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 \(S=0\) \(M_{s}=0\)

Step by step solution

01

Verify the claim for \(\Phi_{1}\) and \(\Phi_{4}\)

Evaluate the total spin operator on the Slater determinant \(\Phi_{1}\) and \(\Phi_{4}\). The total spin operator \(S^{2}\) is \(S^{2} = S_1^2 +S_2^2 + 2S_1\cdot S_2\). Since the total spin quantum number \(s\) for an electron is 1/2 and the only non-zero term when applying \(S^{2}\) is \(2S_1\cdot S_2\), it can be concluded that both \(\Phi_{1}\) and \(\Phi_{4}\) are eigenfunctions of \(S^{2}\) with eigenvalue of \(2 \hbar^{2}\)

02

Find linear combinations for \(\Phi_{2}\) and \(\Phi_{3}\)

The linear combination of \(\Phi_{2}\) and \(\Phi_{3}\) which is itself an eigenfunction of \(S^{2}\) with \(S=1, M_{s}=0\) (eigenvalue \(2\hbar^{2}\)) is \(\Phi_{+}=\frac{1}{\sqrt{2}}(\Phi_{2}+\Phi_{3})\). The orthogonal combination, which is itself an eigenfunction of \( S^{2}\) with \(S=0, M_{s}=0\) (eigenvalue \(0\)) is \(\Phi_{-}=\frac{1}{\sqrt{2}}(\Phi_{2}-\Phi_{3})\).

03

Confirm Spin Values

To confirm the claim, it is needed to compute the expectation values of \(S^{2}\) for these combinations. For \(\Phi_{+}\), \(S=1, M_{s}=0\) and for \(\Phi_{-}\), \(S=0, M_{s}=0\). This shows that the linear combinations \(\Phi_{+}\) and \(\Phi_{-}\) correspond to \(S=1, M_{s}=0\) and \(S=0, M_{s}=0\) respectively.

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