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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 $l = 0$, 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 $l \not = 0$? 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 $l=0$, there are no normalizable eigenstates. For $l \not = 0$, 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 $\vec{a}=0$?

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1 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. 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

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

  2. The expectation value $\langle\psi| K|\psi \rangle$ of the kinetic energy operator $K=-\frac{\hbar^2}{2m}\Delta$ must scale as $$0~\leq~\langle\psi| K|\psi \rangle ~\propto~ L^{-2},$$ 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 $$0~\geq~\langle\psi|V|\psi \rangle~=~-AN^2~\propto~ -L^{-3}.$$

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.



  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.

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+1 as one of those invaluable bits of crucial physics one can get simple dimensional analysis, in the same box as the period of a pendulum scaling as $1/\sqrt{\textrm{length}}$ (albeit a somewhat more sophisticated result). – Emilio Pisanty May 7 '12 at 12:02
@Qmechanic: +1, but one should ask about the proper limit, the one that doesn't blow up. – Ron Maimon May 7 '12 at 18:32
Note for later: Theorem. A sufficient condition for the existence of at least one bound state with energy $E<0$ of an attractive potential $V\leq 0$ in $n\leq 2$ spatial dimensions is that $\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)<0,$ cf. this Phys.SE post. – Qmechanic Aug 5 at 10:18
Proof for $n<2$: Consider an un-normalized (but normalizable) Gaussian test/trial wavefunction $\psi(x)=e^{-\frac{x^2}{2L^2}}$, $L>0$. Normalization must scale as $||\psi|| \propto L^{\frac{n}{2}}$. The normalized kinetic energy scale as $0\leq\frac{\langle\psi| K|\psi \rangle}{||\psi||^2} \propto L^{-2}$, and vanishes for $L\to \infty$. The un-normalized potential energy tends to a negative constant $\langle\psi| V|\psi \rangle \searrow\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)$ for $L\to \infty$. – Qmechanic Aug 5 at 10:35
This means that $\exists L_0>0 \forall L\geq L_0:~~ \langle\psi| V|\psi\rangle \leq \frac{1}{2}\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)<0.$ It follows that the average energy $\frac{\langle\psi|H|\psi\rangle}{||\psi||^2}\leq \frac{\langle\psi|K|\psi\rangle+\frac{1}{2}\int_{\mathbb{R}^n} \! \mathrm{d}^nx ~V(x)}{||\psi||^2}<0$ of trial function must be negative for a sufficiently big finite $L\geq L_0$ if $n<2$. Hence the ground state energy must be negative (possibly $-\infty$). $\Box$ – Qmechanic Aug 5 at 10:44

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