2
$\begingroup$

Let $V\in L^{\infty}(\mathbb{R}^3)$ be a radial, compactly supported potential, and consider the Schrodinger operator $H:=-\Delta + V$ on $L^2(\mathbb{R}^3)$. Let $\psi$ be a resonance for $H$, i.e. a function $\psi\in L^2(\mathbb{R}^3,\langle x\rangle^{-1-\varepsilon}dx)\setminus L^2(\mathbb{R}^3)$ which satisfies $(-\Delta + V)\psi=0$.

Is it true that $\psi$ is radial? If not, is it at least true that the orthogonal projection of $\psi$ into the space of radial functions is a resonance?

$\endgroup$
  • $\begingroup$ $\psi$ will factor into a radial function times an angular dependence, but it is not solely a function of the radial coordinate. $\endgroup$ – Carlo Beenakker Apr 1 at 11:20
  • $\begingroup$ Ok thanks, but if I take the projection of $\psi$ into the space of radial function at least I get a radial function $\psi_r$ that solves $(-\Delta+V)\psi_r=0$. The point is to understand wheter we actually have $\psi_r\not\in L^2$. $\endgroup$ – Capublanca Apr 1 at 11:27
  • $\begingroup$ you require $\psi\in L^2$, doesn't this imply $\psi_r\in L^2$? $\endgroup$ – Carlo Beenakker Apr 1 at 12:02
  • $\begingroup$ No, I require $\psi\not\in L^2$. $\endgroup$ – Capublanca Apr 1 at 12:10
4
$\begingroup$

Expand your resonance in spherical harmonics: $\psi = \sum_{\ell=0}^\infty \sum_m \psi_{\ell m}(r) Y_{\ell m}(\theta,\phi)$. Then each coefficient satisfies the radial Schrödinger equation $$ -\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial \psi_{\ell m}}{\partial r} + \frac{\ell(\ell+1)}{r^2}\psi_{\ell m} + V(r) \psi_{\ell m} = 0 . $$ Since $V(r)$ is compactly supported, for sufficiently large $r$, you must have $\psi_{\ell m} = A r^\ell + B r^{-\ell-1}$. Your asymptotic condition forces $A=0$ for all $\ell$. On the other hand, the asymptotics of the remaining term mean that, for $\ell \ge 1$, $\psi_{\ell m}(r) Y_{\ell m}(\theta,\phi)$ will be in $L^2(\mathbb{R}^3)$. So the component of $\psi$ orthogonal to radial functions is necessarily in $L^2(\mathbb{R}^3)$.

Only the $\ell=0$ case escapes $L^2(\mathbb{R}^3)$. And of course, when $\psi_{\ell=0}$ is non-vanishing, it is $O(r^{-1})$ at infinity and hence a resonance by your definition. So the answer to your second question is Yes, but if there are any solutions at higher $\ell$ (they will be normalizable eigenfunctions), then mixing them with an $\ell=0$ resonance will give you non-radial resonances.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.