understand that the value of $\nu$ is preserved by $\mathfrak{F}$. For a
while, however, we treat $\nu$ as another parameter and express the energy
$E_{a}$ as a function of $\lambda_{a}$, $L_{a}$ and $\nu$,

$E_{a}=E_{a}(\lambda_{a},L_{a},\nu).$ (199)

This corresponds to the energy function $E_{a}(\lambda_{a},L_{a},N)$ in the
semiclassical case. We convert this energy function to the energy spectrum of
system $A$ by replacing the parameters $L_{a}$ and $\nu$ to their quantum
counterparts. If we restrict our interest to bound state solutions, the
parameter $\nu$ is to be replaced by a set of discrete numbers
$\nu=0,1,2,\ldots$. Furthermore, putting the angular parameter $L_{a}$ into
the Langer form (167), we obtain the discrete energy spectrum of system $A$,

$E_{a}(\ell_{a},\nu)=E_{a}(\lambda_{a},\ell_{a}+D/2-1,\nu),$ (200)

where $\ell_{a}\in\mathbb{N}_{0}$ and $\nu\in\mathbb{N}_{0}$.

Since the energy functions $E_{a}(\lambda_{a},L_{a},\nu)$ and
$E_{b}(\lambda_{b},L_{b},\nu)$ are related by the classical energy formulas,
(60) and (64)–(65), the corresponding energy spectra $E_{a}(\ell_{a},\nu)$ and
$E_{b}(\ell_{b},\nu)$ can be related by the same formulas provided the angular
parameter and the quantum parameter are properly expressed in terms of quantum
numbers. Knowing the energy spectrum of the form
$E_{a}(\ell_{a},\nu)=\mathcal{E}(\lambda_{a},L_{a},\nu)$ for system $A$, we
can determine the energy spectrum $E_{b}$ of system $B$ by

$E_{b}(\ell_{b},\nu)=-\eta^{2}C^{a+2}\mathcal{E}^{-1}(-\lambda_{b}/(\eta^{2}C^{2}),L_{b}/\eta,\nu),$
(201)

where $L_{b}=\ell_{b}+D/2-1$ with $\ell_{b}\in\mathbb{N}_{0}$. For the bound
state spectrum, $\nu=0,1,2,\ldots$.

If the energy spectrum of system $A$ is given in the form

$E_{a}(\ell_{a},\nu)=\pm\frac{1}{4}(a+2)^{2}|\lambda_{a}|^{2/(a+2)}\left[\mathcal{F}\Bigl{(}\sqrt{2/(a+2)\,}(\ell_{a}+D/2-1),\nu\Bigr{)}\right]^{1/a}$
(202)

then the energy spectrum of system $B$ is given by

$E_{b}(\ell_{b},\nu)=\pm\frac{1}{4}(b+2)^{2}|\lambda_{b}|^{2/(b+2)}\left[\mathcal{F}\Bigl{(}\sqrt{2/(b+2)\,}(\ell_{b}+D/2-1),\nu\Bigr{)}\right]^{1/b}.$
(203)

These relations are the same as the semiclassical relations (104 and (105)
where the signs are determined by the signs of the coupling constants, ${\rm
sgn}\,E_{a}=-{\rm sgn}\,\lambda_{b}$ and ${\rm sgn}\,E_{b}=-{\rm
sgn}\,\lambda_{a}$.

####  IV.2.2 Wave Functions

The wave function transforms as
$\psi_{a}(r;L_{a},\nu)=h(\rho)\psi_{b}(\rho;L_{b},\nu)$. Therefore, if an
eigenfunction of system $A$ is given, then the corresponding eigenfunction of
system $B$ can be determined by

$\psi_{b}(\rho;L_{b},\nu)=\frac{1}{h(\rho)}\psi_{a}(C\rho^{\eta};L_{b}/\eta,\nu),$
(204)

where $L_{b}=\ell_{b}+D/2-1$ with $\ell_{b}\in\mathbb{N}_{0}$. Both
$\psi_{a}(r)$ and $\psi_{b}(\rho)$ as eigenfunctions are supposed to be
square-integrable, and each of them must be normalizable to unity. However,
even if $\psi_{a}(r)$ is normalized to unity, it is unlikely that
$\psi_{b}(\rho)$ constructed by (204) is normalized to unity. This is because

$\int_{0}^{\infty}\textrm{d}r\,|\psi_{a}(r)|^{2}=\int_{0}^{\infty}\textrm{d}\rho\,g(\rho)\,|\psi_{b}(\rho)|^{2}=1$
(205)

where
$g(\rho)=[f^{\prime}(\rho)]^{2}=[h(\rho)]^{4}=C^{2}\eta^{2}\rho^{2(\eta-1)}$.
In this regard, if system $A$ and system $B$ are power-dual to each other, the
formula (204) determines $\psi_{b}(\rho)$ of system $B$ out of $\psi_{a}(r)$
of system $A$ except for the normalization.

####  IV.2.3 Green Functions

The Green function $G(r,r^{\prime};z)=\langle r|\hat{G}(z)|r^{\prime}\rangle$
is the $r$-representation of the resolvent $\hat{G}(z)=(z-\hat{H})^{-1}$ where
$z\in\mathbb{C}\backslash{\rm spec}\,\hat{H}$ and $\hat{H}$ is the Hamiltonian
operator of the system in question. Let $E(\nu)$ and $|\psi(\nu)\rangle$ be
the eigenvalue of $\hat{H}$ and the corresponding eigenstate, respectively, so
that $\hat{H}|\psi(\nu)\rangle=E(\nu)|\psi(\nu)\rangle$. For simplicity, we
consider the case where $\nu\in\mathbb{N}_{0}$. Assume the eigenstates are
orthonormalized and form a complete set, that is,

$\langle\psi(\nu)|\psi(\nu^{\prime})\rangle=\delta_{\nu,\nu^{\prime}}\,,\,\,\quad\,\,\sum_{\nu\in\mathbb{N}_{0}}|\psi(\nu)\rangle\langle\psi(\nu)|=1.$
(206)

From the completeness condition in (206), it is obvious that

$\hat{G}(z)=\sum_{\nu\in\mathbb{N}_{0}}\frac{|\psi(\nu)\rangle\langle\psi(\nu)|}{z-E(\nu)}.$
(207)

Hence, the Green function can be written as

$G(r,r^{\prime};z)=\sum_{\nu\in\mathbb{N}_{0}}\frac{\psi^{\ast}(r^{\prime};\nu)\psi(r;\nu)}{z-E(\nu)}.$
(208)

Use of Cauchy’s integral formula leads us to the expression,

$\psi^{\ast}(r,\nu)\psi(r^{\prime};\nu)=\frac{1}{2\pi
i}\oint_{C_{\nu}}\textrm{d}z\,G(r,r^{\prime};z),$ (209)

where the closed contour $C_{\nu}$ counterclockwise encloses only the simple
pole $z=E(\nu)$ for a fixed value of $\nu$. Note that we will deal only with
radial, hence one-dimensional, problems where no degeneracies can occur.
Multiplying both sides of (209) by two factors $v(r)$ and $v(r^{\prime})$
yields

$\tilde{\psi}^{\ast}(r,\nu)\tilde{\psi}(r^{\prime};\nu)=\frac{1}{2\pi
i}\oint_{C_{\nu}}\textrm{d}z\,\tilde{G}(r,r^{\prime};z),$ (210)

where $\tilde{\psi}(r;\nu)=v(r)\psi(r;\nu)$ and
$\tilde{G}(r,r^{\prime};z)=v(r)v(r^{\prime})G(r,r^{\prime};z)$.

For instance, the Green function $\mathcal{G}(r,r^{\prime};E)$ for the radial
Schrödinger Equation (165) is related to the Green function
$G(r,r^{\prime};E)$ for the simplified radial Equation (166) by

$\mathcal{G}(r,r^{\prime};E)=(r\,r^{\prime})^{(1-D)/2}G(r,r^{\prime};E)$ (211)

as the wave functions of (165) and (166) are connected by
$R_{\ell}(r)=r^{(1-D)/2}\psi_{\ell}(r)$.

