Welcome to visit Communications in Theoretical Physics,
Quantum Physics and Quantum Information

An indirect approach for quantum-mechanical eigenproblems: duality transforms

  • Yu-Jie Chen 1 ,
  • Shi-Lin Li 1 ,
  • Wen-Du Li , 2 ,
  • Wu-Sheng Dai , 1,
Expand
  • 1Department of Physics, Tianjin University, Tianjin 300350, China
  • 2College of Physics and Materials Science, Tianjin Normal University, Tianjin 300387, China

Author to whom any correspondence should be addressed.

Received date: 2021-12-20

  Revised date: 2022-02-15

  Accepted date: 2022-02-16

  Online published: 2022-05-03

Supported by

National Natural Science Foundation of China(11575125)

National Natural Science Foundation of China(11675119)

National Natural Science Foundation of China(11947124)

Copyright

© 2022 Institute of Theoretical Physics CAS, Chinese Physical Society and IOP Publishing

Abstract

We suggest an indirect approach for solving eigenproblems in quantum mechanics. Unlike the usual method, this method is not a technique for solving differential equations. There exists a duality among potentials in quantum mechanics. The first example is the Newton–Hooke duality revealed by Newton in Principia. Potentials that are dual to each other form a duality family consisting of infinite numbers of family members. If one potential in a duality family is solved, the solutions of all other potentials in the family can be obtained by duality transforms. Instead of directly solving the eigenequation of a given potential, we turn to solve one of its dual potentials which is easier to solve. The solution of the given potential can then be obtained from the solution of this dual potential by a duality transform. The approach is as follows: first to construct the duality family of the given potential, then to find a dual potential which is easier to solve in the family and solve it, and finally to obtain the solution of the given potential by the duality transform. In this paper, as examples, we solve exact solutions for general polynomial potentials.

Cite this article

Yu-Jie Chen , Shi-Lin Li , Wen-Du Li , Wu-Sheng Dai . An indirect approach for quantum-mechanical eigenproblems: duality transforms[J]. Communications in Theoretical Physics, 2022 , 74(5) : 055103 . DOI: 10.1088/1572-9494/ac5585

1. Introduction

Eigenproblems are important in quantum mechanics and exact solutions are rare. In this paper, we seek exact solutions by a duality relation in mechanical systems. The first example of this duality is the Newton–Hooke duality in classical mechanics revealed by Newton in Principia (Corollary III of proposition VII) [1]. The Newton–Hooke duality is a duality between the Newtonian gravitational potential and the Hookian harmonic-oscillator potential. E. Kasner and V. I. Arnold independently generalized the Newton–Hooke duality to general power potentials, called the Kasner–Arnold theorem [26] . Recently, the duality between arbitrary potentials in quantum mechanics was reported [7].
In classical and quantum mechanics there exist duality families [7]. In a duality family, potentials are related by a duality relation. Once a family member is solved, the solutions of all other family members can be obtained by a duality transform. That is, in a duality family, we in principle only need to solve one potential. This inspires us to develop an indirect approach for solving eigenproblems.
Suppose the eigenequation of the potential V is difficult. According to the duality, the potential V must belong to a duality family which consists of an infinite number of potentials. All potentials in V's family are dual potentials of V. If one potential in V's family is solved, the solution of the eigenequation of V can be obtained by the duality transform.
In the approach, the first step is to construct the duality family of the potential V by the duality relation. The second step is to look for an ‘easy' potential in V's family. The third step is to achieve the solution of V by the duality transformation from the solution of the ‘easy' potential.
The eigenproblem has been studied for a long time. Some exact solutions are obtained, such as the one- and three-dimensional inverse square root potentials [811], the pseudo-Gaussian potential [12], the quasi-exactly solvable potential [13], and the Coulomb-type potential [14], and some methods are developed [1519]. However, due to the difficulty of seeking exact solutions, the research mainly focuses on solving isolated examples one by one. The method in this paper turns to finding exactly solvable families rather than directly solving a certain potential.
In this paper, we illustrate this approach by general polynomial central potentials. We first derive the duality relation for general polynomial central potentials. A series of examples are considered, including solving the potentials $\tfrac{\eta }{r}+\tfrac{\lambda }{{r}^{3/2}}$ and ηr2 + λr6 from their dual potential $\xi {r}^{2}+\tfrac{\mu }{r}$, solving the potentials $\tfrac{\eta }{r}+\tfrac{\lambda }{\sqrt{r}}$ and $\tfrac{\eta }{{r}^{2/3}}+\lambda {r}^{2/3}$ from their dual potential ξr2 + μr, solving the potentials $\eta {r}^{2/3}+\tfrac{\lambda }{{r}^{4/3}}$ and ηr6 + λr4 from their dual potential $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$, and solving the potentials $\tfrac{\eta }{r}+\tfrac{\nu }{{r}^{3/2}}+\tfrac{\lambda }{{r}^{1/2}}$, ηr2 + νr6 + λr4 and $\tfrac{\eta }{{r}^{2/3}}+\nu {r}^{2/3}+\tfrac{\lambda }{{r}^{4/3}}$ from their dual potential $\xi {r}^{2}+\tfrac{\mu }{r}+\kappa r$.
In section 2, we derive the duality relation for three-dimensional polynomial central potentials. In section 3, we construct duality families. In sections 47, we give examples. In the appendices, we work out some exact solutions of ‘easy' potentials.

2. Duality of general polynomial central potential

In this section, we consider the duality between general polynomial central potentials. The general polynomial is a polynomial with arbitrary real-number powers. Reference [7] gives a duality relation between arbitrary-dimensional central potentials:
The radial wave functions of the three-dimensional central potentials $U\left(r\right)$ and $V\left(\rho \right)$ that satisfy the duality relation
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{r}^{2}}{{\left(l+\tfrac{1}{2}\right)}^{2}}\left[U\left(r\right)-E\right]=\displaystyle \frac{{\rho }^{2}}{{\left({\ell }+\tfrac{1}{2}\right)}^{2}}\left[V\left(\rho \right)-{ \mathcal E }\right],\\ r\to {\rho }^{\sigma }\end{array}\end{eqnarray}$
are related by the duality transform
$\begin{eqnarray}{u}_{l}\left(r\right)\to {\rho }^{\left(\sigma -1\right)/2}{v}_{{\ell }}\left(\rho \right),\end{eqnarray}$
$\begin{eqnarray}l+\displaystyle \frac{1}{2}\to \displaystyle \frac{1}{\sigma }\left({\ell }+\displaystyle \frac{1}{2}\right),\end{eqnarray}$
where E and ${ \mathcal E }$ are eigenvalues and l and are angular quantum numbers. Here σ is an arbitrary constant.
Based on this duality relation, we derive the duality relation for the general polynomial central potential
$\begin{eqnarray}U\left(r\right)=\xi {r}^{a}+\sum _{n}^{N-1}{\mu }_{n}{r}^{{b}_{n}}.\end{eqnarray}$
Here, for convenience, we deliberately arbitrarily write a term in the polynomial, say ξra, separately.
Substituting the general polynomial potential (4) into the duality relation (1) gives
$\begin{eqnarray}V\left(\rho \right)={\sigma }^{2}\xi {\rho }^{2\sigma -2+\sigma a}+\sum _{n}{\sigma }^{2}{\mu }_{n}{\rho }^{2\sigma -2+\sigma {b}_{n}}-{\sigma }^{2}E{\rho }^{2\sigma -2}+{ \mathcal E }.\end{eqnarray}$
If requiring that the dual potential (5) is still an N-term general polynomial central potential, i.e.,
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{A}+\sum _{n}^{N-1}{\lambda }_{n}{\rho }^{{B}_{n}},\end{eqnarray}$
we have N choices. This means that an N-term general polynomial potential has N dual general polynomial potentials. Without loss of generality, we choose
$\begin{eqnarray}\begin{array}{l}2\sigma -2=A,\\ 2\sigma -2+\sigma a=0,\\ 2\sigma -2+\sigma {b}_{n}={B}_{n},\\ {\sigma }^{2}\xi =-{ \mathcal E },\\ -{\sigma }^{2}E=\eta ,\\ {\sigma }^{2}{\mu }_{n}={\lambda }_{n}.\end{array}\end{eqnarray}$
This gives
$\begin{eqnarray}\begin{array}{l}\sigma =\displaystyle \frac{2}{a+2}=\displaystyle \frac{A+2}{2},\\ \sqrt{\displaystyle \frac{2}{a+2}}\left({b}_{n}+2\right)=\sqrt{\displaystyle \frac{2}{A+2}}\left({B}_{n}+2\right),\\ \xi =-{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{ \mathcal E },\\ E=-{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\eta ,\\ \displaystyle \frac{2}{a+2}{\mu }_{n}=\displaystyle \frac{2}{A+2}{\lambda }_{n}.\end{array}\end{eqnarray}$
Then we arrive at the following conclusion.
Two general polynomial central potentials
$\begin{eqnarray}\begin{array}{rcl}U\left(r\right) & = & \xi {r}^{a}+\sum _{n}{\mu }_{n}{r}^{{b}_{n}},\\ V\left(\rho \right) & = & \eta {\rho }^{A}+\sum _{n}{\lambda }_{n}{\rho }^{{B}_{n}}\end{array}\end{eqnarray}$
are dual to each other, if
$\begin{eqnarray}\displaystyle \frac{a+2}{2}=\displaystyle \frac{2}{A+2},\end{eqnarray}$
$\begin{eqnarray}\sqrt{\displaystyle \frac{2}{a+3}}\left({b}_{n}+2\right)=\sqrt{\displaystyle \frac{2}{A+3}}\left({B}_{n}+2\right).\end{eqnarray}$
Their eigenfunctions are related by the duality transform
$\begin{eqnarray}u\left(r\right)\to {\rho }^{A/4}v\left(\rho \right)\end{eqnarray}$
with
$\begin{eqnarray}r\to {\rho }^{\tfrac{A+2}{2}}.\end{eqnarray}$
Their eigenvalues and their angular momenta are related by
$\begin{eqnarray}E\to -\eta {\left(\displaystyle \frac{2}{A+2}\right)}^{2},\end{eqnarray}$
$\begin{eqnarray}\xi \to -{ \mathcal E }{\left(\displaystyle \frac{2}{A+2}\right)}^{2},\end{eqnarray}$
$\begin{eqnarray}{\mu }_{n}\to {\left(\displaystyle \frac{2}{A+2}\right)}^{2}{\lambda }_{n},\end{eqnarray}$
and
$\begin{eqnarray}l+\displaystyle \frac{1}{2}\to \displaystyle \frac{2}{A+2}\left({\ell }+\displaystyle \frac{1}{2}\right).\end{eqnarray}$
More concretely, substituting the duality relations (14)–(17) into the eigenvalue of the potential $U\left(r\right)$,
$\begin{eqnarray}E=f\left({n}_{r},l,\xi ,{\mu }_{n}\right),\end{eqnarray}$
gives
$\begin{eqnarray}\begin{array}{l}-\eta {\left(\displaystyle \frac{2}{A+2}\right)}^{2}=f\left({n}_{r},\displaystyle \frac{2}{A+2}\left({\ell }+\displaystyle \frac{1}{2}\right)\right.\\ \left.-\displaystyle \frac{1}{2},-{ \mathcal E }{\left(\displaystyle \frac{2}{A+2}\right)}^{2},{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{\lambda }_{n}\right).\end{array}\end{eqnarray}$
Then the eigenvalue ${ \mathcal E }$ of the potential $V\left(\rho \right)$ reads
$\begin{eqnarray}\begin{array}{rcl}{ \mathcal E } & = & -{\left(\displaystyle \frac{A+2}{2}\right)}^{2}{f}^{-1}\left({n}_{r},\displaystyle \frac{2}{A+2}\left({\ell }+\displaystyle \frac{1}{2}\right)\right.\\ & & \left.-\displaystyle \frac{1}{2},-\eta {\left(\displaystyle \frac{2}{A+2}\right)}^{2},{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{\lambda }_{n}\right),\end{array}\end{eqnarray}$
where f−1 denotes the inverse function of f.

3. Solving an eigenproblem by constructing a duality family

All potentials that are dual to each other form a duality family. The family member is labeled by a parameter σ. In a duality family, once a potential is solved, the solutions of all other potentials in the family can be obtained by a duality transform. This inspires the idea that if a potential is difficult to solve, we may first construct its duality family and then see if there are any potentials in the family that are easy to solve. If there is a family member that can be solved, then the potential we want to solve can be solved by the duality transform. This approach requires that we must find a solved potential in the duality family.
Note that not all dual potentials of a physical potential are physical potentials. Some dual potentials of a physical potential are unphysical lower-unbounded potentials. These unphysical potentials still make sense in solving eigenvalue problems. A bound-state solution after a duality transform is still a bound-state solution. If a potential has only bound states, but its dual potential has scattering states, the scattering-state solution can be obtained by the duality transform from the scattering-state solution of its unphysical lower-unbounded dual potential.
A duality family consists of infinite family members. However, an N-term polynomial potential only has N dual N-term polynomial potentials.
In the following, we solve some general polynomial potentials by virtue of the duality transform.

