Hydrogen atom

  • Thread starter eku_girl83
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I solved the differential equation for theta portion of the hydrogen wave function using a power series solution. I got a sub n+2 = a sub n ((n(n+1)-C)/(n+2)(n+1)). I then truncated the power series at n = l to get
C= l(l+1).

I know need to use the recursion formula I found to find the l = 0, 1, 2, and 3 solutions to the differential equation. Do I simply plug l in for n? If so, I get for l = 0, a2 = -Ca0/2. Is this the SOLUTION to the D.E. for
l = 0, or do I need to do something else?
Similarly, for l = 1, I get a3= a1 (2-C)/6.

Any help appreciated!
 

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  • #2
Meir Achuz
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Put in C=L(L+1) and do the arith.
The last step is to require P_L(1)=1, which is the normalization conditon for Legendre polynomials (the name of the theta solutions).
 
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If I sub in C= L(L+1), then I get a2 = -l(l+1)a0/2. But this is the solution for l = 0...so if I let l = 0, then I get a2 =0. This is also true for a3, a4, a5, .... Is this ok?

Also...when I normalize...Do I do the integral of a2^2 from -1 to 1 = 1 (since x = cos theta) and solve for a0??

I'm just a bit confused...thanks!
 
  • #4
Meir Achuz
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eku_girl83 said:
If I sub in C= L(L+1), then I get a2 = -l(l+1)a0/2. But this is the solution for l = 0...so if I let l = 0, then I get a2 =0. This is also true for a3, a4, a5, .... Is this ok?
Yes. For any L, the solution is a poynomial of degree L.
For L=0, P_0=1. For L_1. P_1=1, etc.
 
  • #5
Meir Achuz
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eku_girl83 said:
Also...when I normalize...Do I do the integral of a2^2 from -1 to 1 = 1 (since x = cos theta) and solve for a0??
QUOTE]
The usual Legendre polynomials are normalized so that P_L(1)=1 for each value of L. No integral is involved. This is different (and easier) than the usual normalization of functions. For your problem, the normalization may not be necessary.
 
  • #6
Meir Achuz
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eku_girl83 said:
I'm just a bit confused...thanks!
It may be time to look in a Math Physics book under
Legendre Polynomials in the index.
 

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