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Bound state for a Dirac delta function potential

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the bound state energy for a particle in a Dirac delta function potential.


    2. Relevant equations
    [tex]\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } - \frac{\hbar^2}{2 m} \ \pd{\psi}{x}{2} - \alpha \delta (x) \psi (x) = E\psi (x)[/tex]
    where [tex]\alpha[/tex] > 0 is the strength of the delta function potential.


    3. The attempt at a solution
    This is solved in Griffith's QM book and other places but I'm having an issue with the energy that is given there. The wave function is:
    [tex]\psi(x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m \alpha |x|\hbar^{2}}[/tex]

    Now if we plug this into the Schoedinger question above, times everything by [tex]\psi^{*}[/tex], and integrate over all x, we should get the bound energy.

    The kinetic energy energy term gives [tex]\frac{-m\alpha^{2}}{2\hbar^{2}}[/tex] which is the listed as the total energy in Griffith's, but what about the term from the potential energy? Integrating over that will give a contribution of [tex]\frac{-m\alpha^{2}}{\hbar^{2}} [/tex] so that the total energy is [tex]E=\frac{-m\alpha^{2}}{2\hbar^{2}}\frac{-m\alpha^{2}}{\hbar^{2}}=\frac{-3m\alpha^{2}}{2\hbar^{2}}[/tex].

    Any ideas why the potential energy is never included in the energy people list? It is listed as being the total energy in Griffith's QM book for instance. Am I correct in my evaluation of the potential energy? Any thoughts?
     
    Last edited: Jul 12, 2008
  2. jcsd
  3. Jul 12, 2008 #2
    Nevermind, I think I realized my problem. I was just excluding the origin in my integration of the kinetic energy. It contributes a [tex]\frac{-m \alpha}{\hbar^{2}}[/tex] which cancels out the potential term (of course).
     
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