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In this series on solid state chemistry by MIT, 11:00-13:00 in this lecture, LCAO for molecular orbitals is justified by the fact that Schrodinger's equation is linear (and therefore the superposition principle is used)

However, I have a vague recollection, either from high school or from reading somewhere that expressing the wave function of an MO as a linear combination of the atomic orbital MOs is an approximation. However, there's no such mention of it being so in the lecture. He seems to be implying that this type of a thing is exact.

Am mixing two different ideas up or misinterpreting something?

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LCAO is an approximation, partly because we usually only add up a finite number of orbitals. E.g. for the molecule LiH, we could combine the H 1s orbitals with the Li 2s and 2p to get some wavefunction. But we have ignored the Li 3s for example, because it's high in energy and won't overlap well with H 1s.

Another reason it's an approximation is we're using Hydrogen-like atomic orbitals (which are the exact solutions to the Hydrogen Schrodinger equation), and using a variational method to solve the actual Schrodinger equation which has more than one electron for which we don't have an exact solution.

The "exactness" here is just the fact that if 1s is a solution to the (Hydrogen) Schrodinger equation, and 2s is also a solution, then 1s+2s is a valid solution too.

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  • $\begingroup$ The last part is what I'm confused about. I agree with your statement, but don't see how that's applicable in the context of MOs as was being implied in the lecture. $\endgroup$
    – xasthor
    Commented Feb 10, 2023 at 7:46
  • $\begingroup$ To elaborate on that a bit: I can see how if 1s is a solution to the hydrogen Schrodinger equation, and 2s is, then 1s+2s is. But he seems to be implying that can be extended in the case of MOs and LCAO. He's an MIT professor so I'm sure he knows LCAO is an approximation, as you have confirmed. So I don't get what he's trying to say or why he he's implying some sort of exactness $\endgroup$
    – xasthor
    Commented Feb 10, 2023 at 7:51
  • $\begingroup$ I think all he's trying to point out is that when making up trial wavefunctions for molecules, it is appropriate to add them linearly because each orbital on its own is a solution to the Hydrogen Schrodinger equation, which itself is linear. It would be less appropriate to, say, use a trial wavefunction with the product of two orbitals. Such a wavefunction could still be used in a variational way to get another approximate answer, but it would probably be less good than if we used a wavefunction with addition. $\endgroup$
    – Garf
    Commented Feb 10, 2023 at 12:25
  • $\begingroup$ It's important to remember that when finding MOs for molecules, we do not have the exact answer. The best we can do is "guess" a wavefunction, and a good guess will usually be a linear combination. $\endgroup$
    – Garf
    Commented Feb 10, 2023 at 12:26
  • $\begingroup$ "when making up trial wavefunctions for molecules, it is appropriate to add them linearly because each orbital on its own is a solution to the Hydrogen Schrodinger equation, which itself is linear" Why does the latter mean that when making up trial wavefunctions for molecules, it is appropriate to add the orbitals of the composing atoms linearly? $\endgroup$
    – xasthor
    Commented Feb 10, 2023 at 13:26

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