# First-order perturbation for a simple harmonic potential well

1. Dec 17, 2014

### rwooduk

1. The problem statement, all variables and given/known data
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

2. Relevant equations
E = E0 + V00

3. 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?

2. Dec 17, 2014

### Staff: Mentor

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

3. Dec 17, 2014

### rwooduk

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

4. Dec 17, 2014

### Staff: Mentor

Try $u = x^2$.

5. Dec 17, 2014

### rwooduk

thanks, yes thats 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.

6. Dec 17, 2014

### TeethWhitener

7. Dec 17, 2014

### Staff: Mentor

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

8. Dec 17, 2014

### rwooduk

ahhhh, thats a clever little trick! i see now, many thanks for your help!!

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