On-site repulsion and Pauli exclusion

Been studying hopping conduction and something that everyone is taking for granted is bothering me.

Let's say we have a bunch of sites that are either unoccupied, singly occupied, or doubly occupied. Due to on-site Coulomb repulsion the two electron levels are separated by U energy at a doubly occupied site. Now everyone is saying that the two electrons on the double site are in the spin singlet state due to, I assume, Pauli exclusion. However the two electrons are not in the same energy level - they are separated by U so why is there a restriction on their spins?

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4tnemele's answer is great, but I thought I'd try to give a simpler explanation of the main point of confusion.

Let's say (as in 4tnemele's answer) that there's only one relevant orbital (or "energy level") at each site. That means there's two states a single electron can occupy: spin up and spin down.

Also, let's say there's no magnetic field.

You say "Due to on-site Coulomb repulsion the two electron levels are separated by U energy at a doubly occupied site." This is wrong, because there is only one "level" (orbital), not two. The Coulomb energy penalty U only applies if there are two electrons in that level, one spin up and one spin down.

At a doubly occupied site, the overall energy is increased by U, but that energy doesn't belong to one electron or the other. It comes from the interaction between the electrons. At a doubly occupied site, the two electrons still have exactly the same spatial wavefunction. The many-body wavefunction is spatially symmetric. This is why the spins have to be opposite.

So the basic problem is that you shouldn't say "the two electron levels are separated by U". Instead you should say "the state with two electrons at the same site is increased in energy by U relative to what you'd otherwise expect (twice the single-electron energy)".

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Thanks Keenan. That is really insightful. It makes a lot of sense now that the overall energy is increases by U and not the energy of one electron (b/c which electron would it be anyway?). To follow up on one thing you said though, why would the two electrons necessarily be in a symmetric spatial state (which then of course leads to the singlet spin state)? Why not let them be in antisymmetric spatial state and then a triplet spin state? – BeauGeste Feb 19 '11 at 19:43
Now, with what you've said (@Keenan), I must really wonder about the diagrams I'm seeing in these papers. They are showing one electron at an energy level εa. Then they show another electron hopping to that first electron's site. The new electron is placed at εa+U. With what you said, this seems to be incorrect b/c it suggests that there are two levels. Is there not a better way to draw this? – BeauGeste Feb 19 '11 at 20:55
@BeauGeste With regard to your second comment, the problem in drawing the "correct" scenario is how to communicate the change in energy level of the first electron. It could be shown with two side-by-side depictions with the energy level altered by the appropriate amount, but in a single diagram, the best that could be done would be a dotted line overlay picture, which would get messy. And side-by-side diagrams take up a lot of extra space. Just my thoughts. – Mitchell Feb 20 '11 at 18:20
An antisymmetric spatial state only exists if there is more than one orbital available. Then you can have something like $|1\rangle|2\rangle−|2\rangle|1\rangle$. If all you have is $|1\rangle$, then you can try to make $|1\rangle|1\rangle−|1\rangle|1\rangle$, but that equals zero, which means it just doesn't exist in the two-particle Hilbert space. – Keenan Pepper Feb 20 '11 at 20:39

Maybe I am misunderstanding your question, but let me try to explain.

For simplicity suppose that we have only one site with only one relevant orbital (a big energy gap to other orbitals). Furthermore electrons have internal degrees of freedom, which introduces the index $\sigma = \uparrow, \downarrow$. By Pauli exclusion principle there can only be one electron at each state, so the many-body Hilbert space is four-dimensional $\mathcal H = \{a|0\rangle + b|\uparrow\rangle + c|\downarrow\rangle + d|\uparrow\downarrow\rangle | a,b,c,d\in\mathbb C\}.$

I think you are confused because you have forgotten that there is not one, but two singly occupied states. Now suppose you have a Hamiltonian of the form $H = \sum_{\sigma}\xi c^{\dagger}_{\sigma}c_{\sigma} + Un_{\uparrow}n_{\downarrow} + B\sum_{\sigma\sigma'}c^{\dagger}_{\sigma}\tau^z_{\sigma\sigma'}c_{\sigma'},$ where the second term is the "on-site" repulsion and the last term $\sum_{\sigma\sigma'}c^{\dagger}_{\sigma}\tau^z_{\sigma\sigma'}c_{\sigma'} = c^{\dagger}_{\uparrow}c_{\uparrow} - c^{\dagger}_{\downarrow}c_{\downarrow}$ is a Zeeman-term and represents a magnetic field in the $z$-direction. So the on-site repulsion gives a energy penalty $U$ for occupying the orbital with two electrons, while the magnetic field introduces a gap $B$ between single occupation with spin-up and single occupation with spin-down. It is not hard to find the eigenvectors and energies

$E_{|0\rangle} = 0$, $E_{|\downarrow\rangle} = \xi - B$, $E_{|\uparrow\rangle} = \xi + B$ and $E_{|\uparrow\downarrow\rangle} = 2\xi + U$.

I hope this helped.

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