# boundary conditions for Schr$\ddot{\textrm{o}}$dinger equation in 2D polars?

What are the boundary conditions at $r=0$ for Schr$\ddot{\textrm{o}}$dinger's equation for a quantum particle in 2D polars $(r,\theta)$, with potential $U=0$ for $r<a$ and $U=\infty$ for $r>a$?

I've shown that the angular part of the wave-function is $e^{im\theta}$ where $m\in\mathbb{Z}$, and I've got an ODE for the radial part $R(r)$, with boundary condition $R(a)=0$.

Since $e^{im\theta}$ is not well-defined as $(r,\theta)$ approaches the origin, it looks as though we should have $R(0)=0\;$ to get a well-defined wave-function. But according to the question, the series expansion for $R(r)$ has arbitrary constant term and all odd-order terms zero. What sort of boundary conditions give this??

Many thanks for any help with this!

-

Schrödinger's equation takes the form $$(\nabla^2 + k^2)\psi = 0,$$ where $k = \sqrt{2 m E/\hbar^2}$ inside the well. This is the Helmholtz equation. The solutions in polar coordinates are well-known, $${\psi}_{n}(r,\theta) = \left\{ \begin{array}[c]{l} {J_n}(k r) \\ {Y_n}(k r) \end{array} \right\} \{ e^{\pm i n\theta} \},$$ where $J_n$ and $Y_n$ are the Bessel functions of the first and second kind. (The braces are shorthand for "linear combinations of.") The boundary conditions are that $\psi$ be finite, vanish at $r=a$, and be single-valued. Therefore, the eigenfunctions are $$\psi_{n i}(r,\theta) = J_n\left(\alpha_{n i} \frac{r}{a}\right) (a_{n i}\cos n\theta + b_{n i}\sin n\theta),$$ where $n = 0,1,2,\ldots$ and where $\alpha_{n i}$ is the $i$th zero of $J_n$. (Energy quantization follows since $k = k_{n i} = \alpha_{n i}/a$.)
Notice that $J_n(0) = 0$ for $n\ge 1$. Furthermore, $\psi_0$ has no angular dependence. Thus, we need not impose any conditions at $r=0$. The solutions already "know" about this possible trouble and have taken care of it!