If one wants to assess how an electron orbits in a hydrogen atom one uses the Schrodinger equation. Ditto for an electron in a magnetic well. However if one wants to assess how particles interact or decay one uses Feyman diagrams, amplitudehrons, or various formulae from Feynman diagrams. The latter is what we usually talk about in CERN or RHC experimental results.

Question: where does Schrodinger stop and Feynman diagrams start? Do we need both? Can I derive the hydrogen s,p orbital shapes from a Feynman diagram(s)?

An electron orbiting a proton is still an interaction; I guess the salient difference is the electron is fundamental and cannot decay. Here Schrodinger typically applies. But could one go Feynman? Both are dynamic over time ... Both seem to calculate probabilities of interaction, and yet have different purposes.

  • $\begingroup$ @johnrennie Is interpreting the question as being "when do we need QFT as opposed to Schroedinger's equation?". I think this makes sense but you may want to reformulate the question. Note also that between the two there's Diraq equation. One partial answer would then be: when relativistic effects become important. $\endgroup$
    – lcv
    Oct 10, 2019 at 11:07

1 Answer 1


There are two aspects to this.

Firstly, as long as the interaction energies are well below the energies necessary to create new particles the Schrodinger equation is usually an excellent approximation. For the majority of atoms and molecules it gives an excellent description of the electronic structure. (There is also the pragmatic consideration that bound states are hard to handle in QFT). We would generally only resort to QFT when the scattering energies get high enough that new particles are created. The Schrodinger equation does not and cannot describe the creation of new particles.

However the Schrodinger equation just takes for granted the classical form of the electrostatic potential energy, which brings to the second aspect I want to mention. If we just assume the PE is proportional to $r^{-1}$ then the Schrodinger equation gives us results that match experiment. However to explain why the PE has this form requires QFT. That is, even though we may not need to resort to QFT to describe the electronic structure of a molecule, QFT helps explain why the Schrodinger equation works.

I should also note that in practice when computing the properties of bound states like molecules we generally use the SE then extend it as necessary with perturbations that originate from QFT - for example the Lamb shift.


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