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Overlap integrals in chemical molecules

  1. Sep 20, 2010 #1
    In the explicitly-unsolvable Hamiltonian for a chemical system, approximations are made to solve the energy of the system. In particular we see CNDO (ZDO), INDO and MNDO theories.

    CNDO - complete neglect of differential overlap; two-center electron integrals are zero.

    INDO - intermediate neglect of differential overlap

    MNDO - modified neglect of differential overlap

    So CNDO is easy, if orbitals arent on the same atom their integral is zero. What about INDO and MNDO? What integrals are included in INDO that arent included in CNDO? What integrals are included in MNDO that arent included in INDO? How are these integrals evaluated (parametrized)?
  2. jcsd
  3. Sep 21, 2010 #2


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    You're not starting with an "explicitly-unsolvable Hamiltonian". You're starting with the very solvable Roothaan-Hall equations solved by SCF-LCAO. After which you are then making semi-empirical approximations by throwing out various integrals you know to be small, in order to do less calculating.

    I don't know the exact differences between CNDO/INDO/MNDO offhand. I don't really feel compelled to; these methods have been obsolete for about 20 years.

    The integrals are evaluated the same way as SCF integrals are. (i.e. analytically as far as possible and numerically for the rest)
    Have you learned the Hartree-Fock/SCF method? A thorough understanding of it is necessary if you want to understand post-HF methods.
    Last edited: Sep 21, 2010
  4. Sep 21, 2010 #3
    Excuse me for mis-typing then. The Roothan-Hall equations are merely a representation of the HF method, which is an approximation for the determination of the ground-state energy. I didn't realize you needed me to state all that information before asking my actual question. For the record, yes I am familiar with HF theory both RHF and UHF.

    No offense but if you do not know the answer than why did you bother responding? Just because you don't feel compelled to know the answer doesn't mean it is irrelevant as it is applicable to a question stemming from my current academic pursuits. Furthermore if you took the time to punch MNDO into google scholar, limiting search results to papers published after 2000, you will see these methods are QUITE relevant still and are frequently reparametrized.

    I don't know what your intentions were in responding to my question but to be honest I found your response arrogant, dismissive and unhelpful. If you don't know the answer than move on. Do not tell me the question is not worth asking.
  5. Sep 21, 2010 #4


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    Good! As you noticed, I didn't get that impression from your original post. The impression I got was that you were trying to jump from the Schrödinger equation straight to semi-empirical methodology without first having a solid grasp of the intermediate levels of theory. But to me, a thorough understanding of HF theory does include knowing how it's implemented, which means knowing how the two-electron integrals are evaluated in practice. (And yes, I also know that the Roothan-Hall equations and HF method are the same thing. I don't get offended being told that they are, either.)
    I bothered responding because I suspected you might be wasting your time, either learning the details of methods you would likely never use, or attempting to implement these methods without first understanding the ones they build on. The only reasons I can see for knowing this, and asking how integrals are evaluated, is because you're interested in either implementing these methods or developing them further. Telling people they have to learn to walk before they can run is being helpful, even if it appears to have injured your pride.

    I could not (and still can't) understand what set of circumstances there'd be where someone would know how to calculate the Hartree-Fock two-electron integrals but have problems understanding the simplified ones used by the semi-empirical methods. Also, if method implementation/development was your interest, you'd know or wouldn't have any problems finding out, because you'd have a basic textbook on quantum chemistry. For instance, it's in section 3.9 of Jensen's "Introduction to Computational Chemistry", sec 16.5 of Levine's "Quantum Chemistry", sec 9.5 of Mueller's "Fundamentals of Quantum Chemistry".

    Whereas if your interests were in just doing calculations, I'd suggest you learn about more modern semi-empirical methods such as PM3 or AM1.
    A parametrization makes it a different method. (e.g. PM3 and AM1) Stock MNDO isn't used for calculations anymore. Nor could you (in most contexts) publish a paper having done straight-up Hartree-Fock calculations either. That doesn't mean dozens of papers aren't still being published that have something to do with Hartree-Fock. The fact that a method is obsolete for doing calculations doesn't make it obsolete from the method-development point of view. For reasons stated, I assumed you were interested in practical calculations, not method development.
    Sorry if you feel that way, but hey - you get what you paid for.
    Last edited: Sep 21, 2010
  6. Sep 28, 2010 #5
    Admittedly I may have been a little irritated with your initial response and I apologize. I'm sure you understand, quantum chemistry can sometimes make one's head spin.

    I still seek an answer to my question. I don't care about types of parametrization (e.g. PM3 or AM1) nor method development or any other perspectives on what is worth knowing.

    All I am interested in is which integrals are parametrized or approximated in each method. In other words, the Hamiltonian for each basic method is separated into a core Hamiltonian and then exchange integrals and coulomb integrals.

    In ZDO or CNDO, the coulomb integrals are ignored. In INDO, they are added back in but only when they are on the same atom (one-centered). In MNDO they are all added back in. Is this correct?
  7. Oct 5, 2010 #6


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    I can answer on the "how to evaluate the integrals'-part. Semi-empirical methods typically require only two-center integrals. Such two-center integrals are traditionally evaluated using Slater-Koster tables, and better using the methodology given in http://dx.doi.org/10.1063/1.2945897 and http://dx.doi.org/10.1063/1.2821745 .

    As an introduction into the general procedure of evaluating semi-empirical methods (including the now-obsolete Slater-Koster tables), I recommend reading http://doi:10.1016/j.commatsci.2009.07.013 [Broken] . This article is on "Density functional tight binding", which is another semi-empirical method. Contrary to most other semi-empirical methods, DFTB is rather popular, mainly due to its clever choice of name.
    Last edited by a moderator: May 5, 2017
  8. Oct 5, 2010 #7
    wow great info. Thank you very much.
    Last edited by a moderator: May 5, 2017
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