)
( )
(
)
( ) ( )
(
)
(
)
2 2 2 0 0 1 0 0
2 2 2 1 4 2 2 4 1 2
k k k
k k p kg q n f q n n k f q n p
+ −
+ + = − + + + −
Setting
( )
(
)
( )
(
)
( ) ( )
(
)
(
)
( )( ) ( )
(
)
2 0 0 1 0 1 0
4 2 2 4 1 2 2 2 2 1 2 ,
k k k
kg q n f q n n k f q n k k f q n
− +
− + + + − = + +
the previous expression of
2 2
k
p
+
is written as
( )
(
)
(
)
1 0 0
2 1 2
k
k
p f q n p
+
+=,
with
(
)
(
)
1 0
2
k
f q n
+ a polynomial of degree
1
k
+
in
(
)
0
2
q n
.
Then, by induction,
(
)
(
)
2 0 0
2
k k
p f q n p
= (16)

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
23
for
1, 2, ...
k n
=
,
0,1, ...
n
=
, with
(
)
(
)
0
2
k
f q n
a polynomial of degree
k
in
(
)
0
2
q n
, and
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − . From (16), we see that if
0
p
vanishes, all coefficients of
(
)
2n
p x
vanish,
which is impossible, since
(
)
2n
p x
is monic. Thus 0
0
p
≠
, and then (16) holds for
0
k
=
if
(
)
(
)
0 0
2 1
f q n
=
. Thus, (16) holds for
0,1, ...
k n
=
, with
(
)
(
)
0 0
2 1
f q n
=
and
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − .
Using (16), the recursion relation (14) is written as
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1 0 2 0 0 1 0
2 2 2 1 2 4 2 2 4 1 2 (17)
k k k
k k f q n kg q n f q n n k f q n
+ −
+ + = − + − + −
where we divided both members by 0
0
p
≠
. The recursion relation (17) holds for
0,1, ...,
k n
=
,
0,1, ...
n
=
, with
(
)
(
)
0 0
2 1
f q n
=
and
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − . For
k n
=
, we drop
(
)
(
)
1 0
2
k
f q n
+,
as it is related to the coefficient
( )
2 1
k
p
+
whose index exceeds the degree of
(
)
2n
p x
, and we obtain
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0 0 2 0 1 0
2 2 4 2 4 2 0
n n n
q n f q n ng f q n f q n
−
+ + =
(18)
The equation (18), which is a
(
)
1
n
+
-degree polynomial equation, gives the values of
(
)
0
2
q n
and
then, from (10), the energies of the known eigenstates of the sextic anharmonic oscillator (12). The
equation (18) is equivalently written as
(
)
(
)
1 0
2 0
n
f q n
+
=
(19)
We’ll refer to (18) or (19) as the energy equation.
Next, we’ll show that the polynomial
(
)
(
)
0
2
k
f q n
has
k
(real) zeros.
Since
(
)
(
)
0 0
2 1
f q n
=
(no zeros) and
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − (one zero, at zero), it is enough
to show that
(
)
(
)
0
2
k
f q n
has
k
zeros for
2
k
≥
.
From (17), we see that the leading coefficients of
(
)
(
)
0
2
k
f q n
and
(
)
(
)
1 0
2
k
f q n
+ have opposite
signs. Then, since the leading coefficient of
(
)
(
)
0 0
2
f q n
is positive, we derive that
(
)
(
)
0
2
k
f q n
has
positive/negative leading coefficient if (and only if)
k
is even/odd. As a consequence, the leading
coefficients of
(
)
(
)
1 0
2
k
f q n
− and
(
)
(
)
1 0
2
k
f q n
+ have the same sign. Also, since
(
)
(
)
0
2
k
f q n
is a
polynomial of degree
k
in
(
)
0
2
q n
, if
k
is even,
(
)
k
f
±∞ = ∞
, and thus
(
)
(
)
sgn 1
k
f
±∞ =
, while if
k
is odd,
(
)
k
f
±∞ = ∞
m
, and thus
(
)
(
)
sgn 1
k
f
±∞ =
m
. We observe that in both cases,
(
)
(
)
sgn 1
k
f
−∞ =
.

