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Atomic many-electron configurations and the use of the orbital quantum number

  1. Jul 20, 2012 #1
    Hi,

    For single-electron atomic systems, the electron can be specified by four quantum numbers n, l, m_l, m_s (principal, orbital, z-orbital, z-spin). The orbital quantum numbers are well defined since the problem is spherically symmetric.

    However, for many-electron systems, the spherical symmetry is broken due to the inter-electronic interaction. It seems to me that the orbital quantum numbers would then lose their meaning. Still, the configurations are labelled as 1s1s2p, as if the electrons didn't interact at all. This might be a good approximation for highly stripped positive ions such as Lithium-like Uranium, but surely this can't be the case for neutral Lithium?
     
  2. jcsd
  3. Jul 20, 2012 #2

    cgk

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    The orbitals are solutions of an effective mean-field one-particle problem. When you read "orbital", you should think of "Hartree-Fock". In this approximation, the spherical symmetry of atoms is almost preserved (or even exactly for spherical atoms like N, Ne, etc), so you still end up with a spherical symmetric one-particle problem. But in difference to the hydrogen case, this potential is now not the bare nuclear interaction, but also includes the mean-field potential emitted by the other electrons (in the form of their coulomb and exchange interactions).

    That is, in such cases the L/m quantum numbers for orbitals are still good. The n-quantum number however is only used to enumerate solutions in a certain L subspace and does not carry any specific meaning. That is, 1s is the first orbital of s symmetry, 2s is the second orbital of s symmetry, etc. But 2s and 2p are not degenerate (and in fact, not really related at all, except through the mean field which couples all the orbitals together).

    The actual (correlated) atomic states are then labeled by their dominant HF configuration. E.g., if you talk of a nitrogen state of 1s2 2s2 2p3, you mean the state that is dominated by this configuration. E.g., in the nitrogen case, this confiugration would carry somewhere around 99% of the total weight of the total wave function.That approach works fine for many light elements, but there are atoms where the picture is not so clear (e.g., Be, 3d transition metals).
     
  4. Jul 21, 2012 #3
    Ok, got it, this is what I suspected. Thanks a lot!
     
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