To better understand the effect   ${\hat{\mathbf{S}}}^2$   has on Slater determinants, it will be useful to consider   ${\hat{\mathbf{S}}}^2$   written in terms of the ladder operators [Eq. (2.5)]

Since a single Slater determinant is always an eigenstate of   ${\hat{S}}_z$,   we can focus our attention on understanding what effect the operators   ${\hat{S}}_{\mp }{\hat{S}}_{\pm }$   have on a Slater determinant. It will be sufficient to show that when the ladder operators  (${\hat{S}}_{\pm }$)  are applied to a closed shell Slater determinant, the result is zero. But first, it will be shown that

which is a weaker statement than   ${\hat{S}}_{\pm }|{\displaystyle \Psi }\rangle =0$,   since, in general, for an operator   $\hat{A}$  

and   $\langle {\displaystyle \Psi }|{\displaystyle \Psi \mathbf{\text{'}}}\rangle =\lambda$   does not necessarily imply that   $|{\displaystyle \Psi }\rangle$   is an eigenstate of   $\hat{A}$   (i.e., $|{\displaystyle \Psi \text{'}}\rangle =\lambda |{\displaystyle \Psi }\rangle$),  where   $\lambda$   is a constant. In fact,   $\Psi$   will not be an eigenstate of   ${\hat{S}}_{\mp }{\hat{S}}_{\pm }$   and   ${\hat{\mathbf{S}}}^2$   if   $\Psi$   is an open shell Slater determinant in which the spins of the open shell electrons are not aligned.

To prove Eq. (B.2), the operator   ${\hat{S}}_{\mp }{\hat{S}}_{\pm }$   will be written as the sum of a one-particle operator   $\hat{U}$   and a two-particle operator   $\hat{V}$, i.e.,

where

The expectation value of Eq. (B.4) with respect to a many-electron state is given by

Next, we can use Eq. (5.10) to evaluate   $\langle {\displaystyle \Psi }|\hat{U}|{\displaystyle \Psi }\rangle$,   and Eq. (5.14), times a factor of two (since   ${\hat{s}}_{i\mp }{\hat{s}}_{j\pm }\text{\hspace{0.17em}and\hspace{0.17em}}{\hat{s}}_{j\mp }{\hat{s}}_{i\pm }$   must both be included in the summation) to evaluate   $\langle {\displaystyle \Psi }|\hat{V}|{\displaystyle \Psi }\rangle$;   this gives us

In our previous discussions of spin, we used the notation of Eq. (3.30), which maps a spatial wavefunction, ψ_i, to two spin-orbitals, χ_2i-1 and χ_2i. With this definition of the spin-orbital labels there is a one-to-one correspondence between spin-orbital label and the z-projection of the angular momentum quantum number, and the spin-up orbital follows the spin-down orbital sequentially for each spatial wavefunction. In this appendix, we will redefine the spin-orbital labels so our discussion can be easily generalized to orbital angular momentum. In ket notation, our restricted spin-orbitals are defined by

where   $|{\psi }_i\rangle =|n_i{l}_i{m}_{l_i}\rangle$, and we have introduced the superscript label (m_s) so the discussion that follows will be valid for orbital angular momentum with the appropriate definition of the spin-orbital indices and the substitutions   ${\chi }_i^{\left(m_s\right)}\to {\chi }_i^{\left(m_l\right)}$,     $\hat{\mathbf{S}}\to \hat{\mathbf{L}}$, etc. The superscript labels, (m_s) and (m_l), only serve to remind us whether the spin-orbital indices are defined to have a one-to-one correspondence with the quantum number m_s (or in the next section m_l). In the case of spin angular momentum, the redefinition is trivial

The main difference between this notation and the notation introduced in Chapter 2 [see Eqs. (2.29), (3.30), and (3.31)] is that here j odd labels a spin-down state   $\left(m_s=-\frac{1}{2}\right)$,   j even labels a spin-up state   $\left(m_s=+\frac{1}{2}\right)$,   and the ladder operators have the same effect on the spin-orbital subscript as the quantum number m_s (or in the next section, m_l). The effect the single-particle ladder operators have on the spin eigenstates can be summarized by the following equation

Since we are dealing with spin-half particles, Equation (B.12) can be written as

Equation (B.13) can be used to evaluate the matrix elements in Eqs. (B.7) and (B.8). The matrix elements of the single-particle operator in Eq. (B.7) are

where we have used the fact that   ${\hat{S}}_{-}={\hat{S}}_{+}^{†}$. The matrix elements of the two-particle operator in Eq. (B.8) are

and

If we substitute Eqs. (B.14), (B.15), and (B.16) into Eqs. (B.7) and (B.8), we obtain the following for a restricted closed-shell Slater determinant

Finally, from Eqs. (B.6) and (B.17) it is easy to see that for a restricted closed-shell Slater determinant we have

and hence   $\langle {\displaystyle \Psi }|{\hat{S}}^2|{\displaystyle \Psi }\rangle =0$,   as expected.

If   $\Psi$   includes open shells, then Eq. (B.16) is equal to zero if   ${\chi }_i^{\left(m_s\right)}\text{\hspace{0.17em}or\hspace{0.17em}}{\chi }_j^{\left(m_s\right)}$   corresponds to an open shell spin-orbital, and therefore the contribution to Eq. (B.8) from open shells is zero. Furthermore, the sum in Eqs. (B.7) can be split into two sums: one over closed shell orbitals and one over open-shell orbitals. But it was just shown that the net contribution from closed shells is zero, therefore only the open shells contribute to the total spin angular momentum of a many-electron wavefunction.

It is simple to repeat the above the above discussion to include open shell Slater determinants. The matrix elements of the operator   ${\hat{\mathbf{S}}}^2$   can be calculated using

where

and

By combining Eqs. (B.20), (B.21), and (B.22) one obtains

If all the open shell electrons have the same spin, then   $m_{s_i}=\pm s$,   and Eq. (B.23) becomes

where

and N_o is the number of open shell electrons,   $N_{\uparrow }$   is the total number of spin-up electrons, and   $N_{\downarrow }$   is the total number of spin-down electrons. If the spins of the open shell electrons are not aligned, then   $\Psi$   will not be an eigenstate of   ${\hat{S}}_{\mp }{\hat{S}}_{\pm }$   and, therefore, neither will it be an eigenstate of   ${\hat{\mathbf{S}}}^2$.