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
24
For
2
k
=
, we have
(
)
(
)
2
sgn 1
f
±∞ =
. Also, (17) gives for
1
k
=
, at
(
)
0
2 0
q n
=
,
(
)
2
12 0 4 0
f n
= − <
, since
(
)
1
0 0
f
=
,
(
)
0
0 1
f
=
, and
0
n
>
, because if
0
n
=
,
(
)
(
)
2 0
2
f q n
vanishes.
Thus
(
)
2
0 0
f
<
, and then
(
)
(
)
2 0
2
f q n
changes sign in the interval
(
)
,0
−∞ and in the interval
(
)
0,
∞
. Then, since
(
)
(
)
2 0
2
f q n
is a second degree polynomial, it has exactly one zero in each of
the two previous intervals. Thus,
(
)
(
)
2 0
2
f q n
has exactly one zero in each of the two intervals that
the zero of
(
)
(
)
1 0
2
f q n
splits the real line.
We’ll now show, by induction, that, for
2
k
≥
,
(
)
(
)
0
2
k
f q n
has exactly one zero in each of the
k
intervals that the
1
k
−
zeros of
(
)
(
)
1 0
2
k
f q n
− split the real line.
Proof
For
2
k
=
, we showed that the previous statement holds.
We assume that
(
)
(
)
1 0
2
k
f q n
− has
1
k
−
simple zeros, with
1 1
k
− ≥
, which thus split the real
line into
k
intervals, and in each of these intervals,
(
)
(
)
0
2
k
f q n
has a zero. Since
(
)
(
)
0
2
k
f q n
is a
polynomial of
k
degree, it has exactly one zero in each of the previous
k
intervals. Denoting the
zeros of
(
)
(
)
1 0
2
k
f q n
− by
(
)
(
)
1,
0
2
k i
q n
−, with
1, 2, ..., 1
i k
= −
, without loss of generality, we arrange
them as
(
)
(
)
(
)
(
)
(
)
(
)
1,1 1, 2 1, 1
0 0 0
2 2 ... 2
k k k k
q n q n q n
− − − −
−∞ < < < < < ∞
Then, in each of the
k
intervals
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
( )
(
)
1,1 1,1 1,2 1, 2 1, 1 1, 1
0 0 0 0 0 0
, 2 , 2 , 2 ,..., 2 , 2 , 2 , ,
k k k k k k k k k
q n q n q n q n q n q n
− − − − − − − − −
−∞ ∞
the polynomial
(
)
(
)
1 0
2
k
f q n
− has constant sign. Since the zeros are simple, the derivative
(
)
(
)
(
)
1 0 0
2 2
k
df q n dq n
− (with respect to
0
q
) does not vanish at the zeros of
(
)
(
)
1 0
2
k
f q n
−, and then
the zeros of
(
)
(
)
1 0
2
k
f q n
− are not extreme points of
(
)
(
)
1 0
2
k
f q n
−. Thus, in every two consecutive
intervals of the above
k
intervals, the polynomial
(
)
(
)
1 0
2
k
f q n
− has opposite signs, and thus if
0
q
′
and
0
q
′′
belong, respectively, to any two consecutive intervals of the above
k
intervals, then
(
)
(
)
1 0 1 0
0
k k
f q f q
− −
′ ′′
<
.
Arranging, without loss of generality, the
k
zeros of
(
)
(
)
0
2
k
f q n
as

