Appendix. Appendix A
The radial equation becomes
$$\begin{aligned}{} & {} \frac{\hbar ^2}{2m_0} \left[ \frac{\textrm{d}^2}{\textrm{d}r^2}+\frac{2}{r}\frac{\textrm{d}}{\textrm{d}r}\right] R(r) \nonumber \\{} & {} \quad +\left[ E -\frac{e^4}{4(l+1)^2} -V(r) \right] R(r) = 0, \end{aligned}$$
(A.1)
where
$$\begin{aligned} V(r)= -\frac{e^2}{r} + \frac{\hbar ^2 l(l+1)}{2m_0 r^2}. \end{aligned}$$
(A.2)
The radial coordinate r ranges from 0 to \(\infty \). Now putting
$$\begin{aligned} E=\frac{e^4}{4(l + 1)^2} - \frac{e^4}{4(n + l+ 1)^2} \end{aligned}$$
in (A.1) one gets
$$\begin{aligned}{} & {} -\left[ \frac{\textrm{d}^2}{\textrm{d}r^2}+\frac{2}{r}\frac{\textrm{d}}{\textrm{d}r}\right] R(r)\nonumber \\{} & {} \quad +\frac{2m_0}{\hbar ^2}\left[ \frac{e^4}{4(n + l+ 1)^2} +V(r) \right] R(r) = 0 . \end{aligned}$$
(A.3)
Let
$$\begin{aligned}{} & {} \alpha ^2 = \frac{2m_0}{\hbar ^2}\frac{e^4}{(n + l + 1)^2}-\left[ \frac{\textrm{d}^2}{\textrm{d}r^2}+\frac{2}{r}\frac{\textrm{d}}{\textrm{d}r}\right] R(r)\nonumber \\{} & {} \quad + \left[ \frac{\alpha ^2}{4} -\frac{2m_0}{\hbar ^2}\frac{e^2}{r} + \frac{ l(l+1)}{r^2} \right] R(r) = 0 \end{aligned}$$
(A.4)
or
$$\begin{aligned}{} & {} -\left[ \frac{1}{\alpha ^2}\frac{\textrm{d}^2}{\textrm{d}r^2}+\frac{1}{\alpha }\frac{2}{\alpha r}\frac{\textrm{d}}{\textrm{d}r}\right] R(r) \nonumber \\{} & {} \quad + \left[ \frac{1}{4} -\frac{2m_0}{\hbar ^2}\frac{e^2}{\alpha ^2r} + \frac{ l(l+1)}{\alpha ^2r^2} \right] R(r) = 0 . \end{aligned}$$
(A.5)
On the change of variable \(y=\alpha r\),
$$\begin{aligned}{} & {} \frac{\textrm{d}R}{\textrm{d}r}=\frac{\textrm{d}y}{\textrm{d}r}\frac{\textrm{d}R}{\textrm{d}y} =\alpha \frac{\textrm{d}R}{\textrm{d}y} \end{aligned}$$
(A.6)
$$\begin{aligned}{} & {} \frac{\textrm{d}^2R}{\textrm{d}r^2}=\frac{\textrm{d}}{\textrm{d}r}\frac{\textrm{d}R}{\textrm{d}y} =\alpha \frac{\textrm{d}R}{\textrm{d}y}\left( \alpha \frac{\textrm{d}R}{\textrm{d}y}\right) =\alpha ^2\frac{\textrm{d}^2R}{\textrm{d}y^2}. \end{aligned}$$
(A.7)
Taking
$$\begin{aligned} \lambda =\frac{2m_0}{\hbar ^2}\frac{1}{\alpha }, \end{aligned}$$
we obtain
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2}+\frac{2}{y}\frac{\textrm{d}}{\textrm{d}y}\right] R(y)\nonumber \\{} & {} \quad + \left[ \frac{\lambda }{y}-\frac{1}{4} - \frac{ l(l+1)}{y^2} \right] R(y) = 0 . \end{aligned}$$
(A.8)
Let the solution to this equation be
$$\begin{aligned} R(y)=y^{l} \exp \left( -\frac{y}{2}\right) L_n^{2l+1} (y) \end{aligned}$$
(A.9)
so that, we obtain the following equation
$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}y}R(y)= & {} \left[ \frac{l}{y}-\frac{1}{2}\right] y^{l} \exp \left( -\frac{y}{2}\right) L_n^{2l+1} (y) \nonumber \\{} & {} + y^{l} \exp \left( -\frac{y}{2}\right) \frac{\textrm{d}}{\textrm{d}y} L_n^{2l+1} (y) \end{aligned}$$
