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A seeming contrdiction in deriving wave function for delta function potential

  1. Nov 2, 2009 #1
    First of all, let me copy the standard solution from Griffiths, section 2.5, just for the sake of clarity.
    Potential[tex]V(x) = - \alpha \delta (x)[/tex]

    The bound state eigenfunction:
    [tex]\psi (x) = \left\{ \begin{array}{l}
    B{e^{\kappa x}}{\rm{ (}}x \le 0{\rm{)}} \\
    B{e^{ - \kappa x}}{\rm{ (}}x \ge 0{\rm{)}} \\
    \end{array} \right.{\rm{ where }}\kappa = \frac{{\sqrt { - 2mE} }}{\hbar }[/tex]
    Integrate time independent schrodinger's equation from -infinitesimal to +infinitesimal to get an imposed condition
    [tex] - \frac{{{\hbar ^2}}}{{2m}}\int_{ - \varepsilon }^{ + \varepsilon } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx + \int_{ - \varepsilon }^{ + \varepsilon } {V(x)} \psi (x)dx = E\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx} [/tex]
    For right hand side, since the we're integrating a finite function in a infinitesimal region
    [tex]\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx} = 0[/tex]
    So we have
    [tex] - \frac{{{\hbar ^2}}}{{2m}}\mathop {\lim }\limits_{\varepsilon \to 0} ({\left. {\frac{{d\psi }}{{dx}}} \right|_{ + \varepsilon }} - {\left. {\frac{{d\psi }}{{dx}}} \right|_{ - \varepsilon }}) = \alpha \psi (0) \Rightarrow - 2B\kappa = - \frac{{2m\alpha }}{{{\hbar ^2}}}B[/tex]
    so [tex]\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}[/tex]
    and consequently[tex]E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}[/tex] and [tex]B = \frac{{\sqrt {m\alpha } }}{\hbar }(normalization)[/tex]

    And here comes my problem: instead of integrating a infinitesimal region, I tried to integrate the whole x-axis just to check the consistency
    [tex] - \frac{{{\hbar ^2}}}{{2m}}\int_{ - \infty }^{ + \infty } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx + \int_{ - \infty }^{ + \infty } {V(x)} \psi (x)dx = E\int_{ - \infty }^{ + \infty } {\psi (x)dx} [/tex]
    in which:
    [tex]\int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2}{\kappa }[/tex]
    [tex] - \frac{{{\hbar ^2}}}{{2m}}\int_{ - \infty }^{ + \infty } {\frac{{{d^2}\psi }}{{d{x^2}}}} dx = - \frac{{{\hbar ^2}}}{{2m}}({\left. {\frac{{d\psi }}{{dx}}} \right|_{ + \infty }} - {\left. {\frac{{d\psi }}{{dx}}} \right|_{ - \infty }}) = 0[/tex]
    Finally,
    [tex] - \alpha \psi (0) = \frac{{2E}}{\kappa } \Leftrightarrow - \alpha B = \frac{{2E}}{\kappa }[/tex]
    Now if we sub in what we've got in the first part, e.g.[tex]\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}[/tex], [tex]E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}[/tex], [tex]B = \frac{{\sqrt {m\alpha } }}{\hbar }[/tex]
    we find
    [tex]\frac{{\sqrt {m\alpha } }}{\hbar } = 1[/tex]

    This is obviously incorrect, since m and alpha are arbitrary. So where did I get wrong?
     
  2. jcsd
  3. Nov 2, 2009 #2
    You forgot the normalization of the wavefunction. The integral you quote should be:
    [tex]
    \int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2B}{\kappa }
    [/tex]
    If you do the same trick, you'll just get 1=1 instead of what you have.
     
  4. Nov 2, 2009 #3
    Ah,you're right. Thanks
     
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