DrDu said:
I just wanted to add that even in strongly interacting systems (which are not well described by one determinant), one can define so called "natural orbitals" which are the eigenstates of the 1-particle density matrix. If the wavefunction is well represented through one determinant, they become equal to the canonical Hartree Fock orbitals. However, in general, the occupation probability ranges continuously somewhere between 0 and 1.
Aw, I saw this earlier and was going to say that. Oh well, I'll contribute a link to Löwdin's classic http://link.aps.org/doi/10.1103/PhysRev.97.1474" (which is not obvious). So the partial occupations come about through the CI expansion coefficients. (So IMO it makes more sense to think about it as configurational occupations rather than orbital occupations)
Anyway, to address the original poster, as cgk pointed out, a real, interacting, system cannot be described in terms of orbitals. Orbitals describe individual electronic states, and since electrons interact, you (strictly speaking) cannot separate the system into individual electronic states like that.
To make an analogy, consider the solar system (this is a purely mathematical analogy, not physical, I am not suggesting a planetary model of the atom!). The orbit of every planet influences that of every other planet. It's a many-body problem. So you cannot describe the motion of a single planet around the sun without taking into account the motion of every other planet. But to a good approximation, you can regard the system as if the planets were orbiting independently, and treat the interdependence of the motion (the
correlation) as a correction to that picture. This is of course how we visualize the solar system in practice.
The same goes for electrons in an atom. You can only view its state as composed of independent orbitals if you neglect the correlation (the Hartree-Fock/single Slater determinant picture), or if you allow the electrons' state to be described by many such orbitals (CI). Just as the picture of planets orbiting the sun, the HF picture is qualitatively more-or-less correct (energy-wise more than 95%), and you can view the correlation as a correction to this. So orbitals are important conceptually as they give a very simple yet mostly-correct way of visualizing the electronic behavior, and hence the properties of atoms and molecules. And they are important computationally, since it's easier to begin with the Hartree-Fock method and find ways to correct the last 5% or so, rather than to try to attack the (normally unsolvable) many-electron Schrödinger equation directly.
There is one case where you can solve the two-electron Schrödinger equation, namely "Harmonium" also known as http://en.wikipedia.org/wiki/Hooke%27s_atom" . This is a two-electron 'atom' where you have replaced the Coulomb potential of the nucleus with a harmonic potential. So there is Coulomb repulsion between the electrons, which are connected by harmonic 'springs' to a point (the 'nucleus'). This happens to have an analytical solution, and the ground-state wave function is then:
\Psi(\mathbf{r}_{1},\mathbf{r}_{2}) = N\left(1+\frac{1}{2}|\mathbf{r}_{1}-\mathbf{r}_{2}|\right)\exp\left(-\frac{1}{4}\big(r_{1}^{2}+r_{2}^{2}\big)\right)
Where \mathbf{r}_{1},\mathbf{r}_{2} are the coordinates of the two electrons, and N is just a normalization factor. Note that the interelectronic distance is a factor - hence the real wave function
cannot be separated into single-electron functions (orbitals) as so:
\Psi(\mathbf{r}_{1},\mathbf{r}_{2}) = \phi_1(\mathbf{r}_{1})\phi_2(\mathbf{r}_{2})
But also note that if you eliminated that factor, it would become separable, and you would have such a product, each electron described by a Gaussian. In other words, the exact solution for two entirely non-interacting particles in a harmonic potential. But we can do slightly better than this and still have "orbitals". Rather than neglect the interaction and hence the |r
1-r
2| factor, you could include the electron-electron interaction as an average. Hence, substituting the factor for some average value. That is in essence what the Hartree-Fock method does, and it's the most accurate description you can get while insisting your system being described by functions of a single electron's coordinates. (Real example: Helium's ground state is -2.903 a.u., the best possible orbital description gives -2.86 a.u. The resulting energy is always higher because correlation lowers the energy.)
Finally, there's of course the case of the single-electron (hydrogenic) atom. If you only have a
single electron, then the entire electronic wave function can be described in terms of that one function. So in that one case, the orbitals are the true electronic wave functions, and the orbital energies are those of the full atom.