[Physics] When is the spectrum of the Hamiltonian (purely) continuous

mathematical physicsquantum mechanics

Given a quantum hamiltonian $H = \frac{1}{2m}\vec{p}^2 +V(\vec{x})$ in $n$-dimensions, the general rule-of-thumb is that the energy will be discrete for energies $E$ for which $\{ \vec{x} | V(\vec{x})\leq E\}$ has finite volume, and continuous otherwise.

Question: What mathematical theorems back-up / contradict this rule-of-thumb?

I know of a few (which I will list in an answer below), but they all of the form of sufficient conditions for the spectrum to be discrete (possibly below some energy). Therefore I would be particularly interested in sufficient conditions for the spectrum to be purely continuous (possibly above some energy).

Best Answer

Sufficient conditions for a discrete energy spectrum:

  • In one-dimension Sturm-Louville theory implies that the spectrum is purely discrete provided that the system is restricted to a finite interval [a,b] (with appropriate boundary conditions). I assume (but don't know) this can be extended to the case of a system restricted to finite volume in higher dimensions.

The next two conditions I found in this paper:

  • In dimension $n\geq 3$ the Cwickel-Lieb-Rosenbljum bound implies that the spectrum will be discrete for energies for which the classical phase-space volume of $\{(\vec{x},\vec{p})|\vec{p}^2+V(\vec{x})<E\}$ is finite.
  • As a consequence of the Golden-Thompson inequality the spectrum will be purely discrete in any system where the classical partition function is finite (at some value of temperature)

Conditions for the spectrum to have a continuous part or to be purely continuous seem a lot more complicated. The main purpose of the paper cited above was to provide a counter-example to the 'rule-of-thumb' I gave in the question. Namely, it shows that the Hamiltonian $$ H = - \frac{\partial^2}{\partial x^2} -\frac{\partial^2}{\partial y^2} + x^2 y^2 $$ has a purely discrete spectrum even though the volume of $\{\vec{x}|V(\vec{x})<E \}$ is infinite for any $E>0$.

A reference suggested by yuggib (thanks!), and references therein give some more general criteria for continuous spectra. It looks like a difficult problem and research on classes of Hamiltonian with continous spectra seems to be ongoing. Here are a few results:

Consider the Hamiltonian $$ H = -\nabla^2 +V(\vec{x}) $$ and suppose that $|V(\vec{x})|<C(1+|\vec{x}|)^{-\alpha}$ for some constants $C$ and $\alpha>0$ (i.e. the potential decays to zero at infinity). Then:

  • If $\alpha >1$ then the Hamiltonian has purely continuous spectrum on $E\in [0,\infty)$.
  • On the other hand there exist $V(\vec{x})$ satisfying the above with $\alpha \leq \frac{1}{2}$ for which the spectrum is purely discrete.
  • The interesting stuff seems to happen for $\frac{1}{2}< \alpha \leq 1$. Here there are examples of continous spectra on $E\in [0,\infty)$ with embedded discrete Eigenvalues. The most well known seem to be the Wigner von-Neumann potentials (e.g. $V(r) \sim \frac{\sin 2 r}{r}$ which has a discrete Eigenvalue at $E=1$).
  • It looks like for $\frac{1}{2}< \alpha \leq 1$, while the spectrum may no longer be purely continous, the continous part on $E\in [0,\infty)$ is always preserved. However, the proofs I have found of this always make additional assumptions, so I'm not sure if this has been proved in general. (For example, yuggib's reference proves this result in 1D, and references therein based on scattering theory prove it in any dimension if $V(\vec{x})$ is differentiable and $|\frac{\partial V}{\partial x^i}|<C_1 (1+|\vec{x}|)^{-3/2-\epsilon}$ for some $C_1>0$ and all $i=1,\ldots,n$.)
  • Finally, the assumption that the potential must decay to zero as $|\vec{x}|\to \infty$ can be relaxed (in some cases) to the assumption that it decays in this way only on some cone of $\mathbb{R}^n$. Here 'cone' is defined in the linear algebra sense. See, for example, this paper.
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