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Normalising phi for the Hydrogen atom.

  1. Aug 19, 2009 #1
    This is most likely very simple, but I can't figure it out.

    http://www.sussex.ac.uk/physics/teaching/btv/Lect02_2006.pdf

    Step 5 they've got an equation for [tex]\Phi[/tex]. They then normalise it to get A = [tex]\frac{1}{\sqrt{2\pi}}[/tex]. Every time I do the integral I get:

    [tex]A^2.^{2\pi}_{0}[ \frac{exp(2i\sqrt{\Lambda}\Phi)}{2i\sqrt{\Lambda}}] = 1 [/tex]

    Which makes the integral go to zero when you rewrite the exp using Euler's and and take into account [tex]\sqrt{\Lambda}[/tex] must be an integer?
     
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 19, 2009 #2
    What is the absolute square of phi?
     
  4. Aug 19, 2009 #3
    The probability density function. So?
     
  5. Aug 19, 2009 #4
    No, I mean write down phi and then compute |phi|^2. What do you get for |phi|^2?
     
  6. Aug 19, 2009 #5
    [tex]A^2.exp(2i\sqrt{\Lambda}\Phi)[/tex]
     
  7. Aug 19, 2009 #6
    No, look again. It is the absolute square.
     
  8. Aug 19, 2009 #7
    So I can't take the two inside the exp? I don't really see why not? So i'm guessing you're saying that I have to do the integral of phi multiplied by its conjugate?

    Thanks anyway, that's solved it :)
     
    Last edited: Aug 19, 2009
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