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Basic Gaussian question - What do stability scans mean?

  1. Sep 29, 2012 #1
    I was assigned a multistep task by my professor so I first made the molecule with ChemDraw, then I ran an optimisation using B3LYP/6-31G. I then ran a frequency scan and checked for any imaginary frequencies. There were no negative frequencies and my understanding is that this means that Gaussian optimised the structure to a global minimum, as opposed to a local minimum or saddle point etc.

    The next step of my task was to run a stability=opt scan on the structure to "check the stability of the ground state wavefunction". Unfortunately, it turns out that the wavefunction isn't stable. Problem is I don't know what that means, let alone how to fix it. Maths isn't my strongest field so the Schrodinger equation confuses the hell out of me but my understanding is that a wavefunction represents the probability of finding electrons in 3D space. What exactly is a "ground state" wavefunction? Or a better question: what is a non ground state wave function? I'm guessing an example of an excited state wavefunction would represent the electron distribution of a molecule after I beam it with UV-Vis radiation so that some of its electrons jump to higher energy levels, have I got the right idea?

    If the ground state wavefunction for my molecule is instable, does that mean the optimisation failed? If so, how can I correct it?
     
  2. jcsd
  3. Oct 1, 2012 #2
    Does the professor know that they're being so vague and confusing? Telling a student to just include a keyword in an input and not telling them anything about what it means is at best useless and at worst confusing (as it has confused you).

    In Gaussian, the stability keyword checks to see if the wave function is at the lowest possible spin multiplicity for that arrangement of nuclei. What that means is, say we have a system with an even number of electrons. They can all be paired up (as many "up" as "down", though in the output file you'll see them called "alpha" and "beta") which is called a "singlet" (because when you pass a beam of atoms in a singlet state through a magnetic field it will yield a single beam at the other end) or you can have two more ups than downs which is called a "triplet" or three more ups than downs (called a quintet) etc. This is something that you specify when you ask Gaussian to do a quantum mechanical calculation. You tell it how many electrons you have and how to pair them up and it then optimizes a set of orbitals and fills them in the way you asked (pairing the number you specified). You can then check to see if there's a better way of pairing them to get lower energy.

    You should really, REALLY press your prof on this kind of thing when he/she gives you the assignment. Presumably they're getting paid to teach you, and if you don't see to it that they do then no one will.
     
  4. Oct 9, 2012 #3
    Brilliant explanation! Thanks a lot. I didn't check back here earlier because I didn't think anyone was going to answer the question. I'm struggling to learn all this stuff because all the tutorials I can find (i.e. the Gaussian manual) don't provide explanations that I can understand. I've been looking for a tutorial that gives proper in-depth explanations like your one that give me an idea of what the calculations actually do in terms of the real life molecule.

    So a stability scan determines if the molecule is at its most stable electron pairing configuration, is that it? Now its starting to make sense, the molecule I was running calculations on was a tetrahedral Mn complex so I understand why I'd need to perform this kinda calculation. Don't know why my professor couldn't just explain that. He explains nothing, he seems to assume that I already have good knowledge of all this. Either that or hes just too busy to put time into explaining things.
     
  5. Oct 10, 2012 #4
    Yes, that's it exactly.

    You should really, really think about finding another professor to work with. With the effort that you appear to be putting into this, another would certainly be happy to have you and you would learn a lot more with a lot less frustration with even an "average" professor than the one you currently have.
     
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