sanalsprasad
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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 y = \sin \theta and then assumed that Y is a polynomial i.e. Y(y) = \sum b_n x^n and then arrived at the recursion formula :
b_{n-2} = - \frac{n^2 - m^2}{l(l+1) - (n-1)(n-2) }b_{n} , where l is maximum value of n for which b_n 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 b_l and then finding b_{l-2} and then b_{l-4} and so on in terms of b_l.
Thanks for help!
http://skisickness.com/2009/11/22/
To solve the angular part they gave the substitution of y = \sin \theta and then assumed that Y is a polynomial i.e. Y(y) = \sum b_n x^n and then arrived at the recursion formula :
b_{n-2} = - \frac{n^2 - m^2}{l(l+1) - (n-1)(n-2) }b_{n} , where l is maximum value of n for which b_n 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 b_l and then finding b_{l-2} and then b_{l-4} and so on in terms of b_l.
Thanks for help!
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