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
25
(
)
(
)
(
)
(
)
(
)
(
)
,1 ,2 ,
0 0 0
2 2 ... 2
k k k k
q n q n q n
−∞ < < < < < ∞
,
we have
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
,1 1,1 ,2 1,2
0 0 0 0
2 2 2 2 ...
k k k k
q n q n q n q n
− −
−∞ < < < < <
(
)
(
)
(
)
(
)
(
)
(
)
, 1 1, 1 ,
0 0 0
... 2 2 2
k k k k k k
q n q n q n
− − −
< < < < ∞
Then
( )
( )
(
)
( )
( )
(
)
, , 1
1 0 1 0
2 2 0
k i k i
k k
f q n f q n
+
− −
<
,
for every
1, 2, ... 1
i k
= −
.
Besides, for
1
n k
≥ +
,
(
)
(
)
1 0
2
k
f q n
+ is given by the recursion relation (17), which at
(
)
(
)
,
0
2
k i
q n
and at
(
)
(
)
, 1
0
2
k i
q n
+ is respectively written as
( ) ( )
( )
( )
(
)
( )
( )
( )
(
)
, ,
1 0 1 0
2 2 2 1 2 4 1 2
k i k i
k k
k k f q n n k f q n
+ −
+ + = − + −
( ) ( )
( )
( )
(
)
( )
( )
( )
(
)
, 1 , 1
1 0 1 0
2 2 2 1 2 4 1 2
k i k i
k k
k k f q n n k f q n
+ +
+ −
+ + = − + − ,
since
( )
( )
(
)
( )
( )
(
)
, , 1
0 0
2 2 0
k i k i
k k
f q n f q n
+
= =
. Multiplying side-by-side the two equations, we obtain
( ) ( )
(
)
( )
( )
(
)
( )
( )
(
)
2, , 1
1 0 1 0
2 2 2 1 2 2
k i k i
k k
k k f q n f q n
+
+ +
+ + =
( )
( )
( )
(
)
( )
( )
(
)
2, , 1
1 0 1 0
16 1 2 2
k i k i
k k
n k f q n f q n
+
− −
= + −
For
1
n k
≥ +
,
1 0
n k
+ − ≠
, and since
( )
( )
(
)
( )
( )
(
)
, , 1
1 0 1 0
2 2 0
k i k i
k k
f q n f q n
+
− −
<
, from the previous
equation we derive that
( )
( )
(
)
( )
( )
(
)
, , 1
1 0 1 0
2 2 0
k i k i
k k
f q n f q n
+
+ +
<
,
for every
1, 2, ...., 1
i k
= −
. Thus,
(
)
(
)
1 0
2
k
f q n
+ has at least one zero in each of the
1
k
−
intervals
( )
( )
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
( )
( )
( )
(
)
,1 ,2 ,2 , 3 , 1 ,
0 0 0 0 0 0
2 , 2 , 2 , 2 ,...., 2 , 2
k k k k k k k k
q n q n q n q n q n q n
−
Besides, at
(
)
(
)
,1
0
2
k
q n
, the recursion relation (17) is written as
( ) ( )
( )
( )
(
)
( )
( )
( )
(
)
,1 ,1
1 0 1 0
2 2 2 1 2 4 1 2
k k
k k
k k f q n n k f q n
+ −
+ + = − + −
For
1
n k
≥ +
,
1 0
n k
+ − >
, and from the previous equation we derive that
( )
( )
(
)
(
)
( )
( )
(
)
(
)
,1 ,1
1 0 1 0
sgn 2 sgn 2
k k
k k
f q n f q n
+ −
= − (20)

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
26
But
(
)
(
)
,1
0
2
k
q n
belongs to the interval
( )
( )
(
)
1,1
0
, 2
k
q n
−
−∞ in which
(
)
(
)
1 0
2
k
f q n
− has constant
sign, and thus
( )
( )
(
)
(
)
( )
(
)
,1
1 0 1
sgn 2 sgn 1
k
k k
f q n f
− −
= −∞ =
,
because, as shown above,
(
)
(
)
sgn 1
k
f
−∞ =
for all
k
. Then (20) gives
( )
( )
(
)
(
)
,1
1 0
sgn 2 1
k
k
f q n
+
= −
,
and thus
( )
(
)
( )
( )
(
)
(
)
,1
1 1 0
sgn sgn 2 1
k
k k
f f q n
+ +
−∞ = −
Then, in the interval
( )
( )
(
)
,1
0
, 2
k
q n
−∞ , the polynomial
(
)
(
)
1 0
2
k
f q n
+ changes sign, and
particularly, it becomes from positive, negative, and thus
(
)
(
)
1 0
2
k
f q n
+ has at least one zero in this
interval.
Similarly, at
(
)
(
)
,
0
2
k k
q n
, the recursion relation (17) is written as
( ) ( )
( )
( )
(
)
( )
( )
( )
(
)
, ,
1 0 1 0
2 2 2 1 2 4 1 2
k k k k
k k
k k f q n n k f q n
+ −
+ + = − + − ,
and thus
( )
( )
(
)
(
)
( )
( )
(
)
(
)
, ,
1 0 1 0
sgn 2 sgn 2
k k k k
k k
f q n f q n
+ −
= − (21)
But
(
)
(
)
,
0
2
k k
q n
belongs to the interval
( )
( )
(
)
1, 1
0
2 ,
k k
q n
− −
∞
in which
(
)
(
)
1 0
2
k
f q n
− has constant
sign, and thus
( )
( )
(
)
(
)
( )
(
)
,
1 0 1
sgn 2 sgn
k k
k k
f q n f
− −
= ∞
The numbers
1
k
−
and
1
k
+
are of the same parity, and thus
(
)
(
)
(
)
(
)
1 1
sgn sgn
k k
f f
− +
∞ = ∞
,
because, as shown above, the leading coefficients of
(
)
(
)
1 0
2
k
f q n
− and
(
)
(
)
1 0
2
k
f q n
+ have the
same sign. Comparing the last two equations yields
( )
( )
(
)
(
)
( )
(
)
,
1 0 1
sgn 2 sgn
k k
k k
f q n f
− +
= ∞
,
and comparing (21) with the last equation yields
( )
( )
(
)
(
)
( )
(
)
,
1 0 1
sgn 2 sgn
k k
k k
f q n f
+ +
= − ∞
,

