1
$\begingroup$

I am confused about the symmetry of the electronic structure of open shell atoms, especially when it comes to calculating their response properties (polarizability, to name an example).

Let's take the free oxygen atom as an example. The electronic state of the atom should be triplet with $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^0$. I am using the real $p$ orbitals here, which I think is customary, instead of the hydrogenic imaginary orbitals. This, if I understand right, is permissible because the three $p$ orbitals are degenerate, therefore I can use any linear combination of them as a basis.

If this statement is true, eg. the $p$ orbitals are degenerate, then does it make sense to talk about the $p_x$ and $p_y$ having different occupations then $p_z$, or is it better to say that they are all $2/3$ occupied?

My intuition tells me that if we are talking about a free atom, then there's nothing to break the symmetry. Without any symmetry breaking, I can't define a coordinate system with cardinal direction, so I can't even say which orbital is oriented along the $z$ axis - further proving the degeneracy of all orbitals.

However, if I am to calculate the polarizability tensor of the oxygen atom, no matter what method I use, I clearly get a different value for one of the diagonal elements of the tensor. This can be seen for example in the NIST tabulated polarizability tensor of the oxygen atom, showing that only two diagonal elements are equal. This seems to contradict the idea of the free atom having a complete spherical symmetry.

I could try to say that the individual elements of the polarizability tensor are not observable quantities anyways, so any symmetry breaking is an artifact of the simulation. However, by digging deeper into the question of the polarizability of the atoms, the statement that only closed shell atoms are symmetric appears quite often. This paper, for example, explicitly uses this distinction in the derivations

enter image description here

Looking at the expectation values of the different spherical harmonics as well as their radial averages for free atoms, it does indeed seem that the symmetry is broken, even if there's in principle no reason to have a difference between the cardinal directions.

How can it be true that the $p$ orbitals are degenerate due to a spherical symmetry, but when an observable response quantity is actually evaluated, there seems to be a difference between the directions?

$\endgroup$
1
  • $\begingroup$ FYI: The cardinal directions are North, South, East, West. $\endgroup$
    – JEB
    Commented Jul 4, 2022 at 13:12

1 Answer 1

0
$\begingroup$

It is an example of spontaneous symmetry breaking. There is also one in the spin channel.

Due to these two breakings (spin and spherical symmetry) (i) spin up and spin down orbitals will have different energy (ii) occupied p orbitals will have different energy compared to the empty ones

The spin direction as well as the direction associated to the orbital angular momentum breaking are random

$\endgroup$
4
  • $\begingroup$ Do you happen to know any textbook or chapter that goes into some details regarding this effect in atoms? I think I should learn a bit more about this effect to be comfortable with it. $\endgroup$
    – Szgoger
    Commented Jul 5, 2022 at 8:43
  • $\begingroup$ I do not have a good reference for isolated systems. Spontaneous symmetry breaking is defined when the ground state of an Hamilton is invariant under a smaller symmetry group compared to the one under which the Hamilton itself is invariant. It has far reaching consequences in extended systems (like the existence of Goldstone modes). Instead it is not particularly important in isolated systems. $\endgroup$ Commented Jul 7, 2022 at 22:25
  • $\begingroup$ I understand that this is probably correct, and that's why I want to learn more about it, because so far it sounds as a circular "proof": the ground state has less symmetry than the Hamiltonian, due to spontaneous symmetry breaking, defined as the ground have having less symmetry than the Hamiltonian. $\endgroup$
    – Szgoger
    Commented Jul 8, 2022 at 8:47
  • $\begingroup$ You are right, it is not a proof, just a definition. Indeed, there is no proof of the opposite, i.e. that any eigenstate of an Hamilton must have the same symmetries of the Hamiltonian. On the contrary, it can be proven that the eigenstates belong to the irreps of the symmetry group. For SO(3) the irreps are the spherical harmonics: math.stackexchange.com/questions/78802/… . States of a given irrep transform into one another under the symmetry operations, but independently are not invariant. Like px, py and pz. $\endgroup$ Commented Jul 8, 2022 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.