Quantum Physics Question - Second Year University level.

AI Thread Summary
The discussion focuses on solving a normalization problem related to the antisymmetric eigenfunction of the helium atom. Participants clarify that to show the normalization constant C equals 1/root(2), one must prove that (uab - uba)(uab - uba)* equals 2. The calculation involves expanding the brackets and recognizing that UabUba equals zero, leading to the conclusion that (Uab - Uba)^2 is 2. To normalize this to 1, it is confirmed that the eigenfunction must be divided by the square root of 2. The conversation emphasizes the importance of understanding eigenfunctions and their normalization in quantum physics.
coffeem
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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).
 
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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:
 
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