According to the Q-Chem manual [1], the singlet-triplet spin-orbit couplings between a singlet excited state $I$ and a triplet excited state state $J$ is: $$\langle\Phi_\text{singlet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_s=0,\pm1}|\langle\Phi_\text{singlet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_s}\rangle|^2}~.$$

The further elaborate on the triplet-triplet spin-orbit coupling to be: $$\langle\Phi_\text{triplet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_s=0,\pm1}|\langle\Phi_\text{triplet}^{I,m_s}|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_s}\rangle|^2}~.$$

But shouldn't the latter actually be a double sum like: $$\langle\Phi_\text{triplet}^I|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^J\rangle = \sqrt{\sum_{m_{s,i}=0,\pm1}\sum_{m_{s,j}=0,\pm1}|\langle\Phi_\text{triplet}^{I,m_{s,i}}|\hat{\mathrm H}_{SO}|\Phi_\text{triplet}^{J,m_{s,j}}\rangle|^2}~?$$

As of porphyrin's comment, I'd like to add, why I think so.

The triplet-triplet SOC matrix looks like the following: \begin{array}{c|ccc} \hline m_{s,i}\text{\\} m_{s,j} & 0 & -1 & 1 \\\hline 0 & 0 & a & a \\ -1 & a & b & 0\\ 1 & a & 0 & b\\ \hline \end{array}

If the rules from the "single sum" would apply, only $b$ would count into the sum and $a$ would be neglected, because the summation only accounts for the diagonal elements.

[1] http://www.q-chem.com/qchem-website/manual/qchem50_manual/sect-tddft.html

  • 2
    $\begingroup$ I would agree if the total from all $m$ is needed then this would seem logical. $\endgroup$
    – porphyrin
    Feb 4, 2018 at 15:36


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