When trying to solve for a molecule's molecular orbitals, we start off with the total molecular hamiltonian. However, given the difficulty of solving the Schrödinger Equation for this complicated hamiltonian we can introduce the Born-Oppenheimer approximation. By assuming that the nuclei of molecules are relatively fixed, our molecular hamiltonian can be reduced to the electronic hamiltonian. Then, we can use the electronic hamiltonian and solve for the electronic wavefunctions of our molecule.

This calculation for the wavefunctions is significantly easier. What is confusing to me is what the solution to the electronic Schrödinger equation represents.

Are these solutions considered the molecular orbitals?


1 Answer 1


I think you are close to the right idea, but let me try to clarify some points.

The full Schrodinger equation for a molecule should depend on the $3N$ nuclear coordinates, as well as the $3n$ electronic coordinates, where $N$ and $n$ are the number of nuclei and electrons respectively. This is challenging both due to the number of coordinates and, in particular, due to the coupling between electronic and nuclear coordinates. However, with the Born-Oppenheimer approximation, we assume they are decoupled and solve for the electronic part of the wavefunction at a fixed nuclear geometry.

The issue here is that we still can't solve the Schrodinger equation for an $n$ electron system with $n>1$. To simplify further, we can make the Hartree-Fock approximation, which (in one way of phrasing it) assumes that the $n$ electron wavefunction can be formed from the antisymmetrized product of $n$ one electron wavefunctions/orbitals. These orbitals can be determined by solving a nonlinear system of equations and then stitched together in the form of a Slater Determinant to make an approximate $n$ electron wavefunction.

While chemists like to frame discussions in terms of orbitals, strictly speaking orbitals are always an approximation (except for hydrogen) and the electronic structure of a molecule is determined by a $n$ electron wavefunction. Even though orbitals aren't quite real, they still can give a pretty good qualitative description of a molecule's behavior. For a bit more about the meaning of orbitals, you might want to check out this previous question: Is it correct to talk about an empty orbital?

  • $\begingroup$ Just to be clear, when you say "𝑛 electron wavefunction can be formed from the antisymmetrized product of 𝑛 one electron wavefunctions, which we call orbitals", is the molecular orbital = n electron wave function? $\endgroup$
    – LamGyro
    Commented Apr 8, 2020 at 19:34
  • 1
    $\begingroup$ The molecular orbitals are one-electron wavefunctions that are combined together to make an $n$ electron wavefunction. $\endgroup$
    – Tyberius
    Commented Apr 8, 2020 at 19:37
  • $\begingroup$ Ok I think I understand your answer now. It was difficult for me to understand initially because my professor is constructing MOs via the LCAO method. Looking through my notes, it seems the "one-electron wavefunctions" that you mention, are they MOs we obtain via LCAO? Also, for a given electronic state, the overall electronic wave function involves some product of those MOs (and a spin part), and it represents the OVERALL state of our molecule. $\endgroup$
    – LamGyro
    Commented Apr 8, 2020 at 20:13
  • 1
    $\begingroup$ LCAO adds another layer to it. Basically, we can't solve for the MOs analytically, but if we define a basis of atomic orbitals(AOs), we can approximately express the MOs as linear combinations of the AOs. But yes, the overall state of the molecule is particular product of the MOs, which have a spatial and a spin component. $\endgroup$
    – Tyberius
    Commented Apr 8, 2020 at 20:21
  • $\begingroup$ Since the is some confusion about the many- and one-electron wave function and which one the actual solution to the Schrödinger-Equation is: I would add the Slater-Determinant to the explanation. $\endgroup$
    – Feodoran
    Commented Apr 9, 2020 at 11:11

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