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Why should the quantum number m be less than equal to l.

  1. Jul 10, 2013 #1
    I was reading the following article regarding solution of wavefunction of hydrogen :
    http://skisickness.com/2009/11/22/

    To solve the angular part they gave the substitution of [itex] y = \sin \theta [/itex] and then assumed that Y is a polynomial i.e. [itex] Y(y) = \sum b_n x^n [/itex] and then arrived at the recursion formula :
    [itex] b_{n-2} = - \frac{n^2 - m^2}{l(l+1) - (n-1)(n-2) }b_{n} [/itex] , where [itex]l[/itex] is maximum value of [itex]n[/itex] for which [itex]b_n[/itex] is non-zero.

    Then they say that :
    " There must be a minimum value of n; otherwise, the series will diverge at y=0. Given l, for the series to converge, it is necessary that |m|=l-2k, with k greater than or equal to zero and less than or equal to l/2. For even or odd l, this series gives l+1 solutions. This solution gives the eigenfunctions with both odd or both even m and l. "

    I did not understand why the series converges only for this particular condition. Is it something to do with starting with [itex]b_l[/itex] and then finding [itex]b_{l-2}[/itex] and then [itex]b_{l-4}[/itex] and so on in terms of [itex]b_l[/itex].
    Thanks for help!
     
    Last edited: Jul 10, 2013
  2. jcsd
  3. Jul 10, 2013 #2
    I don't care to post the mathematical reasoning because the physical reason is easy to understand.

    How can a projection of a vector on any arbitrary axis (angular momentum in the case) be larger than the magnitude of the vector?
     
  4. Jul 10, 2013 #3

    hilbert2

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    When you use that recursive formula to generate polynomials (so called Legendre polynomials), every term in the resulting polynomial has coefficient larger than 1, and the coefficients only grow when going to higher-power terms. There's no way how that kind of a series could converge unless it terminates at some point. And the series terminates only if m ≤ l (with that condition, all coefficients are zero after some term).
     
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