The number of eigenstates of a finite circular well in 2D

Question

Consider the following potential in two dimensions: $$ V(r,\theta) = -V_0\,\theta(r_0-r), $$ i.e., $V(r,\theta) = 0$ if $r>r_0$, with a circular well of radius $r_0$ and depth $V_0$ around the origin. (For convenience, write $V_0 = \hbar^2 k_0^2/2m$.)

I have a couple of questions to which I would like rigorous and well-referenced answers:

1. Presumably there is always a ground state, even in the limit where $0<k_0 r_0 \ll 1$ and the well is very shallow and/or very wide. Is there a rigorous theorem that guarantees it? Is there a rigorous estimate for the ground-state energy in this limit? (I have code that solves for the eigen-energies using Bessel functions, but below a certain threshold around $k_0r_0 \approx 0.6$ the root is too close to the continuum for my root finder to locate it.)

2. As we walk away from the long-and-shallow limit and $k_0r_0$ increases to the region of, say, 10 or 20, several excited states will enter the bound-state region at specific threshold values of the well-shape parameter $k_0 r_0$. Is there a formal name for these events? For the specific case of a circular well, are the specific values of $k_0r_0$ for the transitions known rigorously? (A rough look indicates that they happen at around $(n+\tfrac14)\pi$ for $n=0,1,2,\ldots$, but not exactly there.)

More generally, the eigenstates happen at energies $E = \hbar^2 k^2/2m$ for wavenumbers that satisfy $$ k J_\ell'(kr_0)K_\ell(\kappa r_0) - \kappa J_\ell(kr_0)K_\ell'(\kappa r_0) = 0, $$ for $\kappa^2 = k_0^2 - k^2$. This equation is very close to that of the so-called cross-product Bessel zeros, but I don't find it in that class in the DLMF. Is there a formal name for these objects that makes them more discoverable in the literature? Are their properties (e.g. asymptotics, or rough estimates that can be used as seeds for a root finder) known and explored somewhere? I would expect that the problem is basic enough that it's been done to death, but I'm surprised at how tricky it is to find resources for it.

Answer 1

Sketch of a proof for your first question:

Let $\mathcal{H}=L^2(\mathbb{R}^n)$ be our Hilbert space, $V:\mathbb{R}^n\to \mathbb{R}$ be a measurable and bounded function ($V\in L^{\infty}(\mathbb{R}^n)$) and the Hamiltonian $\hat{H}= -\frac{\hbar^2}{2m}\triangle+\hat{V}$ where $\hat{V}$ is the operator related to multiplication by $V(x)$ on $\mathcal{H}$ as usual.

Now by Kato- Rellich, on the domain of the Lapacian ($D(\triangle)$) $\hat{H}$ is self adjoint. Note that: $$D(\triangle)= \left\{\phi\in\mathcal{H}~|~ \mathcal{F}^{-1}(|k|^2\mathcal{F}(\phi))\in\mathcal{H} \right\}= \left\{\phi\in\mathcal{H}~|~ |\left\langle \phi,\phi''\right\rangle|<\infty \right\}$$ The second definition is probably the more familiar one, but using the unitarity of the Fourier transform, $\mathcal{F}$, they are indeed equivalent,looking at the first definition it is clear that the spectrum of the Laplacian is negative (left to show it is indeed self adjoint on this Domain)

Now since $V$ is bounded, $\exists c>0:~||\hat{V}||_{op}=||V ||_{\infty}<c$ and we can calculate $\forall \psi\in D(\triangle)$

\begin{align} \left\langle \psi,\hat{H}\psi\right\rangle&= -\frac{\hbar^2}{2m}\left\langle \psi,\triangle \psi\right\rangle +\left\langle \psi,\hat{V} \psi\right\rangle\geq \left\langle \psi,\hat{V} \psi\right\rangle \end{align}

In your case since you consider $V$ compactly supported, and is constant when it's none zero,then this is where it ends. Either way this integral converges under our assumptions (V is bounded, i.e. $D(\hat{V})=\mathcal{H}$)

Thus the spectrum of $\hat{H}$ is bounded from below, hence there is a ground state.

For the rest of your questions, your guess for the transitions look reasonable (that if I understood correctly- excited states become bounded) but I can't give an actual answer, I would suggest to solve the problem at least symbolically, with boundary conditions and conditioning that it's a bounded state for fixed values of well's parameters I don't think that would take too long and my guess is that it will give you some self consistency equation that you can check numerically.

Answer 2

In rough logical order:

• There is indeed a rigorous theorem that guarantees the existence of a bound ground state for wells in 1D and 2D; for a proof (and an explanation of why the result fails in 3D) see the thread Why the statement "there exist at least one bound state for negative/attractive potential" doesn't hold for 3D case?, linked in the comments above.

• A suitable asymptotic in the $\kappa\to 0$ limit to the solutions of $$ k J_\ell'(kr_0)K_\ell(\kappa r_0) - \kappa J_\ell(kr_0)K_\ell'(\kappa r_0) = 0, $$ is the asymptotic $$ k \approx k_0 \left( 1 - \frac{2e^{-2\gamma}}{(k_0r_0)^2} \exp\mathopen{}\left[ -2\frac{J_0(k_0r_0)}{k_0r_0 J_1(k_0r_0)} \right] \right) , $$ where $\gamma$ is the Euler-Mascheroni constant. This holds for the ground state, where it is useful in the region $0<k_0r_0<\pi/4$, and it also holds for the excited states, where it is useful for $k_0r_0$ just above threshold.

• A good name for the events where eigenstates reach the continuum and disappear (or possibly go off somewhere?) is that they "reach threshold". This is particularly useful in scattering theory, where those events coincide in singularities of the scattering length, which goes off to $-\infty$ and returns from $+\infty$ at the threshold transition.

• For this specific system, the threshold events happen at the zeros of $J_1(k_0r_0) \propto J_0'(k_0r_0)$, which coincides with the energies at which the eigenfunction is flat at $r_0$, thus continuing off into a roughly-constant function. The zeros of $J_1$, often denoted $j_{1,n}$ are slightly below the stated asymptote, $(n+\frac14)\pi$.

• Somewhat surprisingly, as $k_0r_0$ approaches a continuum threshold transition, the energy of the bound state which is about to disappear becomes (super)exponentially close to zero, a behaviour which comes from the term of the form $$\exp(-\mathrm{const}/J_1(k_0r_0)) \approx \exp\mathopen{}\left(\frac{\pm\mathrm{const}}{k_0r_0-j_{1,n}}\right).$$ I suspect this has physical significance but I'm not sure what the implications or connections are.