(adsbygoogle = window.adsbygoogle || []).push({}); 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Bound state for a Dirac delta function potential

**Physics Forums | Science Articles, Homework Help, Discussion**