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Usually the first step in describing the electronic wave function of a molecule or atom is to describe it with a single Slater determinant. I get that this is an independent particle approximation, and that in the context of Hartree-Fock, each electron moves in the average field of the other electrons. That's fine.

But when we correlate the electrons, this is done by generating excitations out of the Slater determinant, and then taking weighted combinations of them. For example, configuration interaction mixes the ground state Slater determinant with singly, doubly, triply, etc excited determinants. A fancier way of doing this is coupled cluster, where cluster operators generate excitations out of the Slater determinant.

Hermann Kummel says that we can think about electron correlation this way:

"The first thing one may imagine happening is that two particles mutually interact, thereby lifting themselves out of the Fermi sea, so that after the interaction both are in unoccupied orbitals"

Kümmel, Hermann. "Origins of the coupled cluster method." Theoretica chimica acta 80.2-3 (1991): 81-89.

Why can we (or should we) think about electron correlation that way?

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In what way? In terms of linear combinations of Slater determinants or in terms of 'interacting electrons being in unoccupied orbitals'? – LordStryker Feb 25 '14 at 19:49
@LordStryker Both, really. The standard mathematical way to treat dynamic electron correlation is to "mix in" excited Slater determinants, which is equivalent to saying that when two electron interact, they "excite" each other out of their ground state occupied orbitals and into previously unoccupied virtual orbitals. In other words, an excitation from occupied to unoccupied. – jjgoings Feb 25 '14 at 22:23
@jjgoings, for a simple physical reasoning watch this video by professor Jack Simons. You could also read almost identical material in his book "Quantum Mechanics in Chemistry" (Chapter 8) freely available here. – Wildcat Oct 22 '14 at 13:03

It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!

The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.

(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)

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Unfortunately I have to agree with you -- it seems to be one of those places that we have to rely on math instead of intuition. The HF Slater determinant forms a basis for an N-electron wavefunction, just like our single electron basis functions form a basis for an independent particle wavefunction. – jjgoings Feb 25 '14 at 22:26
Regarding your point about single electron excitations: they do show up in many correlated methods (CCSD and CISD, for example)! While you are correct in saying that single excitations don't mix with the ground state directly (e.g. Brillouin's theorem), they do mix with higher excitations. According to Thouless, single excitations create any single Slater determinant out of any other single Slater determinant. Basically it optimizes your reference in response to including dynamic electron correlation. It's a big reason CCSD is pretty insensitive to reference (which makes it a robust method). – jjgoings Feb 25 '14 at 22:31
Of course you are correct that my statement in the final paragraph only applies to the ground state, and I’ve edited to make this clear. Apologies if the level of explanation was rather too basic for you! – Aant Mar 2 '14 at 18:17
The concept of 'mixing' these determinants (i.e. ground and some excited state) is fine with me mathematically. However, the physical interpretation of this mathematical approach is very dissatisfying to me. – LordStryker Mar 28 '14 at 19:47
@LordStryker, there is some simple physical interpretation. Please, watch this video. – Wildcat Oct 22 '14 at 13:07

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