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Physical meaning of orbital polarization?

  1. Oct 21, 2014 #1
    In this paper, the Orbital Polarization (OP) is defined as:

    $$OP=\frac{n_{x^2-y^2}-n_{z^2}}{n_{x^2-y^2}+n_{z^2}}$$

    where $$n_i$$ is the occupancy of that given orbital. I would like to understand the physical meaning of this. Also, is there a difference between OP and Orbital Hybridization?
     
  2. jcsd
  3. Oct 23, 2014 #2

    DrClaude

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    I am not an expert in the field, so do not take what I write here with any authority. It is simply my interpretation.

    From what I get from the article, I take it that under normal conditions, the ##d_{x^2-y^2}## and ##d_{z^2}## orbitals would be equally populated by electrons. The presence of strain will lift that degeneracy, and one orbital will end up with a higher electronic occupancy (probability). This is what orbital polarization "measures."

    It does not seem to be related to hybridization, in that it does not consider that the orbitals themselves are modified, only their occupancy. Also, hybrid orbitals are usually constructed from equal proportions of the base orbitals.

    I think you should keep in mind that orbitals are basically single electron solutions, and do not take into account the presence of other electrons in the atom, or chemical bonding. It is expected that the actual electronic wave function will be only approximated by considering that electrons occupy disctinct orbitals. Orbital polarization appears as one way to take this into account.
     
  4. Oct 23, 2014 #3
    "The presence of strain will lift that degeneracy, and one orbital will end up with a higher electronic occupancy (probability)"

    Does this mean the two orbitals will have different energy? This is what first come to my mind when I see "degeneracy lifting".
     
  5. Oct 23, 2014 #4

    DrClaude

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    I was paraphrasing the article:
    I interpreted that as meaning that in the absence of strain, the ##d_{x^2-y^2}## and ##d_{z^2}## orbitals would have the same energy, and thus be equally populated.
     
  6. Oct 23, 2014 #5
    There must be something changed that causes them to have different occupancy. This thing should be the energy of each orbital. Do you agree with this interpretation?
     
  7. Oct 24, 2014 #6

    DrClaude

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    Sounds reasonable.
     
  8. Oct 25, 2014 #7
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