EDIT 1: Many people are recommending that the question Do electrons really perform instantaneous quantum leaps? is very similar to mine. However, that question is very specific to quantum leaps. I am also asking about a possible relation to QFT and the all-fields model.

Having learned that electrons can move from one energy level to another by transmitting or taking in energy, I am interested in knowing how exactly do they "jump" from one level to another. Do the electrons move as a particle or as a wave through energy levels. Because if they move as a particle, the distance and space between the electron shells must be covered, so do electrons move through the orbitals as a whole particle? Or do they move like a wave, because electrons exhibit wave-particle duality? Also, is the "jump" from one energy level to another instantaneous or is there a transmission delay? Because from what I have read, it seems that an instantaneous transmission has not been proved, whereas a delayed transmission by direct action-at-a-distance theory has been proven by experiments. Also, in a quantum mechanics explanation of electron transmission, I learned that the electron is destroyed in the original orbital, then recreated in another orbital, and the result also releases a photon. I thought that matter could not be created nor destroyed.

Furthermore, could it be possible that if the field model is used, it could possibly simplify the matter? If the field model is used, then would it be right to assume that the electron field is part of the shells/energy level where the electrons are, because electrons are just vibrations of the electron field and for them to be "created" they must come from a field? If we do assume that the electron field is present, then can we not say that since the field is able to occupy the space in which the distinct energy levels are in, then an electron's energy in one energy level can be absorbed by the field and then "moved"/transmitted to another energy level, through the use of the field? And this would be done by the energy traveling from one part of the field which is at a certain energy level to another, and then the energy can cause a vibration which will "pop-out" an electron at the new place in the field which is the new energy level.

Please let me know if any of my logic or understanding of the topic is flawed. I am very interested in getting to know everyone's take on this.

  • 3
    $\begingroup$ Does this answer your question? Do electrons really perform instantaneous quantum leaps? $\endgroup$ Dec 4, 2020 at 4:51
  • $\begingroup$ Did the question linked by Nihar help you? The top answer there is excellent, IMHO. $\endgroup$
    – PM 2Ring
    Dec 4, 2020 at 14:53
  • $\begingroup$ It was pretty helpful in understanding the theoretical part. The only difficulty was understanding the mathematical concepts, but that is not due to the writer of the answer, but due to my lack of experience and knowledge in higher-level maths. Is there any way the math can be simplified? $\endgroup$
    – Samarth
    Dec 4, 2020 at 15:39
  • $\begingroup$ I guess you're talking about this equation: $$|\psi(t) \rangle = c_1(t) |2p \rangle + c_2(t) | 1s \rangle$$ It tells you the probability of detecting the electron in either the $2p$ state or the $1s$ state at any time $t$. The functions $c_1(t)$ and $c_2(t)$ vary between 0 & 1, and add up to 1. So if $c_1(t)=0.1$ there's a 0.1 probability of detecting the electron in the $2p$ state and 0.9 probability of detecting it in the $1s$ state. $\endgroup$
    – PM 2Ring
    Dec 4, 2020 at 17:46
  • $\begingroup$ Thank you so much. That clears my doubts and helps me understand the rest of the puzzle which I was missing. $\endgroup$
    – Samarth
    Dec 4, 2020 at 18:11

2 Answers 2


There are misconceptions in your learning experience, (you do not give your physics background).

So having learned that electrons can move from one energy level to another by transmitting or taking in energy,

It is not the electrons that are moving in the classical sense. It is the whole atom which has quantum mechanical solutions with energy levels and orbitals for the electron. The atom absorbs a photon of an energy between two energy levels and the electron goes to the higher energy orbital. Orbitals are not orbits. They are probability loci, of where in (x,y,z) the given electron would be if measured. See this calculation of the orbitals available to the electron of the hydrogen atom.

As for the time taken, the energy levels have a width, and that corresponds to a lifetime, so there is nothing instantaneous.

the electron is destroyed in the original orbital, then recreated in another orbital, and the result also releases a photon. I thought that matter could not be created nor destroyed.

This is a confused use of quantum field theory, which is not very usefull for atomic spectra calculations. QFT describes well high energy scatterings, not the bound states of atoms and molecules.

There is nothing sacred about matter at the level of quantum mechanics and special relativity. Mass is not a conserved quantity, it is the four vector algebra that describes the creation and annihilation of particles . If there is enough energy in an interaction to create a particle antiparticle pair , quantum mechanics gives a probability for this to happen.

As for the last paragraph, Quantum field theory is not very useful for bound states as atoms and molecules, as its calculations involve a series expansion. Atoms can have complete solutions so to use the model of creation and annihilation operators on the corresponding fields makes no sense. There are such calculations but as the answer here says, still in a research stage,

  • 1
    $\begingroup$ Thank you so much for your response. I am currently a sophomore in high school who is taking chemistry and have an interest in physics. I seemed to have gotten some of my understandings of what I learned in chemistry and my research of physics mixed together. Your explanation and the links you provided have answered my questions. Also, for your last sentence, the calculations you mentioned and about how it is still in a research stage, is the research showing any hopeful signs, or is it most likely a dead end, since the answer refers back to a statement from 1995? $\endgroup$
    – Samarth
    Dec 4, 2020 at 12:41
  • $\begingroup$ In the answer to te last link, Arnold states that the situation is nor resolved since then, so it is probable that someone is trying to resolve it. $\endgroup$
    – anna v
    Dec 4, 2020 at 18:44
  • $\begingroup$ Ok. Thanks a lot. $\endgroup$
    – Samarth
    Dec 4, 2020 at 18:58

You may be horrified or delighted to know that this question at least partially falls into the domain of interpretation.

Electrons are not particles that move in classical trajectories (unless you are a devotee of Bohm). They generally don't even have a well defined position, so they don't "jump" in any physical sense.

Experimentally the energy levels are discrete (which match the QM calculations) so the electron's energy does undergo a "jump".

An atom may find itself in a situation where the probability of electron emitting or absorbing a photon steadily increases, in which case the waveform steadily changes shape, but measurement shows that the energy levels are always at the before or after levels and never in between. In that sense, the energy jump is "instantaneous".

Many eminent physicists make comments like "particles are excitations of the quantum field" because that helps them think about the physics, but that is somewhat glib. The reality is more complex. Similarly many believe that the waveform always changes smoothly but that is really inconsistent with phenomena like the photo-electric effect or double slit diffraction. These debates have no winners so it is much easier to follow Feynman and just "shut up and calculate".

  • 1
    $\begingroup$ Didn't Feynman also say "Nobody really understands quantum mechanics"? $\endgroup$
    – wyphan
    Dec 4, 2020 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.