$S$ 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 \begin{equation} 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 ) \end{equation} All one need is to evaluate the expectation value of these operators over the wavefunction of interest.

| improve this answer | |
  • 2
    $\begingroup$ I don't think this formalism is relevant to the question. $\endgroup$ – tobiuchiha Jun 17 '19 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.