(A.10)
$$\begin{aligned} \frac{\textrm{d}^2}{\textrm{d}y^2}R(y)= & {} \left[ \left[ \frac{l}{y}-\frac{1}{2}\right] ^2-\frac{l}{y^2}\right] y^{l} \exp \left( -\frac{y}{2}\right) L_n^{2l+1} (y)\nonumber \\{} & {} +2\left[ \frac{l}{y}-\frac{1}{2}\right] y^{l} \exp \left( -\frac{y}{2}\right) \frac{\textrm{d}}{\textrm{d}y} L_n^{2l+1} (y)\nonumber \\{} & {} +y^{l} \exp \left( -\frac{y}{2}\right) \frac{\textrm{d}^2}{\textrm{d}y^2} L_n^{2l+1} (y). \end{aligned}$$
(A.11)
After performing a series of calculations
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2} + \left[ 2\left[ \frac{l}{y}-\frac{1}{2}\right] +\frac{2}{y}\right] \frac{\textrm{d}}{\textrm{d}y} \right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad +\left[ \left[ \frac{l}{y}-\frac{1}{2}\right] ^2-\frac{l}{y^2}+\frac{2}{y}\left[ \frac{l}{y}-\frac{1}{2}\right] \right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad +\left[ \frac{\lambda }{y}-\frac{1}{4} - \frac{ l(l+1)}{y^2} \right] L_n^{2l+1} (y) = 0. \end{aligned}$$
(A.12)
On simplification
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2} +\frac{1}{y} \left[ 2l+2-y\right] \frac{\textrm{d}}{\textrm{d}y} \right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad +\left[ \frac{l^2}{y^2}+\frac{1}{4}-\frac{l}{y}-\frac{l}{y^2}+\frac{2l}{y^2}\right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad -\left[ \frac{l}{y}+\frac{\lambda }{y}-\frac{1}{4} - \frac{ l(l+1)}{y^2} \right] L_n^{2l+1} (y) = 0 \end{aligned}$$
(A.13)
which reduces to
$$\begin{aligned}{} & {} \left[ y\frac{\textrm{d}^2}{\textrm{d}y^2} + (2l+2-y) \frac{\textrm{d}}{\textrm{d}y} +(\lambda -l-1)\right] \nonumber \\{} & {} \quad \times L_n^{2l+1} (y) = 0. \end{aligned}$$
(A.14)
To verify the supersymmetric partner, we modify eq. (A.8) to
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2}+\frac{2}{y}\frac{\textrm{d}}{\textrm{d}y}\right] R(y) + \left[ \frac{\lambda }{y}-\frac{1}{4}-\frac{V(y)}{y} - \frac{ l(l+1)}{y^2} \right] \nonumber \\{} & {} \quad \times R(y) = 0 . \end{aligned}$$
(A.15)
Let the solution to this equation be
$$\begin{aligned} R(y)=\frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)} L_n^{2l+1} (y), \end{aligned}$$
(A.16)
where \(k=2l+1\). We obtain the following equation
$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}y}R(y)= & {} \left[ \frac{l}{y}-\frac{1}{2}-\frac{1}{(y+k)}\right] \frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)} L_n^{2l+1} (y)\nonumber \\{} & {} +\frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)}\frac{\textrm{d}}{\textrm{d}y} L_n^{2l+1} (y) \end{aligned}$$
(A.17)
