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Schrödinger eq problem

  1. Aug 17, 2008 #1
    The problem is the picture below. The thing I dont understand, since I also have the solution, is the fact that the "radial wavefunction is normalized to 1". And all the constants before it aswell. Why cant I move out the constants in front of the integral, normalize it and then get a competely different answear? And why shall this only equal to 1?

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Aug 17, 2008 #2

    G01

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    I don't know if I understand your question correctly, but I'll try to answer what I think your asking.

    Just because you can pull the normalization constant out of the integral, does not mean it is now separate from the probability function or that you can just drop it. It is still there. You can't just normalize the radial function again. Doing that would then give you the wrong constant (since now you would have two normalization constants multiplied together) and the function would no longer be normalized to one. You would be effectively "un-normalizing" the function. Does this address your first question?

    In regards to your second question,

    [tex]\int_{allspace}\psi^*\psi d\nu = 1[/tex]

    because [itex]\psi^* \psi[/itex] is a probability density and the integral gives a probability. For instance, if this integral was equal to 2, we would be saying that the particle has a 200% chance of being found anywhere. This makes no sense. The particles chance of being found "anywhere" has to be 100%. (It has be somewhere, right?) So, this is why we have to normalize our probability densities to 1. If we don't we get nonsense probabilities as answers.
     
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