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Associated vs. Non-associated Laguerre Polynomials

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Could someone pls clarify if the value of x changes from just Laguerre polynomial to associated one? I am confused about the role of variable x.

    2. Relevant equations
    From what I have learned in the class, I understand that L1n(x) = d/dx Ln(x), n = 1, 2, 3...

    3. The attempt at a solution
    Because L1(x) = 1 - x and L2(x) = 2 - 4x + x2
    I did:
    L11(x) = d/dx L1(x) = d/dx (1 - x) = -1
    L12(x) = d/dx L2(x) = d/dx (2-4x+x2) = 2x - 4 = 2(x - 2)....I wonder if x is a different function of radius in L11(x) (as in 1s orbital) and L12(x) (as in 2s orbital)? I am assuming the orbital polynomial on the basis of node....as in:
    L11(x)= 1 because of 0 nodes...hence 1s
    L12(x) = 2x -4 because of 1 node.....hence 2s
     
  2. jcsd
  3. Dec 9, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Yes. The associated Laguerre polynomial found in hydrogenic wave functions is expressed as
    $$
    L_{n+l}^{2l+1}(\rho)
    $$
    with
    $$
    \rho \equiv \frac{2 Z}{n a} r
    $$
    where ##n## the principal quantum number and ##l## the angular momentum quantum number. You see that ##n## enters as a scaling factor in the argument of the associated Laguerre polynomial. Note also that the association between ##n## and ##l## and the associated Laguerre polynomial is a bit more complicated than your simple assumption based on the number of nodes, as angular nodes have to be considered also.
     
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