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Series Solutions to TISE

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    The eigenvalue problem H[itex]\psi[/itex]=E[itex]\psi[/itex] for [itex]\phi[/itex] becomes

    -[itex]\phi[/itex]''+2x[itex]\phi[/itex]'+((a(a-1))/x2)[itex]\phi[/itex]+(1-2E)=0

    assume that [itex]\phi[/itex](x)=[itex]\sum[/itex]anxn+B, determine B.

    2. The attempt at a solution

    As a first step I took the first and second derivatives of [itex]\phi[/itex]:

    [itex]\phi[/itex]'=[itex]\sum[/itex](n+B)anxn+B-1
    [itex]\phi[/itex]''=[itex]\sum[/itex](n+B-1)(n+B)anxn+B-2

    and then substituted these back into -[itex]\phi[/itex]''+2x[itex]\phi[/itex]'+((a(a-1))/x2)[itex]\phi[/itex]+(1-2E)=0; which is

    -[itex]\sum[/itex](n+B-1)(n+B)anxn+B-2+2x([itex]\sum[/itex](n+B)anxn+B-1)+((a(a-1))/x2)([itex]\sum[/itex]anxn+B)+(1-2E)=0

    And it's at this point (assuming I'm working correctly up to here) that I stop-short mentally; how do I go about solving this monster for B?
     
  2. jcsd
  3. Sep 26, 2011 #2

    kreil

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    Try computing the terms for n=0, then n=1, n=2, and n=3 then grouping the terms by powers of x to see if any noticeable patterns emerge
     
  4. Sep 26, 2011 #3
    I'm certainly getting the impression there's either a (x-1)2 or a(a-1) in the denominator...
     
    Last edited: Sep 26, 2011
  5. Sep 27, 2011 #4

    vela

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    Are you sure the (1-2E) term isn't multiplied by [itex]\phi[/itex]?
     
  6. Sep 27, 2011 #5
    Oh there is! Good catch!
     
  7. Sep 27, 2011 #6

    vela

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    Look up the method of Frobenius in your math methods book. That's what you're doing here.

    To find B, find the relation a0 must satisfy. This is called the indicial equation. By assumption, a0 is not equal to 0, so the relation will only hold for certain values of B.
     
  8. Sep 27, 2011 #7
    Thanks for the help! I'll go look Frobenius up in Arfken.
     
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