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?

  • $\begingroup$ $\psi$ will factor into a radial function times an angular dependence, but it is not solely a function of the radial coordinate. $\endgroup$ Apr 1, 2020 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, 2020 at 11:27
  • $\begingroup$ you require $\psi\in L^2$, doesn't this imply $\psi_r\in L^2$? $\endgroup$ Apr 1, 2020 at 12:02
  • $\begingroup$ No, I require $\psi\not\in L^2$. $\endgroup$
    – Capublanca
    Apr 1, 2020 at 12:10

1 Answer 1


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.


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