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Let $S$ be a nonnegative self-adjoint operator on a complex Hilbert space $X$. (For example, $X$ consists of functions on $\mathbb R^d$; it could be $L^2(\mathbb R^d), \dot{H}^2(\mathbb R^d)$, etc.)

We consider the abstract Cauchy problem for the Schrödinger equation (SE): $$ i\frac{\partial}{\partial t} u(x,t) + Su(x,t)=0, u(x,0)=u_0(x). $$

Formally, we may write the solution of $(SE)$, $$u(x,t)= e^{itS}u_0(x).$$

Question: If $u_0(x)\in X$ is radial, then is it true that $e^{itS}u_0(x)$ is also radial?

Note:
(1) I know that if $S=- \Delta$, then $e^{itS}u_{0}$ is radial whenever $u_0$ is radial.
(2) Specifically, I am interested in $S=-\Delta+ \frac{a}{|x|^2}$, see Section 1.1 for details (with $d\geq 3, a\geq (\frac{d-2}{2})^2$).

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  • $\begingroup$ Any operator which takes radial functions into radial functions should do. I can not see any other reasonable requirement. I.e., S should commute with rotations on the underlying function space $\endgroup$
    – Piero D'Ancona
    Commented Apr 25, 2017 at 9:02
  • $\begingroup$ @PieroD'Ancona: Thanks: What can we say about $S=-\Delta + \frac{a}{|x|^2}$? (I think, $-\Delta + \frac{a}{|x|^2}$ takes radial function to radial, please correct me if I am wrong...) $\endgroup$
    – abcd
    Commented Apr 25, 2017 at 9:10
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    $\begingroup$ Yes of course, any operator like $-\Delta+V(|x|)$ takes radial functions into radial functions $\endgroup$
    – Piero D'Ancona
    Commented Apr 25, 2017 at 13:30

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