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Normalizing Wave Functions

  1. Nov 17, 2014 #1
    Hi, so I'm having a bit of trouble understanding the normalization of radial waves. I understand that the equation is the integral of ((R^2)r^2)=1 but I'm not understanding how the process works. I need the normalization constant on R32. I got the function to come out to be (((r^2)(Co)(e^(-r/3a))/(27a^3)) so I can take that and plug it into the normalization equation (with r/a=z) to get ((Co)/(27a^3)^2) (a^4) integral of (z^2)(e^(-z))(z^2)dz then I combined both the z^2's so it's now the stuff out front integral of ((z^4)(e^(-z))dz and this is where I'm stuck. From what I keep seeing is that people are getting actual numerical values and I don't understand how that part works. Please help.
     
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  3. Nov 17, 2014 #2

    Simon Bridge

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    The process is the same as when you normalize any wavefunction - the extra r^2 comes from the volume element for spherical-polar coordinates.
    You get the numerical values because it is a definite integral. What are the limits of the integration?
     
  4. Nov 17, 2014 #3
    Well the initial equation states that it's from 0 to infinity. That's were I don't see an actual value coming into place. Unless the bounds somehow change?
     
  5. Nov 17, 2014 #4

    Simon Bridge

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    Lets makes sure I follow you - you are trying to evaluate $$\int_0^\infty z^4e^{-z}\;dz$$ ... with a bunch of constant terms out the front?
    If so - then what do you get for the indefinite integral?
     
  6. Nov 17, 2014 #5
    Yes. And I just plugged it in on mathematica and got an answer of 24. I was trying to do it by hand because I hate taking the easy way out, but thanks so much for responding!
     
  7. Nov 17, 2014 #6

    Orodruin

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    There is a particular trick for solving integrals of this type. Try computing
    $$
    \int_0^\infty e^{-st} dt
    $$
    and then differentiate wrt ##s## a few times.
     
  8. Nov 17, 2014 #7

    Simon Bridge

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    ... hint: integration by parts.
    You need to do the "by parts" trick more than once.
     
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