# About the quantum spectrum of a certain potential.

+ 3 like - 0 dislike
2966 views
• 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,

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

$+ 2\vert \sigma \vert ^2 (\sum _i \vert s_i \vert ^2 + n^2 \vert p \vert^2)$

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

This post imported from StackExchange MathOverflow at 2014-09-01 11:18 (UCT), posted by SE-user Anirbit

retagged Aug 28, 2016
I added backticks to your latex code involving \{ \} - these will not display without some workaround.

This post imported from StackExchange MathOverflow at 2014-09-01 11:18 (UCT), posted by SE-user David Roberts
Small nitpick: you mention $D$, but I think that you already eliminated it via its equation of motion.

This post imported from StackExchange MathOverflow at 2014-09-01 11:18 (UCT), posted by SE-user José Figueroa-O'Farrill
@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? :)

This post imported from StackExchange MathOverflow at 2014-09-01 11:18 (UCT), posted by SE-user Anirbit

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$y$\varnothing$icsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.