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
27
and then
( )
( )
(
)
(
)
( )
(
)
( )
(
)
,2
1 0 1 1
sgn 2 sgn sgn 1
k k
k k k
f q n f f
+ + +
∞ = − ∞ = −
Then, in the interval
( )
( )
(
)
,
0
2 ,
k k
q n
∞
, the polynomial
(
)
(
)
1 0
2
k
f q n
+ changes sign, and thus it has
at least one zero in this interval.
We have thus showed that the polynomial
(
)
(
)
1 0
2
k
f q n
+ has at least one zero in each of the
following
1
k
+
intervals
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
( )
( )
( )
(
)
( )
( )
(
)
,1 ,1 ,2 , 1 , ,
0 0 0 0 0 0
, 2 , 2 , 2 ,...., 2 , 2 , 2 ,
k k k k k k k k k
q n q n q n q n q n q n
−
−∞ ∞
,
i.e. it has at least
1
k
+
zeros. But, since it is a
(
)
1
k
+
-degree polynomial, it can have up to
1
k
+
zeros. Thus, the polynomial
(
)
(
)
1 0
2
k
f q n
+ has exactly one zero in each of the above intervals and,
in total, it has
1
k
+
zeros, and this completes the proof.
For every
0,1,...
n
=
, solving (19), we obtain
1
n
+
different values of
(
)
0
2
q n
, and then from
(10) we obtain
1
n
+
different energies of the sextic anharmonic oscillator (12). Since the potential
is one-dimensional, its bound states are non-degenerate1 [8], and thus the
1
n
+
energies correspond
to
1
n
+
eigenstates. To locate the eigenstates, we use the node theorem [9], which, in this case,
where the potential is regular, implies that the ground-state wave function has no zeros, while the
n
th-excited-state wave function has
n
zeros.
Besides, as we showed, the constant term
0
p
of the polynomial
(
)
2n
p x
does not vanish, and
thus
(
)
2n
p x
does not have a zero at zero. Then, since it is an even-parity polynomial, it can have an
even number of zeros, and since it is of
2
n
-degree, it can have up to
2
n
zeros. Thus,
(
)
2n
p x
can
have 0,2,…,
2
n
zeros, which are also the zeros of the wave function (1), which thus can be the
ground-state wave function, or the second-excited-state wave function,…, or the
2
n
-th-excited-
state wave function of the oscillator (12). Therefore, the
1
n
+
eigenstates we find are, respectively,
the ground state, the second-excited state,…, the
2
n
-th-excited state of the sextic anharmonic
oscillator (12).
To conclude this section, we’ll quasi-exactly solve the oscillator (12) for
0,1
n
=
. For
0
n
=
, we
calculate the ground-state energy and wave function of the sextic anharmonic oscillator
1 See ref. [10] for some interesting exceptions.

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
28
(
)
(
)
6 4 2 2
2 2
2 3
V x x g x g x
= − + −
In this case, the energy equation is, from (19),
(
)
(
)
1 0
0 0
f q
=
,
and since
(
)
(
)
(
)
1 0 0
0 0 2
f q q= − , we obtain
(
)
0
0 0
q
=
Then, from (10), the ground-state energy is
0 2
E g
= −
For
0
n
=
,
(
)
0
1
p x
=
and from (1), the ground-state wave function is
( )
4 2
2
0 0
1
exp 4 2
g
x A x x
ψ
 
