CCL:Summary: Are atoms spherical?

[Alan.Shusterman #*at*# directory.reed.edu (Alan Shusterman)]

 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
 woon #*at*# hecla.molres.org
 
 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
 schaft #*at*# cmbi.kun.nl
 
 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
 scot_wherland #*at*# wsu.edu
 
 (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
 Christoph.van.Wuellen #*at*# ruhr-uni-bochum.de
 
 (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
 Christoph.van.Wuellen #*at*# ruhr-uni-bochum.de