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Help normalizing a wave function

  1. Nov 8, 2016 #1
    1. The problem statement, all variables and given/known data
    I don't see how the author normalizes ##u(r)=Asin(kr)##. From Griffiths, Introduction to Quantum Mechanics, 2nd edition, page 141-142:

    http://imgur.com/a/bo8v6

    2. Relevant equations
    ##\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi}|A|^2 \sin^2(\frac{n\pi r}{a})r^2 \sin \theta drd\theta d\phi=1##

    3. The attempt at a solution
    My integral was
    ##\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi}|A|^2 \sin^2(\frac{n\pi r}{a})r^2 \sin \theta drd\theta d\phi=1##

    Mathematica simplifies the integral (without the ##A## for simplicity) to
    ##=\int_0^{\infty}4\pi r^2 \sin^2(\frac{n\pi r}{a})dr##

    but it stops there. I don't think this integral converges. Did I make a mistake somewhere?
     
  2. jcsd
  3. Nov 8, 2016 #2
    As Griffiths himself makes clear - the radial part of the wavefunction is ##R(r) = u(r) /r##, so that is what you should be putting in the integral instead. When he says normalising ##u(r)##, what he really means is finding the coefficient ##A##
     
  4. Nov 8, 2016 #3

    TSny

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    What is the value of the wavefunction for r > a?

    The normalization condition for the radial part of the wavefunction R(r) is given in Griffith's equation [4.31] (not shown in your picture).
     
  5. Nov 8, 2016 #4
    The wavefunction is ##0## for ##r>a##.
     
  6. Nov 8, 2016 #5
    Ah I see. Ok then but normalizing according to that equation yields ##\int_0^{\infty} |A|^2 sin^2 (n*\pi *r/a)dr##, which does not converge.
     
  7. Nov 8, 2016 #6

    TSny

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    Did you take account of the fact that the wavefunction vanishes for r > a?
     
  8. Nov 8, 2016 #7
    But I thought whenever you normalize, you integrate over all space?
     
  9. Nov 8, 2016 #8

    TSny

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    Yes, you integrate over all space. But you have to use the correct wavefunction for all of space.
     
  10. Nov 8, 2016 #9

    TSny

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    To put it another way, is there a need to integrate over regions of space where the wavefunction is zero?
     
  11. Nov 8, 2016 #10
    So you mean the integral should technically be ##\int_0^a |A|^2 sin^2 (n\pi r/a)dr + \int_a^{\infty} 0 dr## ??
     
  12. Nov 8, 2016 #11

    TSny

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    Yes.
     
  13. Nov 8, 2016 #12
    facepalm.jpg


    Ugh....makes sense. Sorry for being so dense. Thanks!
     
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