3D Delta Potential Well

Question

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,

1. 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?

2. 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 $\vec{a}=0$?

Answer 1

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.

Theorem. The attractive delta potential has infinitely many bound states, and the spectrum is not bounded from below, if the spatial dimension $\color{Red}{n\geq 2}$.

Simpler proof for $\color{Red}{n>2}$: The unboundedness from below is rigorously proven via the variational method. Consider a normalized Gaussian test/trial wavefunction

$$\psi(r)~=~Ne^{-\frac{r^2}{2L^2}}, \qquad \int d^nr~|\psi(r)|^2 ~=~\langle\psi|\psi \rangle~=~1,$$

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 $(\text{length})^n$. It follows that

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

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

3. The expectation value $\langle\psi| V|\psi \rangle$ of the potential energy $V=-A\delta^3(\vec{r})$ must scale as $$0~\geq~\langle\psi|V|\psi \rangle~=~-AN^2~\propto~ -L^{-\color{Red}{n}}.$$

Now assume that $\color{Red}{n>2}.$ 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. (This proof also shows the $\color{Red}{n=2}$ case for a sufficiently large $A$.) $\Box$

Proof for $\color{Red}{n=2}$: If there exists a normalized energy eigenfunction $\vec{r}\mapsto \psi_1(\vec{r})$ to the TISE $H\psi_1=E_1\psi_1$ with negative energy $E_1<0$, then it is easy to see by a scaling argument that there exists a whole 1-parameter family

$$\begin{align} \psi_{\lambda}(\vec{r})~=~&\lambda\psi_1(\lambda\vec{r}),\qquad H\psi_{\lambda}~=~E_{\lambda}\psi_{\lambda}, \cr E_{\lambda}~=~&\lambda^2E_1~<~0, \qquad \lambda~>~0, \end{align}$$

of normalized energy eigenfunctions. Together with theorem 1 from my Phys.SE answer here, we conclude that the spectrum is unbounded from below. $\Box$

References:

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

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