Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wave Function from ground to excited

  1. Oct 1, 2014 #1
    I now understand how the wavefunction graphs look from the hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc5.html.

    However, what I fail to understand, is how the wavefunction equation was derived for the 1st excited state, from:

    ground_wavelength0 = (a/pi)1/4 e(-ax2/2), where a = alpha


    excited_wavelength1 = (a/pi)1/4 21/2e(-ax2/2), where a = alpha

    If someone could explain, merci beaucoup!!!!!!!
  2. jcsd
  3. Oct 1, 2014 #2


    User Avatar

    Staff: Mentor

  4. Oct 1, 2014 #3
    Awesome, thank you!!!!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook