Recursion relation in the hydrogen atom

[sunrah]

Homework Statement

Given the following recursion formula:

$b_{j} = 2 \frac{kj - a}{j(j+1)-l(l+1)} \cdot b_{j-1}$

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

$b_{j = l} \neq 0$ if $b_{j - 1} = 0$.

Homework Equations

The Attempt at a Solution

This is a part question and I really can't see why. if b_l-1 = 0 then b_l = 0 also. But then all subsequent b_j as well !

 

[grindfreak]

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.

 

[sunrah]

you could try l'hopital's rule on the first equation with respect to l.

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 b_j are constants not dependent on j so the factor b_j-1=0 would remain after differentiating.

 

[grindfreak]

Perhaps not l'hospital's 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.

 

[sunrah]

second question, how would you use this recursion formula? If b_j-1:= 0, then when moving forward with increasing j all subsequent b_j 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.

 

[grindfreak]

Hmmm, I'm not sure about moving backwards since, like you've stated, every b_j is zero due to the fact that every numerator of the recursion is zero. The one exception being b_j=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.

 

[sunrah]

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

 

[grindfreak]

Yes I noticed this, most texts use:

$$c_{j+1}=\frac{2(j+l+1-n}{(j+1)(j+2(l+1))}c_{j}\$$

I believe.