Molecular material ground state

[hokhani]

Why the ground state HOMO level of a molecule with two opposite-spin electrons, is singlet while we know that a level with two opposite-spin electrons may be singlet(s=0,m=0) or triplet(s=1,m=0)?

 

[DrDu]

The total wavefunction has to be antisymmetric. For two electrons in one orbital $\phi$ the Slater determinant can be written as
$\begin{vmatrix} \phi(1)s_+(1)& \phi(1) s_-(1)\\ \phi(2)s_+(2) &\phi(2) s_-(2) \end{vmatrix}= N\phi(1)\phi(2)(s_+(1)s_-(2)-s_-(1)s_+(2))$
where N is a normalization constant and s_+ and s_- are up or down spin eigenfunctions.
So as the orbital part of the wavefunction can only be symmetric, the spin function has to be antisymmetric.
With two orbital functions, you can form two Slater determinants and also create triplet eigenfunctions, e.g.
$N(\phi_a(1)\phi_b(2)-\phi_b(1)\phi_a(2))(s_+(1)s_-(2)+s_-(1)s_+(2))$

 

[hokhani]

The total wavefunction has to be antisymmetric. For two electrons in one orbital $\phi$ the Slater determinant can be written as
$\begin{vmatrix} \phi(1)s_+(1)& \phi(1) s_-(1)\\ \phi(2)s_+(2) &\phi(2) s_-(2) \end{vmatrix}= N\phi(1)\phi(2)(s_+(1)s_-(2)-s_-(1)s_+(2))$
where N is a normalization constant and s_+ and s_- are up or down spin eigenfunctions.
So as the orbital part of the wavefunction can only be symmetric, the spin function has to be antisymmetric.
With two orbital functions, you can form two Slater determinants and also create triplet eigenfunctions, e.g.
$N(\phi_a(1)\phi_b(2)-\phi_b(1)\phi_a(2))(s_+(1)s_-(2)+s_-(1)s_+(2))$

Thanks for your excellent and exact answer. But it remains another question;
writing the slater determinant this way:
$\begin{vmatrix} \phi_a(1)s_+(1)& \phi_b(1) s_-(1)\\ \phi_a(2)s_+(2) &\phi_b(2) s_-(2) \end{vmatrix}$
how can you separate spin and orbital parts as
$N(\phi_a(1)\phi_b(2)-\phi_b(1)\phi_a(2))(s_+(1)s_-(2)+s_-(1)s_+(2))$?

 

[DrDu]

As I said with two orbitals you can form two different Slater determinants. The wavefunction with definite spins I have written down are combinations of the two.
The second one being
$\begin{vmatrix} \phi_a(1)s_-(1)& \phi_b(1) s_+(1)\\ \phi_a(2)s_-(2) &\phi_b(2) s_+(2) \end{vmatrix}$

 

[hokhani]

As I said with two orbitals you can form two different Slater determinants. The wavefunction with definite spins I have written down are combinations of the two.
The second one being
$\begin{vmatrix} \phi_a(1)s_-(1)& \phi_b(1) s_+(1)\\ \phi_a(2)s_-(2) &\phi_b(2) s_+(2) \end{vmatrix}$

Excuse me. I didn't underestand what you mean by "with two orbitals you can form two different Slater determinants". Do you mean that
$\begin{vmatrix} \phi_a(1)s_-(1)& \phi_b(1) s_+(1)\\ \phi_a(2)s_-(2) &\phi_b(2) s_+(2) \end{vmatrix}$=$\begin{vmatrix} \phi_a(1)& \phi_b(1)\\ \phi_a(2) &\phi_b(2) \end{vmatrix}$$\begin{vmatrix} s_-(1)& \ s_+(1)\\ s_-(2) &\ s_+(2) \end{vmatrix}$?

 

[DrDu]

No. In post #3 you wrote down one of the two possible Slater determinants. In post #4 I posted the second one.
You should try to convince yourself that from these two determinants you can construct both the singlet and the triplet state.