3.1. Two-term polynomial central potential

As an example, consider the two-term polynomial central potential
$\begin{eqnarray}U\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}.\end{eqnarray}$
The general form of a family member in a duality family by the duality relation (1) is
$\begin{eqnarray}\begin{array}{rcl}{U}_{\sigma }\left(r\right) & = & {\sigma }^{2}\eta {r}^{2\sigma -2+\sigma A}\\ & & +{\sigma }^{2}\lambda {r}^{2\sigma -2+\sigma B}-{\sigma }^{2}{{Er}}^{2\sigma -2}+{ \mathcal E }.\end{array}\end{eqnarray}$
The family member is labeled by the parameter σ.
There is an infinite number of family members in a duality family; here we only focus on a special case: two-term polynomial potentials.
If requiring the dual potential of a two-term polynomial potential is still a two-term polynomial potential, there are only three choices of the parameter σ: σ = 1, $\sigma =\tfrac{2}{A+2}$, and $\sigma =\tfrac{2}{B+2}$.
For σ = 1,
$\begin{eqnarray}{U}_{1}\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}-E+{ \mathcal E }.\end{eqnarray}$
The requirement that the dual potential has only two terms requires that
$\begin{eqnarray}E={ \mathcal E },\end{eqnarray}$
i.e.,
$\begin{eqnarray}{U}_{1}\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}.\end{eqnarray}$
For $\sigma =\tfrac{2}{A+2}$, corresponding to 2σ − 2 + σA = 0 in equation (22),
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{A+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}\\ +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)}+{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\eta +{ \mathcal E }.\end{array}\end{eqnarray}$
Requiring only two power terms to be retained requires
$\begin{eqnarray}{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\eta +{ \mathcal E }=0\end{eqnarray}$
and we have
$\begin{eqnarray}\begin{array}{rcl}{U}_{\tfrac{2}{A+2}}\left(r\right) & = & -{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}\\ & & +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)}.\end{array}\end{eqnarray}$
For $\sigma =\tfrac{2}{B+2}$, corresponding to 2σ − 2 + σB = 0 in equation (22),
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{B+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}\\ +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}+{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\lambda +{ \mathcal E }.\end{array}\end{eqnarray}$
Similarly, choosing
$\begin{eqnarray}{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\lambda +{ \mathcal E }=0\end{eqnarray}$
gives
$\begin{eqnarray}\begin{array}{rcl}{U}_{\tfrac{2}{B+2}}\left(r\right) & = & -{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}\\ & & +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}.\end{array}\end{eqnarray}$
The duality family of two-term general polynomial central potentials consists of the following three family members:
$\begin{eqnarray}\begin{array}{l}\{\eta {r}^{A}+\lambda {r}^{B},\\ -{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}+{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)},\\ -{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}+{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}\}.\end{array}\end{eqnarray}$

3.2. Three-term polynomial central potential

Similarly, we consider the three-term polynomial central potential
$\begin{eqnarray}U\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}+\mu {r}^{C}.\end{eqnarray}$
The family member is of the form
$\begin{eqnarray}\begin{array}{l}{U}_{\sigma }\left(r\right)={\sigma }^{2}\eta {r}^{2\sigma -2+\sigma A}+{\sigma }^{2}\lambda {r}^{2\sigma -2+\sigma B}\\ +{\sigma }^{2}\mu {r}^{2\sigma -2+\sigma C}-{\sigma }^{2}{{Er}}^{2\sigma -2}+{ \mathcal E }.\end{array}\end{eqnarray}$
If requiring the dual potential of a three-term polynomial potential is still a three-term polynomial potential, there are only four choices: σ = 1, $\sigma =\tfrac{2}{A+2}$, $\sigma =\tfrac{2}{B+2}$, and $\sigma =\tfrac{2}{C+2}$.
For σ = 1,
$\begin{eqnarray}{U}_{1}\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}+\mu {r}^{C}-E+{ \mathcal E }.\end{eqnarray}$
Choosing
$\begin{eqnarray}E={ \mathcal E }\end{eqnarray}$
gives a three-term polynomial potential,
$\begin{eqnarray}{U}_{1}\left(r\right)=\eta {r}^{A}+\lambda {r}^{B}+\mu {r}^{C}.\end{eqnarray}$
For $\sigma =\tfrac{2}{A+2}$,
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{A+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}\ +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)}\\ +\,{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-A}{A+2}\right)}+{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\eta +{ \mathcal E }.\end{array}\end{eqnarray}$
Choosing
$\begin{eqnarray}{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\eta +{ \mathcal E }=0\end{eqnarray}$
gives
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{A+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}\\ +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)}\\ +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-A}{A+2}\right)}.\end{array}\end{eqnarray}$
For $\sigma =\tfrac{2}{B+2}$,
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{B+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}\\ +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}\\ +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-B}{B+2}\right)}+{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\lambda +{ \mathcal E }.\end{array}\end{eqnarray}$
Choosing
$\begin{eqnarray}{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\lambda =-{ \mathcal E }\end{eqnarray}$
gives
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{B+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}\\ +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}+{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-B}{B+2}\right)}.\end{array}\end{eqnarray}$
For $\sigma =\tfrac{2}{C+2}$,
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{C+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{C+2}\right)}^{2}{{Er}}^{\tfrac{4}{C+2}-2}+{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-C}{C+2}\right)}\\ +\,{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-C}{C+2}\right)}+{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\mu +{ \mathcal E }.\end{array}\end{eqnarray}$
Choosing
$\begin{eqnarray}{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\mu =-{ \mathcal E }\end{eqnarray}$
gives
$\begin{eqnarray}\begin{array}{l}{U}_{\tfrac{2}{C+2}}\left(r\right)=-{\left(\displaystyle \frac{2}{C+2}\right)}^{2}{{Er}}^{\tfrac{4}{C+2}-2}\\ +\,{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-C}{C+2}\right)}+{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-C}{C+2}\right)}.\end{array}\end{eqnarray}$
The duality family of three-term general polynomial central potentials consists of four family members:
$\begin{eqnarray}\begin{array}{l}\{\eta {r}^{A}+\lambda {r}^{B}+\mu {r}^{C},\\ -{\left(\displaystyle \frac{2}{A+2}\right)}^{2}{{Er}}^{\tfrac{4}{A+2}-2}+{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-A}{A+2}\right)}\\ +{\left(\displaystyle \frac{2}{A+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-A}{A+2}\right)},\\ -{\left(\displaystyle \frac{2}{B+2}\right)}^{2}{{Er}}^{\tfrac{4}{B+2}-2}+{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-B}{B+2}\right)}\\ +{\left(\displaystyle \frac{2}{B+2}\right)}^{2}\mu {r}^{2\left(\tfrac{C-B}{B+2}\right)},\\ -{\left(\displaystyle \frac{2}{C+2}\right)}^{2}{{Er}}^{\tfrac{4}{C+2}-2}+{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\eta {r}^{2\left(\tfrac{A-C}{C+2}\right)}\\ +{\left(\displaystyle \frac{2}{C+2}\right)}^{2}\lambda {r}^{2\left(\tfrac{B-C}{C+2}\right)}\}.\end{array}\end{eqnarray}$

4. Solving $\displaystyle \frac{\eta }{r}+\displaystyle \frac{\lambda }{{r}^{3/2}}$ and $\eta {r}^{2}+\lambda {r}^{6}$ from $\xi {r}^{2}+\displaystyle \frac{\mu }{r}$

From equation (32), we can see that the potentials $\tfrac{\eta }{r}+\tfrac{\lambda }{{r}^{3/2}}$, ηr2 + λr6, and $\xi {r}^{2}+\tfrac{\mu }{r}$ belong to the same duality family. Their solutions are related by a duality transform.
This example contains a special case: two dual polynomial potentials are dual to each other term-to-term. The potential $\tfrac{\mu }{r}+\tfrac{\eta }{{r}^{3/2}}$ is dual to λr2 + ξr6, while $\tfrac{\mu }{r}$ is dual to λr2 and $\tfrac{\eta }{{r}^{3/2}}$ is dual to ξr6. A two-term potential has only one term-to-term dual potential.
In the following, we solve the potentials
$\begin{eqnarray}V\left(\rho \right)=\displaystyle \frac{\eta }{\rho }+\displaystyle \frac{\lambda }{{\rho }^{3/2}},\end{eqnarray}$
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{2}+\lambda {\rho }^{6}\end{eqnarray}$
from their dual potential
$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\displaystyle \frac{\mu }{r}.\end{eqnarray}$
The exact solution of the potential (50) is given in appendix A.
The radial equation of $U\left(r\right)$ is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}u\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\xi {r}^{2}-\displaystyle \frac{\mu }{r}\right]u\left(r\right)=0.\end{eqnarray}$
The radial eigenfunction of the potential (50) is
$u(r)=A_{l} r^{l+1} \mathrm{e}^{\left(\sqrt{\xi} r^{2}\right) / 2} \\\\ \times \mathrm{N}\left(2 l+1,0,-\frac{E}{\sqrt{\xi}},-\frac{2 \mathrm{i} \mu}{\xi^{1 / 4}}, \mathrm{i} \xi^{1 / 4} r\right)$
and the eigenvalue of bound states can be expressed by an implicit expression:
$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,0,-\displaystyle \frac{E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)=0,\end{eqnarray}$
i.e., the eigenvalue is the zero of ${{\rm{K}}}_{2}\left(2l+1,0,-\tfrac{E}{\sqrt{\xi }},-\tfrac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)$. Here
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(\alpha ,\beta ,\gamma ,\delta \right)=\displaystyle \frac{{\rm{\Gamma }}\left(1+\alpha \right)}{{\rm{\Gamma }}\left(1+\tfrac{\alpha }{2}+\tfrac{\gamma }{2}\right)}\\ \times {\displaystyle \int }_{0}^{\infty }{x}^{\tfrac{\alpha }{2}+\tfrac{\gamma }{2}}{{\rm{e}}}^{-\beta x-{x}^{2}}{\rm{N}}\left(\displaystyle \frac{\alpha }{2}+\displaystyle \frac{\gamma }{2},\beta ,\right.\\ \left.\displaystyle \frac{3\alpha }{2}-\displaystyle \frac{\gamma }{2},\delta +\displaystyle \frac{\beta \gamma }{2}-\displaystyle \frac{\beta \alpha }{2},x\right){\rm{d}}x,\end{array}\end{eqnarray}$
where N(α, β, γ, δ, z) is the biconfluent Heun function [9, 20, 21].

4.1.$V(\rho )=\tfrac{\eta }{\rho }+\tfrac{\lambda }{{\rho }^{3/2}}$

The potential (48) is a dual potential of the potential (50) with a = 2, b = −1 and A = −1, B = −3/2 in the duality relations (10) and (11). We can solve it from the solution of the potential $U\left(r\right)$.
The radial equation of $V\left(\rho \right)=\tfrac{\eta }{\rho }+\tfrac{\lambda }{{\rho }^{3/2}}$ is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\displaystyle \frac{\eta }{\rho }-\displaystyle \frac{\lambda }{{\rho }^{3/2}}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = 2 and b = −1 read
$\begin{eqnarray}\begin{array}{l}E\to -4\eta ,\qquad \xi \to -4{ \mathcal E },\qquad \mu \to 4\lambda ,\\ l+\displaystyle \frac{1}{2}\to 2\left({\ell }+\displaystyle \frac{1}{2}\right)\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{1/2},\\ u\left(r\right)\to {\rho }^{-1/4}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the above dual relations into the eigenfunction of the potential (50), equation (52), gives
$\begin{eqnarray}\begin{array}{l}v\left(\rho \right)={A}_{{\ell }}{\rho }^{{\ell }+1}{{\rm{e}}}^{\sqrt{-{ \mathcal E }}\rho }\\ \times \,{\rm{N}}\left(4{\ell }+2,0,\displaystyle \frac{2\eta }{\sqrt{-{ \mathcal E }}},-\displaystyle \frac{4\sqrt{2}{\rm{i}}\lambda }{{\left(-{ \mathcal E }\right)}^{1/4}},{\rm{i}}{\left(-{ \mathcal E }\right)}^{1/4}\sqrt{2\rho }\right).\end{array}\end{eqnarray}$
Substituting the above replacements into equation (53) gives an implicit expression of the eigenvalue of the potential (48):
$\begin{eqnarray}{{\rm{K}}}_{2}\left(4{\ell }+2,0,\displaystyle \frac{2\eta }{\sqrt{-{ \mathcal E }}},-\displaystyle \frac{4\sqrt{2}{\rm{i}}\lambda }{{\left(-{ \mathcal E }\right)}^{1/4}}\right)=0.\end{eqnarray}$
It should be noted that the potential $\xi {r}^{2}+\tfrac{\mu }{r}$ with ξ > 0 has only bound states for which E > 0. The duality relation (56) gives the solution of its dual potential $\tfrac{\eta }{\rho }+\tfrac{\lambda }{{\rho }^{3/2}}$ with η < 0 and ${ \mathcal E }\lt 0$, which is also a bound state. That is, the duality relation transforms bound states only to bound states.
If we want to seek the scattering-state solution of the potential $\tfrac{\eta }{\rho }+\tfrac{\lambda }{{\rho }^{3/2}}$ for which η < 0 and ${ \mathcal E }\gt 0$, we should start with the scattering-state solution of the dual potential $\xi {r}^{2}+\tfrac{\mu }{r}$ with ξ < 0. The potential $\xi {r}^{2}+\tfrac{\mu }{r}$ with ξ < 0 is not lower bounded and is not a physical potential, but we can use it to solve physical potentials.

4.2. V(ρ) = ηρ2 + λρ6

The potential (49) is a dual potential of the potential (50) with a = −1, b = 2 and A = 2, B = 6 in the duality relations (10) and (11). We can also solve it from the solution of the potential (50).
The radial equation of $V\left(\rho \right)=\eta {\rho }^{2}+\lambda {\rho }^{6}$ is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\eta {\rho }^{2}-\lambda {\rho }^{6}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = −1 and b = 2 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{\eta }{4},\qquad \mu \to -\displaystyle \frac{{ \mathcal E }}{4},\qquad \xi \to \displaystyle \frac{\lambda }{4},\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{1}{2}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{2},\\ u\left(r\right)\to {\rho }^{1/2}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (50), equation (52), gives
$\begin{eqnarray}\begin{array}{l}v\left(\rho \right)={A}_{{\ell }}{\rho }^{{\ell }+1}{{\rm{e}}}^{\tfrac{\sqrt{\lambda }}{4}{\rho }^{4}}\\ \times \,{\rm{N}}\left({\ell }+\displaystyle \frac{1}{2},0,\displaystyle \frac{\eta }{2\sqrt{\lambda }},\displaystyle \frac{{\rm{i}}{ \mathcal E }}{\sqrt{2}{\lambda }^{1/4}},\displaystyle \frac{{\rm{i}}}{\sqrt{2}}{\lambda }^{1/4}{\rho }^{2}\right).\end{array}\end{eqnarray}$
Substituting the replacements into equation (53) gives an implicit expression of the eigenvalue of the potential (49):
$\begin{eqnarray}{{\rm{K}}}_{2}\left({\ell }+\displaystyle \frac{1}{2},0,\displaystyle \frac{\eta }{2\sqrt{\lambda }},\displaystyle \frac{{\rm{i}}{ \mathcal E }}{\sqrt{2}{\lambda }^{1/4}}\right)=0.\end{eqnarray}$

5. Solving $\displaystyle \frac{\eta }{r}+\displaystyle \frac{\lambda }{\sqrt{r}}$ and $\displaystyle \frac{\eta }{{r}^{2/3}}+\lambda {r}^{2/3}$ from $\xi {r}^{2}+\mu {r}^{}$

From equation (32), we can see that the potentials $\tfrac{\eta }{r}+\tfrac{\lambda }{\sqrt{r}}$, $\tfrac{\eta }{{r}^{2/3}}+\lambda {r}^{2/3}$, and ξr2 + μr belong to the same duality family.
In this section, we solve the potentials
$\begin{eqnarray}V\left(\rho \right)=\displaystyle \frac{\eta }{\rho }+\displaystyle \frac{\lambda }{\sqrt{\rho }},\end{eqnarray}$
$\begin{eqnarray}V\left(\rho \right)=\displaystyle \frac{\eta }{{\rho }^{2/3}}+\lambda {\rho }^{2/3}\end{eqnarray}$
from their dual potential
$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\mu r.\end{eqnarray}$
The exact solution of the potential (67) is given in appendix B.
The radial equation of the potential (67) is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}u\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\xi {r}^{2}-\mu r\right]u\left(r\right)=0.\end{eqnarray}$
The radial eigenfunction is
$\begin{eqnarray}\begin{array}{l}u\left(r\right)={A}_{l}{r}^{l+1}\exp \left(\displaystyle \frac{\sqrt{\xi }}{2}{r}^{2}+\displaystyle \frac{\mu }{2\sqrt{\xi }}r\right)\\ \times \,{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{\sqrt{\xi }}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,{\rm{i}}{\xi }^{1/4}r\right).\end{array}\end{eqnarray}$
The eigenvalue of the bound state can be expressed by the implicit expression
$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{\sqrt{\xi }}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)=0,\end{eqnarray}$
i.e., the eigenvalue of $U\left(r\right)$ is the zero of ${{\rm{K}}}_{2}\left(2l+1,\tfrac{{\rm{i}}\mu }{{\xi }^{3/4}},-\tfrac{E}{\sqrt{\xi }}-\tfrac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)$.

