Is This Calculation of Sz for Molecular Orbitals Correct?

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Homework Help Overview

The discussion revolves around the calculation of the total spin operator Sz for molecular orbitals, specifically examining the contributions from individual spin states (alpha and beta) and their representation in a Slater determinant framework.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of Sz(total) by summing Sz1 and Sz2, questioning the signs and contributions of the alpha and beta states in the wavefunction. There is a focus on the interpretation of terms in the determinant and the implications of the signs in the calculations.

Discussion Status

Some participants express uncertainty about the correctness of the calculations, particularly regarding the signs of the terms in the wavefunction. Clarifications are sought, and there is acknowledgment of mistakes in reasoning, indicating an ongoing exploration of the topic.

Contextual Notes

There is mention of the difficulty in reading the calculations without LaTeX formatting, which may impact the clarity of the mathematical expressions being discussed.

gazepdapi1
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[SOLVED] molecular orbital wavefunctions

http://img255.imageshack.us/img255/7012/37626561fb6.jpg

starting with

Sz(total) = Sz1 + Sz2
Sz(alpha) = +(1/2)hbar(alpha)
Sz(beta) = -(1/2)hbar(beta)

I first found the determinant and then simplified

psi = 1/sqrt(2)psi(b)^2[alpha(1)beta(2) - alpha(2)beta(1)]

So,

Sz1psi = 1/sqrt(2)psi(b)^2[beta(2)(1/2)h(bar)alpha(1) + alpha(2)(1/2)h(bar)beta(1)]

Sz1psi = 1/sqrt(2)psi(b)^2[-alpha(1)(1/2)h(bar)beta(2) - beta(1)(1/2)h(bar)alpha(2)]

So...
Sztotal = Sz1 + Sz2 = 0

Can someone please verify if this is correct?
I would appreciate it.
thank you
 
Last edited by a moderator:
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I realize it's hard to read without the LaTeX, but no one can check if it's correct or not?
 
yes it seems to be ok,

learn LaTeX, it is worth it :)
 
nertil1 said:
http://img255.imageshack.us/img255/7012/37626561fb6.jpg

starting with

Sz(total) = Sz1 + Sz2
Sz(alpha) = +(1/2)hbar(alpha)
Sz(beta) = -(1/2)hbar(beta)

I first found the determinant and then simplified

psi = 1/sqrt(2)psi(b)^2[alpha(1)beta(2) - alpha(2)beta(1)]

So,
[tex] Sz_1 \psi = 1/\sqrt{2} ~\psi_b^2 (\beta_2 ~ (1/2) \hbar \alpha_1 + \alpha_2 ~(1/2) \hbar \beta_1)[/tex]
I don't understand that result. Shouldn't the second term be negative since it's a beta?

[tex]Sz_1 \psi = 1/\sqrt{2} ~\psi_b^2 ~(-\alpha_1 ~ (1/2) \hbar \beta_2 - \beta_1~(1/2) \hbar \alpha_2)[/tex]
You means Sz2 I think here. But the second term should be positive since it's an alpha, no?
 
Last edited by a moderator:
No because when you multiply by the -(1/2)hbar, the beta portion (the second term) becomes a positive, the opposite goes for the alpha, I think, and yes it should be an Sz2.
 
nertil1 said:
No because when you multiply by the -(1/2)hbar, the beta portion (the second term) becomes a positive, the opposite goes for the alpha, I think, and yes it should be an Sz2.

you are absolutely right, my mistake. For some reason, when i replied to your post, I forgot about the Slater determinant and I was thinking about a symmetric wavefunction. Sorry about that.
 
Thank you all for your help
 

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