Suppose the Green functions of system $A$ and system $B$ are given,
respectively, by

$\psi_{a}^{\ast}(r;\nu)\psi_{a}(r^{\prime};\nu)=\frac{1}{2\pi
i}\oint_{C_{\nu}}\textrm{d}z\,G_{a}(r,r^{\prime};z),$ (212)

and

$\psi_{b}^{\ast}(\rho;\nu)\psi_{b}(\rho^{\prime};\nu)=\frac{1}{2\pi
i}\oint_{C_{\nu}}\textrm{d}z\,G_{b}(\rho,\rho^{\prime};z).$ (213)

By comparing these two expressions, we see that if
$\psi_{a}(r;\nu)=h(\rho)\psi_{b}(\rho;\nu)$ then

$G_{a}(r,r^{\prime};E_{a}(\nu))=h(\rho)h(\rho^{\prime})G_{b}(\rho,\rho^{\prime};E_{b}(\nu)).$
(214)

The above result is obtained without considering the detail of the
Hamiltonian. In the following, an alternative account is provided for deriving
the same result by using the Hamiltonian explicitly. Let $\hat{H}_{a}$ be the
Hamiltonian operator of system $A$ such that $\langle
r|\hat{H}_{a}-E_{a}|r^{\prime}\rangle=(H_{a}(r)-E_{a})\langle
r|r^{\prime}\rangle$. Then it is obvious that

$\\{H_{a}(r)-E_{a}\\}G_{a}(r,r^{\prime};E_{a})=-\delta(r-r^{\prime}).$ (215)

According to (194), Equation (215) implies

$\\{H_{b}(\rho)-E_{b}\\}\frac{1}{h}G_{a}(f(\rho),f(\rho^{\prime});E_{a})=-h^{3}(\rho)\delta(f(\rho)-f(\rho^{\prime})).$
(216)

From the relations,

$\int\,\textrm{d}r\,|r\rangle\langle
r|=\int\,\textrm{d}\rho\,f^{\prime}(\rho)|f(\rho)\rangle\langle
f(\rho^{\prime})|=\int\,\textrm{d}\rho\,|\rho\rangle\langle\rho^{\prime}|=1,$
(217)

there follows $|\rho\rangle=h(\rho)|f(\rho)\rangle$. Hence we have,
$\langle\rho|\rho^{\prime}\rangle=h(\rho)h(\rho^{\prime})\langle
f(\rho)|f(\rho^{\prime})\rangle$, that is,
$\delta(r-r^{\prime})=\delta(f(\rho)-f(\rho^{\prime}))=[h(\rho)h(\rho^{\prime})]^{-1}\delta(\rho-\rho^{\prime})$.
Thus, we arrive at the radial equation satisfied by the Green function of
system $B$,

$\\{H_{b}(\rho)-E_{b}\\}G_{b}(\rho,\rho^{\prime};E_{b})=-\delta(\rho-\rho^{\prime}),$
(218)

if the Green function transforms as

$G_{a}(r,r_{0};E_{a},L_{a})=h(\rho)h(\rho^{\prime})G_{b}(\rho,\rho^{\prime};E_{b},L_{b}).$
(219)

Substitution of $L_{a}=\ell_{a}+D/2-1$ with $\ell_{a}\in\mathbb{N}_{0}$ and
$L_{b}=\ell_{b}+D/2-1$ with $\ell_{b}\in\mathbb{N}_{0}$ into (219) results in

$G_{a}(r,r^{\prime};E_{a},\ell_{a}+D/2-1)\doteq
h(\rho)h(\rho_{0})G_{b}(\rho,\rho^{\prime};E_{b},\ell_{b}+D/2-1),$ (220)

which is not an equality as $\ell_{a}\in\mathbb{N}_{0}$ and
$\ell_{b}\in\mathbb{N}_{0}$ are assumed. Insofar as system $B$ is power-dual
to system $A$, the Green function of system $B$ can be expressed in terms of
the Green function of system $A$ as

$G_{b}(\rho,\rho^{\prime};E_{b},\ell_{b}+D/2-1,\lambda_{b},\lambda_{b^{\prime}})=[(f^{\prime}(\rho)f^{\prime}(\rho^{\prime})]^{-1/2}G_{a}\left(f(\rho),f(\rho^{\prime});E_{a},(\ell_{b}+D/2-1)/\eta,\lambda_{a},\lambda_{a^{\prime}}\right)$
(221)

where $f(\rho)=C\rho^{\eta}$ and the parameters $E_{a}$, $\lambda_{a}$ and
$\lambda_{a^{\prime}}$ are given via the relations (177) and (179) in terms of
$E_{b}$, $\lambda_{b}$ and $\lambda_{b^{\prime}}$. This relation is an
equality even though (220) is a dot equality. An expression similar to but
slightly different from (221) has been obtained by Johnson [36] in much the
same way.

###  IV.3 The Coulomb–Hooke Dual Pair

Again, we take up the Coulomb–Hooke dual pair to test the transformation
properties shown in Section III.1. Let system $A$ be the hydrogen atom with
$\lambda_{a}=-e^{2}<0$ and system $B$ a radial oscillator with
$\lambda_{b}=\frac{1}{2}m\omega^{2}>0$. So $(a,b)=(-1,2)$ and $\eta=-b/a=2$.
Both systems are assumed to be in $D$ dimensional space. The Coulomb system
has the scattering states $(E_{a}>0)$ as well as the bound states $(E_{a}<0)$.
However, the exchange relations (177) prohibits the process
$(E_{a}>0,\lambda_{a}<0)\Rightarrow(E_{b}>0,\lambda_{b}>0)$. The Coulomb–Hooke
duality occurs only when the Coulomb system is in bound states.

The energy relations: Suppose we know that the energy spectrum of system $A$
has the form,

$E_{a}(\lambda_{a},L_{a},\nu)=-\frac{me^{4}}{2\hbar^{2}(\nu+L_{a}+1/2)^{2}},$
(222)

where $\lambda_{a}=-e^{2}$, $\nu\in\mathbb{N}_{0}$ and $L_{a}=\ell_{b}+D/2-1$
with $\ell_{a}\in\mathbb{N}_{0}$. Then the formula (202) leads to

$\mathcal{F}\Bigl{(}\sqrt{2}L_{a},\nu\Bigr{)}=\frac{\hbar^{2}}{2m}\left[\nu+(\sqrt{2}L_{a})/\sqrt{2}+1/2\right]^{2}.$
(223)

Careful use of this result in the formula (203) enables us to determine the
energy spectrum of system $B$. Namely,

$E_{b}(\lambda_{b},L_{b},\nu)=4\sqrt{\lambda_{b}}\sqrt{\frac{\hbar^{2}}{2m}}\left[\nu+(L_{b}/\sqrt{2})/\sqrt{2}+1/2\right]^{1/2}.$
(224)

Substituting $\lambda_{b}=m\omega^{2}/2$ and $L_{b}=\ell_{b}+D/2-1$ in (224),
we reach the standard expression for the energy spectrum of the isotropic
harmonic oscillator in $D$-dimensional space,

$E_{b}(\ell_{b},\nu)=\hbar\omega(2\nu+\ell_{b}+D/2)\,\quad\,(\ell_{b},\,\nu\in\mathbb{N}_{0}).$
(225)

Wave functions: The radial Equation (166) for the Coulomb potential
$V(r)=-e^{2}/r$ can easily be converted to the Whittaker Equation [69]

