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Recursion relation in the hydrogen atom

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the following recursion formula:

    [itex]b_{j} = 2 \frac{kj - a}{j(j+1)-l(l+1)} \cdot b_{j-1}[/itex]

    (where a, k and l are constants) how can

    [itex]b_{j = l} \neq 0[/itex] if [itex]b_{j - 1} = 0[/itex].

    2. Relevant equations

    3. The attempt at a solution

    This is a part question and I really can't see why. if bl-1 = 0 then bl = 0 also. But then all subsequent bj as well !!!
     
  2. jcsd
  3. Apr 24, 2012 #2
    Well firstly bj=l = (2(kl-a))/0*bj-1 thus if bj-1 = 0 we come to the indeterminant form 0/0 . This implies that bj=l could possibly be nonzero. I know this isn't very mathematically rigorous, but I suppose you could try l'hopital's rule on the first equation with respect to l.
     
  4. Apr 24, 2012 #3
    hi, I don't see how that would help. surely you'd have to differentiate w.r.t. j and take the limit as j → l but my understanding is that the bj are constants not dependent on j so the factor bj-1=0 would remain after differentiating.
     
  5. Apr 24, 2012 #4
    Perhaps not l'Hopitals rule, but the only way that bl-1 can equal zero is if kl-a=0 or l=a/k. This also makes the numerator of bl=j = 0 as well as it's denominator. Since you end up with the indeterminate form of 0/0, we don't know for sure that bl=j = 0 and thus might actually be nonzero. I believe your problem was looking for speculation and not necessarily a formal proof.
     
  6. Apr 25, 2012 #5
    second question, how would you use this recursion formula? If bj-1:= 0, then when moving forward with increasing j all subsequent bj would also equal zero. Is it possible to move back from a non zero upper limit perhaps j = n, where n is the main quantum number.
     
  7. Apr 25, 2012 #6
    Hmmm, I'm not sure about moving backwards since, like you've stated, every bj is zero due to the fact that every numerator of the recursion is zero. The one exception being bj=l where both the denominator and the numerator are zero. Since this is the case, we end up with an indeterminate form. Now where to go from there I'm not sure, since it is neither zero, nor infinite, there is definitely a chance that it is nonzero. But, judging by how the question is worded I don't think they're looking for a rigorous proof, but how it is possible that it is nonzero.
     
  8. Apr 25, 2012 #7
    hi, this formula is used to calculate the eigenstates of the hydrogen atom (at least that's the context of how we derived it in class) so it should produce a range of non-zero results. It's weird because I can't find this formula in the literature, all the books I know use a different one without the pole at j = l
     
  9. Apr 25, 2012 #8
    Yes I noticed this, most texts use:

    [tex]c_{j+1}=\frac{2(j+l+1-n}{(j+1)(j+2(l+1))}c_{j}\[/tex]

    I believe.
     
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