\phi(\rho) - 4 \alpha \phi(\rho)=0~,$$ if $r=\rho^2$, $u(r)=\rho^{1/2}\phi(\rho)$, and $L=2\ell+1/2$. The modified angular momentum can be interpreted as relevant in a higher-dimensional world [@Mavromatis:1998]. But Eq. (\[eq:KS2\]) is precisely the Schr[" o]{}dinger equation for the three-dimensional oscillator with (fixed) energy $4\alpha$ and oscillator strength $4(-E)$ (which is positive), i.e., $$\label{eq:KS4}
4\alpha =\sqrt{-4E}(3+4n+2 L)~,\quad n=0,\,1, \ldots~,$$ which is equivalent to the Bohr formula $$\label{eq:KS5}
E=-{\alpha^2\over 4(1+n+\ell)^2}~,$$ where $1+n+\ell$ is the usual principal quantum number of atomic physics.

Now, an additional potential $\lambda v(r)$ in the Coulomb equation results into a short-range term $4 \lambda \rho^2v(\rho^2)$ added to the harmonic oscillator, and all results obtained for exotic atoms translate into the properties listed for a narrow hole added to an harmonic well.

Note that the $n$ dependence is also explained. In the KS transformation, the energy $E$ ($E<0$) in the Coulomb system becomes the strength $-4E$ of the oscillator, while four times the fine structure constant, i.e., $4\alpha$ ($\alpha>0$ for attraction) becomes the energy eigenvalue of the oscillator with angular momentum $L$. If $n$ increases, the oscillator deduced from the KS transformation becomes looser, and hence less sensitive to the short range attraction $4\lambda\rho^2v(\rho^2)$. To maintain a fixed oscillator strength, one should 