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Molecule of hydrogen

  1. Jun 10, 2008 #1
    A molecule of hydrogen. We assume that the only possible states of
    its two electrons (indistinguishable) are [tex] |A\uparrow\rangle,
    |A\downarrow\rangle, |B\uparrow\rangle[/tex] and [tex]|B\downarrow\rangle[/tex]
    (A=1s-orbital at atom A, B=1s-orbital at atom B).
    Formulate the 6 basis functions using the four possible single
    particle states above. (Don't forget the Pauli-principle!)


    Here is my solution:

    Total spin S=1 or 0.
    [tex]|1,1\rangle=|A\uparrow\rangle|B\uparrow\rangle[/tex],
    [tex]|1,0\rangle=\frac{1}{\sqrt{2}}(|A\uparrow\rangle|B\downarrow\rangle+|A\downarrow\rangle|B\uparrow\rangle)[/tex],
    [tex]|1,-1\rangle=|A\downarrow\rangle|B\downarrow\rangle[/tex],
    [tex]|0,0\rangle=\frac{1}{\sqrt{2}}(|A\uparrow\rangle|B\downarrow\rangle-|A\downarrow\rangle|B\uparrow\rangle)[/tex].

    Are they basis functions? There should be 6 but I got only 4.
     
  2. jcsd
  3. Jun 10, 2008 #2

    Mute

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    Homework Helper

    You've written down spin triplet and spin singlet states, and those would be all you had if your single particle states were [itex]\left|\uparrow\right>[/itex] and [itex]\left|\downarrow\right>[/itex], but your states aren't labelled by just the spin - you have labels A and B too. However, all of the states you have written are pairs A and B. What about AA or BB pairs?
     
  4. Jun 11, 2008 #3
    Hi, Mute. I think there should be no AA or BB pairs. Because at the same time A or B has only one state.
     
  5. Jun 12, 2008 #4
    Why do you think that? It's not given anywhere in the problem statement that each atom can only have one electron.

    Remember that the Pauli exclusion principle only rules out two indist. particles in the *exact* same state...
     
  6. Jun 13, 2008 #5
    2Tesla and Mute, thanks. I understand now.

    |0,0>=|A up>|A down>
    |0,0>=|B up>|B down>

    Thanks u for ur tips.
     
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