5.1. $V(\rho )=\tfrac{\eta }{\rho }+\tfrac{\lambda }{\sqrt{\rho }}$

The potential (65) is a dual potential of the potential (67) with a = 2, b = 1 and A = −1, B = −1/2 in the duality relations (10) and (11).
The radial equation of $V\left(\rho \right)=\tfrac{\eta }{\rho }+\tfrac{\lambda }{\sqrt{\rho }}$ is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\displaystyle \frac{\eta }{\rho }-\displaystyle \frac{\lambda }{\sqrt{\rho }}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = 2 and b = 1 read
$\begin{eqnarray}\begin{array}{l}E\to -4\eta ,\qquad \xi \to -4{ \mathcal E },\qquad \mu \to 4\lambda ,\\ l+\displaystyle \frac{1}{2}\to 2\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{1/2},\\ u\left(r\right)\to {\rho }^{-1/4}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (67), equation (52), gives
$\begin{eqnarray}\begin{array}{rcl}v\left(\rho \right) & = & {A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\sqrt{-{ \mathcal E }}\rho +\displaystyle \frac{\lambda }{\sqrt{-{ \mathcal E }}}\sqrt{\rho }\right)\\ & & \times {\rm{N}}\left(4{\ell }+2,\displaystyle \frac{\sqrt{2}{\rm{i}}\lambda }{{\left(-{ \mathcal E }\right)}^{3/4}},\displaystyle \frac{2\eta }{\sqrt{-{ \mathcal E }}}\right.\\ & & \left.-\displaystyle \frac{{\lambda }^{2}}{2{\left(-{ \mathcal E }\right)}^{3/2}},0,{\rm{i}}{\left(-{ \mathcal E }\right)}^{1/4}\sqrt{2\rho }\right).\end{array}\end{eqnarray}$
Substituting the dual relations into (70) gives an implicit expression of the eigenvalue of the potential (65):
$\begin{eqnarray}{{\rm{K}}}_{2}\left(4{\ell }+2,\displaystyle \frac{\sqrt{2}{\rm{i}}\lambda }{{\left(-{ \mathcal E }\right)}^{3/4}},\displaystyle \frac{2\eta }{\sqrt{-{ \mathcal E }}}-\displaystyle \frac{{\lambda }^{2}}{2{\left(-{ \mathcal E }\right)}^{3/2}},0\right)=0.\end{eqnarray}$
The potential ξr2 + μr with ξ > 0 has only bound states (E > 0). The duality transform can only give bound-state solutions for its dual potential $\tfrac{\eta }{\rho }+\tfrac{\lambda }{\sqrt{\rho }}$ with η < 0 and ${ \mathcal E }\lt 0$. The scattering-state solution of the potential $\tfrac{\eta }{\rho }+\tfrac{\lambda }{\sqrt{\rho }}$ with η < 0 and ${ \mathcal E }\gt 0$ is given by the scattering-state solution of its lower-unbounded unphysical dual potential ξr2 + μr with ξ < 0.

5.2. $V(\rho )=\frac{\eta }{{\rho }^{2/3}}\,+\lambda {\rho }^{2/3}$

The potential (66) is another dual potential of the potential (67) with a = 1, b = 2 and A = −2/3, B = 2/3 in the duality relations (10) and (11).
The radial equation of $V\left(\rho \right)=\tfrac{\eta }{{\rho }^{2/3}}+\lambda {\rho }^{2/3}$ is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\displaystyle \frac{\eta }{{\rho }^{2/3}}-\lambda {\rho }^{2/3}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = 1 and b = 2 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{9}{4}\eta ,\qquad \mu \to -\displaystyle \frac{9}{4}{ \mathcal E },\qquad \xi \to \displaystyle \frac{9}{4}\lambda ,\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{3}{2}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{2/3},\\ u\left(r\right)\to {\rho }^{-1/6}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (67), equation (69), gives
$\begin{eqnarray}\begin{array}{l}v\left(\rho \right)={A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\displaystyle \frac{{\left(\tfrac{9}{4}\lambda \right)}^{1/2}}{2}{\rho }^{4/3}+\displaystyle \frac{-\tfrac{9}{4}{ \mathcal E }}{2{\left(\tfrac{9}{4}\lambda \right)}^{1/2}}{\rho }^{2/3}\right)\\ \times \,{\rm{N}}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\lambda }^{3/4}},\displaystyle \frac{3\eta }{2\sqrt{\lambda }}\right.\\ \left.-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\lambda }^{3/2}},0,\displaystyle \frac{{\rm{i}}\sqrt{6}}{2}{\lambda }^{1/4}{\rho }^{2/3}\right).\end{array}\end{eqnarray}$
Substituting the dual relations into equation (70) gives an implicit expression of the eigenvalue of the potential (66):
$\begin{eqnarray}{{\rm{K}}}_{2}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\lambda }^{3/4}},\displaystyle \frac{3\eta }{2\sqrt{\lambda }}-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\lambda }^{3/2}},0\right)=0.\end{eqnarray}$

6. Solving $\eta {r}^{2/3}+\displaystyle \frac{\lambda }{{r}^{4/3}}$ and $\eta {r}^{6}+\lambda {r}^{4}$ from $\displaystyle \frac{\xi }{\sqrt{r}}+\displaystyle \frac{\mu }{{r}^{3/2}}$

From equation (32), we can see that the potentials $\eta {r}^{2/3}+\tfrac{\lambda }{{r}^{4/3}}$, ηr6 + λr4, and $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$ belong to the same duality family. In the following, we solve the potentials
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{2/3}+\displaystyle \frac{\lambda }{{\rho }^{4/3}},\end{eqnarray}$
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{6}+\lambda {\rho }^{4}\end{eqnarray}$
from their dual potential
$\begin{eqnarray}U\left(r\right)=\displaystyle \frac{\xi }{\sqrt{r}}+\displaystyle \frac{\mu }{{r}^{3/2}}.\end{eqnarray}$
The exact solution of the potential (83) is given in appendix C.
The radial equation of the potential (83) is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}u\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\displaystyle \frac{\xi }{\sqrt{r}}-\displaystyle \frac{\mu }{{r}^{3/2}}\right]u\left(r\right)=0.\end{eqnarray}$
The radial eigenfunction is
$\begin{eqnarray}\begin{array}{l}u\left(r\right)={A}_{l}{r}^{l+1}\exp \left(\sqrt{-E}r+\displaystyle \frac{\xi }{\sqrt{-E}}\sqrt{r}\right)\\ \times {\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\xi \sqrt{2}}{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\right.\\ \left.\displaystyle \frac{-4\sqrt{2}{\rm{i}}\mu }{{\left(-E\right)}^{1/4}},{\rm{i}}\sqrt{2}{\left(-E\right)}^{1/4}\sqrt{r}\right).\end{array}\end{eqnarray}$
The eigenvalue of the bound state can be expressed by the implicit expression [22]
$\begin{eqnarray}{{\rm{K}}}_{2}\left(4l+2,\displaystyle \frac{{\rm{i}}\xi \sqrt{2}}{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-4\sqrt{2}{\rm{i}}\mu }{{\left(-E\right)}^{1/4}}\right)=0,\end{eqnarray}$
i.e., the eigenvalue is the zero of ${{\rm{K}}}_{2}\left(4l+2,\tfrac{{\rm{i}}\xi \sqrt{2}}{{\left(-E\right)}^{3/4}},-\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\tfrac{-4\sqrt{2}{\rm{i}}\mu }{{\left(-E\right)}^{1/4}}\right)$.

6.1. $V(\rho )=\eta {\rho }^{2/3}+\tfrac{\lambda }{{\rho }^{2/3}}$

The potential (81) is a dual potential of the potential (83) with a = −1/2, b = −3/2 and A = 2/3, B = −4/3 in the duality relations (10) and (11).
The radial equation of the potential (81) is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\eta {\rho }^{2/3}-\displaystyle \frac{\lambda }{{\rho }^{4/3}}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = −1/2 and b = −3/2 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{9}{16}\eta ,\qquad \xi \to -\displaystyle \frac{9}{16}{ \mathcal E },\qquad \mu \to \displaystyle \frac{9}{16}\lambda ,\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{3}{4}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{4/3},\\ u\left(r\right)\to {\rho }^{1/6}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (83), equation (85), gives the eigenfunction of the potential (81):
$\begin{eqnarray}\begin{array}{rcl}v\left(\rho \right) & = & {A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\displaystyle \frac{3\sqrt{\eta }}{4}{\rho }^{4/3}-\displaystyle \frac{3{ \mathcal E }}{4\sqrt{\eta }}{\rho }^{2/3}\right)\\ & & \times {\rm{N}}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\eta }^{3/4}},-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\eta }^{3/2}},\right.\\ & & \left.-\displaystyle \frac{{\rm{i}}3\sqrt{6}\lambda }{2{\eta }^{1/4}},\displaystyle \frac{{\rm{i}}\sqrt{6}}{2}{\eta }^{1/4}{\rho }^{2/3}\right).\end{array}\end{eqnarray}$
Substituting the dual relations into equation (86) gives an implicit expression of the eigenvalue of the potential (81):
$\begin{eqnarray}{{\rm{K}}}_{2}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\eta }^{3/4}},-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\eta }^{3/2}},-\displaystyle \frac{{\rm{i}}3\sqrt{6}\lambda }{2{\eta }^{1/4}}\right)=0.\end{eqnarray}$
The potential $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$ with ξ < 0 has both bound states and scattering states. The bound-state solution of the potential $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$ (ξ < 0 and E < 0) gives the bound-state solution of the potential $\eta {\rho }^{2/3}+\tfrac{\lambda }{{\rho }^{4/3}}$ (η > 0) which has only bound states.

6.2. V(ρ) = ηρ6 + λρ4

The potential (82)
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{6}+\lambda {\rho }^{4}\end{eqnarray}$
is a dual potential of the potential (83) with a = −3/2, b = −1/2 and A = 6, B = 4 in the duality relations (10) and (11).
The radial equation of the potential (82) is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\eta {\rho }^{6}-\lambda {\rho }^{4}\right]v\left(\rho \right)=0.\end{eqnarray}$
The dual relations (12)–(17) with a = −3/2 and b = −1/2 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{1}{16}\eta ,\qquad \mu \to -\displaystyle \frac{1}{16}{ \mathcal E },\qquad \xi \to \displaystyle \frac{1}{16}\lambda ,\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{1}{4}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{4},\\ u\left(r\right)\to {\rho }^{3/2}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (83), equation (85), gives
$\begin{eqnarray}\begin{array}{rcl}v\left(\rho \right) & = & {A}_{l}{\rho }^{{\ell }+1}\exp \left(\displaystyle \frac{\sqrt{\eta }}{4}{\rho }^{4}+\displaystyle \frac{\lambda }{4\sqrt{\eta }}{\rho }^{2}\right)\\ & & \times {\rm{N}}\left({\ell }+\displaystyle \frac{1}{2},\displaystyle \frac{{\rm{i}}\lambda \sqrt{2}}{2{\eta }^{3/4}},-\displaystyle \frac{{\lambda }^{2}}{8{\eta }^{3/2}},\right.\\ & & \left.\displaystyle \frac{{\rm{i}}\sqrt{2}{ \mathcal E }}{2{\eta }^{1/4}},\displaystyle \frac{{\rm{i}}\sqrt{2}}{2}{\eta }^{1/4}{\rho }^{2}\right).\end{array}\end{eqnarray}$
Substituting the dual relations into (86) gives the eigenvalue of the potential (82):
$\begin{eqnarray}{{\rm{K}}}_{2}\left({\ell }+\displaystyle \frac{1}{2},\displaystyle \frac{{\rm{i}}\lambda \sqrt{2}}{2{\eta }^{3/4}},-\displaystyle \frac{{\lambda }^{2}}{8{\eta }^{3/2}},\displaystyle \frac{{\rm{i}}\sqrt{2}{ \mathcal E }}{2{\eta }^{1/4}}\right)=0.\end{eqnarray}$
The potential $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$ with ξ < 0 has both bound states and scattering states. The bound-state solution of the potential $\tfrac{\xi }{\sqrt{r}}+\tfrac{\mu }{{r}^{3/2}}$ (ξ < 0 and E < 0) gives the bound-state solution of the potential ηρ6 + λρ4 (η > 0) which has only bound states.

