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-independent perturbation.

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

with V₀= 2eV and L= 5.10^-10 m

Homework Equations

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

<br /> \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?

 

[DrClaude]

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

Thanks for the reply.

 

[DrClaude]

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.

 

[TeethWhitener]

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

 

[DrClaude]

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!