# Are there Exchange integrals in Semiempirical methods?

One of the major features of semiempirical methods is that they significantly reduce the cost of evaluating 2-electron integrals.

In general, they follow the formula for HF, but, introduce many simplifying assumptions. In HF for same spin electrons you need to calculate both coulomb and exchange integrals.

When applying the simplification in, for instance, CNDO and INDO, that $$(\mu \nu | \lambda \sigma ) = \delta_{\mu \nu} \delta_{\lambda \sigma}(\mu \mu | \lambda \lambda )$$, this seems to de facto rule out calculating an exchange integral.

Is there a scenario where the exchange integral is computed in semiempirical methods (for simplicity, in the case of CNDO/INDO methods)?

Textbooks do not discuss the exchange integral when going over semiempirical, so the answer must be obvious... I just don't know if it is obvious that they cannot be performed, or if they can, and I am missing something.

• As far as I can tell from Jensen, CNDO and INDO neglect all exchange integrals, but, NDDO does not – B. Kelly Mar 25 '20 at 5:55
• If I remember correctly, the exchange integrals become parameters based on the atomic orbital energies - but it's been a long while and my copy of Pople's Approximate Molecular Orbital Theory is on campus. If you can find a copy, the answer's in there - he explains the derivations of both CNDO and INDO in detail. – Geoff Hutchison Mar 27 '20 at 3:10
• I will check it out. As the saying goes... Answer someones stack exchange question and you nourish their knowledge for a day, give them a good book and you nourish their knowledge for a lifetime – B. Kelly Mar 27 '20 at 5:51

Semiempirical quantum chemistry models do carry out proper Hartree-Fock calculations, including both Hartree (electron-electron electrostatics) and Fock exchange contributions to the total electronic energy. As noted in the question, the tensor of 4-center Coulomb integrals is heavily sparsified, approximated, and parameterized to simplify and accelerate semiempirical calculations. Even in the most approximate models (CNDO), the exchange interaction between molecular orbitals does not vanish in general. For example, the exchange integral between a pair of molecular orbitals $$i$$ and $$j$$ with real atomic-orbital coefficients $$\phi_{i \mu}$$ is simplified by the zero-differential-overlap (ZDO) approximation stated in the question as
$$(ij|ji) = \sum_{\mu,\nu,\lambda,\sigma} \phi_{i \mu} \phi_{j \nu} \phi_{j \lambda} \phi_{i \sigma} (\mu \nu | \lambda \sigma) = \sum_{\mu,\lambda} \phi_{i \mu} \phi_{j \mu} \phi_{j \lambda} \phi_{i \lambda} (\mu \mu | \lambda \lambda)$$.
This answer might be confusing precisely what the original question means by "exchange integral". For example, no explicit integrals are involved in calculating the Coulomb matrix elements used by most semiempirical models, and there may be some class of Coulomb matrix elements in the atomic-orbital basis that are considered to be "exchange-like" that are indeed neglected in most semiempirical models (I'm not familiar with all variants of this nomenclature). The neglect of diatomic differential overlap (NDDO) approximation that eventually superseded ZDO does allow for nonzero $$(\mu \nu | \lambda \sigma)$$ values when $$\mu$$ and $$\nu$$ (similarly, $$\lambda$$ and $$\sigma$$) are different atomic orbitals on the same atom.