7. Solving $\tfrac{\eta }{\rho }+\tfrac{\nu }{{\rho }^{3/2}}+\,\tfrac{\lambda }{{\rho }^{1/2}},$ $\eta {\rho }^{2}+\nu {\rho }^{6}+\lambda {\rho }^{4},$ and $\tfrac{\eta }{{\rho }^{2/3}}+\,\nu {\rho }^{2/3}+\,\tfrac{\lambda }{{\rho }^{4/3}}$ from $\xi {r}^{2}+\,\tfrac{\mu }{r}+\kappa r$

From equation (47), we can see that the potentials $\tfrac{\eta }{r}+\tfrac{\nu }{{r}^{3/2}}+\tfrac{\lambda }{{r}^{1/2}}$, ηr2 + νr6 + λr4, $\tfrac{\eta }{{r}^{2/3}}+\nu {r}^{2/3}+\tfrac{\lambda }{{r}^{4/3}}$, and $\xi {r}^{2}+\tfrac{\mu }{r}+\kappa r$ belong to the same duality family. In the following, we solve the potentials
$\begin{eqnarray}V\left(\rho \right)=\displaystyle \frac{\eta }{\rho }+\displaystyle \frac{\nu }{{\rho }^{3/2}}+\displaystyle \frac{\lambda }{{\rho }^{1/2}},\end{eqnarray}$
$\begin{eqnarray}V\left(\rho \right)=\eta {\rho }^{2}+\nu {\rho }^{6}+\lambda {\rho }^{4},\end{eqnarray}$
$\begin{eqnarray}V\left(\rho \right)=\displaystyle \frac{\eta }{{\rho }^{2/3}}+\nu {\rho }^{2/3}+\displaystyle \frac{\lambda }{{\rho }^{4/3}}\end{eqnarray}$
from their dual potential
$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\displaystyle \frac{\mu }{r}+\kappa r.\end{eqnarray}$
The exact solution of the potential (101) is given in appendix D.
The radial equation of the potential (101) is
$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}u\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\xi {r}^{2}-\displaystyle \frac{\mu }{r}-\kappa r\right]u\left(r\right)=0.\end{eqnarray}$
The radial eigenfunction is
$\begin{eqnarray}\begin{array}{rcl}u\left(r\right) & = & {A}_{l}{r}^{l+1}\exp \left(\displaystyle \frac{\sqrt{\xi }}{2}{r}^{2}+\displaystyle \frac{\kappa }{2\sqrt{\xi }}r\right)\\ & & \times {\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}\right.\\ & & \left.-\displaystyle \frac{E}{{\xi }^{1/2}},-{\rm{i}}\displaystyle \frac{2\mu }{{\xi }^{1/4}},{\rm{i}}{\xi }^{1/4}r\right).\end{array}\end{eqnarray}$
The eigenvalue of the bound states can be expressed by the implicit expression [22]
$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-\displaystyle \frac{E}{{\xi }^{1/2}},-{\rm{i}}\displaystyle \frac{2\mu }{{\xi }^{1/4}}\right)=0.\end{eqnarray}$

7.1. $V(\rho )=\tfrac{\eta }{\rho }+\,\tfrac{\nu }{{\rho }^{3/2}}+\,\tfrac{\lambda }{{\rho }^{1/2}}$

The potential (98) is a dual potential of the potential (101) with a = 2, b1 = − 1, b2 = 1 and A = −1, B1 = − 3/2, B2 = − 1/2 in the duality relations (10) and (11).
The radial equation is
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}-\displaystyle \frac{\eta }{\rho }\right.\\ \left.-\displaystyle \frac{\nu }{{\rho }^{3/2}}-\displaystyle \frac{\lambda }{{\rho }^{1/2}}\right]v\left(\rho \right)=0.\end{array}\end{eqnarray}$
The dual relations (12)–(17) with a = 2, b1 = − 1, and b2 = 1 read
$\begin{eqnarray}\begin{array}{l}E\to -4\eta ,\qquad \xi \to -4{ \mathcal E },\qquad \mu \to 4\nu ,\\ \kappa \to 4\lambda ,\\ l+\displaystyle \frac{1}{2}\to 2\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{1/2},\\ u\left(r\right)\to {\rho }^{-1/4}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (101), equation (103), gives the eigenfunction
$\begin{eqnarray}\begin{array}{rcl}v\left(\rho \right) & = & {A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\sqrt{-{ \mathcal E }}\rho +\displaystyle \frac{\lambda }{\sqrt{-{ \mathcal E }}}\sqrt{\rho }\right)\\ & & \times {\rm{N}}\left(4{\ell }+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\lambda }{{\left(-{ \mathcal E }\right)}^{3/4}},-\displaystyle \frac{{\lambda }^{2}}{2{\left(-{ \mathcal E }\right)}^{3/2}}+\displaystyle \frac{2\eta }{{\left(-{ \mathcal E }\right)}^{1/2}},\right.\\ & & \left.-{\rm{i}}\displaystyle \frac{4\sqrt{2}\nu }{{\left(-{ \mathcal E }\right)}^{1/4}},{\rm{i}}{\left(-{ \mathcal E }\right)}^{1/4}\sqrt{2\rho }\right).\end{array}\end{eqnarray}$
Substituting the dual relations into (104) gives an implicit expression of the eigenvalue of the potential (98):
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(4{\ell }+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\lambda }{{\left(-{ \mathcal E }\right)}^{3/4}},-\displaystyle \frac{{\lambda }^{2}}{2{\left(-{ \mathcal E }\right)}^{3/2}}\right.\\ \left.+\displaystyle \frac{2\eta }{{\left(-{ \mathcal E }\right)}^{1/2}},-{\rm{i}}\displaystyle \frac{4\sqrt{2}\nu }{{\left(-{ \mathcal E }\right)}^{1/4}}\right)=0.\end{array}\end{eqnarray}$
The potential $\xi {r}^{2}+\tfrac{\mu }{r}+\kappa r$ with ξ > 0 has only bound states (E > 0). The duality transform can only give bound-state solutions for its dual potential $\tfrac{\eta }{\rho }+\tfrac{\nu }{{\rho }^{3/2}}+\tfrac{\lambda }{{\rho }^{1/2}}$ with η < 0 and ${ \mathcal E }\lt 0$. The scattering-state solution of the potential $\tfrac{\eta }{\rho }+\tfrac{\nu }{{\rho }^{3/2}}+\tfrac{\lambda }{{\rho }^{1/2}}$ with η < 0 and ${ \mathcal E }\gt 0$ is given by the scattering-state solution of its lower-unbounded unphysical dual potential $\xi {r}^{2}+\tfrac{\mu }{r}+\kappa r$ with ξ < 0.

7.2. V(ρ) = ηρ2 + νρ6 + λρ4

The potential (99) is a dual potential of the potential (101) with a = −1, b1 = 2, and b2 = 1 and A = 2, B1 = 6, B2 = 4 in the duality relations (10) and (11).
The radial equation of the potential (99) is
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}\right.\\ \left.-\eta {\rho }^{2}-\nu {\rho }^{6}-\lambda {\rho }^{4}\Space{0ex}{3.19ex}{0ex}\right]v\left(\rho \right)=0.\end{array}\end{eqnarray}$
The dual relations (12)–(17) with a = −1, b1 = 2, and b2 = 1 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{\eta }{4},\qquad \mu \to -\displaystyle \frac{{ \mathcal E }}{4},\qquad \xi \to \displaystyle \frac{\nu }{4},\qquad \kappa \to \displaystyle \frac{\lambda }{4},\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{1}{2}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{2},\\ u\left(r\right)\to {\rho }^{1/2}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (101), equation (103), gives the eigenfunction of the potential (99):
$\begin{eqnarray}\begin{array}{l}v\left(\rho \right)={A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\displaystyle \frac{\sqrt{\nu }}{4}{\rho }^{4}+\displaystyle \frac{\lambda }{4\sqrt{\nu }}{\rho }^{2}\right)\\ \times \,{\rm{N}}\left({\ell }+\displaystyle \frac{1}{2},\displaystyle \frac{{\rm{i}}\sqrt{2}\lambda }{2{\nu }^{3/4}},\displaystyle \frac{\eta }{2{\nu }^{1/2}}\right.\\ \left.-\displaystyle \frac{{\lambda }^{2}}{8{\nu }^{3/2}},{\rm{i}}\displaystyle \frac{\sqrt{2}{ \mathcal E }}{2{\nu }^{1/4}},{\rm{i}}\displaystyle \frac{\sqrt{2}}{2}{\nu }^{1/4}{\rho }^{2}\right).\end{array}\end{eqnarray}$
Substituting the dual relations into (104) gives an implicit expression of the eigenvalue of the potential (99):
$\begin{eqnarray}{{\rm{K}}}_{2}\left({\ell }+\displaystyle \frac{1}{2},\displaystyle \frac{{\rm{i}}\sqrt{2}\lambda }{2{\nu }^{3/4}},\displaystyle \frac{\eta }{2{\nu }^{1/2}}-\displaystyle \frac{{\lambda }^{2}}{8{\nu }^{3/2}},{\rm{i}}\displaystyle \frac{\sqrt{2}{ \mathcal E }}{2{\nu }^{1/4}}\right)=0.\end{eqnarray}$

7.3.$V(\rho )=\tfrac{\eta }{{\rho }^{2/3}}+\,\nu {\rho }^{2/3}+\,\tfrac{\lambda }{{\rho }^{4/3}}$

The potential (100) is a dual potential of the potential (101) with a = 1, b1 = 2, b2 = −1 and A = −2/3, B1 = 2/3, B2 = −4/3 in the duality relations (10) and (11).
The radial equation of the potential (100) is
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{{\rm{d}}}^{2}v\left(\rho \right)}{{\rm{d}}{\rho }^{2}}+\left[{ \mathcal E }-\displaystyle \frac{{\ell }\left({\ell }+1\right)}{{\rho }^{2}}\right.\\ \left.-\displaystyle \frac{\eta }{{\rho }^{2/3}}-\nu {\rho }^{2/3}-\displaystyle \frac{\lambda }{{\rho }^{4/3}}\right]v\left(\rho \right)=0.\end{array}\end{eqnarray}$
The dual relations (12)–(17) with a = 1, b1 = 2, and b2 = −1 read
$\begin{eqnarray}\begin{array}{l}E\to -\displaystyle \frac{9}{4}\eta ,\qquad \kappa \to -\displaystyle \frac{9}{4}{ \mathcal E },\qquad \xi \to \displaystyle \frac{9}{4}\nu ,\qquad \mu \to \displaystyle \frac{9}{4}\lambda ,\\ l+\displaystyle \frac{1}{2}\to \displaystyle \frac{3}{2}\left({\ell }+\displaystyle \frac{1}{2}\right),\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}r\to {\rho }^{2/3},\\ u\left(r\right)\to {\rho }^{-1/6}v\left(\rho \right).\end{array}\end{eqnarray}$
Substituting the dual relations into the eigenfunction of the potential (101), equation (103), gives the eigenfunction of the potential (100):
$\begin{eqnarray}\begin{array}{rcl}v\left(\rho \right) & = & {A}_{{\ell }}{\rho }^{{\ell }+1}\exp \left(\displaystyle \frac{3{\nu }^{1/2}}{4}{\rho }^{4/3}-\displaystyle \frac{3{ \mathcal E }}{4{\nu }^{1/2}}{\rho }^{2/3}\right)\\ & & \times {\rm{N}}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\nu }^{3/4}},\displaystyle \frac{3\eta }{2{\nu }^{1/2}}-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\nu }^{3/2}},\right.\\ & & \left.-{\rm{i}}\displaystyle \frac{3\sqrt{6}\lambda }{2{\nu }^{1/4}},\displaystyle \frac{{\rm{i}}\sqrt{6}}{2}{\nu }^{1/4}{\rho }^{2/3}\right).\end{array}\end{eqnarray}$
Substituting the dual relations into (104) gives an implicit expression of the eigenvalue of the potential (100):
$\begin{eqnarray}{{\rm{K}}}_{2}\left(3{\ell }+\displaystyle \frac{3}{2},-\displaystyle \frac{{\rm{i}}\sqrt{6}{ \mathcal E }}{2{\nu }^{3/4}},\displaystyle \frac{3\eta }{2{\nu }^{1/2}}-\displaystyle \frac{3{{ \mathcal E }}^{2}}{8{\nu }^{3/2}},-{\rm{i}}\displaystyle \frac{3\sqrt{6}\lambda }{2{\nu }^{1/4}}\right)=0.\end{eqnarray}$

8. Conclusion

We suggest a method for solving eigenproblems in quantum mechanics. This is an indirect method. Unlike the usual method, this method is not a technique of solving differential equations. The example given by Newton himself demonstrates how to obtain the solution of the harmonic oscillator potential from the solution of the Newtonian gravitational potential, or vice versa.
The key of the method is to first construct the dual family of the potential to be solved, and then to look for the solvable member in the duality family. If there is a solvable member, all family members, including the potential we want, can be solved by the duality transform.
It is not only the Schrödinger equation that has such a kind of duality. This duality also exists in, e.g., the scalar field equation [7] and the Gross–Pitaevskii equation [23]. The duality can also be applied to long-range potential scattering [24]. The duality relation can be further used to solve other equations. For example, in gravity problems, we also need to solve eigenproblems in curved space [2527]. This method can also be used to seek the exact solutions of eigenproblems in curved space.
It is worth noting that the exact solutions obtained in this paper contain some long-range potentials. For short-range potentials which have only scattering states, there is a uniform treatment since for short-range potentials the scattering boundary conditions are the same [22, 2830]. Long-range potentials may have both scattering and bound states or only bound states. For long-range potential scattering, the scattering boundary conditions depend on potentials and different potentials have different scattering boundary conditions [3137] . Exact solutions play an important role in long-range potential scattering. The exact solution obtained in the paper can also be used to study bound states, e.g., the existence of bound states [38, 39] and the number of bound states [4042, 42, 43] .

We are very indebted to Dr G Zeitrauman for his encouragement. This work is supported in part by the Special Funds for Theoretical Physics Research Program of the NSFC under Grant No. 11947124, and NSFC under Grant Nos. 11575125 and 11675119.

In this appendix, we provide an exact solution of the eigenproblem of the potential

$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\displaystyle \frac{\mu }{r},\end{eqnarray}$
by solving the radial equation directly. This potential has only bound states.

