Series Solutions to TISE

  • #1
208
7

Homework Statement



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?
 

Answers and Replies

  • #2
kreil
Insights Author
Gold Member
668
68
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
 
  • #3
208
7
I'm certainly getting the impression there's either a (x-1)2 or a(a-1) in the denominator...
 
Last edited:
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629

Homework Statement



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
Are you sure the (1-2E) term isn't multiplied by [itex]\phi[/itex]?
 
  • #5
208
7
Oh there is! Good catch!
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629
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.
 
  • #7
208
7
Thanks for the help! I'll go look Frobenius up in Arfken.
 

Related Threads on Series Solutions to TISE

  • Last Post
Replies
2
Views
2K
Replies
7
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
7
Views
2K
Replies
2
Views
730
Top