$\left\\{\frac{\textrm{d}^{2}}{\textrm{d}x^{2}}-\frac{L^{2}-1/4}{x^{2}}+\frac{k}{x}-\frac{1}{4}\right\\}w(x)=0,$
(226)

where $L=\ell+D/2-1\,\,(\ell\in\mathbb{N}_{0})$. In the conversion, we have
let $x=2\kappa r$, $k=me^{2}/(\hbar^{2}\kappa)=k_{a}$,
$\hbar\kappa=\sqrt{-2mE}$, $L=L_{a}$ and $w(x)=\psi_{a}(x/(2\kappa))$. This
set of replacements is indeed a duality map for the self-dual pair
$(a,a)=(-1,-1)$. The Whittaker functions, $M_{k,L}(x)$ and $W_{k,L}(x)$, are
two linearly independent solutions of the Whittaker Equation (226). For $|x|$
small, $M_{k,L}(x)\sim x^{L+\,1/2}$ and
$W_{k,L}(x)\sim-\frac{\Gamma(2L)}{\Gamma(L-k+1/2)}x^{-L+\,1/2}$. If
$-\pi/2<\arg x<3\pi/2$ and $|x|$ is large, then

$M_{k,L}(x)\sim\Gamma(2L+1)\left\\{\frac{\textrm{e}^{i\pi(L-k+\frac{1}{2})}\textrm{e}^{-x/2}x^{k}}{\Gamma(L+k+\frac{1}{2})}+\frac{\textrm{e}^{x/2}x^{-k}}{\Gamma(L-k+\frac{1}{2})}\right\\},$
(227)

and, if $x\notin\mathbb{R}^{-}$ and $|x|$ is large,

$W_{k,L}(x)\sim\textrm{e}^{-x/2}x^{k}[1+O(x^{-1})].$ (228)

The first solution $M_{k,L}(x)$ vanishes at $x=0$ as $L>-1/2$ but diverges as
$|x|\rightarrow\infty$ unless $k-L-\frac{1}{2}\in\mathbb{N}_{0}$, whereas the
second solution $W_{k,L}(x)$ diverges at $x=0$ but converges to zero as
$|x|\rightarrow\infty$.

The solution for the Coulomb problem is given in terms of the Whittaker
function,

$\psi_{a}(r;L_{a},\nu)=\mathcal{N}_{a}(L_{a})\,M_{\nu+L_{a}+\frac{1}{2},\,L_{a}}(2\kappa
r),$ (229)

where $k_{a}$ is replaced by $\nu+L_{a}+\,1/2$. For the bound state solution
which vanishes at infinity, we have to let $\nu=0,1,2,\ldots$. In this case,
$k_{a}=\nu+L_{a}+\,1/2$ implies the discrete spectrum
$E_{a}(\lambda_{a},L_{a},\nu)$ in (222).

Since the Whittaker function $M_{k,\mu}(z)$ is related to the Laguerre
function $L_{\nu}^{2\mu}(z)$ as

$M_{\mu+\nu+\frac{1}{2},\,\mu}(z)=\frac{\Gamma(2\mu+1)\Gamma(\nu+1)}{\Gamma(2\mu+\nu+1)}\,\textrm{e}^{-z/2}z^{\mu+\frac{1}{2}}\,L_{\nu}^{2\mu}(z),$
(230)

the eigenfunction may also be expressed in terms of the Laguerre function as

$\psi_{a}(r;L_{a},\nu)=\mathcal{N}_{a}(L_{a})\,\frac{\Gamma(2L_{a}+1)\Gamma(\nu+1)}{\Gamma(\nu+2L_{a}+1)}\,\textrm{e}^{-\kappa
r}(2\kappa r)^{L_{a}+\frac{1}{2}}\,L_{\nu}^{2L_{a}}(2\kappa r),$ (231)

which is normalized to unity with

$\mathcal{N}_{a}(L_{a})=\frac{\hbar\kappa/\sqrt{me^{2}}}{\Gamma(2L_{a}+1)}\sqrt{\frac{\Gamma(\nu+2L_{a}+1)}{\Gamma(\nu+1)}}.$
(232)

The radial equation for the Hooke system with
$V_{b}(\rho)=\frac{1}{2}m\omega^{2}\rho^{2}$, too, can be reduced to the
Whittaker equation by letting

$y=(m\omega/\hbar)\rho^{2},\,\quad\,L=L_{b}/2,\,\quad\,k=E_{b}/(2\hbar\omega)=k_{b},\,\quad\,w(y)=y^{1/4}\psi_{b}(y),$
(233)

which form a duality map for $(b,c)=(2,-1)$. Here $L_{b}=\ell_{b}+D/2-1$ with
$\ell_{b}\in\mathbb{N}_{0}$. The bound state solution for the radial
oscillator is given by

$\psi_{b}(\rho;L_{b},\nu)=\mathcal{N}_{b}(L_{b})\,\frac{1}{\sqrt{\rho}}\,M_{\nu+\frac{1}{2}L_{b}+\frac{1}{2},\,\frac{1}{2}L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}\right).$
(234)

The choice $k_{b}=(2\nu+L_{b}+1)/2$ with $\nu\in\mathbb{N}_{0}$ makes the
solution (234) the eigenfunction belonging to the energy $E_{b}(\nu,\ell_{b})$
in (224). In terms of the Laguerre function, it reads

$\psi_{b}(\rho;L_{b},\nu)=\mathcal{N}_{b}(L_{b})\,\frac{\Gamma(L_{b}+1)\Gamma(\nu+1)}{\Gamma(\nu+L_{b}+1)}\textrm{e}^{-(m\omega/2\hbar)\rho^{2}}\left(\frac{m\omega}{\hbar}\rho^{2}\right)^{(L_{b}+\frac{1}{2})/2}L_{\nu}^{L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}\right),$
(235)

which is normalized to unity with

$\mathcal{N}_{b}(L_{b})=\frac{(4m\omega/\hbar)^{1/4}}{\Gamma(L_{b}+1)}\,\sqrt{\frac{\Gamma(\nu+L_{b}+1)}{\Gamma(\nu+1)}}.$
(236)

The process of going from (229) to (234) is rather straightforward. First we
notice that $\eta=-b/a=2$ for the Coulomb–Hooke pair $(a,b)=(-1,2)$. Then we
use the relation $\lambda_{b}=-\eta^{2}C^{2}E_{a}$,
$\lambda_{b}=m\omega^{2}/2$ and $\hbar\kappa=\sqrt{-2mE_{a}}$ to get
$C=m\omega/(2\hbar\kappa)$. Hence operation $\mathfrak{R}:\,r=C\rho^{\eta}$
with $\eta=2$ yields $2\kappa r=(m\omega/\hbar)\rho^{2}$. In addition, we
apply $\mathfrak{L}:\,L_{a}=L_{b}/2$. Consequently, we have the right hand
side of (204) for $a=-1$, $\eta=2$ and
$h(\rho)=\sqrt{m\omega/(\hbar\kappa)}\rho^{1/2}$ in the form,

$\sqrt{\hbar\kappa/m\omega}\,\frac{1}{\sqrt{\rho}}\,\psi_{a}((m\omega/2\hbar\kappa)\rho^{2};L_{b}/2,\nu)=\tilde{\mathcal{N}}_{b}(L_{b})\,\frac{1}{\sqrt{\rho}}\,M_{\nu+\frac{1}{2}L_{b}+\frac{1}{2},\,\frac{1}{2}L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}\right),$
(237)

which coincides with the eigenfunction for the radial oscillator in (234)
except for the normalization factor. In (237),

$\tilde{\mathcal{N}}_{b}(L_{b})=\sqrt{\hbar\kappa/m\omega}\mathcal{N}_{a}(L_{b}/2),$
(238)

which differs from $\mathcal{N}_{b}(L_{b})$ of (236) due to the difference of
factors,$\sqrt{\hbar^{2}\kappa^{3}/(me^{2})}(m\omega/\hbar)^{-1/2}\neq\sqrt{2}(m\omega/\hbar)^{1/4}$.
The wave function of the radial oscillator can be determined by the radial
wave function of the hydrogen atom except for its normalization.

