# Series Solutions to TISE

## 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)=$\sum$anxn+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)anxn+B-1
$\phi$''=$\sum$(n+B-1)(n+B)anxn+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)anxn+B-2+2x($\sum$(n+B)anxn+B-1)+((a(a-1))/x2)($\sum$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?

kreil
Gold Member
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

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

Last edited:
vela
Staff Emeritus
Homework Helper

## 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 $\phi$?

Oh there is! Good catch!

vela
Staff Emeritus