The radial equation of the potential (A1) reads

$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}{u}_{l}\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\xi {r}^{2}-\displaystyle \frac{\mu }{r}\right]{u}_{l}\left(r\right)=0.\end{eqnarray}$
Using the variable substitution
$\begin{eqnarray}z={\rm{i}}{\xi }^{1/4}r\end{eqnarray}$
and introducing ${f}_{l}\left(z\right)$ by
$\begin{eqnarray}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}\right){z}^{l+1}{f}_{l}\left(z\right)\end{eqnarray}$
with Al a constant, we convert the radial equation (A2) into an equation of ${f}_{l}\left(z\right)$:
$\begin{eqnarray}\begin{array}{l}{{zf}}_{l}^{{\prime\prime} }\left(z\right)+\left(-2{z}^{2}+2l+2\right){f}_{l}^{{\prime} }\left(z\right)\\ +\left[\left(\displaystyle \frac{-E}{\sqrt{\xi }}-2l-3\right)z+\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}}\right]{f}_{l}\left(z\right)=0.\end{array}\end{eqnarray}$
This is a biconfluent Heun equation [20].

The choice of the boundary condition has been discussed in [9].

A.1. The regular solution

The regular solution is a solution satisfying the boundary condition at r = 0 [9]. The regular solution at r = 0 should satisfy the boundary condition ${\mathrm{lim}}_{r\to 0}{u}_{l}\left(r\right)/{r}^{l\,+\,1}=1$. In this section, we provide the regular solution of equation (A5).

The biconfluent Heun equation (A5) has two linearly independent solutions [20],

$\begin{eqnarray}{y}_{l}^{\left(1\right)}\left(z\right)={\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{y}_{l}^{\left(2\right)}\left(z\right)=c{\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\\ \mathrm{ln}z+\sum _{n\geqslant 0}{d}_{n}{z}^{n-2l-1},\end{array}\end{eqnarray}$
where
$\begin{eqnarray}c=\displaystyle \frac{1}{2l+1}\left[\displaystyle \frac{-{\rm{i}}\mu }{{\xi }^{1/4}}{d}_{2l}-\left(\displaystyle \frac{-E}{\sqrt{\xi }}-2l-3\right){d}_{2l-1}\right]\end{eqnarray}$
is a constant with the coefficient dν given by the following recurrence relation:
$\begin{eqnarray}\begin{array}{l}{d}_{-1}=0,\qquad {d}_{0}=1,\\ \left(\nu +2\right)\left(\nu +1-2l\right){d}_{\nu +2}+\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}}{d}_{\nu +1}\\ +\left(\displaystyle \frac{-E}{\sqrt{\xi }}-2v-1+2l\right){d}_{\nu }=0,\end{array}\end{eqnarray}$
and N(α, β, γ, δ, z) is the biconfluent Heun function [9, 20, 21].

The biconfluent Heun function ${\rm{N}}\left(2l+1,0,\tfrac{-E}{\sqrt{\xi }},\tfrac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)$ has an expansion at z = 0 [20]:

$\begin{eqnarray}{\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)=\sum _{n\geqslant 0}\displaystyle \frac{{A}_{n}}{{\left(2l+2\right)}_{n}}\displaystyle \frac{{z}^{n}}{n!},\end{eqnarray}$
where the expansion coefficients is determined by the recurrence relation
$\begin{eqnarray}\begin{array}{l}{A}_{0}=1,\qquad {A}_{1}=-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}},\\ {A}_{n+2}=-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}}{A}_{n+1}-\left(n+1\right)\\ \times \left(n+2l+2\right)\left(\displaystyle \frac{-E}{\sqrt{\xi }}-3-2l-2n\right){A}_{n},\end{array}\end{eqnarray}$
and ${\left(a\right)}_{n}={\rm{\Gamma }}\left(a+n\right)/{\rm{\Gamma }}\left(a\right)$ is Pochhammer's symbol.

Only ${y}_{l}^{\left(1\right)}\left(z\right)$ satisfies the boundary condition for the regular solution at r = 0, so the radial eigenfunction reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}\right){z}^{l+1}{y}_{l}^{\left(1\right)}\left(z\right)\\ =\,{A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}\right){z}^{l+1}\\ \times \,{\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right).\end{array}\end{eqnarray}$
By equation (A3), we obtain the regular solution
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}\exp \left(\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}\right){r}^{l+1}\\ \times {\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},{\rm{i}}{\xi }^{1/4}r\right).\end{array}\end{eqnarray}$

A.2. The irregular solution

The irregular solution is a solution satisfying the boundary condition at r → ∞ [9].

The biconfluent Heun equation (A5) has two linearly independent irregular solutions [20]:

$\begin{eqnarray}\begin{array}{l}{{\rm{B}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\\ =\,{{\rm{e}}}^{{z}^{2}}{{\rm{B}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{E}{\sqrt{\xi }},\displaystyle \frac{2\mu }{{\xi }^{1/4}},-{\rm{i}}z\right)\\ =\,{{\rm{e}}}^{{z}^{2}}{\left(-{\rm{i}}z\right)}^{\tfrac{1}{2}\left(\tfrac{E}{\sqrt{\xi }}-2l-3\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left(-{\rm{i}}z\right)}^{n}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{H}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{-2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\\ =\,{{\rm{e}}}^{{z}^{2}}{{\rm{H}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{E}{\sqrt{\xi }},\displaystyle \frac{2\mu }{{\xi }^{1/4}},-{\rm{i}}z\right)\\ =\,{\left(-{\rm{i}}z\right)}^{-\tfrac{1}{2}\left(\tfrac{E}{\sqrt{\xi }}+2l+3\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-{\rm{i}}z\right)}^{n}}\end{array}\end{eqnarray}$
with the expansion coefficients given by the recurrence relation
$\begin{eqnarray}{a}_{0}=1,\qquad {a}_{1}=\displaystyle \frac{\mu }{2{\xi }^{1/4}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){a}_{n+2}-\displaystyle \frac{\mu }{{\xi }^{1/4}}{a}_{n+1}\\ +\left[\displaystyle \frac{{E}^{2}}{4\xi }-\displaystyle \frac{{\left(2l+1\right)}^{2}}{4}+1-\displaystyle \frac{E}{\sqrt{\xi }}\right.\\ \left.+n\left(n+2-\displaystyle \frac{E}{\sqrt{\xi }}\right)\right]{a}_{n}=0\end{array}\end{eqnarray}$
and
$\begin{eqnarray}{e}_{0}=1,\qquad {e}_{1}=-\displaystyle \frac{\mu }{2{\xi }^{1/4}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){e}_{n+2}+\displaystyle \frac{\mu }{{\xi }^{1/4}}{e}_{n+1}\\ -\left[\displaystyle \frac{{E}^{2}}{4\xi }-\displaystyle \frac{{\left(2l+1\right)}^{2}}{4}+1+\displaystyle \frac{E}{\sqrt{\xi }}\right.\\ \left.+n\left(n+2+\displaystyle \frac{E}{\sqrt{\xi }}\right)\right]{e}_{n}=0.\end{array}\end{eqnarray}$

A.3. Eigenfunctions and eigenvalues

To construct the solution, we first express the regular solution (A13) as a linear combination of the two irregular solutions (A14) and (A15).

The regular solution (A13), with the relation [9, 20]

$\begin{eqnarray}\begin{array}{c}{\rm{N}}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\\ ={{\rm{K}}}_{1}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ \times {{\rm{B}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\\ +{{\rm{K}}}_{2}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ \times {{\rm{H}}}_{l}^{+}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}},z\right)\end{array}\end{eqnarray}$
and the expansions (A14) and (A15), becomes
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}\right)\exp \left(\left(\displaystyle \frac{E}{2\sqrt{\xi }}-\displaystyle \frac{1}{2}\right)\right.\\ \times \left.\mathrm{ln}\left({\xi }^{1/4}r\right)\right)\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}\\ +{A}_{l}{{\rm{K}}}_{2}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ \times \exp \left(\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}\right)\exp \left(\left(-\displaystyle \frac{E}{2\sqrt{\xi }}-\displaystyle \frac{1}{2}\right)\right.\\ \times \left.\mathrm{ln}\left({\xi }^{1/4}r\right)\right)\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}},\end{array}\end{eqnarray}$
where ${{\rm{K}}}_{1}\left(2l+1,0,\tfrac{-E}{\sqrt{\xi }},-\tfrac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)$ and ${{\rm{K}}}_{2}\left(2l+1,0,\tfrac{-E}{\sqrt{\xi }},-\tfrac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)$ are combination coefficients and z = iξ1/4r.

The boundary condition of bound states, ${\left.u\left(r\right)\right|}_{r\to \infty }\to 0$, requires that the coefficient of the second term must vanish since this term diverges when r → ∞ , i.e.,

$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)=0,\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ =\displaystyle \frac{{\rm{\Gamma }}\left(2l+2\right)}{{\rm{\Gamma }}\left(l+\tfrac{1}{2}+\tfrac{E}{2\sqrt{\xi }}\right){\rm{\Gamma }}\left(l+\tfrac{3}{2}-\tfrac{E}{2\sqrt{\xi }}\right)}\\ \times {{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2\sqrt{\xi }}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2\sqrt{\xi }},0,3l+\displaystyle \frac{3}{2}+\displaystyle \frac{E}{2\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\end{array}\end{eqnarray}$
with
$\begin{eqnarray}\begin{array}{l}{{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2\sqrt{\xi }}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2\sqrt{\xi }},0,3l\right.\\ \left.+\displaystyle \frac{3}{2}+\displaystyle \frac{E}{2\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ ={\displaystyle \int }_{0}^{\infty }{x}^{\tfrac{1}{2}\left(2l+1-\tfrac{E}{\sqrt{\xi }}\right)}{{\rm{e}}}^{-{x}^{2}}{\rm{N}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2\sqrt{\xi }},0,3l\right.\\ \left.+\displaystyle \frac{3}{2}+\displaystyle \frac{E}{2\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}},x\right){\rm{d}}x.\end{array}\end{eqnarray}$

Equation (A22) is an implicit expression of the eigenvalue.

The eigenfunction, from equations (A21) and (A22), reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(2l+1,0,\displaystyle \frac{-E}{\sqrt{\xi }},-\displaystyle \frac{2{\rm{i}}\mu }{{\xi }^{1/4}}\right)\\ \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}\right)\exp \left(\left(\displaystyle \frac{E}{2\sqrt{\xi }}-\displaystyle \frac{1}{2}\right)\right.\\ \times \left.\mathrm{ln}\left({\xi }^{1/4}r\right)\right)\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}.\end{array}\end{eqnarray}$

In this appendix, we provide an exact solution of the eigenproblem of the potential

$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\mu r\end{eqnarray}$
by solving the radial equation directly. This potential has only bound states.

The radial equation reads

$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}{u}_{l}\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\xi {r}^{2}-\mu r\right]{u}_{l}\left(r\right)=0.\end{eqnarray}$
Using the variable substitution
$\begin{eqnarray}z={\rm{i}}{\xi }^{1/4}r\end{eqnarray}$
and introducing ${f}_{l}\left(z\right)$ by
$\begin{eqnarray}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}{f}_{l}\left(z\right)\end{eqnarray}$
with Al a constant, we convert the radial equation (A27) into an equation of ${f}_{l}\left(z\right)$:
$\begin{eqnarray}\begin{array}{l}{{zf}}_{l}^{{\prime\prime} }\left(z\right)+\left(2l+2-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}z-2{z}^{2}\right){f}_{l}^{{\prime} }\left(z\right)\\ +\left[\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2l-3\right)z\right.\\ \left.-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}\left(l+1\right)\right]{f}_{l}\left(z\right)=0.\end{array}\end{eqnarray}$
This is a biconfluent Heun equation [20].

The choice of the boundary condition has been discussed in [9].

B.1. The regular solution

The regular solution is a solution satisfying the boundary condition at r = 0 [9]. The regular solution at r = 0 should satisfy the boundary condition ${\mathrm{lim}}_{r\to 0}{u}_{l}\left(r\right)/{r}^{l\,+\,1}=1$. In this section, we provide the regular solution of equation (A30).

The biconfluent Heun equation (A30) has two linearly independent solutions [20]:

$\begin{eqnarray}\begin{array}{l}{y}_{l}^{\left(1\right)}\left(z\right)={\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right),\\ {y}_{l}^{\left(2\right)}\left(z\right)=c{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ \times \mathrm{ln}z+\displaystyle \sum _{n\geqslant 0}{d}_{n}{z}^{n-2l-1},\end{array}\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{rcl}c & = & \displaystyle \frac{1}{2l+1}\left[\displaystyle \frac{{\rm{i}}l\mu }{{\xi }^{3/4}}{d}_{2l}-{d}_{2l-1}\right.\\ & & \times \left.\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}+1-2l\right)\right]\end{array}\end{eqnarray}$
is a constant with the coefficient dν given by the following recurrence relation:
$\begin{eqnarray}\begin{array}{l}{d}_{-1}=0,\qquad {d}_{0}=1,\\ \left(v+2\right)\left(v+1-2l\right){d}_{v+2}-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}\left(v+1-l\right){d}_{v+1}\\ +\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2v-1+2l\right){d}_{v}=0\end{array}\end{eqnarray}$
and N(α, β, γ, δ, z) is the biconfluent Heun function [9, 20, 21].

The biconfluent Heun function ${\rm{N}}\left(2l+1,\tfrac{{\rm{i}}\mu }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)$ has an expansion at z = 0 [20]:

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ =\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{A}_{n}}{{\left(2l+2\right)}_{n}}\displaystyle \frac{{z}^{n}}{n!},\end{array}\end{eqnarray}$
where the expansion coefficients are determined by the recurrence relation
$\begin{eqnarray}\begin{array}{l}{A}_{0}=1,\qquad {A}_{1}=\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}\left(l+1\right),\\ {A}_{n+2}=\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}\left(n+l+2\right){A}_{n+1}-\left(n+1\right)\left(n+2l+2\right)\\ \times \left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-3-2l-2n\right){A}_{n},\end{array}\end{eqnarray}$
and ${\left(a\right)}_{n}={\rm{\Gamma }}\left(a+n\right)/{\rm{\Gamma }}\left(a\right)$ is Pochhammer's symbol.

