# First-order perturbation for a simple harmonic potential well

• rwooduk

## Homework Statement

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

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

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

$$V(x)= V_{0} x^{3}$$

with V0= 2eV and L= 5.10^-10 m

E = E0 + V00

## 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 V00

$$V_{00}= \left \langle \Psi _{0}| V_{0} x^{3} | \Psi _{0} \right \rangle$$

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

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?

You should be able to get that by integrating by parts, provided you choose the correct parts! What did you use?

• rwooduk
You should be able to get that by integrating by parts, provided you choose the correct parts! What did you use?

Using $$\int u dv = uv - \int vdu$$

I set u = x^3 in order to reduce it to a lower power of x (and continue to so with the next by parts integrals) and dv = exp (-ax^2) where a is a constant.

The problem is when I integrate with 0 to L limits using substitution of y^2, it comes out as ERF.

http://www.integral-calculator.com/#expr=e^(-y^2)&intvar=y&ubound=l&lbound=0

Try ##u = x^2##.

• rwooduk
Try ##u = x^2##.

thanks, yes that's what I did but this gives an exponential of a square for dv, which when I integrate with 0 to L limits using substitution, namely exp (-y^2), it comes out as ERF.

• rwooduk
thanks, yes that's what I did but this gives an exponential of a square for dv, which when I integrate with 0 to L limits using substitution, namely exp (-y^2), it comes out as ERF.
If you set ##u = x^2##, then ##dv = x e^{-a x^2} dx##, which can be easily integrated.

• rwooduk
If you set ##u = x^2##, then ##dv = x e^{-a x^2} dx##, which can be easily integrated.

ahhhh, that's a clever little trick! i see now, many thanks for your help!

ERF is the error function: http://en.wikipedia.org/wiki/Error_function

it's okay, I've got it now, but thanks!