# Multi-electron wavefunctions

1. Apr 29, 2009

### sokrates

What is the basis to say that the wavefunction of a multi-electron system is the product of individual wavefunctions of the electrons that form the system?

In other words, how does theory ensure that the multi-electron wavefunction is seperable into variables r1 and r2?

Even in Hartree form (where things like exhange interaction and exclusion principle is not readily captured) -- the way we write down Psi(r1,r2) is:

$$\Psi(r1,r2) = \phi (r1)\phi(r2)$$

I learned from Wikipedia that this is really an ansatz in the HF theory. But is there any convincing reason that the well-versed quantum camp could deliver here?

2. Apr 29, 2009

### alxm

It follows from the antisymmetry requirement/Pauli exclusion principle for fermions. It's follows fairly easily that the simplest way to satisfy that requirement is to describe the wave function in the term of a Slater determinant.

It's not. It's the assumption that the wave function can be described using a single determinant of one-electron functions which is the essence of the Hartree-Fock approximation.

That would be a Hartree product. (which doesn't satisfy antisymmetry) Since the Hartree-Fock method uses a Slater determinant, so it'd actually be:
$$\Psi(r1,r2) = \phi_1(r_1)\phi_2(r_2) - \phi_1(r_2)\phi_2(r_1)$$
(where the coordinate r is assumed to include spin)

Hartree-Fock does get exchange right. (Within the constraint of a single-determinant description). It's implicit in the derivation. It's the correlation energy that's completely neglected.

Well, as stated above, it's an easy way to satisfy the antisymmetry requirement. Second, it's a fairly decent approximation, given the relative simplicity of it.

The biggest problem is getting the kinetic energy of the electrons. That's the single biggest energy term. Due to correlated motion, that's a many-body problem, and quite difficult to solve. So how would you, for instnace, calculate the motion of the planets around the sun? You'd probably go with first describing them as moving independently of each other and then include their effect on each other as an averaged effect. That's essentially what Hartree-Fock does. IOW: Its kinetic energy is exact for non-interacting or infinitely-slowly-moving electrons. That's a good approximation, because it gives most of the kinetic energy. The effects of correlated motion are very small in comparison.

Improving on it is of course the main deal with QC. For instance there's Møller-Plesset perturbation theory, which works by treating the correlation energy as a perturbation, and Configuration Interaction, which works by using multiple determinants.

Now you don't have to do it this way. For instance, Hylleraas used the Ritz variational method very early, 1929, to find the ground-state energy of Helium, using a total wave function parametricized in terms of $$r_1, r_2, r_{12}$$ (no spin coordinates!) directly, by re-expressing the Hamiltonian in these terms and solving the variational problem, with the symmetry requirement built in via constraints and integration limits. Very clever and extremely accurate. But that method is notoriously difficult to extend to systems of more than two electrons.

'Orbital-less' DFT methods (starting with the old Thomas-Fermi model) don't use single-electron descriptions either. But they tend to fail badly due to the difficulty of getting the kinetic energy. It was Kohn-Sham's Nobel-prize winning method that made DFT accurate enough to be usable, and that entailed re-using the Hartree-Fock approach to get the 'non-interacting' kinetic energy, and treating exchange and correlation as a pseudopotential. (and it's usually that single part that's referred to as a 'functional' now) Exchange was lost because you're not solving the Schrödinger equation; Kohn-Sham orbitals are not 'real' orbitals. (although debate rages over whether or not they can be ascribed some physical significance or not)

3. Apr 30, 2009

### sokrates

What is the anti-symmetry requirement? Is it a universal requirement?

And it seems like the multi-electron wavefunction is seperable into r1 and r2. I think what you are writing as a Slater determinant also supports that.

It has to be separable. But still, I can't see the exact reason. If it's an assumption, what is the basis?

4. May 2, 2009

### alxm

That the wave function changes sign under the interchange of coordinates for two particles:
$$\Psi(r_1,r_2) = -\Psi(r_2,r_1)$$ (where r is the particle coordinate, including spin)
This is a requirement for fermions, or rather, it's the definition of a fermion. (Bosons are the particles which are symmetric under interchange)

This is a boundary condition which has to be imposed on the Schrödinger equation. For fermions, it follows then that the wave functions of two single particles with the same coordinates would cancel out, making the overall solution invalid. Hence the Pauli exclusion principle.

Well, no. It's fairly obvious from the electron-electron repulsion term that the many-electron S.E. is inseparable. (It's less immediately obvious of the kinetic-energy operator, but isn't that hard to find out http://digitalcommons.uconn.edu/chem_educ/8/" [Broken] - the result is pretty nasty)

Since the e-e potential only depends on the coordinates r1 and r2, the separation would be okay if that was the only thing involved (again, HF would be exact for infinitely-slow-moving electrons). The interdependence would be taken care of through demanding self-consistency, (i.e. applying the SCF method).

But the kinetic-energy operator contains cross terms, that is, terms that depend on partial derivatives of both r1 and r2. They essentially represent the correlation energy and are essentially neglected in HF. (If you're aquainted with the Born-Oppenheimer approximation, that also essentialy amounts to neglecting such cross terms, the nuclear-electronic ones). This is valid because they're much smaller than the terms that depend only on one particle ($$\frac{\partial^2\psi}{\partial r_1^2}$$ and $$\frac{\partial^2\psi}{\partial r_2^2}$$)

Is it a good approximation? Well yes. That'd be borne out by the fact that Hartree-Fock is, in most cases, a pretty decent method, especially considering its relative simplicity.

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