very arge enough $n$, the first nodes of the radial function come in the range of $v(r)$, and then $\delta E$ decreases with $n$. Moreover, for very large $n$, the radial Schr[ö]{}dinger equation is dominated at short distance by the energy term.

-   For a Coulomb interaction, $|\phi_n(0)|^2\propto n^{-3}$, and hence $\delta E_n\propto n^{-3}$, a well-known property of exotic atoms. As explained, e.g., in a review article on protonium [@Klempt:2002ap] and briefly explained in Appendix, the first node of the $n$S radial function, as $n$ increases, does not go to $0$. In the case $\hbar^2/(2\mu)=e^2=1$, the node of the 2S level is at $r=4$, while the first node of $n$S at large $n$ is at $r\simeq 3.67$. Hence the Coulomb wave function never exhibits nodes within the range of the nuclear potential. Moreover, the energy term is always negligible in comparison with $\lambda v(r)$ at short distances.

From Coulomb to harmonic rearrangement
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The KS transformation [@Mavromatis:1998] relates Coulomb and harmonic-oscillator potentials. The radial equation for a Coulomb system in three dimension (with $\hbar=2\mu=1$) reads $$\label{eq:KS1}
-u''(r)+{\ell(\ell+1)\over r^2} u(r)-{\alpha\over r} u(r) - E u(r)=0~,$$ with $u(0)=0$ and $u(r)\to 0$ as $r\to \infty$ becomes $$\label{eq:KS2}
-\phi''(\rho)+{L(L+1)\over \rho^2}\phi(\rho) + 4 (-E) \rho^2 