The Green functions: The Green function of interest,
$G_{a}(r,r^{\prime};E,L)$, obeys the radial equation,

$\left\\{\frac{\textrm{d}^{2}}{\textrm{d}r^{2}}-\frac{L^{2}-1/4}{r^{2}}-\frac{2m}{\hbar^{2}}V(r)+\frac{2m}{\hbar^{2}}E\right\\}G_{a}(r,r^{\prime};E,L)=-\frac{2m}{\hbar^{2}}\delta(r-r^{\prime}),$
(239)

where $V_{a}(r)=\lambda_{a}r^{a}+\lambda_{a^{\prime}}r^{a^{\prime}}$. The
boundary conditions we impose on it are

$\lim_{r\rightarrow
0}G(r,r^{\prime};E,L)=0\qquad\mbox{and}\qquad\lim_{r\rightarrow\infty}G(r,r^{\prime};E,L)<\infty.$
(240)

Let $\psi^{(1)}(r)$ and $\psi^{(2)}(r)$ be two independent solutions of the
radial Equation (166). Let us assume that $\psi^{(1)}(r)$ remains finite as
$r\rightarrow\infty$ while the second solution obeys $\psi^{(2)}(0)=0$. With
these solutions, following the standard procedure [70], we can construct the
Green function $G(r,r^{\prime};E,L)$ as

$G(r,r^{\prime};E,L)=\frac{2m}{\hbar^{2}\mathcal{W}[\psi^{(1)},\psi^{(2)}]}\left\\{\begin{array}[]{ll}\psi^{(1)}(r)\,\psi^{(2)}(r^{\prime}),&\,\quad\,r>r^{\prime}\\\
\psi^{(1)}(r^{\prime})\,\psi^{(2)}(r),&\,\quad\,r^{\prime}>r\end{array}\right.$
(241)

where $\mathcal{W}[\cdot,\cdot]$ signifies the Wronskian.

For the Coulomb problem with $V_{a}(r)=-e^{2}r^{-1}$, we let
$\psi^{(1)}(r)=W_{k_{a},\,L_{a}}(2\kappa r)$ and
$\psi^{(2)}(r)=M_{k_{a},\,L_{a}}(2\kappa r)$. Then we calculate the Wronskian
to get

$(2\kappa)^{-1}\mathcal{W}[W_{k,L}(2\kappa r),M_{k,L}(2\kappa
r)]=\mathcal{W}[W_{k,L}(x),M_{k,L}(x)]=-\frac{\Gamma(2L+1)}{\Gamma(L-k+\frac{1}{2})},$
(242)

where we have use the property,

$\mathcal{W}[W_{k,L}(x),\,M_{k,L}(x)]=(dy/dx)\mathcal{W}[W_{k,L}(y),\,M_{k,L}(y)].$
(243)

Substituting this result in the formula (241), we obtain the radial Green
function for the Coulomb problem,

$G_{a}(r,r^{\prime};E_{a},L_{a})=-\frac{m}{\hbar^{2}\kappa}\frac{\Gamma(L_{a}-k_{a}+\frac{1}{2})}{\Gamma(2L_{a}+1)}W_{k_{a},\,L_{a}}(2\kappa
r_{>})\,M_{k_{a},\,L_{a}}(2\kappa r_{<}),$ (244)

where $r_{>}=\max\\{r,r^{\prime}\\}$ and $r_{<}=\min\\{r,r^{\prime}\\}$. We
have also set $\kappa=\sqrt{-2mE_{a}}\,\hbar$ and
$k_{a}=me^{2}/(\hbar\sqrt{-2mE_{a}})$, both of which are in general complex
numbers. The resultant Green function is a double-valued function of $E_{a}$.
It contains the contribution from the continuous states (corresponding to the
branch-cut along the positive real line on $E_{a}$) as well as the bound
states (corresponding to the poles on the negative real axis). The poles of
$G(r,r^{\prime};L_{a},E_{a})$ on the $E_{a}$-plane occur when
$L_{a}-k_{a}+\frac{1}{2}=-\nu$ with $\nu\in\mathbb{N}_{0}$, yielding the
discrete energy spectrum (222).

Similarly, for the radial oscillator with
$V_{b}(\rho)=(m/2)\omega^{2}\rho^{2}$, we
let$\psi^{(1)}(\rho)=W_{k_{b},L_{b}}((m\omega/\hbar)\rho^{2})$ and
$\psi^{(2)}(\rho)=M_{k_{b},L_{b}}((m\omega/\hbar)\rho^{2})$. Use of the
property,

$\mathcal{W}[\chi(y)W_{k,L}(y),\,\chi(y)M_{k,L}(y)]=[\chi(y)]^{2}\mathcal{W}[W_{k,L}(y),\,M_{k,L}(y)],$
(245)

for a differentiable function $\chi(y)$, together with (243) and (242),
enables us to evaluate the Wronskian and to get to the Green function for the
radial oscillator,

$G_{b}(\rho,\rho^{\prime};L_{b},E_{b})=-\frac{1}{\hbar\omega\,\sqrt{\rho\rho^{\prime}}}\frac{\Gamma(\frac{1}{2}L_{b}-k_{b}+\frac{1}{2})}{\Gamma(L_{b}+1)}\,W_{k_{b},\frac{1}{2}L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}_{{}_{>}}\right)\,M_{k_{b},\frac{1}{2}L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}_{{}_{<}}\right),$
(246)

where $k_{b}=E_{b}/(2\hbar\omega)$. Since $G(\rho,\rho^{\prime};l_{b},E_{b})$
is not a multi-valued function of $E_{b}$, it has no branch point on the
$E_{b}$-plane and contains no contribution corresponding to a continuous
spectrum, but has poles at $k_{b}=\nu+\frac{1}{2}L_{b}+\frac{1}{2}$ with
$\nu\in\mathbb{N}_{0}$ yielding the discrete energy spectrum (225).

Finally, we compare the Green function for the bound state of the Coulomb
problem (244) and the Green function for the radial oscillator (246). The
Gamma functions and the Whittaker functions in (244) are brought to those in
(246) by transformations $r=C\rho^{2}$ with $C=m\omega/(2\hbar\kappa)$,
$L_{a}=L_{b}/\eta$ with $\eta=2$, and $k_{a}=k_{b}$. Although the first two
transformations are two of the dual operations, the last one must be verified.
Since $k_{a}=me^{2}/(\hbar^{2}\kappa)=-m\lambda_{a}/(\hbar^{2}\kappa)$ and
$\lambda_{a}=-E_{b}/(4C)$, it immediately follows that
$k_{a}=E_{b}/(2\hbar\omega)=k_{b}$ provided $C=m\omega/(2\hbar\kappa)$. For
the bound state problem, $k_{a}=\nu+L_{a}+\frac{1}{2}$ and
$k_{b}=\nu+\frac{1}{2}L_{b}+\frac{1}{2}$. Hence, it is apparent that
$k_{a}=k_{b}$ when $L_{a}=L_{b}/2$. The extra function in (219) is now given
by
$h(\rho)h(\rho^{\prime})=\sqrt{m\omega/(\hbar\kappa)}\sqrt{\rho\rho^{\prime}}$.
Hence the prefactor $m/(\hbar^{2}\kappa)$ in (244) divided by the extra
function gives rise to the prefactor
$(\hbar\omega\,\sqrt{\rho\rho^{\prime}})^{-1}$ in (246). In this fashion,
$G_{a}(r,r^{\prime};L_{a},E_{a})$ of (244) is completely transformed into
$G_{b}(\rho,\rho^{\prime};L_{b},E_{b})$ by the duality procedures with
$C=(m\omega/2\hbar\kappa)$. By letting $L_{b}=\ell_{b}+D/2-1$ with
$\ell_{b}\in\mathbb{N}_{0}$, we can see that the formula (221) works well for
the Coulomb–Hooke pair.

###  IV.4 A Confinement Potential as a Multi-Term Power-Law Example

One of the motivations that urged the study of power-law potentials was the
quark-antiquark confinement problem. See, for instance, references [39, 40,
36]. Here we examine a two-term power potential as a model of the confinement
potential.

