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I understand the vanishing integral rule for a one-electron integral to be:

$$\langle i | \hat{O} | j \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma_O \neq \Gamma_i \otimes \Gamma_j$$

For electronic repulsion integrals, the Coulomb operator is totally symmetry, so the product must contain the totally symmetric irrep (TSIR). However, I am not sure what form the vanishing integral rule will have. Is it simply

$$\langle ij | kl \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma_\mathrm{TSIR} \neq \Gamma_i \otimes \Gamma_j \otimes \Gamma_k \otimes \Gamma_l$$

or does it have a different form? Additionally, will a Schwarz-inequality matrix composed of:

$$Q_{ij} = \sqrt{\langle ii | jj \rangle}$$

have an identical block-diagonal structure to the one-electron matrices constructed in this fashion?

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  • $\begingroup$ I would agree that the product of irreps should contain the TSIR if it is not to be zero. Will not your Q always be totally symmetric for all i and j. $\endgroup$ – porphyrin Dec 3 '18 at 16:26
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    $\begingroup$ this should be somewhat more complex, then just the product of all 4 irreps, since there are 2 different electronic coordinates involved. The argument with the vanishing integral always goes over just one coordinate at a time. (I can try for a more detailed answer later) $\endgroup$ – Feodoran Apr 22 at 6:53
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    $\begingroup$ Ok, it actually seems to be as simple as taking the product of the 4 IRREPs. However, I could not find a decent reference with a detailed derivation. There is though this wikipedia article and a paper mentioned there (although in a slightly different context) DOI: 10.1002/qua.560050606 $\endgroup$ – Feodoran Apr 24 at 8:22

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