3D Delta Potential Well The 1D delta potential well $V(x) = -A\delta(x - a)$ always has exactly one bound state. The same is true for the 3D delta potential well $V(\vec{r}) = -A\delta(\vec{r}-\vec{a})$. I can show this for $\ell = 0$, I don't know how to do the calculations otherwise.
So two questions,


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*Can I conclude that there is only one bound state for the 3D potential well for $\ell \not = 0$? I've seen that the energies of the eigenstates for the hydrogen atom depend only on $n$, but I am wondering whether this is an instance of a more general result?

*When $\vec{a} = 0$ and $\ell=0$, there are no normalizable eigenstates. For $\ell \not = 0$, the effective potential in the radial equation becomes large at the origin, can I use this to conclude that there are no bound states when $\vec{a}=0$?
 A: Since the energy spectrum does not depend on the absolute position $\vec{r}=\vec{a}$ of the delta potential, we may assume that $\vec{a}=\vec{0}$. Therefore, in its current formulation (v1), OP is effectively saying that 

The attractive 1D delta potential $V(x) = -A\delta(x)$, $A>0$, has exactly one bound state. The same is true for the 3D delta potential $V(\vec{r}) = -A\delta^3(\vec{r})$. 

No, the bare 3D delta potential does not constitute a well-posed mathematical problem without some kind of regularization/renormalization, see e.g. Ref. 1 and Ref. 2. The bare spectrum has infinitely many bound states, and it is not bounded from below. 
The latter can be rigorously proven via e.g. the variational method. Proof: Consider a normalized Gaussian test/trial wavefunction 
$$\psi(r)~=~Ne^{-\frac{r^2}{2L^2}}~=~Ne^{-\frac{x^2+y^2+z^2}{2L^2}}, \qquad \int d^3r~|\psi(r)|^2 ~=~\langle\psi|\psi \rangle~=~1,$$
where $N,L>0$ are two constants. For dimensional reasons, the constant $L$ must have dimension of length, and $1/N^2$ must have dimension of volume. It follows that 


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*The normalization constant $N$ must scale as
$$N ~\propto~ L^{-\frac{3}{2}}.$$

*The expectation value $\langle\psi| K|\psi \rangle$ of the kinetic energy operator $K=-\frac{\hbar^2}{2m}\Delta$ must scale as
$$0~\leq~\langle\psi| K|\psi \rangle ~\propto~ L^{-2},$$
essentially because the Laplacian $\Delta=\vec{\nabla}^2$ contains two position derivatives.

*The expectation value $\langle\psi| V|\psi \rangle$ of the potential energy $V=-A\delta^3(\vec{r})$ must scale as
$$0~\geq~\langle\psi|V|\psi \rangle~=~-AN^2~\propto~ -L^{-3}.$$
Thus by choosing $L\to 0^{+}$ smaller and smaller, the negative potential energy $\langle\psi| V|\psi \rangle\leq 0$ beats the positive kinetic energy $\langle\psi| K|\psi \rangle\geq 0$, so that the average energy $\langle\psi| H|\psi \rangle$ becomes more and more negative,
$$ \langle\psi| H|\psi \rangle ~=~\langle\psi| K|\psi \rangle + \langle\psi| V|\psi \rangle ~\to~ -\infty \qquad \text{for}\qquad L\to 0^{+}. $$
Hence, the spectrum is unbounded from below. 
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References:


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*S. Geltman, Bound States in Delta Function Potentials, Journal of Atomic, Molecular, and Optical Physics, Volume 2011, Article ID 573179.

*R.J. Henderson and S.G. Rajeev, Renormalized Path Integral in Quantum Mechanics, arXiv:hep-th/9609109.