Let system $A$ consist of a particle of mass $m$ confined in a two-term power
potential,

$V_{a}(r)=\lambda_{a}r^{a}+\lambda_{a^{\prime}}r^{a^{\prime}},$ (247)

where $\lambda_{a}\neq 0$, $\lambda_{a^{\prime}}\neq 0$, $a\neq a^{\prime}$,
$a\neq 0$, and $a^{\prime}\neq 0$. Let system $B$ be power-dual to system $A$
and quantum-mechanically solvable. Then we expect that some quantum-mechanical
information can be obtained concerning the confined system $A$ by analyzing
the properties of system $B$. As we have seen earlier, when system $A$ and
system $B$ are dual to each other, the shifted potential of system $A$,

$U_{a}(r)=\lambda_{a}r^{a}+\lambda_{a^{\prime}}r^{a^{\prime}}-E_{a},$ (248)

transforms to that of system $B$,

$U_{b}(\rho)=\lambda_{b}\rho^{b}+\lambda_{b^{\prime}}\rho^{b^{\prime}}-E_{b},$
(249)

by

$U_{b}(\rho)=g(\rho)U_{a}(f(\rho)).$ (250)

Here $r=f(\rho)=C\rho^{\eta}$, $g(\rho)=C^{2}\eta^{2}\rho^{2\eta-2}$,
$\eta=2/(a+2)=-b/a$, and

$b^{\prime}=2(a^{\prime}-a)/(a+2)\,\,\quad\,\,\lambda_{b^{\prime}}=\lambda_{a^{\prime}}\eta^{2}C^{a^{\prime}+2}.$
(251)

Note also that the exchange relations,

$E_{b}=-\eta^{2}C^{a+2}\lambda_{a},\,\quad\,\lambda_{b}=-\eta^{2}C^{2}E_{a},$
(252)

play an essential role in verifying the equality (250).

First, we wish to tailor the potential of system $A$ to be a confinement
potential. To this end, we set the following conditions.

(i) System $B$ behaves as a radial harmonic oscillator
$\,(\lambda_{b}=0,\,\lambda_{b^{\prime}}>0,\,b^{\prime}=2)$

(ii) System $A$ has a bound state with $E_{a}=0$ and its potential is
asymptotically linearly-increasing
$\,(\lambda_{a^{\prime}}>0,\,a^{\prime}=1)$.

Since we are unable to solve analytically the Schrödinger equation for system
$B$ with (249) in general, we consider the limiting case for which
$\lambda_{b}\rightarrow 0$, that is, we employ for the potential of system $B$

$\mathcal{U}_{b}(\rho)=\lim_{\lambda_{b}\rightarrow
0}U_{b}(\rho)=\lambda_{b^{\prime}}\rho^{b^{\prime}}-E_{b}.$ (253)

According to the second relation of (252), the limit $\lambda_{b}\rightarrow
0$ implies $E_{a}\rightarrow 0$. Hence we study only the zero-energy state of
system $A$ by assuming that it exists and is characterized by an integral
number $\nu_{0}$. We denote the zero-energy by $E_{a}(\nu_{0})$. There are
only a few exactly soluble nontrivial examples with $\mathcal{U}_{b}$ of
(253). Our choice is the one for the radial harmonic oscillator with
$b^{\prime}=2$ and $\lambda_{b^{\prime}}>0$,

$\mathcal{U}_{b}(\rho)=\lambda_{b^{\prime}}\rho^{2}-E_{b}\,\,\,(\lambda_{b^{\prime}}>0).$
(254)

Namely, we consider that system $B$ behaves as the radial harmonic oscillator
with frequency $\Omega=\sqrt{2\lambda_{b^{\prime}}/m}$ and angular momentum
$L_{b}$. Since $b^{\prime}=2$ implies $2(a^{\prime}-a)/(a+2)=2$ as obvious
from (251), the corresponding potential of system $A$ is

$V_{a}(r)=\lambda_{a}r^{(a^{\prime}-2)/2}+\lambda_{a^{\prime}}r^{a^{\prime}}.$
(255)

Next we assume that a possible confinement potential behaves asymptotically as
a linearly increasing function. Thus, letting $a^{\prime}=1$ and
$\lambda_{a^{\prime}}>0$ in (255), we have

$V_{a}(r)=\lambda_{a}r^{-1/2}+\lambda_{a^{\prime}}r,\,\,\quad\,(\lambda_{a}<0,\,\lambda_{a^{\prime}}>0).$
(256)

If $\lambda_{a}>0$, then $V_{a}(r)>0$ for all $r$, and the assumed zero-energy
state cannot exist. For $\lambda_{a}<0$, the effective potential of system
$A$,

$V_{a}^{eff}(r)=\frac{(L_{a}^{2}-\frac{1}{4})\hbar^{2}}{2mr^{2}}-|\lambda_{a}|r^{-1/2}+|\lambda_{a^{\prime}}|r,$
(257)

can accommodate the zero-energy state provided that $\lambda_{a}$ and
$\lambda_{a^{\prime}}$ are so selected that $V_{a}^{eff}(r_{1})<0$ where
$r_{1}$ is a positive root of $\textrm{d}V_{a}^{eff}(r)/\textrm{d}r=0$. Here
$L_{a}=L_{b}/\eta$ and $L_{a}=\ell_{a}+D/2-1$ with
$\ell_{a}\in\mathbb{N}_{0}$. In this manner, we are able to obtain the
confinement potential (256) which is asymptotically linearly increasing and
may accommodate at least the assumed zero-energy state. Figure 2 shows the
effective potential (257) of system $A$ for $\ell_{a}=1$, $D=3$,
$\lambda_{a^{\prime}}=1$ and $\nu_{0}=0,1,2,3,4$ in units $2m=\hbar=1$.

Figure 2: The effective potential (257) related to the eigenfunctions (266)
for $\nu_{0}=0,1,2,3,4$ from top to bottom. The parameters and units are set
to $\ell_{a}=\lambda_{a^{\prime}}=1$, $D=3$ and $2m=\hbar=1$, respectivly.

Since $a^{\prime}=1$, we have $a=(a^{\prime}-2)/2=-1/2$, $\eta=2/(a+2)=4/3$
and $b=-a\eta=2/3$. The last information concerning $b$ is unimportant insofar
as $\lambda_{b}\rightarrow 0$ is assumed. The second relation of (251) demands
that

$C=\left(9\lambda_{b^{\prime}}/16\lambda_{a^{\prime}}\right)^{1/3}.$ (258)

Therefore, the first relation of (252) yields

$E_{b}=-\frac{4}{3}\lambda_{a}\,\sqrt{\frac{\lambda_{b^{\prime}}}{\lambda_{a^{\prime}}}}.$
(259)

On the other hand, since system $B$ behaves as a radial harmonic oscillator
with frequency $\Omega=\sqrt{2\lambda_{b^{\prime}}/m}$ and angular momentum
$L_{b}$, its energy spectrum is given by

$E_{b}(\nu_{0},\ell_{b})=\hbar\Omega\,(2\nu_{0}+L_{b}+1),$ (260)

where $\nu=\nu_{0}$ is fixed by $E_{a}(\nu_{0})$ and $L_{b}=\ell_{b}+D/2-1$
with $\ell_{b}\in\mathbb{N}_{0}$. Letting $L_{b}=(4/3)L_{a}$ in (260) and
interpreting that $E_{b}$ of (259) represents an allowed value in the spectrum
(260), we observe that the coupling constant $\lambda_{a}$ may take one of the
values specified by the set of $(\nu_{0},\ell_{a})$ via

$\lambda_{a}=-\frac{3}{4}\sqrt{\frac{2\lambda_{a^{\prime}}\hbar^{2}}{m}}(2\nu_{0}+(4/3)L_{a}+1),$
(261)

where $L_{a}=\ell_{a}+D/2-1$ with $\ell_{a}\in\mathbb{N}_{0}$.

