Determinantal wave function of Li using LCAO?

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magnesium12
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Homework Statement



The wave function for lithium can be written as:

| 1sα(1) 1sβ(1) 2sα(1) | ##\frac{1}{\sqrt(3)} ## = ψ(1,2,3)
| 1sα(2) 1sβ(2) 2sα(2) |
| 1sα(1) 1sβ(2) 2sα(1) |

How can each row be a linear combination of atomic orbitals that makes a new orbital in which the electron actually exists if these new functions are made up of the old functions which all have different spin states?

Like in the first row, |1sα(1) 1sβ(1) 2sα(1)| represents a wave function made up of 3 other wave functions. But two of these wave functions are functions of α (positive spin) and one has β (negative spin) in it. So how can we combine all three orbitals to make a new orbital? You can only have +1/2 or -1/2 spin, nothing in between. So what would the spin state of the new orbital be?

Could someone clarify this for me? I'm really sorry if I didn't word the question clearly.

Homework Equations

The Attempt at a Solution

 
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magnesium12 said:
| 1sα(1) 1sβ(1) 2sα(1) | ##\frac{1}{\sqrt(3)} ## = ψ(1,2,3)
| 1sα(2) 1sβ(2) 2sα(2) |
| 1sα(1) 1sβ(2) 2sα(1) |
That last row should be 1sα(3) 1sβ(3) 2sα(3)

magnesium12 said:
How can each row be a linear combination of atomic orbitals that makes a new orbital in which the electron actually exists if these new functions are made up of the old functions which all have different spin states?
What you get from the Slater determinant is a 3-electron wave function. Using that particular set of atomic orbitals, you get a 3-electron orbital in which 2 electrons are spin-up and one spin-down.

magnesium12 said:
So how can we combine all three orbitals to make a new orbital? You can only have +1/2 or -1/2 spin, nothing in between. So what would the spin state of the new orbital be?
The total spin of 3 electrons can be 1/2 or 3/2. In the case at hand, the total spin is S = 1/2, and MS = +1/2.