Series Solutions for TISE: Finding B in the Eigenvalue Problem H\psi=E\psi

[atomicpedals]

Homework Statement

The eigenvalue problem H\psi=E\psi for \phi becomes

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

assume that \phi(x)=\suma_nx^n+B, determine B.

2. The attempt at a solution

As a first step I took the first and second derivatives of \phi:

\phi'=\sum(n+B)a_nx^n+B-1
\phi''=\sum(n+B-1)(n+B)a_nx^n+B-2

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

-\sum(n+B-1)(n+B)a_nx^n+B-2+2x(\sum(n+B)a_nx^n+B-1)+((a(a-1))/x²)(\suma_nx^n+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?

 

[kreil]

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

 

[atomicpedals]

I'm certainly getting the impression there's either a (x-1)² or a(a-1) in the denominator...

 

[vela]

Homework Statement

The eigenvalue problem H\psi=E\psi for \phi becomes

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

Are you sure the (1-2E) term isn't multiplied by \phi?

 

[atomicpedals]

Oh there is! Good catch!

 

[vela]

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.

 

[atomicpedals]

Thanks for the help! I'll go look Frobenius up in Arfken.