Only ${y}_{l}^{\left(1\right)}\left(z\right)$ satisfies the boundary condition for the regular solution at r = 0, so the radial eigenfunction reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}{y}_{l}^{\left(1\right)}\left(z\right)\\ =\,{A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}\\ \times \,{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right).\end{array}\end{eqnarray}$
From equation (A28), we obtain the regular solution
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}\exp \left(\displaystyle \frac{{\xi }^{1/2}}{2}{r}^{2}+\displaystyle \frac{\mu }{2{\xi }^{1/2}}r\right){r}^{l+1}\\ \times \,{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,{\rm{i}}{\xi }^{1/4}r\right).\end{array}\end{eqnarray}$

B.2. The irregular solution

The irregular solution is a solution satisfying the boundary condition at r → ∞ [9].

The biconfluent Heun equation (A30) has two linearly independent irregular solutions [20]:

$\begin{eqnarray}\begin{array}{l}{{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ =\exp \left(\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}z+{z}^{2}\right){{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{\mu }{{\xi }^{3/4}},\displaystyle \frac{E}{{\xi }^{1/2}}\right.\\ \left.+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,-{\rm{i}}z\right)\\ =\exp \left(\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}z+{z}^{2}\right){\left(-{\rm{i}}z\right)}^{\tfrac{1}{2}\left(\tfrac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2l-3\right)}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left(-{\rm{i}}z\right)}^{n}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}}z+{z}^{2}\right){{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{\mu }{{\xi }^{3/4}},\displaystyle \frac{E}{{\xi }^{1/2}}\right.\\ \left.+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,-{\rm{i}}z\right)\\ =\,{\left(-{\rm{i}}z\right)}^{-\tfrac{1}{2}\left(\tfrac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}+2l+3\right)}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-{\rm{i}}z\right)}^{n}}\end{array}\end{eqnarray}$
with the expansion coefficients given by the recurrence relation
$\begin{eqnarray}{a}_{0}=1,\qquad {a}_{1}=\displaystyle \frac{\mu }{4{\xi }^{3/4}}\left(\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-1\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){a}_{n+2}+\displaystyle \frac{\mu }{{\xi }^{3/4}}\left(\displaystyle \frac{3}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}+n\right){a}_{n+1}\\ +\left[\displaystyle \frac{1}{4}{\left(\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}\right)}^{2}-\displaystyle \frac{1}{4}{\left(2l+1\right)}^{2}+1\right.\\ \left.-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}+n\left(n+2-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}\right)\right]{a}_{n}=0\end{array}\end{eqnarray}$
and
$\begin{eqnarray}{e}_{0}=1,\qquad {e}_{1}=-\displaystyle \frac{\mu }{4{\xi }^{3/4}}\left(\gamma +1\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){e}_{n+2}+\displaystyle \frac{\mu }{{\xi }^{3/4}}\left(\displaystyle \frac{3}{2}+\displaystyle \frac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}+n\right){e}_{n+1}\\ -\left[\displaystyle \frac{1}{4}{\left(\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}\right)}^{2}-\displaystyle \frac{1}{4}{\left(2l+1\right)}^{2}+1+\displaystyle \frac{E}{{\xi }^{1/2}}\right.\\ \left.+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}+n\left(n+2+\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}\right)\right]{e}_{n}=0.\end{array}\end{eqnarray}$

B.3. Eigenfunctions and eigenvalues

To construct the solution, we first express the regular solution (A37) as a linear combination of the two irregular solutions (A38) and (A39).

The regular solution (A37), with the relation [9, 20]

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ ={{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ \times {{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\\ +{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ \times {{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0,z\right)\end{array}\end{eqnarray}$
and the expansions (A38) and (A39), becomes
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}}{2}{r}^{2}-\displaystyle \frac{\mu }{2{\xi }^{1/2}}r\right){r}^{\tfrac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}\\ +{A}_{l}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ \times \exp \left(\displaystyle \frac{{\xi }^{1/2}}{2}{r}^{2}+\displaystyle \frac{\mu }{2{\xi }^{1/2}}r\right){r}^{-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}},\end{array}\end{eqnarray}$
where ${{\rm{K}}}_{1}\left(2l+1,\tfrac{{\rm{i}}\mu }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)$ and ${{\rm{K}}}_{2}\left(2l+1,\tfrac{{\rm{i}}\mu }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)$ are combination coefficients and z = iξ1/4r.

The boundary condition of bound states, ${\left.u\left(r\right)\right|}_{r\to \infty }\to 0$, requires that the coefficient of the second term must vanish since this term diverges when r → ∞ , i.e.,

$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)=0,\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ =\displaystyle \frac{{\rm{\Gamma }}\left(2l+2\right)}{{\rm{\Gamma }}\left(l+\tfrac{1}{2}+\tfrac{E}{2{\xi }^{1/2}}+\tfrac{{\mu }^{2}}{8{\xi }^{3/2}}\right){\rm{\Gamma }}\left(l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\tfrac{{\mu }^{2}}{8{\xi }^{3/2}}\right)}\\ \times {{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}\right.\\ -\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)+\displaystyle \frac{E}{2{\xi }^{1/2}}\\ \left.+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{2{\xi }^{3/4}}\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2l-1\right)\right)\end{array}\end{eqnarray}$
with
$\begin{eqnarray}\begin{array}{l}{{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)\right.\\ \left.+\displaystyle \frac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{2{\xi }^{3/4}}\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2l-1\right)\right)\\ ={\displaystyle \int }_{0}^{\infty }{\rm{d}}{{xx}}^{l+\tfrac{1}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}}{{\rm{e}}}^{-{x}^{2}}\\ \times {\rm{N}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)+\displaystyle \frac{E}{2{\xi }^{1/2}}\right.\\ \left.+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\mu }{2{\xi }^{3/4}}\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}}-2l-1\right),x\right).\end{array}\end{eqnarray}$

Equation (A46) is an implicit expression of the eigenvalue.

The eigenfunction, from equations (A45) and (A46), reads

$\begin{eqnarray}\begin{array}{rcl}{u}_{l}\left(r\right) & = & {A}_{l}{{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\mu }^{2}}{4{\xi }^{3/2}},0\right)\\ & & \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}}{2}{r}^{2}-\displaystyle \frac{\mu }{2{\xi }^{1/2}}r\right){r}^{\tfrac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\mu }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\\ & & \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}.\end{array}\end{eqnarray}$

In this appendix, we provide an exact solution of the eigenproblem of the potential

$\begin{eqnarray}U\left(r\right)=\displaystyle \frac{\xi }{\sqrt{r}}+\displaystyle \frac{\mu }{{r}^{3/2}}\end{eqnarray}$
by solving the radial equation directly. This potential has both bound states and scattering states.

The radial equation reads

$\begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}{u}_{l}\left(r\right)}{{\rm{d}}{r}^{2}}+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}-\displaystyle \frac{\xi }{\sqrt{r}}-\displaystyle \frac{\mu }{{r}^{3/2}}\right]{u}_{l}\left(r\right)=0.\end{eqnarray}$
Using the variable substitution
$\begin{eqnarray}z={\rm{i}}\sqrt{2r}{\left(-E\right)}^{1/4}\end{eqnarray}$
and introducing ${f}_{l}\left(z\right)$ by
$\begin{eqnarray}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{2\left(l+1\right)}{f}_{l}\left(z\right)\end{eqnarray}$
with Al a constant, we convert the radial equation (A51) into an equation of ${f}_{l}\left(z\right)$:
$\begin{eqnarray}\begin{array}{l}{{zf}}_{l}^{{\prime\prime} }\left(z\right)+\left(4l+3-\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}z-2{z}^{2}\right){f}_{l}^{{\prime} }\left(z\right)\\ +\left[\left(-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-4l-4\right)z\right.\\ \left.-\displaystyle \frac{1}{2}\left(\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}+\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(4l+3\right)\right)\right]\\ \times {f}_{l}\left(z\right)=0.\end{array}\end{eqnarray}$
This is a biconfluent Heun equation [20].

The choice of the boundary condition has been discussed in [9].

C.1. The regular solution

The regular solution is a solution satisfying the boundary condition at r = 0 [9]. The regular solution at r = 0 should satisfy the boundary condition ${\mathrm{lim}}_{r\to 0}{u}_{l}\left(r\right)/{r}^{l\,+\,1}=1$ for both bound states and scattering states. In this section, we provide the regular solution of equation (A54).

The biconfluent Heun equation (A54) has two linearly independent solutions [20]:

$\begin{eqnarray}\begin{array}{l}{y}_{l}^{\left(1\right)}\left(z\right)={\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right),\\ {y}_{l}^{\left(2\right)}\left(z\right)=c{\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ \times \mathrm{ln}z+\displaystyle \sum _{n\geqslant 0}{d}_{n}{z}^{n-4l-2},\end{array}\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{rcl}c & = & \displaystyle \frac{1}{4l+2}\left[{d}_{4l+1}\displaystyle \frac{1}{2}\left(\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}+\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(4l+1\right)\right)\right.\\ & & \left.-{d}_{4l}\left(-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-4l\right)\right]\end{array}\end{eqnarray}$
is a constant with the coefficient dν given by the following recurrence relation:
$\begin{eqnarray}\begin{array}{l}{d}_{-1}=0,\qquad {d}_{0}=1,\\ \left(v+2\right)\left(v-4l\right){d}_{v+2}-\displaystyle \frac{1}{2}\left[\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right.\\ \left.+\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(2v+1-4l\right)\right]{d}_{v+1}\\ +\left[-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-2v+4l\right]{d}_{v}=0\end{array}\end{eqnarray}$
and N(α, β, γ, δ, z) is the biconfluent Heun function [9, 20, 21].

The biconfluent Heun function ${\rm{N}}\left(4l+2,\tfrac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\tfrac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)$ has an expansion at z = 0 [20]:

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ =\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{A}_{n}}{{\left(4l+3\right)}_{n}}\displaystyle \frac{{z}^{n}}{n!},\end{array}\end{eqnarray}$
where the expansion coefficients are determined by the recurrence relation
$\begin{eqnarray}\begin{array}{l}{A}_{0}=1,\qquad {A}_{1}=\displaystyle \frac{1}{2}\left[\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}+\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(4l+3\right)\right],\\ {A}_{n+2}=\left\{\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(n+1\right)\right.\\ \left.+\displaystyle \frac{1}{2}\left[\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}+\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(4l+3\right)\right]\right\}{A}_{n+1}-\left(n+1\right)\\ \times \left(n+4l+3\right)\left[-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-4-4l-2n\right]{A}_{n},\end{array}\end{eqnarray}$
and ${\left(a\right)}_{n}={\rm{\Gamma }}\left(a+n\right)/{\rm{\Gamma }}\left(a\right)$ is Pochhammer's symbol.

Only ${y}_{l}^{\left(1\right)}\left(z\right)$ satisfies the boundary condition for the regular solution at r = 0, so the radial eigenfunction reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(z\right)={A}_{l}\left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{2\left(l+1\right)}{y}_{l}^{\left(1\right)}\left(z\right)\\ =\,{A}_{l}\left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{2\left(l+1\right)}\\ \times \,{\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right).\end{array}\end{eqnarray}$
From equation (A3), we obtain the regular solution
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}\left[{\left(-E\right)}^{1/2}r+\displaystyle \frac{\xi }{{\left(-E\right)}^{1/2}}\sqrt{r}\right]{r}^{l+1}\\ \times \,{\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\right.\\ \left.\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},{\rm{i}}\sqrt{2r}{\left(-E\right)}^{1/4}\right).\end{array}\end{eqnarray}$

C.2. Irregular solution

The irregular solution is a solution satisfying the boundary condition at r → ∞ [9]. The boundary conditions for bound states and scattering states at r → ∞ are different.

The biconfluent Heun equation (A54) has two linearly independent irregular solutions [20]:

$\begin{eqnarray}\begin{array}{l}{{\rm{B}}}_{l}^{+}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}z+{z}^{2}\right){{\rm{B}}}_{l}^{+}\left(4l+2,\displaystyle \frac{\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},\right.\\ \left.\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},-{\rm{i}}z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}z+{z}^{2}\right){\left(-{\rm{i}}z\right)}^{\tfrac{1}{2}\left(\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-4l-4\right)}\\ \displaystyle \mathop{\times \,\sum }\limits_{\,n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left(-{\rm{i}}z\right)}^{n}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{H}}}_{l}^{+}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ =\exp \left(\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}z+{z}^{2}\right)\\ \times {{\rm{H}}}_{l}^{+}\left(4l+2,\displaystyle \frac{\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},-{\rm{i}}z\right)\\ ={\left(-{\rm{i}}z\right)}^{-\tfrac{1}{2}\left(\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+4l+4\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-{\rm{i}}z\right)}^{n}}\end{array}\end{eqnarray}$
with the expansion coefficients given by the recurrence relation
$\begin{eqnarray}\begin{array}{l}{a}_{0}=1,\qquad {a}_{1}=\displaystyle \frac{1}{4}\left[\displaystyle \frac{4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right.\\ \left.+\displaystyle \frac{\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}-1\right)\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){a}_{n+2}+\left[\displaystyle \frac{\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(\displaystyle \frac{3}{2}\right.\right.\\ \left.\left.-\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}+n\right)-\displaystyle \frac{2\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right]{a}_{n+1}\\ +\left[\displaystyle \frac{{\xi }^{4}}{16{\left(-E\right)}^{3}}-\displaystyle \frac{{\left(4l+2\right)}^{2}}{4}+1-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}\right.\\ \left.+n\left(n+2-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}\right)\right]{a}_{n}=0\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}{e}_{0}=1,\qquad {e}_{1}=-\displaystyle \frac{1}{4}\left[\displaystyle \frac{4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right.\\ \left.+\displaystyle \frac{\sqrt{2}\xi }{{\left(-E\right)}^{3/4}}\left(\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+1\right)\right],\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){e}_{n+2}+\left[\beta \left(\displaystyle \frac{3}{2}+\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}+n\right)+\displaystyle \frac{\delta }{2}\right]{e}_{n+1}\\ -\left[\displaystyle \frac{{\xi }^{4}}{16{\left(-E\right)}^{3}}-\displaystyle \frac{{\left(4l+2\right)}^{2}}{4}+1\right.\\ \left.+\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+n\left(n+2+\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}\right)\right]{e}_{n}=0.\end{array}\end{eqnarray}$

C.3. Bound states and scattering states C.3.1. The bound state

To construct the solution, we first express the regular solution (A61) as a linear combination of the two irregular solutions (A62) and (A63).

