# Roothaan matrix equation equation orbital coefficients

To go from the Hartree_Fock

$$f\psi{_i}=\epsilon\psi{_i}$$

to the Roothaan equation

$$FC=SC\epsilon$$.

equation we expand the orbitals as

$$\psi{_i}=\sum C_{\mu i}\phi_i$$

But for Helium atom we have just one orbital function so the coefficient $$C_{\mu i}$$ should be a vector not a matrix. My question is what is the interpretation of the matrix coefficient $$C$$ in the Roothaan equation $$FC=SC\epsilon$$ for the helium atom since we've got just one spatial orbital function?

The $$\mathbf{C}$$ matrix will always be $$K\times K$$, where $$K$$ is the size of the finite basis set you are using. That is to say, with a basis set of size $$K$$, you will produce exactly $$K$$ MOs. This means you will never wind up with $$\mathbf{C}$$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $$\mathbf{C}$$ matrix is $$2\times 2$$. While helium has only one occupied spatial orbital, it will have $$K-1$$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.
• So what you mean is that if $\mathbf{C}$ is $2\times 2$ matrix, one column of the matrix will be the coefficients of the occupied spatial orbital(the one with smaller $\epsilon$ ) and the other coefficients of a virtual excited state ? Jan 10 '19 at 17:45
• @amiltonmoreira exactly. The only way you wouldn't wind up with a $\mathbf{C}$ matrix per se is if you only used a single basis function, though you could obviously still treat it as a $1\times 1$ matrix. Jan 10 '19 at 17:52