# How to calculate S² value of a broken-symmetry wave function?

$$S$$ represents spin, signifies the number of unpaired electrons in the system. For example, if the number of unpaired electrons is $$1$$, then $$S=1/2$$. $$S^2$$ is calculated as $$S(S+1)$$.

From what I have read, the $$S^2$$ value of a broken-symmetry singlet (contaminated by a triplet) is $$1.0$$, which is calculated to be the average of the singlet and triplet $$S^2$$. Similarly, the $$S^2$$ of broken-symmetry doublet (contaminated by the quartet state) turns out to be $$1.75$$ (average of a doublet and a quartet). I would like to know how these average values are being calculated. I understand that these are weighted averages (as $$1.75 \neq 0.5\cdot(0.75+3.75)$$, where $$0.75$$ and $$3.75$$ are $$S^2$$ values of doublet and quartet, respectively), but I don't know how that weighting is being done. I would be grateful if someone could provide a detailed explanation (mathematical derivation) of this.

The spin operators for the $$p^\textrm{th}$$ MO are defined as follows: \begin{align} S_{x,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^x_{\tau,\tau'} c_{p,\tau'}, \\ S_{y,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^y_{\tau,\tau'} c_{p,\tau'}, \\ S_{z,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^z_{\tau,\tau'} c_{p,\tau'}, \\ \end{align} where $$\sigma_{\tau,\tau'}^{x}$$ denotes the matrix elements of the Pauli matrix $$\sigma^x$$ and so on. Any spin-observable can then be constructed from these basic blocks. For instance, $$S^2$$ operator takes the form $$$$S^2 = \sum_{p \in MO} \frac{3}{4}\left( c_{p,\uparrow}^\dagger c_{p,\uparrow} + c^\dagger_{p,\downarrow} c_{p,\downarrow} - 2c^\dagger_{p,\uparrow} c^\dagger_{p, \downarrow} c_{p,\downarrow} c_{p,\uparrow} \right )$$$$ All one need is to evaluate the expectation value of these operators over the wavefunction of interest.