On-site repulsion and Pauli exclusion

Question

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?

Answer 1

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)".

Answer 2

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.