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This source is showing that solving the Schrödinger equation for a triatmoic linear molecule yields the same formula for the rotationaI quantum states $BJ(J+1)$ as for dipoles.

For dipoles, the total rotational energy can be expressed as the total length of the molecule $R$ and a particle with reduced mass $\mu$ which makes it possible to express the coordinates in the Schrodinger equation specifically for only one particle with that reduced mass $\mu$.

I am clueless how the Schrödinger equation is solved for triatomic linear molecules because there are more variables that can not be reduced the same way. I would therefore expect to be seperate coordinates for each atom in the Schrodinger equation, as well as more than one bond length.

How is it derived and is it actually possible to do so?

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  • $\begingroup$ See this answer chemistry.stackexchange.com/questions/135539/… . The 'trick' is to change to polar coordinates and then in your case to make the potential energy $V(r)=0$ and then to solve the Schroedinger equation. Its complicated but standard maths. $\endgroup$
    – porphyrin
    Jul 30, 2020 at 7:53
  • $\begingroup$ @porphyrin I indeed know that polar coordinates has to be used just like with dipoles. But for dipoles, there's only one particle with reduced mass and one bond length $R$. In the case of a triatomic linear molecule, one would have to use polar coordinates for each of these atoms, each having a different distance to the center of mass. Is there a derivation that shows how all these variables can be reduced to solve the Schrödinger equation? $\endgroup$
    – Phy
    Jul 30, 2020 at 17:01
  • $\begingroup$ do you mean diatomics in the place of dipoles? $\endgroup$
    – Antonio de Oliveira-Filho
    Jul 30, 2020 at 17:33
  • $\begingroup$ @AntoniodeOliveira-Filho I mean triatomics, a linear molecule with three atoms (or more). $\endgroup$
    – Phy
    Jul 30, 2020 at 17:37
  • $\begingroup$ You use still use the reduced mass to form the moment of inertia as the molecule rotates about it s centre of mass. $\endgroup$
    – porphyrin
    Jul 30, 2020 at 18:21

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