How to show the ground state is orthagnol with the first in a quantum harmonic os

[leoflindall]

How to show the ground state is orthagnol with the first in a quantum harmonic os...

Homework Statement

This is a revision question for an upcoming quantum mechanics exam, that i am stuck on. Any help or idea at how to approach the question would be greatly appreciated, i get the feeling it is quite easy and I am just missing something!

Harmonic Oscillator.

Show that the ground state is orthagonol to the first excited state wave function.

\psi_{0} (x) = \sqrt{\frac{1}{b\sqrt{\Pi}}} e ^{\frac{-x^{2}}{2b^{2}}}

\psi_{1} (x) = \sqrt{}\frac{2}{b^{3} \sqrt{\Pi}} e ^{\frac{-x^{2}}{2b^{2}}}

The Attempt at a Solution

I think that you have to work out the overlap by intergrating between the limits of +/- infinity of the complex conjugate of one wavefunction multiplied by the other. I don't get zero when i do this.

In this case is \psi_{1} (x) * = \psi_{1} (x) ?

Or is since the wave functions are in a harmonic osscilator that the limits of the intergral should be [0,L]?

Kind regards

Leo

 

[badphysicist]

You have the wrong equation for \psi^1. See http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html#c1" for the correct wave functions. You should be integrating from -infinity to infinity though because the standard quantum harmonic oscillator doesn't have finite bounds on it. And yes, you should be integrating one function versus the Hermitian conjugate of the other.