The regular solution (A61), with the relation [9, 20]

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ ={{\rm{K}}}_{1}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ \times {{\rm{B}}}_{l}^{+}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\\ +{{\rm{K}}}_{2}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ \times {{\rm{H}}}_{l}^{+}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}},z\right)\end{array}\end{eqnarray}$
and the expansions (A62) and (A63), become
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ \times \exp \left(-{\left(-E\right)}^{1/2}r-\displaystyle \frac{\xi }{{\left(-E\right)}^{1/2}}\sqrt{r}\right){\left[{\rm{i}}{\left(-E\right)}^{1/4}\sqrt{2r}\right]}^{{\xi }^{2}/\left[4{\left(-E\right)}^{3/2}\right]}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left[\sqrt{2r}{\left(-E\right)}^{1/4}\right]}^{n}}\\ +{A}_{l}{{\rm{K}}}_{2}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ \times \exp \left({\left(-E\right)}^{1/2}r+\displaystyle \frac{\xi }{{\left(-E\right)}^{1/2}}\sqrt{r}\right){\left[{\rm{i}}{\left(-E\right)}^{1/4}\sqrt{2r}\right]}^{-{\xi }^{2}/\left[4{\left(-E\right)}^{3/2}\right]}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left[\sqrt{2r}{\left(-E\right)}^{1/4}\right]}^{n}},\end{array}\end{eqnarray}$
where ${{\rm{K}}}_{1}\left(4l+2,\tfrac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\tfrac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)$ and ${{\rm{K}}}_{2}\left(4l+2,\tfrac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\tfrac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\tfrac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)$ are combination coefficients and $z={\rm{i}}\sqrt{2r}{\left(-E\right)}^{1/4}$.

The boundary condition of bound states, ${\left.u\left(r\right)\right|}_{r\to \infty }\to 0$, requires that the coefficient of the second term must vanish since this term diverges when r → ∞ , i.e.,

$\begin{eqnarray}{{\rm{K}}}_{2}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)=0,\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ =\displaystyle \frac{{\rm{\Gamma }}\left(4l+3\right)}{{\rm{\Gamma }}\left(2l+1+\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}\right){\rm{\Gamma }}\left(2l+2-\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}\right)}\\ \times {{\rm{J}}}_{2l+2-\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}}\left(2l+1-\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},6l+3\right.\\ +\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\\ \left.-\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{2{\left(-E\right)}^{3/4}}\left(\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+4l+2\right)\right)\end{array}\end{eqnarray}$
with
$\begin{eqnarray}\begin{array}{l}{{\rm{J}}}_{2l+2-\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}}\left(2l+1-\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},6l+3\right.\\ +\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\\ \left.-\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{2{\left(-E\right)}^{3/4}}\left(\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+4l+2\right)\right)\\ ={\displaystyle \int }_{0}^{\infty }{\rm{d}}{{xx}}^{2l+1-\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}}{{\rm{e}}}^{-{x}^{2}}\\ \times {\rm{N}}\left(2l+1-\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},6l+3\right.\\ +\displaystyle \frac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}-\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{2{\left(-E\right)}^{3/4}}\\ \left.\left(\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}}+4l+2\right),x\right).\end{array}\end{eqnarray}$

Equation (A70) is an implicit expression of the eigenvalue.

The eigenfunction, from equations (A69) and (A70), reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(4l+2,\displaystyle \frac{{\rm{i}}\sqrt{2}\xi }{{\left(-E\right)}^{3/4}},-\displaystyle \frac{{\xi }^{2}}{2{\left(-E\right)}^{3/2}},\displaystyle \frac{-{\rm{i}}4\sqrt{2}\mu }{{\left(-E\right)}^{1/4}}\right)\\ \times \exp \left(-{\left(-E\right)}^{1/2}r-\displaystyle \frac{\xi }{{\left(-E\right)}^{1/2}}\sqrt{r}\right)\\ \times {\left[{\rm{i}}{\left(-E\right)}^{1/4}\sqrt{2r}\right]}^{\tfrac{{\xi }^{2}}{4{\left(-E\right)}^{3/2}}}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left[\sqrt{2r}{\left(-E\right)}^{1/4}\right]}^{n}}.\end{array}\end{eqnarray}$

C.3.2. The scattering state

For scattering states, E > 0, we introduce

$\begin{eqnarray}E={k}^{2}.\end{eqnarray}$

The singularity of the S-matrix on the positive imaginary axis corresponds to the eigenvalues of bound states [44], so the zero of ${{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)$ on the positive imaginary is just the singularity of the S-matrix. Considering that the S-matrix is unitary, i.e.,

$\begin{eqnarray}{S}_{l}={{\rm{e}}}^{2{\rm{i}}{\delta }_{l}},\end{eqnarray}$
we have
$\begin{eqnarray}\begin{array}{l}{S}_{l}\left(k\right)=\displaystyle \frac{{{\rm{K}}}_{2}^{* }\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)}{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)}\\ =\displaystyle \frac{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1-{\rm{i}}\right)\xi }{{k}^{3/2}},-\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{2{\rm{i}}}\mu }{\sqrt{k}}\right)}{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)}.\end{array}\end{eqnarray}$

The scattering wave function can be constructed with the help of the S-matrix. The scattering wave function can be expressed as a linear combination of the radially ingoing wave ${u}_{\mathrm{in}}\left(r\right)$ and the radially outgoing wave ${u}_{\mathrm{out}}\left(r\right)$, which are conjugate to each other, i.e., [44]

$\begin{eqnarray}{u}_{l}\left(r\right)={A}_{l}\left[{\left(-1\right)}^{l+1}{u}_{\mathrm{in}}\left(r\right)+{S}_{l}\left(k\right){u}_{\mathrm{out}}\left(r\right)\right].\end{eqnarray}$
From equation (A69), we have
$\begin{eqnarray}\begin{array}{l}{u}_{\mathrm{in}}\left(r\right)=\exp \left(-{\rm{i}}{kr}+{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right)\\ \times {\left(2{\rm{i}}{kr}\right)}^{-{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-2{\rm{i}}{kr}\right)}^{n/2}},\\ {u}_{\mathrm{out}}\left(r\right)=\exp \left({\rm{i}}{kr}-{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right)\\ \times {\left(-2{\rm{i}}{kr}\right)}^{{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}^{* }}{{\left(2{\rm{i}}{kr}\right)}^{n/2}}.\end{array}\end{eqnarray}$
Then, from equation (A76), we obtain the scattering wave function,
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}\left[{\left(-1\right)}^{l+1}\exp \left(-{\rm{i}}{kr}+{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right)\right.\\ \times {\left(2{\rm{i}}{kr}\right)}^{-{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-2{\rm{i}}{kr}\right)}^{n/2}}\\ +\displaystyle \frac{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1-{\rm{i}}\right)\xi }{{k}^{3/2}},-\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{2{\rm{i}}}\mu }{\sqrt{k}}\right)}{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)}\\ \times \exp \left({\rm{i}}{kr}-{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right){\left(-2{\rm{i}}{kr}\right)}^{{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\\ \times \left.\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}^{* }}{{\left(2{\rm{i}}{kr}\right)}^{n/2}}\right].\end{array}\end{eqnarray}$
Taking r → ∞ , we have
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)\mathop{\sim }\limits^{r\to \infty }{A}_{l}\left[{\left(-1\right)}^{l+1}\exp \left(-{\rm{i}}{kr}+{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right){\left(2{kr}\right)}^{-{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\right.\\ +\displaystyle \frac{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1-{\rm{i}}\right)\xi }{{k}^{3/2}},-\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{2{\rm{i}}}\mu }{\sqrt{k}}\right)}{{{\rm{K}}}_{2}\left(4l+2,-\tfrac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\tfrac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\tfrac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right)}\\ \times \left.\exp \left({\rm{i}}{kr}-{\rm{i}}\displaystyle \frac{\xi }{k}\sqrt{r}\right){\left(2{kr}\right)}^{{\rm{i}}{\xi }^{2}/\left(8{k}^{3}\right)}\right]\\ ={A}_{l}{{\rm{e}}}^{{\rm{i}}{\delta }_{l}}\sin \left({kr}-\displaystyle \frac{\xi }{k}\sqrt{r}+\displaystyle \frac{{\xi }^{2}}{8{k}^{3}}\mathrm{ln}2{kr}+{\delta }_{l}-\displaystyle \frac{l\pi }{2}\right).\end{array}\end{eqnarray}$
From equations (A75) and (A76), we obtain the scattering phase shift
$\begin{eqnarray}{\delta }_{l}=-{\mathrm{argK}}_{2}\left(4l+2,-\displaystyle \frac{\left(1+{\rm{i}}\right)\xi }{{k}^{3/2}},\displaystyle \frac{{\rm{i}}{\xi }^{2}}{2{k}^{3}},\displaystyle \frac{4\sqrt{-2{\rm{i}}}\mu }{\sqrt{k}}\right).\end{eqnarray}$

In this appendix, we provide an exact solution of the eigenproblem of the potential

$\begin{eqnarray}U\left(r\right)=\xi {r}^{2}+\displaystyle \frac{\mu }{r}+\kappa r\end{eqnarray}$
by solving the radial equation directly. This potential has only bound states.

The radial equation reads

$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{{\rm{d}}}^{2}}{{\rm{d}}{r}^{2}}{u}_{l}\left(r\right)+\left[E-\displaystyle \frac{l\left(l+1\right)}{{r}^{2}}\right.\\ \left.-\xi {r}^{2}-\displaystyle \frac{\mu }{r}-\kappa r\right]{u}_{l}\left(r\right)=0.\end{array}\end{eqnarray}$
Using the variable substitution
$\begin{eqnarray}z={\rm{i}}{\xi }^{1/4}r\end{eqnarray}$
and introducing ${f}_{l}\left(z\right)$ by
$\begin{eqnarray}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}{f}_{l}\left(z\right)\end{eqnarray}$
with Al a constant, we convert the radial equation (A83) into an equation of ${f}_{l}\left(z\right)$:
$\begin{eqnarray}\begin{array}{l}{{zf}}_{l}^{{\prime\prime} }\left(z\right)+\left(2l+2-\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}z-2{z}^{2}\right){f}_{l}^{{\prime} }\left(z\right)\\ +\left[\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-3\right)z\right.\\ \left.+\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}}-\left(l+1\right)\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}\right]{f}_{l}\left(z\right)=0.\end{array}\end{eqnarray}$
This is a biconfluent Heun equation [20].

The choice of the boundary condition has been discussed in [9].

D.1. The regular solution

The regular solution is a solution satisfying the boundary condition at r = 0 [9]. The regular solution at r = 0 should satisfy the boundary condition ${\mathrm{lim}}_{r\to 0}{u}_{l}\left(r\right)/{r}^{l\,+\,1}=1$. In this section, we provide the regular solution of equation (A86).

The biconfluent Heun equation (A86) has two linearly independent solutions [20]:

$\begin{eqnarray}\begin{array}{l}{y}_{l}^{\left(1\right)}\left(z\right)={\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right),\\ {y}_{l}^{\left(2\right)}\left(z\right)=c{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ \times \mathrm{ln}z+\displaystyle \sum _{n\geqslant 0}{d}_{n}{z}^{n-2l-1},\end{array}\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}c=\displaystyle \frac{1}{2l+1}\left[{d}_{2l}\left(-\displaystyle \frac{{\rm{i}}\mu }{{\xi }^{1/4}}+\displaystyle \frac{{\rm{i}}l\kappa }{{\xi }^{3/4}}\right)\right.\\ \left.-{d}_{2l-1}\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}+1-2l\right)\right]\end{array}\end{eqnarray}$
is a constant with the coefficient dν given by the following recurrence relation:
$\begin{eqnarray}\begin{array}{l}{d}_{-1}=0,\qquad {d}_{0}=1,\\ \left(v+2\right)\left(v+1-2l\right){d}_{v+2}-\left(\displaystyle \frac{-{\rm{i}}\mu }{{\xi }^{1/4}}+\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}\left(v+1-l\right)\right){d}_{v+1}\\ +\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2v-1+2l\right){d}_{v}=0\end{array}\end{eqnarray}$
and N(α, β, γ, δ, z) is the biconfluent Heun function [9, 20, 21].

The biconfluent Heun function ${\rm{N}}\left(2l+1,\tfrac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\kappa }^{2}}{4{\xi }^{3/2}},\tfrac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)$ has an expansion at z = 0 [20]:

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ =\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{A}_{n}}{{\left(2l+2\right)}_{n}}\displaystyle \frac{{z}^{n}}{n!},\end{array}\end{eqnarray}$
where the expansion coefficients are determined by the recurrence relation
$\begin{eqnarray}\begin{array}{l}{A}_{0}=1,\qquad {A}_{1}=\displaystyle \frac{-{\rm{i}}\mu }{{\xi }^{1/4}}+\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}\left(l+1\right),\\ {A}_{n+2}=\left[\left(n+l+2\right)\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}+\displaystyle \frac{-{\rm{i}}\mu }{{\xi }^{1/4}}\right]{A}_{n+1}\\ -\left(n+1\right)\left(n+2l+2\right)\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-3-2n\right){A}_{n},\end{array}\end{eqnarray}$
and ${\left(a\right)}_{n}={\rm{\Gamma }}\left(a+n\right)/{\rm{\Gamma }}\left(a\right)$ is Pochhammer's symbol.

Only ${y}_{l}^{\left(1\right)}\left(z\right)$ satisfies the boundary condition for the regular solution at r = 0, so the radial eigenfunction reads

$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(z\right)={A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}{y}_{l}^{\left(1\right)}\left(z\right)\\ =\,{A}_{l}\exp \left(-\displaystyle \frac{{z}^{2}}{2}-\displaystyle \frac{\beta }{2}z\right){z}^{l+1}\\ \times \,{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right).\end{array}\end{eqnarray}$
From equation (A84), we obtain the regular solution
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}\exp \left(\displaystyle \frac{{\xi }^{1/2}}{2}{r}^{2}+\displaystyle \frac{\kappa }{2{\xi }^{1/2}}r\right){r}^{l+1}\\ \times {\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},{\rm{i}}{\xi }^{1/4}r\right).\end{array}\end{eqnarray}$

D.2. The irregular solution

The irregular solution is a solution satisfying the boundary condition at r → ∞ [9].

