Non-normalizable QM bound state in 4 spatial dimensions?

Edit 26/Sept/13: Fixed Typo in potential

I'm solving the following (seemingly simple) quantum-mechanical problem in four spatial dimensions. In natural units ($\hbar^2/2m=1$), the Schrödinger equation reads:

$$\Big[-\nabla^2-\frac{24 R^2}{(\mathbf{x}^2+R^2)^2}\Big]\psi(\mathbf{x})=E\,\psi(\mathbf{x})\,,$$

where $R>0$ is a parameter simultaneously characterizing the depth and range of the potential. The potential depends only on the distance away from the origin, so I can separate variables $\psi=R_{nl}(r)\,Y_l(\vec{\theta})$ and the radial equation then reads:

$$\Big[-\frac{\partial^2}{\partial r^2}-\frac{3}{r}\frac{\partial}{\partial r}+\frac{l(l+2)}{r^2}-\frac{24 R^2}{(r^2+R^2)^2}\Big]\,R_{nl}(r)=E_{nl}\,R_{nl}(r)\,.$$

Problem: I seem to have found an s-wave ($l=0$) non-scattering state with zero energy $E_{nl}=0$ that appears to be localized:

$$R_{n,l=0}(r)=\mathcal{N}\frac{r^2-R^2}{(r^2+R^2)^2}\qquad E_{nl}=0.$$

But, I am unable to normalize this "bound" state since the integral $\int_0^\infty dr\, r^3 |R(r)|^2$ does not converge. What is the nature of this state? Or am I just totally screwing something up?

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After the editing the potential is a bounded operator so the Hamiltonian is well-defined and bounded from below.

It seems to me that $E_{nl}=0$ is simply not a legitimate discrete eigenvalue, the associated eigenfunction not being square integrable.
It may well be that the potential does not support any bound states at all.

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Thank you for the answer. If it is not a discrete eigenvalue, then is it a scattering state? – QuantumDot Oct 23 '13 at 14:13

I) Comments to the question (v1) with potential $V(r)=-\frac{24 R^2}{(r^2{\color{Red}- }R^2)^2}$ before OP's correction:

There is a bigger problem with OP's model: We claim that the Hamiltonian $H$ is unbounded from below. The idea of the proof is similar to this Phys.SE answer: It is possible to find a one-parameter family of normalized trial wavefunctions $\psi_{\epsilon}(r)$, which are localized very near the 3-sphere $r=R$, in such a manner, that the positive kinetic energy remains bounded, but the negative potential energy diverges to $-\infty$, as $\epsilon\to 0^{+}$, cf. the variational method.

II) Comments to the question (v2) with potential $V(r)=-\frac{24 R^2}{(r^2{\color{Red}+}R^2)^2}$ after OP's correction:

The potential $V(r)$ is now bounded

$$-\frac{24}{R^2}~\leq~ V(r) ~< ~0,$$

and since $-\nabla^2\geq 0$ is semipositive, the Hamiltonian

$$H~=~-\nabla^2+V~\geq ~ -\frac{24}{R^2}$$

is now bounded from below. Indeed OP's solution satisfies the TISE with $E=0$ and $\ell=0$, and indeed it is not normalizable, as OP claims, so it does not belong to the Hilbert space $H=L^2(\mathbb{R}^4)$ of square integrable wavefunctions.

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Uh-oh! I made a typo in my original question: the denominator in the potential should read $(r^2+R^2)$, so that the binding region is near the origin. What is the protocol now that you have given an answer to a question with a typo? – QuantumDot Sep 26 '13 at 7:28
I updated the answer. – Qmechanic Sep 26 '13 at 20:19
Thank you for the answer. If it is not a discrete eigenvalue, then is it a scattering state? – QuantumDot Oct 23 '13 at 14:14

The spectrum of a self-adjoint operator acting in a Hilbert space can be decomposed into three types, discrete eigenvalues associated with square integrable eigenvectors, absolutely continuous spectrum and singular continuous spectrum. The discrete spectrum can be continuum-embedded but usually is not and singular continuous spectrum is mainly restricted to random systems. In the case at hand 0 is part of the continuous spectrum. One can define it to be a scattering state but there are plenty of situations where the continuous spectrum has nothing to do with scattering.

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