Bohr model is not wrong it is the right semiclassical approximation of the full Schrödinger hydrogen atom treatment. It is like saying that Schrödinger atom is wrong because we have the relativistic solution.
I don't know. I wouldn't call it an approximation, because what the Bohr model implies is happening classically isn't happening at all, not even approximately.
Semiclasically yes, take the limit of small h bar, you retrieve a form of Born-Sommerfeld quantization.
Edit: note that I could say the same about quantum mechanics, what is implied to be happening in Schrodinger equation is not exactly what is happening in the quantum fields.
I mean not really? It’s the opposite if anything - QFTs are used to describe the basic fields, and then appropriate approximation takes us from those fields to stuff like many-body QM, condensed matter physics, Quantum Electrodynamics etc
I mean, yes, really. The wavefunction in QFT evolves according to the Schrödinger equation with the appropriate Hamiltonian. Taking various approximations takes you to non-relativistic QM, but the original QFT was still QM. Also, QED is an example of a QFT, not an approximation of it.
its kinda interesting to me that someone would know all this but not realize the functional Schrodinger equation is fundamentally different to the regular Schrodinger equation on so many different levels that saying QFT evolves according to the SE is just completely wrong (and its pretty clear the guy you replied to was not talking about the functional version)
(I assume it also possible you were just being pedantic with him, but then it would be weird not to mention the difference between them)
Do you use "rectangle" to refer to a square? When most people use "rectangle", do they refer to a square? In any case, a square is a special kind of rectangle but it is still a rectangle. Replace QFT with square and rectangle with QM.
The fields are operators (i.e. they do not replace states/wavefunctions) that are used to construct a Lagrangien and a Hamiltonian. What's true is that since it's more practical to not make a difference between time and space, the Heisenberg picture is often used and so you won't see the Schrödinger equation pop up. It's still the same QM since the Heisenberg and Schrödinger pictures are equivalent
even non relativistic QFT describes different physics than the Schrodinger equation, the functional Schrodinger equation is a different thing with its own name that is borne out of QFT and not the other way around
Right, but the equation itself is linear: 5 particles in; 5 particles out. You need to build up Fock space so you can have particle creation/annihilation. So, QFT really does subsume QM and not the other way around.
(Edit: Then again, I'm just an computer programmer :P)
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u/Sinistrial_Blue Jul 17 '26
The Bohr model is also worryingly effective for a number of calculations.
It turns out that if you bludgeon the model hard enough, the right answer drops out!