The energy eigenfunction of the radial oscillator has been given in (234).
Replacing $(m\omega/\hbar)$ in the previous result by
$\beta=m\Omega/\hbar=\sqrt{2m\lambda_{b^{\prime}}\,}/\hbar$, we write down the
eigenfunction of the present oscillator as

$\phi_{b}(\rho;L_{b},\nu_{0})=\mathcal{N}_{b}(L_{b},\nu_{0},\beta)\,{\left(\beta\rho^{2}\right)^{-1/4}}\,M_{\nu_{0}+\frac{1}{2}L_{b}+\frac{1}{2},\,\frac{1}{2}L_{b}}\left(\beta\rho^{2}\right),$
(262)

which is normalized to unity with

$\mathcal{N}_{b}(L_{b},\nu_{0},\beta)=\frac{(4\beta)^{1/4}}{\Gamma(L_{b}+1)}\,\sqrt{\frac{\Gamma(\nu_{0}+L_{b}+1)}{\Gamma(\nu_{0}+1)}}.$
(263)

Moreover, utilizing the eigenfunction just obtained, we construct the
eigenfunction for the zero-energy state in the confinement potential (256) by
following the simple prescription $\phi_{a}(r)=h(\rho)\phi_{b}(\rho)$. For the
pair $(a,b)=(-1/2,2/3)$, the two variables $r$ and $\rho$ are related by
$r=C\rho^{4/3}$ with $C$ given in (258). Since $\rho^{2}=C^{-3/2}r^{3/2}$ and
$C^{-3/2}=(4/3)\sqrt{\lambda_{a^{\prime}}/\lambda_{b^{\prime}}}$, we let

$\alpha=\frac{4}{3}\beta\sqrt{\frac{\lambda_{a^{\prime}}}{\lambda_{b^{\prime}}}}=\frac{4}{3}\frac{\sqrt{2m\lambda_{a^{\prime}}}}{\hbar},\,\,\quad\,\beta=\frac{\sqrt{2m\lambda_{b^{\prime}}}}{\hbar},$
(264)

and

$\beta\rho^{2}=\alpha r^{3/2}.$ (265)

Multiplying $\phi_{b}(\rho)$ of (262) by
$h(\rho)=\sqrt{dr/d\rho}=\sqrt{4C/3}\rho^{1/6}$, and substituting (264) and
$L_{b}=(4/3)L_{a}$ into $\phi_{b}(\rho)$, we arrive at the eigenfunction for
the zero-energy state of system $A$,

$\phi_{a}(\rho;L_{a},\nu_{0})=\mathcal{N}_{a}(L_{a},\nu_{0},\alpha)\,{\left(\alpha
r^{3/2}\right)^{-1/6}}\,M_{\nu_{0}+\frac{2}{3}L_{a}+\frac{1}{2},\,\frac{2}{3}L_{a}}\left(\alpha
r^{3/2}\right),$ (266)

where $L_{a}=\ell_{a}+D/2-1$ with $\ell_{a}\in\mathbb{N}_{0}$. Here the factor
$\mathcal{N}_{a}(L_{a},\nu_{0},\alpha)$ that normalizes $\phi_{a}(\rho)$ to
unity cannot be determined by
$\mathcal{N}_{b}((4/3)L_{a},\nu_{0},(3/4)\alpha\sqrt{\lambda_{b^{\prime}}/\lambda_{a^{\prime}}})$.
Corresponding to the value of $\lambda_{a}$ specified in (261) by the set
$(\nu_{0},\ell_{a})$, the eigenfunction $\phi_{a}(\rho;\ell_{a},\nu_{0})$ is
characterized by the same set $(\nu_{0},\ell_{a})$ of numbers.

The Green function of system $A$ obeys the inhomogeneous radial equation,

$\left\\{\frac{\textrm{d}^{2}}{\textrm{d}r^{2}}-\frac{L_{a}^{2}-1/4}{r^{2}}-\frac{2me^{2}}{\hbar^{2}}\left(\lambda_{a}r^{-1/2}+\lambda_{a^{\prime}}r\right)+\frac{2me^{2}}{\hbar^{2}}E_{a}\right\\}G_{a}(r,r^{\prime};E_{a},L_{a})=-\frac{2m}{\hbar^{2}}\delta(r-r^{\prime}).$
(267)

Since the Green function for the radial oscillator has been given in (246), we
can write down the Green function $G_{b}(\rho,\rho^{\prime};E_{b}(\nu_{0}))$
of system $B$ with $\lambda_{b}=0$ as

$G_{b}(\rho,\rho^{\prime};E_{b},L_{b})=-\frac{m}{\hbar^{2}\beta}\,\frac{1}{\sqrt{\rho\rho^{\prime}}}\frac{\Gamma(\frac{1}{2}L_{b}-k_{b}+\frac{1}{2})}{\Gamma(L_{b}+1)}\,W_{k_{b},\frac{1}{2}L_{b}}(\beta\rho_{{}_{>}}^{2})\,M_{k_{b},\frac{1}{2}L_{b}}(\beta\rho^{2}_{{}_{<}}),$
(268)

where $k_{b}=E_{b}/(2\hbar\Omega)$. The pole of
$G_{b}(\rho,\rho^{\prime};E_{b})$ that corresponds to $E_{b}(\nu_{0})$ occurs
when $k_{b}(L_{b},\nu_{0})=\nu_{0}+\frac{1}{2}L_{b}+\frac{1}{2}$ where
$\nu_{0}$ is a non-negative integer.

The Green function $G_{a}(r,r^{\prime};E_{a},L_{a})$ of system $A$ at
$E_{a}=0$ can be found by substituting (265) together with

$h(\rho)=\sqrt{4/3}C^{3/8}r^{1/8},\,\,\quad\frac{1}{2}L_{b}=\frac{2}{3}L_{a},\,\,\quad$

into $h(\rho)h(\rho^{\prime})G_{b}(\rho,\rho^{\prime};E_{b},L_{b})$. Namely,

$G_{a}(r,r^{\prime};E_{a}=0,L_{a})=\frac{4}{3}C^{3/4}(rr^{\prime})^{1/8}\,G_{b}\left((r/C)^{3/4},(r^{\prime}/C)^{3/4};E_{b}=\frac{16}{9}|\lambda_{a}|,\frac{4}{3}L_{a}\right),$
(269)

where $C$ has been given in (258). Explicitly, we have

$G_{a}(r,r^{\prime};E_{a},L_{a})=-\frac{4m}{3\hbar^{2}\alpha}\,(rr^{\prime})^{-1/4}\frac{\Gamma(\frac{2}{3}L_{a}-k_{a}+\frac{1}{2})}{\Gamma(\frac{4}{3}L_{a}+1)}\,W_{k_{a},\frac{2}{3}L_{a}}(\alpha
r_{{}_{>}}^{3/2})\,M_{k_{a},\frac{3}{2}L_{a}}(\alpha r^{3/2}_{{}_{<}}).$ (270)

where $\alpha$ and $\beta$ have been given by (264). The pole corresponding to
$E_{a}(\nu_{0})=0$ occurs when $k_{a}=\nu_{0}+(2/3)L_{a}+\,1/2$ and
$L_{a}=\ell_{a}+D/2-1$. We have to remember that the Green function (269) is
meaningful only in the vicinity of $E_{a}=0$.

Remark 25: The angular momentum $L$ in (167) is identical in form to that used
in the semiclassical case (86). However, no Langer-like ad hoc treatment has
been made in the Schrödinger equation. The angular contribution
$\ell(\ell+D-2)$ and an additional contribution $(D-1)(D-3)/4$ from the
kinetic term due to the transformation of base function, $R_{\ell}(r)$ to
$\psi_{\ell}(r)$, make up the term $L^{2}-1/4$ in the effective centrifugal
potential term of (169).

