- #1

- 747

- 55

## Homework Statement

The ground state of the wavefunction for an electron in a simple one-dimensional harmonic potential well is

[tex]\Psi _{0}(x)= \left ( \frac{m\omega }{\pi \hbar} \right )^{1/4} exp(-\frac{m\omega x^{2}}{2\hbar})[/tex]

By employing first-order perturbation theory calculate the energy shift [tex]E_{0}= \frac{\hbar \omega}{2} = 2eV[/tex] caused by the time-independant perturbation.

[tex]V(x)= V_{0} x^{3}[/tex]

with V

_{0}= 2eV and L= 5.10^-10 m

## Homework Equations

E = E

_{0}+ V

_{00}

## The Attempt at a Solution

The problem I'm having is that no integrals are given for this question (it's a past exam question). If I follow the process of finding V

_{00}

[tex]V_{00}= \left \langle \Psi _{0}| V_{0} x^{3} | \Psi _{0} \right \rangle [/tex]

[tex]

\int_{0}^{L} (\frac{m\omega }{\pi \hbar} \right ))^{1/2} x^{3} exp (-\frac{m\omega x^{2}}{\hbar})[/tex]

if I integrate by parts between 0 and L I get an integral of exp (-y^2) which ordinarily between minus infinity and infinity would be root pi, but between 0 and L gives ERF. Any Ideas of what to do in this situation?