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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?

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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.

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  • $\begingroup$ 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 ? $\endgroup$ Jan 10 '19 at 17:45
  • $\begingroup$ @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. $\endgroup$
    – Tyberius
    Jan 10 '19 at 17:52

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