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Why does bringing N 1-orbital atoms together yield N levels?

  1. Oct 3, 2013 #1
    A common example of this is that when bringing N hydrogen atoms together into a ring. Far apart, assume each electron exists in the 1s state. As we bring them together, instead of each electron staying at the original 1s level, or all of them changing by the same amount, the 1s level fans out into N.

    For the case of 2 atoms, I can understand this as bonding or anti-bonding of the atoms. i.e., do the wavefunctions add between the protons, meaning each electron can share in the potential of both protons (bonding) or do the wavefunctions destructively interfere between the protons (anti-bonding).

    With 3 atoms, I can't find 3 levels. Assuming Gaussian shaped wavefunctions, note that the sign of each wavefunction between any two atoms defines the wavefunction on the rest of the ring. Since the signs of the wavefunction are independent, there should be 2^3=8 possibilities since each wavefunction can be + or -. Yet, there are really only 2 energetically distinct arrangements that I see: all have the same sign (two cases) or 2 of 3 have the same sign (2*(3 choose 2)), to account for both sign cases). So I get 3 atoms yield 2 levels.

    Can somebody shed light on what I've done incorrectly? Or is 3 too small to work correctly? Is there an argument about the shape of the orbitals I've neglected?

    Thank you.
  2. jcsd
  3. Oct 4, 2013 #2

    Simon Bridge

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    Hint: Pauli exclusion principle.
  4. Oct 4, 2013 #3


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    You are right in that you only get two distinct energy levels. However there are two independent wavefunctions giving the same energy in case of the upper level.
    In the case of a ring with N atoms, you get (N+1)/2 distinct energy levels.
  5. Oct 4, 2013 #4
    Yeah, I suppose I'm unclear about what defines a state in this context. If I assume all of the electrons remain in the 1s state, while sort of forming a metal, then I have more electrons than available states. So, the creation of more states solves this problem, but isn't obviously the answer.

    By the two independent wavefunctions in the upper (anti-bonding) level, do you mean the sign of them? Then this should yield 4 total levels since you could argue the same thing for the constructively interfering wavefunctions.

    Also, the (N+1)/2 argument seems contrary to what I remember reading in Kittel: N atoms yield N levels. I assumed the levels were distinct.

    Thanks for your replies, please follow up!
  6. Oct 4, 2013 #5

    Simon Bridge

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    Lets restrict talk to energy eigenstates of the system then.

    When you bring a lot of atoms together, the electron-electron interactions mean that there is no longer a well-defined 1s state. The new system has different energy levels. The old 1s state is a superposition of the new system ... and the electrons decay into the new energy levels creating the new ground state.

    You can model this by putting two finite square wells at different distances from each other and look what happens to the first few bound states when they are different distances apart. Also start with just one square well, with one particle, and "switch on" a second one.

    Another way of thinking about it: no two electrons in the universe are ever in exactly the same discrete state - its just that the energy levels for far-separated wells are so very similar our instruments cannot tell the difference. When the wells get closer, their interactions force the states to split and you can see the difference.

    Whatever - once the combined system has formed, the inital 1s states no longer exist and their energy levels may even be forbidden.

    When lots of atoms are close together, like in a solid, the energy levels merge to form bands.

    Note: I was wondering if DrDu was thinking of molecular energy or, maybe, configuration states.
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