Consider the potential below:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]

V(x)=\left\{ \begin{array}{cc} -\frac{e^2}{4\pi\varepsilon_0 x} &x>0 \\ \infty &x\leq 0 \end{array} \right.

[/itex]

The time independent Schrodinger equation becomes:

[itex]

\frac{d^2X}{dx^2}=-\frac{2m}{\hbar^2} (E+\frac{e^2}{4\pi\varepsilon_0 x})X

[/itex]

I wanna find the ground state wave function.This is how I did it:

[itex]

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

[/itex]

But because bound states of this potential are for small x and the ground state has a very very small x,I assumed [itex] x\to 0 [/itex] and considered the approximated equation below:

[itex]2\frac{dY}{dx}+\frac{me^2}{2\pi \varepsilon_0 \hbar^2}Y=0[/itex],whose answer is [itex] Y=A\exp{ (-\frac{me^2}{4\pi \varepsilon_0 \hbar^2}x)} [/itex] and so [itex] X=Ax\exp{ (-\frac{me^2}{4\pi \varepsilon_0 \hbar^2}x)}[/itex]

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

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# One dimensional Coulomb potential

Loading...

Similar Threads - dimensional Coulomb potential | Date |
---|---|

I Operators and vectors in infinite dimensional vector spaces | Mar 2, 2018 |

A Can zero dimensional QFT be real? | Sep 21, 2017 |

A Photon propagator in Coulomb gauge | Jul 20, 2017 |

B How can something be "Zero Dimensional?" | May 20, 2017 |

**Physics Forums - The Fusion of Science and Community**