Physical meaning of orbital polarization?

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SUMMARY

The discussion centers on the concept of Orbital Polarization (OP), defined mathematically as $$OP=\frac{n_{x^2-y^2}-n_{z^2}}{n_{x^2-y^2}+n_{z^2}}$$, where $$n_i$$ represents the occupancy of specific orbitals. It is established that OP measures the difference in electronic occupancy between the ##d_{x^2-y^2}## and ##d_{z^2}## orbitals, particularly under conditions of strain that lift their degeneracy. Unlike Orbital Hybridization, OP focuses solely on occupancy rather than modifications to the orbitals themselves. The discussion concludes that the lifting of degeneracy due to strain results in differing energy levels for these orbitals, leading to unequal occupancy.

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  • Understanding of quantum mechanics and atomic orbitals
  • Familiarity with concepts of degeneracy and energy levels
  • Knowledge of electronic occupancy in quantum systems
  • Basic grasp of strain effects in materials science
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Researchers in condensed matter physics, materials scientists, and students studying quantum mechanics who are interested in the electronic properties of materials and the effects of strain on orbital occupancy.

phys_student1
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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?
 
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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.
 
"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".
 
phys_student1 said:
"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".
I was paraphrasing the article:
One of the major effects of epitaxial strain or heterostructuring is the degeneracy lifting of the ##e_g## states, resulting in ‘orbital polarization’ (OP) of the electronic structure.
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.
 
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?
 
phys_student1 said:
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?
Sounds reasonable.
 
Thanks.
 

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