Why an electron bound to a nucleus does not emit photons when accelerated?

Question

I think electrically neutral materials do not generate electromagnetic radiation/photons when accelerated, but I might be wrong. If I am correct though, why is it that accelerated ions generate electromagnetic radiation but not electrically neutral stuff? Do their electrical fields cancel each other (like a positive proton cancelling the negative electron fields)? Even so, a perfect cancellation would assume a faster than light propagation of those fields otherwise when the entire atom proton plus electron is shaken, there should be some lag between the two and thus a non perfect cancellation, letting the differential field escape...? Even if the effect is extremely small and boring for you, I am still interested.

Answer 1

Why an electron bound to a nucleus does not emit photons when accelerated?

Because bound electrons have no individual identity, no orbits, otherwise they would neutralize on the nucleus, but orbitals which are probability loci. The whole atom/molecule is the quantum mechanical entity, i.e. obeys the rules of quantum mechanics.

If an atom molecule has an electric dipole then it will behave while accelerating according to the theory of classical Maxwell's equations, similar to an accelerating electron, which emits synchrotron radiation, for example, that is calculated classically.

Quantum electrodynamics solutions and Maxwell's equation solutions are compatible because the classical fields emerge from the quantum ones, as shown here. Only at the level of individual quantum entities one has to be careful to distinguish between probability distributions and predictions, from the deterministic predictions of maxwell equations.

why is it that accelerated ions generate electromagnetic radiation but not electrically neutral stuff?

Because the quantum mechanical entities are considered as a whole, not a la cart, their constituents are quantum mechanically bound and not seperable. A neutral atom/molecule with no dipoles is invisible to Maxwell equation solutions.