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Hydrogen probability distribution

  1. Mar 1, 2014 #1
    I have found that:

    For l = 1:

    [tex]\sum_{m=-l}^l |Y_l^m|^2 = \frac{3}{4\pi}[/tex]

    For l = 2:

    [tex]\sum_{m=-l}^l |Y_l^m|^2 = \frac{5}{4\pi}[/tex]

    What significance does this have for the probability distribution in an hydrogen atom?
     
  2. jcsd
  3. Mar 1, 2014 #2

    jfizzix

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    the significance is that there are five-thirds as many states of orbital angular momentum l=2, than for l=1. This makes sense because there are five orbitals for l=2, and three orbitals for l=1, for all the values of m.
     
    Last edited: Mar 1, 2014
  4. Mar 1, 2014 #3

    Simon Bridge

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    Do you know what those terms mean?
    Do you know why you did the sum in each case?
     
  5. Mar 2, 2014 #4
    I think the numerator gives the degeneracy for a certain l?

    since -l ≤ m ≤ l, m can take on 2l+1 possible values.

    Like possible states for |n,l,m>:

    For l = 1:

    Possible states are |n,1,-1> and |n,1,0> and |n,1,1>
     
  6. Mar 2, 2014 #5

    Simon Bridge

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    Each spin/orbit state contributes a factor of ##1/4\pi## to the overall thingy being calculated.

    But I meant the terms on the RHS.

    For a particular |n,l,m> state, what does |Ylm|2 mean?

    Now - what was your question?
     
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