Subject: CCL:Summary: Are atoms spherical?

From: [email-masked] (Alan Shusterman)
Date: 19 Jul 2000 16:42:56 PDT

Here is a summary of responses. Thanks to all who responded.

-Alan

Alan Shusterman

Department of Chemistry

Reed College

Portland, OR, USA

----

My original question:

Programs like Spartan produce nonspherical isodensity surfaces for many atoms
(for example, any of the halogens). I would like to know if this outcome is
physically reliable, i.e., are only some atoms spherical?

I can see why the computational programs yield nonspherical electron density
distributions. This is because a UHF or UDFT calculation on fluorine (5 valence
p electrons) assigns 2 electrons to p_x, 2 electrons to p_y, but only 1
electron to p_z. But is this physically realistic?

----

Responses:

Several of the responses I received were to a question I did not ask, namely,
the shapes of atoms in molecules. My question concerns isolated atoms. Atoms in
molecules do not need to be spherical because their external environment does
not have spherical symmetry (but thanks anyway to all who pointed this out).

As far as isolated atoms are concerned, the concensus is, "yes, atoms are
spherical, even if most quantum chemical computations lead to solutions of
lower symmetry". Responses are reproduced below.

Thanks again to David Woon, Fred Arnold, Gijs Schaftenaar, Carlos Faerman, Scot
Wherland, Artem Masunov, Christoph van Wuellen, and Douglas Smith for helping
me out.

1. From David Woon:

To generate the proper spherical charge distribution for an open-shell atom,
one must use a multireference method that allows equal weighting of degenerate
occupations. This is also true for molecules: to properly describe a Pi state
in a linear molecule, one must include and equally weight the pi_x and pi_y
occupations.

Practically, however, the energy differences are usually small, much smaller
than other sources of error present in the calculation.

Dave Woon

[email-masked]

2. From Gijs Schaftenaar:

Atom groundstates yield spherically symmetric electron densities. However the
density is a linear combination of px(1)py(2)pz(2), px(2)py(1)pz(2),
px(2)py(2)pz(1) contributions, these solutions are degenerate. However this
degeneracy is lifted even at very long range proximity of other atoms. This has
its manifestation in difference density plots.

Gijs Schaftenaar

[email-masked]

3. From Scot Wherland:

In support of the "all (isoolated) atoms are spherical" position I refer
students to an old letter in J Chem Ed. Irwin Cohen J. Chem. Ed. 42 397-8
(1965). The letter is in response to what he calls a "readily understandable
misunderstanding"

-Scot Wherland

[email-masked]

(Alan: the first half of the letter that Scot refers to is exactly what I was
looking for. Here is, in part, what I. Cohen wrote:

"In contrast to the idea we may well get from the shapes of the orbitals, all
isolated atoms are spherical. To explain this perhaps surprising conclusion,
consider first the simplest case - an excited H atom with one 2p electron. Let
us, with Johnson and Rettew [p. 145 in this Journal], place the atom in an
indifferent environment, one with no selection or preference for any direction.
The location of the electron in such an environment must necessarily be equally
probable in any direction from the nucleus.")

4. From Christoph van Wuellen:

Are atoms spherical?

For an isolated atom, since the external potential is spherical, the "density"
has to be spherical as well. For degenerate states, the "density" thus is the
ensemble density.

For an atom in an environment, there is no such argument. E.g. the density of a
hydrogen atom will be deformed by the nearby presence of another hydrogen atom
-- they form a bond.

So, if atoms in VanDerWaals situations are nonspherical, this does not mean the
"atom itself" is non-spherical.

Nonspherical atoms as the result of a quantum-chemical calculations are a
different topic, they mostly aris through symmetry breaking, that mostly is a
defect of the chosen method. If, e.g. a fluorine atom is treated at the
single-configuration level and if pz is chosen to be singly occupied, then

the radial wave function of pz will be different from px and py --- no
spherical density can be obtained even for an ensemble. This will change if one
goes to a state-averaged calculation.

-Christoph van Wuellen

[email-masked]

(Alan: I wrote some additional questions to Christoph, and his replies shed
some more light on this question)

5. From Christoph van Wuellen:

for the electronic system, the electron-nuclear attraction is the "external
potential" (i.e. external to the electrons), and that is spherical. The
electron-electron interaction is "internal".

So if you have a nonspherical density, you can just rotate the wave function a
little bit and you get a rotated density. The rotated wave function must be
degenerate with the original one. In the whole set of degenerate wave
functions, you can find linear combinations which are angular momentum
eigenfunctions and so forth...

In many-electron atoms, you will only find that the rotated wave functions form
an irreducible representation of the rotation group with dimension 2L+1. Such
"states" are then called S P D F .... states for L=0,1,2,3,....

If you do a calculation, you might encounter "symmetry breaking".  This means
that all the rotated images of your original wave function do not form an irrep
of the rotation group (i.e. states higher L are mixed in). This does not happen
for "true" atomic wave functions since configurations with different L are
non-degenerate except for one-electron systems.

Christoph van Wuellen

[email-masked]