kof9595995
Nov2-09, 11:56 AM
First of all, let me copy the standard solution from Griffiths, section 2.5, just for the sake of clarity.
PotentialV(x) = - \alpha \delta (x)
The bound state eigenfunction:
\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 }
Integrate time independent schrodinger's equation from -infinitesimal to +infinitesimal to get an imposed condition
- \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}
For right hand side, since the we're integrating a finite function in a infinitesimal region
\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx} = 0
So we have
- \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
so \kappa = \frac{{m\alpha }}{{{\hbar ^2}}}
and consequentlyE = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}} and B = \frac{{\sqrt {m\alpha } }}{\hbar }(normalization)
And here comes my problem: instead of integrating a infinitesimal region, I tried to integrate the whole x-axis just to check the consistency
- \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}
in which:
\int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2}{\kappa }
- \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
Finally,
- \alpha \psi (0) = \frac{{2E}}{\kappa } \Leftrightarrow - \alpha B = \frac{{2E}}{\kappa }
Now if we sub in what we've got in the first part, e.g.\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}, E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}, B = \frac{{\sqrt {m\alpha } }}{\hbar }
we find
\frac{{\sqrt {m\alpha } }}{\hbar } = 1
This is obviously incorrect, since m and alpha are arbitrary. So where did I get wrong?
PotentialV(x) = - \alpha \delta (x)
The bound state eigenfunction:
\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 }
Integrate time independent schrodinger's equation from -infinitesimal to +infinitesimal to get an imposed condition
- \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}
For right hand side, since the we're integrating a finite function in a infinitesimal region
\int_{ - \varepsilon }^{ + \varepsilon } {\psi (x)dx} = 0
So we have
- \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
so \kappa = \frac{{m\alpha }}{{{\hbar ^2}}}
and consequentlyE = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}} and B = \frac{{\sqrt {m\alpha } }}{\hbar }(normalization)
And here comes my problem: instead of integrating a infinitesimal region, I tried to integrate the whole x-axis just to check the consistency
- \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}
in which:
\int_{ - \infty }^{ + \infty } {\psi (x)dx} = \frac{2}{\kappa }
- \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
Finally,
- \alpha \psi (0) = \frac{{2E}}{\kappa } \Leftrightarrow - \alpha B = \frac{{2E}}{\kappa }
Now if we sub in what we've got in the first part, e.g.\kappa = \frac{{m\alpha }}{{{\hbar ^2}}}, E = - \frac{{m{\alpha ^2}}}{{2{\hbar ^2}}}, B = \frac{{\sqrt {m\alpha } }}{\hbar }
we find
\frac{{\sqrt {m\alpha } }}{\hbar } = 1
This is obviously incorrect, since m and alpha are arbitrary. So where did I get wrong?