Bound states of a quartic and sextic inverse-power-law potential for all angular momenta

Abstract

We use the tridiagonal representation approach to solve the radial Schrödinger equation for an inverse-power-law potential of a combined quartic and sextic degrees and for all angular momenta. The amplitude of the quartic singularity is larger than that of the sextic but the signs are negative and positive, respectively. It turns out that the system has a finite number of bound states, which is determined by the larger ratio of the two singularity amplitudes. The solution is written as a finite series of square integrable functions written in terms of the Bessel polynomial.

Appendix: Bessel polynomial on the real line

The Bessel polynomial on the positive real line is defined in terms of the hypergeometric or confluent hypergeometric functions as follows (see section 9.13 of the book by Koekoek et al. [19] but make the replacement $x \mapsto 2x$ and $a \mapsto 2\mu$)

$$ Y_{n}^{\mu } (x) = {}_{2}F_{0} \left( {\left. {_{{\_\_}}^{{ - n,n + 2\mu + 1}} } \right| - x} \right) = \left( {n + 2\mu + 1} \right)_{n} x^{n} {}_{1}F_{1} \left( {\left. {_{{ - 2(n + \mu )}}^{{ - n}} } \right|{1 \mathord{\left/ {\vphantom {1 x}} \right. \kern-\nulldelimiterspace} x}} \right), $$

(A1)

where x ≥ 0, $n = 0,1,2, \ldots ,N$ and N is a non-negative integer. The real parameter μ is negative such that $\mu < - N - \tfrac{1}{2}$. The Pochhammer symbol $(a)_{n}$ (a.k.a. shifted factorial) is defined as $\left( a \right)_{n} = a(a + 1)(a + 2) \ldots (a + n - 1) = \tfrac{\varGamma (n + a)}{{\varGamma (a)}}$. The Bessel polynomial could also be written in terms of the associated Laguerre polynomial as: $Y_{n}^{\mu } (x) = n!\left( { - x} \right)^{n} L_{n}^{ - (2n + 2\mu + 1)} \left( {{1 \mathord{\left/ {\vphantom {1 x}} \right. \kern-\nulldelimiterspace} x}} \right)$. The three-term recursion relation reads as follows:

$$ \begin{aligned} & 2x\,Y_{n}^{\mu } (x) = \frac{ - \mu }{{(n + \mu )(n + \mu + 1)}}Y_{n}^{\mu } (x) \\ & \quad - \frac{n}{(n + \mu )(2n + 2\mu + 1)}\,Y_{n - 1}^{\mu } (x) + \frac{n + 2\mu + 1}{{(n + \mu + 1)(2n + 2\mu + 1)}}\,Y_{n + 1}^{\mu } (x) \\ \end{aligned} $$

(A2)

Note that the constraints on μ and on the maximum polynomial degree make this recursion definite (i.e., the signs of the two recursion coefficients multiplying $Y_{n \pm 1}^{\mu } (x)$ are the same). Otherwise, these polynomials could not be defined on the real line but on the unit circle in the complex plane. The orthogonality relation reads as follows

$$ \int_{0}^{\infty } {x^{2\mu } {\text{e}}^{{ - {1 \mathord{\left/ {\vphantom {1 x}} \right. \kern-\nulldelimiterspace} x}}} Y_{n}^{\mu } (x)Y_{m}^{\mu } (x)\,{\text{d}}x} = - \frac{n!\varGamma ( - n - 2\mu )}{{2n + 2\mu + 1}}\delta_{nm} . $$

(A3)

The differential equation is

$$ \left\{ {x^{2} \frac{{{\text{d}}^{2} }}{{{\text{d}}x^{2} }} + \left[ {1 + 2x\left( {\mu + 1} \right)} \right]\frac{{\text{d}}}{{{\text{d}}x}} - n\left( {n + 2\mu + 1} \right)} \right\}Y_{n}^{\mu } (x) = 0. $$

(A4)

The forward and backward shift differential relations read as follows:

$$ \frac{{\text{d}}}{{{\text{d}}x}}Y_{n}^{\mu } (x) = n\left( {n + 2\mu + 1} \right)Y_{n - 1}^{\mu + 1} (x). $$

(A5)

$$ x^{2} \frac{{\text{d}}}{{{\text{d}}x}}Y_{n}^{\mu } (x) = - \left( {2\mu x + 1} \right)Y_{n}^{\mu } (x) + Y_{n + 1}^{\mu - 1} (x). $$

(A6)

We can write $Y_{n + 1}^{\mu - 1} (x)$ in terms of $Y_{n}^{\mu } (x)$ and $Y_{n \pm 1}^{\mu } (x)$ as follows:

$$ \begin{aligned} & 2Y_{n + 1}^{\mu - 1} (x) = \frac{(n + 1)(n + 2\mu )}{{(n + \mu )(n + \mu + 1)}}\,Y_{n}^{\mu } (x) \\ & \quad + \frac{n\,(n + 1)}{{(n + \mu )(2n + 2\mu + 1)}}\,Y_{n - 1}^{\mu } (x) + \frac{(n + 2\mu )(n + 2\mu + 1)}{{(n + \mu + 1)(2n + 2\mu + 1)}}\,Y_{n + 1}^{\mu } (x) \\ \end{aligned} $$

(A7)

Using this identity and the recursion relation (A2), we can rewrite the backward shift differential relation as follows:

$$ \begin{aligned} & 2x^{2} \frac{{\text{d}}}{{{\text{d}}x}}Y_{n}^{\mu } (x) = n\,(n + 2\mu + 1) \\ & \quad \times \left[ { - \frac{{Y_{n}^{\mu } (x)}}{(n + \mu )(n + \mu + 1)} + \frac{{Y_{n - 1}^{\mu } (x)}}{(n + \mu )(2n + 2\mu + 1)} + \frac{{Y_{n + 1}^{\mu } (x)}}{(n + \mu + 1)(2n + 2\mu + 1)}} \right] \\ \end{aligned} $$

(A8)

The generating function is

$$ \sum\limits_{n = 0}^{\infty } {Y_{n}^{\mu } (x)\frac{{t^{n} }}{n!}} = \frac{{2^{2\mu } }}{{\sqrt {1 - 4xt} }}\left( {1 + \sqrt {1 - 4xt} } \right)^{ - 2\mu } \exp \left[ {{{2t} \mathord{\left/ {\vphantom {{2t} {(1 + \sqrt {1 - 4xt} )}}} \right. \kern-\nulldelimiterspace} {(1 + \sqrt {1 - 4xt} )}}} \right]. $$

(A9)

The polynomial $B_{n}^{\mu } (z;\gamma )$ is defined in [15] by its three-term recursion relation Eq. (16) therein, which reads

$$ \begin{aligned} & z\,B_{n}^{\mu } (z;\gamma ) = \left[ {\frac{ - 2\mu }{{(n + \mu )(n + \mu + 1)}} + \gamma \left( {n + \mu + \tfrac{1}{2}} \right)^{2} } \right]B_{n}^{\mu } (z;\gamma ) \\ & \quad - \frac{n}{{(n + \mu )\left( {n + \mu + \tfrac{1}{2}} \right)}}\,B_{n - 1}^{\mu } (z;\gamma ) + \frac{n + 2\mu + 1}{{(n + \mu + 1)\left( {n + \mu + \tfrac{1}{2}} \right)}}\,B_{n + 1}^{\mu } (z;\gamma ) \\ \end{aligned} $$

(A10)

where $B_{0}^{\mu } (z;\gamma ) = 1$ and $B_{ - 1}^{\mu } (z;\gamma ): = 0$.