= − +
 
 
For
1
n
=
, we calculate the ground-state and the second-excited-state energies and wave
functions of the sextic anharmonic oscillator
(
)
(
)
6 4 2 2
2 2
2 7
V x x g x g x
= − + −
In this case, the energy equation is, from (18),
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0 1 0 2 1 0 0 0
2 2 4 2 4 2 0
q f q g f q f q
+ + =
,
and using that
(
)
(
)
0 0
2 1
f q
=
and
(
)
(
)
(
)
1 0 0
2 2 2
f q q= − , we end up to the equation
(
)
(
)
2
0 2 0
2 4 2 8 0
q g q
+ − =
,
which has two real roots, namely
( )
2
0 2 2
2 2 2 2
q g g
= − ± +
Then, from (10), we obtain the ground-state energy and the second-excited-state energy, which
are, respectively,
2
0 2 2
3 2 2
E g g
= − − +
2
2 2 2
3 2 2
E g g
= − + +
The polynomial
(
)
2
p x
has the form
(
)
2
2 0
p x x p
= +
To calculate
0
p
, we use (16), which for
1
k n
= =
is written as

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
29
(
)
(
)
1 0 0
2 1
f q p
=
Then, since
(
)
(
)
(
)
1 0 0
2 2 2
f q q= − , we obtain for the two values of
(
)
0
2
q, respectively,
02
2 2
1
2
p
g g
=
− +
and 02
2 2
1
2
p
g g
=
+ +
The first one corresponds to the value of
(
)
0
2
q with the plus sign, i.e. it corresponds to the
second-excited state energy, while the second one corresponds to the ground-state energy. Then,
using (1), the ground-state wave function is
( )
2 4 2
2
0 0 2
2 2
1 1
exp 4 2
2
g
x A x x x
g g
ψ
 
 
 
= + − +
 
 
 
+ +
 
and the second-excited-state wave function is
( )
2 4 2
2
2 2 2
2 2
1 1
exp 4 2
2
g
x A x x x
g g
ψ
 
 
 
= + − +
 
 
 
− +
 
Observe that the ground-state wave function has no zeros, while the second-excited-state wave
function has two zeros2, as expected.
4. The odd-parity case
This case is similar to the even-parity case and we’ll examine it in brief.
Since the polynomial
(
)
n
p x
is of
n
degree,
n
must be odd in this case, and from (1) we see
that the ansatz eigenfunction is also of odd parity. Setting
2 1
n n
→ +
, with
0,1, ...
n
=
, the oscillator
(11) takes the form
( ) ( )
(
)
(
)
6 4 2 2
2 2
2 3 2 2 1
V x x g x g n x
= − + − + + (22)
while the differential equation (8) becomes
( )
(
)
( ) ( ) ( )
(
)
( )
3 2
2 1 2 2 1 0 2 1
2 2 2 1 2 1
n n n
p x x g x p x n x q n p x
+ + +
′′ ′
+ − + = − + + + (23)
where
(
)
2 1 2 1
2 1 2 1 1
...
n n
n n
p x x p x p x
+ −
+ −
= + + +
2 Since 2
2 2 2
2
g g g
+ > = ±
, we obtain 2
2 2
2 0
g g
+ + >
(from the second inequality) and 2
2 2
2 0
g g
− + <
(from
the first inequality).

