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  • Intuitively one understands that if one is solving the Schroedinger's equation for energies $E$ such that $\{ x \vert U(x)\leq E \}$ is compact (..is there a weaker criteria?..) then the spectrum will turn out to be discrete and the wave-functions will decay exponentially for large values of $x$. What is the most rigorous statement and proof of this?

I want to focus on one potential,

$U = \vert G(s_i)\vert ^2 + \vert p \vert ^2 \sum _ {i} \vert \frac{\partial G}{\partial s_i} \vert ^2 + \frac{e^2}{2}(\sum _i \vert s_i \vert ^2 - n^2 \vert p \vert^2 - r ) + 2\vert \sigma \vert ^2 (\sum _i \vert s_i \vert ^2 + n^2 \vert p \vert^2) $

where $e$ is a real constant, $s_i$, $p$ and $\sigma$ are complex and $r$ is a real field and $G$ is a degree $n$ transverse homogeneous function in $s_i$.

Now apparently the following claims are true,

  • If $r = 0$ then for any value of $\sigma$, the range of $s_i$, $p$ where $U(s_i,p) \leq E$ is true is compact and hence the spectrum is discrete.

  • If $\sigma \neq 0$ then for any fixed non-zero value of $r$, the region of $s_i$ and p where $U(s_i,p;r) \leq E$ is true is compact for small enough values of $E$ and hence the spectrum is discrete for low-lying values of $E$ below some $E_{critical}$.

(..the above is apparently motivated by the fact that at $s_i=p=0$, $U$ becomes constant and equal to $\frac{e^2 r^2}{2}$ and hence independent of $\sigma$..so apparently if one goes above some critical value of $E$ the spectrum is continuous thanks to field configurations with large $\vert \sigma \vert$ but at $s_i = p =0$... )

  • My reading of the literature is that the above two claims are true independent of the topology of the space on which the fields are valued though in a case of interest one wants the theory to be on a circle and hence I guess one wants to think of $s_i$, $p$, $\sigma$ to be maps from $S^1$ to $\mathbb{C}$ or $\mathbb{R}$. If the theory is on a circle then semiclassically apparently the estimate for $E_{critical}$ is $\frac{e^2 r^2}{2} 2\pi R$ where $R$ is the radius of the circle.

I would be glad if someone can help justify the above three claims.

This paper is the reference for this question

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I added backticks to your latex code involving \{ \} - these will not display without some workaround. –  David Roberts May 7 '12 at 4:24
    
Small nitpick: you mention $D$, but I think that you already eliminated it via its equation of motion. –  José Figueroa-O'Farrill May 7 '12 at 12:39
    
@David Thanks! @Jose I meant $r$ and not $D$. Thanks for pointing out the typo. Now I have corrected it. You can may be write out an answer to this? :) –  Anirbit May 8 '12 at 22:23

1 Answer 1

This question (and a partial converse with counterexamples) is answered in

B. Simon, Some quantum operators with discrete spectrum but classically continuous spectrum, Annals of physics 46 (1983), 209-220. http://www.math.caltech.edu/SimonPapers/158.pdf

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