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Schrödinger equation: P(r)>1 ?

  1. Nov 19, 2005 #1
    Schrödinger equation: P(r)>1 ???

    I have the solution of the Schrödinger equation for the ground state of the hydrogen electron. The solution ist:

    u100(r)=sqrt(1/(pi*a^3))*exp(-r/a)

    If I want to calculate some probabilty values I do this with:

    P(r)=4*pi*r^2*|u100|^2

    If I set r=10^-13 I get a value that is greater than 1, I get P(10^-13)=10^34.

    This cannot be. Whats wrong with my probability formula?
     
  2. jcsd
  3. Nov 19, 2005 #2

    Hurkyl

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    P is a probability... density. To get the probability that it lies in a region, you need to integrate over that region.
     
  4. Nov 19, 2005 #3
    With what function are the graphs plotted? (I mean one like this:
    http://panda.unm.edu/courses/finley/P262/Hydrogen/img106.gif)

    But to calculate the charge distribution of the electron I can take:

    Charge(r)=-e*abs(u100)^2

    thanks for your help
     
  5. Nov 19, 2005 #4

    Hurkyl

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    I'm not familiar with those graphs -- I might look at the Bernoulli or Airy functions, but if it's not them, I don't think I could guess. (Also, I think I've read that the Airy functions appear in dealing with the hydrogen atom)
     
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