$$\begin{aligned} \frac{\textrm{d}^2}{\textrm{d}y^2}R(y)= & {} \left[ \left[ \frac{l}{y}-\frac{1}{2}-\frac{1}{(y+k)}\right] ^2\right] \frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)}\nonumber \\{} & {} -\left[ \frac{l}{y^2}+\frac{1}{(y+k)^2}\right] \frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)}\nonumber \\{} & {} +2\left[ \frac{l}{y}-\frac{1}{2}-\frac{1}{(y+k)}\right] \frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)}\nonumber \\{} & {} \times \frac{\textrm{d}}{\textrm{d}y} L_n^{2l+1} (y)\nonumber \\{} & {} +\frac{y^{l} \exp ( -\frac{y}{2})}{(y+k)}\frac{\textrm{d}^2}{\textrm{d}y^2} L_n^{2l+1} (y). \end{aligned}$$
(A.18)
One can see that the only extra contribution comes from \({1}/{(y+k)}\). If this term is not present, it will revert back to the associated Lagrange equation. After performing a series of calculations
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2} +\bigg (\frac{k+1}{y} -1-\frac{2}{(y+k)}\bigg ) \frac{\textrm{d}}{\textrm{d}y}\right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad + \left[ \frac{1}{y}(\lambda -l-1)-\frac{V(y)}{y}+\frac{2}{(y+k)^2}\right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad - \left[ 2\left[ \frac{l}{y}-\frac{1}{2}\right] \frac{1}{(y+k)}-\frac{2}{y(y+k)}\right] \nonumber \\{} & {} \quad \times L_n^{2l+1} (y) = 0 \end{aligned}$$
(A.19)
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2} +\bigg (\frac{(k+1)(y+k)-y(y+k)-2y}{y(y+k)}\bigg ) \frac{\textrm{d}}{\textrm{d}y}\right] \nonumber \\{} & {} \quad \times L_n^{2l+1} (y)\nonumber \\{} & {} \quad + \frac{1}{y}\left[ (\lambda -l-1)-V(y)+\frac{2y}{(y+k)^2}\right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad -\frac{1}{y}\left[ \frac{2l}{(y+k)}+\frac{y}{(y+k)}-\frac{2}{(y+k)}\right] L_n^{2l+1} (y) = 0 \end{aligned}$$
(A.20)
$$\begin{aligned}{} & {} \left[ \frac{\textrm{d}^2}{\textrm{d}y^2} +\bigg (\frac{ky+k+y+k^2-y^2+yk-2y}{y(y+k)}\bigg ) \frac{\textrm{d}}{\textrm{d}y}\right] \nonumber \\{} & {} \quad \times L_n^{2l+1} (y)\nonumber \\{} & {} \quad + \frac{1}{y}\left[ (\lambda -l-1)+\mathcal {V}(y)\right] L_n^{2l+1} (y) = 0 \end{aligned}$$
(A.21)
which reduce to
$$\begin{aligned}{} & {} \left[ y \frac{\textrm{d}^2}{\textrm{d}y^2} +\bigg (\frac{(y-k)(k+y+1)}{(y+k)}\bigg ) \frac{\textrm{d}}{\textrm{d}y}\right] L_n^{2l+1} (y)\nonumber \\{} & {} \quad +\left[ (\lambda -l-1)+\mathcal {V}(y)\right] L_n^{2l+1} (y) = 0, \end{aligned}$$
(A.22)
where
$$\begin{aligned} \mathcal {V}(y)=-V(y)+\frac{2y}{(y+k)^2} -\frac{1}{(y+k)}+\frac{y-k}{(y+k)}.\nonumber \\ \end{aligned}$$
(A.23)
Since the eigenvalues are the same, the last term is equal to
$$\begin{aligned}{} & {} -V(y)+\frac{2y}{(y+k)^2}-\frac{1}{(y+k)}+\frac{y-k}{(y+k)} = \frac{(y-k)}{(y+k)} \end{aligned}$$
(A.24)
$$\begin{aligned}{} & {} V(y)=\frac{2y}{(y+k)^2}-\frac{1}{(y+k)} \end{aligned}$$
(A.25)
It should be clear from eq. (A.15) we have added
$$\begin{aligned} \frac{V(y)}{y}=\frac{2}{(y+k)^2}-\frac{1}{y(y+k)}. \end{aligned}$$
(A.26)
