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Are atoms perfect spheres?

  1. Dec 4, 2006 #1
    One of my students Chris wanted to ask this question: are atoms perfect spheres? I could have offered one of my own answers to this but I think he would prefer to hear what you guys have to say. The only thing I would say is that I like the question because of its apparent simplicity. I guess my first responce would be: "define perfection!"
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  3. Dec 4, 2006 #2
    I would respond to someone random asking this question like this -- there is no way to know for sure. At that level, it becomes a quantum physics deal. The Uncertainty Principle would say you never really know where the electrons are, so each electron could be spread out to be a sphere around the nucleus. But electrons have likely positions, and thus can be defined more like a particle orbiting the nucleus.

    In the end -- I would say no, but there is no way to be 100% sure.
  4. Dec 4, 2006 #3
    I would give an answer stating the following two things:

    (1) The electrons orbitting an atom are described by a wave function, which is a probability amplitude. So you can't say "where" anything is to begin with. This is a more fundamental statement than the uncertainty principle.

    (2) After I explained that, I would say that even the probability amplitudes aren't spherically symmetric. Just look up p, d, f, orbitals, etc., and they come in a variety of not spherically symmetric shapes. While this may sound surprising, since the Coulomb potential goes like 1/r and is spherically symmetric, just look at Kepler problem. It's the same kind of problem, an attractive 1/r potential, but it permits elliptic, parabolic, and hyperbolic orbits, none of which are spherically symmetric.

    This is a good way of illustrating that symmetries don't always manifest themselves in the most obvious manner.
  5. Dec 4, 2006 #4


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    Well, I'm not sure that there is spontaneous symmetry breaking in atoms (as long as they aren't in any asymmetrical environment such as a crystal, or a magnetic field or so). Consider an Argon atom. Sure, it is composed of s and p orbitals, but these are just mathematical components of the overall wavefunction. I would suppose that the overall wavefunction IS spherically symmetric, in that if you have the total solution (not for one orbital, but for the entire electronic structure), and you apply a rotation to it, that you find the same structure, up to a phase factor.

    The question is in fact if in genuine atoms, the ground state of the entire electronic structure is degenerate or not. If it is degenerate, there is a possibility for spontaneous symmetry breaking, but if the ground state is unique, there isn't.
  6. Dec 4, 2006 #5
    There isn't spontaneous symmetry breaking in atoms, and I never said that. What does the fact that the potential is spherically symmetric mean in terms of conserved quantities?

    Edit: Allow me to clarify -- In my admittedly limited knowledge of quantum atomic structure, I am not aware of the presence of spontaneous symmetry breaking. There might be. The point is that there is a spherical symmetry in the atom, and it does manifest itself, but not as "well, the solutions have to be symmetric".

    This is actually a good way of introducing Noether's theorem to a high school class.
  7. Dec 4, 2006 #6


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    The answer is no. See any book on atomic physics, particularly check out the physics of the periodic table. The so-called Noble gases are spherical, most others are not. It all has to do with the basic energy-level structure of atomic electrons. (See vanesch's post above.)
    Reilly Atkinson
    Last edited: Dec 4, 2006
  8. Dec 4, 2006 #7
    Define perfection ? Well perfection as in the sense of a perfect sphere, so i don't see how asking about perfection would give this student the correct picture as described by QM. Just my $$

    Anyhow, my first answer would be this : "you are asking the wrong question"

    I mean, we cannot talk about the "structure" (in the same way we talk about the structure of a material for example) of atoms because that violates the HUP. The question should be directed towards the structure of electronic orbitals !!! But that's ofcourse something totally different than talking about structure in the "classical" way. Since not all orbitals (eg p orbitals) are spheres (like the s orbital) the answer is NO. But again, this "NO" needs to be viewed at within the context of orbitals (so not just "the structure" of atoms)!!!

  9. Dec 4, 2006 #8
    Atoms are composed of lots of smaller particles in motion. How can you ask if it is a perfect sphere?

    If I had a collection of golf balls, all moving around, the only configuration that would reasonably correspond to a sphere is if the outermost balls moved on a sphere of some constant radius (or perhaps varying, to have a pulsing sphere).

    A first note is that (as others above have noted) electrons in all but the s shell (or equivalently electrons with [itex]l=0[/itex]) do not have spherically symmetric wavefunctions.

    The second note, and this is crucial in my opinion, is that the radial part of the wavefunction is usually a polynomial multiplied by a decaying exponential. There is no reason why any of these electrons, even if they were to be in the s shell, should all be at the same distance away from the centre at any one time.

    The answer therefore is no, atoms are not, in general, perfect spheres.
  10. Dec 5, 2006 #9
    An easy answer is that nothing is a perfect sphere. Keep zooming in and eventually you find an imperfection or an uncertainty - atom or not.
  11. Dec 5, 2006 #10
  12. Dec 5, 2006 #11


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    I was wondering if other atoms, when alone, and outside of any non-spherical influence, aren't also "spherical" (in the sence that their ground state is spherically symmetric).
    In order for them NOT to be spherically symmetric, one would need a degenerate ground state (because the ground state solution space has to be a representation of the symmetry group of the hamiltonian - which is spherically symmetric, and if that solution space is one-dimensional, then it is automatically a singlet representation which is another way of saying that it is invariant under rotations).
  13. Dec 7, 2006 #12
    Thanks for all your interesting responces. There is enough food for thought here to keep us going for years!


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