Quantum Physics Question - Second Year University level.

Click For Summary

Homework Help Overview

The discussion revolves around the normalization of an antisymmetric eigenfunction of the helium atom in quantum physics. Participants are exploring the mathematical properties of the eigenfunction and its components.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the form of the antisymmetric eigenfunction and the requirement to show that the normalization constant is C = 1/root(2). There are attempts to prove the normalization condition by expanding the expression and checking orthonormality.

Discussion Status

Participants are actively engaging with the problem, providing hints and exploring the implications of the normalization condition. There is a collaborative effort to clarify the steps needed to arrive at the normalization constant, with some guidance offered regarding the mathematical expansion.

Contextual Notes

There is an emphasis on proving mathematical relationships without providing complete solutions, reflecting the forum's homework help guidelines.

coffeem
Messages
91
Reaction score
0
Hi there,

I have some questions which I cannot solve from my Quantum book. I would appretiate some hints as to how they should be solved. Thanks.

Question

The antisymmetric eigenfunction of the helium atom can be written in the form UA - C(uab-uba), where uab and uba are orthonormal. Show that the normalisation constant C = 1/root(2).
 
Physics news on Phys.org
Hi coffeem! :smile:
coffeem said:
The antisymmetric eigenfunction of the helium atom can be written in the form UA - C(uab-uba), where uab and uba are orthonormal. Show that the normalisation constant C = 1/root(2).

well, that means you have to prove that (uab-uba)(uab-uba)* = 2 :wink:
 
tiny-tim said:
Hi coffeem! :smile:


well, that means you have to prove that (uab-uba)(uab-uba)* = 2 :wink:

Thanks for your help, I really do appreciate it. I can expand the brackets and get:

UabUab + UbaUba = 2 (since UabUba = 0)

So would a reasonable answer be:

Since (Uab-Uba)^2 is 2, to normalise this to 1, we must divide the eigenfunction by the square root of this?

Thanks.
 
coffeem said:
Thanks for your help, I really do appreciate it. I can expand the brackets and get:

UabUab + UbaUba = 2 (since UabUba = 0)

So would a reasonable answer be:

Since (Uab-Uba)^2 is 2, to normalise this to 1, we must divide the eigenfunction by the square root of this?

Thanks.

Yes, except I'd say "we must divide it by the square root" …

any multiple of (Uab-Uba) is an eigenfunction. :wink:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
579
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
46
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K