Is This Calculation of Sz for Molecular Orbitals Correct?

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

Sz_1 \psi = 1/\sqrt{2} ~\psi_b^2 ~(-\alpha_1 ~ (1/2) \hbar \beta_2 - \beta_1~(1/2) \hbar \alpha_2)
You means Sz2 I think here. But the second term should be positive since it's an alpha, no?
 
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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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