# Slater Determinant for simple covalent bond

1. Jun 9, 2012

### nigelscott

I have read that when 2 Hydrogen atoms come together their individual spacial wavefunctions overlap in the following way:

ψsymmetric = ψa + ψb ... bonding case
ψasymmetric = ψa - ψb ... antibonding case

How do you express this in terms of the Slater Determinant?

2. Jun 10, 2012

### cgk

You are mixing up something. The bonding and anti-bonding wave functions you wrote down are one-particle wave functions ("molecular orbitals"). A Slater-Determinant is a N-electron wave function; in particular, it is a N-electron wave function you get by taking a number of molecular orbitals and putting them into a determiant. From the bonding and anti-bonding MOs you wrote down, one can form the following 2-electron Slater determinants with Sz = 0:
$$\Psi(x_1, x_2) = \phi_+(r_1)A(s_1)\cdot \phi_+(r_2)B(s_2) - \phi_+(r_2)A(s_2)\cdot\phi_+(r_1)B(s_1)$$
$$\Psi(x_1, x_2) = \phi_+(r_1)A(s_1)\cdot \phi_-(r_2)B(s_2) - \phi_+(r_2)A(s_2)\cdot\phi_-(r_1)B(s_1)$$
$$\Psi(x_1, x_2) = \phi_-(r_1)A(s_1)\cdot \phi_+(r_2)B(s_2) - \phi_-(r_2)A(s_2)\cdot\phi_+(r_1)B(s_1)$$
$$\Psi(x_1, x_2) = \phi_-(r_1)A(s_1)\cdot \phi_-(r_2)B(s_2) - \phi_-(r_2)A(s_2)\cdot\phi_-(r_1)B(s_1)$$
where i used $$\phi_\pm$$ for the bonding/antibonding MO, xi = (ri,si) are the combined space/spin coordinates of the two electrons, and A/B are the alpha/beta spin functions (or up/down, if you prefer that). In this case the first determinant (in which both electrons sit in the bonding MO) would be the lowest energy single determinant close to equilibrium (i.e., the Hartree-Fock determinant). If you look at large separation, however, you will have substantial weight in all determinants if you express the wave function in the bonding/antibonding basis instead of the localized basis.