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
30
In (23), we have only odd-degree terms in
x
, and equating the coefficients of the
(
)
2 1
k
+
-
degree terms in
x
on both sides of (23) yields
(
)
(
)
(
)
(
)
(
)
(
)
2 3 2 0 2 1 2 1
2 3 2 2 2 2 1 2 1 4 1
k k k
k k p k g q n p n k p
+ + −
+ + = − + + + − + − (24)
As in the even-parity case, the recursion relation (24) holds non-trivially for
0,1, ...,
k n
=
.
Using the same reasoning as in the even-parity case, we derive the following relation
(
)
(
)
2 1 0 1
2 1
k k
p f q n p
+= + (25)
for
0,1, ...
k n
=
, with
(
)
(
)
0
2 1
k
f q n
+
a polynomial of degree
k
in
(
)
0
2 1
q n
+
, and
(
)
(
)
0 0
2 1 1
f q n
+ =
and
(
)
(
)
(
)
(
)
1 0 0 2
2 1 2 1 2 6
f q n q n g+ = − + + .
Using (25), the recursion relation (24) is written in terms of
(
)
(
)
0
2 1
k
f q n
+
as
(
)
(
)
(
)
(
)
1 0
2 3 2 2 2 1
k
k k f q n
+
+ + + =
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2 0 0 1 0
2 2 1 2 1 2 1 4 1 2 1
k k
k g q n f q n n k f q n
−
= − + + + + − + − +
(26)
for
0,1, ...,
k n
=
. As in the even-parity case, for
k n
=
we obtain the energy equation
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0 0 2 0 1 0
2 1 2 1 2 2 1 2 1 4 2 1 0
n n n
q n f q n n g f q n f q n
−
+ + + + + + + =
(27)
or, equivalently,
(
)
(
)
1 0
2 1 0
n
f q n
+
+ =
(28)
In the same way as in the even-parity case, we can show that the polynomial
(
)
(
)
0
2 1
k
f q n
+
has
k
(real) zeros, and thus from (28) we obtain
1
n
+
different values of
(
)
0
2 1
q n
+
, and then from
(10) we take
1
n
+
different energies, which are non-degenerate and correspond to
1
n
+
eigenstates
of the sextic anharmonic oscillator (22).
As seen from (25), 1
0
p
≠
, otherwise all coefficients of
(
)
2 1n
p x
+ vanish, which is impossible,
as
(
)
2 1n
p x
+ is monic. Then,
(
)
(
)
2 1 2n n
p x xQ x
+=, where
(
)
2n
Q x
an even-parity,
2
n
-degree
polynomial, which does not vanish at zero, as
(
)
2 1
0 0
n
Q p
= ≠
, and thus it has an even number of
zeros. Then, the polynomial
(
)
2 1n
p x
+ has an odd number of zeros, and since it is of
(
)
2 1
n
+
-
degree, it can have up to
2 1
n
+
zeros, i.e. it can have 1,3,…,
2 1
n
+
zeros, which are also the zeros
of the wave function (1), which thus can be, according to the node theorem [9], the first-excited-
state wave function, or the third-excited-state wave function,…, or the
(
)
2 1
n
+
-th-excited-state
wave function of the oscillator (22). Therefore, the
1
n
+
eigenstates we find are, respectively, the

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
31
first-excited state, the third-excited state,..., the
(
)
2 1
n
+
-th-excited state of the sextic anharmonic
oscillator (22).
We’ll conclude this section by quasi-exactly solving the oscillator (22) for
0,1
n
=
.
For
0
n
=
, we calculate the first-excited-state energy and wave function of the sextic
anharmonic oscillator
(
)
(
)
6 4 2 2
2 2
2 5
V x x g x g x
= − + −
In this case, the energy equation is, from (28),
(
)
(
)
1 0
1 0
f q
=
,
and using that
(
)
(
)
(
)
(
)
1 0 0 2
1 1 2 6
f q q g= − + , we obtain
(
)
0 2
1 2
q g
= −
Then, from (10), the first-excited-state energy is
1 2
3
E g
= −
For
0
n
=
,
(
)
1
p x x
=
and from (1), the first-excited-state wave function is
( )
4 2
2
1 1
1
exp 4 2
g
x A x x x
ψ
 
= − +
 
 
For
1
n
=
, we calculate the first and third-excited-state energies and wave functions of the
sextic anharmonic oscillator
(
)
(
)
6 4 2 2
2 2
2 9
V x x g x g x
= − + −
In this case, the energy equation is, from (27),
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0 1 0 2 1 0 0 0
3 3 6 3 4 3 0
q f q g f q f q
+ + =
,
and using that
(
)
(
)
0 0
3 1
f q
=
and
(
)
(
)
(
)
(
)
1 0 0 2
3 3 2 6
f q q g= − + , we end up to the equation
(
)
(
)
2 2
0 2 0 2
3 8 3 12 24 0
q g q g
+ + − =
,
which has two real roots, namely
( )
2
0 2 2
3 4 2 6
q g g
= − ± +
Then, from (10), we obtain the first and third-excited-state energies, which are, respectively,
2
1 2 2
5 2 6
E g g
= − − +
2
3 2 2
5 2 6
E g g
= − + +

