One dimensional Coulomb potential

[ShayanJ]

Consider the potential below:
<br /> V(x)=\left\{ \begin{array}{cc} -\frac{e^2}{4\pi\varepsilon_0 x} &amp;x&gt;0 \\ \infty &amp;x\leq 0 \end{array} \right.<br />
The time independent Schrodinger equation becomes:
<br /> \frac{d^2X}{dx^2}=-\frac{2m}{\hbar^2} (E+\frac{e^2}{4\pi\varepsilon_0 x})X<br />
I want to find the ground state wave function.This is how I did it:
<br /> Y=\frac{X}{x} \Rightarrow x\frac{d^2Y}{dx^2}+2\frac{dY}{dx}+(\frac{2mE}{\hbar^2}x+\frac{me^2}{2 \pi \varepsilon_0\hbar^2})Y=0<br />
But because bound states of this potential are for small x and the ground state has a very very small x,I assumed x\to 0 and considered the approximated equation below:
2\frac{dY}{dx}+\frac{me^2}{2\pi \varepsilon_0 \hbar^2}Y=0,whose answer is Y=A\exp{ (-\frac{me^2}{4\pi \varepsilon_0 \hbar^2}x)} and so X=Ax\exp{ (-\frac{me^2}{4\pi \varepsilon_0 \hbar^2}x)}
My problems are:
1-There is noting in X that indicates it is the ground state.What should I do about it? Is it an issue at all?
2-How can I find energy levels?
Thanks

 

[Hawkwind]

You have to solve the exact equation. Finally, applying the boundary condition
X(0) = 0
(because of the unfinite potential step) should deliver the discrete spectrum of solutions.