Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hydrogen probability distribution
Loading...