Let $S$ be a nonnegative self- adjoint operator in a complex Hilbert Space $X$. (Say $X$ consists of functions on $\mathbb R^d$, e.g., $L^2(\mathbb R^d), \dot{H}^1(\mathbb R^d)$, 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 $(SE)$ as: $$u(x,t)= e^{itS}u_0(x).$$

Question: If $u_0(x)\in X$ is radial, then can we expect $e^{itS}u_0(x)$ also a radial?

Note:
(1) I know 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 here Section 1.1 for detail (with $d\geq 3, a\geq (\frac{d-2}{2})^2$).

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 $d\geq 3, a\geq (\frac{d-2}{2})^2$).

Let $S$ be a nonnegative self- adjoint operator in a complex Hilbert Space $X$ (say $X$ consists of functions on $\mathbb R^d$, e.g., $L^2(\mathbb R^d), \dot{H}^1(\mathbb R^d)$, etc.. )

We consider abstract Cauchy problem for the Schr"odinger 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 $(SE)$ as $$u(x,t)= e^{itS}u_0(x).$$

My Question: If $u_0(x)\in X$ is radial, then can we expect $e^{itS}u_0(x)$ also a radial?

Note: (1) I know 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 here Section 1.1 for detail (with $d\geq 3, a\geq (\frac{d-2}{2})^2$)

Let $S$ be a nonnegative self- adjoint operator in a complex Hilbert Space $X$. (Say $X$ consists of functions on $\mathbb R^d$, e.g., $L^2(\mathbb R^d), \dot{H}^1(\mathbb R^d)$, 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 $(SE)$ as: $$u(x,t)= e^{itS}u_0(x).$$

Question: If $u_0(x)\in X$ is radial, then can we expect $e^{itS}u_0(x)$ also a radial?

Note:
(1) I know 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 here Section 1.1 for detail (with $d\geq 3, a\geq (\frac{d-2}{2})^2$).