Solutions to Schrödinger's equation in 2D polar coordinates when the potential is zero For Schrödinger’s equation, in polar coordinates $(r, \theta)$, when the potential in the Hamiltonian is $0$ (free particle), I think a solution is $r^{-1} e^{-i(kr - \omega t)}$. This radial wave is centered at the origin of coordinates.
Can another solution, for the above described Hamiltonian, also be a radial wave function (with amplitude, at any point, that is again the inverse of the distance from the wave’s center), but not centered at the origin of coordinates?
 A: The radial solutions are Bessel functions and the angular part are waves. You start with the free Hamiltonian in polar coordinates,
\begin{equation} 
H \psi(r,\varphi) = - \left(\frac{\partial ^2}{\partial r^2} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \varphi^2} \right) \psi(r, \varphi) = E \psi(r, \varphi).
\end{equation}
As so often with these types of differential equations, we can make the Ansatz
\begin{equation} 
\phi(r, \varphi) = R(r) \Phi(\varphi).
\end{equation}
Inserting this into the differential equation and dividing by $R(r) \Phi(\varphi)$ yields
\begin{equation} 
- \frac{r^2}{R} \frac{d^2 R}{dr^2} - \frac{r}{R} \frac{dR}{dr} - Er^2 = \frac{1}{\Phi} \frac{d^2 \Phi}{d\varphi}.
\end{equation}
The left hand side depends only on $r$, while the right hand side only depends on $\varphi$, thus both sides must be constant. We call the constant $-A^2$. Then the angular differential equation reads,
\begin{equation} 
\frac{d^2 \Phi}{d\varphi^2} + A^2 \Phi = 0 
\end{equation}
which is easily solved by
\begin{equation} 
\Phi(\varphi) = C_1 e^{i A \varphi} + C_2 e^{-i A \varphi}. 
\end{equation}
Since $\Phi(0) = \Phi(2 \pi)$, we conclude that $A \in \mathbb{N}$. The radial part is then given by
\begin{equation} 
r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} + (Er^2 - A^2) R = 0.
\end{equation}
After substituting $\tilde{r} =  \sqrt{E} r$, this becomes the Bessel differential equation and the solutions are the Bessel fuctions.
So to answer your question, yes there are more solutions given by the Bessel functions modulated by trig' functions. As usual, you can represent any smooth function by a sum of suitable Eigenfunctions. But I guess your question is if we can also find Eigenfunctions that are displaced from the origin. And there the answer is no as this would break the symmetry of the problem.