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
32
The polynomial
(
)
3
p x
has the form
(
)
3
3 1
p x x p x
= +
To calculate
1
p
, we use (25), which for
1
k n
= =
is written as
(
)
(
)
1 0 1
3 1
f q p
=
Then, since
(
)
(
)
(
)
(
)
1 0 0 2
3 3 2 6
f q q g= − + , we obtain for the two values of
(
)
0
3
q, respectively,
12
2 2
3
6
p
g g
=
− +
and 12
2 2
3
6
p
g g
=
+ +
The first one corresponds to the value of
(
)
0
3
q with the plus sign, i.e. it corresponds to the third-
excited-state energy, while the second one corresponds to the first-excited-state energy. Then, using
(1), the first-excited-state wave function is
( )
2 4 2
2
1 1 2
2 2
3 1
exp 4 2
6
g
x A x x x x
g g
ψ
 
 
 
= + − +
 
 
 
+ +
 
and the third-excited-state wave function is
( )
2 4 2
2
3 3 2
2 2
3 1
exp 4 2
6
g
x A x x x x
g g
ψ
 
 
 
= + − +
 
 
 
− +
 
Observe that the first-excited-state wave function has one zero, while the third-excited-state wave
function has three zeros, as expected.
5. Energy reflection symmetry if the coupling constant vanishes
The coupling constant of the sextic anharmonic oscillator (11) is
2
2
g
−, and thus it vanishes if
(and only if)
2
g
vanishes, and then the oscillator becomes
(
)
(
)
6 2
3 2
V x x n x
= − + (29)
The oscillator (29) has energy reflection symmetry, i.e. its known non-zero energies (10) come
in symmetric pairs
(
)
,
E E
−
.
Proof
If
2
g
vanishes, the recursion relations (17) and (26), for the polynomial
k
f
in the even and
odd-parity cases, become, respectively,

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
33
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1 0 0 0 1 0
2 2 2 1 2 2 2 4 1 2
k k k
k k f q n q n f q n n k f q n
+ −
+ + = − − + − ,
with
(
)
(
)
0 0
2 1
f q n
=
and
(
)
(
)
(
)
1 0 0
2 2 2
f q n q n= − , and
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1 0 0 0 1 0
2 3 2 2 2 1 2 1 2 1 4 1 2 1 ,
k k k
k k f q n q n f q n n k f q n
+ −
+ + + = − + + − + − +
with
(
)
(
)
0 0
2 1 1
f q n
+ =
and
(
)
(
)
(
)
1 0 0
2 1 2 1 6
f q n q n+ = − + . In both cases,
0
f
is of even parity
and
1
f
is of odd parity, and if
1
k
f
−
and
k
f
have different parities,
1
k
f
+
has the same parity as
1
k
f
−
.
Thus, in both cases,
k
f
is of even/odd parity if
k
is even/odd. Then, the polynomial
1
n
f
+
in (19)
and (28) has definite parity, and thus if
0
q
is a root,
0
q
−
is also a root. Then, since in this case
0
q
is
equal to the energy (10), if
E
is an energy,
E
−
is also an energy, and thus the known non-zero
energies (10) of the sextic anharmonic oscillator (29) come in symmetric pairs.
6. Concluding remarks
Using as basic parameter, the constant term of a quotient polynomial, we have shown that the
sextic anharmonic oscillator (11) is quasi-exactly solvable, in the sense that for every value of the
non-negative integer
n
,
1
2
n
 
+
 
 
(see foot note 3) energies and the respective eigen functions can be
exactly calculated by the relation (10) and the ansatz (1), respectively.
The method we have presented can be straightforwardly used to study not only real and analytic
polynomial potentials, but also piecewise-analytic or complex PT-symmetric polynomial potentials,
such as the symmetrized quartic and sextic anharmonic oscillators that have been introduced in [11,
12], or the complex PT-symmetric quartic and sextic anharmonic oscillators that have been studied
in [13, 14]. In the cases of symmetrized or complex PT-symmetric sextic oscillators, the quotient
polynomial has, respectively, a non-analytic linear term or a linear term with an imaginary
coefficient and if this term vanishes, the quotient polynomial becomes analytic or real, respectively.
Then, the use of our approach opens up the possibility to search for symmetrized (non-analytic) or
complex PT-symmetric sextic oscillators resulting, respectively, from analytic or real quotient
polynomials, and to identify a possible dynamical symmetry which distinguishes those oscillators
from the other of the same class.
3 The symbol
 
 
denotes the floor function, i.e.
x
 
 
is the greatest integer less than or equal to
x
.

Spiros Konstantogiannis Journal for Foundations and Applications of Physics, vol. 6, No. 1 (2019)
34
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