# Does a relaxing electron really accelerate?

This is not a duplicate. I am not asking about quantum leaps or quantum jumps or whether the transition is instantaneous (yes I asked that question before here Do electrons really perform instantaneous quantum leaps?).

Where cmaster says:

When an electron falls from one shell to a lower one, it's wave function transitions smoothly from one eigenvector to another, creating a superposition of the two eigenvectors with changing amplitudes. This superposition wave function has the property of oscillating precisely with the frequency of the emitted photon. It's the expected location of the electron that oscillates. So, electron relaxation actually falls within the accelerated charge case.

where Kyle Oman says:

So what does the electron do between observations? I don't think anyone can answer that question. All we can say is that at one time the electron was observed at point A, and at a later time it was observed at point B. It got from A to B... somehow. This leads to a different way of thinking about where an electron (or other particle) is.

where annav says:

As far as the mathematics goes there is only the probability and no continuity between two space points so that a classical velocity could be defined.

So basically one says that electron relaxation is an accelerating charge.

The other ones say, that the electron's position is not even classically defined during the transition, it is in a superposition and not even velocity could be calculated, thus I believe we cannot talk about the acceleration of the electron during relaxation.

This is a contradiction, because as per QM, the wavefunction transition goes smoothly, creating a superposition. Now I believe it does not make sense to talk about classical speed or acceleration in this superposition of states.

Question:

1. Is the relaxing electron really accelerating or not?

Let us be clear, acceleration , velocity, force are classical definitions. What exists in the quantum mechanical and, generally, quantum field theory is four vectors and interactions. It is simple to think of Feynman diagrams, which have a $$\mathrm{d}p/\mathrm{d}t$$ at each vertex, being given or taken away by the interaction, that is the connection with $$F=ma$$, acceleration.

When coming to the bound states of atoms, one has to see how the quantum field theory calculations ( Feynman diagrams) can apply to bound states. Please see the answer by Arnold Neumaier here for links and theoretical explanation.

QFT has been used to get the fine structure of atomic levels, one has to stop using classical analogues for the quantum mechanical level. The atom is an entangled system of the electron and the nucleus , the calculations give probability distributions so it has little meaning, imo, to talk of accelerating an electron as if it can be separated from the total solution.

The answer by cmaster is correct. As an example, consider the probability $$|\psi(x)|^2$$ of a superposition of the two lowest eigenstates of a particle in a box:

That charge density is oscillating, sloshing back and forth in the box, with a frequency $$\omega = (E_1-E_2)/\hbar$$, which is the frequency of the emitted photon.

While it is radiating, the expectation value of the energy of the mixed state goes down, the relative weight of the excited state becomes smaller, the amplitude of the oscillation becomes smaller, until it is essential equal to the ground state with its stationary probability distribution.

• My take on this is that it is not the electron that is doing this, but the total (electron-nucleus) bound state, and it is misleading to talk of acceleration of one of the entangled components in a quantum mechanical system, which needs probability distributions to be modeled. A given electron does not have a continuous distribution as the plot above , the transition is a probability for a given atom, happens once, and is just a point adding to this plot. Acceleration is a continuous variable by definition on one particle. Oct 6, 2019 at 5:07