A single Slater determinant is not necessarily an cigenfunction of the total
clectron spin operator. Therefore, even within the Hartree-Fock approximation,
for the wavefunction \(\Phi_{0}\) to be an eigenfunction of \(S^{2},\) it might
have to be expressed as a linear combination of Slater determinants. The
linear combination is referred to as a spin-adapted configuration. As a simple
example, consider a two-electron system with four possible Slater
determinants:
$$\begin{array}{l}
\Phi_{1}=\left(\frac{1}{2}\right)^{1 / 2}
\operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1)
\psi_{2}\left(r_{2}\right) \alpha(2)\right| \\
\Phi_{2}=\left(\frac{1}{2}\right)^{1 / 2}
\operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1)
\psi_{2}\left(r_{2}\right) \beta(2)\right| \\
\Phi_{3}=\left(\frac{1}{2}\right)^{1 / 2}
\operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1)
\psi_{2}\left(r_{2}\right) \alpha(2)\right| \\
\Phi_{4}=\left(\frac{1}{2}\right)^{1 / 2}
\operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1)
\psi_{2}\left(r_{2}\right) \beta(2)\right|
\end{array}$$
(a) Show that the Slater determinants \(\Phi_{1}\) and \(\phi_{4}\) are themselves
cigcnfunctions of \(S^{2}\) with cigenvaluc \(2 \hbar^{2}\) (corresponding to
\(S=1\) ). (b) From \(\Phi_{2}\) and \(\Phi_{3}\), determine two linear
combinations, one of which corresponds to \(S=1, M_{s}=0\) and the other of
which corresponds to \(S=0\) \(M_{s}=0\)