Bound state for a Dirac delta function potential

[badphysicist]

Homework Statement

Find the bound state energy for a particle in a Dirac delta function potential.

Homework Equations

\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)
where \alpha > 0 is the strength of the delta function potential.

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:
\psi(x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m \alpha |x|\hbar^{2}}

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

The kinetic energy energy term gives \frac{-m\alpha^{2}}{2\hbar^{2}} 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 \frac{-m\alpha^{2}}{\hbar^{2}} so that the total energy is E=\frac{-m\alpha^{2}}{2\hbar^{2}}\frac{-m\alpha^{2}}{\hbar^{2}}=\frac{-3m\alpha^{2}}{2\hbar^{2}}.

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?

 

[badphysicist]

Nevermind, I think I realized my problem. I was just excluding the origin in my integration of the kinetic energy. It contributes a \frac{-m \alpha}{\hbar^{2}} which cancels out the potential term (of course).