# One dimensional Coulomb potential

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## Main Question or Discussion Point

Consider the potential below:
$V(x)=\left\{ \begin{array}{cc} -\frac{e^2}{4\pi\varepsilon_0 x} &x>0 \\ \infty &x\leq 0 \end{array} \right.$
The time independent Schrodinger equation becomes:
$\frac{d^2X}{dx^2}=-\frac{2m}{\hbar^2} (E+\frac{e^2}{4\pi\varepsilon_0 x})X$
I wanna find the ground state wave function.This is how I did it:
$Y=\frac{X}{x} \Rightarrow x\frac{d^2Y}{dx^2}+2\frac{dY}{dx}+(\frac{2mE}{\hbar^2}x+\frac{me^2}{2 \pi \varepsilon_0\hbar^2})Y=0$
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

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