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One dimensional Coulomb potential

  1. Jan 22, 2014 #1


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    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 [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?
    Last edited: Jan 23, 2014
  2. jcsd
  3. Jan 24, 2014 #2
    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.
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