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Wavefunction normalization

  1. Dec 31, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm trying to determine a normalization value, A, for the following wavefunction:

    [tex]\Psi = Ax{^2}exp(-\alpha x)}, x>0[/tex]
    [tex]\Psi = 0, x<0[/tex]

    In the past, I've had an i term in my exponential, so when applying the Normalization Condition:

    [tex]\int|\Psi(x)|^2 dx = \int\Psi{^*}(x) \Psi(x) dx[/tex]

    the exponentials always multiply to equal one, leaving me with an easy route to getting the normalization factor.

    However in this case, I'm left with the following integral:

    [tex]\int|\Psi(x)|^2 dx = \int A{^2}x{^4}exp(-2\alpha x)} dx[/tex]

    ...which seems horrible!

    Can anyone advise what I'm doing wrong here? I'm sure there's a simpler way...
     
  2. jcsd
  3. Dec 31, 2007 #2

    George Jones

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    This looks OK.

    Have you ever seen tabular integration by parts?
     
  4. Dec 31, 2007 #3
    No, you're doing in the right way. Just do your integral by parts now.

    You can use the gamma function to do it faster.
     
    Last edited: Dec 31, 2007
  5. Dec 31, 2007 #4

    Never...

    I tried it in an online integrator and got a horrendous answer so assumed I'd gone drastically wrong somewhere...
     
  6. Dec 31, 2007 #5

    George Jones

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    I tried (maybe not very successfully) to explain it here.

    Most of the terms will be zero; remember, as [itex]x \rightarrow \infty[/itex], an exponential dominates any power of [itex]x[/itex].
     
  7. Dec 31, 2007 #6

    George Jones

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    Sorry, you need to see the whole thread.

    It looks a little complicated,but it's very easy and useful once you get the hang of it.
     
  8. Dec 31, 2007 #7
    OK, I think I've followed that George Jones;

    [tex]\int|\Psi(x)|^2 dx = \int A{^2}x{^4}exp(-2\alpha x)} dx[/tex]

    Which, using your method, I get to be:

    [tex]exp(-2\alpha x) \left(-\frac{x{^4}}{2\alpha} + \frac{x{^3}}{\alpha{^2}} - \frac{3x{^2}}{2\alpha{^3}} + \frac{3x}{2\alpha{^4}} - \frac{3}{4\alpha{^5}} \right)[/tex]


    Which I *think* leads to [tex]\frac{3A{^2}}{4\alpha{^5}} = 1[/tex] or [tex]A = \sqrt{4/3}\alpha{^\frac{5}{2}}[/tex]

    Would someone be able to verify that? I'm paranoid about this being wrong now!!
     
  9. Dec 31, 2007 #8
    Seems good, from my calculation, as I arrive, too, at
    [itex]1=\frac{3A^2}{4\alpha ^5}[/itex]

    by two differents methods (that is, by the use of the gamma function and with the long method of integration by parts).
     
    Last edited: Dec 31, 2007
  10. Dec 31, 2007 #9
    Awesome, thanks for your help George Jones and Erythro73!

    That technique is awesome - I'll definitely be using it again!!
     
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