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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

    to

    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

    jtbell

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    Staff: Mentor

  4. Oct 1, 2014 #3
    Awesome, thank you!!!!!!
     
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