Remark 26: The time transformation $\mathfrak{T}$ needed in classical
mechanics takes no part in the power duality of the stationary Schrödinger
equation. Instead, the change of the base function plays an essential role.
While $\mathfrak{T}$ assumes $\textrm{d}t=g(\rho)\textrm{d}s$, the state
function changes as $\psi_{a}(r)=[g(\rho)]^{1/4}\psi_{b}(\rho)$. The possible
connection between the time transformation and the change of state function
has been discussed in the context of path integration for the Green function
in [41]. So long as the stationary Schrödinger equation is concerned, there is
no clue to draw any causal relation between $\mathfrak{T}$ and $\mathfrak{F}$.
However, one might expect that $\mathfrak{T}$ would play a role in the time-
dependent Schödinger equation. If the energy-coupling exchange operation
$\mathfrak{E}$ of (93) is formally modified as

$\mathfrak{E}^{\prime}:gV_{a}(r)\rightarrow-i\hbar\frac{\partial}{\partial\bar{s}},\,\,\quad\,gi\hbar\frac{\partial}{\partial\bar{t}}\rightarrow-
V_{b}(\rho),$ (271)

then the time-dependent radial Schrödinger equation,

$\left[-\frac{\hbar^{2}}{2m}\frac{\textrm{d}^{2}}{\textrm{d}r^{2}}+\frac{\hbar^{2}(L_{a}^{2}-1/4)}{2mr^{2}}+V_{a}(r)\right]\psi_{a}(r)=i\hbar\frac{\partial\psi_{a}(r)}{\partial\bar{t}},$
(272)

transforms into

$\left[-\frac{\hbar^{2}}{2m}\frac{\textrm{d}^{2}}{\textrm{d}\rho^{2}}+\frac{\hbar^{2}(L_{b}^{2}-1/4)}{2m\rho^{2}}+V_{b}(\rho)\right]\psi_{b}(\rho)=i\hbar\frac{\partial\psi_{b}(\rho)}{\partial\bar{s}},$
(273)

under the set of
$\\{\mathfrak{R},\mathfrak{L},\mathfrak{E}^{\prime},\mathfrak{F}\\}$. It is
important that $\bar{t}$ and $\bar{s}$ are not necessarily connected by
$\mathfrak{T}$; they are basically independent time-like parameters. In
conclusion, the time transformation $\mathfrak{T}$ has no role in the time-
dependent Schrödinger equation.

Remark 27: More on time transformations. Since we are dealing with the action
integral (170) rather than the Schrödinger equation, it is easy to observe
that the time transformation $\mathfrak{T}$ in the classical action in Section
II is closely related to the transformation $\mathfrak{F}$ of wave functions
in the quantum action (170). Recall that
$\mathfrak{T}:\,{(\textrm{d}t/\textrm{d}\varphi)}=g(\rho){(\textrm{d}s/\textrm{d}\varphi)}$
where $g=f^{\prime 2}$ with $f=C\rho^{\eta}$, and that

$\textrm{d}t\,U_{a}=\textrm{d}s\,gU_{a}=\textrm{d}s\,U_{b}.$ (274)

From (IV.1) and (190), we have

$\textrm{d}r\,U_{a}\psi_{a}^{\ast}\psi_{a}=\textrm{d}\rho\,f^{\prime}h^{2}U_{a}\psi_{b}^{\ast}\psi_{b}=\textrm{d}\rho\,U_{b}\psi_{b}^{\ast}\psi_{b},$
(275)

where $g=f^{\prime}h^{2}=f^{\prime 2}$. Comparing (274) and (275), we see that
$\textrm{d}t=g\textrm{d}s$ in classical mechanics corresponds to
$\textrm{d}r\,\psi_{a}^{\ast}\psi_{a}=g\,\textrm{d}\rho\,\psi_{b}^{\ast}\psi_{b}$
in quantum mechanics. In other words, $\textrm{d}r\,\psi_{a}^{\ast}\psi_{a}$
has the same transformation behavior that $\textrm{d}t$ does. In this respect,
we may say that the role of $\mathfrak{T}$ in classical mechanics is replaced
by $\mathfrak{F}$ in quantum mechanics.

##  V Summary and Outlook

In the present paper we have revisited the Newton–Hooke power-law duality and
its generalizations from the symmetry point of view.

(1) We have stipulated the power-dual symmetry in classical mechanics by form-
invariance and reciprocity of the classical action in the form of Hamilton’s
characteristic function, and clarified the roles of duality operations
$\\{\mathfrak{C},\mathfrak{R},\mathfrak{T},\mathfrak{E},\mathfrak{L}\\}$. The
exchange operation $\mathfrak{E}$ has a double role; it may decide the
constant $C$ appearing in the transformation $r=C\rho^{\eta}$, while it leads
to an energy formula that relates the new energy to the old energy.

(2) We have shown that the semiclassical action is symmetric under the set of
duality operations $\\{\mathfrak{C},\mathfrak{R},\mathfrak{E},\mathfrak{L}\\}$
without $\mathfrak{T}$ insofar as angular momentum $L$ is treated as a
continuous parameter, and observed that the power-duality is essentially a
classical notion and breaks down at the level of angular quantization. To
preserve the basic spirit of power-duality in the semiclassical action, we
have proposed an ad hoc procedure in which angular momentum transforms as
$L_{b}=\eta L_{a}$, as the classical case, rather than
$\ell_{b}=\eta\ell_{a}$; after that each of $L$ is quantized as
$L=\ell+\,D/2-1$ with $\ell\in\mathbb{N}_{0}$. As an example, we have solved
by the WKB formula a simple problem for a linear motion in a fractional power
potential.

(3) We have failed to verify the dual symmetry of the supersymmetric (SUSY)
semiclassical action for an arbitrary power potential, but have succeeded to
reveal the Coulomb–Hooke duality in the SUSY action.

(4) To study the power-dual symmetry in quantum mechanics, we have chosen the
action in which the variables are the wave function $\psi(r)$ and its complex
conjugate $\psi^{\ast}(r)$ and from which the radial Schrödinger equation can
be derived. The potential appearing in the action is a two-term power
potential. We have shown that the action is symmetric under the set of
operations $\\{\mathfrak{C},\mathfrak{R},\mathfrak{E},\mathfrak{L}\\}$ plus
the transformation of wave function $\mathfrak{F}$ provided that angular
momentum $L$ is a continuous parameter. Again the ad hoc procedure introduced
for the semiclassical case must be used in quantum mechanics. Associated with
$\mathfrak{F}$ is the transformation of Green functions from which we have
derived a formula that relates the new Green function and the old one. We have
studied the Coulomb–Hooke duality to verify the energy formula and the formula
for the Green functions. We also discussed a confinement potential and the
Coulomb–Hooke–Morse triality.

There are more topics that we considered important but left out for the future
work. They include the power-dual symmetry in the path integral formulation of
quantum mechanics, the Coulomb–Hooke duality in Dirac’s equation, and the
confinement problem in Witten’s framework of supersymetric quantum mechanics.
Feynman’s path integral is defined for the propagator (or the transition
probability) with the classical action in the form of Hamilton’s principal
function, whereas the path integral pertinent to the duality discussion is
based on the classical action in the form of Hamilton’s characteristic
function. Since the power-dual symmetry of the characteristic action has been
shown, it seems obvious that the path integral remains form-invariant under
the duality operations, but the verification of it is tedious. As is well-
known, Dirac’s equation is exactly solvable for the hydrogen atom. There are
also solutions of Dirac’s equation for the harmonic oscillator. However, the
Coulomb–Hooke duality of Dirac’s equation has never been established. The
situation is similar to Witten’s model of SUSYQM. Using the same
superpotential as that used for the semiclassical case in Section IV, we may
be able to show the Coulomb–Hooke symmetry and handle the confinement problem
in Witten’s framework.

