A seeming contrdiction in deriving wave function for delta function potential

  • Thread starter kof9595995
  • Start date
  • #1
679
2
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?
 

Answers and Replies

  • #2
525
7
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.
 
  • #3
679
2
Ah,you're right. Thanks
 

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