What are the exact relations between bound states, discrete spectra, and negative energies in quantum mechanics?

Question

Consider the nonrelativistic quantum mechanics of one particle in one dimension ("NRQMOPOD") with the time-independent Schrodinger equation

$$ \left( -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E\ \psi(x), $$ where we assume that the potential energy function $V(x)$ is continuous and at large distances approaches a constant value $$ V_\infty = \lim_{|x| \to \infty} V(x) $$ in the extended real line (which is often set to 0 if $V_\infty$ is finite). We do not assume that the eigenfunction $\psi(x)$ is necessarily normalizable.

"Folk wisdom" says that the following three statements are equivalent:

1. $\psi(x)$ is normalizable, i.e. $$\int \limits_{-\infty}^\infty dx\ |\psi(x)|^2 < \infty.$$ Or equivalently, $\psi(x)$ represents a bound state, i.e. $$\lim_{R \to \infty} \int \limits_{|x| > R} dx\ |\psi(x)|^2 = 0.$$ (The equivalence of these two statements is a straightforward exercise in real analysis.)
2. $E < V_\infty$, and
3. $E$ lies in a discrete part of the energy spectrum, i.e. there exists a proper energy interval such that $E$ is the only eigenvalue in the Hamiltonian's spectrum that lies within that interval.

But this folk wisdom is incorrect. This answer gives an explicit example of a potential energy function $V(x)$ and a normalizable energy eigenstate $\psi(x)$ with energy $E > V_\infty$. Therefore, statement #1 above does not imply statement #2. (I do not know whether the spectrum for the particular Hamiltonian given in that example is discrete or continuous around the relevant eigenvalue $\lambda = 1$, so I don't know the status of claim #3 for this example.)

What are the exact implications between the three statements above? Of the six possible implications, which have been proven to be true, which have explicit known counterexamples, and which are still open problems?

I'd also like to know about the case of multiple spatial dimensions, although I assume that the answers are probably the same as for the 1D case.

Answer 1

Part of the answer is in Reed and Simon but it is somewhat technical (it is, after all, Reed and Simon). Basically your intuition is correct as long as the potential $V(x)$ is sufficiently nice. I believe that it suffices that $\int (1+|x|) V(x) dx$ be finite. If this is the case then the operator $$ -\frac{d^2\psi}{dx^2} + V(x) \psi = E \psi $$
has

• A finite number of bound states $\psi_{E_i}(x)$ with $E<0$ which belong to $L^2(\mathbb{R})$
• no "imbedded" eigenvalues in the essential spectrum $E>0$

In the borderline case where $V(x)$ decays like $x^{-2}$ there is a surprisingly sharp result: if $V(x)>-\frac{1}{4 x^2}$ for large $x$ then you still have a finite number of eigenvalues. The proof of this is actually quite elementary. I think that if $V(x)< -\frac{C}{x^2}$ for $C>\frac14$ then you always get an infinite number of bound states but I could not find a reference.

The example you cite has a fairly odd looking potential: it is only algebraically decaying and is oscillatory. The conventional wisdom is that it is trapped by multiple reflections off of the peaks of the potential.

What happens when this condition fails is harder. It is fairly easy to construct long-range potentials that have an infinite number of bound states. I suspect that having an imbedded eigenvalue for $E>0$ is very non-generic and is unlikely to persist if you tweak the potential (blah blah blah coupling to radiation, Fermi golden rule, blah blah blah) but I am not aware of any rigorous results to that effect.

There is a result here https://link.springer.com/article/10.1007/BF02099252 that analyzes a system with bound state coupled to a radiation field and shows that generically the bound state goes away but I don't know if it can be applied in this context.

Answer 2

To shed light in this shadowed folk wisdom, let's stablish some facts.

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The most general argument that I know, and you could find it in this wiki article is based on WKB approximation. The thing is that, since we have

$$ \Psi(x) \approx \frac{ C_{+} e^{+ \frac{i}{\hbar} \int \sqrt{2m \left( E - V(x) \right)}\,dx} + C_{-} e^{- \frac{i}{\hbar} \int \sqrt{2 m \left( E - V(x) \right)}\,dx} }{ \sqrt[4]{2m \mid E - V(x) \mid} } $$

For $E< V_{\infty}$ , we have exponentially decaying eigenfunctions, so bound states. It stablish at least $2\to 1$ in your chain of equivalences. The answer that you mention provides a counter-example of the back, so $1 \not \to 2$.

The matching condition of bound states in a region where the WKB approximation could be trusted implies relations like

$$ \int_a^b dx \sqrt{E-V(x)} = \left (n+ \frac 12\right )\hbar \pi. $$

This result is applied to one dimensional case, and it strongly suggest $1\to 3$. This somehow recovers the Bohr-Sommerfeld quantization condition.

In the book The Quantum theory for Mathematicians by Brian Hall, page 320, it is showed that for well-like potentials (i.e., potentials that goes to infinity at infinity), the Bohr-Sommerfeld quantization condition works as a very good approximation for the real eigenvalues of the Hamiltonian operator (Theorem 15.8 in the book). Actually, he also shows that it enough that the wave functions vanishes at infinity in order to match this quantization rule (Claim 15.7 pg 316-317).

It proves that $1\to 3$. By transitivity, we have $2\to 3$.

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If we suppose $3\to 2$, by transitivity we would have $1\to 3 \to 2$, which is a contradiction, so $3\not \to 2$. Breaking another equivalence in the folk wisdom.

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The only remaining relation we should check is $3\to 1$. I think I found the concrete counter-example of this relation. Consider an electron moving in a unit circle around an infinite solenoid with magnetic flux $\Phi$. The Hamiltonian of such system, in cylindrical coordinates, is given by

$$ H =\frac{-\hbar^2}{2m}\left(-i {\partial\over \partial \theta} + e {\Phi\over hc}\right)^2 - {\hbar^2\over 2m}{\partial^2\over \partial r^2} - {\hbar^2\over 2m}{\partial^2\over \partial z^2} $$

The eigenstates are given by

$$ \Psi_n(r,\theta,z) = e^{in\theta} $$

Which are not bounded in space, but the spectrum is given by

$$ E_n = {\hbar^2\over 2m}\left (n + e{\Phi\over hc}\right)^2 $$

Which constitute a discrete spectrum. So $3\not \to 1$.

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In resume, we have $1\leftarrow 2 \rightarrow 3$ and $1\to 3$.

Answer 3

Negative energies become when the particle is unbonded and falls in at least finite potential well.

Stationary waves of finite potential wells for one dimension are available on https://en.wikipedia.org/wiki/Finite_potential_well

Here there is already the case of spherical potential well for 3D. In this case the radial part of the wave vanishes when going far because amplitude is well spread at spherical surface as a function of the radius.