##  Appendix A The Coulomb–Hooke–Morse Triality

In this Appendix A, we wish to present the Coulomb–Hooke–Morse triality that
relates the Morse oscillator to the Coulomb–Hooke duality. Specifically,
letting system $A$ be the hydrogen atom (for the Coulomb system), system $B$
be the radial harmonic oscillator (for the Hooke system) and system $C$ be the
Morse oscillator, we deal with their triangular relation. The Morse oscillator
is a system obeying the one-dimensional Schrödinger Equation [71],

$-\frac{\hbar^{2}}{2m}\frac{\textrm{d}^{2}\psi_{c}(\xi)}{\textrm{d}^{2}\xi}+\left(V_{c}(\xi)-E_{c}\right)\psi_{c}(\xi)=0,\,\quad\,\xi\in\mathbb{R},$
(276)

where

$V_{c}(\xi)=D_{1}\,{\rm e}^{-2\alpha\xi}-2D_{2}\,{\rm
e}^{-\alpha\xi},\,\qquad\alpha,\,D_{1},\,D_{2}>0,$ (277)

which is the Morse potential in a slightly modified form. The potential (277),
being not a power-law potential, is beyond the scope of the main text. It is
yet interesting to observe how the Morse oscillator is related to the
Coulomb–Hooke duality. It is straightforward, if one follows the general
transformation procedure [41] for the Schrödinger equation, to transform (276)
directly to the Schrödinger equation for each of the hydrogen atom and the
radial harmonic oscillation. Here, to focus our attention on their trial
nature, we place the Whittaker function at the center of the triangular
relation. In fact, the Schrödinger Equation (276) is easily transformed to the
Whittaker Equation (226) under the substitutions

$\displaystyle
x=\gamma\,\textrm{e}^{-\alpha\xi},\,\quad\gamma=\frac{\sqrt{8mD_{1}}}{\hbar\alpha}$
(278) $\displaystyle L_{c}=\frac{\sqrt{-2mE_{c}}}{\hbar\alpha},\,\quad
k_{c}=\sqrt{\frac{2mD_{2}^{2}}{\hbar^{2}\alpha^{2}D_{1}}},$ (279)
$\displaystyle w(x)=x^{1/2}\psi_{c}(\xi).$ (280)

Hence the bound state solution of (276) can be expressed in terms of the
Whittaker function as

$\psi_{c}(\xi)=\mathcal{N}_{c}\,\textrm{e}^{\alpha\xi/2}M_{k_{c},L_{c}}\left(\gamma\,\textrm{e}^{-\alpha\xi}\right),$
(281)

subject to the condition

$k_{c}=\nu+L_{c}+\frac{1}{2}\,,\,\quad\nu\in\mathbb{N}_{0}.$ (282)

The last condition yields the energy spectrum,

$E_{c}=-\frac{\hbar^{2}\alpha^{2}}{2m}\left\\{\sqrt{\frac{2mD_{2}^{2}}{\hbar^{2}\alpha^{2}D_{1}}}-\left(\nu+\frac{1}{2}\right)\right\\}^{2},\,\,\nu=0,1,2,\ldots<\sqrt{\frac{2mD_{2}^{2}}{\hbar^{2}\alpha^{2}D_{1}}}-\frac{1}{2}\,.$
(283)

The Morse oscillator solution $\psi_{c}(\xi)$ in (281) may be compared with
the Coulomb bound state solution $\psi_{a}(r)$ and the Hooke oscillator
solution $\psi_{b}(\rho)$ given, respectively, by

$\psi_{a}(r)=\mathcal{N}_{a}\,M_{k_{a},L_{a}}\left(2\kappa r\right),$ (284)

with

$k_{a}=\nu+L_{a}+\frac{1}{2}\,\,\quad\nu\in\mathbb{N}_{0},$ (285)

and

$\psi_{b}(\rho)=\mathcal{N}_{b}\,{\left(\frac{m\omega}{\hbar}\rho^{2}\right)^{-1/4}}M_{k_{b},\frac{1}{2}L_{b}}\left(\frac{m\omega}{\hbar}\rho^{2}\right),$
(286)

with

$k_{b}=\nu+\frac{1}{2}L_{b}+\frac{1}{2}\,\,\quad\nu\in\mathbb{N}_{0}.$ (287)

The bound state conditions (285) and (287) lead to the energy spectrum of the
Coulomb system ($A$) and that of the Hooke system ($B$), respectively, when

$k_{a}=me^{2}/(\hbar^{2}\kappa)\,,\quad\,\hbar\kappa=\sqrt{-2mE_{a}}\,,\quad\,L_{a}=\ell+\,1/2\,,\quad\ell\in\mathbb{N}_{0}\,,$
(288)
$k_{b}=E_{b}/(\hbar\omega)\,,\quad\,L_{b}=\ell+\,1/2\,,\quad\ell\in\mathbb{N}_{0}\,.$
(289)

The triality relations are schematically shown below,

$\begin{array}[]{ccccccc}&{\rm
Morse}&{}{}{}{}{}{}\hfil&&{}{}{}{}{}{}{}\hfil&{\rm Morse}&\\\
{}^{CA}\,\swarrow&&\nwarrow\,{}^{BC}&{}{}{}{}{}\hfil&{}^{AC}\,\nearrow&&\searrow\,{}^{CB}\\\
{\rm Coulomb}&\longrightarrow&{\rm Hooke}&{}{}{}{}{}\hfil&{\rm
Coulomb}&\longleftarrow&{\rm Hooke}\\\
{}\hfil&{}^{AB}&{}\hfil&&{}\hfil&{}^{BA}&{}\hfil\end{array}$

and the dual transformations $AC$, $CB$ and $BA$ are given by

$\begin{array}[]{lllll}AC:&2\kappa
r=\gamma\textrm{e}^{-\alpha\xi},&k_{a}=k_{c},&L_{a}=L_{c},&\psi_{a}(r)=\textrm{e}^{-\alpha\xi/2}\psi_{c}(\xi)\\\
CB:&\gamma\textrm{e}^{-\alpha\xi}=(m\omega/\hbar)\rho^{2},&k_{c}=k_{b},&L_{c}=(1/2)L_{b},&\textrm{e}^{-\alpha\xi/2}\psi_{c}(\xi)=\rho^{-1/2}\psi_{b}(\rho)\\\
BA:&(m\omega/\hbar)\rho^{2}=2\kappa
r,&k_{b}=k_{a},&(1/2)L_{b}=L_{a},&\rho^{-1/2}\psi_{b}(\rho)=\psi_{a}(r)\end{array}$

which are all invertible. Although none of the energy formulas discussed
earlier for the power-duality works when the Morse (non-power-law) potential
is involved, transforming one of the bound state conditions to another
suffices as each condition generates an energy spectrum. Let
$\chi(k_{s},\eta_{s}L_{s})$ represent the condition
$k_{s}-\eta_{s}L_{s}-\frac{1}{2}=\nu$ where $s=a,b,c$, and
$\eta_{a}=\eta_{c}=1$ and $\eta_{b}=1/2$. The map
$\,\chi(k_{s},\eta_{s}L_{s})\,\Rightarrow\,\chi(k_{s^{\prime}},\eta_{s^{\prime}}L_{s^{\prime}})\,$
induces $E_{s}\,\Rightarrow\,E_{s^{\prime}}$.

$\begin{array}[]{ccccccc}&~{}~{}~{}\chi(k_{c},L_{c})&{}{}{}{}\hfil&&{}{}{}{}{}{}{}\hfil&E_{c}&\\\
~{}~{}{}^{CA}\,\swarrow&&\nwarrow\,{}^{BC}&~{}~{}\Rightarrow{}{}&{}^{CA}\,\swarrow&&\nwarrow\,{}^{BC}\\\
\chi(k_{a},L_{a})&\longrightarrow&\chi(k_{b},\frac{1}{2}L_{b})&{}{}{}{}{}\hfil&E_{a}&\longrightarrow&E_{b}\\\
{}\hfil&{}^{AB}&{}\hfil&&{}\hfil&{}^{AB}&{}\hfil\end{array}$

Finally, it must be mentioned that this triangular relation has been discussed
in the context of so-called shape invariant potentials in supersymmetric
quantum mechanics [72]. It may also be worth pointing out that the three
systems share the $SU(1,1)$ dynamical group [50, 56].

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