(adsbygoogle = window.adsbygoogle || []).push({}); How to show the ground state is orthagnol with the first in a quantum harmonic os....

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

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 im just missing something!

Harmonic Oscillator.

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

[tex]\psi[/tex][tex]_{0}[/tex] (x) = [tex]\sqrt{\frac{1}{b\sqrt{\Pi}}}[/tex] e [tex]^{\frac{-x^{2}}{2b^{2}}}[/tex]

[tex]\psi[/tex][tex]_{1}[/tex] (x) = [tex]\sqrt{}\frac{2}{b^{3} \sqrt{\Pi}}[/tex] e [tex]^{\frac{-x^{2}}{2b^{2}}}[/tex]

3. 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 dont get zero when i do this.

In this case is [tex]\psi[/tex][tex]_{1}[/tex] (x) * = [tex]\psi[/tex][tex]_{1}[/tex] (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

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

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