The biconfluent Heun equation (A86) has two linearly independent irregular solutions [20]:

$\begin{eqnarray}\begin{array}{l}{{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}z+{z}^{2}\right)\\ \times \,{{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{\kappa }{{\xi }^{3/4}},\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{2\mu }{{\xi }^{1/4}},-{\rm{i}}z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}z+{z}^{2}\right){\left(-{\rm{i}}z\right)}^{\tfrac{1}{2}\left(\tfrac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-3\right)}\\ \displaystyle \times \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left(-{\rm{i}}z\right)}^{n}},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ =\,\exp \left(\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}}z+{z}^{2}\right){{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{\kappa }{{\xi }^{3/4}},\displaystyle \frac{E}{{\xi }^{1/2}}\right.\\ \left.+\,\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{2\mu }{{\xi }^{1/4}},-{\rm{i}}z\right)\\ =\,{\left(-{\rm{i}}z\right)}^{-\tfrac{1}{2}\left(\tfrac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}+2l+3\right)}\displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left(-{\rm{i}}z\right)}^{n}}\end{array}\end{eqnarray}$
with the expansion coefficients given by the recurrence relation
$\begin{eqnarray}{a}_{0}=1,\qquad {a}_{1}=\displaystyle \frac{\mu }{2{\xi }^{1/4}}+\displaystyle \frac{\kappa }{4{\xi }^{3/4}}\left(\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-1\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){a}_{n+2}+\left[\displaystyle \frac{\kappa }{{\xi }^{3/4}}\left(\displaystyle \frac{3}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}+n\right)\right.\\ \left.-\displaystyle \frac{\mu }{{\xi }^{1/4}}\right]{a}_{n+1}\\ +\left[{\left(\displaystyle \frac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}-1\right)}^{2}-\displaystyle \frac{{\left(2l+1\right)}^{2}}{4}\right.\\ \left.+n\left(n+2-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}\right)\right]{a}_{n}=0\end{array}\end{eqnarray}$
and
$\begin{eqnarray}\begin{array}{l}{e}_{0}=1,\qquad {e}_{1}=-\displaystyle \frac{\mu }{2{\xi }^{1/4}}-\displaystyle \frac{\kappa }{4{\xi }^{3/4}}\\ \times \left(\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}+1\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}2\left(n+2\right){e}_{n+2}+\left[\displaystyle \frac{\kappa }{{\xi }^{3/4}}\left(\displaystyle \frac{3}{2}+\displaystyle \frac{E}{2{\xi }^{1/2}}\right.\right.\\ \left.\left.+\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}+n\right)+\displaystyle \frac{\mu }{{\xi }^{1/4}}\right]{e}_{n+1}\\ -\left[{\left(\displaystyle \frac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}+1\right)}^{2}-\displaystyle \frac{{\left(2l+1\right)}^{2}}{4}\right.\\ \left.+n\left(n+2+\displaystyle \frac{E}{{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}\right)\right]{e}_{n}=0.\end{array}\end{eqnarray}$

D.3. Eigenfunctions and eigenvalues

To construct the solution, we first express the regular solution (A93) as a linear combination of the two irregular solutions (A94) and (A95).

The regular solution (A93), with the relation [9, 20]

$\begin{eqnarray}\begin{array}{l}{\rm{N}}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ =\,{{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ \times \,{{\rm{B}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\\ +\,{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ \times \,{{\rm{H}}}_{l}^{+}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}},z\right)\end{array}\end{eqnarray}$
and the expansions (A94) and (A95), becomes
$\begin{eqnarray}\begin{array}{l}{u}_{l}\left(r\right)={A}_{l}{{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}-\displaystyle \frac{\kappa }{2{\xi }^{1/2}}r\right){r}^{\tfrac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}\\ +{A}_{l}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ \times \exp \left(\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}+\displaystyle \frac{\kappa }{2{\xi }^{1/2}}r\right){r}^{-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\\ \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{e}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}},\end{array}\end{eqnarray}$
where ${{\rm{K}}}_{1}\left(2l+1,\tfrac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\kappa }^{2}}{4{\xi }^{3/2}},\tfrac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)$ and ${{\rm{K}}}_{2}\left(2l+1,\tfrac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\tfrac{E}{{\xi }^{1/2}}-\tfrac{{\kappa }^{2}}{4{\xi }^{3/2}},\tfrac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)$ are combination coefficients and z = iξ1/4r.

The boundary condition of bound states, ${\left.u\left(r\right)\right|}_{r\to \infty }\to 0$, requires that the coefficient of the second term must vanish since this term diverges when r → ∞ , i.e.,

$\begin{eqnarray}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)=0,\end{eqnarray}$
where
$\begin{eqnarray}\begin{array}{l}{{\rm{K}}}_{2}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ =\displaystyle \frac{{\rm{\Gamma }}\left(2l+2\right)}{{\rm{\Gamma }}\left(l+\tfrac{1}{2}+\tfrac{E}{2{\xi }^{1/2}}+\tfrac{{\kappa }^{2}}{8{\xi }^{3/2}}\right){\rm{\Gamma }}\left(l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\tfrac{{\kappa }^{2}}{8{\xi }^{3/2}}\right)}\\ \times {{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}\right.\\ -\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)+\displaystyle \frac{E}{2{\xi }^{1/2}}\\ +\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}+\displaystyle \frac{{\rm{i}}\kappa }{2{\xi }^{3/4}}\left(-\displaystyle \frac{E}{{\xi }^{1/2}}\right.\\ \left.\left.-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-1\right)\right)\end{array}\end{eqnarray}$
with
$\begin{eqnarray}\,\begin{array}{c}{{\rm{J}}}_{l+\tfrac{3}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\right.\\ \displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)+\displaystyle \frac{E}{2{\xi }^{1/2}}\\ +\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}+\displaystyle \frac{i\kappa }{2{\xi }^{3/4}}\\ \times \left.\left.\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-1\right.\right)\right)\\ ={\displaystyle \int }_{0}^{\infty }{\rm{d}}{{xx}}^{l+\tfrac{1}{2}-\tfrac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}}{{\rm{e}}}^{-{x}^{2}}\\ {\rm{N}}\left(l+\displaystyle \frac{1}{2}-\displaystyle \frac{E}{2{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},\displaystyle \frac{3}{2}\left(2l+1\right)+\displaystyle \frac{E}{2{\xi }^{1/2}}\right.\\ +\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}+\displaystyle \frac{{\rm{i}}\kappa }{2{\xi }^{3/4}}\\ \times \left.\left(-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}}-2l-1\right),x\right).\end{array}\end{eqnarray}$

Equation (A102) is an implicit expression of the eigenvalue.

The eigenfunction, from equations (A101) and (A102), reads

$\begin{eqnarray}\begin{array}{rcl}{u}_{l}\left(r\right) & = & {A}_{l}{{\rm{K}}}_{1}\left(2l+1,\displaystyle \frac{{\rm{i}}\kappa }{{\xi }^{3/4}},-\displaystyle \frac{E}{{\xi }^{1/2}}-\displaystyle \frac{{\kappa }^{2}}{4{\xi }^{3/2}},\displaystyle \frac{-{\rm{i}}2\mu }{{\xi }^{1/4}}\right)\\ & & \times \exp \left(-\displaystyle \frac{{\xi }^{1/2}{r}^{2}}{2}-\displaystyle \frac{\kappa }{2{\xi }^{1/2}}r\right){r}^{\tfrac{E}{2{\xi }^{1/2}}+\displaystyle \frac{{\kappa }^{2}}{8{\xi }^{3/2}}-\displaystyle \frac{1}{2}}\\ & & \times \displaystyle \sum _{n\geqslant 0}\displaystyle \frac{{a}_{n}}{{\left({\xi }^{1/4}r\right)}^{n}}.\end{array}\end{eqnarray}$

1
Chandrasekhar S 1995 Newton's Principia for the Common Reader Oxford Clarendon

2
Arnold V Vogtmann K Weinstein A 2013 Mathematical methods of classical mechanics Graduate Texts in Mathematics New York Springer

3
Arnold V 1990 Huygens and Barrow, Newton and Hooke: Pioneers in Mathematical Analysis and Catastrophe Theory from Evolvents to Quasicrystals Basel: Birkhäuser Basel

4
Needham T 1998 Visual Complex Analysis Oxford Oxford University Press

5
Needham T 1993 Newton and the transmutation of force Am. Math. Monthly 100 119 137

DOI

6
Hall R W Josic K 2000 Planetary motion and the duality of force laws SIAM Rev. 42 115 124

DOI

7
Li W-D Dai W-S 2021 Duality family of scalar field Nuclear Physics. B 972 115569

DOI

8
Ishkhanyan A 2015 Exact solution of the Schrödinger equation for the inverse square root potential Europhys. Lett. 112 10006

DOI

9
Li W-D Dai W-S 2016 Exact solution of inverse-square-root potential $V\left(r\right)=-\tfrac{\alpha }{\sqrt{r}}$ Ann. Phys. 373 207 215

DOI

10
Grant A K Rosner J L 1994 Classical orbits in power-law potentials Am. J. Phys. 62 310 314

DOI

11
Wu Z-B Zeng J-Y 2000 Dynamical symmetry of screened Coulomb potential and isotropic harmonic oscillator Phys. Rev. A 62 032509

DOI

12
Iacob F Lute M 2015 Exact solution to the Schrödinger's equation with pseudo-Gaussian potential J. Math. Phys. 56 121501

DOI

13
Ushveridze A 2017 Quasi-Exactly Solvable Models in Quantum Mechanics Boca Raton, FL CRC Press

14
Ran Y Xue L Hu S Su R-K 2000 On the Coulomb-type potential of the one-dimensional Schrödinger equation J. Phys. A: Math. Gen. 33 9265

DOI

15
Karayer H Demirhan D Büyükkılıç F 2018 Solution of Schrödinger equation for two different potentials using extended Nikiforov–Uvarov method and polynomial solutions of biconfluent Heun equation J. Math. Phys. 59 053501

DOI

16
Karayer H Demirhan D 2021 Analytical eigenstate solutions of Schrödinger equation with noncentral generalized oscillator potential by extended Nikiforov–Uvarov method Phys. Lett. A 413 127608

DOI

17
Chen C-Y You Y Wang X-H Dong S-H 2013 Exact solutions of the Schrödinger equation with double ring-shaped oscillator Phys. Lett. A 377 1521 1525

DOI

18
Sun G-H Chen C-Y Taud H Yáñez-Márquez C Dong S-H 2020 Exact solutions of the 1D Schrödinger equation with the Mathieu potential Phys. Lett. A 384 126480

DOI

19
Ciftci H Hall R L Saad N 2005 Construction of exact solutions to eigenvalue problems by the asymptotic iteration method J. Phys. A: Math. Gen. 38 1147

DOI

20
Ronveaux A Arscott F M 1995 Heun's Differential Equations Oxford Oxford University Press

21
Slavyanov S Lay W 2000 Special Functions: A Unified Theory Based on Singularities Oxford Oxford University Press

22
Li W-D Dai W-S 2016 Scattering theory without large-distance asymptotics in arbitrary dimensions J. Phys. A: Math. Theor. 49 465202

DOI

23
Liu Y-Y Li W-D Dai W 2021 Exactly solvable Gross–Pitaevskii type equations J. Phys. Commun. 015011

DOI

24
Li W-D Dai W-S 2021 Long-range potential scattering: Converting long-range potential to short-range potential by tortoise coordinate J. Math. Phys. 62 122102

DOI

25
Li W-D Chen Y-Z Dai W-S 2019 Scattering state and bound state of scalar field in Schwarzschild spacetime: exact solution Ann. Phys. 409 167919

DOI

26
Li W-D Chen Y-Z Dai W-S 2018 Scalar scattering in Schwarzschild spacetime: integral equation method Phys. Lett. B 786 300 304

DOI

27
Li S-L Liu Y-Y Li W-D Dai W-S 2021 Scalar field in Reissner–Nordström spacetime: bound state and scattering state (with appendix on eliminating oscillation in partial sum approximation of periodic function) Ann. Phys. 432 168578

DOI

28
Pike E Sabatier P 2008 Scattering: Scattering and Inverse Scattering in Pure and Applied Science New York Academic

29
Burke P G 2011 R-Matrix Theory of Atomic Collisions: Application to Atomic, Molecular and Optical Processes vol 61 New York Springer

30
Liu T Li W-D Dai W-S 2014 Scattering theory without large-distance asymptotics J. High Energy Phys. 2014 1 12

DOI

31
Enss V 1979 Asymptotic completeness for quantum-mechanical potential scattering: II. Singular and long-range potentials Ann. Phys. 119 117 132

DOI

32
Levy B R Keller J B 1963 Low-energy expansion of scattering phase shifts for long-range potentials J. Math. Phys. 4 54 64

DOI

33
Hinckelmann O Spruch L 1971 Low-energy scattering by long-range potentials Phys. Rev. A 3 642

DOI

34
Barford T Birse M C 2003 Renormalization group approach to two-body scattering in the presence of long-range forces Phys. Rev. C 67 064006

DOI

35
Hod S 2013 Scattering by a long-range potential J. High Energy Phys. 2013 1 11

DOI

36
Yafaev D 1998 The scattering amplitude for the Schrödinger equation with a long-range potential Commun. Math. Phys. 191 183 218

DOI

37
Romo W Valluri S 1998 A study of the momentum dependence of the phase shift for finite range and Coulomb potentials and its possible applications Nucl. Phys. A 636 467 484

DOI

38
Brau F 2003 Necessary and sufficient conditions for existence of bound states in a central potential J. Phys. A: Math. Gen. 36 9907

DOI

39
Brau F 2004 Sufficient conditions for the existence of bound states in a central potential J. Phys. A: Math. Gen. 37 6687

DOI

40
Dai W-S Xie M 2009 The number of eigenstates: counting function and heat kernel J. High Energy Phys. 2009 033

DOI

41
Dai W-S Xie M 2010 An approach for the calculation of one-loop effective actions, vacuum energies, and spectral counting functions J. High Energy Phys. 2010 1 29

DOI

42
Bargmann V 1952 On the number of bound states in a central field of force Proc. Natl Acad. Sci. 38 961 966

DOI

43
Calogero F 1965 Upper and lower limits for the number of bound states in a given central potential Commun. Math. Phys. 1 80 88

DOI

44
Joachain C J 1975 Quantum Collision Theory Amsterdam North-Holland Publishing Company

Outlines

/