Spin expectation values for RHF

We can derive the spin expectation values for RHF from the UHF spin expectation values. From the UHF case, we already have: \[\begin{align} \require{physics} & \ev{\hat{S}_x}{\text{core}} = 0 \\ % & \ev{\hat{S}_y}{\text{core}} = 0 \thinspace . \end{align}\] Furthermore, since in the RHF wave function model: \[\begin{equation} N_\alpha = N_\beta = N_P \thinspace , \end{equation}\] the expectation value for \(\hat{S}_z\) also vanishes: \[\begin{equation} \ev{\hat{S}_z}{\text{core}} = 0 \thinspace . \end{equation}\]

Since the \(\alpha\)- and \(\beta\)-spinors are required to be equal, the expectation value of \(\hat{S}_- \hat{S}_+\) also vanishes: \[\begin{equation} \ev{\hat{S}_- \hat{S}_+}{\text{core}} = 0 \thinspace , \end{equation}\] which means that the expectation value for \(\hat{S}^2\) is zero: \[\begin{equation} \ev{\hat{S}^2}{\text{core}} = 0 \thinspace . \end{equation}\]